3.227 \(\int \frac{\coth ^6(x)}{(i+\sinh (x))^2} \, dx\)

Optimal. Leaf size=48 \[ \frac{\coth ^5(x)}{5}-\frac{2 \coth ^3(x)}{3}-\frac{1}{4} i \tanh ^{-1}(\cosh (x))+\frac{1}{2} i \coth (x) \text{csch}^3(x)+\frac{1}{4} i \coth (x) \text{csch}(x) \]

[Out]

(-I/4)*ArcTanh[Cosh[x]] - (2*Coth[x]^3)/3 + Coth[x]^5/5 + (I/4)*Coth[x]*Csch[x] + (I/2)*Coth[x]*Csch[x]^3

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Rubi [A]  time = 0.0905616, antiderivative size = 48, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 5, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.385, Rules used = {2709, 3767, 8, 3768, 3770} \[ \frac{\coth ^5(x)}{5}-\frac{2 \coth ^3(x)}{3}-\frac{1}{4} i \tanh ^{-1}(\cosh (x))+\frac{1}{2} i \coth (x) \text{csch}^3(x)+\frac{1}{4} i \coth (x) \text{csch}(x) \]

Antiderivative was successfully verified.

[In]

Int[Coth[x]^6/(I + Sinh[x])^2,x]

[Out]

(-I/4)*ArcTanh[Cosh[x]] - (2*Coth[x]^3)/3 + Coth[x]^5/5 + (I/4)*Coth[x]*Csch[x] + (I/2)*Coth[x]*Csch[x]^3

Rule 2709

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*tan[(e_.) + (f_.)*(x_)]^(p_), x_Symbol] :> Dist[a^p, Int[Expan
dIntegrand[(Sin[e + f*x]^p*(a + b*Sin[e + f*x])^(m - p/2))/(a - b*Sin[e + f*x])^(p/2), x], x], x] /; FreeQ[{a,
b, e, f}, x] && EqQ[a^2 - b^2, 0] && IntegersQ[m, p/2] && (LtQ[p, 0] || GtQ[m - p/2, 0])

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[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))^2} \, dx &=\int \left (\text{csch}^2(x)-2 i \text{csch}^3(x)-2 i \text{csch}^5(x)-\text{csch}^6(x)\right ) \, dx\\ &=-\left (2 i \int \text{csch}^3(x) \, dx\right )-2 i \int \text{csch}^5(x) \, dx+\int \text{csch}^2(x) \, dx-\int \text{csch}^6(x) \, dx\\ &=i \coth (x) \text{csch}(x)+\frac{1}{2} i \coth (x) \text{csch}^3(x)+i \int \text{csch}(x) \, dx-i \operatorname{Subst}(\int 1 \, dx,x,-i \coth (x))+i \operatorname{Subst}\left (\int \left (1+2 x^2+x^4\right ) \, dx,x,-i \coth (x)\right )+\frac{3}{2} i \int \text{csch}^3(x) \, dx\\ &=-i \tanh ^{-1}(\cosh (x))-\frac{2 \coth ^3(x)}{3}+\frac{\coth ^5(x)}{5}+\frac{1}{4} i \coth (x) \text{csch}(x)+\frac{1}{2} i \coth (x) \text{csch}^3(x)-\frac{3}{4} i \int \text{csch}(x) \, dx\\ &=-\frac{1}{4} i \tanh ^{-1}(\cosh (x))-\frac{2 \coth ^3(x)}{3}+\frac{\coth ^5(x)}{5}+\frac{1}{4} i \coth (x) \text{csch}(x)+\frac{1}{2} i \coth (x) \text{csch}^3(x)\\ \end{align*}

