3.54 \(\int \frac{\text{csch}^3(x)}{(i+\sinh (x))^2} \, dx\)

Optimal. Leaf size=58 \[ \frac{16}{3} i \coth (x)-\frac{7}{2} \tanh ^{-1}(\cosh (x))+\frac{7}{2} \coth (x) \text{csch}(x)-\frac{8 i \coth (x) \text{csch}(x)}{3 (\sinh (x)+i)}+\frac{\coth (x) \text{csch}(x)}{3 (\sinh (x)+i)^2} \]

[Out]

(-7*ArcTanh[Cosh[x]])/2 + ((16*I)/3)*Coth[x] + (7*Coth[x]*Csch[x])/2 + (Coth[x]*Csch[x])/(3*(I + Sinh[x])^2) -
(((8*I)/3)*Coth[x]*Csch[x])/(I + Sinh[x])

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Rubi [A]  time = 0.139687, antiderivative size = 58, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.538, Rules used = {2766, 2978, 2748, 3768, 3770, 3767, 8} \[ \frac{16}{3} i \coth (x)-\frac{7}{2} \tanh ^{-1}(\cosh (x))+\frac{7}{2} \coth (x) \text{csch}(x)-\frac{8 i \coth (x) \text{csch}(x)}{3 (\sinh (x)+i)}+\frac{\coth (x) \text{csch}(x)}{3 (\sinh (x)+i)^2} \]

Antiderivative was successfully verified.

[In]

Int[Csch[x]^3/(I + Sinh[x])^2,x]

[Out]

(-7*ArcTanh[Cosh[x]])/2 + ((16*I)/3)*Coth[x] + (7*Coth[x]*Csch[x])/2 + (Coth[x]*Csch[x])/(3*(I + Sinh[x])^2) -
(((8*I)/3)*Coth[x]*Csch[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 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 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]

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 8

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

Rubi steps

\begin{align*} \int \frac{\text{csch}^3(x)}{(i+\sinh (x))^2} \, dx &=\frac{\coth (x) \text{csch}(x)}{3 (i+\sinh (x))^2}-\frac{1}{3} \int \frac{\text{csch}^3(x) (5 i-3 \sinh (x))}{i+\sinh (x)} \, dx\\ &=\frac{\coth (x) \text{csch}(x)}{3 (i+\sinh (x))^2}-\frac{8 i \coth (x) \text{csch}(x)}{3 (i+\sinh (x))}+\frac{1}{3} \int \text{csch}^3(x) (-21-16 i \sinh (x)) \, dx\\ &=\frac{\coth (x) \text{csch}(x)}{3 (i+\sinh (x))^2}-\frac{8 i \coth (x) \text{csch}(x)}{3 (i+\sinh (x))}-\frac{16}{3} i \int \text{csch}^2(x) \, dx-7 \int \text{csch}^3(x) \, dx\\ &=\frac{7}{2} \coth (x) \text{csch}(x)+\frac{\coth (x) \text{csch}(x)}{3 (i+\sinh (x))^2}-\frac{8 i \coth (x) \text{csch}(x)}{3 (i+\sinh (x))}+\frac{7}{2} \int \text{csch}(x) \, dx-\frac{16}{3} \operatorname{Subst}(\int 1 \, dx,x,-i \coth (x))\\ &=-\frac{7}{2} \tanh ^{-1}(\cosh (x))+\frac{16}{3} i \coth (x)+\frac{7}{2} \coth (x) \text{csch}(x)+\frac{\coth (x) \text{csch}(x)}{3 (i+\sinh (x))^2}-\frac{8 i \coth (x) \text{csch}(x)}{3 (i+\sinh (x))}\\ \end{align*}

Mathematica [B]  time = 0.311149, size = 131, normalized size = 2.26 \[ \frac{1}{24} \left (24 i \tanh \left (\frac{x}{2}\right )+24 i \coth \left (\frac{x}{2}\right )+3 \text{csch}^2\left (\frac{x}{2}\right )+3 \text{sech}^2\left (\frac{x}{2}\right )+84 \log \left (\tanh \left (\frac{x}{2}\right )\right )+\frac{160 i \sinh \left (\frac{x}{2}\right )}{\cosh \left (\frac{x}{2}\right )-i \sinh \left (\frac{x}{2}\right )}+\frac{8}{\left (\cosh \left (\frac{x}{2}\right )-i \sinh \left (\frac{x}{2}\right )\right )^2}+\frac{16 \sinh \left (\frac{x}{2}\right )}{\left (\sinh \left (\frac{x}{2}\right )+i \cosh \left (\frac{x}{2}\right )\right )^3}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Csch[x]^3/(I + Sinh[x])^2,x]

[Out]

((24*I)*Coth[x/2] + 3*Csch[x/2]^2 + 84*Log[Tanh[x/2]] + 3*Sech[x/2]^2 + 8/(Cosh[x/2] - I*Sinh[x/2])^2 + ((160*
I)*Sinh[x/2])/(Cosh[x/2] - I*Sinh[x/2]) + (16*Sinh[x/2])/(I*Cosh[x/2] + Sinh[x/2])^3 + (24*I)*Tanh[x/2])/24

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Maple [A]  time = 0.047, size = 76, normalized size = 1.3 \begin{align*} i\tanh \left ({\frac{x}{2}} \right ) -{\frac{1}{8} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}}+{i \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{-1}}+{\frac{1}{8} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{-2}}+{\frac{7}{2}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) }+{8\,i \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) ^{-1}}-{{\frac{4\,i}{3}} \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) ^{-3}}+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)^3/(I+sinh(x))^2,x)

[Out]

