3.55 \(\int \frac{\text{csch}^4(x)}{(i+\sinh (x))^2} \, dx\)

Optimal. Leaf size=64 \[ 4 \coth ^3(x)-12 \coth (x)-5 i \tanh ^{-1}(\cosh (x))+5 i \coth (x) \text{csch}(x)-\frac{10 i \coth (x) \text{csch}^2(x)}{3 (\sinh (x)+i)}+\frac{\coth (x) \text{csch}^2(x)}{3 (\sinh (x)+i)^2} \]

[Out]

(-5*I)*ArcTanh[Cosh[x]] - 12*Coth[x] + 4*Coth[x]^3 + (5*I)*Coth[x]*Csch[x] + (Coth[x]*Csch[x]^2)/(3*(I + Sinh[
x])^2) - (((10*I)/3)*Coth[x]*Csch[x]^2)/(I + Sinh[x])

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Rubi [A]  time = 0.130932, antiderivative size = 64, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.462, Rules used = {2766, 2978, 2748, 3767, 3768, 3770} \[ 4 \coth ^3(x)-12 \coth (x)-5 i \tanh ^{-1}(\cosh (x))+5 i \coth (x) \text{csch}(x)-\frac{10 i \coth (x) \text{csch}^2(x)}{3 (\sinh (x)+i)}+\frac{\coth (x) \text{csch}^2(x)}{3 (\sinh (x)+i)^2} \]

Antiderivative was successfully verified.

[In]

Int[Csch[x]^4/(I + Sinh[x])^2,x]

[Out]

(-5*I)*ArcTanh[Cosh[x]] - 12*Coth[x] + 4*Coth[x]^3 + (5*I)*Coth[x]*Csch[x] + (Coth[x]*Csch[x]^2)/(3*(I + Sinh[
x])^2) - (((10*I)/3)*Coth[x]*Csch[x]^2)/(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 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}^4(x)}{(i+\sinh (x))^2} \, dx &=\frac{\coth (x) \text{csch}^2(x)}{3 (i+\sinh (x))^2}-\frac{1}{3} \int \frac{\text{csch}^4(x) (6 i-4 \sinh (x))}{i+\sinh (x)} \, dx\\ &=\frac{\coth (x) \text{csch}^2(x)}{3 (i+\sinh (x))^2}-\frac{10 i \coth (x) \text{csch}^2(x)}{3 (i+\sinh (x))}+\frac{1}{3} \int \text{csch}^4(x) (-36-30 i \sinh (x)) \, dx\\ &=\frac{\coth (x) \text{csch}^2(x)}{3 (i+\sinh (x))^2}-\frac{10 i \coth (x) \text{csch}^2(x)}{3 (i+\sinh (x))}-10 i \int \text{csch}^3(x) \, dx-12 \int \text{csch}^4(x) \, dx\\ &=5 i \coth (x) \text{csch}(x)+\frac{\coth (x) \text{csch}^2(x)}{3 (i+\sinh (x))^2}-\frac{10 i \coth (x) \text{csch}^2(x)}{3 (i+\sinh (x))}+5 i \int \text{csch}(x) \, dx-12 i \operatorname{Subst}\left (\int \left (1+x^2\right ) \, dx,x,-i \coth (x)\right )\\ &=-5 i \tanh ^{-1}(\cosh (x))-12 \coth (x)+4 \coth ^3(x)+5 i \coth (x) \text{csch}(x)+\frac{\coth (x) \text{csch}^2(x)}{3 (i+\sinh (x))^2}-\frac{10 i \coth (x) \text{csch}^2(x)}{3 (i+\sinh (x))}\\ \end{align*}

Mathematica [B]  time = 1.5293, size = 143, normalized size = 2.23 \[ \frac{1}{24} \left (-44 \coth \left (\frac{x}{2}\right )+6 i \text{csch}^2\left (\frac{x}{2}\right )+\frac{1}{2} \sinh (x) \text{csch}^4\left (\frac{x}{2}\right )+2 \left (-\frac{4}{\sinh (x)+i}-22 \tanh \left (\frac{x}{2}\right )+3 i \text{sech}^2\left (\frac{x}{2}\right )+60 i \log \left (\sinh \left (\frac{x}{2}\right )\right )-60 i \log \left (\cosh \left (\frac{x}{2}\right )\right )+\frac{8 (13 \sinh (x)+14 i) \sinh \left (\frac{x}{2}\right )}{\left (\sinh \left (\frac{x}{2}\right )+i \cosh \left (\frac{x}{2}\right )\right )^3}-4 \sinh ^4\left (\frac{x}{2}\right ) \text{csch}^3(x)\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Csch[x]^4/(I + Sinh[x])^2,x]

