3.71 \(\int \frac{\sinh ^2(x)}{1+\tanh (x)} \, dx\)

Optimal. Leaf size=38 \[ -\frac{x}{8}+\frac{1}{8 (1-\tanh (x))}+\frac{1}{4 (\tanh (x)+1)}-\frac{1}{8 (\tanh (x)+1)^2} \]

[Out]

-x/8 + 1/(8*(1 - Tanh[x])) - 1/(8*(1 + Tanh[x])^2) + 1/(4*(1 + Tanh[x]))

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Rubi [A]  time = 0.0587289, antiderivative size = 38, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.364, Rules used = {3516, 848, 88, 207} \[ -\frac{x}{8}+\frac{1}{8 (1-\tanh (x))}+\frac{1}{4 (\tanh (x)+1)}-\frac{1}{8 (\tanh (x)+1)^2} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[x]^2/(1 + Tanh[x]),x]

[Out]

-x/8 + 1/(8*(1 - Tanh[x])) - 1/(8*(1 + Tanh[x])^2) + 1/(4*(1 + Tanh[x]))

Rule 3516

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

Rule 848

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)
^(m + p)*(f + g*x)^n*(a/d + (c*x)/e)^p, x] /; FreeQ[{a, c, d, e, f, g, m, n}, x] && NeQ[e*f - d*g, 0] && EqQ[c
*d^2 + a*e^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && EqQ[m + p, 0]))

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.0371192, size = 24, normalized size = 0.63 \[ \frac{1}{32} (-4 x+\sinh (4 x)+4 \cosh (2 x)-\cosh (4 x)) \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[x]^2/(1 + Tanh[x]),x]

[Out]

(-4*x + 4*Cosh[2*x] - Cosh[4*x] + Sinh[4*x])/32

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Maple [B]  time = 0.027, size = 68, normalized size = 1.8 \begin{align*} -{\frac{1}{2} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-4}}+ \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-3}-{\frac{1}{2} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-2}}-{\frac{1}{8}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) }+{\frac{1}{4} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-2}}+{\frac{1}{4} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-1}}+{\frac{1}{8}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(x)^2/(1+tanh(x)),x)

[Out]

-1/2/(tanh(1/2*x)+1)^4+1/(tanh(1/2*x)+1)^3-1/2/(tanh(1/2*x)+1)^2-1/8*ln(tanh(1/2*x)+1)+1/4/(tanh(1/2*x)-1)^2+1
/4/(tanh(1/2*x)-1)+1/8*ln(tanh(1/2*x)-1)

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Maxima [A]  time = 1.13915, size = 30, normalized size = 0.79 \begin{align*} -\frac{1}{8} \, x + \frac{1}{16} \, e^{\left (2 \, x\right )} + \frac{1}{16} \, e^{\left (-2 \, x\right )} - \frac{1}{32} \, e^{\left (-4 \, x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)^2/(1+tanh(x)),x, algorithm="maxima")

[Out]

-1/8*x + 1/16*e^(2*x) + 1/16*e^(-2*x) - 1/32*e^(-4*x)

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Fricas [A]  time = 2.31414, size = 176, normalized size = 4.63 \begin{align*} \frac{\cosh \left (x\right )^{3} + 3 \, \cosh \left (x\right ) \sinh \left (x\right )^{2} + 3 \, \sinh \left (x\right )^{3} - 2 \,{\left (2 \, x - 1\right )} \cosh \left (x\right ) +{\left (9 \, \cosh \left (x\right )^{2} - 4 \, x - 2\right )} \sinh \left (x\right )}{32 \,{\left (\cosh \left (x\right ) + \sinh \left (x\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)^2/(1+tanh(x)),x, algorithm="fricas")

[Out]

1/32*(cosh(x)^3 + 3*cosh(x)*sinh(x)^2 + 3*sinh(x)^3 - 2*(2*x - 1)*cosh(x) + (9*cosh(x)^2 - 4*x - 2)*sinh(x))/(
cosh(x) + sinh(x))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sinh ^{2}{\left (x \right )}}{\tanh{\left (x \right )} + 1}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)**2/(1+tanh(x)),x)

[Out]

Integral(sinh(x)**2/(tanh(x) + 1), x)

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Giac [A]  time = 1.22711, size = 41, normalized size = 1.08 \begin{align*} \frac{1}{32} \,{\left (3 \, e^{\left (4 \, x\right )} + 2 \, e^{\left (2 \, x\right )} - 1\right )} e^{\left (-4 \, x\right )} - \frac{1}{8} \, x + \frac{1}{16} \, e^{\left (2 \, x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)^2/(1+tanh(x)),x, algorithm="giac")

[Out]

1/32*(3*e^(4*x) + 2*e^(2*x) - 1)*e^(-4*x) - 1/8*x + 1/16*e^(2*x)