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

Optimal. Leaf size=27 \[ -e^{-4+e^{2 x}} x \log (x \log (5))+\log \left (-e^x+\log (x)\right ) \]

________________________________________________________________________________________

Rubi [F]  time = 1.93, 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-e^x x+e^{-4+e^{2 x}} \left (e^x x-x \log (x)+\left (e^x x+2 e^{3 x} x^2+\left (-x-2 e^{2 x} x^2\right ) \log (x)\right ) \log (x \log (5))\right )}{-e^x x+x \log (x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

x - ExpIntegralEi[E^(2*x)]/(2*E^4) - (ExpIntegralEi[E^(2*x)]*Log[x*Log[5]])/(2*E^4) - 2*Log[x*Log[5]]*Defer[In
t][E^(-4 + E^(2*x) + 2*x)*x, x] + Defer[Int][ExpIntegralEi[E^(2*x)]/x, x]/(2*E^4) - Defer[Int][1/(x*(E^x - Log
[x])), x] + Defer[Int][Log[x]/(E^x - Log[x]), x] + 2*Defer[Int][Defer[Int][E^(-4 + E^(2*x) + 2*x)*x, x]/x, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-1+e^x x-e^{-4+e^{2 x}} x \left (e^x-\log (x)\right ) \left (1+\left (1+2 e^{2 x} x\right ) \log (x \log (5))\right )}{x \left (e^x-\log (x)\right )} \, dx\\ &=\int \left (\frac {-1+x \log (x)}{x \left (e^x-\log (x)\right )}-2 e^{-4+e^{2 x}+2 x} x \log (x \log (5))+\frac {e^4-e^{e^{2 x}}-e^{e^{2 x}} \log (x \log (5))}{e^4}\right ) \, dx\\ &=-\left (2 \int e^{-4+e^{2 x}+2 x} x \log (x \log (5)) \, dx\right )+\frac {\int \left (e^4-e^{e^{2 x}}-e^{e^{2 x}} \log (x \log (5))\right ) \, dx}{e^4}+\int \frac {-1+x \log (x)}{x \left (e^x-\log (x)\right )} \, dx\\ &=x+2 \int \frac {\int e^{-4+e^{2 x}+2 x} x \, dx}{x} \, dx-\frac {\int e^{e^{2 x}} \, dx}{e^4}-\frac {\int e^{e^{2 x}} \log (x \log (5)) \, dx}{e^4}-(2 \log (x \log (5))) \int e^{-4+e^{2 x}+2 x} x \, dx+\int \left (-\frac {1}{x \left (e^x-\log (x)\right )}+\frac {\log (x)}{e^x-\log (x)}\right ) \, dx\\ &=x-\frac {\text {Ei}\left (e^{2 x}\right ) \log (x \log (5))}{2 e^4}+2 \int \frac {\int e^{-4+e^{2 x}+2 x} x \, dx}{x} \, dx-\frac {\operatorname {Subst}\left (\int \frac {e^x}{x} \, dx,x,e^{2 x}\right )}{2 e^4}+\frac {\int \frac {\text {Ei}\left (e^{2 x}\right )}{2 x} \, dx}{e^4}-(2 \log (x \log (5))) \int e^{-4+e^{2 x}+2 x} x \, dx-\int \frac {1}{x \left (e^x-\log (x)\right )} \, dx+\int \frac {\log (x)}{e^x-\log (x)} \, dx\\ &=x-\frac {\text {Ei}\left (e^{2 x}\right )}{2 e^4}-\frac {\text {Ei}\left (e^{2 x}\right ) \log (x \log (5))}{2 e^4}+2 \int \frac {\int e^{-4+e^{2 x}+2 x} x \, dx}{x} \, dx+\frac {\int \frac {\text {Ei}\left (e^{2 x}\right )}{x} \, dx}{2 e^4}-(2 \log (x \log (5))) \int e^{-4+e^{2 x}+2 x} x \, dx-\int \frac {1}{x \left (e^x-\log (x)\right )} \, dx+\int \frac {\log (x)}{e^x-\log (x)} \, dx\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.57, size = 27, normalized size = 1.00 \begin {gather*} -e^{-4+e^{2 x}} x \log (x \log (5))+\log \left (e^x-\log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(E^(-4 + E^(2*x))*x*Log[x*Log[5]]) + Log[E^x - Log[x]]

________________________________________________________________________________________

fricas [A]  time = 0.67, size = 28, normalized size = 1.04 \begin {gather*} -{\left (x \log \relax (x) + x \log \left (\log \relax (5)\right )\right )} e^{\left (e^{\left (2 \, x\right )} - 4\right )} + \log \left (-e^{x} + \log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(x*log(x) + x*log(log(5)))*e^(e^(2*x) - 4) + log(-e^x + log(x))

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x e^{x} - {\left (x e^{x} + {\left (2 \, x^{2} e^{\left (3 \, x\right )} + x e^{x} - {\left (2 \, x^{2} e^{\left (2 \, x\right )} + x\right )} \log \relax (x)\right )} \log \left (x \log \relax (5)\right ) - x \log \relax (x)\right )} e^{\left (e^{\left (2 \, x\right )} - 4\right )} - 1}{x e^{x} - x \log \relax (x)}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((x*e^x - (x*e^x + (2*x^2*e^(3*x) + x*e^x - (2*x^2*e^(2*x) + x)*log(x))*log(x*log(5)) - x*log(x))*e^(
e^(2*x) - 4) - 1)/(x*e^x - x*log(x)), x)

________________________________________________________________________________________

maple [A]  time = 0.10, size = 30, normalized size = 1.11

method	result	size
risch	\(\ln \left (\ln \relax (x )-{\mathrm e}^{x}\right )+\left (-x \ln \left (\ln \relax (5)\right )-x \ln \relax (x )\right ) {\mathrm e}^{{\mathrm e}^{2 x}-4}\)	\(30\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

ln(ln(x)-exp(x))+(-x*ln(ln(5))-x*ln(x))*exp(exp(2*x)-4)

________________________________________________________________________________________

maxima [A]  time = 0.75, size = 28, normalized size = 1.04 \begin {gather*} -{\left (x \log \relax (x) + x \log \left (\log \relax (5)\right )\right )} e^{\left (e^{\left (2 \, x\right )} - 4\right )} + \log \left (e^{x} - \log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(x*log(x) + x*log(log(5)))*e^(e^(2*x) - 4) + log(e^x - log(x))

________________________________________________________________________________________

mupad [B]  time = 1.55, size = 28, normalized size = 1.04 \begin {gather*} \ln \left (\ln \relax (x)-{\mathrm {e}}^x\right )-{\mathrm {e}}^{{\mathrm {e}}^{2\,x}-4}\,\left (x\,\ln \left (\ln \relax (5)\right )+x\,\ln \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(log(x) - exp(x)) - exp(exp(2*x) - 4)*(x*log(log(5)) + x*log(x))

________________________________________________________________________________________

sympy [A]  time = 158.64, size = 29, normalized size = 1.07 \begin {gather*} \left (- x \log {\relax (x )} - x \log {\left (\log {\relax (5 )} \right )}\right ) e^{e^{2 x} - 4} + \log {\left (e^{x} - \log {\relax (x )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-x*log(x) - x*log(log(5)))*exp(exp(2*x) - 4) + log(exp(x) - log(x))

________________________________________________________________________________________