3.111 \(\int \frac{\coth ^5(x)}{i+\text{csch}(x)} \, dx\)

Optimal. Leaf size=30 \[ -\frac{1}{3} \text{csch}^3(x)+\frac{1}{2} i \text{csch}^2(x)-\text{csch}(x)-i \log (\sinh (x)) \]

[Out]

-Csch[x] + (I/2)*Csch[x]^2 - Csch[x]^3/3 - I*Log[Sinh[x]]

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Rubi [A]  time = 0.0459022, antiderivative size = 30, 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 = {3879, 75} \[ -\frac{1}{3} \text{csch}^3(x)+\frac{1}{2} i \text{csch}^2(x)-\text{csch}(x)-i \log (\sinh (x)) \]

Antiderivative was successfully verified.

[In]

Int[Coth[x]^5/(I + Csch[x]),x]

[Out]

-Csch[x] + (I/2)*Csch[x]^2 - Csch[x]^3/3 - I*Log[Sinh[x]]

Rule 3879

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_.), x_Symbol] :> Dist[1/(a^(m - n
- 1)*b^n*d), Subst[Int[((a - b*x)^((m - 1)/2)*(a + b*x)^((m - 1)/2 + n))/x^(m + n), x], x, Sin[c + d*x]], x] /
; FreeQ[{a, b, c, d}, x] && IntegerQ[(m - 1)/2] && EqQ[a^2 - b^2, 0] && IntegerQ[n]

Rule 75

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && EqQ[b*e + a*f, 0] &&  !(ILtQ[n
+ p + 2, 0] && GtQ[n + 2*p, 0])

Rubi steps

\begin{align*} \int \frac{\coth ^5(x)}{i+\text{csch}(x)} \, dx &=\operatorname{Subst}\left (\int \frac{(i-i x)^2 (i+i x)}{x^4} \, dx,x,i \sinh (x)\right )\\ &=\operatorname{Subst}\left (\int \left (-\frac{i}{x^4}+\frac{i}{x^3}+\frac{i}{x^2}-\frac{i}{x}\right ) \, dx,x,i \sinh (x)\right )\\ &=-\text{csch}(x)+\frac{1}{2} i \text{csch}^2(x)-\frac{\text{csch}^3(x)}{3}-i \log (\sinh (x))\\ \end{align*}

Mathematica [A]  time = 0.0145455, size = 30, normalized size = 1. \[ -\frac{1}{3} \text{csch}^3(x)+\frac{1}{2} i \text{csch}^2(x)-\text{csch}(x)-i \log (\sinh (x)) \]

Antiderivative was successfully verified.

[In]

Integrate[Coth[x]^5/(I + Csch[x]),x]

[Out]

-Csch[x] + (I/2)*Csch[x]^2 - Csch[x]^3/3 - I*Log[Sinh[x]]

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Maple [B]  time = 0.068, size = 78, normalized size = 2.6 \begin{align*}{\frac{3}{8}\tanh \left ({\frac{x}{2}} \right ) }+{\frac{1}{24} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{3}}+{\frac{i}{8}} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+i\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) -{\frac{1}{24} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{-3}}-i\ln \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) +{{\frac{i}{8}} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{-2}}-{\frac{3}{8} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{-1}}+i\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(x)^5/(I+csch(x)),x)

[Out]

3/8*tanh(1/2*x)+1/24*tanh(1/2*x)^3+1/8*I*tanh(1/2*x)^2+I*ln(tanh(1/2*x)+1)-1/24/tanh(1/2*x)^3-I*ln(tanh(1/2*x)
)+1/8*I/tanh(1/2*x)^2-3/8/tanh(1/2*x)+I*ln(tanh(1/2*x)-1)

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Maxima [B]  time = 1.02183, size = 101, normalized size = 3.37 \begin{align*} -i \, x + \frac{6 \, e^{\left (-x\right )} - 6 i \, e^{\left (-2 \, x\right )} - 4 \, e^{\left (-3 \, x\right )} + 6 i \, e^{\left (-4 \, x\right )} + 6 \, e^{\left (-5 \, x\right )}}{3 \,{\left (3 \, e^{\left (-2 \, x\right )} - 3 \, e^{\left (-4 \, x\right )} + e^{\left (-6 \, x\right )} - 1\right )}} - i \, \log \left (e^{\left (-x\right )} + 1\right ) - i \, \log \left (e^{\left (-x\right )} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(x)^5/(I+csch(x)),x, algorithm="maxima")

[Out]

-I*x + 1/3*(6*e^(-x) - 6*I*e^(-2*x) - 4*e^(-3*x) + 6*I*e^(-4*x) + 6*e^(-5*x))/(3*e^(-2*x) - 3*e^(-4*x) + e^(-6
*x) - 1) - I*log(e^(-x) + 1) - I*log(e^(-x) - 1)

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Fricas [B]  time = 1.72243, size = 292, normalized size = 9.73 \begin{align*} \frac{3 i \, x e^{\left (6 \, x\right )} +{\left (-9 i \, x + 6 i\right )} e^{\left (4 \, x\right )} +{\left (9 i \, x - 6 i\right )} e^{\left (2 \, x\right )} +{\left (-3 i \, e^{\left (6 \, x\right )} + 9 i \, e^{\left (4 \, x\right )} - 9 i \, e^{\left (2 \, x\right )} + 3 i\right )} \log \left (e^{\left (2 \, x\right )} - 1\right ) - 3 i \, x - 6 \, e^{\left (5 \, x\right )} + 4 \, e^{\left (3 \, x\right )} - 6 \, e^{x}}{3 \,{\left (e^{\left (6 \, x\right )} - 3 \, e^{\left (4 \, x\right )} + 3 \, e^{\left (2 \, x\right )} - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(x)^5/(I+csch(x)),x, algorithm="fricas")

[Out]

1/3*(3*I*x*e^(6*x) + (-9*I*x + 6*I)*e^(4*x) + (9*I*x - 6*I)*e^(2*x) + (-3*I*e^(6*x) + 9*I*e^(4*x) - 9*I*e^(2*x
) + 3*I)*log(e^(2*x) - 1) - 3*I*x - 6*e^(5*x) + 4*e^(3*x) - 6*e^x)/(e^(6*x) - 3*e^(4*x) + 3*e^(2*x) - 1)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\coth ^{5}{\left (x \right )}}{\operatorname{csch}{\left (x \right )} + i}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(x)**5/(I+csch(x)),x)

[Out]

Integral(coth(x)**5/(csch(x) + I), x)

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Giac [B]  time = 1.15493, size = 92, normalized size = 3.07 \begin{align*} -\frac{11 \,{\left (e^{\left (-x\right )} - e^{x}\right )}^{3} - 12 i \,{\left (e^{\left (-x\right )} - e^{x}\right )}^{2} + 12 \, e^{\left (-x\right )} - 12 \, e^{x} - 16 i}{6 \,{\left (-i \, e^{\left (-x\right )} + i \, e^{x}\right )}^{3}} - i \, \log \left (-i \, e^{\left (-x\right )} + i \, e^{x}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(x)^5/(I+csch(x)),x, algorithm="giac")

[Out]

-1/6*(11*(e^(-x) - e^x)^3 - 12*I*(e^(-x) - e^x)^2 + 12*e^(-x) - 12*e^x - 16*I)/(-I*e^(-x) + I*e^x)^3 - I*log(-
I*e^(-x) + I*e^x)