3.21.68 \(\int \frac {-x+2 x^3+e^x (-1+2 x^2)+(e^x+x) \log (x)+(6 x^3+e^x (4 x^2+2 x^3)+(e^x (-2-x)-3 x) \log (x)) \log (\frac {x}{-10 x^2+5 \log (x)})}{-6 e^{2 x} x^5-12 e^x x^6-6 x^7+(3 e^{2 x} x^3+6 e^x x^4+3 x^5) \log (x)} \, dx\)

Optimal. Leaf size=30 \[ \frac {\log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right )}{3 x^2 \left (e^x+x\right )} \]

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Rubi [F]  time = 9.75, 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 {-x+2 x^3+e^x \left (-1+2 x^2\right )+\left (e^x+x\right ) \log (x)+\left (6 x^3+e^x \left (4 x^2+2 x^3\right )+\left (e^x (-2-x)-3 x\right ) \log (x)\right ) \log \left (\frac {x}{-10 x^2+5 \log (x)}\right )}{-6 e^{2 x} x^5-12 e^x x^6-6 x^7+\left (3 e^{2 x} x^3+6 e^x x^4+3 x^5\right ) \log (x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-x + 2*x^3 + E^x*(-1 + 2*x^2) + (E^x + x)*Log[x] + (6*x^3 + E^x*(4*x^2 + 2*x^3) + (E^x*(-2 - x) - 3*x)*Lo
g[x])*Log[x/(-10*x^2 + 5*Log[x])])/(-6*E^(2*x)*x^5 - 12*E^x*x^6 - 6*x^7 + (3*E^(2*x)*x^3 + 6*E^x*x^4 + 3*x^5)*
Log[x]),x]

[Out]

-1/3*(Log[-1/5*x/(2*x^2 - Log[x])]*Defer[Int][1/(x^2*(E^x + x)^2), x]) + (Log[-1/5*x/(2*x^2 - Log[x])]*Defer[I
nt][1/(x*(E^x + x)^2), x])/3 + Defer[Int][1/(x^3*(E^x + x)*(2*x^2 - Log[x])), x]/3 - (2*Defer[Int][1/(x*(E^x +
x)*(2*x^2 - Log[x])), x])/3 - Defer[Int][Log[x]/(x^3*(E^x + x)*(2*x^2 - Log[x])), x]/3 - (2*Defer[Int][Log[x/
(5*(-2*x^2 + Log[x]))]/((E^x + x)*(2*x^2 - Log[x])), x])/3 - (4*Defer[Int][Log[x/(5*(-2*x^2 + Log[x]))]/(x*(E^
x + x)*(2*x^2 - Log[x])), x])/3 + (2*Defer[Int][(Log[x]*Log[x/(5*(-2*x^2 + Log[x]))])/(x^3*(E^x + x)*(2*x^2 -
Log[x])), x])/3 + Defer[Int][(Log[x]*Log[x/(5*(-2*x^2 + Log[x]))])/(x^2*(E^x + x)*(2*x^2 - Log[x])), x]/3 + De
fer[Int][Defer[Int][1/(x^2*(E^x + x)^2), x]/(x*(2*x^2 - Log[x])), x]/3 - (2*Defer[Int][(x*Defer[Int][1/(x^2*(E
^x + x)^2), x])/(2*x^2 - Log[x]), x])/3 + Defer[Int][(Log[x]*Defer[Int][1/(x^2*(E^x + x)^2), x])/(x*(-2*x^2 +
Log[x])), x]/3 - Defer[Int][Defer[Int][1/(x*(E^x + x)^2), x]/(x*(2*x^2 - Log[x])), x]/3 + (2*Defer[Int][(x*Def
er[Int][1/(x*(E^x + x)^2), x])/(2*x^2 - Log[x]), x])/3 - Defer[Int][(Log[x]*Defer[Int][1/(x*(E^x + x)^2), x])/
(x*(-2*x^2 + Log[x])), x]/3

