3.208 \(\int \frac {\text {csch}(c+d x)}{a+i a \sinh (c+d x)} \, dx\)

Optimal. Leaf size=41 \[ -\frac {\tanh ^{-1}(\cosh (c+d x))}{a d}+\frac {\cosh (c+d x)}{d (a+i a \sinh (c+d x))} \]

[Out]

-arctanh(cosh(d*x+c))/a/d+cosh(d*x+c)/d/(a+I*a*sinh(d*x+c))

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Rubi [A]  time = 0.06, antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {2747, 3770, 2648} \[ -\frac {\tanh ^{-1}(\cosh (c+d x))}{a d}+\frac {\cosh (c+d x)}{d (a+i a \sinh (c+d x))} \]

Antiderivative was successfully verified.

[In]

Int[Csch[c + d*x]/(a + I*a*Sinh[c + d*x]),x]

[Out]

-(ArcTanh[Cosh[c + d*x]]/(a*d)) + Cosh[c + d*x]/(d*(a + I*a*Sinh[c + d*x]))

Rule 2648

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Simp[Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2747

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Dist[b/(
b*c - a*d), Int[1/(a + b*Sin[e + f*x]), x], x] - Dist[d/(b*c - a*d), Int[1/(c + d*Sin[e + f*x]), x], x] /; Fre
eQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin {align*} \int \frac {\text {csch}(c+d x)}{a+i a \sinh (c+d x)} \, dx &=-\left (i \int \frac {1}{a+i a \sinh (c+d x)} \, dx\right )+\frac {\int \text {csch}(c+d x) \, dx}{a}\\ &=-\frac {\tanh ^{-1}(\cosh (c+d x))}{a d}+\frac {\cosh (c+d x)}{d (a+i a \sinh (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 52, normalized size = 1.27 \[ -\frac {\text {sech}(c+d x) \left (i \sinh (c+d x)+\sqrt {\cosh ^2(c+d x)} \tanh ^{-1}\left (\sqrt {\cosh ^2(c+d x)}\right )-1\right )}{a d} \]

Antiderivative was successfully verified.

[In]

Integrate[Csch[c + d*x]/(a + I*a*Sinh[c + d*x]),x]

[Out]

-((Sech[c + d*x]*(-1 + ArcTanh[Sqrt[Cosh[c + d*x]^2]]*Sqrt[Cosh[c + d*x]^2] + I*Sinh[c + d*x]))/(a*d))

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fricas [A]  time = 1.60, size = 57, normalized size = 1.39 \[ -\frac {{\left (e^{\left (d x + c\right )} - i\right )} \log \left (e^{\left (d x + c\right )} + 1\right ) - {\left (e^{\left (d x + c\right )} - i\right )} \log \left (e^{\left (d x + c\right )} - 1\right ) - 2}{a d e^{\left (d x + c\right )} - i \, a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(d*x+c)/(a+I*a*sinh(d*x+c)),x, algorithm="fricas")

[Out]

-((e^(d*x + c) - I)*log(e^(d*x + c) + 1) - (e^(d*x + c) - I)*log(e^(d*x + c) - 1) - 2)/(a*d*e^(d*x + c) - I*a*
d)

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giac [A]  time = 0.27, size = 48, normalized size = 1.17 \[ -\frac {\frac {\log \left (e^{\left (d x + c\right )} + 1\right )}{a} - \frac {\log \left (e^{\left (d x + c\right )} - 1\right )}{a} - \frac {2}{a {\left (e^{\left (d x + c\right )} - i\right )}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(d*x+c)/(a+I*a*sinh(d*x+c)),x, algorithm="giac")

[Out]

-(log(e^(d*x + c) + 1)/a - log(e^(d*x + c) - 1)/a - 2/(a*(e^(d*x + c) - I)))/d

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maple [A]  time = 0.08, size = 42, normalized size = 1.02 \[ -\frac {2 i}{d a \left (-i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csch(d*x+c)/(a+I*a*sinh(d*x+c)),x)

[Out]

-2*I/d/a/(-I+tanh(1/2*d*x+1/2*c))+1/d/a*ln(tanh(1/2*d*x+1/2*c))

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maxima [A]  time = 0.50, size = 62, normalized size = 1.51 \[ -\frac {\log \left (e^{\left (-d x - c\right )} + 1\right )}{a d} + \frac {\log \left (e^{\left (-d x - c\right )} - 1\right )}{a d} + \frac {2}{{\left (a e^{\left (-d x - c\right )} + i \, a\right )} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(d*x+c)/(a+I*a*sinh(d*x+c)),x, algorithm="maxima")

[Out]

-log(e^(-d*x - c) + 1)/(a*d) + log(e^(-d*x - c) - 1)/(a*d) + 2/((a*e^(-d*x - c) + I*a)*d)

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mupad [B]  time = 0.88, size = 56, normalized size = 1.37 \[ -\frac {2\,\mathrm {atan}\left (\frac {{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\,\sqrt {-a^2\,d^2}}{a\,d}\right )}{\sqrt {-a^2\,d^2}}+\frac {2}{a\,d\,\left ({\mathrm {e}}^{c+d\,x}-\mathrm {i}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sinh(c + d*x)*(a + a*sinh(c + d*x)*1i)),x)

[Out]

2/(a*d*(exp(c + d*x) - 1i)) - (2*atan((exp(d*x)*exp(c)*(-a^2*d^2)^(1/2))/(a*d)))/(-a^2*d^2)^(1/2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {i \int \frac {\operatorname {csch}{\left (c + d x \right )}}{\sinh {\left (c + d x \right )} - i}\, dx}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(d*x+c)/(a+I*a*sinh(d*x+c)),x)

[Out]

-I*Integral(csch(c + d*x)/(sinh(c + d*x) - I), x)/a

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