3.156 \(\int \tanh ^2(a+2 \log (x)) \, dx\)

Optimal. Leaf size=165 \[ \frac{x}{e^{2 a} x^4+1}+\frac{e^{-a/2} \log \left (e^a x^2-\sqrt{2} e^{a/2} x+1\right )}{4 \sqrt{2}}-\frac{e^{-a/2} \log \left (e^a x^2+\sqrt{2} e^{a/2} x+1\right )}{4 \sqrt{2}}+\frac{e^{-a/2} \tan ^{-1}\left (1-\sqrt{2} e^{a/2} x\right )}{2 \sqrt{2}}-\frac{e^{-a/2} \tan ^{-1}\left (\sqrt{2} e^{a/2} x+1\right )}{2 \sqrt{2}}+x \]

[Out]

x + x/(1 + E^(2*a)*x^4) + ArcTan[1 - Sqrt[2]*E^(a/2)*x]/(2*Sqrt[2]*E^(a/2)) - ArcTan[1 + Sqrt[2]*E^(a/2)*x]/(2
*Sqrt[2]*E^(a/2)) + Log[1 - Sqrt[2]*E^(a/2)*x + E^a*x^2]/(4*Sqrt[2]*E^(a/2)) - Log[1 + Sqrt[2]*E^(a/2)*x + E^a
*x^2]/(4*Sqrt[2]*E^(a/2))

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Rubi [F]  time = 0.0100514, antiderivative size = 0, normalized size of antiderivative = 0., number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0., Rules used = {} \[ \int \tanh ^2(a+2 \log (x)) \, dx \]

Verification is Not applicable to the result.

[In]

Int[Tanh[a + 2*Log[x]]^2,x]

[Out]

Defer[Int][Tanh[a + 2*Log[x]]^2, x]

Rubi steps

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

Mathematica [A]  time = 0.531036, size = 146, normalized size = 0.88 \[ \frac{1}{4} \left (\frac{4 x}{e^{2 a} x^4+1}+\sqrt [4]{-1} e^{-a/2} \log \left (\sqrt [4]{-1} e^{-a/2}-x\right )+(-1)^{3/4} e^{-a/2} \log \left ((-1)^{3/4} e^{-a/2}-x\right )-\sqrt [4]{-1} e^{-a/2} \log \left (\sqrt [4]{-1} e^{-a/2}+x\right )-(-1)^{3/4} e^{-a/2} \log \left ((-1)^{3/4} e^{-a/2}+x\right )+4 x\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Tanh[a + 2*Log[x]]^2,x]

[Out]

(4*x + (4*x)/(1 + E^(2*a)*x^4) + ((-1)^(1/4)*Log[(-1)^(1/4)/E^(a/2) - x])/E^(a/2) + ((-1)^(3/4)*Log[(-1)^(3/4)
/E^(a/2) - x])/E^(a/2) - ((-1)^(1/4)*Log[(-1)^(1/4)/E^(a/2) + x])/E^(a/2) - ((-1)^(3/4)*Log[(-1)^(3/4)/E^(a/2)
+ x])/E^(a/2))/4

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Maple [C]  time = 0.02, size = 47, normalized size = 0.3 \begin{align*} x+{\frac{x}{1+{{\rm e}^{2\,a}}{x}^{4}}}-{\frac{{{\rm e}^{-2\,a}}}{4}\sum _{{\it \_R}={\it RootOf} \left ({{\rm e}^{2\,a}}{{\it \_Z}}^{4}+1 \right ) }{\frac{\ln \left ( x-{\it \_R} \right ) }{{{\it \_R}}^{3}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(a+2*ln(x))^2,x)

[Out]

x+x/(1+exp(2*a)*x^4)-1/4*exp(-2*a)*sum(1/_R^3*ln(x-_R),_R=RootOf(exp(2*a)*_Z^4+1))

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Maxima [A]  time = 1.70158, size = 186, normalized size = 1.13 \begin{align*} -\frac{1}{4} \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (2 \, x e^{a} + \sqrt{2} e^{\left (\frac{1}{2} \, a\right )}\right )} e^{\left (-\frac{1}{2} \, a\right )}\right ) e^{\left (-\frac{1}{2} \, a\right )} - \frac{1}{4} \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (2 \, x e^{a} - \sqrt{2} e^{\left (\frac{1}{2} \, a\right )}\right )} e^{\left (-\frac{1}{2} \, a\right )}\right ) e^{\left (-\frac{1}{2} \, a\right )} - \frac{1}{8} \, \sqrt{2} e^{\left (-\frac{1}{2} \, a\right )} \log \left (x^{2} e^{a} + \sqrt{2} x e^{\left (\frac{1}{2} \, a\right )} + 1\right ) + \frac{1}{8} \, \sqrt{2} e^{\left (-\frac{1}{2} \, a\right )} \log \left (x^{2} e^{a} - \sqrt{2} x e^{\left (\frac{1}{2} \, a\right )} + 1\right ) + x + \frac{x}{x^{4} e^{\left (2 \, a\right )} + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*sqrt(2)*arctan(1/2*sqrt(2)*(2*x*e^a + sqrt(2)*e^(1/2*a))*e^(-1/2*a))*e^(-1/2*a) - 1/4*sqrt(2)*arctan(1/2*
sqrt(2)*(2*x*e^a - sqrt(2)*e^(1/2*a))*e^(-1/2*a))*e^(-1/2*a) - 1/8*sqrt(2)*e^(-1/2*a)*log(x^2*e^a + sqrt(2)*x*
e^(1/2*a) + 1) + 1/8*sqrt(2)*e^(-1/2*a)*log(x^2*e^a - sqrt(2)*x*e^(1/2*a) + 1) + x + x/(x^4*e^(2*a) + 1)

