3.12.58 \(\int \frac {e^{\frac {1+x \log (x)}{(-25 x+5 e^x x-5 x^2) \log (x)}} (5-e^x+x+(5+e^x (-1-x)+2 x) \log (x)+(x^2-e^x x^2) \log ^2(x))}{e^{\frac {1+x \log (x)}{(-25 x+5 e^x x-5 x^2) \log (x)}} (125 x^2+5 e^{2 x} x^2+50 x^3+5 x^4+e^x (-50 x^2-10 x^3)) \log ^2(x)+(-250 x^2-10 e^{2 x} x^2-100 x^3-10 x^4+e^x (100 x^2+20 x^3)) \log ^2(x)} \, dx\)

Optimal. Leaf size=30 \[ \log \left (4 \left (-2+e^{\frac {x+\frac {1}{\log (x)}}{5 \left (-5+e^x-x\right ) x}}\right )\right ) \]

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Rubi [F]  time = 75.42, 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 {\exp \left (\frac {1+x \log (x)}{\left (-25 x+5 e^x x-5 x^2\right ) \log (x)}\right ) \left (5-e^x+x+\left (5+e^x (-1-x)+2 x\right ) \log (x)+\left (x^2-e^x x^2\right ) \log ^2(x)\right )}{\exp \left (\frac {1+x \log (x)}{\left (-25 x+5 e^x x-5 x^2\right ) \log (x)}\right ) \left (125 x^2+5 e^{2 x} x^2+50 x^3+5 x^4+e^x \left (-50 x^2-10 x^3\right )\right ) \log ^2(x)+\left (-250 x^2-10 e^{2 x} x^2-100 x^3-10 x^4+e^x \left (100 x^2+20 x^3\right )\right ) \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((1 + x*Log[x])/((-25*x + 5*E^x*x - 5*x^2)*Log[x]))*(5 - E^x + x + (5 + E^x*(-1 - x) + 2*x)*Log[x] + (x
^2 - E^x*x^2)*Log[x]^2))/(E^((1 + x*Log[x])/((-25*x + 5*E^x*x - 5*x^2)*Log[x]))*(125*x^2 + 5*E^(2*x)*x^2 + 50*
x^3 + 5*x^4 + E^x*(-50*x^2 - 10*x^3))*Log[x]^2 + (-250*x^2 - 10*E^(2*x)*x^2 - 100*x^3 - 10*x^4 + E^x*(100*x^2
+ 20*x^3))*Log[x]^2),x]

[Out]

-1/5*Defer[Int][1/((-1 + 2*E^(1/(5*(5 - E^x + x)) + 1/(5*x*(5 - E^x + x)*Log[x])))*(-5 + E^x - x)^2), x] + Def
er[Int][E^x/((-1 + 2*E^(1/(5*(5 - E^x + x)) + 1/(5*x*(5 - E^x + x)*Log[x])))*(-5 + E^x - x)^2), x]/5 + Defer[I
nt][E^x/((-1 + 2*E^(1/(5*(5 - E^x + x)) + 1/(5*x*(5 - E^x + x)*Log[x])))*(-5 + E^x - x)^2*x^2*Log[x]^2), x]/5
- Defer[Int][1/((-1 + 2*E^(1/(5*(5 - E^x + x)) + 1/(5*x*(5 - E^x + x)*Log[x])))*x^2*(5 - E^x + x)^2*Log[x]^2),
x] - Defer[Int][1/((-1 + 2*E^(1/(5*(5 - E^x + x)) + 1/(5*x*(5 - E^x + x)*Log[x])))*x*(5 - E^x + x)^2*Log[x]^2
), x]/5 + Defer[Int][E^x/((-1 + 2*E^(1/(5*(5 - E^x + x)) + 1/(5*x*(5 - E^x + x)*Log[x])))*(-5 + E^x - x)^2*x^2
*Log[x]), x]/5 + Defer[Int][E^x/((-1 + 2*E^(1/(5*(5 - E^x + x)) + 1/(5*x*(5 - E^x + x)*Log[x])))*(-5 + E^x - x
)^2*x*Log[x]), x]/5 - Defer[Int][1/((-1 + 2*E^(1/(5*(5 - E^x + x)) + 1/(5*x*(5 - E^x + x)*Log[x])))*x^2*(5 - E
^x + x)^2*Log[x]), x] - (2*Defer[Int][1/((-1 + 2*E^(1/(5*(5 - E^x + x)) + 1/(5*x*(5 - E^x + x)*Log[x])))*x*(5
- E^x + x)^2*Log[x]), x])/5

