3.224 \(\int \frac{\coth ^3(x)}{(i+\sinh (x))^2} \, dx\)

Optimal. Leaf size=29 \[ \frac{\text{csch}^2(x)}{2}+2 i \text{csch}(x)+2 \log (\sinh (x))-2 \log (\sinh (x)+i) \]

[Out]

(2*I)*Csch[x] + Csch[x]^2/2 + 2*Log[Sinh[x]] - 2*Log[I + Sinh[x]]

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Rubi [A]  time = 0.0480786, antiderivative size = 29, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {2707, 77} \[ \frac{\text{csch}^2(x)}{2}+2 i \text{csch}(x)+2 \log (\sinh (x))-2 \log (\sinh (x)+i) \]

Antiderivative was successfully verified.

[In]

Int[Coth[x]^3/(I + Sinh[x])^2,x]

[Out]

(2*I)*Csch[x] + Csch[x]^2/2 + 2*Log[Sinh[x]] - 2*Log[I + Sinh[x]]

Rule 2707

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p_.), x_Symbol] :> Dist[1/f, Subst[I
nt[(x^p*(a + x)^(m - (p + 1)/2))/(a - x)^((p + 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& EqQ[a^2 - b^2, 0] && IntegerQ[(p + 1)/2]

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps

\begin{align*} \int \frac{\coth ^3(x)}{(i+\sinh (x))^2} \, dx &=-\operatorname{Subst}\left (\int \frac{i-x}{x^3 (i+x)} \, dx,x,\sinh (x)\right )\\ &=-\operatorname{Subst}\left (\int \left (\frac{1}{x^3}+\frac{2 i}{x^2}-\frac{2}{x}+\frac{2}{i+x}\right ) \, dx,x,\sinh (x)\right )\\ &=2 i \text{csch}(x)+\frac{\text{csch}^2(x)}{2}+2 \log (\sinh (x))-2 \log (i+\sinh (x))\\ \end{align*}

Mathematica [A]  time = 0.016163, size = 29, normalized size = 1. \[ \frac{\text{csch}^2(x)}{2}+2 i \text{csch}(x)+2 \log (\sinh (x))-2 \log (\sinh (x)+i) \]

Antiderivative was successfully verified.

[In]

Integrate[Coth[x]^3/(I + Sinh[x])^2,x]

[Out]

(2*I)*Csch[x] + Csch[x]^2/2 + 2*Log[Sinh[x]] - 2*Log[I + Sinh[x]]

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Maple [A]  time = 0.06, size = 51, normalized size = 1.8 \begin{align*} -i\tanh \left ({\frac{x}{2}} \right ) +{\frac{1}{8} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}}+{i \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{-1}}+{\frac{1}{8} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{-2}}+2\,\ln \left ( \tanh \left ( x/2 \right ) \right ) -4\,\ln \left ( \tanh \left ( x/2 \right ) +i \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(x)^3/(I+sinh(x))^2,x)

[Out]

-I*tanh(1/2*x)+1/8*tanh(1/2*x)^2+I/tanh(1/2*x)+1/8/tanh(1/2*x)^2+2*ln(tanh(1/2*x))-4*ln(tanh(1/2*x)+I)

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Maxima [B]  time = 1.14238, size = 86, normalized size = 2.97 \begin{align*} \frac{-4 i \, e^{\left (-x\right )} - 2 \, e^{\left (-2 \, x\right )} + 4 i \, e^{\left (-3 \, x\right )}}{2 \, e^{\left (-2 \, x\right )} - e^{\left (-4 \, x\right )} - 1} + 2 \, \log \left (e^{\left (-x\right )} + 1\right ) - 4 \, \log \left (e^{\left (-x\right )} - i\right ) + 2 \, \log \left (e^{\left (-x\right )} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(x)^3/(I+sinh(x))^2,x, algorithm="maxima")

[Out]

(-4*I*e^(-x) - 2*e^(-2*x) + 4*I*e^(-3*x))/(2*e^(-2*x) - e^(-4*x) - 1) + 2*log(e^(-x) + 1) - 4*log(e^(-x) - I)
+ 2*log(e^(-x) - 1)

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Fricas [B]  time = 2.06189, size = 207, normalized size = 7.14 \begin{align*} \frac{2 \,{\left (e^{\left (4 \, x\right )} - 2 \, e^{\left (2 \, x\right )} + 1\right )} \log \left (e^{\left (2 \, x\right )} - 1\right ) - 4 \,{\left (e^{\left (4 \, x\right )} - 2 \, e^{\left (2 \, x\right )} + 1\right )} \log \left (e^{x} + i\right ) + 4 i \, e^{\left (3 \, x\right )} + 2 \, e^{\left (2 \, x\right )} - 4 i \, e^{x}}{e^{\left (4 \, x\right )} - 2 \, e^{\left (2 \, x\right )} + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(x)^3/(I+sinh(x))^2,x, algorithm="fricas")

[Out]

(2*(e^(4*x) - 2*e^(2*x) + 1)*log(e^(2*x) - 1) - 4*(e^(4*x) - 2*e^(2*x) + 1)*log(e^x + I) + 4*I*e^(3*x) + 2*e^(
2*x) - 4*I*e^x)/(e^(4*x) - 2*e^(2*x) + 1)

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Sympy [A]  time = 0.688085, size = 53, normalized size = 1.83 \begin{align*} \frac{4 i e^{3 x} + 2 e^{2 x} - 4 i e^{x}}{e^{4 x} - 2 e^{2 x} + 1} - 4 \log{\left (e^{x} + i \right )} + 2 \log{\left (e^{2 x} - 1 \right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(x)**3/(I+sinh(x))**2,x)

[Out]

(4*I*exp(3*x) + 2*exp(2*x) - 4*I*exp(x))/(exp(4*x) - 2*exp(2*x) + 1) - 4*log(exp(x) + I) + 2*log(exp(2*x) - 1)

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Giac [B]  time = 1.12132, size = 72, normalized size = 2.48 \begin{align*} \frac{4 i \, e^{\left (3 \, x\right )} + 2 \, e^{\left (2 \, x\right )} - 4 i \, e^{x}}{{\left (e^{x} + 1\right )}^{2}{\left (e^{x} - 1\right )}^{2}} + 2 \, \log \left (e^{x} + 1\right ) - 4 \, \log \left (e^{x} + i\right ) + 2 \, \log \left ({\left | e^{x} - 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(x)^3/(I+sinh(x))^2,x, algorithm="giac")

[Out]

(4*I*e^(3*x) + 2*e^(2*x) - 4*I*e^x)/((e^x + 1)^2*(e^x - 1)^2) + 2*log(e^x + 1) - 4*log(e^x + I) + 2*log(abs(e^
x - 1))