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

Optimal. Leaf size=26 \[ \frac{1}{3} i \coth ^3(x)-\frac{1}{2} \tanh ^{-1}(\cosh (x))-\frac{1}{2} \coth (x) \text{csch}(x) \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0779558, antiderivative size = 26, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.385, Rules used = {2706, 2607, 30, 2611, 3770} \[ \frac{1}{3} i \coth ^3(x)-\frac{1}{2} \tanh ^{-1}(\cosh (x))-\frac{1}{2} \coth (x) \text{csch}(x) \]

Antiderivative was successfully verified.

[In]

Int[Coth[x]^4/(I + Sinh[x]),x]

[Out]

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

Rule 2706

Int[((g_.)*tan[(e_.) + (f_.)*(x_)])^(p_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[1/a, Int[S
ec[e + f*x]^2*(g*Tan[e + f*x])^p, x], x] - Dist[1/(b*g), Int[Sec[e + f*x]*(g*Tan[e + f*x])^(p + 1), x], x] /;
FreeQ[{a, b, e, f, g, p}, x] && EqQ[a^2 - b^2, 0] && NeQ[p, -1]

Rule 2607

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(b*x)
^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] &&  !(IntegerQ[(n
- 1)/2] && LtQ[0, n, m - 1])

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2611

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a*Sec[e
+ f*x])^m*(b*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] - Dist[(b^2*(n - 1))/(m + n - 1), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 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)} \, dx &=-\left (i \int \coth ^2(x) \text{csch}^2(x) \, dx\right )+\int \coth ^2(x) \text{csch}(x) \, dx\\ &=-\frac{1}{2} \coth (x) \text{csch}(x)+\frac{1}{2} \int \text{csch}(x) \, dx-\operatorname{Subst}\left (\int x^2 \, dx,x,i \coth (x)\right )\\ &=-\frac{1}{2} \tanh ^{-1}(\cosh (x))+\frac{1}{3} i \coth ^3(x)-\frac{1}{2} \coth (x) \text{csch}(x)\\ \end{align*}

Mathematica [B]  time = 0.0366139, size = 100, normalized size = 3.85 \[ \frac{1}{6} i \tanh \left (\frac{x}{2}\right )+\frac{1}{6} i \coth \left (\frac{x}{2}\right )-\frac{1}{8} \text{csch}^2\left (\frac{x}{2}\right )-\frac{1}{8} \text{sech}^2\left (\frac{x}{2}\right )+\frac{1}{2} \log \left (\tanh \left (\frac{x}{2}\right )\right )+\frac{1}{24} i \coth \left (\frac{x}{2}\right ) \text{csch}^2\left (\frac{x}{2}\right )-\frac{1}{24} i \tanh \left (\frac{x}{2}\right ) \text{sech}^2\left (\frac{x}{2}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Coth[x]^4/(I + Sinh[x]),x]

[Out]

(I/6)*Coth[x/2] - Csch[x/2]^2/8 + (I/24)*Coth[x/2]*Csch[x/2]^2 + Log[Tanh[x/2]]/2 - Sech[x/2]^2/8 + (I/6)*Tanh
[x/2] - (I/24)*Sech[x/2]^2*Tanh[x/2]

________________________________________________________________________________________

Maple [B]  time = 0.046, size = 59, normalized size = 2.3 \begin{align*}{\frac{i}{8}}\tanh \left ({\frac{x}{2}} \right ) +{\frac{i}{24}} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{3}+{\frac{1}{8} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}}-{\frac{1}{8} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{-2}}+{{\frac{i}{8}} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{-1}}+{{\frac{i}{24}} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{-3}}+{\frac{1}{2}\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)),x)

[Out]

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

________________________________________________________________________________________

Maxima [B]  time = 1.05202, size = 82, normalized size = 3.15 \begin{align*} \frac{3 \, e^{\left (-x\right )} - 6 i \, e^{\left (-4 \, x\right )} - 3 \, e^{\left (-5 \, x\right )} - 2 i}{3 \,{\left (3 \, e^{\left (-2 \, x\right )} - 3 \, e^{\left (-4 \, x\right )} + e^{\left (-6 \, x\right )} - 1\right )}} - \frac{1}{2} \, \log \left (e^{\left (-x\right )} + 1\right ) + \frac{1}{2} \, \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)),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [B]  time = 2.01305, size = 263, normalized size = 10.12 \begin{align*} -\frac{3 \,{\left (e^{\left (6 \, x\right )} - 3 \, e^{\left (4 \, x\right )} + 3 \, e^{\left (2 \, x\right )} - 1\right )} \log \left (e^{x} + 1\right ) - 3 \,{\left (e^{\left (6 \, x\right )} - 3 \, e^{\left (4 \, x\right )} + 3 \, e^{\left (2 \, x\right )} - 1\right )} \log \left (e^{x} - 1\right ) + 6 \, e^{\left (5 \, x\right )} - 12 i \, e^{\left (4 \, x\right )} - 6 \, e^{x} - 4 i}{6 \,{\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)),x, algorithm="fricas")

[Out]

-1/6*(3*(e^(6*x) - 3*e^(4*x) + 3*e^(2*x) - 1)*log(e^x + 1) - 3*(e^(6*x) - 3*e^(4*x) + 3*e^(2*x) - 1)*log(e^x -
1) + 6*e^(5*x) - 12*I*e^(4*x) - 6*e^x - 4*I)/(e^(6*x) - 3*e^(4*x) + 3*e^(2*x) - 1)

________________________________________________________________________________________

Sympy [B]  time = 0.478784, size = 58, normalized size = 2.23 \begin{align*} \frac{- e^{5 x} + 2 i e^{4 x} + e^{x} + \frac{2 i}{3}}{e^{6 x} - 3 e^{4 x} + 3 e^{2 x} - 1} + \frac{\log{\left (e^{x} - 1 \right )}}{2} - \frac{\log{\left (e^{x} + 1 \right )}}{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-exp(5*x) + 2*I*exp(4*x) + exp(x) + 2*I/3)/(exp(6*x) - 3*exp(4*x) + 3*exp(2*x) - 1) + log(exp(x) - 1)/2 - log
(exp(x) + 1)/2

________________________________________________________________________________________

Giac [B]  time = 1.15295, size = 59, normalized size = 2.27 \begin{align*} -\frac{3 \, e^{\left (5 \, x\right )} - 6 i \, e^{\left (4 \, x\right )} - 3 \, e^{x} - 2 i}{3 \,{\left (e^{\left (2 \, x\right )} - 1\right )}^{3}} - \frac{1}{2} \, \log \left (e^{x} + 1\right ) + \frac{1}{2} \, \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)),x, algorithm="giac")

[Out]

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