Rubi steps

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

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Mathematica [F]  time = 1.23, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {e^{\frac {1+x \log (x)}{\left (-25 x+5 e^x x-5 x^2\right ) \log (x)}} \left (5-e^x+x+\left (5+e^x (-1-x)+2 x\right ) \log (x)+\left (x^2-e^x x^2\right ) \log ^2(x)\right )}{e^{\frac {1+x \log (x)}{\left (-25 x+5 e^x x-5 x^2\right ) \log (x)}} \left (125 x^2+5 e^{2 x} x^2+50 x^3+5 x^4+e^x \left (-50 x^2-10 x^3\right )\right ) \log ^2(x)+\left (-250 x^2-10 e^{2 x} x^2-100 x^3-10 x^4+e^x \left (100 x^2+20 x^3\right )\right ) \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[(E^((1 + x*Log[x])/((-25*x + 5*E^x*x - 5*x^2)*Log[x]))*(5 - E^x + x + (5 + E^x*(-1 - x) + 2*x)*Log[x
] + (x^2 - E^x*x^2)*Log[x]^2))/(E^((1 + x*Log[x])/((-25*x + 5*E^x*x - 5*x^2)*Log[x]))*(125*x^2 + 5*E^(2*x)*x^2
+ 50*x^3 + 5*x^4 + E^x*(-50*x^2 - 10*x^3))*Log[x]^2 + (-250*x^2 - 10*E^(2*x)*x^2 - 100*x^3 - 10*x^4 + E^x*(10
0*x^2 + 20*x^3))*Log[x]^2),x]

[Out]

Integrate[(E^((1 + x*Log[x])/((-25*x + 5*E^x*x - 5*x^2)*Log[x]))*(5 - E^x + x + (5 + E^x*(-1 - x) + 2*x)*Log[x
] + (x^2 - E^x*x^2)*Log[x]^2))/(E^((1 + x*Log[x])/((-25*x + 5*E^x*x - 5*x^2)*Log[x]))*(125*x^2 + 5*E^(2*x)*x^2
+ 50*x^3 + 5*x^4 + E^x*(-50*x^2 - 10*x^3))*Log[x]^2 + (-250*x^2 - 10*E^(2*x)*x^2 - 100*x^3 - 10*x^4 + E^x*(10
0*x^2 + 20*x^3))*Log[x]^2), x]

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fricas [A]  time = 0.69, size = 30, normalized size = 1.00 \begin {gather*} \log \left (e^{\left (-\frac {x \log \relax (x) + 1}{5 \, {\left (x^{2} - x e^{x} + 5 \, x\right )} \log \relax (x)}\right )} - 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-exp(x)*x^2+x^2)*log(x)^2+((-x-1)*exp(x)+5+2*x)*log(x)+x-exp(x)+5)*exp((x*log(x)+1)/(5*exp(x)*x-5*
x^2-25*x)/log(x))/((5*exp(x)^2*x^2+(-10*x^3-50*x^2)*exp(x)+5*x^4+50*x^3+125*x^2)*log(x)^2*exp((x*log(x)+1)/(5*
exp(x)*x-5*x^2-25*x)/log(x))+(-10*exp(x)^2*x^2+(20*x^3+100*x^2)*exp(x)-10*x^4-100*x^3-250*x^2)*log(x)^2),x, al
gorithm="fricas")

