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+exp(2*a)*x^4)-1/4*arctan(-1+exp(1/2*a)*x*2^(1/2))/exp(1/2*a)*2^(1/2)-1/4*arctan(1+exp(1/2*a)*x*2^(1/2))
/exp(1/2*a)*2^(1/2)+1/8*ln(1+exp(a)*x^2-exp(1/2*a)*x*2^(1/2))/exp(1/2*a)*2^(1/2)-1/8*ln(1+exp(a)*x^2+exp(1/2*a
)*x*2^(1/2))/exp(1/2*a)*2^(1/2)

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Rubi [F]  time = 0.01, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, 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*}

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Mathematica [A]  time = 0.58, 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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fricas [B]  time = 0.64, size = 228, normalized size = 1.38 \[ \frac {8 \, x^{5} e^{\left (2 \, a\right )} + 4 \, {\left (\sqrt {2} x^{4} e^{\left (2 \, a\right )} + \sqrt {2}\right )} \arctan \left (-\sqrt {2} x e^{\left (\frac {1}{2} \, a\right )} + \sqrt {2} \sqrt {\sqrt {2} x e^{\left (-\frac {1}{2} \, a\right )} + x^{2} + e^{\left (-a\right )}} e^{\left (\frac {1}{2} \, a\right )} - 1\right ) e^{\left (-\frac {1}{2} \, a\right )} + 4 \, {\left (\sqrt {2} x^{4} e^{\left (2 \, a\right )} + \sqrt {2}\right )} \arctan \left (-\sqrt {2} x e^{\left (\frac {1}{2} \, a\right )} + \sqrt {2} \sqrt {-\sqrt {2} x e^{\left (-\frac {1}{2} \, a\right )} + x^{2} + e^{\left (-a\right )}} e^{\left (\frac {1}{2} \, a\right )} + 1\right ) e^{\left (-\frac {1}{2} \, a\right )} - {\left (\sqrt {2} x^{4} e^{\left (2 \, a\right )} + \sqrt {2}\right )} e^{\left (-\frac {1}{2} \, a\right )} \log \left (\sqrt {2} x e^{\left (-\frac {1}{2} \, a\right )} + x^{2} + e^{\left (-a\right )}\right ) + {\left (\sqrt {2} x^{4} e^{\left (2 \, a\right )} + \sqrt {2}\right )} e^{\left (-\frac {1}{2} \, a\right )} \log \left (-\sqrt {2} x e^{\left (-\frac {1}{2} \, a\right )} + x^{2} + e^{\left (-a\right )}\right ) + 16 \, x}{8 \, {\left (x^{4} e^{\left (2 \, a\right )} + 1\right )}} \]

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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giac [A]  time = 0.15, size = 133, normalized size = 0.81 \[ -\frac {1}{4} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} e^{\left (-\frac {1}{2} \, a\right )} + 2 \, x\right )} e^{\left (\frac {1}{2} \, a\right )}\right ) e^{\left (-\frac {1}{2} \, a\right )} - \frac {1}{4} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} e^{\left (-\frac {1}{2} \, a\right )} - 2 \, x\right )} e^{\left (\frac {1}{2} \, a\right )}\right ) e^{\left (-\frac {1}{2} \, a\right )} - \frac {1}{8} \, \sqrt {2} e^{\left (-\frac {1}{2} \, a\right )} \log \left (\sqrt {2} x e^{\left (-\frac {1}{2} \, a\right )} + x^{2} + e^{\left (-a\right )}\right ) + \frac {1}{8} \, \sqrt {2} e^{\left (-\frac {1}{2} \, a\right )} \log \left (-\sqrt {2} x e^{\left (-\frac {1}{2} \, a\right )} + x^{2} + e^{\left (-a\right )}\right ) + x + \frac {x}{x^{4} e^{\left (2 \, a\right )} + 1} \]

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)

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maple [C]  time = 0.10, size = 47, normalized size = 0.28 \[ x +\frac {x}{1+{\mathrm e}^{2 a} x^{4}}-\frac {{\mathrm e}^{-2 a} \left (\munderset {\textit {\_R} =\RootOf \left ({\mathrm e}^{2 a} \textit {\_Z}^{4}+1\right )}{\sum }\frac {\ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{3}}\right )}{4} \]

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 = 0.44, size = 138, normalized size = 0.84 \[ -\frac {1}{4} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x e^{a} + \sqrt {2} e^{\left (\frac {1}{2} \, a\right )}\right )} e^{\left (-\frac {1}{2} \, a\right )}\right ) e^{\left (-\frac {1}{2} \, a\right )} - \frac {1}{4} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x e^{a} - \sqrt {2} e^{\left (\frac {1}{2} \, a\right )}\right )} e^{\left (-\frac {1}{2} \, a\right )}\right ) e^{\left (-\frac {1}{2} \, a\right )} - \frac {1}{8} \, \sqrt {2} e^{\left (-\frac {1}{2} \, a\right )} \log \left (x^{2} e^{a} + \sqrt {2} x e^{\left (\frac {1}{2} \, a\right )} + 1\right ) + \frac {1}{8} \, \sqrt {2} e^{\left (-\frac {1}{2} \, a\right )} \log \left (x^{2} e^{a} - \sqrt {2} x e^{\left (\frac {1}{2} \, a\right )} + 1\right ) + x + \frac {x}{x^{4} e^{\left (2 \, a\right )} + 1} \]

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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mupad [B]  time = 1.09, size = 61, normalized size = 0.37 \[ x-\frac {\mathrm {atan}\left (x\,{\left (-{\mathrm {e}}^{2\,a}\right )}^{1/4}\right )}{2\,{\left (-{\mathrm {e}}^{2\,a}\right )}^{1/4}}+\frac {x}{{\mathrm {e}}^{2\,a}\,x^4+1}+\frac {\mathrm {atan}\left (x\,{\left (-{\mathrm {e}}^{2\,a}\right )}^{1/4}\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2\,{\left (-{\mathrm {e}}^{2\,a}\right )}^{1/4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(a + 2*log(x))^2,x)

[Out]

x - atan(x*(-exp(2*a))^(1/4))/(2*(-exp(2*a))^(1/4)) + (atan(x*(-exp(2*a))^(1/4)*1i)*1i)/(2*(-exp(2*a))^(1/4))
+ x/(x^4*exp(2*a) + 1)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \tanh ^{2}{\left (a + 2 \log {\relax (x )} \right )}\, dx \]

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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