3.3.99 \(\int \frac {-4 e^x x \log (x)+e^x (8 x+2 x^2) \log ^2(x)+e^x (-8 x-4 x^2) \log ^3(x)+e^x (-2+2 x+2 x^2) \log ^4(x)}{5 e^{2 x} x^4-20 e^{2 x} x^4 \log (x)+(10 e^x x^2+e^{2 x} (-10 x^3+30 x^4)) \log ^2(x)+(-20 e^x x^2+e^{2 x} (20 x^3-20 x^4)) \log ^3(x)+(5+e^x (-10 x+10 x^2)+e^{2 x} (5 x^2-10 x^3+5 x^4)) \log ^4(x)} \, dx\)

Optimal. Leaf size=32 \[ \frac {2 x}{5 \left (-x+e^x x \left (x-\left (-x+\frac {x}{\log (x)}\right )^2\right )\right )} \]

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Rubi [F]  time = 10.34, 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 {-4 e^x x \log (x)+e^x \left (8 x+2 x^2\right ) \log ^2(x)+e^x \left (-8 x-4 x^2\right ) \log ^3(x)+e^x \left (-2+2 x+2 x^2\right ) \log ^4(x)}{5 e^{2 x} x^4-20 e^{2 x} x^4 \log (x)+\left (10 e^x x^2+e^{2 x} \left (-10 x^3+30 x^4\right )\right ) \log ^2(x)+\left (-20 e^x x^2+e^{2 x} \left (20 x^3-20 x^4\right )\right ) \log ^3(x)+\left (5+e^x \left (-10 x+10 x^2\right )+e^{2 x} \left (5 x^2-10 x^3+5 x^4\right )\right ) \log ^4(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-4*E^x*x*Log[x] + E^x*(8*x + 2*x^2)*Log[x]^2 + E^x*(-8*x - 4*x^2)*Log[x]^3 + E^x*(-2 + 2*x + 2*x^2)*Log[x
]^4)/(5*E^(2*x)*x^4 - 20*E^(2*x)*x^4*Log[x] + (10*E^x*x^2 + E^(2*x)*(-10*x^3 + 30*x^4))*Log[x]^2 + (-20*E^x*x^
2 + E^(2*x)*(20*x^3 - 20*x^4))*Log[x]^3 + (5 + E^x*(-10*x + 10*x^2) + E^(2*x)*(5*x^2 - 10*x^3 + 5*x^4))*Log[x]
^4),x]

[Out]

(-4*Defer[Int][(E^x*x*Log[x])/(E^x*x^2 - 2*E^x*x^2*Log[x] + Log[x]^2 - E^x*x*Log[x]^2 + E^x*x^2*Log[x]^2)^2, x
])/5 + (8*Defer[Int][(E^x*x*Log[x]^2)/(E^x*x^2 - 2*E^x*x^2*Log[x] + Log[x]^2 - E^x*x*Log[x]^2 + E^x*x^2*Log[x]
^2)^2, x])/5 + (2*Defer[Int][(E^x*x^2*Log[x]^2)/(E^x*x^2 - 2*E^x*x^2*Log[x] + Log[x]^2 - E^x*x*Log[x]^2 + E^x*
x^2*Log[x]^2)^2, x])/5 - (8*Defer[Int][(E^x*x*Log[x]^3)/(E^x*x^2 - 2*E^x*x^2*Log[x] + Log[x]^2 - E^x*x*Log[x]^
2 + E^x*x^2*Log[x]^2)^2, x])/5 - (4*Defer[Int][(E^x*x^2*Log[x]^3)/(E^x*x^2 - 2*E^x*x^2*Log[x] + Log[x]^2 - E^x
*x*Log[x]^2 + E^x*x^2*Log[x]^2)^2, x])/5 - (2*Defer[Int][(E^x*Log[x]^4)/(E^x*x^2 - 2*E^x*x^2*Log[x] + Log[x]^2
- E^x*x*Log[x]^2 + E^x*x^2*Log[x]^2)^2, x])/5 + (2*Defer[Int][(E^x*x*Log[x]^4)/(E^x*x^2 - 2*E^x*x^2*Log[x] +
Log[x]^2 - E^x*x*Log[x]^2 + E^x*x^2*Log[x]^2)^2, x])/5 + (2*Defer[Int][(E^x*x^2*Log[x]^4)/(E^x*x^2 - 2*E^x*x^2
*Log[x] + Log[x]^2 - E^x*x*Log[x]^2 + E^x*x^2*Log[x]^2)^2, x])/5

