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

Optimal. Leaf size=32 \[ -e^{5+x+\frac {4}{\log (3)}} \log (x)+\frac {\log \left (x^4\right )}{e^{2-x}+x} \]

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Rubi [F]  time = 6.24, 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 {4 e^{2-x}+4 x+e^{\frac {4+(5+x) \log (3)}{\log (3)}} \left (-e^{4-2 x}-2 e^{2-x} x-x^2\right )+e^{\frac {4+(5+x) \log (3)}{\log (3)}} \left (-e^{4-2 x} x-2 e^{2-x} x^2-x^3\right ) \log (x)+\left (-x+e^{2-x} x\right ) \log \left (x^4\right )}{e^{4-2 x} x+2 e^{2-x} x^2+x^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

E^(5 + 4/Log[3])*ExpIntegralEi[x] - E^(5 + x + 4/Log[3])*Log[x] + Log[x^4]*Defer[Int][E^(2 + x)/(E^2 + E^x*x)^
2, x] + Log[x^4]*Defer[Int][E^(2 + x)/(x*(E^2 + E^x*x)^2), x] - Defer[Int][E^(5 + 2*x + 4/Log[3])/(E^2 + E^x*x
), x] + (4 - E^(7 + 4/Log[3]))*Defer[Int][E^x/(x*(E^2 + E^x*x)), x] - Log[x^4]*Defer[Int][E^x/(x*(E^2 + E^x*x)
), x] - 4*Defer[Int][Defer[Int][E^(2 + x)/(E^2 + E^x*x)^2, x]/x, x] - 4*Defer[Int][Defer[Int][E^(2 + x)/(x*(E^
2 + E^x*x)^2), x]/x, x] + 4*Defer[Int][Defer[Int][E^x/(x*(E^2 + E^x*x)), x]/x, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int e^x \left (-e^{5+\frac {4}{\log (3)}} \log (x)-\frac {\left (e^2+e^x x\right ) \left (-4+e^{7+\frac {4}{\log (3)}}+e^{5+x+\frac {4}{\log (3)}} x\right )+\left (-e^2+e^x\right ) x \log \left (x^4\right )}{x \left (e^2+e^x x\right )^2}\right ) \, dx\\ &=\int \left (-e^{5+x+\frac {4}{\log (3)}} \log (x)+\frac {e^x \left (4 e^2 \left (1-\frac {1}{4} e^{7+\frac {4}{\log (3)}}\right )+4 e^x \left (1-\frac {1}{2} e^{7+\frac {4}{\log (3)}}\right ) x-e^{5+2 x+\frac {4}{\log (3)}} x^2+e^2 x \log \left (x^4\right )-e^x x \log \left (x^4\right )\right )}{x \left (e^2+e^x x\right )^2}\right ) \, dx\\ &=-\int e^{5+x+\frac {4}{\log (3)}} \log (x) \, dx+\int \frac {e^x \left (4 e^2 \left (1-\frac {1}{4} e^{7+\frac {4}{\log (3)}}\right )+4 e^x \left (1-\frac {1}{2} e^{7+\frac {4}{\log (3)}}\right ) x-e^{5+2 x+\frac {4}{\log (3)}} x^2+e^2 x \log \left (x^4\right )-e^x x \log \left (x^4\right )\right )}{x \left (e^2+e^x x\right )^2} \, dx\\ &=-e^{5+x+\frac {4}{\log (3)}} \log (x)+\int \frac {e^{5+x+\frac {4}{\log (3)}}}{x} \, dx+\int \frac {e^x \left (-\left (\left (e^2+e^x x\right ) \left (-4+e^{7+\frac {4}{\log (3)}}+e^{5+x+\frac {4}{\log (3)}} x\right )\right )-\left (-e^2+e^x\right ) x \log \left (x^4\right )\right )}{x \left (e^2+e^x x\right )^2} \, dx\\ &=e^{5+\frac {4}{\log (3)}} \text {Ei}(x)-e^{5+x+\frac {4}{\log (3)}} \log (x)+\int \left (-\frac {e^{5+2 x+\frac {4}{\log (3)}}}{e^2+e^x x}+\frac {e^x \left (4 e^2 \left (1-\frac {1}{4} e^{7+\frac {4}{\log (3)}}\right )+4 