3.217 \(\int \frac{\coth ^6(x)}{i+\sinh (x)} \, dx\)

Optimal. Leaf size=36 \[ \frac{1}{5} i \coth ^5(x)-\frac{3}{8} \tanh ^{-1}(\cosh (x))-\frac{1}{4} \coth ^3(x) \text{csch}(x)-\frac{3}{8} \coth (x) \text{csch}(x) \]

[Out]

(-3*ArcTanh[Cosh[x]])/8 + (I/5)*Coth[x]^5 - (3*Coth[x]*Csch[x])/8 - (Coth[x]^3*Csch[x])/4

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Rubi [A]  time = 0.0969328, antiderivative size = 36, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.385, Rules used = {2706, 2607, 30, 2611, 3770} \[ \frac{1}{5} i \coth ^5(x)-\frac{3}{8} \tanh ^{-1}(\cosh (x))-\frac{1}{4} \coth ^3(x) \text{csch}(x)-\frac{3}{8} \coth (x) \text{csch}(x) \]

Antiderivative was successfully verified.

[In]

Int[Coth[x]^6/(I + Sinh[x]),x]

[Out]

(-3*ArcTanh[Cosh[x]])/8 + (I/5)*Coth[x]^5 - (3*Coth[x]*Csch[x])/8 - (Coth[x]^3*Csch[x])/4

Rule 2706

Int[((g_.)*tan[(e_.) + (f_.)*(x_)])^(p_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[1/a, Int[S
ec[e + f*x]^2*(g*Tan[e + f*x])^p, x], x] - Dist[1/(b*g), Int[Sec[e + f*x]*(g*Tan[e + f*x])^(p + 1), x], x] /;
FreeQ[{a, b, e, f, g, p}, x] && EqQ[a^2 - b^2, 0] && NeQ[p, -1]

Rule 2607

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(b*x)
^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] &&  !(IntegerQ[(n
- 1)/2] && LtQ[0, n, m - 1])

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2611

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a*Sec[e
+ f*x])^m*(b*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] - Dist[(b^2*(n - 1))/(m + n - 1), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

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{\coth ^6(x)}{i+\sinh (x)} \, dx &=-\left (i \int \coth ^4(x) \text{csch}^2(x) \, dx\right )+\int \coth ^4(x) \text{csch}(x) \, dx\\ &=-\frac{1}{4} \coth ^3(x) \text{csch}(x)+\frac{3}{4} \int \coth ^2(x) \text{csch}(x) \, dx+\operatorname{Subst}\left (\int x^4 \, dx,x,i \coth (x)\right )\\ &=\frac{1}{5} i \coth ^5(x)-\frac{3}{8} \coth (x) \text{csch}(x)-\frac{1}{4} \coth ^3(x) \text{csch}(x)+\frac{3}{8} \int \text{csch}(x) \, dx\\ &=-\frac{3}{8} \tanh ^{-1}(\cosh (x))+\frac{1}{5} i \coth ^5(x)-\frac{3}{8} \coth (x) \text{csch}(x)-\frac{1}{4} \coth ^3(x) \text{csch}(x)\\ \end{align*}

Mathematica [B]  time = 0.0402153, size = 164, normalized size = 4.56 \[ \frac{1}{10} i \tanh \left (\frac{x}{2}\right )+\frac{1}{10} i \coth \left (\frac{x}{2}\right )-\frac{1}{64} \text{csch}^4\left (\frac{x}{2}\right )-\frac{5}{32} \text{csch}^2\left (\frac{x}{2}\right )+\frac{1}{64} \text{sech}^4\left (\frac{x}{2}\right )-\frac{5}{32} \text{sech}^2\left (\frac{x}{2}\right )+\frac{3}{8} \log \left (\tanh \left (\frac{x}{2}\right )\right )+\frac{1}{160} i \coth \left (\frac{x}{2}\right ) \text{csch}^4\left (\frac{x}{2}\right )+\frac{7}{160} i \coth \left (\frac{x}{2}\right ) \text{csch}^2\left (\frac{x}{2}\right )+\frac{1}{160} i \tanh \left (\frac{x}{2}\right ) \text{sech}^4\left (\frac{x}{2}\right )-\frac{7}{160} i \tanh \left (\frac{x}{2}\right ) \text{sech}^2\left (\frac{x}{2}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Coth[x]^6/(I + Sinh[x]),x]

