3.104 \(\int \frac{\tanh ^3(x)}{i+\text{csch}(x)} \, dx\)

Optimal. Leaf size=77 \[ -\frac{i}{8 (1-i \sinh (x))}-\frac{3 i}{4 (1+i \sinh (x))}+\frac{i}{8 (1+i \sinh (x))^2}-\frac{11}{16} i \log (-\sinh (x)+i)-\frac{5}{16} i \log (\sinh (x)+i) \]

[Out]

((-11*I)/16)*Log[I - Sinh[x]] - ((5*I)/16)*Log[I + Sinh[x]] - (I/8)/(1 - I*Sinh[x]) + (I/8)/(1 + I*Sinh[x])^2
- ((3*I)/4)/(1 + I*Sinh[x])

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Rubi [A]  time = 0.0672505, antiderivative size = 77, 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, 88} \[ -\frac{i}{8 (1-i \sinh (x))}-\frac{3 i}{4 (1+i \sinh (x))}+\frac{i}{8 (1+i \sinh (x))^2}-\frac{11}{16} i \log (-\sinh (x)+i)-\frac{5}{16} i \log (\sinh (x)+i) \]

Antiderivative was successfully verified.

[In]

Int[Tanh[x]^3/(I + Csch[x]),x]

[Out]

((-11*I)/16)*Log[I - Sinh[x]] - ((5*I)/16)*Log[I + Sinh[x]] - (I/8)/(1 - I*Sinh[x]) + (I/8)/(1 + I*Sinh[x])^2
- ((3*I)/4)/(1 + I*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 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rubi steps

\begin{align*} \int \frac{\tanh ^3(x)}{i+\text{csch}(x)} \, dx &=\operatorname{Subst}\left (\int \frac{x^4}{(i-i x)^2 (i+i x)^3} \, dx,x,i \sinh (x)\right )\\ &=\operatorname{Subst}\left (\int \left (-\frac{i}{8 (-1+x)^2}-\frac{5 i}{16 (-1+x)}-\frac{i}{4 (1+x)^3}+\frac{3 i}{4 (1+x)^2}-\frac{11 i}{16 (1+x)}\right ) \, dx,x,i \sinh (x)\right )\\ &=-\frac{11}{16} i \log (i-\sinh (x))-\frac{5}{16} i \log (i+\sinh (x))-\frac{i}{8 (1-i \sinh (x))}+\frac{i}{8 (1+i \sinh (x))^2}-\frac{3 i}{4 (1+i \sinh (x))}\\ \end{align*}

Mathematica [A]  time = 0.115431, size = 61, normalized size = 0.79 \[ \frac{1}{16} \left (-\frac{2 \left (5 \sinh ^2(x)+3 i \sinh (x)+6\right )}{(\sinh (x)-i)^2 (\sinh (x)+i)}-11 i \log (-\sinh (x)+i)-5 i \log (\sinh (x)+i)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Tanh[x]^3/(I + Csch[x]),x]

[Out]

((-11*I)*Log[I - Sinh[x]] - (5*I)*Log[I + Sinh[x]] - (2*(6 + (3*I)*Sinh[x] + 5*Sinh[x]^2))/((-I + Sinh[x])^2*(
I + Sinh[x])))/16

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Maple [A]  time = 0.055, size = 109, normalized size = 1.4 \begin{align*} i\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) -{\frac{11\,i}{8}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -i \right ) +{{\frac{i}{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) -i \right ) ^{-4}}+{{\frac{i}{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) -i \right ) ^{-2}}+ \left ( \tanh \left ({\frac{x}{2}} \right ) -i \right ) ^{-3}+ \left ( \tanh \left ({\frac{x}{2}} \right ) -i \right ) ^{-1}+i\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) -{\frac{5\,i}{8}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) +{{\frac{i}{4}} \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) ^{-2}}-{\frac{1}{4} \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(x)^3/(I+csch(x)),x)

[Out]

I*ln(tanh(1/2*x)+1)-11/8*I*ln(tanh(1/2*x)-I)+1/2*I/(tanh(1/2*x)-I)^4+1/2*I/(tanh(1/2*x)-I)^2+1/(tanh(1/2*x)-I)
^3+1/(tanh(1/2*x)-I)+I*ln(tanh(1/2*x)-1)-5/8*I*ln(tanh(1/2*x)+I)+1/4*I/(tanh(1/2*x)+I)^2-1/4/(tanh(1/2*x)+I)

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Maxima [B]  time = 1.04781, size = 130, normalized size = 1.69 \begin{align*} -i \, x + \frac{5 \, e^{\left (-x\right )} + 6 i \, e^{\left (-2 \, x\right )} + 14 \, e^{\left (-3 \, x\right )} - 6 i \, e^{\left (-4 \, x\right )} + 5 \, e^{\left (-5 \, x\right )}}{8 i \, e^{\left (-x\right )} - 4 \, e^{\left (-2 \, x\right )} + 16 i \, e^{\left (-3 \, x\right )} + 4 \, e^{\left (-4 \, x\right )} + 8 i \, e^{\left (-5 \, x\right )} + 4 \, e^{\left (-6 \, x\right )} - 4} - \frac{5}{8} i \, \log \left (e^{\left (-x\right )} - i\right ) - \frac{11}{8} i \, \log \left (i \, e^{\left (-x\right )} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)^3/(I+csch(x)),x, algorithm="maxima")

