3.22.30 \(\int \frac {e^{-\frac {1}{-2 \log (\frac {3}{2})+2 e^{2 x} \log (2)}} (2 x \log ^2(\frac {3}{2})+e^{2 x} (x^2-4 x \log (\frac {3}{2})) \log (2)+2 e^{4 x} x \log ^2(2))}{4 \log ^2(\frac {3}{2})-8 e^{2 x} \log (\frac {3}{2}) \log (2)+4 e^{4 x} \log ^2(2)} \, dx\)

Optimal. Leaf size=30 \[ \frac {1}{4} e^{-\frac {1}{2 \left (-\log \left (\frac {3}{2}\right )+e^{2 x} \log (2)\right )}} x^2 \]

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Rubi [F]  time = 5.17, 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 = {} \begin {gather*} \int \frac {e^{-\frac {1}{-2 \log \left (\frac {3}{2}\right )+2 e^{2 x} \log (2)}} \left (2 x \log ^2\left (\frac {3}{2}\right )+e^{2 x} \left (x^2-4 x \log \left (\frac {3}{2}\right )\right ) \log (2)+2 e^{4 x} x \log ^2(2)\right )}{4 \log ^2\left (\frac {3}{2}\right )-8 e^{2 x} \log \left (\frac {3}{2}\right ) \log (2)+4 e^{4 x} \log ^2(2)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(2*x*Log[3/2]^2 + E^(2*x)*(x^2 - 4*x*Log[3/2])*Log[2] + 2*E^(4*x)*x*Log[2]^2)/(E^(-2*Log[3/2] + 2*E^(2*x)*
Log[2])^(-1)*(4*Log[3/2]^2 - 8*E^(2*x)*Log[3/2]*Log[2] + 4*E^(4*x)*Log[2]^2)),x]

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{-\frac {1}{-2 \log \left (\frac {3}{2}\right )+2 e^{2 x} \log (2)}} \left (2 x \log ^2\left (\frac {3}{2}\right )+e^{2 x} \left (x^2-4 x \log \left (\frac {3}{2}\right )\right ) \log (2)+2 e^{4 x} x \log ^2(2)\right )}{4 \left (\log \left (\frac {3}{2}\right )-e^{2 x} \log (2)\right )^2} \, dx\\ &=\frac {1}{4} \int \frac {e^{-\frac {1}{-2 \log \left (\frac {3}{2}\right )+2 e^{2 x} \log (2)}} \left (2 x \log ^2\left (\frac {3}{2}\right )+e^{2 x} \left (x^2-4 x \log \left (\frac {3}{2}\right )\right ) \log (2)+2 e^{4 x} x \log ^2(2)\right )}{\left (\log \left (\frac {3}{2}\right )-e^{2 x} \log (2)\right )^2} \, dx\\ &=\frac {1}{4} \int \left (2 e^{-\frac {1}{-2 \log \left (\frac {3}{2}\right )+2 e^{2 x} \log (2)}} x+\frac {e^{-\frac {1}{-2 \log \left (\frac {3}{2}\right )+2 e^{2 x} \log (2)}} x^2 \log \left (\frac {3}{2}\right )}{\left (\log \left (\frac {3}{2}\right )-e^{2 x} \log (2)\right )^2}-\frac {e^{-\frac {1}{-2 \log \left (\frac {3}{2}\right )+2 e^{2 x} \log (2)}} x^2}{\log \left (\frac {3}{2}\right )-e^{2 x} \log (2)}\right ) \, dx\\ &=-\left (\frac {1}{4} \int \frac {e^{-\frac {1}{-2 \log \left (\frac {3}{2}\right )+2 e^{2 x} \log (2)}} x^2}{\log \left (\frac {3}{2}\right )-e^{2 x} \log (2)} \, dx\right )+\frac {1}{2} \int e^{-\frac {1}{-2 \log \left (\frac {3}{2}\right )+2 e^{2 x} \log (2)}} x \, dx+\frac {1}{4} \log \left (\frac {3}{2}\right ) \int \frac {e^{-\frac {1}{-2 \log \left (\frac {3}{2}\right )+2 e^{2 x} \log (2)}} x^2}{\left (\log \left (\frac {3}{2}\right )-e^{2 x} \log (2)\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [F]  time = 1.39, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {e^{-\frac {1}{-2 \log \left (\frac {3}{2}\right )+2 e^{2 x} \log (2)}} \left (2 x \log ^2\left (\frac {3}{2}\right )+e^{2 x} \left (x^2-4 x \log \left (\frac {3}{2}\right )\right ) \log (2)+2 e^{4 x} x \log ^2(2)\right )}{4 \log ^2\left (\frac {3}{2}\right )-8 e^{2 x} \log \left (\frac {3}{2}\right ) \log (2)+4 e^{4 x} \log ^2(2)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[(2*x*Log[3/2]^2 + E^(2*x)*(x^2 - 4*x*Log[3/2])*Log[2] + 2*E^(4*x)*x*Log[2]^2)/(E^(-2*Log[3/2] + 2*E^
(2*x)*Log[2])^(-1)*(4*Log[3/2]^2 - 8*E^(2*x)*Log[3/2]*Log[2] + 4*E^(4*x)*Log[2]^2)),x]