[Out]

(-44*Coth[x/2] + (6*I)*Csch[x/2]^2 + (Csch[x/2]^4*Sinh[x])/2 + 2*((-60*I)*Log[Cosh[x/2]] + (60*I)*Log[Sinh[x/2
]] + (3*I)*Sech[x/2]^2 - 4*Csch[x]^3*Sinh[x/2]^4 - 4/(I + Sinh[x]) + (8*Sinh[x/2]*(14*I + 13*Sinh[x]))/(I*Cosh
[x/2] + Sinh[x/2])^3 - 22*Tanh[x/2]))/24

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Maple [A]  time = 0.046, size = 92, normalized size = 1.4 \begin{align*} -{\frac{15}{8}\tanh \left ({\frac{x}{2}} \right ) }+{\frac{1}{24} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{3}}-{\frac{i}{4}} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+{{\frac{i}{4}} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{-2}}+5\,i\ln \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) +{\frac{1}{24} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{-3}}-{\frac{15}{8} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{-1}}+{2\,i \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) ^{-2}}+{\frac{4}{3} \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) ^{-3}}-10\, \left ( \tanh \left ( x/2 \right ) +i \right ) ^{-1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csch(x)^4/(I+sinh(x))^2,x)

[Out]

-15/8*tanh(1/2*x)+1/24*tanh(1/2*x)^3-1/4*I*tanh(1/2*x)^2+1/4*I/tanh(1/2*x)^2+5*I*ln(tanh(1/2*x))+1/24/tanh(1/2
*x)^3-15/8/tanh(1/2*x)+2*I/(tanh(1/2*x)+I)^2+4/3/(tanh(1/2*x)+I)^3-10/(tanh(1/2*x)+I)

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Maxima [B]  time = 1.27862, size = 174, normalized size = 2.72 \begin{align*} -\frac{16 \,{\left (57 \, e^{\left (-x\right )} + 99 i \, e^{\left (-2 \, x\right )} - 155 \, e^{\left (-3 \, x\right )} - 153 i \, e^{\left (-4 \, x\right )} + 135 \, e^{\left (-5 \, x\right )} + 85 i \, e^{\left (-6 \, x\right )} - 45 \, e^{\left (-7 \, x\right )} - 15 i \, e^{\left (-8 \, x\right )} - 24 i\right )}}{72 \, e^{\left (-x\right )} + 144 i \, e^{\left (-2 \, x\right )} - 240 \, e^{\left (-3 \, x\right )} - 288 i \, e^{\left (-4 \, x\right )} + 288 \, e^{\left (-5 \, x\right )} + 240 i \, e^{\left (-6 \, x\right )} - 144 \, e^{\left (-7 \, x\right )} - 72 i \, e^{\left (-8 \, x\right )} + 24 \, e^{\left (-9 \, x\right )} - 24 i} - 5 i \, \log \left (e^{\left (-x\right )} + 1\right ) + 5 i \, \log \left (e^{\left (-x\right )} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(x)^4/(I+sinh(x))^2,x, algorithm="maxima")

[Out]

-16*(57*e^(-x) + 99*I*e^(-2*x) - 155*e^(-3*x) - 153*I*e^(-4*x) + 135*e^(-5*x) + 85*I*e^(-6*x) - 45*e^(-7*x) -
15*I*e^(-8*x) - 24*I)/(72*e^(-x) + 144*I*e^(-2*x) - 240*e^(-3*x) - 288*I*e^(-4*x) + 288*e^(-5*x) + 240*I*e^(-6
*x) - 144*e^(-7*x) - 72*I*e^(-8*x) + 24*e^(-9*x) - 24*I) - 5*I*log(e^(-x) + 1) + 5*I*log(e^(-x) - 1)

