3.97.76 \(\int \frac {1}{3} e^{-2 x} x^{2 x} (96 x+(-9 x^2+4 x^3) \log (4)+(96 x^2+(-6 x^3+2 x^4) \log (4)) \log (x)) \, dx\)

Optimal. Leaf size=26 \[ e^{-2 x} x^{2+2 x} \left (16+\left (-1+\frac {x}{3}\right ) x \log (4)\right ) \]

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Rubi [F]  time = 0.74, 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 {1}{3} e^{-2 x} x^{2 x} \left (96 x+\left (-9 x^2+4 x^3\right ) \log (4)+\left (96 x^2+\left (-6 x^3+2 x^4\right ) \log (4)\right ) \log (x)\right ) \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(x^(2*x)*(96*x + (-9*x^2 + 4*x^3)*Log[4] + (96*x^2 + (-6*x^3 + 2*x^4)*Log[4])*Log[x]))/(3*E^(2*x)),x]

[Out]

32*Defer[Int][x^(1 + 2*x)/E^(2*x), x] - 3*Log[4]*Defer[Int][x^(2 + 2*x)/E^(2*x), x] + 32*Log[x]*Defer[Int][x^(
2 + 2*x)/E^(2*x), x] + (4*Log[4]*Defer[Int][x^(3 + 2*x)/E^(2*x), x])/3 - 2*Log[4]*Log[x]*Defer[Int][x^(3 + 2*x
)/E^(2*x), x] + (2*Log[4]*Log[x]*Defer[Int][x^(4 + 2*x)/E^(2*x), x])/3 - 32*Defer[Int][Defer[Int][x^(2 + 2*x)/
E^(2*x), x]/x, x] + 2*Log[4]*Defer[Int][Defer[Int][x^(3 + 2*x)/E^(2*x), x]/x, x] - (2*Log[4]*Defer[Int][Defer[
Int][x^(4 + 2*x)/E^(2*x), x]/x, x])/3

