\(\int \frac {1}{4} e^{e^x-x} (-3 x^2+2 x^3+(-9 x^2+11 x^3-2 x^4+e^x (-3 x^3+2 x^4)) \log (x)) \, dx\) [8174]

Optimal result

Integrand size = 58, antiderivative size = 23 \[ \int \frac {1}{4} e^{e^x-x} \left (-3 x^2+2 x^3+\left (-9 x^2+11 x^3-2 x^4+e^x \left (-3 x^3+2 x^4\right )\right ) \log (x)\right ) \, dx=\frac {1}{2} e^{e^x-x} \left (-\frac {3}{2}+x\right ) x^3 \log (x) \]

[Out]

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

Rubi [F]

\[ \int \frac {1}{4} e^{e^x-x} \left (-3 x^2+2 x^3+\left (-9 x^2+11 x^3-2 x^4+e^x \left (-3 x^3+2 x^4\right )\right ) \log (x)\right ) \, dx=\int \frac {1}{4} e^{e^x-x} \left (-3 x^2+2 x^3+\left (-9 x^2+11 x^3-2 x^4+e^x \left (-3 x^3+2 x^4\right )\right ) \log (x)\right ) \, dx \]

[In]

Int[(E^(E^x - x)*(-3*x^2 + 2*x^3 + (-9*x^2 + 11*x^3 - 2*x^4 + E^x*(-3*x^3 + 2*x^4))*Log[x]))/4,x]

[Out]

(-3*Defer[Int][E^(E^x - x)*x^2, x])/4 - (9*Log[x]*Defer[Int][E^(E^x - x)*x^2, x])/4 - (3*Log[x]*Defer[Int][E^E
^x*x^3, x])/4 + Defer[Int][E^(E^x - x)*x^3, x]/2 + (11*Log[x]*Defer[Int][E^(E^x - x)*x^3, x])/4 + (Log[x]*Defe
r[Int][E^E^x*x^4, x])/2 - (Log[x]*Defer[Int][E^(E^x - x)*x^4, x])/2 + (9*Defer[Int][Defer[Int][E^(E^x - x)*x^2
, x]/x, x])/4 + (3*Defer[Int][Defer[Int][E^E^x*x^3, x]/x, x])/4 - (11*Defer[Int][Defer[Int][E^(E^x - x)*x^3, x
]/x, x])/4 - Defer[Int][Defer[Int][E^E^x*x^4, x]/x, x]/2 + Defer[Int][Defer[Int][E^(E^x - x)*x^4, x]/x, x]/2

