3.50.50 \(\int \frac {e^{-x} (x-x^2+e^x (x-60 x^2)-22 e^x x^2 \log (x)-2 e^x x^2 \log ^2(x)+e^{e^x} (10 e^x+25 e^{2 x} x+(2 e^x+10 e^{2 x} x) \log (x)+e^{2 x} x \log ^2(x)))}{x} \, dx\)

Optimal. Leaf size=30 \[ 2+x+e^{-x} x+\log (5)+\left (e^{e^x}-x^2\right ) (5+\log (x))^2 \]

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Rubi [F]  time = 2.24, 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 {e^{-x} \left (x-x^2+e^x \left (x-60 x^2\right )-22 e^x x^2 \log (x)-2 e^x x^2 \log ^2(x)+e^{e^x} \left (10 e^x+25 e^{2 x} x+\left (2 e^x+10 e^{2 x} x\right ) \log (x)+e^{2 x} x \log ^2(x)\right )\right )}{x} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(x - x^2 + E^x*(x - 60*x^2) - 22*E^x*x^2*Log[x] - 2*E^x*x^2*Log[x]^2 + E^E^x*(10*E^x + 25*E^(2*x)*x + (2*E
^x + 10*E^(2*x)*x)*Log[x] + E^(2*x)*x*Log[x]^2))/(E^x*x),x]

[Out]

25*E^E^x + x + x/E^x - 25*x^2 + 10*E^E^x*Log[x] - 10*x^2*Log[x] - x^2*Log[x]^2 + 2*Log[x]*Defer[Int][E^E^x/x,
x] + Defer[Int][E^(E^x + x)*Log[x]^2, x] - 2*Defer[Int][Defer[Int][E^E^x/x, x]/x, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (e^{-x}-e^{-x} x+e^{e^x+x} (5+\log (x))^2-\frac {-10 e^{e^x}-x+60 x^2-2 e^{e^x} \log (x)+22 x^2 \log (x)+2 x^2 \log ^2(x)}{x}\right ) \, dx\\ &=\int e^{-x} \, dx-\int e^{-x} x \, dx+\int e^{e^x+x} (5+\log (x))^2 \, dx-\int \frac {-10 e^{e^x}-x+60 x^2-2 e^{e^x} \log (x)+22 x^2 \log (x)+2 x^2 \log ^2(x)}{x} \, dx\\ &=-e^{-x}+e^{-x} x-\int e^{-x} \, dx+\int \left (25 e^{e^x+x}+10 e^{e^x+x} \log (x)+e^{e^x+x} \log ^2(x)\right ) \, dx-\int \left (-1+60 x+22 x \log (x)+2 x \log ^2(x)-\frac {2 e^{e^x} (5+\log (x))}{x}\right ) \, dx\\ &=x+e^{-x} x-30 x^2-2 \int x \log ^2(x) \, dx+2 \int \frac {e^{e^x} (5+\log (x))}{x} \, dx+10 \int e^{e^x+x} \log (x) \, dx-22 \int x \log (x) \, dx+25 \int e^{e^x+x} \, dx+\int e^{e^x+x} \log ^2(x) \, dx\\ &=x+e^{-x} x-\frac {49 x^2}{2}+10 e^{e^x} \log (x)-11 x^2 \log (x)-x^2 \log ^2(x)+2 \int x \log (x) \, dx+2 \int \left (\frac {5 e^{e^x}}{x}+\frac {e^{e^x} \log (x)}{x}\right ) \, dx-10 \int \frac {e^{e^x}}{x} \, dx+25 \operatorname {Subst}\left (\int e^x \, dx,x,e^x\right )+\int e^{e^x+x} \log ^2(x) \, dx\\ &=25 e^{e^x}+x+e^{-x} x-25 x^2+10 e^{e^x} \log (x)-10 x^2 \log (x)-x^2 \log ^2(x)+2 \int \frac {e^{e^x} \log (x)}{x} \, dx+\int e^{e^x+x} \log ^2(x) \, dx\\ &=25 e^{e^x}+x+e^{-x} x-25 x^2+10 e^{e^x} \log (x)-10 x^2 \log (x)-x^2 \log ^2(x)-2 \int \frac {\int \frac {e^{e^x}}{x} \, dx}{x} \, dx+(2 \log (x)) \int \frac {e^{e^x}}{x} \, dx+\int e^{e^x+x} \log ^2(x) \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.13, size = 42, normalized size = 1.40 \begin {gather*} x+e^{-x} x-25 x^2-10 x^2 \log (x)-x^2 \log ^2(x)+e^{e^x} (5+\log (x))^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x - x^2 + E^x*(x - 60*x^2) - 22*E^x*x^2*Log[x] - 2*E^x*x^2*Log[x]^2 + E^E^x*(10*E^x + 25*E^(2*x)*x
+ (2*E^x + 10*E^(2*x)*x)*Log[x] + E^(2*x)*x*Log[x]^2))/(E^x*x),x]

[Out]

x + x/E^x - 25*x^2 - 10*x^2*Log[x] - x^2*Log[x]^2 + E^E^x*(5 + Log[x])^2

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fricas [B]  time = 1.02, size = 64, normalized size = 2.13 \begin {gather*} -{\left (x^{2} e^{x} \log \relax (x)^{2} + 10 \, x^{2} e^{x} \log \relax (x) + {\left (25 \, x^{2} - x\right )} e^{x} - {\left (e^{x} \log \relax (x)^{2} + 10 \, e^{x} \log \relax (x) + 25 \, e^{x}\right )} e^{\left (e^{x}\right )} - x\right )} e^{\left (-x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*exp(x)^2*log(x)^2+(10*x*exp(x)^2+2*exp(x))*log(x)+25*x*exp(x)^2+10*exp(x))*exp(exp(x))-2*x^2*exp
(x)*log(x)^2-22*x^2*exp(x)*log(x)+(-60*x^2+x)*exp(x)-x^2+x)/exp(x)/x,x, algorithm="fricas")