[Out]

(I/10)*Coth[x/2] - (5*Csch[x/2]^2)/32 + ((7*I)/160)*Coth[x/2]*Csch[x/2]^2 - Csch[x/2]^4/64 + (I/160)*Coth[x/2]
*Csch[x/2]^4 + (3*Log[Tanh[x/2]])/8 - (5*Sech[x/2]^2)/32 + Sech[x/2]^4/64 + (I/10)*Tanh[x/2] - ((7*I)/160)*Sec
h[x/2]^2*Tanh[x/2] + (I/160)*Sech[x/2]^4*Tanh[x/2]

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Maple [B]  time = 0.07, size = 93, normalized size = 2.6 \begin{align*}{\frac{i}{16}}\tanh \left ({\frac{x}{2}} \right ) +{\frac{i}{160}} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{5}+{\frac{1}{64} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{4}}+{\frac{i}{32}} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{3}+{\frac{1}{8} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}}-{\frac{1}{8} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{-2}}+{{\frac{i}{16}} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{-1}}+{{\frac{i}{160}} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{-5}}+{{\frac{i}{32}} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{-3}}+{\frac{3}{8}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) }-{\frac{1}{64} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{-4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(x)^6/(I+sinh(x)),x)

[Out]

1/16*I*tanh(1/2*x)+1/160*I*tanh(1/2*x)^5+1/64*tanh(1/2*x)^4+1/32*I*tanh(1/2*x)^3+1/8*tanh(1/2*x)^2-1/8/tanh(1/
2*x)^2+1/16*I/tanh(1/2*x)+1/160*I/tanh(1/2*x)^5+1/32*I/tanh(1/2*x)^3+3/8*ln(tanh(1/2*x))-1/64/tanh(1/2*x)^4

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Maxima [B]  time = 1.08917, size = 123, normalized size = 3.42 \begin{align*} \frac{25 \, e^{\left (-x\right )} - 10 \, e^{\left (-3 \, x\right )} - 80 i \, e^{\left (-4 \, x\right )} + 10 \, e^{\left (-7 \, x\right )} - 40 i \, e^{\left (-8 \, x\right )} - 25 \, e^{\left (-9 \, x\right )} - 8 i}{20 \,{\left (5 \, e^{\left (-2 \, x\right )} - 10 \, e^{\left (-4 \, x\right )} + 10 \, e^{\left (-6 \, x\right )} - 5 \, e^{\left (-8 \, x\right )} + e^{\left (-10 \, x\right )} - 1\right )}} - \frac{3}{8} \, \log \left (e^{\left (-x\right )} + 1\right ) + \frac{3}{8} \, \log \left (e^{\left (-x\right )} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(x)^6/(I+sinh(x)),x, algorithm="maxima")

[Out]

1/20*(25*e^(-x) - 10*e^(-3*x) - 80*I*e^(-4*x) + 10*e^(-7*x) - 40*I*e^(-8*x) - 25*e^(-9*x) - 8*I)/(5*e^(-2*x) -
10*e^(-4*x) + 10*e^(-6*x) - 5*e^(-8*x) + e^(-10*x) - 1) - 3/8*log(e^(-x) + 1) + 3/8*log(e^(-x) - 1)