I*tanh(1/2*x)-1/8*tanh(1/2*x)^2+I/tanh(1/2*x)+1/8/tanh(1/2*x)^2+7/2*ln(tanh(1/2*x))+8*I/(tanh(1/2*x)+I)-4/3*I/
(tanh(1/2*x)+I)^3+2/(tanh(1/2*x)+I)^2

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Maxima [B]  time = 1.23323, size = 142, normalized size = 2.45 \begin{align*} -\frac{8 \,{\left (-75 i \, e^{\left (-x\right )} + 97 \, e^{\left (-2 \, x\right )} + 126 i \, e^{\left (-3 \, x\right )} - 98 \, e^{\left (-4 \, x\right )} - 63 i \, e^{\left (-5 \, x\right )} + 21 \, e^{\left (-6 \, x\right )} - 32\right )}}{72 \, e^{\left (-x\right )} + 120 i \, e^{\left (-2 \, x\right )} - 168 \, e^{\left (-3 \, x\right )} - 168 i \, e^{\left (-4 \, x\right )} + 120 \, e^{\left (-5 \, x\right )} + 72 i \, e^{\left (-6 \, x\right )} - 24 \, e^{\left (-7 \, x\right )} - 24 i} - \frac{7}{2} \, \log \left (e^{\left (-x\right )} + 1\right ) + \frac{7}{2} \, \log \left (e^{\left (-x\right )} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(x)^3/(I+sinh(x))^2,x, algorithm="maxima")

[Out]

-8*(-75*I*e^(-x) + 97*e^(-2*x) + 126*I*e^(-3*x) - 98*e^(-4*x) - 63*I*e^(-5*x) + 21*e^(-6*x) - 32)/(72*e^(-x) +
120*I*e^(-2*x) - 168*e^(-3*x) - 168*I*e^(-4*x) + 120*e^(-5*x) + 72*I*e^(-6*x) - 24*e^(-7*x) - 24*I) - 7/2*log
(e^(-x) + 1) + 7/2*log(e^(-x) - 1)

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Fricas [B]  time = 2.10982, size = 578, normalized size = 9.97 \begin{align*} -\frac{{\left (21 \, e^{\left (7 \, x\right )} + 63 i \, e^{\left (6 \, x\right )} - 105 \, e^{\left (5 \, x\right )} - 147 i \, e^{\left (4 \, x\right )} + 147 \, e^{\left (3 \, x\right )} + 105 i \, e^{\left (2 \, x\right )} - 63 \, e^{x} - 21 i\right )} \log \left (e^{x} + 1\right ) -{\left (21 \, e^{\left (7 \, x\right )} + 63 i \, e^{\left (6 \, x\right )} - 105 \, e^{\left (5 \, x\right )} - 147 i \, e^{\left (4 \, x\right )} + 147 \, e^{\left (3 \, x\right )} + 105 i \, e^{\left (2 \, x\right )} - 63 \, e^{x} - 21 i\right )} \log \left (e^{x} - 1\right ) - 42 \, e^{\left (6 \, x\right )} - 126 i \, e^{\left (5 \, x\right )} + 196 \, e^{\left (4 \, x\right )} + 252 i \, e^{\left (3 \, x\right )} - 194 \, e^{\left (2 \, x\right )} - 150 i \, e^{x} + 64}{6 \, e^{\left (7 \, x\right )} + 18 i \, e^{\left (6 \, x\right )} - 30 \, e^{\left (5 \, x\right )} - 42 i \, e^{\left (4 \, x\right )} + 42 \, e^{\left (3 \, x\right )} + 30 i \, e^{\left (2 \, x\right )} - 18 \, e^{x} - 6 i} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(x)^3/(I+sinh(x))^2,x, algorithm="fricas")

[Out]

-((21*e^(7*x) + 63*I*e^(6*x) - 105*e^(5*x) - 147*I*e^(4*x) + 147*e^(3*x) + 105*I*e^(2*x) - 63*e^x - 21*I)*log(
e^x + 1) - (21*e^(7*x) + 63*I*e^(6*x) - 105*e^(5*x) - 147*I*e^(4*x) + 147*e^(3*x) + 105*I*e^(2*x) - 63*e^x - 2
1*I)*log(e^x - 1) - 42*e^(6*x) - 126*I*e^(5*x) + 196*e^(4*x) + 252*I*e^(3*x) - 194*e^(2*x) - 150*I*e^x + 64)/(
6*e^(7*x) + 18*I*e^(6*x) - 30*e^(5*x) - 42*I*e^(4*x) + 42*e^(3*x) + 30*I*e^(2*x) - 18*e^x - 6*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)**3/(I+sinh(x))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.36617, size = 80, normalized size = 1.38 \begin{align*} \frac{e^{\left (3 \, x\right )} + 4 i \, e^{\left (2 \, x\right )} + e^{x} - 4 i}{{\left (e^{\left (2 \, x\right )} - 1\right )}^{2}} + \frac{2 \,{\left (9 \, e^{\left (2 \, x\right )} + 21 i \, e^{x} - 10\right )}}{3 \,{\left (e^{x} + i\right )}^{3}} - \frac{7}{2} \, \log \left (e^{x} + 1\right ) + \frac{7}{2} \, \log \left ({\left | e^{x} - 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(x)^3/(I+sinh(x))^2,x, algorithm="giac")

[Out]

(e^(3*x) + 4*I*e^(2*x) + e^x - 4*I)/(e^(2*x) - 1)^2 + 2/3*(9*e^(2*x) + 21*I*e^x - 10)/(e^x + I)^3 - 7/2*log(e^
x + 1) + 7/2*log(abs(e^x - 1))