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Fricas [B]  time = 2.20292, size = 680, normalized size = 4.12 \begin{align*} \frac{8 \, x^{5} e^{\left (2 \, a\right )} + 4 \,{\left (\sqrt{2} x^{4} e^{\left (2 \, a\right )} + \sqrt{2}\right )} \arctan \left (-\sqrt{2} x e^{\left (\frac{1}{2} \, a\right )} + \sqrt{2} \sqrt{\sqrt{2} x e^{\left (-\frac{1}{2} \, a\right )} + x^{2} + e^{\left (-a\right )}} e^{\left (\frac{1}{2} \, a\right )} - 1\right ) e^{\left (-\frac{1}{2} \, a\right )} + 4 \,{\left (\sqrt{2} x^{4} e^{\left (2 \, a\right )} + \sqrt{2}\right )} \arctan \left (-\sqrt{2} x e^{\left (\frac{1}{2} \, a\right )} + \sqrt{2} \sqrt{-\sqrt{2} x e^{\left (-\frac{1}{2} \, a\right )} + x^{2} + e^{\left (-a\right )}} e^{\left (\frac{1}{2} \, a\right )} + 1\right ) e^{\left (-\frac{1}{2} \, a\right )} -{\left (\sqrt{2} x^{4} e^{\left (2 \, a\right )} + \sqrt{2}\right )} e^{\left (-\frac{1}{2} \, a\right )} \log \left (\sqrt{2} x e^{\left (-\frac{1}{2} \, a\right )} + x^{2} + e^{\left (-a\right )}\right ) +{\left (\sqrt{2} x^{4} e^{\left (2 \, a\right )} + \sqrt{2}\right )} e^{\left (-\frac{1}{2} \, a\right )} \log \left (-\sqrt{2} x e^{\left (-\frac{1}{2} \, a\right )} + x^{2} + e^{\left (-a\right )}\right ) + 16 \, x}{8 \,{\left (x^{4} e^{\left (2 \, a\right )} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(8*x^5*e^(2*a) + 4*(sqrt(2)*x^4*e^(2*a) + sqrt(2))*arctan(-sqrt(2)*x*e^(1/2*a) + sqrt(2)*sqrt(sqrt(2)*x*e^
(-1/2*a) + x^2 + e^(-a))*e^(1/2*a) - 1)*e^(-1/2*a) + 4*(sqrt(2)*x^4*e^(2*a) + sqrt(2))*arctan(-sqrt(2)*x*e^(1/
2*a) + sqrt(2)*sqrt(-sqrt(2)*x*e^(-1/2*a) + x^2 + e^(-a))*e^(1/2*a) + 1)*e^(-1/2*a) - (sqrt(2)*x^4*e^(2*a) + s
qrt(2))*e^(-1/2*a)*log(sqrt(2)*x*e^(-1/2*a) + x^2 + e^(-a)) + (sqrt(2)*x^4*e^(2*a) + sqrt(2))*e^(-1/2*a)*log(-
sqrt(2)*x*e^(-1/2*a) + x^2 + e^(-a)) + 16*x)/(x^4*e^(2*a) + 1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(a+2*ln(x))**2,x)

[Out]

Integral(tanh(a + 2*log(x))**2, x)

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Giac [A]  time = 1.45328, size = 180, normalized size = 1.09 \begin{align*} -\frac{1}{4} \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} e^{\left (-\frac{1}{2} \, a\right )} + 2 \, x\right )} e^{\left (\frac{1}{2} \, a\right )}\right ) e^{\left (-\frac{1}{2} \, a\right )} - \frac{1}{4} \, \sqrt{2} \arctan \left (-\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} e^{\left (-\frac{1}{2} \, a\right )} - 2 \, x\right )} e^{\left (\frac{1}{2} \, a\right )}\right ) e^{\left (-\frac{1}{2} \, a\right )} - \frac{1}{8} \, \sqrt{2} e^{\left (-\frac{1}{2} \, a\right )} \log \left (\sqrt{2} x e^{\left (-\frac{1}{2} \, a\right )} + x^{2} + e^{\left (-a\right )}\right ) + \frac{1}{8} \, \sqrt{2} e^{\left (-\frac{1}{2} \, a\right )} \log \left (-\sqrt{2} x e^{\left (-\frac{1}{2} \, a\right )} + x^{2} + e^{\left (-a\right )}\right ) + x + \frac{x}{x^{4} e^{\left (2 \, a\right )} + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*e^(-1/2*a) + 2*x)*e^(1/2*a))*e^(-1/2*a) - 1/4*sqrt(2)*arctan(-1/2*sqr
t(2)*(sqrt(2)*e^(-1/2*a) - 2*x)*e^(1/2*a))*e^(-1/2*a) - 1/8*sqrt(2)*e^(-1/2*a)*log(sqrt(2)*x*e^(-1/2*a) + x^2
+ e^(-a)) + 1/8*sqrt(2)*e^(-1/2*a)*log(-sqrt(2)*x*e^(-1/2*a) + x^2 + e^(-a)) + x + x/(x^4*e^(2*a) + 1)