\(\int \frac {e^{3+x-e^{\frac {1+(4+x) \log (x)}{\log (x)}} x} (\log ^2(x)+e^{\frac {1+(4+x) \log (x)}{\log (x)}} (1+(-1-x) \log ^2(x)))}{\log (5) \log ^2(x)} \, dx\) [1462]

Optimal result

Integrand size = 65, antiderivative size = 29 \[ \int \frac {e^{3+x-e^{\frac {1+(4+x) \log (x)}{\log (x)}} x} \left (\log ^2(x)+e^{\frac {1+(4+x) \log (x)}{\log (x)}} \left (1+(-1-x) \log ^2(x)\right )\right )}{\log (5) \log ^2(x)} \, dx=\frac {5+e^{x-\left (e^{4+x+\frac {1}{\log (x)}}-\frac {3}{x}\right ) x}}{\log (5)} \]

[Out]

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

Rubi [F]

\[ \int \frac {e^{3+x-e^{\frac {1+(4+x) \log (x)}{\log (x)}} x} \left (\log ^2(x)+e^{\frac {1+(4+x) \log (x)}{\log (x)}} \left (1+(-1-x) \log ^2(x)\right )\right )}{\log (5) \log ^2(x)} \, dx=\int \frac {e^{3+x-e^{\frac {1+(4+x) \log (x)}{\log (x)}} x} \left (\log ^2(x)+e^{\frac {1+(4+x) \log (x)}{\log (x)}} \left (1+(-1-x) \log ^2(x)\right )\right )}{\log (5) \log ^2(x)} \, dx \]

[In]

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

[Out]

Defer[Int][E^(3 + x - E^(4 + x + Log[x]^(-1))*x), x]/Log[5] - Defer[Int][E^(7 + 2*x - E^((1 + (4 + x)*Log[x])/
Log[x])*x + Log[x]^(-1)), x]/Log[5] - Defer[Int][E^(7 + 2*x - E^((1 + (4 + x)*Log[x])/Log[x])*x + Log[x]^(-1))
*x, x]/Log[5] + Defer[Int][E^(7 + 2*x - E^((1 + (4 + x)*Log[x])/Log[x])*x + Log[x]^(-1))/Log[x]^2, x]/Log[5]

Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {e^{3+x-e^{\frac {1+(4+x) \log (x)}{\log (x)}} x} \left (\log ^2(x)+e^{\frac {1+(4+x) \log (x)}{\log (x)}} \left (1+(-1-x) \log ^2(x)\right )\right )}{\log ^2(x)} \, dx}{\log (5)} \\ & = \frac {\int \left (e^{3+x-e^{\frac {1+(4+x) \log (x)}{\log (x)}} x}-\frac {\exp \left (7+2 x-e^{\frac {1+(4+x) \log (x)}{\log (x)}} x+\frac {1}{\log (x)}\right ) \left (-1+\log ^2(x)+x \log ^2(x)\right )}{\log ^2(x)}\right ) \, dx}{\log (5)} \\ & = \frac {\int e^{3+x-e^{\frac {1+(4+x) \log (x)}{\log (x)}} x} \, dx}{\log (5)}-\frac {\int \frac {\exp \left (7+2 x-e^{\frac {1+(4+x) \log (x)}{\log (x)}} x+\frac {1}{\log (x)}\right ) \left (-1+\log ^2(x)+x \log ^2(x)\right )}{\log ^2(x)} \, dx}{\log (5)} \\ & = \frac {\int e^{3+x-e^{4+x+\frac {1}{\log (x)}} x} \, dx}{\log (5)}-\frac {\int \left (\exp \left (7+2 x-e^{\frac {1+(4+x) \log (x)}{\log (x)}} x+\frac {1}{\log (x)}\right )+\exp \left (7+2 x-e^{\frac {1+(4+x) \log (x)}{\log (x)}} x+\frac {1}{\log (x)}\right ) x-\frac {\exp \left (7+2 x-e^{\frac {1+(4+x) \log (x)}{\log (x)}} x+\frac {1}{\log (x)}\right )}{\log ^2(x)}\right ) \, dx}{\log (5)} \\ & = \frac {\int e^{3+x-e^{4+x+\frac {1}{\log (x)}} x} \, dx}{\log (5)}-\frac {\int \exp \left (7+2 x-e^{\frac {1+(4+x) \log (x)}{\log (x)}} x+\frac {1}{\log (x)}\right ) \, dx}{\log (5)}-\frac {\int \exp \left (7+2 x-e^{\frac {1+(4+x) \log (x)}{\log (x)}} x+\frac {1}{\log (x)}\right ) x \, dx}{\log (5)}+\frac {\int \frac {\exp \left (7+2 x-e^{\frac {1+(4+x) \log (x)}{\log (x)}} x+\frac {1}{\log (x)}\right )}{\log ^2(x)} \, dx}{\log (5)} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.79 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76 \[ \int \frac {e^{3+x-e^{\frac {1+(4+x) \log (x)}{\log (x)}} x} \left (\log ^2(x)+e^{\frac {1+(4+x) \log (x)}{\log (x)}} \left (1+(-1-x) \log ^2(x)\right )\right )}{\log (5) \log ^2(x)} \, dx=\frac {e^{3+x-e^{4+x+\frac {1}{\log (x)}} x}}{\log (5)} \]