Rubi steps \begin{align*} \text {integral}& = \frac {1}{4} \int e^{e^x-x} \left (-3 x^2+2 x^3+\left (-9 x^2+11 x^3-2 x^4+e^x \left (-3 x^3+2 x^4\right )\right ) \log (x)\right ) \, dx \\ & = \frac {1}{4} \int e^{e^x-x} x^2 \left (-3+2 x+\left (-9+\left (11-3 e^x\right ) x+2 \left (-1+e^x\right ) x^2\right ) \log (x)\right ) \, dx \\ & = \frac {1}{4} \int \left (e^{e^x} x^3 (-3+2 x) \log (x)-e^{e^x-x} x^2 \left (3-2 x+9 \log (x)-11 x \log (x)+2 x^2 \log (x)\right )\right ) \, dx \\ & = \frac {1}{4} \int e^{e^x} x^3 (-3+2 x) \log (x) \, dx-\frac {1}{4} \int e^{e^x-x} x^2 \left (3-2 x+9 \log (x)-11 x \log (x)+2 x^2 \log (x)\right ) \, dx \\ & = -\left (\frac {1}{4} \int \left (-e^{e^x-x} x^2 (-3+2 x)+e^{e^x-x} x^2 \left (9-11 x+2 x^2\right ) \log (x)\right ) \, dx\right )-\frac {1}{4} \int \frac {-3 \int e^{e^x} x^3 \, dx+2 \int e^{e^x} x^4 \, dx}{x} \, dx+\frac {1}{2} \log (x) \int e^{e^x} x^4 \, dx-\frac {1}{4} (3 \log (x)) \int e^{e^x} x^3 \, dx \\ & = \frac {1}{4} \int e^{e^x-x} x^2 (-3+2 x) \, dx-\frac {1}{4} \int e^{e^x-x} x^2 \left (9-11 x+2 x^2\right ) \log (x) \, dx-\frac {1}{4} \int \left (-\frac {3 \int e^{e^x} x^3 \, dx}{x}+\frac {2 \int e^{e^x} x^4 \, dx}{x}\right ) \, dx+\frac {1}{2} \log (x) \int e^{e^x} x^4 \, dx-\frac {1}{4} (3 \log (x)) \int e^{e^x} x^3 \, dx \\ & = \frac {1}{4} \int \left (-3 e^{e^x-x} x^2+2 e^{e^x-x} x^3\right ) \, dx+\frac {1}{4} \int \frac {9 \int e^{e^x-x} x^2 \, dx-11 \int e^{e^x-x} x^3 \, dx+2 \int e^{e^x-x} x^4 \, dx}{x} \, dx-\frac {1}{2} \int \frac {\int e^{e^x} x^4 \, dx}{x} \, dx+\frac {3}{4} \int \frac {\int e^{e^x} x^3 \, dx}{x} \, dx+\frac {1}{2} \log (x) \int e^{e^x} x^4 \, dx-\frac {1}{2} \log (x) \int e^{e^x-x} x^4 \, dx-\frac {1}{4} (3 \log (x)) \int e^{e^x} x^3 \, dx-\frac {1}{4} (9 \log (x)) \int e^{e^x-x} x^2 \, dx+\frac {1}{4} (11 \log (x)) \int e^{e^x-x} x^3 \, dx \\ & = \frac {1}{4} \int \left (\frac {9 \int e^{e^x-x} x^2 \, dx-11 \int e^{e^x-x} x^3 \, dx}{x}+\frac {2 \int e^{e^x-x} x^4 \, dx}{x}\right ) \, dx+\frac {1}{2} \int e^{e^x-x} x^3 \, dx-\frac {1}{2} \int \frac {\int e^{e^x} x^4 \, dx}{x} \, dx-\frac {3}{4} \int e^{e^x-x} x^2 \, dx+\frac {3}{4} \int \frac {\int e^{e^x} x^3 \, dx}{x} \, dx+\frac {1}{2} \log (x) \int e^{e^x} x^4 \, dx-\frac {1}{2} \log (x) \int e^{e^x-x} x^4 \, dx-\frac {1}{4} (3 \log (x)) \int e^{e^x} x^3 \, dx-\frac {1}{4} (9 \log (x)) \int e^{e^x-x} x^2 \, dx+\frac {1}{4} (11 \log (x)) \int e^{e^x-x} x^3 \, dx \\ & = \frac {1}{4} \int \frac {9 \int e^{e^x-x} x^2 \, dx-11 \int e^{e^x-x} x^3 \, dx}{x} \, dx+\frac {1}{2} \int e^{e^x-x} x^3 \, dx-\frac {1}{2} \int \frac {\int e^{e^x} x^4 \, dx}{x} \, dx+\frac {1}{2} \int \frac {\int e^{e^x-x} x^4 \, dx}{x} \, dx-\frac {3}{4} \int e^{e^x-x} x^2 \, dx+\frac {3}{4} \int \frac {\int e^{e^x} x^3 \, dx}{x} \, dx+\frac {1}{2} \log (x) \int e^{e^x} x^4 \, dx-\frac {1}{2} \log (x) \int e^{e^x-x} x^4 \, dx-\frac {1}{4} (3 \log (x)) \int e^{e^x} x^3 \, dx-\frac {1}{4} (9 \log (x)) \int e^{e^x-x} x^2 \, dx+\frac {1}{4} (11 \log (x)) \int e^{e^x-x} x^3 \, dx \\ & = \frac {1}{4} \int \left (\frac {9 \int e^{e^x-x} x^2 \, dx}{x}-\frac {11 \int e^{e^x-x} x^3 \, dx}{x}\right ) \, dx+\frac {1}{2} \int e^{e^x-x} x^3 \, dx-\frac {1}{2} \int \frac {\int e^{e^x} x^4 \, dx}{x} \, dx+\frac {1}{2} \int \frac {\int e^{e^x-x} x^4 \, dx}{x} \, dx-\frac {3}{4} \int e^{e^x-x} x^2 \, dx+\frac {3}{4} \int \frac {\int e^{e^x} x^3 \, dx}{x} \, dx+\frac {1}{2} \log (x) \int e^{e^x} x^4 \, dx-\frac {1}{2} \log (x) \int e^{e^x-x} x^4 \, dx-\frac {1}{4} (3 \log (x)) \int e^{e^x} x^3 \, dx-\frac {1}{4} (9 \log (x)) \int e^{e^x-x} x^2 \, dx+\frac {1}{4} (11 \log (x)) \int e^{e^x-x} x^3 \, dx \\ & = \frac {1}{2} \int e^{e^x-x} x^3 \, dx-\frac {1}{2} \int \frac {\int e^{e^x} x^4 \, dx}{x} \, dx+\frac {1}{2} \int \frac {\int e^{e^x-x} x^4 \, dx}{x} \, dx-\frac {3}{4} \int e^{e^x-x} x^2 \, dx+\frac {3}{4} \int \frac {\int e^{e^x} x^3 \, dx}{x} \, dx+\frac {9}{4} \int \frac {\int e^{e^x-x} x^2 \, dx}{x} \, dx-\frac {11}{4} \int \frac {\int e^{e^x-x} x^3 \, dx}{x} \, dx+\frac {1}{2} \log (x) \int e^{e^x} x^4 \, dx-\frac {1}{2} \log (x) \int e^{e^x-x} x^4 \, dx-\frac {1}{4} (3 \log (x)) \int e^{e^x} x^3 \, dx-\frac {1}{4} (9 \log (x)) \int e^{e^x-x} x^2 \, dx+\frac {1}{4} (11 \log (x)) \int e^{e^x-x} x^3 \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 2.31 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {1}{4} e^{e^x-x} \left (-3 x^2+2 x^3+\left (-9 x^2+11 x^3-2 x^4+e^x \left (-3 x^3+2 x^4\right )\right ) \log (x)\right ) \, dx=\frac {1}{4} e^{e^x-x} x^3 (-3+2 x) \log (x) \]