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Fricas [B]  time = 2.1101, size = 730, normalized size = 11.41 \begin{align*} \frac{{\left (-15 i \, e^{\left (9 \, x\right )} + 45 \, e^{\left (8 \, x\right )} + 90 i \, e^{\left (7 \, x\right )} - 150 \, e^{\left (6 \, x\right )} - 180 i \, e^{\left (5 \, x\right )} + 180 \, e^{\left (4 \, x\right )} + 150 i \, e^{\left (3 \, x\right )} - 90 \, e^{\left (2 \, x\right )} - 45 i \, e^{x} + 15\right )} \log \left (e^{x} + 1\right ) +{\left (15 i \, e^{\left (9 \, x\right )} - 45 \, e^{\left (8 \, x\right )} - 90 i \, e^{\left (7 \, x\right )} + 150 \, e^{\left (6 \, x\right )} + 180 i \, e^{\left (5 \, x\right )} - 180 \, e^{\left (4 \, x\right )} - 150 i \, e^{\left (3 \, x\right )} + 90 \, e^{\left (2 \, x\right )} + 45 i \, e^{x} - 15\right )} \log \left (e^{x} - 1\right ) + 30 i \, e^{\left (8 \, x\right )} - 90 \, e^{\left (7 \, x\right )} - 170 i \, e^{\left (6 \, x\right )} + 270 \, e^{\left (5 \, x\right )} + 306 i \, e^{\left (4 \, x\right )} - 310 \, e^{\left (3 \, x\right )} - 198 i \, e^{\left (2 \, x\right )} + 114 \, e^{x} + 48 i}{3 \, e^{\left (9 \, x\right )} + 9 i \, e^{\left (8 \, x\right )} - 18 \, e^{\left (7 \, x\right )} - 30 i \, e^{\left (6 \, x\right )} + 36 \, e^{\left (5 \, x\right )} + 36 i \, e^{\left (4 \, x\right )} - 30 \, e^{\left (3 \, x\right )} - 18 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)^4/(I+sinh(x))^2,x, algorithm="fricas")

[Out]

((-15*I*e^(9*x) + 45*e^(8*x) + 90*I*e^(7*x) - 150*e^(6*x) - 180*I*e^(5*x) + 180*e^(4*x) + 150*I*e^(3*x) - 90*e
^(2*x) - 45*I*e^x + 15)*log(e^x + 1) + (15*I*e^(9*x) - 45*e^(8*x) - 90*I*e^(7*x) + 150*e^(6*x) + 180*I*e^(5*x)
- 180*e^(4*x) - 150*I*e^(3*x) + 90*e^(2*x) + 45*I*e^x - 15)*log(e^x - 1) + 30*I*e^(8*x) - 90*e^(7*x) - 170*I*
e^(6*x) + 270*e^(5*x) + 306*I*e^(4*x) - 310*e^(3*x) - 198*I*e^(2*x) + 114*e^x + 48*I)/(3*e^(9*x) + 9*I*e^(8*x)
- 18*e^(7*x) - 30*I*e^(6*x) + 36*e^(5*x) + 36*I*e^(4*x) - 30*e^(3*x) - 18*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)**4/(I+sinh(x))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.36133, size = 113, normalized size = 1.77 \begin{align*} -\frac{2 \,{\left (-15 i \, e^{\left (8 \, x\right )} + 45 \, e^{\left (7 \, x\right )} + 85 i \, e^{\left (6 \, x\right )} - 135 \, e^{\left (5 \, x\right )} - 153 i \, e^{\left (4 \, x\right )} + 155 \, e^{\left (3 \, x\right )} + 99 i \, e^{\left (2 \, x\right )} - 57 \, e^{x} - 24 i\right )}}{3 \,{\left (e^{\left (3 \, x\right )} + i \, e^{\left (2 \, x\right )} - e^{x} - i\right )}^{3}} - 5 i \, \log \left (e^{x} + 1\right ) + 5 i \, \log \left ({\left | e^{x} - 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(x)^4/(I+sinh(x))^2,x, algorithm="giac")

[Out]

-2/3*(-15*I*e^(8*x) + 45*e^(7*x) + 85*I*e^(6*x) - 135*e^(5*x) - 153*I*e^(4*x) + 155*e^(3*x) + 99*I*e^(2*x) - 5
7*e^x - 24*I)/(e^(3*x) + I*e^(2*x) - e^x - I)^3 - 5*I*log(e^x + 1) + 5*I*log(abs(e^x - 1))