\(\int \frac {1}{3} e^{-2 x} x^{2 x} (96 x+(-9 x^2+4 x^3) \log (4)+(96 x^2+(-6 x^3+2 x^4) \log (4)) \log (x)) \, dx\) [9676]

Optimal result

Integrand size = 55, antiderivative size = 26 \[ \int \frac {1}{3} e^{-2 x} x^{2 x} \left (96 x+\left (-9 x^2+4 x^3\right ) \log (4)+\left (96 x^2+\left (-6 x^3+2 x^4\right ) \log (4)\right ) \log (x)\right ) \, dx=e^{-2 x} x^{2+2 x} \left (16+\left (-1+\frac {x}{3}\right ) x \log (4)\right ) \]

[Out]

(16+2*(-1+1/3*x)*x*ln(2))/exp(x)^2*exp(x*ln(x))^2*x^2

Rubi [F]

\[ \int \frac {1}{3} e^{-2 x} x^{2 x} \left (96 x+\left (-9 x^2+4 x^3\right ) \log (4)+\left (96 x^2+\left (-6 x^3+2 x^4\right ) \log (4)\right ) \log (x)\right ) \, dx=\int \frac {1}{3} e^{-2 x} x^{2 x} \left (96 x+\left (-9 x^2+4 x^3\right ) \log (4)+\left (96 x^2+\left (-6 x^3+2 x^4\right ) \log (4)\right ) \log (x)\right ) \, dx \]

[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{align*} \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{align*}

Mathematica [A] (verified)

Time = 1.34 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12 \[ \int \frac {1}{3} e^{-2 x} x^{2 x} \left (96 x+\left (-9 x^2+4 x^3\right ) \log (4)+\left (96 x^2+\left (-6 x^3+2 x^4\right ) \log (4)\right ) \log (x)\right ) \, dx=\frac {1}{3} e^{-2 x} x^{2+2 x} \left (48-3 x \log (4)+x^2 \log (4)\right ) \]

[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))

Maple [A] (verified)

Time = 6.04 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08

method	result	size
risch	\(\frac {2 x^{2} \left (x^{2} \ln \left (2\right )-3 x \ln \left (2\right )+24\right ) {\mathrm e}^{-2 x} x^{2 x}}{3}\)	\(28\)
parallelrisch	\(\frac {\left (2 \ln \left (2\right ) x^{4} {\mathrm e}^{2 x \ln \left (x \right )}-6 \ln \left (2\right ) x^{3} {\mathrm e}^{2 x \ln \left (x \right )}+48 \,{\mathrm e}^{2 x \ln \left (x \right )} x^{2}\right ) {\mathrm e}^{-2 x}}{3}\)	\(48\)

[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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12 \[ \int \frac {1}{3} e^{-2 x} x^{2 x} \left (96 x+\left (-9 x^2+4 x^3\right ) \log (4)+\left (96 x^2+\left (-6 x^3+2 x^4\right ) \log (4)\right ) \log (x)\right ) \, dx=\frac {2}{3} \, {\left (24 \, x^{2} + {\left (x^{4} - 3 \, x^{3}\right )} \log \left (2\right )\right )} x^{2 \, x} e^{\left (-2 \, x\right )} \]

[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)

Sympy [A] (verification not implemented)

Time = 13.17 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.77 \[ \int \frac {1}{3} e^{-2 x} x^{2 x} \left (96 x+\left (-9 x^2+4 x^3\right ) \log (4)+\left (96 x^2+\left (-6 x^3+2 x^4\right ) \log (4)\right ) \log (x)\right ) \, dx=\frac {\left (2 x^{4} e^{- 2 x} \log {\left (2 \right )} - 6 x^{3} e^{- 2 x} \log {\left (2 \right )} + 48 x^{2} e^{- 2 x}\right ) e^{2 x \log {\left (x \right )}}}{3} \]

[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

Maxima [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19 \[ \int \frac {1}{3} e^{-2 x} x^{2 x} \left (96 x+\left (-9 x^2+4 x^3\right ) \log (4)+\left (96 x^2+\left (-6 x^3+2 x^4\right ) \log (4)\right ) \log (x)\right ) \, dx=\frac {2}{3} \, {\left (x^{4} \log \left (2\right ) - 3 \, x^{3} \log \left (2\right ) + 24 \, x^{2}\right )} e^{\left (2 \, x \log \left (x\right ) - 2 \, x\right )} \]

[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)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 50 vs. \(2 (23) = 46\).

Time = 0.34 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.92 \[ \int \frac {1}{3} e^{-2 x} x^{2 x} \left (96 x+\left (-9 x^2+4 x^3\right ) \log (4)+\left (96 x^2+\left (-6 x^3+2 x^4\right ) \log (4)\right ) \log (x)\right ) \, dx=\frac {2}{3} \, x^{4} e^{\left (2 \, x \log \left (x\right ) - 2 \, x\right )} \log \left (2\right ) - 2 \, x^{3} e^{\left (2 \, x \log \left (x\right ) - 2 \, x\right )} \log \left (2\right ) + 16 \, x^{2} e^{\left (2 \, x \log \left (x\right ) - 2 \, x\right )} \]

[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)

Mupad [B] (verification not implemented)

Time = 14.71 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.04 \[ \int \frac {1}{3} e^{-2 x} x^{2 x} \left (96 x+\left (-9 x^2+4 x^3\right ) \log (4)+\left (96 x^2+\left (-6 x^3+2 x^4\right ) \log (4)\right ) \log (x)\right ) \, dx=\frac {2\,x^{2\,x}\,x^2\,{\mathrm {e}}^{-2\,x}\,\left (\ln \left (2\right )\,x^2-3\,\ln \left (2\right )\,x+24\right )}{3} \]

[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