\(\int \frac {e^{\frac {1}{5} (-12+e^{e^3} (-3-5 x)-5 x+25 x^2+(-3-5 x) \log (x))} (-3-10 x-5 e^{e^3} x+50 x^2-5 x \log (x))}{5 x} \, dx\) [7741]

Optimal result

Integrand size = 65, antiderivative size = 28 \[ \int \frac {e^{\frac {1}{5} \left (-12+e^{e^3} (-3-5 x)-5 x+25 x^2+(-3-5 x) \log (x)\right )} \left (-3-10 x-5 e^{e^3} x+50 x^2-5 x \log (x)\right )}{5 x} \, dx=e^{\left (\frac {3}{5}+x\right ) \left (-e^{e^3}-4 (1-x)+x-\log (x)\right )} \]

[Out]

exp((x+3/5)*(5*x-ln(x)-exp(exp(3))-4))

Rubi [F]

\[ \int \frac {e^{\frac {1}{5} \left (-12+e^{e^3} (-3-5 x)-5 x+25 x^2+(-3-5 x) \log (x)\right )} \left (-3-10 x-5 e^{e^3} x+50 x^2-5 x \log (x)\right )}{5 x} \, dx=\int \frac {\exp \left (\frac {1}{5} \left (-12+e^{e^3} (-3-5 x)-5 x+25 x^2+(-3-5 x) \log (x)\right )\right ) \left (-3-10 x-5 e^{e^3} x+50 x^2-5 x \log (x)\right )}{5 x} \, dx \]

[In]

Int[(E^((-12 + E^E^3*(-3 - 5*x) - 5*x + 25*x^2 + (-3 - 5*x)*Log[x])/5)*(-3 - 10*x - 5*E^E^3*x + 50*x^2 - 5*x*L
og[x]))/(5*x),x]

[Out]

-((2 + E^E^3)*Defer[Int][E^(((3 + 5*x)*(-4*(1 + E^E^3/4) + 5*x - Log[x]))/5), x]) - (3*Defer[Int][E^(((3 + 5*x
)*(-4*(1 + E^E^3/4) + 5*x - Log[x]))/5)/x, x])/5 + 10*Defer[Int][E^(((3 + 5*x)*(-4*(1 + E^E^3/4) + 5*x - Log[x
]))/5)*x, x] - Defer[Int][E^(((3 + 5*x)*(-4*(1 + E^E^3/4) + 5*x - Log[x]))/5)*Log[x], x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {\exp \left (\frac {1}{5} \left (-12+e^{e^3} (-3-5 x)-5 x+25 x^2+(-3-5 x) \log (x)\right )\right ) \left (-3+\left (-10-5 e^{e^3}\right ) x+50 x^2-5 x \log (x)\right )}{5 x} \, dx \\ & = \frac {1}{5} \int \frac {\exp \left (\frac {1}{5} \left (-12+e^{e^3} (-3-5 x)-5 x+25 x^2+(-3-5 x) \log (x)\right )\right ) \left (-3+\left (-10-5 e^{e^3}\right ) x+50 x^2-5 x \log (x)\right )}{x} \, dx \\ & = \frac {1}{5} \int \frac {\exp \left (\frac {1}{5} (3+5 x) \left (-4 \left (1+\frac {e^{e^3}}{4}\right )+5 x-\log (x)\right )\right ) \left (-3+\left (-10-5 e^{e^3}\right ) x+50 x^2-5 x \log (x)\right )}{x} \, dx \\ & = \frac {1}{5} \int \left (\frac {\exp \left (\frac {1}{5} (3+5 x) \left (-4 \left (1+\frac {e^{e^3}}{4}\right )+5 x-\log (x)\right )\right ) \left (-3-5 \left (2+e^{e^3}\right ) x+50 x^2\right )}{x}-5 \exp \left (\frac {1}{5} (3+5 x) \left (-4 \left (1+\frac {e^{e^3}}{4}\right )+5 x-\log (x)\right )\right ) \log (x)\right ) \, dx \\ & = \frac {1}{5} \int \frac {\exp \left (\frac {1}{5} (3+5 x) \left (-4 \left (1+\frac {e^{e^3}}{4}\right )+5 x-\log (x)\right )\right ) \left (-3-5 \left (2+e^{e^3}\right ) x+50 x^2\right )}{x} \, dx-\int \exp \left (\frac {1}{5} (3+5 x) \left (-4 \left (1+\frac {e^{e^3}}{4}\right )+5 x-\log (x)\right )\right ) \log (x) \, dx \\ & = \frac {1}{5} \int \left (-5 \exp \left (\frac {1}{5} (3+5 x) \left (-4 \left (1+\frac {e^{e^3}}{4}\right )+5 x-\log (x)\right )\right ) \left (2+e^{e^3}\right )-\frac {3 \exp \left (\frac {1}{5} (3+5 x) \left (-4 \left (1+\frac {e^{e^3}}{4}\right )+5 x-\log (x)\right )\right )}{x}+50 \exp \left (\frac {1}{5} (3+5 x) \left (-4 \left (1+\frac {e^{e^3}}{4}\right )+5 x-\log (x)\right )\right ) x\right ) \, dx-\int \exp \left (\frac {1}{5} (3+5 x) \left (-4 \left (1+\frac {e^{e^3}}{4}\right )+5 x-\log (x)\right )\right ) \log (x) \, dx \\ & = -\left (\frac {3}{5} \int \frac {\exp \left (\frac {1}{5} (3+5 x) \left (-4 \left (1+\frac {e^{e^3}}{4}\right )+5 x-\log (x)\right )\right )}{x} \, dx\right )+10 \int \exp \left (\frac {1}{5} (3+5 x) \left (-4 \left (1+\frac {e^{e^3}}{4}\right )+5 x-\log (x)\right )\right ) x \, dx+\left (-2-e^{e^3}\right ) \int \exp \left (\frac {1}{5} (3+5 x) \left (-4 \left (1+\frac {e^{e^3}}{4}\right )+5 x-\log (x)\right )\right ) \, dx-\int \exp \left (\frac {1}{5} (3+5 x) \left (-4 \left (1+\frac {e^{e^3}}{4}\right )+5 x-\log (x)\right )\right ) \log (x) \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 1.33 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.11 \[ \int \frac {e^{\frac {1}{5} \left (-12+e^{e^3} (-3-5 x)-5 x+25 x^2+(-3-5 x) \log (x)\right )} \left (-3-10 x-5 e^{e^3} x+50 x^2-5 x \log (x)\right )}{5 x} \, dx=e^{-\frac {1}{5} \left (4+e^{e^3}-5 x\right ) (3+5 x)} x^{-\frac {3}{5}-x} \]

