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

Optimal. Leaf size=27 \[ -i x-\frac {1}{2} \tanh ^{-1}(\cosh (x))+\frac {1}{2} \coth (x) (-\text {csch}(x)+2 i) \]

[Out]

-I*x-1/2*arctanh(cosh(x))+1/2*coth(x)*(2*I-csch(x))

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Rubi [A]  time = 0.06, antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {3888, 3881, 3770} \[ -i x-\frac {1}{2} \tanh ^{-1}(\cosh (x))+\frac {1}{2} \coth (x) (-\text {csch}(x)+2 i) \]

Antiderivative was successfully verified.

[In]

Int[Coth[x]^4/(I + Csch[x]),x]

[Out]

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

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3881

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> -Simp[(e*(e*Cot[
c + d*x])^(m - 1)*(a*m + b*(m - 1)*Csc[c + d*x]))/(d*m*(m - 1)), x] - Dist[e^2/m, Int[(e*Cot[c + d*x])^(m - 2)
*(a*m + b*(m - 1)*Csc[c + d*x]), x], x] /; FreeQ[{a, b, c, d, e}, x] && GtQ[m, 1]

Rule 3888

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Dist[a^(2*n
)/e^(2*n), Int[(e*Cot[c + d*x])^(m + 2*n)/(-a + b*Csc[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && E
qQ[a^2 - b^2, 0] && ILtQ[n, 0]

Rubi steps

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

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Mathematica [B]  time = 0.04, size = 65, normalized size = 2.41 \[ -i x+\frac {1}{2} i \tanh \left (\frac {x}{2}\right )+\frac {1}{2} i \coth \left (\frac {x}{2}\right )-\frac {1}{8} \text {csch}^2\left (\frac {x}{2}\right )-\frac {1}{8} \text {sech}^2\left (\frac {x}{2}\right )+\frac {1}{2} \log \left (\tanh \left (\frac {x}{2}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Coth[x]^4/(I + Csch[x]),x]

[Out]

(-I)*x + (I/2)*Coth[x/2] - Csch[x/2]^2/8 + Log[Tanh[x/2]]/2 - Sech[x/2]^2/8 + (I/2)*Tanh[x/2]

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fricas [B]  time = 1.06, size = 85, normalized size = 3.15 \[ \frac {-2 i \, x e^{\left (4 \, x\right )} + {\left (4 i \, x + 4 i\right )} e^{\left (2 \, x\right )} - {\left (e^{\left (4 \, x\right )} - 2 \, e^{\left (2 \, x\right )} + 1\right )} \log \left (e^{x} + 1\right ) + {\left (e^{\left (4 \, x\right )} - 2 \, e^{\left (2 \, x\right )} + 1\right )} \log \left (e^{x} - 1\right ) - 2 i \, x - 2 \, e^{\left (3 \, x\right )} - 2 \, e^{x} - 4 i}{2 \, {\left (e^{\left (4 \, x\right )} - 2 \, e^{\left (2 \, x\right )} + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.13, size = 48, normalized size = 1.78 \[ \frac {e^{\left (3 \, x\right )} - 2 i \, e^{\left (2 \, x\right )} + e^{x} + 2 i}{{\left (i \, e^{\left (2 \, x\right )} - i\right )}^{2}} - i \, \log \left (-i \, e^{x}\right ) - \frac {1}{2} \, \log \left (e^{x} + 1\right ) + \frac {1}{2} \, \log \left ({\left | e^{x} - 1 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.18, size = 61, normalized size = 2.26 \[ \frac {i \tanh \left (\frac {x}{2}\right )}{2}+\frac {\left (\tanh ^{2}\left (\frac {x}{2}\right )\right )}{8}+i \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )-i \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )-\frac {1}{8 \tanh \left (\frac {x}{2}\right )^{2}}+\frac {i}{2 \tanh \left (\frac {x}{2}\right )}+\frac {\ln \left (\tanh \left (\frac {x}{2}\right )\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.31, size = 55, normalized size = 2.04 \[ -i \, x + \frac {e^{\left (-x\right )} + 2 i \, e^{\left (-2 \, x\right )} + e^{\left (-3 \, x\right )} - 2 i}{2 \, e^{\left (-2 \, x\right )} - e^{\left (-4 \, x\right )} - 1} - \frac {1}{2} \, \log \left (e^{\left (-x\right )} + 1\right ) + \frac {1}{2} \, \log \left (e^{\left (-x\right )} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.64, size = 56, normalized size = 2.07 \[ \frac {\ln \left (1-{\mathrm {e}}^x\right )}{2}-\frac {\ln \left (-{\mathrm {e}}^x-1\right )}{2}-\frac {{\mathrm {e}}^x-2{}\mathrm {i}}{{\mathrm {e}}^{2\,x}-1}-\frac {2\,{\mathrm {e}}^x}{{\mathrm {e}}^{4\,x}-2\,{\mathrm {e}}^{2\,x}+1}-x\,1{}\mathrm {i} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(x)^4/(1/sinh(x) + 1i),x)

[Out]

log(1 - exp(x))/2 - log(- exp(x) - 1)/2 - x*1i - (exp(x) - 2i)/(exp(2*x) - 1) - (2*exp(x))/(exp(4*x) - 2*exp(2
*x) + 1)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\coth ^{4}{\relax (x )}}{\operatorname {csch}{\relax (x )} + i}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(coth(x)**4/(csch(x) + I), x)

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