### 3.993 $$\int \frac{\text{sech}^2(x) \tanh ^2(x)}{(2+\tanh ^3(x))^2} \, dx$$

Optimal. Leaf size=12 $-\frac{1}{3 \left (\tanh ^3(x)+2\right )}$

[Out]

-1/(3*(2 + Tanh[x]^3))

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Rubi [A]  time = 0.0857805, antiderivative size = 12, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 17, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.118, Rules used = {4342, 261} $-\frac{1}{3 \left (\tanh ^3(x)+2\right )}$

Antiderivative was successfully veriﬁed.

[In]

Int[(Sech[x]^2*Tanh[x]^2)/(2 + Tanh[x]^3)^2,x]

[Out]

-1/(3*(2 + Tanh[x]^3))

Rule 4342

Int[(u_)*(F_)[(c_.)*((a_.) + (b_.)*(x_))]^2, x_Symbol] :> With[{d = FreeFactors[Tan[c*(a + b*x)], x]}, Dist[d/
(b*c), Subst[Int[SubstFor[1, Tan[c*(a + b*x)]/d, u, x], x], x, Tan[c*(a + b*x)]/d], x] /; FunctionOfQ[Tan[c*(a
+ b*x)]/d, u, x, True]] /; FreeQ[{a, b, c}, x] && NonsumQ[u] && (EqQ[F, Sec] || EqQ[F, sec])

Rule 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rubi steps

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

Mathematica [A]  time = 0.0418217, size = 12, normalized size = 1. $-\frac{1}{3 \left (\tanh ^3(x)+2\right )}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(Sech[x]^2*Tanh[x]^2)/(2 + Tanh[x]^3)^2,x]

[Out]

-1/(3*(2 + Tanh[x]^3))

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Maple [A]  time = 0.041, size = 11, normalized size = 0.9 \begin{align*} -{\frac{1}{6+3\, \left ( \tanh \left ( x \right ) \right ) ^{3}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(sech(x)^2*tanh(x)^2/(2+tanh(x)^3)^2,x)

[Out]

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

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Maxima [A]  time = 0.991249, size = 14, normalized size = 1.17 \begin{align*} -\frac{1}{3 \,{\left (\tanh \left (x\right )^{3} + 2\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^2*tanh(x)^2/(2+tanh(x)^3)^2,x, algorithm="maxima")

[Out]

-1/3/(tanh(x)^3 + 2)

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Fricas [B]  time = 1.99299, size = 250, normalized size = 20.83 \begin{align*} -\frac{8 \,{\left (\cosh \left (x\right )^{2} + \cosh \left (x\right ) \sinh \left (x\right ) + \sinh \left (x\right )^{2}\right )}}{9 \,{\left (3 \, \cosh \left (x\right )^{4} + 12 \, \cosh \left (x\right ) \sinh \left (x\right )^{3} + 3 \, \sinh \left (x\right )^{4} + 2 \,{\left (9 \, \cosh \left (x\right )^{2} + 2\right )} \sinh \left (x\right )^{2} + 4 \, \cosh \left (x\right )^{2} + 4 \,{\left (3 \, \cosh \left (x\right )^{3} + \cosh \left (x\right )\right )} \sinh \left (x\right ) + 9\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^2*tanh(x)^2/(2+tanh(x)^3)^2,x, algorithm="fricas")

[Out]

-8/9*(cosh(x)^2 + cosh(x)*sinh(x) + sinh(x)^2)/(3*cosh(x)^4 + 12*cosh(x)*sinh(x)^3 + 3*sinh(x)^4 + 2*(9*cosh(x
)^2 + 2)*sinh(x)^2 + 4*cosh(x)^2 + 4*(3*cosh(x)^3 + cosh(x))*sinh(x) + 9)

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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(sech(x)**2*tanh(x)**2/(2+tanh(x)**3)**2,x)

[Out]

Timed out

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Giac [B]  time = 1.51189, size = 43, normalized size = 3.58 \begin{align*} -\frac{2 \,{\left (3 \, e^{\left (4 \, x\right )} + 1\right )}}{9 \,{\left (3 \, e^{\left (6 \, x\right )} + 3 \, e^{\left (4 \, x\right )} + 9 \, e^{\left (2 \, x\right )} + 1\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^2*tanh(x)^2/(2+tanh(x)^3)^2,x, algorithm="giac")

[Out]

-2/9*(3*e^(4*x) + 1)/(3*e^(6*x) + 3*e^(4*x) + 9*e^(2*x) + 1)