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

Optimal. Leaf size=151 \[ -\frac {e^{-3 a/2} \log \left (e^a x^2-\sqrt {2} e^{a/2} x+1\right )}{2 \sqrt {2}}+\frac {e^{-3 a/2} \log \left (e^a x^2+\sqrt {2} e^{a/2} x+1\right )}{2 \sqrt {2}}+\frac {e^{-3 a/2} \tan ^{-1}\left (1-\sqrt {2} e^{a/2} x\right )}{\sqrt {2}}-\frac {e^{-3 a/2} \tan ^{-1}\left (\sqrt {2} e^{a/2} x+1\right )}{\sqrt {2}}+\frac {x^3}{3} \]

[Out]

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

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Rubi [F]  time = 0.02, 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 = {} \[ \int x^2 \tanh (a+2 \log (x)) \, dx \]

Verification is Not applicable to the result.

[In]

Int[x^2*Tanh[a + 2*Log[x]],x]

[Out]

Defer[Int][x^2*Tanh[a + 2*Log[x]], x]

Rubi steps

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

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Mathematica [C]  time = 0.26, size = 64, normalized size = 0.42 \[ \frac {1}{6} \left (3 (\cosh (2 a)-\sinh (2 a)) \text {RootSum}\left [\text {$\#$1}^4 \sinh (a)+\text {$\#$1}^4 \cosh (a)-\sinh (a)+\cosh (a)\& ,\frac {\log (x)-\log (x-\text {$\#$1})}{\text {$\#$1}}\& \right ]+2 x^3\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*Tanh[a + 2*Log[x]],x]

[Out]

(2*x^3 + 3*RootSum[Cosh[a] - Sinh[a] + Cosh[a]*#1^4 + Sinh[a]*#1^4 & , (Log[x] - Log[x - #1])/#1 & ]*(Cosh[2*a
] - Sinh[2*a]))/6

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fricas [A]  time = 0.64, size = 158, normalized size = 1.05 \[ \frac {1}{3} \, x^{3} + \sqrt {2} \arctan \left (-\sqrt {2} x e^{\left (\frac {1}{2} \, a\right )} + \sqrt {2} \sqrt {\sqrt {2} x e^{\left (-\frac {1}{2} \, a\right )} + x^{2} + e^{\left (-a\right )}} e^{\left (\frac {1}{2} \, a\right )} - 1\right ) e^{\left (-\frac {3}{2} \, a\right )} + \sqrt {2} \arctan \left (-\sqrt {2} x e^{\left (\frac {1}{2} \, a\right )} + \sqrt {2} \sqrt {-\sqrt {2} x e^{\left (-\frac {1}{2} \, a\right )} + x^{2} + e^{\left (-a\right )}} e^{\left (\frac {1}{2} \, a\right )} + 1\right ) e^{\left (-\frac {3}{2} \, a\right )} + \frac {1}{4} \, \sqrt {2} e^{\left (-\frac {3}{2} \, a\right )} \log \left (\sqrt {2} x e^{\left (-\frac {1}{2} \, a\right )} + x^{2} + e^{\left (-a\right )}\right ) - \frac {1}{4} \, \sqrt {2} e^{\left (-\frac {3}{2} \, a\right )} \log \left (-\sqrt {2} x e^{\left (-\frac {1}{2} \, a\right )} + x^{2} + e^{\left (-a\right )}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.14, size = 123, normalized size = 0.81 \[ \frac {1}{3} \, x^{3} - \frac {1}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} e^{\left (-\frac {1}{2} \, a\right )} + 2 \, x\right )} e^{\left (\frac {1}{2} \, a\right )}\right ) e^{\left (-\frac {3}{2} \, a\right )} - \frac {1}{2} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} e^{\left (-\frac {1}{2} \, a\right )} - 2 \, x\right )} e^{\left (\frac {1}{2} \, a\right )}\right ) e^{\left (-\frac {3}{2} \, a\right )} + \frac {1}{4} \, \sqrt {2} e^{\left (-\frac {3}{2} \, a\right )} \log \left (\sqrt {2} x e^{\left (-\frac {1}{2} \, a\right )} + x^{2} + e^{\left (-a\right )}\right ) - \frac {1}{4} \, \sqrt {2} e^{\left (-\frac {3}{2} \, a\right )} \log \left (-\sqrt {2} x e^{\left (-\frac {1}{2} \, a\right )} + x^{2} + e^{\left (-a\right )}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.10, size = 37, normalized size = 0.25 \[ \frac {x^{3}}{3}-\frac {{\mathrm e}^{-2 a} \left (\munderset {\textit {\_R} =\RootOf \left ({\mathrm e}^{2 a} \textit {\_Z}^{4}+1\right )}{\sum }\frac {\ln \left (x -\textit {\_R} \right )}{\textit {\_R}}\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.45, size = 128, normalized size = 0.85 \[ \frac {1}{3} \, x^{3} - \frac {1}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x e^{a} + \sqrt {2} e^{\left (\frac {1}{2} \, a\right )}\right )} e^{\left (-\frac {1}{2} \, a\right )}\right ) e^{\left (-\frac {3}{2} \, a\right )} - \frac {1}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x e^{a} - \sqrt {2} e^{\left (\frac {1}{2} \, a\right )}\right )} e^{\left (-\frac {1}{2} \, a\right )}\right ) e^{\left (-\frac {3}{2} \, a\right )} + \frac {1}{4} \, \sqrt {2} e^{\left (-\frac {3}{2} \, a\right )} \log \left (x^{2} e^{a} + \sqrt {2} x e^{\left (\frac {1}{2} \, a\right )} + 1\right ) - \frac {1}{4} \, \sqrt {2} e^{\left (-\frac {3}{2} \, a\right )} \log \left (x^{2} e^{a} - \sqrt {2} x e^{\left (\frac {1}{2} \, a\right )} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.10, size = 47, normalized size = 0.31 \[ \frac {\mathrm {atan}\left (x\,{\left (-{\mathrm {e}}^{2\,a}\right )}^{1/4}\right )}{{\left (-{\mathrm {e}}^{2\,a}\right )}^{3/4}}-\frac {\mathrm {atanh}\left (x\,{\left (-{\mathrm {e}}^{2\,a}\right )}^{1/4}\right )}{{\left (-{\mathrm {e}}^{2\,a}\right )}^{3/4}}+\frac {x^3}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*tanh(a + 2*log(x)),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2*tanh(a + 2*log(x)), x)

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