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

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

[Out]

-3/2*I*arctanh(cosh(x))-2*coth(x)+3/2*I*coth(x)*csch(x)+coth(x)*csch(x)/(I+sinh(x))

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Rubi [A]  time = 0.07, antiderivative size = 37, normalized size of antiderivative = 1.00, 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 = {2768, 2748, 3768, 3770, 3767, 8} \[ -2 \coth (x)-\frac {3}{2} i \tanh ^{-1}(\cosh (x))+\frac {3}{2} i \coth (x) \text {csch}(x)+\frac {\coth (x) \text {csch}(x)}{\sinh (x)+i} \]

Antiderivative was successfully verified.

[In]

Int[Csch[x]^3/(I + Sinh[x]),x]

[Out]

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

Rule 8

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

Rule 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 2768

Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(b
^2*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/(a*f*(b*c - a*d)*(a + b*Sin[e + f*x])), x] + Dist[d/(a*(b*c - a*
d)), Int[(c + d*Sin[e + f*x])^n*(a*n - b*(n + 1)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[
b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[n, 0] && (IntegerQ[2*n] || EqQ[c, 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 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}^3(x)}{i+\sinh (x)} \, dx &=\frac {\coth (x) \text {csch}(x)}{i+\sinh (x)}+\int \text {csch}^3(x) (-3 i+2 \sinh (x)) \, dx\\ &=\frac {\coth (x) \text {csch}(x)}{i+\sinh (x)}-3 i \int \text {csch}^3(x) \, dx+2 \int \text {csch}^2(x) \, dx\\ &=\frac {3}{2} i \coth (x) \text {csch}(x)+\frac {\coth (x) \text {csch}(x)}{i+\sinh (x)}+\frac {3}{2} i \int \text {csch}(x) \, dx-2 i \operatorname {Subst}(\int 1 \, dx,x,-i \coth (x))\\ &=-\frac {3}{2} i \tanh ^{-1}(\cosh (x))-2 \coth (x)+\frac {3}{2} i \coth (x) \text {csch}(x)+\frac {\coth (x) \text {csch}(x)}{i+\sinh (x)}\\ \end {align*}

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Mathematica [A]  time = 0.19, size = 49, normalized size = 1.32 \[ \frac {1}{2} i \tanh (x) \left (\text {csch}^3(x)+2 i \text {csch}^2(x)+3 \text {csch}(x)-3 \sqrt {\cosh ^2(x)} \text {csch}(x) \tanh ^{-1}\left (\sqrt {\cosh ^2(x)}\right )+4 i\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Csch[x]^3/(I + Sinh[x]),x]

[Out]

(I/2)*(4*I + 3*Csch[x] - 3*ArcTanh[Sqrt[Cosh[x]^2]]*Sqrt[Cosh[x]^2]*Csch[x] + (2*I)*Csch[x]^2 + Csch[x]^3)*Tan
h[x]

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fricas [B]  time = 0.73, size = 129, normalized size = 3.49 \[ \frac {{\left (-3 i \, e^{\left (5 \, x\right )} + 3 \, e^{\left (4 \, x\right )} + 6 i \, e^{\left (3 \, x\right )} - 6 \, e^{\left (2 \, x\right )} - 3 i \, e^{x} + 3\right )} \log \left (e^{x} + 1\right ) + {\left (3 i \, e^{\left (5 \, x\right )} - 3 \, e^{\left (4 \, x\right )} - 6 i \, e^{\left (3 \, x\right )} + 6 \, e^{\left (2 \, x\right )} + 3 i \, e^{x} - 3\right )} \log \left (e^{x} - 1\right ) + 6 i \, e^{\left (4 \, x\right )} - 6 \, e^{\left (3 \, x\right )} - 10 i \, e^{\left (2 \, x\right )} + 2 \, e^{x} + 8 i}{2 \, e^{\left (5 \, x\right )} + 2 i \, e^{\left (4 \, x\right )} - 4 \, e^{\left (3 \, x\right )} - 4 i \, e^{\left (2 \, x\right )} + 2 \, e^{x} + 2 i} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((-3*I*e^(5*x) + 3*e^(4*x) + 6*I*e^(3*x) - 6*e^(2*x) - 3*I*e^x + 3)*log(e^x + 1) + (3*I*e^(5*x) - 3*e^(4*x) -
6*I*e^(3*x) + 6*e^(2*x) + 3*I*e^x - 3)*log(e^x - 1) + 6*I*e^(4*x) - 6*e^(3*x) - 10*I*e^(2*x) + 2*e^x + 8*I)/(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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giac [A]  time = 0.19, size = 51, normalized size = 1.38 \[ \frac {i \, e^{\left (3 \, x\right )} - 2 \, e^{\left (2 \, x\right )} + i \, e^{x} + 2}{{\left (e^{\left (2 \, x\right )} - 1\right )}^{2}} + \frac {2 i}{e^{x} + i} - \frac {3}{2} i \, \log \left (e^{x} + 1\right ) + \frac {3}{2} i \, \log \left ({\left | e^{x} - 1 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 53, normalized size = 1.43 \[ -\frac {\tanh \left (\frac {x}{2}\right )}{2}-\frac {i \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )}{8}-\frac {2}{\tanh \left (\frac {x}{2}\right )+i}+\frac {i}{8 \tanh \left (\frac {x}{2}\right )^{2}}+\frac {3 i \ln \left (\tanh \left (\frac {x}{2}\right )\right )}{2}-\frac {1}{2 \tanh \left (\frac {x}{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.32, size = 79, normalized size = 2.14 \[ -\frac {8 \, {\left (e^{\left (-x\right )} + 5 i \, e^{\left (-2 \, x\right )} - 3 \, e^{\left (-3 \, x\right )} - 3 i \, e^{\left (-4 \, x\right )} - 4 i\right )}}{8 \, e^{\left (-x\right )} + 16 i \, e^{\left (-2 \, x\right )} - 16 \, e^{\left (-3 \, x\right )} - 8 i \, e^{\left (-4 \, x\right )} + 8 \, e^{\left (-5 \, x\right )} - 8 i} - \frac {3}{2} i \, \log \left (e^{\left (-x\right )} + 1\right ) + \frac {3}{2} i \, \log \left (e^{\left (-x\right )} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-8*(e^(-x) + 5*I*e^(-2*x) - 3*e^(-3*x) - 3*I*e^(-4*x) - 4*I)/(8*e^(-x) + 16*I*e^(-2*x) - 16*e^(-3*x) - 8*I*e^(
-4*x) + 8*e^(-5*x) - 8*I) - 3/2*I*log(e^(-x) + 1) + 3/2*I*log(e^(-x) - 1)

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mupad [B]  time = 0.66, size = 70, normalized size = 1.89 \[ -\frac {\ln \left (-{\mathrm {e}}^x\,3{}\mathrm {i}-3{}\mathrm {i}\right )\,3{}\mathrm {i}}{2}+\frac {\ln \left (-{\mathrm {e}}^x\,3{}\mathrm {i}+3{}\mathrm {i}\right )\,3{}\mathrm {i}}{2}+\frac {2{}\mathrm {i}}{{\mathrm {e}}^x+1{}\mathrm {i}}+\frac {{\mathrm {e}}^x\,2{}\mathrm {i}}{{\mathrm {e}}^{4\,x}-2\,{\mathrm {e}}^{2\,x}+1}+\frac {-2+{\mathrm {e}}^x\,1{}\mathrm {i}}{{\mathrm {e}}^{2\,x}-1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sinh(x)^3*(sinh(x) + 1i)),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(csch(x)**3/(sinh(x) + I), x)

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