[In]

Integrate[(E^(E^x - x)*(-3*x^2 + 2*x^3 + (-9*x^2 + 11*x^3 - 2*x^4 + E^x*(-3*x^3 + 2*x^4))*Log[x]))/4,x]

[Out]

(E^(E^x - x)*x^3*(-3 + 2*x)*Log[x])/4

Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87

method	result	size
risch	\(\frac {\ln \left (x \right ) x^{3} \left (-3+2 x \right ) {\mathrm e}^{{\mathrm e}^{x}-x}}{4}\)	\(20\)
parallelrisch	\(\frac {\left (2 \,{\mathrm e}^{{\mathrm e}^{x}} \ln \left (x \right ) x^{4}-3 \,{\mathrm e}^{{\mathrm e}^{x}} \ln \left (x \right ) x^{3}\right ) {\mathrm e}^{-x}}{4}\)	\(28\)

[In]

int(1/4*(((2*x^4-3*x^3)*exp(x)-2*x^4+11*x^3-9*x^2)*ln(x)+2*x^3-3*x^2)*exp(exp(x))/exp(x),x,method=_RETURNVERBO
SE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {1}{4} e^{e^x-x} \left (-3 x^2+2 x^3+\left (-9 x^2+11 x^3-2 x^4+e^x \left (-3 x^3+2 x^4\right )\right ) \log (x)\right ) \, dx=\frac {1}{4} \, {\left (2 \, x^{4} - 3 \, x^{3}\right )} e^{\left (-x + e^{x}\right )} \log \left (x\right ) \]

[In]

integrate(1/4*(((2*x^4-3*x^3)*exp(x)-2*x^4+11*x^3-9*x^2)*log(x)+2*x^3-3*x^2)*exp(exp(x))/exp(x),x, algorithm="
fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13 \[ \int \frac {1}{4} e^{e^x-x} \left (-3 x^2+2 x^3+\left (-9 x^2+11 x^3-2 x^4+e^x \left (-3 x^3+2 x^4\right )\right ) \log (x)\right ) \, dx=\frac {\left (2 x^{4} \log {\left (x \right )} - 3 x^{3} \log {\left (x \right )}\right ) e^{- x} e^{e^{x}}}{4} \]

[In]

integrate(1/4*(((2*x**4-3*x**3)*exp(x)-2*x**4+11*x**3-9*x**2)*ln(x)+2*x**3-3*x**2)*exp(exp(x))/exp(x),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {1}{4} e^{e^x-x} \left (-3 x^2+2 x^3+\left (-9 x^2+11 x^3-2 x^4+e^x \left (-3 x^3+2 x^4\right )\right ) \log (x)\right ) \, dx=\frac {1}{4} \, {\left (2 \, x^{4} - 3 \, x^{3}\right )} e^{\left (-x + e^{x}\right )} \log \left (x\right ) \]

[In]

integrate(1/4*(((2*x^4-3*x^3)*exp(x)-2*x^4+11*x^3-9*x^2)*log(x)+2*x^3-3*x^2)*exp(exp(x))/exp(x),x, algorithm="
maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.26 \[ \int \frac {1}{4} e^{e^x-x} \left (-3 x^2+2 x^3+\left (-9 x^2+11 x^3-2 x^4+e^x \left (-3 x^3+2 x^4\right )\right ) \log (x)\right ) \, dx=\frac {1}{2} \, x^{4} e^{\left (-x + e^{x}\right )} \log \left (x\right ) - \frac {3}{4} \, x^{3} e^{\left (-x + e^{x}\right )} \log \left (x\right ) \]

[In]

integrate(1/4*(((2*x^4-3*x^3)*exp(x)-2*x^4+11*x^3-9*x^2)*log(x)+2*x^3-3*x^2)*exp(exp(x))/exp(x),x, algorithm="
giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{4} e^{e^x-x} \left (-3 x^2+2 x^3+\left (-9 x^2+11 x^3-2 x^4+e^x \left (-3 x^3+2 x^4\right )\right ) \log (x)\right ) \, dx=\int -\frac {{\mathrm {e}}^{-x}\,{\mathrm {e}}^{{\mathrm {e}}^x}\,\left (\ln \left (x\right )\,\left ({\mathrm {e}}^x\,\left (3\,x^3-2\,x^4\right )+9\,x^2-11\,x^3+2\,x^4\right )+3\,x^2-2\,x^3\right )}{4} \,d x \]

[In]

int(-(exp(-x)*exp(exp(x))*(log(x)*(exp(x)*(3*x^3 - 2*x^4) + 9*x^2 - 11*x^3 + 2*x^4) + 3*x^2 - 2*x^3))/4,x)

[Out]

int(-(exp(-x)*exp(exp(x))*(log(x)*(exp(x)*(3*x^3 - 2*x^4) + 9*x^2 - 11*x^3 + 2*x^4) + 3*x^2 - 2*x^3))/4, x)