[Out]

Integrate[(2*x*Log[3/2]^2 + E^(2*x)*(x^2 - 4*x*Log[3/2])*Log[2] + 2*E^(4*x)*x*Log[2]^2)/(E^(-2*Log[3/2] + 2*E^
(2*x)*Log[2])^(-1)*(4*Log[3/2]^2 - 8*E^(2*x)*Log[3/2]*Log[2] + 4*E^(4*x)*Log[2]^2)), x]

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fricas [A]  time = 1.07, size = 22, normalized size = 0.73 \begin {gather*} \frac {1}{4} \, x^{2} e^{\left (-\frac {1}{2 \, {\left (e^{\left (2 \, x\right )} \log \relax (2) - \log \left (\frac {3}{2}\right )\right )}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*log(2)^2*exp(2*x)^2+(-4*x*log(3/2)+x^2)*log(2)*exp(2*x)+2*x*log(3/2)^2)/(4*log(2)^2*exp(2*x)^2-
8*log(3/2)*log(2)*exp(2*x)+4*log(3/2)^2)/exp(1/(2*log(2)*exp(2*x)-2*log(3/2))),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.36, size = 114, normalized size = 3.80 \begin {gather*} \frac {1}{4} \, x^{2} e^{\left (2 \, x - \frac {4 \, x e^{\left (2 \, x\right )} \log \relax (3) \log \relax (2) - 4 \, x e^{\left (2 \, x\right )} \log \relax (2)^{2} - 4 \, x \log \relax (3)^{2} + 8 \, x \log \relax (3) \log \relax (2) - 4 \, x \log \relax (2)^{2} + e^{\left (2 \, x\right )} \log \relax (2)}{2 \, {\left (e^{\left (2 \, x\right )} \log \relax (3) \log \relax (2) - e^{\left (2 \, x\right )} \log \relax (2)^{2} - \log \relax (3)^{2} + 2 \, \log \relax (3) \log \relax (2) - \log \relax (2)^{2}\right )}} + \frac {1}{2 \, {\left (\log \relax (3) - \log \relax (2)\right )}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*log(2)^2*exp(2*x)^2+(-4*x*log(3/2)+x^2)*log(2)*exp(2*x)+2*x*log(3/2)^2)/(4*log(2)^2*exp(2*x)^2-
8*log(3/2)*log(2)*exp(2*x)+4*log(3/2)^2)/exp(1/(2*log(2)*exp(2*x)-2*log(3/2))),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.25, size = 25, normalized size = 0.83