[In]

Integrate[(E^((-12 + E^E^3*(-3 - 5*x) - 5*x + 25*x^2 + (-3 - 5*x)*Log[x])/5)*(-3 - 10*x - 5*E^E^3*x + 50*x^2 -
5*x*Log[x]))/(5*x),x]

[Out]

x^(-3/5 - x)/E^(((4 + E^E^3 - 5*x)*(3 + 5*x))/5)

Maple [A] (verified)

Time = 0.24 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96

method	result	size
risch	\(x^{-\frac {3}{5}-x} {\mathrm e}^{\frac {\left (5 x +3\right ) \left (5 x -{\mathrm e}^{{\mathrm e}^{3}}-4\right )}{5}}\)	\(27\)
norman	\({\mathrm e}^{\frac {\left (-5 x -3\right ) \ln \left (x \right )}{5}+\frac {\left (-5 x -3\right ) {\mathrm e}^{{\mathrm e}^{3}}}{5}+5 x^{2}-x -\frac {12}{5}}\)	\(31\)
parallelrisch	\({\mathrm e}^{\frac {\left (-5 x -3\right ) \ln \left (x \right )}{5}+\frac {\left (-5 x -3\right ) {\mathrm e}^{{\mathrm e}^{3}}}{5}+5 x^{2}-x -\frac {12}{5}}\)	\(31\)

[In]

int(1/5*(-5*x*ln(x)-5*x*exp(exp(3))+50*x^2-10*x-3)*exp(1/5*(-5*x-3)*ln(x)+1/5*(-5*x-3)*exp(exp(3))+5*x^2-x-12/
5)/x,x,method=_RETURNVERBOSE)

[Out]

x^(-3/5-x)*exp(1/5*(5*x+3)*(5*x-exp(exp(3))-4))

Fricas [A] (verification not implemented)

none

Time = 0.44 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07 \[ \int \frac {e^{\frac {1}{5} \left (-12+e^{e^3} (-3-5 x)-5 x+25 x^2+(-3-5 x) \log (x)\right )} \left (-3-10 x-5 e^{e^3} x+50 x^2-5 x \log (x)\right )}{5 x} \, dx=e^{\left (5 \, x^{2} - \frac {1}{5} \, {\left (5 \, x + 3\right )} e^{\left (e^{3}\right )} - \frac {1}{5} \, {\left (5 \, x + 3\right )} \log \left (x\right ) - x - \frac {12}{5}\right )} \]

[In]

integrate(1/5*(-5*x*log(x)-5*x*exp(exp(3))+50*x^2-10*x-3)*exp(1/5*(-5*x-3)*log(x)+1/5*(-5*x-3)*exp(exp(3))+5*x
^2-x-12/5)/x,x, algorithm="fricas")