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {x-2 x^3-e^x \left (-1+2 x^2\right )-\left (e^x+x\right ) \log (x)-\left (6 x^3+e^x \left (4 x^2+2 x^3\right )+\left (e^x (-2-x)-3 x\right ) \log (x)\right ) \log \left (\frac {x}{-10 x^2+5 \log (x)}\right )}{3 x^3 \left (e^x+x\right )^2 \left (2 x^2-\log (x)\right )} \, dx\\ &=\frac {1}{3} \int \frac {x-2 x^3-e^x \left (-1+2 x^2\right )-\left (e^x+x\right ) \log (x)-\left (6 x^3+e^x \left (4 x^2+2 x^3\right )+\left (e^x (-2-x)-3 x\right ) \log (x)\right ) \log \left (\frac {x}{-10 x^2+5 \log (x)}\right )}{x^3 \left (e^x+x\right )^2 \left (2 x^2-\log (x)\right )} \, dx\\ &=\frac {1}{3} \int \left (\frac {(-1+x) \log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right )}{x^2 \left (e^x+x\right )^2}-\frac {-1+2 x^2+\log (x)+4 x^2 \log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right )+2 x^3 \log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right )-2 \log (x) \log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right )-x \log (x) \log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right )}{x^3 \left (e^x+x\right ) \left (2 x^2-\log (x)\right )}\right ) \, dx\\ &=\frac {1}{3} \int \frac {(-1+x) \log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right )}{x^2 \left (e^x+x\right )^2} \, dx-\frac {1}{3} \int \frac {-1+2 x^2+\log (x)+4 x^2 \log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right )+2 x^3 \log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right )-2 \log (x) \log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right )-x \log (x) \log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right )}{x^3 \left (e^x+x\right ) \left (2 x^2-\log (x)\right )} \, dx\\ &=-\left (\frac {1}{3} \int \frac {-1+2 x^2+\log (x)+2 x^2 (2+x) \log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right )-(2+x) \log (x) \log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right )}{x^3 \left (e^x+x\right ) \left (2 x^2-\log (x)\right )} \, dx\right )+\frac {1}{3} \int \frac {\left (-1+2 x^2+\log (x)\right ) \left (-\int \frac {1}{x^2 \left (e^x+x\right )^2} \, dx+\int \frac {1}{x \left (e^x+x\right )^2} \, dx\right )}{2 x^3-x \log (x)} \, dx-\frac {1}{3} \log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right ) \int \frac {1}{x^2 \left (e^x+x\right )^2} \, dx+\frac {1}{3} \log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right ) \int \frac {1}{x \left (e^x+x\right )^2} \, dx\\ &=-\left (\frac {1}{3} \int \left (-\frac {1}{x^3 \left (e^x+x\right ) \left (2 x^2-\log (x)\right )}+\frac {2}{x \left (e^x+x\right ) \left (2 x^2-\log (x)\right )}+\frac {\log (x)}{x^3 \left (e^x+x\right ) \left (2 x^2-\log (x)\right )}+\frac {2 \log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right )}{\left (e^x+x\right ) \left (2 x^2-\log (x)\right )}+\frac {4 \log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right )}{x \left (e^x+x\right ) \left (2 x^2-\log (x)\right )}-\frac {2 \log (x) \log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right )}{x^3 \left (e^x+x\right ) \left (2 x^2-\log (x)\right )}-\frac {\log (x) \log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right )}{x^2 \left (e^x+x\right ) \left (2 x^2-\log (x)\right )}\right ) \, dx\right )+\frac {1}{3} \int \frac {\left (-1+2 x^2+\log (x)\right ) \left (-\int \frac {1}{x^2 \left (e^x+x\right )^2} \, dx+\int \frac {1}{x \left (e^x+x\right )^2} \, dx\right )}{x \left (2 x^2-\log (x)\right )} \, dx-\frac {1}{3} \log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right ) \int \frac {1}{x^2 \left (e^x+x\right )^2} \, dx+\frac {1}{3} \log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right ) \int \frac {1}{x \left (e^x+x\right )^2} \, dx\\ &=\frac {1}{3} \int \frac {1}{x^3 \left (e^x+x\right ) \left (2 x^2-\log (x)\right )} \, dx-\frac {1}{3} \int \frac {\log (x)}{x^3 \left (e^x+x\right ) \left (2 x^2-\log (x)\right )} \, dx+\frac {1}{3} \int \frac {\log (x) \log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right )}{x^2 \left (e^x+x\right ) \left (2 x^2-\log (x)\right )} \, dx+\frac {1}{3} \int \left (-\frac {\left (-1+2 x^2+\log (x)\right ) \int \frac {1}{x^2 \left (e^x+x\right )^2} \, dx}{x \left (2 x^2-\log (x)\right )}+\frac {\left (-1+2 x^2+\log (x)\right ) \int \frac {1}{x \left (e^x+x\right )^2} \, dx}{x \left (2 x^2-\log (x)\right )}\right ) \, dx-\frac {2}{3} \int \frac {1}{x \left (e^x+x\right ) \left (2 x^2-\log (x)\right )} \, dx-\frac {2}{3} \int \frac {\log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right )}{\left (e^x+x\right ) \left (2 x^2-\log (x)\right )} \, dx+\frac {2}{3} \int \frac {\log (x) \log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right )}{x^3 \left (e^x+x\right ) \left (2 x^2-\log (x)\right )} \, dx-\frac {4}{3} \int \frac {\log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right )}{x \left (e^x+x\right ) \left (2 x^2-\log (x)\right )} \, dx-\frac {1}{3} \log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right ) \int \frac {1}{x^2 \left (e^x+x\right )^2} \, dx+\frac {1}{3} \log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right ) \int \frac {1}{x \left (e^x+x\right )^2} \, dx\\ &=\frac {1}{3} \int \frac {1}{x^3 \left (e^x+x\right ) \left (2 x^2-\log (x)\right )} \, dx-\frac {1}{3} \int \frac {\log (x)}{x^3 \left (e^x+x\right ) \left (2 x^2-\log (x)\right )} \, dx+\frac {1}{3} \int \frac {\log (x) \log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right )}{x^2 \left (e^x+x\right ) \left (2 x^2-\log (x)\right )} \, dx-\frac {1}{3} \int \frac {\left (-1+2 x^2+\log (x)\right ) \int \frac {1}{x^2 \left (e^x+x\right )^2} \, dx}{x \left (2 x^2-\log (x)\right )} \, dx+\frac {1}{3} \int \frac {\left (-1+2 x^2+\log (x)\right ) \int \frac {1}{x \left (e^x+x\right )^2} \, dx}{x \left (2 x^2-\log (x)\right )} \, dx-\frac {2}{3} \int \frac {1}{x \left (e^x+x\right ) \left (2 x^2-\log (x)\right )} \, dx-\frac {2}{3} \int \frac {\log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right )}{\left (e^x+x\right ) \left (2 x^2-\log (x)\right )} \, dx+\frac {2}{3} \int \frac {\log (x) \log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right )}{x^3 \left (e^x+x\right ) \left (2 x^2-\log (x)\right )} \, dx-\frac {4}{3} \int \frac {\log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right )}{x \left (e^x+x\right ) \left (2 x^2-\log (x)\right )} \, dx-\frac {1}{3} \log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right ) \int \frac {1}{x^2 \left (e^x+x\right )^2} \, dx+\frac {1}{3} \log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right ) \int \frac {1}{x \left (e^x+x\right )^2} \, dx\\ &=\frac {1}{3} \int \frac {1}{x^3 \left (e^x+x\right ) \left (2 x^2-\log (x)\right )} \, dx-\frac {1}{3} \int \frac {\log (x)}{x^3 \left (e^x+x\right ) \left (2 x^2-\log (x)\right )} \, dx+\frac {1}{3} \int \frac {\log (x) \log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right )}{x^2 \left (e^x+x\right ) \left (2 x^2-\log (x)\right )} \, dx-\frac {1}{3} \int \left (-\frac {\int \frac {1}{x^2 \left (e^x+x\right )^2} \, dx}{x \left (2 x^2-\log (x)\right )}+\frac {2 x \int \frac {1}{x^2 \left (e^x+x\right )^2} \, dx}{2 x^2-\log (x)}-\frac {\log (x) \int \frac {1}{x^2 \left (e^x+x\right )^2} \, dx}{x \left (-2 x^2+\log (x)\right )}\right ) \, dx+\frac {1}{3} \int \left (-\frac {\int \frac {1}{x \left (e^x+x\right )^2} \, dx}{x \left (2 x^2-\log (x)\right )}+\frac {2 x \int \frac {1}{x \left (e^x+x\right )^2} \, dx}{2 x^2-\log (x)}-\frac {\log (x) \int \frac {1}{x \left (e^x+x\right )^2} \, dx}{x \left (-2 x^2+\log (x)\right )}\right ) \, dx-\frac {2}{3} \int \frac {1}{x \left (e^x+x\right ) \left (2 x^2-\log (x)\right )} \, dx-\frac {2}{3} \int \frac {\log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right )}{\left (e^x+x\right ) \left (2 x^2-\log (x)\right )} \, dx+\frac {2}{3} \int \frac {\log (x) \log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right )}{x^3 \left (e^x+x\right ) \left (2 x^2-\log (x)\right )} \, dx-\frac {4}{3} \int \frac {\log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right )}{x \left (e^x+x\right ) \left (2 x^2-\log (x)\right )} \, dx-\frac {1}{3} \log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right ) \int \frac {1}{x^2 \left (e^x+x\right )^2} \, dx+\frac {1}{3} \log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right ) \int \frac {1}{x \left (e^x+x\right )^2} \, dx\\ &=\frac {1}{3} \int \frac {1}{x^3 \left (e^x+x\right ) \left (2 x^2-\log (x)\right )} \, dx-\frac {1}{3} \int \frac {\log (x)}{x^3 \left (e^x+x\right ) \left (2 x^2-\log (x)\right )} \, dx+\frac {1}{3} \int \frac {\log (x) \log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right )}{x^2 \left (e^x+x\right ) \left (2 x^2-\log (x)\right )} \, dx+\frac {1}{3} \int \frac {\int \frac {1}{x^2 \left (e^x+x\right )^2} \, dx}{x \left (2 x^2-\log (x)\right )} \, dx+\frac {1}{3} \int \frac {\log (x) \int \frac {1}{x^2 \left (e^x+x\right )^2} \, dx}{x \left (-2 x^2+\log (x)\right )} \, dx-\frac {1}{3} \int \frac {\int \frac {1}{x \left (e^x+x\right )^2} \, dx}{x \left (2 x^2-\log (x)\right )} \, dx-\frac {1}{3} \int \frac {\log (x) \int \frac {1}{x \left (e^x+x\right )^2} \, dx}{x \left (-2 x^2+\log (x)\right )} \, dx-\frac {2}{3} \int \frac {1}{x \left (e^x+x\right ) \left (2 x^2-\log (x)\right )} \, dx-\frac {2}{3} \int \frac {\log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right )}{\left (e^x+x\right ) \left (2 x^2-\log (x)\right )} \, dx+\frac {2}{3} \int \frac {\log (x) \log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right )}{x^3 \left (e^x+x\right ) \left (2 x^2-\log (x)\right )} \, dx-\frac {2}{3} \int \frac {x \int \frac {1}{x^2 \left (e^x+x\right )^2} \, dx}{2 x^2-\log (x)} \, dx+\frac {2}{3} \int \frac {x \int \frac {1}{x \left (e^x+x\right )^2} \, dx}{2 x^2-\log (x)} \, dx-\frac {4}{3} \int \frac {\log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right )}{x \left (e^x+x\right ) \left (2 x^2-\log (x)\right )} \, dx-\frac {1}{3} \log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right ) \int \frac {1}{x^2 \left (e^x+x\right )^2} \, dx+\frac {1}{3} \log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right ) \int \frac {1}{x \left (e^x+x\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.18, size = 30, normalized size = 1.00 \begin {gather*} \frac {\log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right )}{3 x^2 \left (e^x+x\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-x + 2*x^3 + E^x*(-1 + 2*x^2) + (E^x + x)*Log[x] + (6*x^3 + E^x*(4*x^2 + 2*x^3) + (E^x*(-2 - x) - 3
*x)*Log[x])*Log[x/(-10*x^2 + 5*Log[x])])/(-6*E^(2*x)*x^5 - 12*E^x*x^6 - 6*x^7 + (3*E^(2*x)*x^3 + 6*E^x*x^4 + 3
*x^5)*Log[x]),x]

