3.158 \(\int \frac{\tanh ^2(a+2 \log (x))}{x^2} \, dx\)

Optimal. Leaf size=190 \[ -\frac{2 e^{2 a} x^3}{e^{2 a} x^4+1}-\frac{1}{x \left (e^{2 a} x^4+1\right )}-\frac{e^{a/2} \log \left (e^a x^2-\sqrt{2} e^{a/2} x+1\right )}{4 \sqrt{2}}+\frac{e^{a/2} \log \left (e^a x^2+\sqrt{2} e^{a/2} x+1\right )}{4 \sqrt{2}}+\frac{e^{a/2} \tan ^{-1}\left (1-\sqrt{2} e^{a/2} x\right )}{2 \sqrt{2}}-\frac{e^{a/2} \tan ^{-1}\left (\sqrt{2} e^{a/2} x+1\right )}{2 \sqrt{2}} \]

[Out]

-(1/(x*(1 + E^(2*a)*x^4))) - (2*E^(2*a)*x^3)/(1 + E^(2*a)*x^4) + (E^(a/2)*ArcTan[1 - Sqrt[2]*E^(a/2)*x])/(2*Sq
rt[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])

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Rubi [F]  time = 0.044585, 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 = {} \[ \int \frac{\tanh ^2(a+2 \log (x))}{x^2} \, dx \]

Verification is Not applicable to the result.

[In]

Int[Tanh[a + 2*Log[x]]^2/x^2,x]

[Out]

Defer[Int][Tanh[a + 2*Log[x]]^2/x^2, x]

Rubi steps

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

Mathematica [A]  time = 0.735635, size = 181, normalized size = 0.95 \[ \frac{1}{4} \left (-\frac{4}{\frac{e^{-2 a}}{x^3}+x}+(-1)^{3/4} e^{a/2} \log \left (\frac{e^{-2 a} \left (\sqrt [4]{-1}-e^{a/2} x\right )}{x^4}\right )+\sqrt [4]{-1} e^{a/2} \log \left (\frac{e^{-2 a} \left ((-1)^{3/4}-e^{a/2} x\right )}{x^4}\right )-(-1)^{3/4} e^{a/2} \log \left (\frac{e^{-2 a} \left (e^{a/2} x+\sqrt [4]{-1}\right )}{x^4}\right )-\sqrt [4]{-1} e^{a/2} \log \left (\frac{e^{-2 a} \left (e^{a/2} x+(-1)^{3/4}\right )}{x^4}\right )-\frac{4}{x}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Tanh[a + 2*Log[x]]^2/x^2,x]

[Out]

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

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Maple [C]  time = 0.026, size = 64, normalized size = 0.3 \begin{align*}{\frac{-2\,{{\rm e}^{2\,a}}{x}^{4}-1}{x \left ( 1+{{\rm e}^{2\,a}}{x}^{4} \right ) }}+{\frac{\sum _{{\it \_R}={\it RootOf} \left ({{\it \_Z}}^{4}+{{\rm e}^{2\,a}} \right ) }{\it \_R}\,\ln \left ( \left ( 5\,{{\it \_R}}^{4}+4\,{{\rm e}^{2\,a}} \right ) x+{{\it \_R}}^{3} \right ) }{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-2*exp(2*a)*x^4-1)/x/(1+exp(2*a)*x^4)+1/4*sum(_R*ln((5*_R^4+4*exp(2*a))*x+_R^3),_R=RootOf(_Z^4+exp(2*a)))

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Maxima [A]  time = 1.93076, size = 197, normalized size = 1.04 \begin{align*} \frac{1}{4} \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} e^{\left (\frac{1}{2} \, a\right )} + \frac{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 )} - \frac{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 (\frac{\sqrt{2} e^{\left (\frac{1}{2} \, a\right )}}{x} + \frac{1}{x^{2}} + e^{a}\right ) - \frac{1}{8} \, \sqrt{2} e^{\left (\frac{1}{2} \, a\right )} \log \left (-\frac{\sqrt{2} e^{\left (\frac{1}{2} \, a\right )}}{x} + \frac{1}{x^{2}} + e^{a}\right ) - \frac{1}{x} - \frac{e^{\left (2 \, a\right )}}{x{\left (\frac{1}{x^{4}} + e^{\left (2 \, a\right )}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

