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

Optimal. Leaf size=58 \[ -\frac{15 i x}{8}+\frac{4 \cosh ^3(x)}{3}-4 \cosh (x)-\frac{5}{4} i \sinh ^3(x) \cosh (x)+\frac{15}{8} i \sinh (x) \cosh (x)-\frac{\sinh ^3(x) \cosh (x)}{\text{csch}(x)+i} \]

[Out]

((-15*I)/8)*x - 4*Cosh[x] + (4*Cosh[x]^3)/3 + ((15*I)/8)*Cosh[x]*Sinh[x] - ((5*I)/4)*Cosh[x]*Sinh[x]^3 - (Cosh
[x]*Sinh[x]^3)/(I + Csch[x])

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Rubi [A]  time = 0.0705064, antiderivative size = 58, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.385, Rules used = {3819, 3787, 2635, 8, 2633} \[ -\frac{15 i x}{8}+\frac{4 \cosh ^3(x)}{3}-4 \cosh (x)-\frac{5}{4} i \sinh ^3(x) \cosh (x)+\frac{15}{8} i \sinh (x) \cosh (x)-\frac{\sinh ^3(x) \cosh (x)}{\text{csch}(x)+i} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[x]^4/(I + Csch[x]),x]

[Out]

((-15*I)/8)*x - 4*Cosh[x] + (4*Cosh[x]^3)/3 + ((15*I)/8)*Cosh[x]*Sinh[x] - ((5*I)/4)*Cosh[x]*Sinh[x]^3 - (Cosh
[x]*Sinh[x]^3)/(I + Csch[x])

Rule 3819

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[(Cot[e + f*
x]*(d*Csc[e + f*x])^n)/(f*(a + b*Csc[e + f*x])), x] - Dist[1/a^2, Int[(d*Csc[e + f*x])^n*(a*(n - 1) - b*n*Csc[
e + f*x]), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[n, 0]

Rule 3787

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 8

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

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rubi steps

\begin{align*} \int \frac{\sinh ^4(x)}{i+\text{csch}(x)} \, dx &=-\frac{\cosh (x) \sinh ^3(x)}{i+\text{csch}(x)}+\int (-5 i+4 \text{csch}(x)) \sinh ^4(x) \, dx\\ &=-\frac{\cosh (x) \sinh ^3(x)}{i+\text{csch}(x)}-5 i \int \sinh ^4(x) \, dx+4 \int \sinh ^3(x) \, dx\\ &=-\frac{5}{4} i \cosh (x) \sinh ^3(x)-\frac{\cosh (x) \sinh ^3(x)}{i+\text{csch}(x)}+\frac{15}{4} i \int \sinh ^2(x) \, dx-4 \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cosh (x)\right )\\ &=-4 \cosh (x)+\frac{4 \cosh ^3(x)}{3}+\frac{15}{8} i \cosh (x) \sinh (x)-\frac{5}{4} i \cosh (x) \sinh ^3(x)-\frac{\cosh (x) \sinh ^3(x)}{i+\text{csch}(x)}-\frac{15}{8} i \int 1 \, dx\\ &=-\frac{15 i x}{8}-4 \cosh (x)+\frac{4 \cosh ^3(x)}{3}+\frac{15}{8} i \cosh (x) \sinh (x)-\frac{5}{4} i \cosh (x) \sinh ^3(x)-\frac{\cosh (x) \sinh ^3(x)}{i+\text{csch}(x)}\\ \end{align*}

Mathematica [A]  time = 0.140282, size = 63, normalized size = 1.09 \[ \frac{1}{96} \left (-180 i x+48 i \sinh (2 x)-3 i \sinh (4 x)-168 \cosh (x)+8 \cosh (3 x)+\frac{192 \sinh \left (\frac{x}{2}\right )}{\sinh \left (\frac{x}{2}\right )-i \cosh \left (\frac{x}{2}\right )}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[x]^4/(I + Csch[x]),x]

[Out]

((-180*I)*x - 168*Cosh[x] + 8*Cosh[3*x] + (192*Sinh[x/2])/((-I)*Cosh[x/2] + Sinh[x/2]) + (48*I)*Sinh[2*x] - (3
*I)*Sinh[4*x])/96