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Fricas [B]  time = 2.13792, size = 437, normalized size = 12.14 \begin{align*} -\frac{15 \,{\left (e^{\left (10 \, x\right )} - 5 \, e^{\left (8 \, x\right )} + 10 \, e^{\left (6 \, x\right )} - 10 \, e^{\left (4 \, x\right )} + 5 \, e^{\left (2 \, x\right )} - 1\right )} \log \left (e^{x} + 1\right ) - 15 \,{\left (e^{\left (10 \, x\right )} - 5 \, e^{\left (8 \, x\right )} + 10 \, e^{\left (6 \, x\right )} - 10 \, e^{\left (4 \, x\right )} + 5 \, e^{\left (2 \, x\right )} - 1\right )} \log \left (e^{x} - 1\right ) + 50 \, e^{\left (9 \, x\right )} - 80 i \, e^{\left (8 \, x\right )} - 20 \, e^{\left (7 \, x\right )} - 160 i \, e^{\left (4 \, x\right )} + 20 \, e^{\left (3 \, x\right )} - 50 \, e^{x} - 16 i}{40 \,{\left (e^{\left (10 \, x\right )} - 5 \, e^{\left (8 \, x\right )} + 10 \, e^{\left (6 \, x\right )} - 10 \, e^{\left (4 \, x\right )} + 5 \, e^{\left (2 \, x\right )} - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(x)^6/(I+sinh(x)),x, algorithm="fricas")

[Out]

-1/40*(15*(e^(10*x) - 5*e^(8*x) + 10*e^(6*x) - 10*e^(4*x) + 5*e^(2*x) - 1)*log(e^x + 1) - 15*(e^(10*x) - 5*e^(
8*x) + 10*e^(6*x) - 10*e^(4*x) + 5*e^(2*x) - 1)*log(e^x - 1) + 50*e^(9*x) - 80*I*e^(8*x) - 20*e^(7*x) - 160*I*
e^(4*x) + 20*e^(3*x) - 50*e^x - 16*I)/(e^(10*x) - 5*e^(8*x) + 10*e^(6*x) - 10*e^(4*x) + 5*e^(2*x) - 1)

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Sympy [B]  time = 1.37699, size = 104, normalized size = 2.89 \begin{align*} \frac{3 \log{\left (e^{x} - 1 \right )}}{8} - \frac{3 \log{\left (e^{x} + 1 \right )}}{8} + \frac{- \frac{5 e^{9 x}}{4} + 2 i e^{8 x} + \frac{e^{7 x}}{2} + 4 i e^{4 x} - \frac{e^{3 x}}{2} + \frac{5 e^{x}}{4} + \frac{2 i}{5}}{e^{10 x} - 5 e^{8 x} + 10 e^{6 x} - 10 e^{4 x} + 5 e^{2 x} - 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(x)**6/(I+sinh(x)),x)

[Out]

3*log(exp(x) - 1)/8 - 3*log(exp(x) + 1)/8 + (-5*exp(9*x)/4 + 2*I*exp(8*x) + exp(7*x)/2 + 4*I*exp(4*x) - exp(3*
x)/2 + 5*exp(x)/4 + 2*I/5)/(exp(10*x) - 5*exp(8*x) + 10*exp(6*x) - 10*exp(4*x) + 5*exp(2*x) - 1)

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Giac [B]  time = 1.12559, size = 84, normalized size = 2.33 \begin{align*} -\frac{25 \, e^{\left (9 \, x\right )} - 40 i \, e^{\left (8 \, x\right )} - 10 \, e^{\left (7 \, x\right )} - 80 i \, e^{\left (4 \, x\right )} + 10 \, e^{\left (3 \, x\right )} - 25 \, e^{x} - 8 i}{20 \,{\left (e^{\left (2 \, x\right )} - 1\right )}^{5}} - \frac{3}{8} \, \log \left (e^{x} + 1\right ) + \frac{3}{8} \, \log \left ({\left | e^{x} - 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(x)^6/(I+sinh(x)),x, algorithm="giac")

[Out]

-1/20*(25*e^(9*x) - 40*I*e^(8*x) - 10*e^(7*x) - 80*I*e^(4*x) + 10*e^(3*x) - 25*e^x - 8*I)/(e^(2*x) - 1)^5 - 3/
8*log(e^x + 1) + 3/8*log(abs(e^x - 1))