3.52 \(\int \frac{\text{csch}(x)}{(i+\sinh (x))^2} \, dx\)

Optimal. Leaf size=34 \[ -\frac{4 i \cosh (x)}{3 (\sinh (x)+i)}+\frac{\cosh (x)}{3 (\sinh (x)+i)^2}+\tanh ^{-1}(\cosh (x)) \]

[Out]

ArcTanh[Cosh[x]] + Cosh[x]/(3*(I + Sinh[x])^2) - (((4*I)/3)*Cosh[x])/(I + Sinh[x])

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Rubi [A]  time = 0.0826279, antiderivative size = 34, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.364, Rules used = {2766, 2978, 12, 3770} \[ -\frac{4 i \cosh (x)}{3 (\sinh (x)+i)}+\frac{\cosh (x)}{3 (\sinh (x)+i)^2}+\tanh ^{-1}(\cosh (x)) \]

Antiderivative was successfully verified.

[In]

Int[Csch[x]/(I + Sinh[x])^2,x]

[Out]

ArcTanh[Cosh[x]] + Cosh[x]/(3*(I + Sinh[x])^2) - (((4*I)/3)*Cosh[x])/(I + Sinh[x])

Rule 2766

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1))/(a*f*(2*m + 1)*(b*c - a*d)), x] + Dis
t[1/(a*(2*m + 1)*(b*c - a*d)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[b*c*(m + 1) - a*d*
(2*m + n + 2) + b*d*(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d,
0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] &&  !GtQ[n, 0] && (IntegersQ[2*m, 2*n] || (IntegerQ
[m] && EqQ[c, 0]))

Rule 2978

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*
x])^(n + 1))/(a*f*(2*m + 1)*(b*c - a*d)), x] + Dist[1/(a*(2*m + 1)*(b*c - a*d)), Int[(a + b*Sin[e + f*x])^(m +
1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b*d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*
(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2
- b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)] &&  !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c,
0])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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

Mathematica [B]  time = 0.0867691, size = 91, normalized size = 2.68 \[ \frac{\cosh \left (\frac{x}{2}\right ) \left (6-9 \log \left (\tanh \left (\frac{x}{2}\right )\right )\right )+\cosh \left (\frac{3 x}{2}\right ) \left (3 \log \left (\tanh \left (\frac{x}{2}\right )\right )-8\right )+6 i \sinh \left (\frac{x}{2}\right ) \left (2 \log \left (\tanh \left (\frac{x}{2}\right )\right )+\cosh (x) \log \left (\tanh \left (\frac{x}{2}\right )\right )-3\right )}{6 \left (\cosh \left (\frac{x}{2}\right )-i \sinh \left (\frac{x}{2}\right )\right )^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Csch[x]/(I + Sinh[x])^2,x]

[Out]

(Cosh[x/2]*(6 - 9*Log[Tanh[x/2]]) + Cosh[(3*x)/2]*(-8 + 3*Log[Tanh[x/2]]) + (6*I)*(-3 + 2*Log[Tanh[x/2]] + Cos
h[x]*Log[Tanh[x/2]])*Sinh[x/2])/(6*(Cosh[x/2] - I*Sinh[x/2])^3)

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Maple [A]  time = 0.039, size = 44, normalized size = 1.3 \begin{align*} -\ln \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) +{{\frac{4\,i}{3}} \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) ^{-3}}-{4\,i \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) ^{-1}}-2\, \left ( \tanh \left ( x/2 \right ) +i \right ) ^{-2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csch(x)/(I+sinh(x))^2,x)

[Out]

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

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Maxima [B]  time = 1.226, size = 74, normalized size = 2.18 \begin{align*} \frac{2 \,{\left (-9 i \, e^{\left (-x\right )} + 3 \, e^{\left (-2 \, x\right )} - 4\right )}}{9 \, e^{\left (-x\right )} + 9 i \, e^{\left (-2 \, x\right )} - 3 \, e^{\left (-3 \, x\right )} - 3 i} + \log \left (e^{\left (-x\right )} + 1\right ) - \log \left (e^{\left (-x\right )} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(x)/(I+sinh(x))^2,x, algorithm="maxima")

[Out]

2*(-9*I*e^(-x) + 3*e^(-2*x) - 4)/(9*e^(-x) + 9*I*e^(-2*x) - 3*e^(-3*x) - 3*I) + log(e^(-x) + 1) - log(e^(-x) -
1)

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Fricas [B]  time = 2.09156, size = 240, normalized size = 7.06 \begin{align*} \frac{{\left (3 \, e^{\left (3 \, x\right )} + 9 i \, e^{\left (2 \, x\right )} - 9 \, e^{x} - 3 i\right )} \log \left (e^{x} + 1\right ) -{\left (3 \, e^{\left (3 \, x\right )} + 9 i \, e^{\left (2 \, x\right )} - 9 \, e^{x} - 3 i\right )} \log \left (e^{x} - 1\right ) - 6 \, e^{\left (2 \, x\right )} - 18 i \, e^{x} + 8}{3 \, e^{\left (3 \, x\right )} + 9 i \, e^{\left (2 \, x\right )} - 9 \, e^{x} - 3 i} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(x)/(I+sinh(x))^2,x, algorithm="fricas")

[Out]

((3*e^(3*x) + 9*I*e^(2*x) - 9*e^x - 3*I)*log(e^x + 1) - (3*e^(3*x) + 9*I*e^(2*x) - 9*e^x - 3*I)*log(e^x - 1) -
6*e^(2*x) - 18*I*e^x + 8)/(3*e^(3*x) + 9*I*e^(2*x) - 9*e^x - 3*I)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(x)/(I+sinh(x))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.3646, size = 46, normalized size = 1.35 \begin{align*} -\frac{2 \,{\left (3 \, e^{\left (2 \, x\right )} + 9 i \, e^{x} - 4\right )}}{3 \,{\left (e^{x} + i\right )}^{3}} + \log \left (e^{x} + 1\right ) - \log \left ({\left | e^{x} - 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(x)/(I+sinh(x))^2,x, algorithm="giac")

[Out]

-2/3*(3*e^(2*x) + 9*I*e^x - 4)/(e^x + I)^3 + log(e^x + 1) - log(abs(e^x - 1))