[Out]

e^(5*x^2 - 1/5*(5*x + 3)*e^(e^3) - 1/5*(5*x + 3)*log(x) - x - 12/5)

Sympy [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.14 \[ \int \frac {e^{\frac {1}{5} \left (-12+e^{e^3} (-3-5 x)-5 x+25 x^2+(-3-5 x) \log (x)\right )} \left (-3-10 x-5 e^{e^3} x+50 x^2-5 x \log (x)\right )}{5 x} \, dx=e^{5 x^{2} - x + \left (- x - \frac {3}{5}\right ) \log {\left (x \right )} + \left (- x - \frac {3}{5}\right ) e^{e^{3}} - \frac {12}{5}} \]

[In]

integrate(1/5*(-5*x*ln(x)-5*x*exp(exp(3))+50*x**2-10*x-3)*exp(1/5*(-5*x-3)*ln(x)+1/5*(-5*x-3)*exp(exp(3))+5*x*
*2-x-12/5)/x,x)

[Out]

exp(5*x**2 - x + (-x - 3/5)*log(x) + (-x - 3/5)*exp(exp(3)) - 12/5)

Maxima [F]

\[ \int \frac {e^{\frac {1}{5} \left (-12+e^{e^3} (-3-5 x)-5 x+25 x^2+(-3-5 x) \log (x)\right )} \left (-3-10 x-5 e^{e^3} x+50 x^2-5 x \log (x)\right )}{5 x} \, dx=\int { \frac {{\left (50 \, x^{2} - 5 \, x e^{\left (e^{3}\right )} - 5 \, x \log \left (x\right ) - 10 \, x - 3\right )} e^{\left (5 \, x^{2} - \frac {1}{5} \, {\left (5 \, x + 3\right )} e^{\left (e^{3}\right )} - \frac {1}{5} \, {\left (5 \, x + 3\right )} \log \left (x\right ) - x - \frac {12}{5}\right )}}{5 \, x} \,d x } \]

[In]

integrate(1/5*(-5*x*log(x)-5*x*exp(exp(3))+50*x^2-10*x-3)*exp(1/5*(-5*x-3)*log(x)+1/5*(-5*x-3)*exp(exp(3))+5*x
^2-x-12/5)/x,x, algorithm="maxima")

[Out]

1/5*integrate((50*x^2 - 5*x*e^(e^3) - 5*x*log(x) - 10*x - 3)*e^(5*x^2 - 1/5*(5*x + 3)*e^(e^3) - 1/5*(5*x + 3)*
log(x) - x - 12/5)/x, x)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.11 \[ \int \frac {e^{\frac {1}{5} \left (-12+e^{e^3} (-3-5 x)-5 x+25 x^2+(-3-5 x) \log (x)\right )} \left (-3-10 x-5 e^{e^3} x+50 x^2-5 x \log (x)\right )}{5 x} \, dx=e^{\left (5 \, x^{2} - x e^{\left (e^{3}\right )} - x \log \left (x\right ) - x - \frac {3}{5} \, e^{\left (e^{3}\right )} - \frac {3}{5} \, \log \left (x\right ) - \frac {12}{5}\right )} \]

[In]

integrate(1/5*(-5*x*log(x)-5*x*exp(exp(3))+50*x^2-10*x-3)*exp(1/5*(-5*x-3)*log(x)+1/5*(-5*x-3)*exp(exp(3))+5*x
^2-x-12/5)/x,x, algorithm="giac")

[Out]

e^(5*x^2 - x*e^(e^3) - x*log(x) - x - 3/5*e^(e^3) - 3/5*log(x) - 12/5)

Mupad [B] (verification not implemented)

Time = 12.73 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.18 \[ \int \frac {e^{\frac {1}{5} \left (-12+e^{e^3} (-3-5 x)-5 x+25 x^2+(-3-5 x) \log (x)\right )} \left (-3-10 x-5 e^{e^3} x+50 x^2-5 x \log (x)\right )}{5 x} \, dx=\frac {{\mathrm {e}}^{-x\,{\mathrm {e}}^{{\mathrm {e}}^3}}\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^{-\frac {12}{5}}\,{\mathrm {e}}^{-\frac {3\,{\mathrm {e}}^{{\mathrm {e}}^3}}{5}}\,{\mathrm {e}}^{5\,x^2}}{x^{x+\frac {3}{5}}} \]

[In]

int(-(exp(5*x^2 - (exp(exp(3))*(5*x + 3))/5 - (log(x)*(5*x + 3))/5 - x - 12/5)*(10*x + 5*x*exp(exp(3)) + 5*x*l
og(x) - 50*x^2 + 3))/(5*x),x)

[Out]

(exp(-x*exp(exp(3)))*exp(-x)*exp(-12/5)*exp(-(3*exp(exp(3)))/5)*exp(5*x^2))/x^(x + 3/5)