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-exp(x)*x^2+x^2)*log(x)^2+((-x-1)*exp(x)+5+2*x)*log(x)+x-exp(x)+5)*exp((x*log(x)+1)/(5*exp(x)*x-5*
x^2-25*x)/log(x))/((5*exp(x)^2*x^2+(-10*x^3-50*x^2)*exp(x)+5*x^4+50*x^3+125*x^2)*log(x)^2*exp((x*log(x)+1)/(5*
exp(x)*x-5*x^2-25*x)/log(x))+(-10*exp(x)^2*x^2+(20*x^3+100*x^2)*exp(x)-10*x^4-100*x^3-250*x^2)*log(x)^2),x, al
gorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:exp(sageVARx)^2=exp(2*sageVARx)exp(sageVARx)^2=exp(2*sageVARx)exp(sageVARx)^2=exp(2*sageVARx)exp(sageVARx)^
2=exp(2*sag

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maple [B]  time = 0.07, size = 87, normalized size = 2.90

method	result	size
risch	\(-\frac {1}{5 \left (x -{\mathrm e}^{x}+5\right )}-\frac {1}{5 x \left (x -{\mathrm e}^{x}+5\right ) \ln \relax (x )}-\frac {x \ln \relax (x )+1}{\left (5 \,{\mathrm e}^{x} x -5 x^{2}-25 x \right ) \ln \relax (x )}+\ln \left ({\mathrm e}^{\frac {x \ln \relax (x )+1}{5 x \left ({\mathrm e}^{x}-5-x \right ) \ln \relax (x )}}-2\right )\)	\(87\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-exp(x)*x^2+x^2)*ln(x)^2+((-x-1)*exp(x)+5+2*x)*ln(x)+x-exp(x)+5)*exp((x*ln(x)+1)/(5*exp(x)*x-5*x^2-25*x)
/ln(x))/((5*exp(x)^2*x^2+(-10*x^3-50*x^2)*exp(x)+5*x^4+50*x^3+125*x^2)*ln(x)^2*exp((x*ln(x)+1)/(5*exp(x)*x-5*x
^2-25*x)/ln(x))+(-10*exp(x)^2*x^2+(20*x^3+100*x^2)*exp(x)-10*x^4-100*x^3-250*x^2)*ln(x)^2),x,method=_RETURNVER
BOSE)

[Out]

-1/5/(x-exp(x)+5)-1/5/x/(x-exp(x)+5)/ln(x)-(x*ln(x)+1)/(5*exp(x)*x-5*x^2-25*x)/ln(x)+ln(exp(1/5*(x*ln(x)+1)/x/
(exp(x)-5-x)/ln(x))-2)

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maxima [B]  time = 0.72, size = 144, normalized size = 4.80 \begin {gather*} \frac {{\left (x \log \relax (x) + 1\right )} e^{x} - 5 \, x \log \relax (x) - x - 5}{5 \, {\left (x e^{\left (2 \, x\right )} \log \relax (x) - {\left (x^{2} + 10 \, x\right )} e^{x} \log \relax (x) + 5 \, {\left (x^{2} + 5 \, x\right )} \log \relax (x)\right )}} + \log \left ({\left (e^{\left (-\frac {1}{5 \, {\left (x - e^{x} + 5\right )}} - \frac {1}{5 \, {\left ({\left (x + 10\right )} e^{x} - 5 \, x - e^{\left (2 \, x\right )} - 25\right )} \log \relax (x)} + \frac {1}{5 \, {\left (x e^{x} - 5 \, x\right )} \log \relax (x)}\right )} - 2\right )} e^{\left (\frac {1}{5 \, {\left (x - e^{x} + 5\right )}} - \frac {1}{5 \, {\left (x e^{x} - 5 \, x\right )} \log \relax (x)}\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-exp(x)*x^2+x^2)*log(x)^2+((-x-1)*exp(x)+5+2*x)*log(x)+x-exp(x)+5)*exp((x*log(x)+1)/(5*exp(x)*x-5*
x^2-25*x)/log(x))/((5*exp(x)^2*x^2+(-10*x^3-50*x^2)*exp(x)+5*x^4+50*x^3+125*x^2)*log(x)^2*exp((x*log(x)+1)/(5*
exp(x)*x-5*x^2-25*x)/log(x))+(-10*exp(x)^2*x^2+(20*x^3+100*x^2)*exp(x)-10*x^4-100*x^3-250*x^2)*log(x)^2),x, al
gorithm="maxima")