[Out]

Log[x/(5*(-2*x^2 + Log[x]))]/(3*x^2*(E^x + x))

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fricas [A]  time = 1.13, size = 30, normalized size = 1.00 \begin {gather*} \frac {\log \left (-\frac {x}{5 \, {\left (2 \, x^{2} - \log \relax (x)\right )}}\right )}{3 \, {\left (x^{3} + x^{2} e^{x}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((-x-2)*exp(x)-3*x)*log(x)+(2*x^3+4*x^2)*exp(x)+6*x^3)*log(x/(5*log(x)-10*x^2))+(exp(x)+x)*log(x)+
(2*x^2-1)*exp(x)+2*x^3-x)/((3*exp(x)^2*x^3+6*exp(x)*x^4+3*x^5)*log(x)-6*x^5*exp(x)^2-12*x^6*exp(x)-6*x^7),x, a
lgorithm="fricas")

[Out]

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

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giac [A]  time = 1.47, size = 30, normalized size = 1.00 \begin {gather*} -\frac {\log \relax (5) + \log \left (-2 \, x^{2} + \log \relax (x)\right ) - \log \relax (x)}{3 \, {\left (x^{3} + x^{2} e^{x}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((-x-2)*exp(x)-3*x)*log(x)+(2*x^3+4*x^2)*exp(x)+6*x^3)*log(x/(5*log(x)-10*x^2))+(exp(x)+x)*log(x)+
(2*x^2-1)*exp(x)+2*x^3-x)/((3*exp(x)^2*x^3+6*exp(x)*x^4+3*x^5)*log(x)-6*x^5*exp(x)^2-12*x^6*exp(x)-6*x^7),x, a
lgorithm="giac")

[Out]

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

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maple [C]  time = 0.22, size = 205, normalized size = 6.83

method	result	size
risch	\(-\frac {\ln \left (x^{2}-\frac {\ln \relax (x )}{2}\right )}{3 x^{2} \left ({\mathrm e}^{x}+x \right )}-\frac {i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i}{-x^{2}+\frac {\ln \relax (x )}{2}}\right ) \mathrm {csgn}\left (\frac {i x}{-x^{2}+\frac {\ln \relax (x )}{2}}\right )-i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i x}{-x^{2}+\frac {\ln \relax (x )}{2}}\right )^{2}+2 i \pi \mathrm {csgn}\left (\frac {i x}{-x^{2}+\frac {\ln \relax (x )}{2}}\right )^{2}+i \pi \,\mathrm {csgn}\left (\frac {i}{-x^{2}+\frac {\ln \relax (x )}{2}}\right ) \mathrm {csgn}\left (\frac {i x}{-x^{2}+\frac {\ln \relax (x )}{2}}\right )^{2}+i \pi \mathrm {csgn}\left (\frac {i x}{-x^{2}+\frac {\ln \relax (x )}{2}}\right )^{3}-2 i \pi +2 \ln \relax (5)+2 \ln \relax (2)-2 \ln \relax (x )}{6 x^{2} \left ({\mathrm e}^{x}+x \right )}\)	\(205\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((((-x-2)*exp(x)-3*x)*ln(x)+(2*x^3+4*x^2)*exp(x)+6*x^3)*ln(x/(5*ln(x)-10*x^2))+(exp(x)+x)*ln(x)+(2*x^2-1)*
exp(x)+2*x^3-x)/((3*exp(x)^2*x^3+6*exp(x)*x^4+3*x^5)*ln(x)-6*x^5*exp(x)^2-12*x^6*exp(x)-6*x^7),x,method=_RETUR
NVERBOSE)

