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

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

[Out]

-1/x/(1+exp(2*a)*x^4)-2*exp(2*a)*x^3/(1+exp(2*a)*x^4)-1/4*exp(1/2*a)*arctan(-1+exp(1/2*a)*x*2^(1/2))*2^(1/2)-1
/4*exp(1/2*a)*arctan(1+exp(1/2*a)*x*2^(1/2))*2^(1/2)-1/8*exp(1/2*a)*ln(1+exp(a)*x^2-exp(1/2*a)*x*2^(1/2))*2^(1
/2)+1/8*exp(1/2*a)*ln(1+exp(a)*x^2+exp(1/2*a)*x*2^(1/2))*2^(1/2)

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Rubi [F]  time = 0.04, 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 = {} \[ \int \frac {\tanh ^2(a+2 \log (x))}{x^2} \, dx \]

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*}

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Mathematica [A]  time = 0.80, size = 181, normalized size = 0.95 \[ \frac {1}{4} \left ((-1)^{3/4} e^{a/2} \log \left (\frac {e^{-2 a} \left (\sqrt [4]{-1}-e^{a/2} x\right )}{x^4}\right )+\sqrt [4]{-1} e^{a/2} \log \left (\frac {e^{-2 a} \left ((-1)^{3/4}-e^{a/2} x\right )}{x^4}\right )-(-1)^{3/4} e^{a/2} \log \left (\frac {e^{-2 a} \left (e^{a/2} x+\sqrt [4]{-1}\right )}{x^4}\right )-\sqrt [4]{-1} e^{a/2} \log \left (\frac {e^{-2 a} \left (e^{a/2} x+(-1)^{3/4}\right )}{x^4}\right )-\frac {4}{\frac {e^{-2 a}}{x^3}+x}-\frac {4}{x}\right ) \]

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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fricas [B]  time = 0.69, size = 274, normalized size = 1.44 \[ -\frac {16 \, x^{4} e^{\left (2 \, a\right )} - 4 \, {\left (\sqrt {2} x^{5} e^{\left (2 \, a\right )} + \sqrt {2} x\right )} \arctan \left (-{\left (\sqrt {2} x e^{\left (\frac {5}{2} \, a\right )} - \sqrt {2} \sqrt {x^{2} e^{\left (4 \, a\right )} + \sqrt {2} x e^{\left (\frac {7}{2} \, a\right )} + e^{\left (3 \, a\right )}} e^{\left (\frac {1}{2} \, a\right )} + e^{\left (2 \, a\right )}\right )} e^{\left (-2 \, a\right )}\right ) e^{\left (\frac {1}{2} \, a\right )} - 4 \, {\left (\sqrt {2} x^{5} e^{\left (2 \, a\right )} + \sqrt {2} x\right )} \arctan \left (-{\left (\sqrt {2} x e^{\left (\frac {5}{2} \, a\right )} - \sqrt {2} \sqrt {x^{2} e^{\left (4 \, a\right )} - \sqrt {2} x e^{\left (\frac {7}{2} \, a\right )} + e^{\left (3 \, a\right )}} e^{\left (\frac {1}{2} \, a\right )} - e^{\left (2 \, a\right )}\right )} e^{\left (-2 \, a\right )}\right ) e^{\left (\frac {1}{2} \, a\right )} - {\left (\sqrt {2} x^{5} e^{\left (2 \, a\right )} + \sqrt {2} x\right )} e^{\left (\frac {1}{2} \, a\right )} \log \left (x^{2} e^{\left (4 \, a\right )} + \sqrt {2} x e^{\left (\frac {7}{2} \, a\right )} + e^{\left (3 \, a\right )}\right ) + {\left (\sqrt {2} x^{5} e^{\left (2 \, a\right )} + \sqrt {2} x\right )} e^{\left (\frac {1}{2} \, a\right )} \log \left (x^{2} e^{\left (4 \, a\right )} - \sqrt {2} x e^{\left (\frac {7}{2} \, a\right )} + e^{\left (3 \, a\right )}\right ) + 8}{8 \, {\left (x^{5} e^{\left (2 \, a\right )} + x\right )}} \]

