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

Optimal. Leaf size=26 \[ -\coth (x)+i \tanh ^{-1}(\cosh (x))-\frac {i \coth (x)}{\text {csch}(x)+i} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {3790, 3789, 3770, 3794} \[ -\coth (x)+i \tanh ^{-1}(\cosh (x))-\frac {i \coth (x)}{\text {csch}(x)+i} \]

Antiderivative was successfully verified.

[In]

Int[Csch[x]^3/(I + Csch[x]),x]

[Out]

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

Rule 3770

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

Rule 3789

Int[csc[(e_.) + (f_.)*(x_)]^2/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[1/b, Int[Csc[e + f*x],
x], x] - Dist[a/b, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, x]

Rule 3790

Int[csc[(e_.) + (f_.)*(x_)]^3/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> -Simp[Cot[e + f*x]/(b*f), x
] - Dist[a/b, Int[Csc[e + f*x]^2/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, x]

Rule 3794

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

Rubi steps

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

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Mathematica [B]  time = 0.11, size = 70, normalized size = 2.69 \[ -\frac {1}{2} \tanh \left (\frac {x}{2}\right )-\frac {1}{2} \coth \left (\frac {x}{2}\right )-i \log \left (\sinh \left (\frac {x}{2}\right )\right )+i \log \left (\cosh \left (\frac {x}{2}\right )\right )-\frac {2 \sinh \left (\frac {x}{2}\right )}{\cosh \left (\frac {x}{2}\right )+i \sinh \left (\frac {x}{2}\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[Csch[x]^3/(I + Csch[x]),x]

[Out]

-1/2*Coth[x/2] + I*Log[Cosh[x/2]] - I*Log[Sinh[x/2]] - (2*Sinh[x/2])/(Cosh[x/2] + I*Sinh[x/2]) - Tanh[x/2]/2

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fricas [B]  time = 0.42, size = 77, normalized size = 2.96 \[ \frac {{\left (i \, e^{\left (3 \, x\right )} + e^{\left (2 \, x\right )} - i \, e^{x} - 1\right )} \log \left (e^{x} + 1\right ) + {\left (-i \, e^{\left (3 \, x\right )} - e^{\left (2 \, x\right )} + i \, e^{x} + 1\right )} \log \left (e^{x} - 1\right ) - 2 i \, e^{\left (2 \, x\right )} - 2 \, e^{x} + 4 i}{e^{\left (3 \, x\right )} - i \, e^{\left (2 \, x\right )} - e^{x} + i} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.12, size = 46, normalized size = 1.77 \[ \frac {2 \, {\left (e^{\left (2 \, x\right )} - i \, e^{x} - 2\right )}}{i \, e^{\left (3 \, x\right )} + e^{\left (2 \, x\right )} - i \, e^{x} - 1} + i \, \log \left (e^{x} + 1\right ) - i \, \log \left ({\left | e^{x} - 1 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.11, size = 35, normalized size = 1.35 \[ -\frac {\tanh \left (\frac {x}{2}\right )}{2}-\frac {2}{\tanh \left (\frac {x}{2}\right )-i}-\frac {1}{2 \tanh \left (\frac {x}{2}\right )}-i \ln \left (\tanh \left (\frac {x}{2}\right )\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*tanh(1/2*x)-2/(tanh(1/2*x)-I)-1/2/tanh(1/2*x)-I*ln(tanh(1/2*x))

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maxima [B]  time = 0.32, size = 55, normalized size = 2.12 \[ -\frac {8 \, {\left (e^{\left (-x\right )} - i \, e^{\left (-2 \, x\right )} + 2 i\right )}}{4 \, e^{\left (-x\right )} - 4 i \, e^{\left (-2 \, x\right )} - 4 \, e^{\left (-3 \, x\right )} + 4 i} + i \, \log \left (e^{\left (-x\right )} + 1\right ) - i \, \log \left (e^{\left (-x\right )} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.64, size = 60, normalized size = 2.31 \[ -\ln \left ({\mathrm {e}}^x\,2{}\mathrm {i}-2{}\mathrm {i}\right )\,1{}\mathrm {i}+\ln \left ({\mathrm {e}}^x\,2{}\mathrm {i}+2{}\mathrm {i}\right )\,1{}\mathrm {i}+\frac {{\mathrm {e}}^{2\,x}\,2{}\mathrm {i}+2\,{\mathrm {e}}^x-4{}\mathrm {i}}{{\mathrm {e}}^{2\,x}\,1{}\mathrm {i}-{\mathrm {e}}^{3\,x}+{\mathrm {e}}^x-\mathrm {i}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sinh(x)^3*(1/sinh(x) + 1i)),x)

[Out]

log(exp(x)*2i + 2i)*1i - log(exp(x)*2i - 2i)*1i + (exp(2*x)*2i + 2*exp(x) - 4i)/(exp(2*x)*1i - exp(3*x) + exp(
x) - 1i)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {csch}^{3}{\relax (x )}}{\operatorname {csch}{\relax (x )} + i}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(csch(x)**3/(csch(x) + I), x)

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