3.214 \(\int \frac {\coth ^3(x)}{i+\sinh (x)} \, dx\)

Optimal. Leaf size=15 \[ -\text {csch}(x)+\frac {1}{2} i \text {csch}^2(x) \]

[Out]

-csch(x)+1/2*I*csch(x)^2

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Rubi [A]  time = 0.06, antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {2706, 2606, 30, 8} \[ -\text {csch}(x)+\frac {1}{2} i \text {csch}^2(x) \]

Antiderivative was successfully verified.

[In]

Int[Coth[x]^3/(I + Sinh[x]),x]

[Out]

-Csch[x] + (I/2)*Csch[x]^2

Rule 8

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2606

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rule 2706

Int[((g_.)*tan[(e_.) + (f_.)*(x_)])^(p_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[1/a, Int[S
ec[e + f*x]^2*(g*Tan[e + f*x])^p, x], x] - Dist[1/(b*g), Int[Sec[e + f*x]*(g*Tan[e + f*x])^(p + 1), x], x] /;
FreeQ[{a, b, e, f, g, p}, x] && EqQ[a^2 - b^2, 0] && NeQ[p, -1]

Rubi steps

\begin {align*} \int \frac {\coth ^3(x)}{i+\sinh (x)} \, dx &=-\left (i \int \coth (x) \text {csch}^2(x) \, dx\right )+\int \coth (x) \text {csch}(x) \, dx\\ &=-(i \operatorname {Subst}(\int 1 \, dx,x,-i \text {csch}(x)))-i \operatorname {Subst}(\int x \, dx,x,-i \text {csch}(x))\\ &=-\text {csch}(x)+\frac {1}{2} i \text {csch}^2(x)\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 15, normalized size = 1.00 \[ -\text {csch}(x)+\frac {1}{2} i \text {csch}^2(x) \]

Antiderivative was successfully verified.

[In]

Integrate[Coth[x]^3/(I + Sinh[x]),x]

[Out]

-Csch[x] + (I/2)*Csch[x]^2

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fricas [B]  time = 1.02, size = 33, normalized size = 2.20 \[ -\frac {2 \, e^{\left (3 \, x\right )} - 2 i \, e^{\left (2 \, x\right )} - 2 \, e^{x}}{e^{\left (4 \, x\right )} - 2 \, e^{\left (2 \, x\right )} + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.16, size = 24, normalized size = 1.60 \[ \frac {2 \, e^{\left (-x\right )} - 2 \, e^{x} + 2 i}{{\left (e^{\left (-x\right )} - e^{x}\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*e^(-x) - 2*e^x + 2*I)/(e^(-x) - e^x)^2

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maple [B]  time = 0.07, size = 34, normalized size = 2.27 \[ \frac {\tanh \left (\frac {x}{2}\right )}{2}+\frac {i \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )}{8}-\frac {1}{2 \tanh \left (\frac {x}{2}\right )}+\frac {i}{8 \tanh \left (\frac {x}{2}\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.31, size = 67, normalized size = 4.47 \[ \frac {2 \, e^{\left (-x\right )}}{2 \, e^{\left (-2 \, x\right )} - e^{\left (-4 \, x\right )} - 1} - \frac {2 i \, e^{\left (-2 \, x\right )}}{2 \, e^{\left (-2 \, x\right )} - e^{\left (-4 \, x\right )} - 1} - \frac {2 \, e^{\left (-3 \, x\right )}}{2 \, e^{\left (-2 \, x\right )} - e^{\left (-4 \, x\right )} - 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.57, size = 25, normalized size = 1.67 \[ \frac {2\,{\mathrm {e}}^x\,\left (1-{\mathrm {e}}^{2\,x}+{\mathrm {e}}^x\,1{}\mathrm {i}\right )}{{\left ({\mathrm {e}}^{2\,x}-1\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(x)^3/(sinh(x) + 1i),x)

[Out]

(2*exp(x)*(exp(x)*1i - exp(2*x) + 1))/(exp(2*x) - 1)^2

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sympy [B]  time = 0.16, size = 32, normalized size = 2.13 \[ \frac {- 2 e^{3 x} + 2 i e^{2 x} + 2 e^{x}}{e^{4 x} - 2 e^{2 x} + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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