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2 e^x \log (x) \left (-2 x+x (4+x) \log (x)-2 x (2+x) \log ^2(x)+\left (-1+x+x^2\right ) \log ^3(x)\right )}{5 \left (e^x x^2-2 e^x x^2 \log (x)+\left (1+e^x (-1+x) x\right ) \log ^2(x)\right )^2} \, dx\\ &=\frac {2}{5} \int \frac {e^x \log (x) \left (-2 x+x (4+x) \log (x)-2 x (2+x) \log ^2(x)+\left (-1+x+x^2\right ) \log ^3(x)\right )}{\left (e^x x^2-2 e^x x^2 \log (x)+\left (1+e^x (-1+x) x\right ) \log ^2(x)\right )^2} \, dx\\ &=\frac {2}{5} \int \left (-\frac {2 e^x x \log (x)}{\left (e^x x^2-2 e^x x^2 \log (x)+\log ^2(x)-e^x x \log ^2(x)+e^x x^2 \log ^2(x)\right )^2}+\frac {4 e^x x \log ^2(x)}{\left (e^x x^2-2 e^x x^2 \log (x)+\log ^2(x)-e^x x \log ^2(x)+e^x x^2 \log ^2(x)\right )^2}+\frac {e^x x^2 \log ^2(x)}{\left (e^x x^2-2 e^x x^2 \log (x)+\log ^2(x)-e^x x \log ^2(x)+e^x x^2 \log ^2(x)\right )^2}-\frac {4 e^x x \log ^3(x)}{\left (e^x x^2-2 e^x x^2 \log (x)+\log ^2(x)-e^x x \log ^2(x)+e^x x^2 \log ^2(x)\right )^2}-\frac {2 e^x x^2 \log ^3(x)}{\left (e^x x^2-2 e^x x^2 \log (x)+\log ^2(x)-e^x x \log ^2(x)+e^x x^2 \log ^2(x)\right )^2}-\frac {e^x \log ^4(x)}{\left (e^x x^2-2 e^x x^2 \log (x)+\log ^2(x)-e^x x \log ^2(x)+e^x x^2 \log ^2(x)\right )^2}+\frac {e^x x \log ^4(x)}{\left (e^x x^2-2 e^x x^2 \log (x)+\log ^2(x)-e^x x \log ^2(x)+e^x x^2 \log ^2(x)\right )^2}+\frac {e^x x^2 \log ^4(x)}{\left (e^x x^2-2 e^x x^2 \log (x)+\log ^2(x)-e^x x \log ^2(x)+e^x x^2 \log ^2(x)\right )^2}\right ) \, dx\\ &=\frac {2}{5} \int \frac {e^x x^2 \log ^2(x)}{\left (e^x x^2-2 e^x x^2 \log (x)+\log ^2(x)-e^x x \log ^2(x)+e^x x^2 \log ^2(x)\right )^2} \, dx-\frac {2}{5} \int \frac {e^x \log ^4(x)}{\left (e^x x^2-2 e^x x^2 \log (x)+\log ^2(x)-e^x x \log ^2(x)+e^x x^2 \log ^2(x)\right )^2} \, dx+\frac {2}{5} \int \frac {e^x x \log ^4(x)}{\left (e^x x^2-2 e^x x^2 \log (x)+\log ^2(x)-e^x x \log ^2(x)+e^x x^2 \log ^2(x)\right )^2} \, dx+\frac {2}{5} \int \frac {e^x x^2 \log ^4(x)}{\left (e^x x^2-2 e^x x^2 \log (x)+\log ^2(x)-e^x x \log ^2(x)+e^x x^2 \log ^2(x)\right )^2} \, dx-\frac {4}{5} \int \frac {e^x x \log (x)}{\left (e^x x^2-2 e^x x^2 \log (x)+\log ^2(x)-e^x x \log ^2(x)+e^x x^2 \log ^2(x)\right )^2} \, dx-\frac {4}{5} \int \frac {e^x x^2 \log ^3(x)}{\left (e^x x^2-2 e^x x^2 \log (x)+\log ^2(x)-e^x x \log ^2(x)+e^x x^2 \log ^2(x)\right )^2} \, dx+\frac {8}{5} \int \frac {e^x x \log ^2(x)}{\left (e^x x^2-2 e^x x^2 \log (x)+\log ^2(x)-e^x x \log ^2(x)+e^x x^2 \log ^2(x)\right )^2} \, dx-\frac {8}{5} \int \frac {e^x x \log ^3(x)}{\left (e^x x^2-2 e^x x^2 \log (x)+\log ^2(x)-e^x x \log ^2(x)+e^x x^2 \log ^2(x)\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 1.24, size = 43, normalized size = 1.34 \begin {gather*} -\frac {2 \log ^2(x)}{5 \left (e^x x^2-2 e^x x^2 \log (x)+\left (1+e^x (-1+x) x\right ) \log ^2(x)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-4*E^x*x*Log[x] + E^x*(8*x + 2*x^2)*Log[x]^2 + E^x*(-8*x - 4*x^2)*Log[x]^3 + E^x*(-2 + 2*x + 2*x^2)
*Log[x]^4)/(5*E^(2*x)*x^4 - 20*E^(2*x)*x^4*Log[x] + (10*E^x*x^2 + E^(2*x)*(-10*x^3 + 30*x^4))*Log[x]^2 + (-20*
E^x*x^2 + E^(2*x)*(20*x^3 - 20*x^4))*Log[x]^3 + (5 + E^x*(-10*x + 10*x^2) + E^(2*x)*(5*x^2 - 10*x^3 + 5*x^4))*
Log[x]^4),x]