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Fricas [B]  time = 2.14456, size = 771, normalized size = 4.06 \begin{align*} -\frac{16 \, x^{4} e^{\left (2 \, a\right )} - 4 \,{\left (\sqrt{2} x^{5} e^{\left (2 \, a\right )} + \sqrt{2} x\right )} \arctan \left (-{\left (\sqrt{2} x e^{\left (\frac{5}{2} \, a\right )} - \sqrt{2} \sqrt{x^{2} e^{\left (4 \, a\right )} + \sqrt{2} x e^{\left (\frac{7}{2} \, a\right )} + e^{\left (3 \, a\right )}} e^{\left (\frac{1}{2} \, a\right )} + e^{\left (2 \, a\right )}\right )} e^{\left (-2 \, a\right )}\right ) e^{\left (\frac{1}{2} \, a\right )} - 4 \,{\left (\sqrt{2} x^{5} e^{\left (2 \, a\right )} + \sqrt{2} x\right )} \arctan \left (-{\left (\sqrt{2} x e^{\left (\frac{5}{2} \, a\right )} - \sqrt{2} \sqrt{x^{2} e^{\left (4 \, a\right )} - \sqrt{2} x e^{\left (\frac{7}{2} \, a\right )} + e^{\left (3 \, a\right )}} e^{\left (\frac{1}{2} \, a\right )} - e^{\left (2 \, a\right )}\right )} e^{\left (-2 \, a\right )}\right ) e^{\left (\frac{1}{2} \, a\right )} -{\left (\sqrt{2} x^{5} e^{\left (2 \, a\right )} + \sqrt{2} x\right )} e^{\left (\frac{1}{2} \, a\right )} \log \left (x^{2} e^{\left (4 \, a\right )} + \sqrt{2} x e^{\left (\frac{7}{2} \, a\right )} + e^{\left (3 \, a\right )}\right ) +{\left (\sqrt{2} x^{5} e^{\left (2 \, a\right )} + \sqrt{2} x\right )} e^{\left (\frac{1}{2} \, a\right )} \log \left (x^{2} e^{\left (4 \, a\right )} - \sqrt{2} x e^{\left (\frac{7}{2} \, a\right )} + e^{\left (3 \, a\right )}\right ) + 8}{8 \,{\left (x^{5} e^{\left (2 \, a\right )} + x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(16*x^4*e^(2*a) - 4*(sqrt(2)*x^5*e^(2*a) + sqrt(2)*x)*arctan(-(sqrt(2)*x*e^(5/2*a) - sqrt(2)*sqrt(x^2*e^(
4*a) + sqrt(2)*x*e^(7/2*a) + e^(3*a))*e^(1/2*a) + e^(2*a))*e^(-2*a))*e^(1/2*a) - 4*(sqrt(2)*x^5*e^(2*a) + sqrt
(2)*x)*arctan(-(sqrt(2)*x*e^(5/2*a) - sqrt(2)*sqrt(x^2*e^(4*a) - sqrt(2)*x*e^(7/2*a) + e^(3*a))*e^(1/2*a) - e^
(2*a))*e^(-2*a))*e^(1/2*a) - (sqrt(2)*x^5*e^(2*a) + sqrt(2)*x)*e^(1/2*a)*log(x^2*e^(4*a) + sqrt(2)*x*e^(7/2*a)
+ e^(3*a)) + (sqrt(2)*x^5*e^(2*a) + sqrt(2)*x)*e^(1/2*a)*log(x^2*e^(4*a) - sqrt(2)*x*e^(7/2*a) + e^(3*a)) + 8
)/(x^5*e^(2*a) + x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(tanh(a + 2*log(x))**2/x**2, x)

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Giac [A]  time = 1.4497, size = 193, normalized size = 1.02 \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 ) - \frac{2 \, x^{4} e^{\left (2 \, a\right )} + 1}{x^{5} e^{\left (2 \, a\right )} + x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(a+2*log(x))^2/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*sqrt
(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)) - (2*x^4*e^(2*a) + 1)/(x^5*e^(2*a) +
x)