[Out]

-I*x + (5*e^(-x) + 6*I*e^(-2*x) + 14*e^(-3*x) - 6*I*e^(-4*x) + 5*e^(-5*x))/(8*I*e^(-x) - 4*e^(-2*x) + 16*I*e^(
-3*x) + 4*e^(-4*x) + 8*I*e^(-5*x) + 4*e^(-6*x) - 4) - 5/8*I*log(e^(-x) - I) - 11/8*I*log(I*e^(-x) - 1)

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Fricas [B]  time = 1.58237, size = 568, normalized size = 7.38 \begin{align*} \frac{8 i \, x e^{\left (6 \, x\right )} + 2 \,{\left (8 \, x - 5\right )} e^{\left (5 \, x\right )} +{\left (8 i \, x - 12 i\right )} e^{\left (4 \, x\right )} + 4 \,{\left (8 \, x - 7\right )} e^{\left (3 \, x\right )} +{\left (-8 i \, x + 12 i\right )} e^{\left (2 \, x\right )} + 2 \,{\left (8 \, x - 5\right )} e^{x} +{\left (-5 i \, e^{\left (6 \, x\right )} - 10 \, e^{\left (5 \, x\right )} - 5 i \, e^{\left (4 \, x\right )} - 20 \, e^{\left (3 \, x\right )} + 5 i \, e^{\left (2 \, x\right )} - 10 \, e^{x} + 5 i\right )} \log \left (e^{x} + i\right ) +{\left (-11 i \, e^{\left (6 \, x\right )} - 22 \, e^{\left (5 \, x\right )} - 11 i \, e^{\left (4 \, x\right )} - 44 \, e^{\left (3 \, x\right )} + 11 i \, e^{\left (2 \, x\right )} - 22 \, e^{x} + 11 i\right )} \log \left (e^{x} - i\right ) - 8 i \, x}{8 \, e^{\left (6 \, x\right )} - 16 i \, e^{\left (5 \, x\right )} + 8 \, e^{\left (4 \, x\right )} - 32 i \, e^{\left (3 \, x\right )} - 8 \, e^{\left (2 \, x\right )} - 16 i \, e^{x} - 8} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)^3/(I+csch(x)),x, algorithm="fricas")

[Out]

(8*I*x*e^(6*x) + 2*(8*x - 5)*e^(5*x) + (8*I*x - 12*I)*e^(4*x) + 4*(8*x - 7)*e^(3*x) + (-8*I*x + 12*I)*e^(2*x)
+ 2*(8*x - 5)*e^x + (-5*I*e^(6*x) - 10*e^(5*x) - 5*I*e^(4*x) - 20*e^(3*x) + 5*I*e^(2*x) - 10*e^x + 5*I)*log(e^
x + I) + (-11*I*e^(6*x) - 22*e^(5*x) - 11*I*e^(4*x) - 44*e^(3*x) + 11*I*e^(2*x) - 22*e^x + 11*I)*log(e^x - I)
- 8*I*x)/(8*e^(6*x) - 16*I*e^(5*x) + 8*e^(4*x) - 32*I*e^(3*x) - 8*e^(2*x) - 16*I*e^x - 8)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tanh ^{3}{\left (x \right )}}{\operatorname{csch}{\left (x \right )} + i}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)**3/(I+csch(x)),x)

[Out]

Integral(tanh(x)**3/(csch(x) + I), x)

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Giac [B]  time = 1.13793, size = 132, normalized size = 1.71 \begin{align*} \frac{5 \, e^{\left (-x\right )} - 5 \, e^{x} - 6 i}{16 \,{\left (-i \, e^{\left (-x\right )} + i \, e^{x} - 2\right )}} + \frac{33 i \,{\left (e^{\left (-x\right )} - e^{x}\right )}^{2} - 84 \, e^{\left (-x\right )} + 84 \, e^{x} - 52 i}{32 \,{\left (e^{\left (-x\right )} - e^{x} + 2 i\right )}^{2}} - \frac{5}{16} i \, \log \left (i \, e^{\left (-x\right )} - i \, e^{x} + 2\right ) - \frac{11}{16} i \, \log \left (i \, e^{\left (-x\right )} - i \, e^{x} - 2\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)^3/(I+csch(x)),x, algorithm="giac")

[Out]

1/16*(5*e^(-x) - 5*e^x - 6*I)/(-I*e^(-x) + I*e^x - 2) + 1/32*(33*I*(e^(-x) - e^x)^2 - 84*e^(-x) + 84*e^x - 52*
I)/(e^(-x) - e^x + 2*I)^2 - 5/16*I*log(I*e^(-x) - I*e^x + 2) - 11/16*I*log(I*e^(-x) - I*e^x - 2)