[Out]

-1/3/x^2/(exp(x)+x)*ln(x^2-1/2*ln(x))-1/6*(I*Pi*csgn(I*x)*csgn(I/(-x^2+1/2*ln(x)))*csgn(I*x/(-x^2+1/2*ln(x)))-
I*Pi*csgn(I*x)*csgn(I*x/(-x^2+1/2*ln(x)))^2+2*I*Pi*csgn(I*x/(-x^2+1/2*ln(x)))^2+I*Pi*csgn(I/(-x^2+1/2*ln(x)))*
csgn(I*x/(-x^2+1/2*ln(x)))^2+I*Pi*csgn(I*x/(-x^2+1/2*ln(x)))^3-2*I*Pi+2*ln(5)+2*ln(2)-2*ln(x))/x^2/(exp(x)+x)

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maxima [A]  time = 0.63, size = 30, normalized size = 1.00 \begin {gather*} -\frac {\log \relax (5) + \log \left (-2 \, x^{2} + \log \relax (x)\right ) - \log \relax (x)}{3 \, {\left (x^{3} + x^{2} e^{x}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((-x-2)*exp(x)-3*x)*log(x)+(2*x^3+4*x^2)*exp(x)+6*x^3)*log(x/(5*log(x)-10*x^2))+(exp(x)+x)*log(x)+
(2*x^2-1)*exp(x)+2*x^3-x)/((3*exp(x)^2*x^3+6*exp(x)*x^4+3*x^5)*log(x)-6*x^5*exp(x)^2-12*x^6*exp(x)-6*x^7),x, a
lgorithm="maxima")

[Out]

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

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mupad [B]  time = 1.48, size = 26, normalized size = 0.87 \begin {gather*} \frac {\ln \left (\frac {x}{5\,\left (\ln \relax (x)-2\,x^2\right )}\right )}{3\,x^2\,\left (x+{\mathrm {e}}^x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(x/(5*log(x) - 10*x^2))*(exp(x)*(4*x^2 + 2*x^3) - log(x)*(3*x + exp(x)*(x + 2)) + 6*x^3) - x + exp(x)
*(2*x^2 - 1) + 2*x^3 + log(x)*(x + exp(x)))/(12*x^6*exp(x) - log(x)*(6*x^4*exp(x) + 3*x^3*exp(2*x) + 3*x^5) +
6*x^5*exp(2*x) + 6*x^7),x)

[Out]

log(x/(5*(log(x) - 2*x^2)))/(3*x^2*(x + exp(x)))

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sympy [A]  time = 0.49, size = 26, normalized size = 0.87 \begin {gather*} \frac {\log {\left (\frac {x}{- 10 x^{2} + 5 \log {\relax (x )}} \right )}}{3 x^{3} + 3 x^{2} e^{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((-x-2)*exp(x)-3*x)*ln(x)+(2*x**3+4*x**2)*exp(x)+6*x**3)*ln(x/(5*ln(x)-10*x**2))+(exp(x)+x)*ln(x)+
(2*x**2-1)*exp(x)+2*x**3-x)/((3*exp(x)**2*x**3+6*exp(x)*x**4+3*x**5)*ln(x)-6*x**5*exp(x)**2-12*x**6*exp(x)-6*x
**7),x)

[Out]

log(x/(-10*x**2 + 5*log(x)))/(3*x**3 + 3*x**2*exp(x))

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