[Out]

(-2*Log[x]^2)/(5*(E^x*x^2 - 2*E^x*x^2*Log[x] + (1 + E^x*(-1 + x)*x)*Log[x]^2))

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fricas [A]  time = 1.11, size = 43, normalized size = 1.34 \begin {gather*} \frac {2 \, \log \relax (x)^{2}}{5 \, {\left (2 \, x^{2} e^{x} \log \relax (x) - x^{2} e^{x} - {\left ({\left (x^{2} - x\right )} e^{x} + 1\right )} \log \relax (x)^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2+2*x-2)*exp(x)*log(x)^4+(-4*x^2-8*x)*exp(x)*log(x)^3+(2*x^2+8*x)*exp(x)*log(x)^2-4*x*exp(x)*l
og(x))/(((5*x^4-10*x^3+5*x^2)*exp(x)^2+(10*x^2-10*x)*exp(x)+5)*log(x)^4+((-20*x^4+20*x^3)*exp(x)^2-20*exp(x)*x
^2)*log(x)^3+((30*x^4-10*x^3)*exp(x)^2+10*exp(x)*x^2)*log(x)^2-20*x^4*exp(x)^2*log(x)+5*exp(x)^2*x^4),x, algor
ithm="fricas")

[Out]

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

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giac [A]  time = 1.73, size = 47, normalized size = 1.47 \begin {gather*} -\frac {2 \, \log \relax (x)^{2}}{5 \, {\left (x^{2} e^{x} \log \relax (x)^{2} - 2 \, x^{2} e^{x} \log \relax (x) - x e^{x} \log \relax (x)^{2} + x^{2} e^{x} + \log \relax (x)^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2+2*x-2)*exp(x)*log(x)^4+(-4*x^2-8*x)*exp(x)*log(x)^3+(2*x^2+8*x)*exp(x)*log(x)^2-4*x*exp(x)*l
og(x))/(((5*x^4-10*x^3+5*x^2)*exp(x)^2+(10*x^2-10*x)*exp(x)+5)*log(x)^4+((-20*x^4+20*x^3)*exp(x)^2-20*exp(x)*x
^2)*log(x)^3+((30*x^4-10*x^3)*exp(x)^2+10*exp(x)*x^2)*log(x)^2-20*x^4*exp(x)^2*log(x)+5*exp(x)^2*x^4),x, algor
ithm="giac")

[Out]

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

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maple [B]  time = 0.09, size = 88, normalized size = 2.75

method	result	size
risch	\(-\frac {2}{5 \left ({\mathrm e}^{x} x^{2}-{\mathrm e}^{x} x +1\right )}-\frac {2 x^{2} {\mathrm e}^{x} \left (2 \ln \relax (x )-1\right )}{5 \left ({\mathrm e}^{x} x^{2}-{\mathrm e}^{x} x +1\right ) \left (x^{2} {\mathrm e}^{x} \ln \relax (x )^{2}-x \,{\mathrm e}^{x} \ln \relax (x )^{2}-2 x^{2} {\mathrm e}^{x} \ln \relax (x )+{\mathrm e}^{x} x^{2}+\ln \relax (x )^{2}\right )}\)	\(88\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x^2+2*x-2)*exp(x)*ln(x)^4+(-4*x^2-8*x)*exp(x)*ln(x)^3+(2*x^2+8*x)*exp(x)*ln(x)^2-4*x*exp(x)*ln(x))/(((
5*x^4-10*x^3+5*x^2)*exp(x)^2+(10*x^2-10*x)*exp(x)+5)*ln(x)^4+((-20*x^4+20*x^3)*exp(x)^2-20*exp(x)*x^2)*ln(x)^3
+((30*x^4-10*x^3)*exp(x)^2+10*exp(x)*x^2)*ln(x)^2-20*x^4*exp(x)^2*ln(x)+5*exp(x)^2*x^4),x,method=_RETURNVERBOS
E)

[Out]