[Out]

-(x^2*e^x*log(x)^2 + 10*x^2*e^x*log(x) + (25*x^2 - x)*e^x - (e^x*log(x)^2 + 10*e^x*log(x) + 25*e^x)*e^(e^x) -
x)*e^(-x)

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giac [B]  time = 0.22, size = 68, normalized size = 2.27 \begin {gather*} -{\left (x^{2} e^{x} \log \relax (x)^{2} + 10 \, x^{2} e^{x} \log \relax (x) + 25 \, x^{2} e^{x} - e^{\left (x + e^{x}\right )} \log \relax (x)^{2} - x e^{x} - 10 \, e^{\left (x + e^{x}\right )} \log \relax (x) - x - 25 \, e^{\left (x + e^{x}\right )}\right )} e^{\left (-x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*exp(x)^2*log(x)^2+(10*x*exp(x)^2+2*exp(x))*log(x)+25*x*exp(x)^2+10*exp(x))*exp(exp(x))-2*x^2*exp
(x)*log(x)^2-22*x^2*exp(x)*log(x)+(-60*x^2+x)*exp(x)-x^2+x)/exp(x)/x,x, algorithm="giac")

[Out]

-(x^2*e^x*log(x)^2 + 10*x^2*e^x*log(x) + 25*x^2*e^x - e^(x + e^x)*log(x)^2 - x*e^x - 10*e^(x + e^x)*log(x) - x
- 25*e^(x + e^x))*e^(-x)

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maple [A]  time = 0.09, size = 50, normalized size = 1.67

method	result	size
risch	\(-x^{2} \ln \relax (x )^{2}-10 x^{2} \ln \relax (x )-x \left (25 \,{\mathrm e}^{x} x -{\mathrm e}^{x}-1\right ) {\mathrm e}^{-x}+\left (25+\ln \relax (x )^{2}+10 \ln \relax (x )\right ) {\mathrm e}^{{\mathrm e}^{x}}\)	\(50\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x*exp(x)^2*ln(x)^2+(10*x*exp(x)^2+2*exp(x))*ln(x)+25*x*exp(x)^2+10*exp(x))*exp(exp(x))-2*x^2*exp(x)*ln(x
)^2-22*x^2*exp(x)*ln(x)+(-60*x^2+x)*exp(x)-x^2+x)/exp(x)/x,x,method=_RETURNVERBOSE)

[Out]

-x^2*ln(x)^2-10*x^2*ln(x)-x*(25*exp(x)*x-exp(x)-1)*exp(-x)+(25+ln(x)^2+10*ln(x))*exp(exp(x))

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maxima [A]  time = 0.49, size = 51, normalized size = 1.70 \begin {gather*} -x^{2} \log \relax (x)^{2} - 10 \, x^{2} \log \relax (x) - 25 \, x^{2} + {\left (x + 1\right )} e^{\left (-x\right )} + {\left (\log \relax (x)^{2} + 10 \, \log \relax (x) + 25\right )} e^{\left (e^{x}\right )} + x - e^{\left (-x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*exp(x)^2*log(x)^2+(10*x*exp(x)^2+2*exp(x))*log(x)+25*x*exp(x)^2+10*exp(x))*exp(exp(x))-2*x^2*exp
(x)*log(x)^2-22*x^2*exp(x)*log(x)+(-60*x^2+x)*exp(x)-x^2+x)/exp(x)/x,x, algorithm="maxima")

[Out]

-x^2*log(x)^2 - 10*x^2*log(x) - 25*x^2 + (x + 1)*e^(-x) + (log(x)^2 + 10*log(x) + 25)*e^(e^x) + x - e^(-x)

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mupad [B]  time = 3.33, size = 43, normalized size = 1.43 \begin {gather*} x+x\,{\mathrm {e}}^{-x}-10\,x^2\,\ln \relax (x)-x^2\,{\ln \relax (x)}^2+{\mathrm {e}}^{{\mathrm {e}}^x}\,\left ({\ln \relax (x)}^2+10\,\ln \relax (x)+25\right )-25\,x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-x)*(x + exp(x)*(x - 60*x^2) + exp(exp(x))*(10*exp(x) + 25*x*exp(2*x) + log(x)*(2*exp(x) + 10*x*exp(2
*x)) + x*exp(2*x)*log(x)^2) - x^2 - 2*x^2*exp(x)*log(x)^2 - 22*x^2*exp(x)*log(x)))/x,x)

[Out]

x + x*exp(-x) - 10*x^2*log(x) - x^2*log(x)^2 + exp(exp(x))*(10*log(x) + log(x)^2 + 25) - 25*x^2

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*exp(x)**2*ln(x)**2+(10*x*exp(x)**2+2*exp(x))*ln(x)+25*x*exp(x)**2+10*exp(x))*exp(exp(x))-2*x**2*
exp(x)*ln(x)**2-22*x**2*exp(x)*ln(x)+(-60*x**2+x)*exp(x)-x**2+x)/exp(x)/x,x)

[Out]

Timed out

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