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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giac [A]  time = 0.13, size = 143, normalized size = 0.75 \[ -\frac {1}{4} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} e^{\left (-\frac {1}{2} \, a\right )} + 2 \, x\right )} e^{\left (\frac {1}{2} \, a\right )}\right ) e^{\left (\frac {1}{2} \, a\right )} - \frac {1}{4} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} e^{\left (-\frac {1}{2} \, a\right )} - 2 \, x\right )} e^{\left (\frac {1}{2} \, a\right )}\right ) e^{\left (\frac {1}{2} \, a\right )} + \frac {1}{8} \, \sqrt {2} e^{\left (\frac {1}{2} \, a\right )} \log \left (\sqrt {2} x e^{\left (-\frac {1}{2} \, a\right )} + x^{2} + e^{\left (-a\right )}\right ) - \frac {1}{8} \, \sqrt {2} e^{\left (\frac {1}{2} \, a\right )} \log \left (-\sqrt {2} x e^{\left (-\frac {1}{2} \, a\right )} + x^{2} + e^{\left (-a\right )}\right ) - \frac {2 \, x^{4} e^{\left (2 \, a\right )} + 1}{x^{5} e^{\left (2 \, a\right )} + x} \]

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)

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maple [C]  time = 0.09, size = 64, normalized size = 0.34 \[ \frac {-2 \,{\mathrm e}^{2 a} x^{4}-1}{x \left (1+{\mathrm e}^{2 a} x^{4}\right )}+\frac {\left (\munderset {\textit {\_R} =\RootOf \left (\textit {\_Z}^{4}+{\mathrm e}^{2 a}\right )}{\sum }\textit {\_R} \ln \left (\left (5 \textit {\_R}^{4}+4 \,{\mathrm e}^{2 a}\right ) x +\textit {\_R}^{3}\right )\right )}{4} \]

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 = 0.44, size = 146, normalized size = 0.77 \[ \frac {1}{4} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} e^{\left (\frac {1}{2} \, a\right )} + \frac {2}{x}\right )} e^{\left (-\frac {1}{2} \, a\right )}\right ) e^{\left (\frac {1}{2} \, a\right )} + \frac {1}{4} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} e^{\left (\frac {1}{2} \, a\right )} - \frac {2}{x}\right )} e^{\left (-\frac {1}{2} \, a\right )}\right ) e^{\left (\frac {1}{2} \, a\right )} + \frac {1}{8} \, \sqrt {2} e^{\left (\frac {1}{2} \, a\right )} \log \left (\frac {\sqrt {2} e^{\left (\frac {1}{2} \, a\right )}}{x} + \frac {1}{x^{2}} + e^{a}\right ) - \frac {1}{8} \, \sqrt {2} e^{\left (\frac {1}{2} \, a\right )} \log \left (-\frac {\sqrt {2} e^{\left (\frac {1}{2} \, a\right )}}{x} + \frac {1}{x^{2}} + e^{a}\right ) - \frac {1}{x} - \frac {e^{\left (2 \, a\right )}}{x {\left (\frac {1}{x^{4}} + e^{\left (2 \, a\right )}\right )}} \]

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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mupad [B]  time = 1.10, size = 68, normalized size = 0.36 \[ \frac {\mathrm {atanh}\left (x\,{\left (-{\mathrm {e}}^{2\,a}\right )}^{1/4}\right )\,{\left (-{\mathrm {e}}^{2\,a}\right )}^{1/4}}{2}-\frac {\mathrm {atan}\left (x\,{\left (-{\mathrm {e}}^{2\,a}\right )}^{1/4}\right )\,{\left (-{\mathrm {e}}^{2\,a}\right )}^{1/4}}{2}-\frac {2\,{\mathrm {e}}^{2\,a}\,x^4+1}{{\mathrm {e}}^{2\,a}\,x^5+x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(a + 2*log(x))^2/x^2,x)

[Out]

(atanh(x*(-exp(2*a))^(1/4))*(-exp(2*a))^(1/4))/2 - (atan(x*(-exp(2*a))^(1/4))*(-exp(2*a))^(1/4))/2 - (2*x^4*ex
p(2*a) + 1)/(x + x^5*exp(2*a))

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

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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