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

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maxima [A]  time = 0.95, size = 44, normalized size = 1.38 \begin {gather*} \frac {2 \, \log \relax (x)^{2}}{5 \, {\left ({\left (2 \, x^{2} \log \relax (x) - {\left (x^{2} - x\right )} \log \relax (x)^{2} - x^{2}\right )} e^{x} - \log \relax (x)^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2+2*x-2)*exp(x)*log(x)^4+(-4*x^2-8*x)*exp(x)*log(x)^3+(2*x^2+8*x)*exp(x)*log(x)^2-4*x*exp(x)*l
og(x))/(((5*x^4-10*x^3+5*x^2)*exp(x)^2+(10*x^2-10*x)*exp(x)+5)*log(x)^4+((-20*x^4+20*x^3)*exp(x)^2-20*exp(x)*x
^2)*log(x)^3+((30*x^4-10*x^3)*exp(x)^2+10*exp(x)*x^2)*log(x)^2-20*x^4*exp(x)^2*log(x)+5*exp(x)^2*x^4),x, algor
ithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int -\frac {-{\mathrm {e}}^x\,\left (2\,x^2+2\,x-2\right )\,{\ln \relax (x)}^4+{\mathrm {e}}^x\,\left (4\,x^2+8\,x\right )\,{\ln \relax (x)}^3-{\mathrm {e}}^x\,\left (2\,x^2+8\,x\right )\,{\ln \relax (x)}^2+4\,x\,{\mathrm {e}}^x\,\ln \relax (x)}{{\ln \relax (x)}^4\,\left ({\mathrm {e}}^{2\,x}\,\left (5\,x^4-10\,x^3+5\,x^2\right )-{\mathrm {e}}^x\,\left (10\,x-10\,x^2\right )+5\right )+5\,x^4\,{\mathrm {e}}^{2\,x}+{\ln \relax (x)}^2\,\left (10\,x^2\,{\mathrm {e}}^x-{\mathrm {e}}^{2\,x}\,\left (10\,x^3-30\,x^4\right )\right )-{\ln \relax (x)}^3\,\left (20\,x^2\,{\mathrm {e}}^x-{\mathrm {e}}^{2\,x}\,\left (20\,x^3-20\,x^4\right )\right )-20\,x^4\,{\mathrm {e}}^{2\,x}\,\ln \relax (x)} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(4*x*exp(x)*log(x) - exp(x)*log(x)^4*(2*x + 2*x^2 - 2) - exp(x)*log(x)^2*(8*x + 2*x^2) + exp(x)*log(x)^3*
(8*x + 4*x^2))/(log(x)^4*(exp(2*x)*(5*x^2 - 10*x^3 + 5*x^4) - exp(x)*(10*x - 10*x^2) + 5) + 5*x^4*exp(2*x) + l
og(x)^2*(10*x^2*exp(x) - exp(2*x)*(10*x^3 - 30*x^4)) - log(x)^3*(20*x^2*exp(x) - exp(2*x)*(20*x^3 - 20*x^4)) -
20*x^4*exp(2*x)*log(x)),x)

[Out]

int(-(4*x*exp(x)*log(x) - exp(x)*log(x)^4*(2*x + 2*x^2 - 2) - exp(x)*log(x)^2*(8*x + 2*x^2) + exp(x)*log(x)^3*
(8*x + 4*x^2))/(log(x)^4*(exp(2*x)*(5*x^2 - 10*x^3 + 5*x^4) - exp(x)*(10*x - 10*x^2) + 5) + 5*x^4*exp(2*x) + l
og(x)^2*(10*x^2*exp(x) - exp(2*x)*(10*x^3 - 30*x^4)) - log(x)^3*(20*x^2*exp(x) - exp(2*x)*(20*x^3 - 20*x^4)) -
20*x^4*exp(2*x)*log(x)), x)

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sympy [B]  time = 0.90, size = 49, normalized size = 1.53 \begin {gather*} - \frac {2 \log {\relax (x )}^{2}}{\left (5 x^{2} \log {\relax (x )}^{2} - 10 x^{2} \log {\relax (x )} + 5 x^{2} - 5 x \log {\relax (x )}^{2}\right ) e^{x} + 5 \log {\relax (x )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x**2+2*x-2)*exp(x)*ln(x)**4+(-4*x**2-8*x)*exp(x)*ln(x)**3+(2*x**2+8*x)*exp(x)*ln(x)**2-4*x*exp(x
)*ln(x))/(((5*x**4-10*x**3+5*x**2)*exp(x)**2+(10*x**2-10*x)*exp(x)+5)*ln(x)**4+((-20*x**4+20*x**3)*exp(x)**2-2
0*exp(x)*x**2)*ln(x)**3+((30*x**4-10*x**3)*exp(x)**2+10*exp(x)*x**2)*ln(x)**2-20*x**4*exp(x)**2*ln(x)+5*exp(x)
**2*x**4),x)

[Out]

-2*log(x)**2/((5*x**2*log(x)**2 - 10*x**2*log(x) + 5*x**2 - 5*x*log(x)**2)*exp(x) + 5*log(x)**2)

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