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

Optimal. Leaf size=173 \[ -\frac {3 e^{-3 a/2} \log \left (e^a x^2-\sqrt {2} e^{a/2} x+1\right )}{4 \sqrt {2}}+\frac {3 e^{-3 a/2} \log \left (e^a x^2+\sqrt {2} e^{a/2} x+1\right )}{4 \sqrt {2}}+\frac {x^3}{e^{2 a} x^4+1}+\frac {3 e^{-3 a/2} \tan ^{-1}\left (1-\sqrt {2} e^{a/2} x\right )}{2 \sqrt {2}}-\frac {3 e^{-3 a/2} \tan ^{-1}\left (\sqrt {2} e^{a/2} x+1\right )}{2 \sqrt {2}}+\frac {x^3}{3} \]

[Out]

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

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Rubi [F]  time = 0.05, 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 = {} \[ \int x^2 \tanh ^2(a+2 \log (x)) \, dx \]

Verification is Not applicable to the result.

[In]

Int[x^2*Tanh[a + 2*Log[x]]^2,x]

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.71, size = 174, normalized size = 1.01 \[ \frac {1}{12} \left (\frac {12 x^3}{e^{2 a} x^4+1}+9 (-1)^{3/4} e^{-3 a/2} \log \left (\sqrt [4]{-1} e^{-3 a/2}-e^{-a} x\right )+9 \sqrt [4]{-1} e^{-3 a/2} \log \left ((-1)^{3/4} e^{-3 a/2}-e^{-a} x\right )-9 (-1)^{3/4} e^{-3 a/2} \log \left (e^{-a} x+\sqrt [4]{-1} e^{-3 a/2}\right )-9 \sqrt [4]{-1} e^{-3 a/2} \log \left (e^{-a} x+(-1)^{3/4} e^{-3 a/2}\right )+4 x^3\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*Tanh[a + 2*Log[x]]^2,x]

[Out]

(4*x^3 + (12*x^3)/(1 + E^(2*a)*x^4) + (9*(-1)^(3/4)*Log[(-1)^(1/4)/E^((3*a)/2) - x/E^a])/E^((3*a)/2) + (9*(-1)
^(1/4)*Log[(-1)^(3/4)/E^((3*a)/2) - x/E^a])/E^((3*a)/2) - (9*(-1)^(3/4)*Log[(-1)^(1/4)/E^((3*a)/2) + x/E^a])/E
^((3*a)/2) - (9*(-1)^(1/4)*Log[(-1)^(3/4)/E^((3*a)/2) + x/E^a])/E^((3*a)/2))/12

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fricas [A]  time = 0.97, size = 231, normalized size = 1.34 \[ \frac {8 \, x^{7} e^{\left (2 \, a\right )} + 32 \, x^{3} + 36 \, {\left (\sqrt {2} x^{4} e^{\left (2 \, a\right )} + \sqrt {2}\right )} \arctan \left (-\sqrt {2} x e^{\left (\frac {1}{2} \, a\right )} + \sqrt {2} \sqrt {\sqrt {2} x e^{\left (-\frac {1}{2} \, a\right )} + x^{2} + e^{\left (-a\right )}} e^{\left (\frac {1}{2} \, a\right )} - 1\right ) e^{\left (-\frac {3}{2} \, a\right )} + 36 \, {\left (\sqrt {2} x^{4} e^{\left (2 \, a\right )} + \sqrt {2}\right )} \arctan \left (-\sqrt {2} x e^{\left (\frac {1}{2} \, a\right )} + \sqrt {2} \sqrt {-\sqrt {2} x e^{\left (-\frac {1}{2} \, a\right )} + x^{2} + e^{\left (-a\right )}} e^{\left (\frac {1}{2} \, a\right )} + 1\right ) e^{\left (-\frac {3}{2} \, a\right )} + 9 \, {\left (\sqrt {2} x^{4} e^{\left (2 \, a\right )} + \sqrt {2}\right )} e^{\left (-\frac {3}{2} \, a\right )} \log \left (\sqrt {2} x e^{\left (-\frac {1}{2} \, a\right )} + x^{2} + e^{\left (-a\right )}\right ) - 9 \, {\left (\sqrt {2} x^{4} e^{\left (2 \, a\right )} + \sqrt {2}\right )} e^{\left (-\frac {3}{2} \, a\right )} \log \left (-\sqrt {2} x e^{\left (-\frac {1}{2} \, a\right )} + x^{2} + e^{\left (-a\right )}\right )}{24 \, {\left (x^{4} e^{\left (2 \, a\right )} + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(8*x^7*e^(2*a) + 32*x^3 + 36*(sqrt(2)*x^4*e^(2*a) + sqrt(2))*arctan(-sqrt(2)*x*e^(1/2*a) + sqrt(2)*sqrt(s
qrt(2)*x*e^(-1/2*a) + x^2 + e^(-a))*e^(1/2*a) - 1)*e^(-3/2*a) + 36*(sqrt(2)*x^4*e^(2*a) + sqrt(2))*arctan(-sqr
t(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^(-3/2*a) + 9*(sqrt(2)*x
^4*e^(2*a) + sqrt(2))*e^(-3/2*a)*log(sqrt(2)*x*e^(-1/2*a) + x^2 + e^(-a)) - 9*(sqrt(2)*x^4*e^(2*a) + sqrt(2))*
e^(-3/2*a)*log(-sqrt(2)*x*e^(-1/2*a) + x^2 + e^(-a)))/(x^4*e^(2*a) + 1)

