3.210 \(\int \frac{\tanh ^2(x)}{i+\sinh (x)} \, dx\)

Optimal. Leaf size=23 \[ -\frac{1}{3} i \tanh ^3(x)+\frac{\text{sech}^3(x)}{3}-\text{sech}(x) \]

[Out]

-Sech[x] + Sech[x]^3/3 - (I/3)*Tanh[x]^3

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Rubi [A]  time = 0.0762514, antiderivative size = 23, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.308, Rules used = {2706, 2607, 30, 2606} \[ -\frac{1}{3} i \tanh ^3(x)+\frac{\text{sech}^3(x)}{3}-\text{sech}(x) \]

Antiderivative was successfully verified.

[In]

Int[Tanh[x]^2/(I + Sinh[x]),x]

[Out]

-Sech[x] + Sech[x]^3/3 - (I/3)*Tanh[x]^3

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 2606

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

Rubi steps

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

Mathematica [B]  time = 0.0590886, size = 67, normalized size = 2.91 \[ \frac{4 i \sinh (x)-\cosh (2 x)+(5-5 i \sinh (x)) \cosh (x)-3}{6 \left (\cosh \left (\frac{x}{2}\right )-i \sinh \left (\frac{x}{2}\right )\right )^3 \left (\cosh \left (\frac{x}{2}\right )+i \sinh \left (\frac{x}{2}\right )\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[Tanh[x]^2/(I + Sinh[x]),x]

[Out]

(-3 - Cosh[2*x] + Cosh[x]*(5 - (5*I)*Sinh[x]) + (4*I)*Sinh[x])/(6*(Cosh[x/2] - I*Sinh[x/2])^3*(Cosh[x/2] + I*S
inh[x/2]))

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Maple [B]  time = 0.045, size = 47, normalized size = 2. \begin{align*}{{\frac{i}{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) -i \right ) ^{-1}}-{{\frac{2\,i}{3}} \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) ^{-3}}-{{\frac{i}{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) ^{-1}}+ \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) ^{-2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(x)^2/(I+sinh(x)),x)

[Out]

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

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Maxima [B]  time = 1.11683, size = 147, normalized size = 6.39 \begin{align*} \frac{2 \, e^{\left (-x\right )}}{-6 i \, e^{\left (-x\right )} - 6 i \, e^{\left (-3 \, x\right )} + 3 \, e^{\left (-4 \, x\right )} - 3} + \frac{6 i \, e^{\left (-2 \, x\right )}}{-6 i \, e^{\left (-x\right )} - 6 i \, e^{\left (-3 \, x\right )} + 3 \, e^{\left (-4 \, x\right )} - 3} - \frac{6 \, e^{\left (-3 \, x\right )}}{-6 i \, e^{\left (-x\right )} - 6 i \, e^{\left (-3 \, x\right )} + 3 \, e^{\left (-4 \, x\right )} - 3} + \frac{2 i}{-6 i \, e^{\left (-x\right )} - 6 i \, e^{\left (-3 \, x\right )} + 3 \, e^{\left (-4 \, x\right )} - 3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.03306, size = 111, normalized size = 4.83 \begin{align*} -\frac{6 \, e^{\left (3 \, x\right )} + 6 i \, e^{\left (2 \, x\right )} - 2 \, e^{x} + 2 i}{3 \, e^{\left (4 \, x\right )} + 6 i \, e^{\left (3 \, x\right )} + 6 i \, e^{x} - 3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B]  time = 0.379415, size = 48, normalized size = 2.09 \begin{align*} \frac{- 2 e^{3 x} - 2 i e^{2 x} + \frac{2 e^{x}}{3} - \frac{2 i}{3}}{e^{4 x} + 2 i e^{3 x} + 2 i e^{x} - 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)**2/(I+sinh(x)),x)

[Out]

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

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Giac [A]  time = 1.12204, size = 39, normalized size = 1.7 \begin{align*} -\frac{1}{2 \,{\left (e^{x} - i\right )}} - \frac{9 \, e^{\left (2 \, x\right )} + 12 i \, e^{x} - 7}{6 \,{\left (e^{x} + i\right )}^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/(e^x - I) - 1/6*(9*e^(2*x) + 12*I*e^x - 7)/(e^x + I)^3