e^x \left (1-\frac {1}{4} e^{7+\frac {4}{\log (3)}}\right ) x+e^2 x \log \left (x^4\right )-e^x x \log \left (x^4\right )\right )}{x \left (e^2+e^x x\right )^2}\right ) \, dx\\ &=e^{5+\frac {4}{\log (3)}} \text {Ei}(x)-e^{5+x+\frac {4}{\log (3)}} \log (x)-\int \frac {e^{5+2 x+\frac {4}{\log (3)}}}{e^2+e^x x} \, dx+\int \frac {e^x \left (4 e^2 \left (1-\frac {1}{4} e^{7+\frac {4}{\log (3)}}\right )+4 e^x \left (1-\frac {1}{4} e^{7+\frac {4}{\log (3)}}\right ) x+e^2 x \log \left (x^4\right )-e^x x \log \left (x^4\right )\right )}{x \left (e^2+e^x x\right )^2} \, dx\\ &=e^{5+\frac {4}{\log (3)}} \text {Ei}(x)-e^{5+x+\frac {4}{\log (3)}} \log (x)-\int \frac {e^{5+2 x+\frac {4}{\log (3)}}}{e^2+e^x x} \, dx+\int \frac {e^x \left (-\left (\left (-4+e^{7+\frac {4}{\log (3)}}\right ) \left (e^2+e^x x\right )\right )-\left (-e^2+e^x\right ) x \log \left (x^4\right )\right )}{x \left (e^2+e^x x\right )^2} \, dx\\ &=e^{5+\frac {4}{\log (3)}} \text {Ei}(x)-e^{5+x+\frac {4}{\log (3)}} \log (x)-\int \frac {e^{5+2 x+\frac {4}{\log (3)}}}{e^2+e^x x} \, dx+\int \left (\frac {e^x \left (4 \left (1-\frac {1}{4} e^{7+\frac {4}{\log (3)}}\right )-\log \left (x^4\right )\right )}{x \left (e^2+e^x x\right )}+\frac {e^{2+x} (1+x) \log \left (x^4\right )}{x \left (e^2+e^x x\right )^2}\right ) \, dx\\ &=e^{5+\frac {4}{\log (3)}} \text {Ei}(x)-e^{5+x+\frac {4}{\log (3)}} \log (x)-\int \frac {e^{5+2 x+\frac {4}{\log (3)}}}{e^2+e^x x} \, dx+\int \frac {e^x \left (4 \left (1-\frac {1}{4} e^{7+\frac {4}{\log (3)}}\right )-\log \left (x^4\right )\right )}{x \left (e^2+e^x x\right )} \, dx+\int \frac {e^{2+x} (1+x) \log \left (x^4\right )}{x \left (e^2+e^x x\right )^2} \, dx\\ &=e^{5+\frac {4}{\log (3)}} \text {Ei}(x)-e^{5+x+\frac {4}{\log (3)}} \log (x)+\log \left (x^4\right ) \int \frac {e^{2+x}}{\left (e^2+e^x x\right )^2} \, dx+\log \left (x^4\right ) \int \frac {e^{2+x}}{x \left (e^2+e^x x\right )^2} \, dx-\int \frac {e^{5+2 x+\frac {4}{\log (3)}}}{e^2+e^x x} \, dx+\int \left (-\frac {e^x \left (-4+e^{7+\frac {4}{\log (3)}}\right )}{x \left (e^2+e^x x\right )}-\frac {e^x \log \left (x^4\right )}{x \left (e^2+e^x x\right )}\right ) \, dx-\int \frac {4 \left (\int \frac {e^{2+x}}{\left (e^2+e^x x\right )^2} \, dx+\int \frac {e^{2+x}}{x \left (e^2+e^x x\right )^2} \, dx\right )}{x} \, dx\\ &=e^{5+\frac {4}{\log (3)}} \text {Ei}(x)-e^{5+x+\frac {4}{\log (3)}} \log (x)-4 \int \frac {\int \frac {e^{2+x}}{\left (e^2+e^x x\right )^2} \, dx+\int \frac {e^{2+x}}{x \left (e^2+e^x x\right )^2} \, dx}{x} \, dx+\left (4-e^{7+\frac {4}{\log (3)}}\right ) \int \frac {e^x}{x \left (e^2+e^x x\right )} \, dx+\log \left (x^4\right ) \int \frac {e^{2+x}}{\left (e^2+e^x x\right )^2} \, dx+\log \left (x^4\right ) \int \frac {e^{2+x}}{x \left (e^2+e^x x\right )^2} \, dx-\int \frac {e^{5+2 x+\frac {4}{\log (3)}}}{e^2+e^x x} \, dx-\int \frac {e^x \log \left (x^4\right )}{x \left (e^2+e^x x\right )} \, dx\\ &=e^{5+\frac {4}{\log (3)}} \text {Ei}(x)-e^{5+x+\frac {4}{\log (3)}} \log (x)-4 \int \left (\frac {\int \frac {e^{2+x}}{\left (e^2+e^x x\right )^2} \, dx}{x}+\frac {\int \frac {e^{2+x}}{x \left (e^2+e^x x\right )^2} \, dx}{x}\right ) \, dx+\left (4-e^{7+\frac {4}{\log (3)}}\right ) \int \frac {e^x}{x \left (e^2+e^x x\right )} \, dx+\log \left (x^4\right ) \int \frac {e^{2+x}}{\left (e^2+e^x x\right )^2} \, dx+\log \left (x^4\right ) \int \frac {e^{2+x}}{x \left (e^2+e^x x\right )^2} \, dx-\log \left (x^4\right ) \int \frac {e^x}{x \left (e^2+e^x x\right )} \, dx-\int \frac {e^{5+2 x+\frac {4}{\log (3)}}}{e^2+e^x x} \, dx+\int \frac {4 \int \frac {e^x}{x \left (e^2+e^x x\right )} \, dx}{x} \, dx\\ &=e^{5+\frac {4}{\log (3)}} \text {Ei}(x)-e^{5+x+\frac {4}{\log (3)}} \log (x)-4 \int \frac {\int \frac {e^{2+x}}{\left (e^2+e^x x\right )^2} \, dx}{x} \, dx-4 \int \frac {\int \frac {e^{2+x}}{x \left (e^2+e^x x\right )^2} \, dx}{x} \, dx+4 \int \frac {\int \frac {e^x}{x \left (e^2+e^x x\right )} \, dx}{x} \, dx+\left (4-e^{7+\frac {4}{\log (3)}}\right ) \int \frac {e^x}{x \left (e^2+e^x x\right )} \, dx+\log \left (x^4\right ) \int \frac {e^{2+x}}{\left (e^2+e^x x\right )^2} \, dx+\log \left (x^4\right ) \int \frac {e^{2+x}}{x \left (e^2+e^x x\right )^2} \, dx-\log \left (x^4\right ) \int \frac {e^x}{x \left (e^2+e^x x\right )} \, dx-\int \frac {e^{5+2 x+\frac {4}{\log (3)}}}{e^2+e^x x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.31, size = 35, normalized size = 1.09 \begin {gather*} -e^{5+x+\frac {4}{\log (3)}} \log (x)+\frac {e^x \log \left (x^4\right )}{e^2+e^x x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.97, size = 82, normalized size = 2.56 \begin {gather*} -\frac {{\left (x e^{\left (\frac {2 \, {\left ({\left (x + 5\right )} \log \relax (3) + 4\right )}}{\log \relax (3)}\right )} + {\left (e^{\left (\frac {7 \, \log \relax (3) + 4}{\log \relax (3)}\right )} - 4\right )} e^{\left (\frac {{\left (x + 5\right )} \log \relax (3) + 4}{\log \relax (3)}\right )}\right )} \log \relax (x)}{x e^{\left (\frac {{\left (x + 5\right )} \log \relax (3) + 4}{\log \relax (3)}\right )} + e^{\left (\frac {7 \, \log \relax (3) + 4}{\log \relax (3)}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*exp(2-x)-x)*log(x^4)+(-x*exp(2-x)^2-2*x^2*exp(2-x)-x^3)*exp(((5+x)*log(3)+4)/log(3))*log(x)+(-ex
p(2-x)^2-2*x*exp(2-x)-x^2)*exp(((5+x)*log(3)+4)/log(3))+4*exp(2-x)+4*x)/(x*exp(2-x)^2+2*x^2*exp(2-x)+x^3),x, a
lgorithm="fricas")