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{3} \int e^{-2 x} x^{2 x} \left (96 x+\left (-9 x^2+4 x^3\right ) \log (4)+\left (96 x^2+\left (-6 x^3+2 x^4\right ) \log (4)\right ) \log (x)\right ) \, dx\\ &=\frac {1}{3} \int \left (96 e^{-2 x} x^{1+2 x}+e^{-2 x} x^{2+2 x} (-9+4 x) \log (4)+2 e^{-2 x} x^{2+2 x} \left (48-3 x \log (4)+x^2 \log (4)\right ) \log (x)\right ) \, dx\\ &=\frac {2}{3} \int e^{-2 x} x^{2+2 x} \left (48-3 x \log (4)+x^2 \log (4)\right ) \log (x) \, dx+32 \int e^{-2 x} x^{1+2 x} \, dx+\frac {1}{3} \log (4) \int e^{-2 x} x^{2+2 x} (-9+4 x) \, dx\\ &=-\left (\frac {2}{3} \int \frac {48 \int e^{-2 x} x^{2+2 x} \, dx+\log (4) \left (-3 \int e^{-2 x} x^{3+2 x} \, dx+\int e^{-2 x} x^{4+2 x} \, dx\right )}{x} \, dx\right )+32 \int e^{-2 x} x^{1+2 x} \, dx+\frac {1}{3} \log (4) \int \left (-9 e^{-2 x} x^{2+2 x}+4 e^{-2 x} x^{3+2 x}\right ) \, dx+(32 \log (x)) \int e^{-2 x} x^{2+2 x} \, dx+\frac {1}{3} (2 \log (4) \log (x)) \int e^{-2 x} x^{4+2 x} \, dx-(2 \log (4) \log (x)) \int e^{-2 x} x^{3+2 x} \, dx\\ &=-\left (\frac {2}{3} \int \left (\frac {3 \left (16 \int e^{-2 x} x^{2+2 x} \, dx-\log (4) \int e^{-2 x} x^{3+2 x} \, dx\right )}{x}+\frac {\log (4) \int e^{-2 x} x^{4+2 x} \, dx}{x}\right ) \, dx\right )+32 \int e^{-2 x} x^{1+2 x} \, dx+\frac {1}{3} (4 \log (4)) \int e^{-2 x} x^{3+2 x} \, dx-(3 \log (4)) \int e^{-2 x} x^{2+2 x} \, dx+(32 \log (x)) \int e^{-2 x} x^{2+2 x} \, dx+\frac {1}{3} (2 \log (4) \log (x)) \int e^{-2 x} x^{4+2 x} \, dx-(2 \log (4) \log (x)) \int e^{-2 x} x^{3+2 x} \, dx\\ &=-\left (2 \int \frac {16 \int e^{-2 x} x^{2+2 x} \, dx-\log (4) \int e^{-2 x} x^{3+2 x} \, dx}{x} \, dx\right )+32 \int e^{-2 x} x^{1+2 x} \, dx-\frac {1}{3} (2 \log (4)) \int \frac {\int e^{-2 x} x^{4+2 x} \, dx}{x} \, dx+\frac {1}{3} (4 \log (4)) \int e^{-2 x} x^{3+2 x} \, dx-(3 \log (4)) \int e^{-2 x} x^{2+2 x} \, dx+(32 \log (x)) \int e^{-2 x} x^{2+2 x} \, dx+\frac {1}{3} (2 \log (4) \log (x)) \int e^{-2 x} x^{4+2 x} \, dx-(2 \log (4) \log (x)) \int e^{-2 x} x^{3+2 x} \, dx\\ &=-\left (2 \int \left (\frac {16 \int e^{-2 x} x^{2+2 x} \, dx}{x}-\frac {\log (4) \int e^{-2 x} x^{3+2 x} \, dx}{x}\right ) \, dx\right )+32 \int e^{-2 x} x^{1+2 x} \, dx-\frac {1}{3} (2 \log (4)) \int \frac {\int e^{-2 x} x^{4+2 x} \, dx}{x} \, dx+\frac {1}{3} (4 \log (4)) \int e^{-2 x} x^{3+2 x} \, dx-(3 \log (4)) \int e^{-2 x} x^{2+2 x} \, dx+(32 \log (x)) \int e^{-2 x} x^{2+2 x} \, dx+\frac {1}{3} (2 \log (4) \log (x)) \int e^{-2 x} x^{4+2 x} \, dx-(2 \log (4) \log (x)) \int e^{-2 x} x^{3+2 x} \, dx\\ &=32 \int e^{-2 x} x^{1+2 x} \, dx-32 \int \frac {\int e^{-2 x} x^{2+2 x} \, dx}{x} \, dx-\frac {1}{3} (2 \log (4)) \int \frac {\int e^{-2 x} x^{4+2 x} \, dx}{x} \, dx+\frac {1}{3} (4 \log (4)) \int e^{-2 x} x^{3+2 x} \, dx+(2 \log (4)) \int \frac {\int e^{-2 x} x^{3+2 x} \, dx}{x} \, dx-(3 \log (4)) \int e^{-2 x} x^{2+2 x} \, dx+(32 \log (x)) \int e^{-2 x} x^{2+2 x} \, dx+\frac {1}{3} (2 \log (4) \log (x)) \int e^{-2 x} x^{4+2 x} \, dx-(2 \log (4) \log (x)) \int e^{-2 x} x^{3+2 x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.28, size = 29, normalized size = 1.12 \begin {gather*} \frac {1}{3} e^{-2 x} x^{2+2 x} \left (48-3 x \log (4)+x^2 \log (4)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^(2*x)*(96*x + (-9*x^2 + 4*x^3)*Log[4] + (96*x^2 + (-6*x^3 + 2*x^4)*Log[4])*Log[x]))/(3*E^(2*x)),x
]

[Out]

(x^(2 + 2*x)*(48 - 3*x*Log[4] + x^2*Log[4]))/(3*E^(2*x))

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fricas [A]  time = 0.60, size = 29, normalized size = 1.12 \begin {gather*} \frac {2}{3} \, {\left (24 \, x^{2} + {\left (x^{4} - 3 \, x^{3}\right )} \log \relax (2)\right )} x^{2 \, x} e^{\left (-2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((2*(2*x^4-6*x^3)*log(2)+96*x^2)*log(x)+2*(4*x^3-9*x^2)*log(2)+96*x)*exp(x*log(x))^2/exp(x)^2,x,
algorithm="fricas")