[In]

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

[Out]

E^(3 + x - E^(4 + x + Log[x]^(-1))*x)/Log[5]

Maple [A] (verified)

Time = 0.35 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.93

method	result	size
parallelrisch	\(\frac {{\mathrm e}^{-x \,{\mathrm e}^{\frac {\left (4+x \right ) \ln \left (x \right )+1}{\ln \left (x \right )}}+3+x}}{\ln \left (5\right )}\)	\(27\)
risch	\(\frac {{\mathrm e}^{-x \,{\mathrm e}^{\frac {x \ln \left (x \right )+4 \ln \left (x \right )+1}{\ln \left (x \right )}}+3+x}}{\ln \left (5\right )}\)	\(29\)

[In]

int((((-1-x)*ln(x)^2+1)*exp(((4+x)*ln(x)+1)/ln(x))+ln(x)^2)*exp(-x*exp(((4+x)*ln(x)+1)/ln(x))+3+x)/ln(5)/ln(x)
^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90 \[ \int \frac {e^{3+x-e^{\frac {1+(4+x) \log (x)}{\log (x)}} x} \left (\log ^2(x)+e^{\frac {1+(4+x) \log (x)}{\log (x)}} \left (1+(-1-x) \log ^2(x)\right )\right )}{\log (5) \log ^2(x)} \, dx=\frac {e^{\left (-x e^{\left (\frac {{\left (x + 4\right )} \log \left (x\right ) + 1}{\log \left (x\right )}\right )} + x + 3\right )}}{\log \left (5\right )} \]

[In]

integrate((((-1-x)*log(x)^2+1)*exp(((4+x)*log(x)+1)/log(x))+log(x)^2)*exp(-x*exp(((4+x)*log(x)+1)/log(x))+3+x)
/log(5)/log(x)^2,x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 1.10 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76 \[ \int \frac {e^{3+x-e^{\frac {1+(4+x) \log (x)}{\log (x)}} x} \left (\log ^2(x)+e^{\frac {1+(4+x) \log (x)}{\log (x)}} \left (1+(-1-x) \log ^2(x)\right )\right )}{\log (5) \log ^2(x)} \, dx=\frac {e^{- x e^{\frac {\left (x + 4\right ) \log {\left (x \right )} + 1}{\log {\left (x \right )}}} + x + 3}}{\log {\left (5 \right )}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.69 \[ \int \frac {e^{3+x-e^{\frac {1+(4+x) \log (x)}{\log (x)}} x} \left (\log ^2(x)+e^{\frac {1+(4+x) \log (x)}{\log (x)}} \left (1+(-1-x) \log ^2(x)\right )\right )}{\log (5) \log ^2(x)} \, dx=\frac {e^{\left (-x e^{\left (x + \frac {1}{\log \left (x\right )} + 4\right )} + x + 3\right )}}{\log \left (5\right )} \]

[In]

integrate((((-1-x)*log(x)^2+1)*exp(((4+x)*log(x)+1)/log(x))+log(x)^2)*exp(-x*exp(((4+x)*log(x)+1)/log(x))+3+x)
/log(5)/log(x)^2,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.69 \[ \int \frac {e^{3+x-e^{\frac {1+(4+x) \log (x)}{\log (x)}} x} \left (\log ^2(x)+e^{\frac {1+(4+x) \log (x)}{\log (x)}} \left (1+(-1-x) \log ^2(x)\right )\right )}{\log (5) \log ^2(x)} \, dx=\frac {e^{\left (-x e^{\left (x + \frac {1}{\log \left (x\right )} + 4\right )} + x + 3\right )}}{\log \left (5\right )} \]

[In]

integrate((((-1-x)*log(x)^2+1)*exp(((4+x)*log(x)+1)/log(x))+log(x)^2)*exp(-x*exp(((4+x)*log(x)+1)/log(x))+3+x)
/log(5)/log(x)^2,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 8.67 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76 \[ \int \frac {e^{3+x-e^{\frac {1+(4+x) \log (x)}{\log (x)}} x} \left (\log ^2(x)+e^{\frac {1+(4+x) \log (x)}{\log (x)}} \left (1+(-1-x) \log ^2(x)\right )\right )}{\log (5) \log ^2(x)} \, dx=\frac {{\mathrm {e}}^{-x\,{\mathrm {e}}^4\,{\mathrm {e}}^{\frac {1}{\ln \left (x\right )}}\,{\mathrm {e}}^x}\,{\mathrm {e}}^3\,{\mathrm {e}}^x}{\ln \left (5\right )} \]

[In]

int((exp(x - x*exp((log(x)*(x + 4) + 1)/log(x)) + 3)*(log(x)^2 - exp((log(x)*(x + 4) + 1)/log(x))*(log(x)^2*(x
+ 1) - 1)))/(log(5)*log(x)^2),x)

[Out]

(exp(-x*exp(4)*exp(1/log(x))*exp(x))*exp(3)*exp(x))/log(5)