Mathematica [B]  time = 0.0582698, size = 175, normalized size = 3.65 \[ -\frac{7}{30} \tanh \left (\frac{x}{2}\right )-\frac{7}{30} \coth \left (\frac{x}{2}\right )+\frac{1}{32} i \text{csch}^4\left (\frac{x}{2}\right )+\frac{1}{16} i \text{csch}^2\left (\frac{x}{2}\right )-\frac{1}{32} i \text{sech}^4\left (\frac{x}{2}\right )+\frac{1}{16} i \text{sech}^2\left (\frac{x}{2}\right )+\frac{1}{4} i \log \left (\sinh \left (\frac{x}{2}\right )\right )-\frac{1}{4} i \log \left (\cosh \left (\frac{x}{2}\right )\right )+\frac{1}{160} \coth \left (\frac{x}{2}\right ) \text{csch}^4\left (\frac{x}{2}\right )-\frac{19}{480} \coth \left (\frac{x}{2}\right ) \text{csch}^2\left (\frac{x}{2}\right )+\frac{1}{160} \tanh \left (\frac{x}{2}\right ) \text{sech}^4\left (\frac{x}{2}\right )+\frac{19}{480} \tanh \left (\frac{x}{2}\right ) \text{sech}^2\left (\frac{x}{2}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Coth[x]^6/(I + Sinh[x])^2,x]

[Out]

(-7*Coth[x/2])/30 + (I/16)*Csch[x/2]^2 - (19*Coth[x/2]*Csch[x/2]^2)/480 + (I/32)*Csch[x/2]^4 + (Coth[x/2]*Csch
[x/2]^4)/160 - (I/4)*Log[Cosh[x/2]] + (I/4)*Log[Sinh[x/2]] + (I/16)*Sech[x/2]^2 - (I/32)*Sech[x/2]^4 - (7*Tanh
[x/2])/30 + (19*Sech[x/2]^2*Tanh[x/2])/480 + (Sech[x/2]^4*Tanh[x/2])/160

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Maple [B]  time = 0.073, size = 74, normalized size = 1.5 \begin{align*} -{\frac{3}{16}\tanh \left ({\frac{x}{2}} \right ) }+{\frac{1}{160} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{5}}-{\frac{i}{32}} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{4}-{\frac{5}{96} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{3}}-{\frac{3}{16} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{-1}}+{\frac{1}{160} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{-5}}-{\frac{5}{96} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{-3}}+{\frac{i}{4}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) +{{\frac{i}{32}} \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))^2,x)

[Out]

-3/16*tanh(1/2*x)+1/160*tanh(1/2*x)^5-1/32*I*tanh(1/2*x)^4-5/96*tanh(1/2*x)^3-3/16/tanh(1/2*x)+1/160/tanh(1/2*
x)^5-5/96/tanh(1/2*x)^3+1/4*I*ln(tanh(1/2*x))+1/32*I/tanh(1/2*x)^4

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Maxima [B]  time = 1.06903, size = 139, normalized size = 2.9 \begin{align*} \frac{-15 i \, e^{\left (-x\right )} - 80 \, e^{\left (-2 \, x\right )} - 90 i \, e^{\left (-3 \, x\right )} + 40 \, e^{\left (-4 \, x\right )} - 240 \, e^{\left (-6 \, x\right )} + 90 i \, e^{\left (-7 \, x\right )} + 60 \, e^{\left (-8 \, x\right )} + 15 i \, e^{\left (-9 \, x\right )} + 28}{30 \,{\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{1}{4} i \, \log \left (e^{\left (-x\right )} + 1\right ) + \frac{1}{4} 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)^6/(I+sinh(x))^2,x, algorithm="maxima")

[Out]

1/30*(-15*I*e^(-x) - 80*e^(-2*x) - 90*I*e^(-3*x) + 40*e^(-4*x) - 240*e^(-6*x) + 90*I*e^(-7*x) + 60*e^(-8*x) +
15*I*e^(-9*x) + 28)/(5*e^(-2*x) - 10*e^(-4*x) + 10*e^(-6*x) - 5*e^(-8*x) + e^(-10*x) - 1) - 1/4*I*log(e^(-x) +
1) + 1/4*I*log(e^(-x) - 1)

