AngouriMath
SolveAFunctionOfTheHyperbolicTangent(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean)
Method (no overloads)
Summary
A function of the hyperbolic tangent of a linear, under u = tanh(a + b x) : the parser
writestanh(L) as (W - 1)/(W + 1) and coth(L) as (W + 1)/(W - 1) with W = e^(2 L) , so an integrand built from those alone is R(u) , and
dx = du/(b (1 - u^2)) . tanh(x)/sqrt(a + b tanh(x)^2 + c tanh(x)^4) is
u/((1 - u^2) sqrt(a + b u^2 + c u^4)) , and sin(tanh(x)) is
sin(u)/(1 - u^2) , onto the sine and cosine integrals. The trigonometric
counterparts are answered underu = tan(x) , which the substitution reads in the
tangent as a node; the hyperbolic functions arrive as exponentials, and nothing read them
as a tangent. Rubi's 6.3 and 6.4.
https://github.com/asc-community/AngouriMath/issues/233
writes
counterparts are answered under
tangent as a node; the hyperbolic functions arrive as exponentials, and nothing read them
as a tangent. Rubi's 6.3 and 6.4.
https://github.com/asc-community/AngouriMath/issues/233
Remarks
Only where R is not a rational function: those the exponential substitution answers,
and so a power ofa + a coth(L) or a - a coth(L) alone, a binomial in W .
Every exponential has to be a whole power ofW inside one of the two quotients;
one anywhere else, an odd power ofe^L from a sinh or cosh among
them, and the rule declines.d/dx F(tanh(L)) = b (1 - tanh(L)^2) F'(tanh(L)) wherever F is differentiable, so the answer written back holds where the integrand
is defined -- and says nothing where the conditions ofF fail for every u in
(-1, 1) , which is the rule's to decline.
and so a power of
Every exponential has to be a whole power of
one anywhere else, an odd power of
them, and the rule declines.
is defined -- and says nothing where the conditions of
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