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Maple [B]  time = 0.048, size = 182, normalized size = 3.1 \begin{align*} -{\frac{15\,i}{8}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) +{\frac{15\,i}{8}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) +{\frac{1}{3} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-3}}+{{\frac{5\,i}{8}} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-2}}-{\frac{1}{2} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-2}}+{{\frac{i}{4}} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-4}}-{\frac{3}{2} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-1}}+{2\,i \left ( \tanh \left ({\frac{x}{2}} \right ) -i \right ) ^{-1}}+{{\frac{7\,i}{8}} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-1}}-{{\frac{i}{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-3}}-{{\frac{5\,i}{8}} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-2}}+{\frac{3}{2} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-1}}-{{\frac{i}{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-3}}-{\frac{1}{2} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-2}}+{{\frac{7\,i}{8}} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-1}}-{\frac{1}{3} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-3}}-{{\frac{i}{4}} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-15/8*I*ln(tanh(1/2*x)+1)+15/8*I*ln(tanh(1/2*x)-1)+1/3/(tanh(1/2*x)+1)^3+5/8*I/(tanh(1/2*x)-1)^2-1/2/(tanh(1/2
*x)+1)^2+1/4*I/(tanh(1/2*x)+1)^4-3/2/(tanh(1/2*x)+1)+2*I/(tanh(1/2*x)-I)+7/8*I/(tanh(1/2*x)+1)-1/2*I/(tanh(1/2
*x)-1)^3-5/8*I/(tanh(1/2*x)+1)^2+3/2/(tanh(1/2*x)-1)-1/2*I/(tanh(1/2*x)+1)^3-1/2/(tanh(1/2*x)-1)^2+7/8*I/(tanh
(1/2*x)-1)-1/3/(tanh(1/2*x)-1)^3-1/4*I/(tanh(1/2*x)-1)^4

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Maxima [A]  time = 1.14486, size = 96, normalized size = 1.66 \begin{align*} -\frac{15}{8} i \, x - \frac{-5 i \, e^{\left (-x\right )} + 40 \, e^{\left (-2 \, x\right )} + 120 i \, e^{\left (-3 \, x\right )} + 552 \, e^{\left (-4 \, x\right )} - 3}{16 \,{\left (12 i \, e^{\left (-4 \, x\right )} + 12 \, e^{\left (-5 \, x\right )}\right )}} - \frac{7}{8} \, e^{\left (-x\right )} - \frac{1}{4} i \, e^{\left (-2 \, x\right )} + \frac{1}{24} \, e^{\left (-3 \, x\right )} + \frac{1}{64} i \, e^{\left (-4 \, x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-15/8*I*x - 1/16*(-5*I*e^(-x) + 40*e^(-2*x) + 120*I*e^(-3*x) + 552*e^(-4*x) - 3)/(12*I*e^(-4*x) + 12*e^(-5*x))
- 7/8*e^(-x) - 1/4*I*e^(-2*x) + 1/24*e^(-3*x) + 1/64*I*e^(-4*x)

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Fricas [A]  time = 1.65756, size = 244, normalized size = 4.21 \begin{align*} \frac{{\left (-360 i \, x + 168 i\right )} e^{\left (5 \, x\right )} - 24 \,{\left (15 \, x + 23\right )} e^{\left (4 \, x\right )} - 3 i \, e^{\left (9 \, x\right )} + 5 \, e^{\left (8 \, x\right )} + 40 i \, e^{\left (7 \, x\right )} - 120 \, e^{\left (6 \, x\right )} + 120 i \, e^{\left (3 \, x\right )} - 40 \, e^{\left (2 \, x\right )} - 5 i \, e^{x} + 3}{192 \,{\left (e^{\left (5 \, x\right )} - i \, e^{\left (4 \, x\right )}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/192*((-360*I*x + 168*I)*e^(5*x) - 24*(15*x + 23)*e^(4*x) - 3*I*e^(9*x) + 5*e^(8*x) + 40*I*e^(7*x) - 120*e^(6
*x) + 120*I*e^(3*x) - 40*e^(2*x) - 5*I*e^x + 3)/(e^(5*x) - I*e^(4*x))

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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(sinh(x)**4/(I+csch(x)),x)

[Out]

Timed out

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Giac [A]  time = 1.18036, size = 89, normalized size = 1.53 \begin{align*} -\frac{{\left (552 \, e^{\left (4 \, x\right )} - 120 i \, e^{\left (3 \, x\right )} + 40 \, e^{\left (2 \, x\right )} + 5 i \, e^{x} - 3\right )} e^{\left (-4 \, x\right )}}{192 \,{\left (e^{x} - i\right )}} - \frac{1}{64} i \, e^{\left (4 \, x\right )} + \frac{1}{24} \, e^{\left (3 \, x\right )} + \frac{1}{4} i \, e^{\left (2 \, x\right )} - \frac{7}{8} \, e^{x} - \frac{15}{8} i \, \log \left (i \, e^{x}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/192*(552*e^(4*x) - 120*I*e^(3*x) + 40*e^(2*x) + 5*I*e^x - 3)*e^(-4*x)/(e^x - I) - 1/64*I*e^(4*x) + 1/24*e^(
3*x) + 1/4*I*e^(2*x) - 7/8*e^x - 15/8*I*log(I*e^x)