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Fricas [B]  time = 2.11256, size = 527, normalized size = 10.98 \begin{align*} \frac{{\left (-15 i \, e^{\left (10 \, x\right )} + 75 i \, e^{\left (8 \, x\right )} - 150 i \, e^{\left (6 \, x\right )} + 150 i \, e^{\left (4 \, x\right )} - 75 i \, e^{\left (2 \, x\right )} + 15 i\right )} \log \left (e^{x} + 1\right ) +{\left (15 i \, e^{\left (10 \, x\right )} - 75 i \, e^{\left (8 \, x\right )} + 150 i \, e^{\left (6 \, x\right )} - 150 i \, e^{\left (4 \, x\right )} + 75 i \, e^{\left (2 \, x\right )} - 15 i\right )} \log \left (e^{x} - 1\right ) + 30 i \, e^{\left (9 \, x\right )} - 120 \, e^{\left (8 \, x\right )} + 180 i \, e^{\left (7 \, x\right )} + 480 \, e^{\left (6 \, x\right )} - 80 \, e^{\left (4 \, x\right )} - 180 i \, e^{\left (3 \, x\right )} + 160 \, e^{\left (2 \, x\right )} - 30 i \, e^{x} - 56}{60 \,{\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))^2,x, algorithm="fricas")

[Out]

1/60*((-15*I*e^(10*x) + 75*I*e^(8*x) - 150*I*e^(6*x) + 150*I*e^(4*x) - 75*I*e^(2*x) + 15*I)*log(e^x + 1) + (15
*I*e^(10*x) - 75*I*e^(8*x) + 150*I*e^(6*x) - 150*I*e^(4*x) + 75*I*e^(2*x) - 15*I)*log(e^x - 1) + 30*I*e^(9*x)
- 120*e^(8*x) + 180*I*e^(7*x) + 480*e^(6*x) - 80*e^(4*x) - 180*I*e^(3*x) + 160*e^(2*x) - 30*I*e^x - 56)/(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 = 1.52104, size = 117, normalized size = 2.44 \begin{align*} \operatorname{RootSum}{\left (16 z^{2} + 1, \left ( i \mapsto i \log{\left (4 i i + e^{x} \right )} \right )\right )} + \frac{\frac{i e^{9 x}}{2} - 2 e^{8 x} + 3 i e^{7 x} + 8 e^{6 x} - \frac{4 e^{4 x}}{3} - 3 i e^{3 x} + \frac{8 e^{2 x}}{3} - \frac{i e^{x}}{2} - \frac{14}{15}}{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))**2,x)

[Out]

RootSum(16*_z**2 + 1, Lambda(_i, _i*log(4*_i*I + exp(x)))) + (I*exp(9*x)/2 - 2*exp(8*x) + 3*I*exp(7*x) + 8*exp
(6*x) - 4*exp(4*x)/3 - 3*I*exp(3*x) + 8*exp(2*x)/3 - I*exp(x)/2 - 14/15)/(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 = 1.14469, size = 100, normalized size = 2.08 \begin{align*} -\frac{-15 i \, e^{\left (9 \, x\right )} + 60 \, e^{\left (8 \, x\right )} - 90 i \, e^{\left (7 \, x\right )} - 240 \, e^{\left (6 \, x\right )} + 40 \, e^{\left (4 \, x\right )} + 90 i \, e^{\left (3 \, x\right )} - 80 \, e^{\left (2 \, x\right )} + 15 i \, e^{x} + 28}{30 \,{\left (e^{\left (2 \, x\right )} - 1\right )}^{5}} - \frac{1}{4} i \, \log \left (e^{x} + 1\right ) + \frac{1}{4} i \, \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))^2,x, algorithm="giac")

[Out]

-1/30*(-15*I*e^(9*x) + 60*e^(8*x) - 90*I*e^(7*x) - 240*e^(6*x) + 40*e^(4*x) + 90*I*e^(3*x) - 80*e^(2*x) + 15*I
*e^x + 28)/(e^(2*x) - 1)^5 - 1/4*I*log(e^x + 1) + 1/4*I*log(abs(e^x - 1))