method	result	size
risch	\(\frac {x^{2} {\mathrm e}^{-\frac {1}{2 \left (\ln \relax (2) {\mathrm e}^{2 x}+\ln \relax (2)-\ln \relax (3)\right )}}}{4}\)	\(25\)
norman	\(\frac {\left (\left (\frac {\ln \relax (2)}{4}-\frac {\ln \relax (3)}{4}\right ) x^{2}+\frac {x^{2} \ln \relax (2) {\mathrm e}^{2 x}}{4}\right ) {\mathrm e}^{-\frac {1}{{\mathrm e}^{2 x} \ln \relax (4)+\ln \left (\frac {4}{9}\right )}}}{\ln \relax (2) {\mathrm e}^{2 x}-\ln \left (\frac {3}{2}\right )}\)	\(59\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x*ln(2)^2*exp(2*x)^2+(-4*x*ln(3/2)+x^2)*ln(2)*exp(2*x)+2*x*ln(3/2)^2)/(4*ln(2)^2*exp(2*x)^2-8*ln(3/2)*l
n(2)*exp(2*x)+4*ln(3/2)^2)/exp(1/(2*ln(2)*exp(2*x)-2*ln(3/2))),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {1}{4} \, \int \frac {{\left (2 \, x e^{\left (4 \, x\right )} \log \relax (2)^{2} + {\left (x^{2} - 4 \, x \log \left (\frac {3}{2}\right )\right )} e^{\left (2 \, x\right )} \log \relax (2) + 2 \, x \log \left (\frac {3}{2}\right )^{2}\right )} e^{\left (-\frac {1}{2 \, {\left (e^{\left (2 \, x\right )} \log \relax (2) - \log \left (\frac {3}{2}\right )\right )}}\right )}}{e^{\left (4 \, x\right )} \log \relax (2)^{2} - 2 \, e^{\left (2 \, x\right )} \log \relax (2) \log \left (\frac {3}{2}\right ) + \log \left (\frac {3}{2}\right )^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*log(2)^2*exp(2*x)^2+(-4*x*log(3/2)+x^2)*log(2)*exp(2*x)+2*x*log(3/2)^2)/(4*log(2)^2*exp(2*x)^2-
8*log(3/2)*log(2)*exp(2*x)+4*log(3/2)^2)/exp(1/(2*log(2)*exp(2*x)-2*log(3/2))),x, algorithm="maxima")

[Out]

1/4*integrate((2*x*e^(4*x)*log(2)^2 + (x^2 - 4*x*log(3/2))*e^(2*x)*log(2) + 2*x*log(3/2)^2)*e^(-1/2/(e^(2*x)*l
og(2) - log(3/2)))/(e^(4*x)*log(2)^2 - 2*e^(2*x)*log(2)*log(3/2) + log(3/2)^2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {{\mathrm {e}}^{\frac {1}{2\,\ln \left (\frac {3}{2}\right )-2\,{\mathrm {e}}^{2\,x}\,\ln \relax (2)}}\,\left (2\,x\,{\ln \left (\frac {3}{2}\right )}^2+2\,x\,{\mathrm {e}}^{4\,x}\,{\ln \relax (2)}^2-{\mathrm {e}}^{2\,x}\,\ln \relax (2)\,\left (4\,x\,\ln \left (\frac {3}{2}\right )-x^2\right )\right )}{4\,{\mathrm {e}}^{4\,x}\,{\ln \relax (2)}^2+4\,{\ln \left (\frac {3}{2}\right )}^2-8\,{\mathrm {e}}^{2\,x}\,\ln \relax (2)\,\ln \left (\frac {3}{2}\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(1/(2*log(3/2) - 2*exp(2*x)*log(2)))*(2*x*log(3/2)^2 + 2*x*exp(4*x)*log(2)^2 - exp(2*x)*log(2)*(4*x*lo
g(3/2) - x^2)))/(4*exp(4*x)*log(2)^2 + 4*log(3/2)^2 - 8*exp(2*x)*log(2)*log(3/2)),x)

[Out]

int((exp(1/(2*log(3/2) - 2*exp(2*x)*log(2)))*(2*x*log(3/2)^2 + 2*x*exp(4*x)*log(2)^2 - exp(2*x)*log(2)*(4*x*lo
g(3/2) - x^2)))/(4*exp(4*x)*log(2)^2 + 4*log(3/2)^2 - 8*exp(2*x)*log(2)*log(3/2)), x)

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sympy [A]  time = 3.87, size = 24, normalized size = 0.80 \begin {gather*} \frac {x^{2} e^{- \frac {1}{2 e^{2 x} \log {\relax (2 )} - 2 \log {\left (\frac {3}{2} \right )}}}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*ln(2)**2*exp(2*x)**2+(-4*x*ln(3/2)+x**2)*ln(2)*exp(2*x)+2*x*ln(3/2)**2)/(4*ln(2)**2*exp(2*x)**2
-8*ln(3/2)*ln(2)*exp(2*x)+4*ln(3/2)**2)/exp(1/(2*ln(2)*exp(2*x)-2*ln(3/2))),x)

[Out]

x**2*exp(-1/(2*exp(2*x)*log(2) - 2*log(3/2)))/4

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