[Out]

1/5*((x*log(x) + 1)*e^x - 5*x*log(x) - x - 5)/(x*e^(2*x)*log(x) - (x^2 + 10*x)*e^x*log(x) + 5*(x^2 + 5*x)*log(
x)) + log((e^(-1/5/(x - e^x + 5) - 1/5/(((x + 10)*e^x - 5*x - e^(2*x) - 25)*log(x)) + 1/5/((x*e^x - 5*x)*log(x
))) - 2)*e^(1/5/(x - e^x + 5) - 1/5/((x*e^x - 5*x)*log(x))))

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mupad [B]  time = 1.48, size = 43, normalized size = 1.43 \begin {gather*} \ln \left ({\mathrm {e}}^{-\frac {1}{5\,x-5\,{\mathrm {e}}^x+25}}\,{\mathrm {e}}^{-\frac {1}{5\,x^2\,\ln \relax (x)+25\,x\,\ln \relax (x)-5\,x\,{\mathrm {e}}^x\,\ln \relax (x)}}-2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-(x*log(x) + 1)/(log(x)*(25*x - 5*x*exp(x) + 5*x^2)))*(x - exp(x) - log(x)^2*(x^2*exp(x) - x^2) + lo
g(x)*(2*x - exp(x)*(x + 1) + 5) + 5))/(log(x)^2*(10*x^2*exp(2*x) - exp(x)*(100*x^2 + 20*x^3) + 250*x^2 + 100*x
^3 + 10*x^4) - exp(-(x*log(x) + 1)/(log(x)*(25*x - 5*x*exp(x) + 5*x^2)))*log(x)^2*(5*x^2*exp(2*x) - exp(x)*(50
*x^2 + 10*x^3) + 125*x^2 + 50*x^3 + 5*x^4)),x)

[Out]

log(exp(-1/(5*x - 5*exp(x) + 25))*exp(-1/(5*x^2*log(x) + 25*x*log(x) - 5*x*exp(x)*log(x))) - 2)

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sympy [A]  time = 4.67, size = 29, normalized size = 0.97 \begin {gather*} \log {\left (e^{\frac {x \log {\relax (x )} + 1}{\left (- 5 x^{2} + 5 x e^{x} - 25 x\right ) \log {\relax (x )}}} - 2 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-exp(x)*x**2+x**2)*ln(x)**2+((-x-1)*exp(x)+5+2*x)*ln(x)+x-exp(x)+5)*exp((x*ln(x)+1)/(5*exp(x)*x-5*
x**2-25*x)/ln(x))/((5*exp(x)**2*x**2+(-10*x**3-50*x**2)*exp(x)+5*x**4+50*x**3+125*x**2)*ln(x)**2*exp((x*ln(x)+
1)/(5*exp(x)*x-5*x**2-25*x)/ln(x))+(-10*exp(x)**2*x**2+(20*x**3+100*x**2)*exp(x)-10*x**4-100*x**3-250*x**2)*ln
(x)**2),x)

[Out]

log(exp((x*log(x) + 1)/((-5*x**2 + 5*x*exp(x) - 25*x)*log(x))) - 2)

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