3.225 \(\int \frac{\coth ^4(x)}{(i+\sinh (x))^2} \, dx\)

Optimal. Leaf size=28 \[ \frac{\coth ^3(x)}{3}-2 \coth (x)-i \tanh ^{-1}(\cosh (x))+i \coth (x) \text{csch}(x) \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.094395, antiderivative size = 28, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.462, Rules used = {2708, 2757, 3767, 8, 3768, 3770} \[ \frac{\coth ^3(x)}{3}-2 \coth (x)-i \tanh ^{-1}(\cosh (x))+i \coth (x) \text{csch}(x) \]

Antiderivative was successfully verified.

[In]

Int[Coth[x]^4/(I + Sinh[x])^2,x]

[Out]

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

Rule 2708

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p_), x_Symbol] :> Dist[a^p, Int[Sin[
e + f*x]^p/(a - b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0] && IntegersQ[m, p] &&
EqQ[p, 2*m]

Rule 2757

Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Int[Expan
dTrig[(a + b*sin[e + f*x])^m*(d*sin[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && EqQ[a^2 - b^2, 0] &
& IGtQ[m, 0] && RationalQ[n]

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]

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

Mathematica [B]  time = 0.0541919, size = 107, normalized size = 3.82 \[ -\frac{5}{6} \tanh \left (\frac{x}{2}\right )-\frac{5}{6} \coth \left (\frac{x}{2}\right )+\frac{1}{4} i \text{csch}^2\left (\frac{x}{2}\right )+\frac{1}{4} i \text{sech}^2\left (\frac{x}{2}\right )+i \log \left (\sinh \left (\frac{x}{2}\right )\right )-i \log \left (\cosh \left (\frac{x}{2}\right )\right )+\frac{1}{24} \coth \left (\frac{x}{2}\right ) \text{csch}^2\left (\frac{x}{2}\right )-\frac{1}{24} \tanh \left (\frac{x}{2}\right ) \text{sech}^2\left (\frac{x}{2}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*Coth[x/2])/6 + (I/4)*Csch[x/2]^2 + (Coth[x/2]*Csch[x/2]^2)/24 - I*Log[Cosh[x/2]] + I*Log[Sinh[x/2]] + (I/4
)*Sech[x/2]^2 - (5*Tanh[x/2])/6 - (Sech[x/2]^2*Tanh[x/2])/24

________________________________________________________________________________________

Maple [B]  time = 0.051, size = 58, normalized size = 2.1 \begin{align*} -{\frac{7}{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}}-{\frac{7}{8} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{-1}}+{\frac{1}{24} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{-3}}+i\ln \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-7/8*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)^2-7/8/tanh(1/2*x)+1/24/tanh(1/2*x)^3
+I*ln(tanh(1/2*x))

________________________________________________________________________________________

Maxima [B]  time = 1.02983, size = 90, normalized size = 3.21 \begin{align*} \frac{-6 i \, e^{\left (-x\right )} - 24 \, e^{\left (-2 \, x\right )} + 6 \, e^{\left (-4 \, x\right )} + 6 i \, e^{\left (-5 \, x\right )} + 10}{3 \,{\left (3 \, e^{\left (-2 \, x\right )} - 3 \, e^{\left (-4 \, x\right )} + e^{\left (-6 \, x\right )} - 1\right )}} - i \, \log \left (e^{\left (-x\right )} + 1\right ) + i \, \log \left (e^{\left (-x\right )} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(-6*I*e^(-x) - 24*e^(-2*x) + 6*e^(-4*x) + 6*I*e^(-5*x) + 10)/(3*e^(-2*x) - 3*e^(-4*x) + e^(-6*x) - 1) - I*
log(e^(-x) + 1) + I*log(e^(-x) - 1)

________________________________________________________________________________________

Fricas [B]  time = 2.03329, size = 302, normalized size = 10.79 \begin{align*} \frac{{\left (-3 i \, e^{\left (6 \, x\right )} + 9 i \, e^{\left (4 \, x\right )} - 9 i \, e^{\left (2 \, x\right )} + 3 i\right )} \log \left (e^{x} + 1\right ) +{\left (3 i \, e^{\left (6 \, x\right )} - 9 i \, e^{\left (4 \, x\right )} + 9 i \, e^{\left (2 \, x\right )} - 3 i\right )} \log \left (e^{x} - 1\right ) + 6 i \, e^{\left (5 \, x\right )} - 6 \, e^{\left (4 \, x\right )} + 24 \, e^{\left (2 \, x\right )} - 6 i \, e^{x} - 10}{3 \,{\left (e^{\left (6 \, x\right )} - 3 \, e^{\left (4 \, x\right )} + 3 \, e^{\left (2 \, x\right )} - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(x)^4/(I+sinh(x))^2,x, algorithm="fricas")

[Out]

1/3*((-3*I*e^(6*x) + 9*I*e^(4*x) - 9*I*e^(2*x) + 3*I)*log(e^x + 1) + (3*I*e^(6*x) - 9*I*e^(4*x) + 9*I*e^(2*x)
- 3*I)*log(e^x - 1) + 6*I*e^(5*x) - 6*e^(4*x) + 24*e^(2*x) - 6*I*e^x - 10)/(e^(6*x) - 3*e^(4*x) + 3*e^(2*x) -
1)

________________________________________________________________________________________

Sympy [B]  time = 0.550075, size = 66, normalized size = 2.36 \begin{align*} \operatorname{RootSum}{\left (z^{2} + 1, \left ( i \mapsto i \log{\left (i i + e^{x} \right )} \right )\right )} + \frac{2 i e^{5 x} - 2 e^{4 x} + 8 e^{2 x} - 2 i e^{x} - \frac{10}{3}}{e^{6 x} - 3 e^{4 x} + 3 e^{2 x} - 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(x)**4/(I+sinh(x))**2,x)

[Out]

RootSum(_z**2 + 1, Lambda(_i, _i*log(_i*I + exp(x)))) + (2*I*exp(5*x) - 2*exp(4*x) + 8*exp(2*x) - 2*I*exp(x) -
10/3)/(exp(6*x) - 3*exp(4*x) + 3*exp(2*x) - 1)

________________________________________________________________________________________

Giac [B]  time = 1.14062, size = 68, normalized size = 2.43 \begin{align*} -\frac{-6 i \, e^{\left (5 \, x\right )} + 6 \, e^{\left (4 \, x\right )} - 24 \, e^{\left (2 \, x\right )} + 6 i \, e^{x} + 10}{3 \,{\left (e^{\left (2 \, x\right )} - 1\right )}^{3}} - i \, \log \left (e^{x} + 1\right ) + i \, \log \left ({\left | e^{x} - 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(-6*I*e^(5*x) + 6*e^(4*x) - 24*e^(2*x) + 6*I*e^x + 10)/(e^(2*x) - 1)^3 - I*log(e^x + 1) + I*log(abs(e^x -
1))