[Out]

2/3*(24*x^2 + (x^4 - 3*x^3)*log(2))*x^(2*x)*e^(-2*x)

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giac [B]  time = 0.26, size = 50, normalized size = 1.92 \begin {gather*} \frac {2}{3} \, x^{4} e^{\left (2 \, x \log \relax (x) - 2 \, x\right )} \log \relax (2) - 2 \, x^{3} e^{\left (2 \, x \log \relax (x) - 2 \, x\right )} \log \relax (2) + 16 \, x^{2} e^{\left (2 \, x \log \relax (x) - 2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((2*(2*x^4-6*x^3)*log(2)+96*x^2)*log(x)+2*(4*x^3-9*x^2)*log(2)+96*x)*exp(x*log(x))^2/exp(x)^2,x,
algorithm="giac")

[Out]

2/3*x^4*e^(2*x*log(x) - 2*x)*log(2) - 2*x^3*e^(2*x*log(x) - 2*x)*log(2) + 16*x^2*e^(2*x*log(x) - 2*x)

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maple [A]  time = 0.04, size = 28, normalized size = 1.08

method	result	size
risch	\(\frac {2 x^{2} \left (x^{2} \ln \relax (2)-3 x \ln \relax (2)+24\right ) {\mathrm e}^{-2 x} x^{2 x}}{3}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/3*((2*(2*x^4-6*x^3)*ln(2)+96*x^2)*ln(x)+2*(4*x^3-9*x^2)*ln(2)+96*x)*exp(x*ln(x))^2/exp(x)^2,x,method=_RE
TURNVERBOSE)

[Out]

2/3*x^2*(x^2*ln(2)-3*x*ln(2)+24)*exp(-2*x)*(x^x)^2

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maxima [A]  time = 0.62, size = 31, normalized size = 1.19 \begin {gather*} \frac {2}{3} \, {\left (x^{4} \log \relax (2) - 3 \, x^{3} \log \relax (2) + 24 \, x^{2}\right )} e^{\left (2 \, x \log \relax (x) - 2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((2*(2*x^4-6*x^3)*log(2)+96*x^2)*log(x)+2*(4*x^3-9*x^2)*log(2)+96*x)*exp(x*log(x))^2/exp(x)^2,x,
algorithm="maxima")

[Out]

2/3*(x^4*log(2) - 3*x^3*log(2) + 24*x^2)*e^(2*x*log(x) - 2*x)

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mupad [B]  time = 6.89, size = 27, normalized size = 1.04 \begin {gather*} \frac {2\,x^{2\,x}\,x^2\,{\mathrm {e}}^{-2\,x}\,\left (\ln \relax (2)\,x^2-3\,\ln \relax (2)\,x+24\right )}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(2*x*log(x))*exp(-2*x)*(log(x)*(2*log(2)*(6*x^3 - 2*x^4) - 96*x^2) - 96*x + 2*log(2)*(9*x^2 - 4*x^3))
)/3,x)

[Out]

(2*x^(2*x)*x^2*exp(-2*x)*(x^2*log(2) - 3*x*log(2) + 24))/3

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sympy [A]  time = 27.55, size = 46, normalized size = 1.77 \begin {gather*} \frac {\left (2 x^{4} e^{- 2 x} \log {\relax (2 )} - 6 x^{3} e^{- 2 x} \log {\relax (2 )} + 48 x^{2} e^{- 2 x}\right ) e^{2 x \log {\relax (x )}}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((2*(2*x**4-6*x**3)*ln(2)+96*x**2)*ln(x)+2*(4*x**3-9*x**2)*ln(2)+96*x)*exp(x*ln(x))**2/exp(x)**2
,x)

[Out]

(2*x**4*exp(-2*x)*log(2) - 6*x**3*exp(-2*x)*log(2) + 48*x**2*exp(-2*x))*exp(2*x*log(x))/3

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