[Out]

-(x*e^(2*((x + 5)*log(3) + 4)/log(3)) + (e^((7*log(3) + 4)/log(3)) - 4)*e^(((x + 5)*log(3) + 4)/log(3)))*log(x
)/(x*e^(((x + 5)*log(3) + 4)/log(3)) + e^((7*log(3) + 4)/log(3)))

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giac [B]  time = 1.43, size = 259, normalized size = 8.09 \begin {gather*} \frac {4 \, x^{4} e^{\left (2 \, x\right )} \log \relax (x)}{x^{5} e^{\left (2 \, x\right )} + x^{4} e^{\left (2 \, x\right )} + 2 \, x^{4} e^{\left (x + 2\right )} + x^{3} e^{4} + 2 \, x^{3} e^{\left (x + 2\right )} + x^{2} e^{4}} + \frac {4 \, x^{3} e^{\left (2 \, x\right )} \log \relax (x)}{x^{5} e^{\left (2 \, x\right )} + x^{4} e^{\left (2 \, x\right )} + 2 \, x^{4} e^{\left (x + 2\right )} + x^{3} e^{4} + 2 \, x^{3} e^{\left (x + 2\right )} + x^{2} e^{4}} + \frac {4 \, x^{3} e^{\left (x + 2\right )} \log \relax (x)}{x^{5} e^{\left (2 \, x\right )} + x^{4} e^{\left (2 \, x\right )} + 2 \, x^{4} e^{\left (x + 2\right )} + x^{3} e^{4} + 2 \, x^{3} e^{\left (x + 2\right )} + x^{2} e^{4}} + \frac {4 \, x^{2} e^{\left (x + 2\right )} \log \relax (x)}{x^{5} e^{\left (2 \, x\right )} + x^{4} e^{\left (2 \, x\right )} + 2 \, x^{4} e^{\left (x + 2\right )} + x^{3} e^{4} + 2 \, x^{3} e^{\left (x + 2\right )} + x^{2} e^{4}} - e^{\left (x + \frac {5 \, \log \relax (3) + 4}{\log \relax (3)}\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*exp(2-x)-x)*log(x^4)+(-x*exp(2-x)^2-2*x^2*exp(2-x)-x^3)*exp(((5+x)*log(3)+4)/log(3))*log(x)+(-ex
p(2-x)^2-2*x*exp(2-x)-x^2)*exp(((5+x)*log(3)+4)/log(3))+4*exp(2-x)+4*x)/(x*exp(2-x)^2+2*x^2*exp(2-x)+x^3),x, a
lgorithm="giac")

[Out]

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

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maple [C]  time = 0.31, size = 241, normalized size = 7.53

method	result	size
risch	\(-\frac {\left (x \,{\mathrm e}^{\frac {4+7 \ln \relax (3)}{\ln \relax (3)}}+{\mathrm e}^{-\frac {x \ln \relax (3)-9 \ln \relax (3)-4}{\ln \relax (3)}}-4 \,{\mathrm e}^{2-x}\right ) {\mathrm e}^{x -2} \ln \relax (x )}{x +{\mathrm e}^{2-x}}-\frac {i \left (\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x \right )^{2}-2 \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x^{3}\right )-\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{3}\right )^{2}+\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{3}\right ) \mathrm {csgn}\left (i x^{4}\right )-\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{4}\right )^{2}+\mathrm {csgn}\left (i x^{2}\right )^{3}-\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x^{3}\right )^{2}+\mathrm {csgn}\left (i x^{3}\right )^{3}-\mathrm {csgn}\left (i x^{3}\right ) \mathrm {csgn}\left (i x^{4}\right )^{2}+\mathrm {csgn}\left (i x^{4}\right )^{3}\right ) \pi }{2 \left (x +{\mathrm e}^{2-x}\right )}\)	\(241\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x*exp(2-x)-x)*ln(x^4)+(-x*exp(2-x)^2-2*x^2*exp(2-x)-x^3)*exp(((5+x)*ln(3)+4)/ln(3))*ln(x)+(-exp(2-x)^2-2
*x*exp(2-x)-x^2)*exp(((5+x)*ln(3)+4)/ln(3))+4*exp(2-x)+4*x)/(x*exp(2-x)^2+2*x^2*exp(2-x)+x^3),x,method=_RETURN
VERBOSE)