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giac [A]  time = 0.15, size = 139, normalized size = 0.80 \[ \frac {1}{3} \, x^{3} - \frac {3}{4} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} e^{\left (-\frac {1}{2} \, a\right )} + 2 \, x\right )} e^{\left (\frac {1}{2} \, a\right )}\right ) e^{\left (-\frac {3}{2} \, a\right )} - \frac {3}{4} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} e^{\left (-\frac {1}{2} \, a\right )} - 2 \, x\right )} e^{\left (\frac {1}{2} \, a\right )}\right ) e^{\left (-\frac {3}{2} \, a\right )} + \frac {3}{8} \, \sqrt {2} e^{\left (-\frac {3}{2} \, a\right )} \log \left (\sqrt {2} x e^{\left (-\frac {1}{2} \, a\right )} + x^{2} + e^{\left (-a\right )}\right ) - \frac {3}{8} \, \sqrt {2} e^{\left (-\frac {3}{2} \, a\right )} \log \left (-\sqrt {2} x e^{\left (-\frac {1}{2} \, a\right )} + x^{2} + e^{\left (-a\right )}\right ) + \frac {x^{3}}{x^{4} e^{\left (2 \, a\right )} + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.08, size = 53, normalized size = 0.31 \[ \frac {x^{3}}{3}+\frac {x^{3}}{1+{\mathrm e}^{2 a} x^{4}}-\frac {3 \,{\mathrm e}^{-2 a} \left (\munderset {\textit {\_R} =\RootOf \left ({\mathrm e}^{2 a} \textit {\_Z}^{4}+1\right )}{\sum }\frac {\ln \left (x -\textit {\_R} \right )}{\textit {\_R}}\right )}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.43, size = 144, normalized size = 0.83 \[ \frac {1}{3} \, x^{3} - \frac {3}{4} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x e^{a} + \sqrt {2} e^{\left (\frac {1}{2} \, a\right )}\right )} e^{\left (-\frac {1}{2} \, a\right )}\right ) e^{\left (-\frac {3}{2} \, a\right )} - \frac {3}{4} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x e^{a} - \sqrt {2} e^{\left (\frac {1}{2} \, a\right )}\right )} e^{\left (-\frac {1}{2} \, a\right )}\right ) e^{\left (-\frac {3}{2} \, a\right )} + \frac {3}{8} \, \sqrt {2} e^{\left (-\frac {3}{2} \, a\right )} \log \left (x^{2} e^{a} + \sqrt {2} x e^{\left (\frac {1}{2} \, a\right )} + 1\right ) - \frac {3}{8} \, \sqrt {2} e^{\left (-\frac {3}{2} \, a\right )} \log \left (x^{2} e^{a} - \sqrt {2} x e^{\left (\frac {1}{2} \, a\right )} + 1\right ) + \frac {x^{3}}{x^{4} e^{\left (2 \, a\right )} + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x^3 - 3/4*sqrt(2)*arctan(1/2*sqrt(2)*(2*x*e^a + sqrt(2)*e^(1/2*a))*e^(-1/2*a))*e^(-3/2*a) - 3/4*sqrt(2)*ar
ctan(1/2*sqrt(2)*(2*x*e^a - sqrt(2)*e^(1/2*a))*e^(-1/2*a))*e^(-3/2*a) + 3/8*sqrt(2)*e^(-3/2*a)*log(x^2*e^a + s
qrt(2)*x*e^(1/2*a) + 1) - 3/8*sqrt(2)*e^(-3/2*a)*log(x^2*e^a - sqrt(2)*x*e^(1/2*a) + 1) + x^3/(x^4*e^(2*a) + 1
)

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mupad [B]  time = 1.12, size = 67, normalized size = 0.39 \[ \frac {x^3}{{\mathrm {e}}^{2\,a}\,x^4+1}+\frac {3\,\mathrm {atan}\left (x\,{\left (-{\mathrm {e}}^{2\,a}\right )}^{1/4}\right )}{2\,{\left (-{\mathrm {e}}^{2\,a}\right )}^{3/4}}+\frac {x^3}{3}+\frac {\mathrm {atan}\left (x\,{\left (-{\mathrm {e}}^{2\,a}\right )}^{1/4}\,1{}\mathrm {i}\right )\,3{}\mathrm {i}}{2\,{\left (-{\mathrm {e}}^{2\,a}\right )}^{3/4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*tanh(a + 2*log(x))^2,x)

[Out]

x^3/(x^4*exp(2*a) + 1) + (3*atan(x*(-exp(2*a))^(1/4)))/(2*(-exp(2*a))^(3/4)) + (atan(x*(-exp(2*a))^(1/4)*1i)*3
i)/(2*(-exp(2*a))^(3/4)) + x^3/3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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