3.69 \(\int \frac{\text{csch}^4(x)}{i+\text{csch}(x)} \, dx\)

Optimal. Leaf size=37 \[ 2 i \coth (x)+\frac{3}{2} \tanh ^{-1}(\cosh (x))+\frac{\coth (x) \text{csch}^2(x)}{\text{csch}(x)+i}-\frac{3}{2} \coth (x) \text{csch}(x) \]

[Out]

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

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Rubi [A]  time = 0.064237, antiderivative size = 37, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.462, Rules used = {3818, 3787, 3767, 8, 3768, 3770} \[ 2 i \coth (x)+\frac{3}{2} \tanh ^{-1}(\cosh (x))+\frac{\coth (x) \text{csch}^2(x)}{\text{csch}(x)+i}-\frac{3}{2} \coth (x) \text{csch}(x) \]

Antiderivative was successfully verified.

[In]

Int[Csch[x]^4/(I + Csch[x]),x]

[Out]

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

Rule 3818

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[(d^2*Cot[e
+ f*x]*(d*Csc[e + f*x])^(n - 2))/(f*(a + b*Csc[e + f*x])), x] - Dist[d^2/(a*b), Int[(d*Csc[e + f*x])^(n - 2)*(
b*(n - 2) - a*(n - 1)*Csc[e + f*x]), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && GtQ[n, 1]

Rule 3787

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

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

Mathematica [B]  time = 0.307058, size = 81, normalized size = 2.19 \[ \frac{1}{8} \left (4 i \tanh \left (\frac{x}{2}\right )+4 i \coth \left (\frac{x}{2}\right )-\text{csch}^2\left (\frac{x}{2}\right )-\text{sech}^2\left (\frac{x}{2}\right )-12 \log \left (\tanh \left (\frac{x}{2}\right )\right )+\frac{16 \sinh \left (\frac{x}{2}\right )}{\sinh \left (\frac{x}{2}\right )-i \cosh \left (\frac{x}{2}\right )}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Csch[x]^4/(I + Csch[x]),x]

[Out]

((4*I)*Coth[x/2] - Csch[x/2]^2 - 12*Log[Tanh[x/2]] - Sech[x/2]^2 + (16*Sinh[x/2])/((-I)*Cosh[x/2] + Sinh[x/2])
+ (4*I)*Tanh[x/2])/8

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Maple [A]  time = 0.025, size = 53, normalized size = 1.4 \begin{align*}{\frac{i}{2}}\tanh \left ({\frac{x}{2}} \right ) +{\frac{1}{8} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}}+{2\,i \left ( \tanh \left ({\frac{x}{2}} \right ) -i \right ) ^{-1}}-{\frac{1}{8} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{-2}}+{{\frac{i}{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{-1}}-{\frac{3}{2}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csch(x)^4/(I+csch(x)),x)

[Out]

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

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Maxima [B]  time = 1.03503, size = 109, normalized size = 2.95 \begin{align*} -\frac{16 \,{\left (-i \, e^{\left (-x\right )} - 5 \, e^{\left (-2 \, x\right )} + 3 i \, e^{\left (-3 \, x\right )} + 3 \, e^{\left (-4 \, x\right )} + 4\right )}}{16 \, e^{\left (-x\right )} - 32 i \, e^{\left (-2 \, x\right )} - 32 \, e^{\left (-3 \, x\right )} + 16 i \, e^{\left (-4 \, x\right )} + 16 \, e^{\left (-5 \, x\right )} + 16 i} + \frac{3}{2} \, \log \left (e^{\left (-x\right )} + 1\right ) - \frac{3}{2} \, \log \left (e^{\left (-x\right )} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(x)^4/(I+csch(x)),x, algorithm="maxima")

[Out]

-16*(-I*e^(-x) - 5*e^(-2*x) + 3*I*e^(-3*x) + 3*e^(-4*x) + 4)/(16*e^(-x) - 32*I*e^(-2*x) - 32*e^(-3*x) + 16*I*e
^(-4*x) + 16*e^(-5*x) + 16*I) + 3/2*log(e^(-x) + 1) - 3/2*log(e^(-x) - 1)

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Fricas [B]  time = 1.60965, size = 381, normalized size = 10.3 \begin{align*} \frac{{\left (3 \, e^{\left (5 \, x\right )} - 3 i \, e^{\left (4 \, x\right )} - 6 \, e^{\left (3 \, x\right )} + 6 i \, e^{\left (2 \, x\right )} + 3 \, e^{x} - 3 i\right )} \log \left (e^{x} + 1\right ) -{\left (3 \, e^{\left (5 \, x\right )} - 3 i \, e^{\left (4 \, x\right )} - 6 \, e^{\left (3 \, x\right )} + 6 i \, e^{\left (2 \, x\right )} + 3 \, e^{x} - 3 i\right )} \log \left (e^{x} - 1\right ) - 6 \, e^{\left (4 \, x\right )} + 6 i \, e^{\left (3 \, x\right )} + 10 \, e^{\left (2 \, x\right )} - 2 i \, e^{x} - 8}{2 \, e^{\left (5 \, x\right )} - 2 i \, e^{\left (4 \, x\right )} - 4 \, e^{\left (3 \, x\right )} + 4 i \, e^{\left (2 \, x\right )} + 2 \, e^{x} - 2 i} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(x)^4/(I+csch(x)),x, algorithm="fricas")

[Out]

((3*e^(5*x) - 3*I*e^(4*x) - 6*e^(3*x) + 6*I*e^(2*x) + 3*e^x - 3*I)*log(e^x + 1) - (3*e^(5*x) - 3*I*e^(4*x) - 6
*e^(3*x) + 6*I*e^(2*x) + 3*e^x - 3*I)*log(e^x - 1) - 6*e^(4*x) + 6*I*e^(3*x) + 10*e^(2*x) - 2*I*e^x - 8)/(2*e^
(5*x) - 2*I*e^(4*x) - 4*e^(3*x) + 4*I*e^(2*x) + 2*e^x - 2*I)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{csch}^{4}{\left (x \right )}}{\operatorname{csch}{\left (x \right )} + i}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(x)**4/(I+csch(x)),x)

[Out]

Integral(csch(x)**4/(csch(x) + I), x)

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Giac [A]  time = 1.16423, size = 68, normalized size = 1.84 \begin{align*} -\frac{e^{\left (3 \, x\right )} - 2 i \, e^{\left (2 \, x\right )} + e^{x} + 2 i}{{\left (e^{\left (2 \, x\right )} - 1\right )}^{2}} - \frac{2 i}{i \, e^{x} + 1} + \frac{3}{2} \, \log \left (e^{x} + 1\right ) - \frac{3}{2} \, \log \left ({\left | e^{x} - 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(x)^4/(I+csch(x)),x, algorithm="giac")

[Out]

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