[Out]

-(x*exp((4+7*ln(3))/ln(3))+exp(-(x*ln(3)-9*ln(3)-4)/ln(3))-4*exp(2-x))*exp(x-2)/(x+exp(2-x))*ln(x)-1/2*I*(csgn
(I*x^2)*csgn(I*x)^2-2*csgn(I*x)*csgn(I*x^2)^2+csgn(I*x)*csgn(I*x^2)*csgn(I*x^3)-csgn(I*x)*csgn(I*x^3)^2+csgn(I
*x)*csgn(I*x^3)*csgn(I*x^4)-csgn(I*x)*csgn(I*x^4)^2+csgn(I*x^2)^3-csgn(I*x^2)*csgn(I*x^3)^2+csgn(I*x^3)^3-csgn
(I*x^3)*csgn(I*x^4)^2+csgn(I*x^4)^3)*Pi/(x+exp(2-x))

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maxima [A]  time = 0.49, size = 44, normalized size = 1.38 \begin {gather*} -\frac {x e^{\left (2 \, x + \frac {4}{\log \relax (3)} + 5\right )} \log \relax (x) + {\left (e^{\left (\frac {4}{\log \relax (3)} + 7\right )} - 4\right )} e^{x} \log \relax (x)}{x e^{x} + e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*exp(2-x)-x)*log(x^4)+(-x*exp(2-x)^2-2*x^2*exp(2-x)-x^3)*exp(((5+x)*log(3)+4)/log(3))*log(x)+(-ex
p(2-x)^2-2*x*exp(2-x)-x^2)*exp(((5+x)*log(3)+4)/log(3))+4*exp(2-x)+4*x)/(x*exp(2-x)^2+2*x^2*exp(2-x)+x^3),x, a
lgorithm="maxima")

[Out]

-(x*e^(2*x + 4/log(3) + 5)*log(x) + (e^(4/log(3) + 7) - 4)*e^x*log(x))/(x*e^x + e^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [B]  time = 0.53, size = 66, normalized size = 2.06 \begin {gather*} - e^{\frac {\left (x + 5\right ) \log {\relax (3 )} + 4}{\log {\relax (3 )}}} \log {\relax (x )} - \frac {4 e^{7} e^{\frac {4}{\log {\relax (3 )}}} \log {\relax (x )}}{x^{2} e^{\frac {\left (x + 5\right ) \log {\relax (3 )} + 4}{\log {\relax (3 )}}} + x e^{7} e^{\frac {4}{\log {\relax (3 )}}}} + \frac {4 \log {\relax (x )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*exp(2-x)-x)*ln(x**4)+(-x*exp(2-x)**2-2*x**2*exp(2-x)-x**3)*exp(((5+x)*ln(3)+4)/ln(3))*ln(x)+(-ex
p(2-x)**2-2*x*exp(2-x)-x**2)*exp(((5+x)*ln(3)+4)/ln(3))+4*exp(2-x)+4*x)/(x*exp(2-x)**2+2*x**2*exp(2-x)+x**3),x
)

[Out]

-exp(((x + 5)*log(3) + 4)/log(3))*log(x) - 4*exp(7)*exp(4/log(3))*log(x)/(x**2*exp(((x + 5)*log(3) + 4)/log(3)
) + x*exp(7)*exp(4/log(3))) + 4*log(x)/x

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