3.596 \(\int x \tanh ^2(x) \, dx\)

Optimal. Leaf size=16 \[ \frac {x^2}{2}-x \tanh (x)+\log (\cosh (x)) \]

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Rubi [A]  time = 0.02, antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3720, 3475, 30} \[ \frac {x^2}{2}-x \tanh (x)+\log (\cosh (x)) \]

Antiderivative was successfully verified.

[In]

Int[x*Tanh[x]^2,x]

[Out]

x^2/2 + Log[Cosh[x]] - x*Tanh[x]

Rule 30

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

Rule 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3720

Int[((c_.) + (d_.)*(x_))^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(c + d*x)^m*(b*Tan[e
+ f*x])^(n - 1))/(f*(n - 1)), x] + (-Dist[(b*d*m)/(f*(n - 1)), Int[(c + d*x)^(m - 1)*(b*Tan[e + f*x])^(n - 1)
, x], x] - Dist[b^2, Int[(c + d*x)^m*(b*Tan[e + f*x])^(n - 2), x], x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n,
1] && GtQ[m, 0]

Rubi steps

\begin {align*} \int x \tanh ^2(x) \, dx &=-x \tanh (x)+\int x \, dx+\int \tanh (x) \, dx\\ &=\frac {x^2}{2}+\log (\cosh (x))-x \tanh (x)\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 16, normalized size = 1.00 \[ \frac {x^2}{2}-x \tanh (x)+\log (\cosh (x)) \]

Antiderivative was successfully verified.

[In]

Integrate[x*Tanh[x]^2,x]

[Out]

x^2/2 + Log[Cosh[x]] - x*Tanh[x]

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x \tanh ^2(x) \, dx \]

Verification is Not applicable to the result.

[In]

IntegrateAlgebraic[x*Tanh[x]^2,x]

[Out]

Could not integrate

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fricas [B]  time = 1.29, size = 93, normalized size = 5.81 \[ \frac {{\left (x^{2} - 4 \, x\right )} \cosh \relax (x)^{2} + 2 \, {\left (x^{2} - 4 \, x\right )} \cosh \relax (x) \sinh \relax (x) + {\left (x^{2} - 4 \, x\right )} \sinh \relax (x)^{2} + x^{2} + 2 \, {\left (\cosh \relax (x)^{2} + 2 \, \cosh \relax (x) \sinh \relax (x) + \sinh \relax (x)^{2} + 1\right )} \log \left (\frac {2 \, \cosh \relax (x)}{\cosh \relax (x) - \sinh \relax (x)}\right )}{2 \, {\left (\cosh \relax (x)^{2} + 2 \, \cosh \relax (x) \sinh \relax (x) + \sinh \relax (x)^{2} + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*tanh(x)^2,x, algorithm="fricas")

[Out]

1/2*((x^2 - 4*x)*cosh(x)^2 + 2*(x^2 - 4*x)*cosh(x)*sinh(x) + (x^2 - 4*x)*sinh(x)^2 + x^2 + 2*(cosh(x)^2 + 2*co
sh(x)*sinh(x) + sinh(x)^2 + 1)*log(2*cosh(x)/(cosh(x) - sinh(x))))/(cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2
+ 1)

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giac [B]  time = 0.63, size = 51, normalized size = 3.19 \[ \frac {x^{2} e^{\left (2 \, x\right )} + x^{2} - 4 \, x e^{\left (2 \, x\right )} + 2 \, e^{\left (2 \, x\right )} \log \left (e^{\left (2 \, x\right )} + 1\right ) + 2 \, \log \left (e^{\left (2 \, x\right )} + 1\right )}{2 \, {\left (e^{\left (2 \, x\right )} + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*tanh(x)^2,x, algorithm="giac")

[Out]

1/2*(x^2*e^(2*x) + x^2 - 4*x*e^(2*x) + 2*e^(2*x)*log(e^(2*x) + 1) + 2*log(e^(2*x) + 1))/(e^(2*x) + 1)

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maple [A]  time = 0.03, size = 28, normalized size = 1.75

method	result	size
risch	\(\frac {x^{2}}{2}-2 x +\frac {2 x}{1+{\mathrm e}^{2 x}}+\ln \left (1+{\mathrm e}^{2 x}\right )\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*tanh(x)^2,x,method=_RETURNVERBOSE)

[Out]

1/2*x^2-2*x+2*x/(1+exp(2*x))+ln(1+exp(2*x))

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maxima [B]  time = 1.12, size = 49, normalized size = 3.06 \[ -\frac {x e^{\left (2 \, x\right )}}{e^{\left (2 \, x\right )} + 1} + \frac {x^{2} + {\left (x^{2} - 2 \, x\right )} e^{\left (2 \, x\right )}}{2 \, {\left (e^{\left (2 \, x\right )} + 1\right )}} + \log \left (e^{\left (2 \, x\right )} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*tanh(x)^2,x, algorithm="maxima")

[Out]

-x*e^(2*x)/(e^(2*x) + 1) + 1/2*(x^2 + (x^2 - 2*x)*e^(2*x))/(e^(2*x) + 1) + log(e^(2*x) + 1)

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mupad [B]  time = 0.31, size = 21, normalized size = 1.31 \[ \ln \left ({\mathrm {e}}^{2\,x}+1\right )-x-x\,\mathrm {tanh}\relax (x)+\frac {x^2}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*tanh(x)^2,x)

[Out]

log(exp(2*x) + 1) - x - x*tanh(x) + x^2/2

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sympy [A]  time = 0.19, size = 17, normalized size = 1.06 \[ \frac {x^{2}}{2} - x \tanh {\relax (x )} + x - \log {\left (\tanh {\relax (x )} + 1 \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*tanh(x)**2,x)

[Out]

x**2/2 - x*tanh(x) + x - log(tanh(x) + 1)

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