3.34.73 \(\int \frac {10+15 x^3+e^x (10 x+5 x^2)+5 \log (\frac {e^{9+e^x (-1-x)-x^3}}{x^2})}{\log ^2(\frac {e^{9+e^x (-1-x)-x^3}}{x^2})} \, dx\)

Optimal. Leaf size=30 \[ \frac {5 x}{\log \left (\frac {e^{9-e^x-e^x x-x^3}}{x^2}\right )} \]

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Rubi [F]  time = 0.93, 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 {10+15 x^3+e^x \left (10 x+5 x^2\right )+5 \log \left (\frac {e^{9+e^x (-1-x)-x^3}}{x^2}\right )}{\log ^2\left (\frac {e^{9+e^x (-1-x)-x^3}}{x^2}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(10 + 15*x^3 + E^x*(10*x + 5*x^2) + 5*Log[E^(9 + E^x*(-1 - x) - x^3)/x^2])/Log[E^(9 + E^x*(-1 - x) - x^3)/
x^2]^2,x]

[Out]

10*Defer[Int][Log[E^(9 - x^3 - E^x*(1 + x))/x^2]^(-2), x] + 10*Defer[Int][(E^x*x)/Log[E^(9 - x^3 - E^x*(1 + x)
)/x^2]^2, x] + 5*Defer[Int][(E^x*x^2)/Log[E^(9 - x^3 - E^x*(1 + x))/x^2]^2, x] + 15*Defer[Int][x^3/Log[E^(9 -
x^3 - E^x*(1 + x))/x^2]^2, x] + 5*Defer[Int][Log[E^(9 - x^3 - E^x*(1 + x))/x^2]^(-1), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {5 \left (2+2 e^x x+e^x x^2+3 x^3+\log \left (\frac {e^{9-x^3-e^x (1+x)}}{x^2}\right )\right )}{\log ^2\left (\frac {e^{9+e^x (-1-x)-x^3}}{x^2}\right )} \, dx\\ &=5 \int \frac {2+2 e^x x+e^x x^2+3 x^3+\log \left (\frac {e^{9-x^3-e^x (1+x)}}{x^2}\right )}{\log ^2\left (\frac {e^{9+e^x (-1-x)-x^3}}{x^2}\right )} \, dx\\ &=5 \int \left (\frac {e^x x (2+x)}{\log ^2\left (\frac {e^{9-x^3-e^x (1+x)}}{x^2}\right )}+\frac {2+3 x^3+\log \left (\frac {e^{9-x^3-e^x (1+x)}}{x^2}\right )}{\log ^2\left (\frac {e^{9-x^3-e^x (1+x)}}{x^2}\right )}\right ) \, dx\\ &=5 \int \frac {e^x x (2+x)}{\log ^2\left (\frac {e^{9-x^3-e^x (1+x)}}{x^2}\right )} \, dx+5 \int \frac {2+3 x^3+\log \left (\frac {e^{9-x^3-e^x (1+x)}}{x^2}\right )}{\log ^2\left (\frac {e^{9-x^3-e^x (1+x)}}{x^2}\right )} \, dx\\ &=5 \int \left (\frac {2 e^x x}{\log ^2\left (\frac {e^{9-x^3-e^x (1+x)}}{x^2}\right )}+\frac {e^x x^2}{\log ^2\left (\frac {e^{9-x^3-e^x (1+x)}}{x^2}\right )}\right ) \, dx+5 \int \left (\frac {2+3 x^3}{\log ^2\left (\frac {e^{9-x^3-e^x (1+x)}}{x^2}\right )}+\frac {1}{\log \left (\frac {e^{9-x^3-e^x (1+x)}}{x^2}\right )}\right ) \, dx\\ &=5 \int \frac {e^x x^2}{\log ^2\left (\frac {e^{9-x^3-e^x (1+x)}}{x^2}\right )} \, dx+5 \int \frac {2+3 x^3}{\log ^2\left (\frac {e^{9-x^3-e^x (1+x)}}{x^2}\right )} \, dx+5 \int \frac {1}{\log \left (\frac {e^{9-x^3-e^x (1+x)}}{x^2}\right )} \, dx+10 \int \frac {e^x x}{\log ^2\left (\frac {e^{9-x^3-e^x (1+x)}}{x^2}\right )} \, dx\\ &=5 \int \left (\frac {2}{\log ^2\left (\frac {e^{9-x^3-e^x (1+x)}}{x^2}\right )}+\frac {3 x^3}{\log ^2\left (\frac {e^{9-x^3-e^x (1+x)}}{x^2}\right )}\right ) \, dx+5 \int \frac {e^x x^2}{\log ^2\left (\frac {e^{9-x^3-e^x (1+x)}}{x^2}\right )} \, dx+5 \int \frac {1}{\log \left (\frac {e^{9-x^3-e^x (1+x)}}{x^2}\right )} \, dx+10 \int \frac {e^x x}{\log ^2\left (\frac {e^{9-x^3-e^x (1+x)}}{x^2}\right )} \, dx\\ &=5 \int \frac {e^x x^2}{\log ^2\left (\frac {e^{9-x^3-e^x (1+x)}}{x^2}\right )} \, dx+5 \int \frac {1}{\log \left (\frac {e^{9-x^3-e^x (1+x)}}{x^2}\right )} \, dx+10 \int \frac {1}{\log ^2\left (\frac {e^{9-x^3-e^x (1+x)}}{x^2}\right )} \, dx+10 \int \frac {e^x x}{\log ^2\left (\frac {e^{9-x^3-e^x (1+x)}}{x^2}\right )} \, dx+15 \int \frac {x^3}{\log ^2\left (\frac {e^{9-x^3-e^x (1+x)}}{x^2}\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.32, size = 27, normalized size = 0.90 \begin {gather*} \frac {5 x}{\log \left (\frac {e^{9-x^3-e^x (1+x)}}{x^2}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(10 + 15*x^3 + E^x*(10*x + 5*x^2) + 5*Log[E^(9 + E^x*(-1 - x) - x^3)/x^2])/Log[E^(9 + E^x*(-1 - x) -
x^3)/x^2]^2,x]

[Out]

(5*x)/Log[E^(9 - x^3 - E^x*(1 + x))/x^2]

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fricas [A]  time = 0.56, size = 25, normalized size = 0.83 \begin {gather*} \frac {5 \, x}{\log \left (\frac {e^{\left (-x^{3} - {\left (x + 1\right )} e^{x} + 9\right )}}{x^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*log(exp((-x-1)*exp(x)-x^3+9)/x^2)+(5*x^2+10*x)*exp(x)+15*x^3+10)/log(exp((-x-1)*exp(x)-x^3+9)/x^2
)^2,x, algorithm="fricas")

[Out]

5*x/log(e^(-x^3 - (x + 1)*e^x + 9)/x^2)

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giac [A]  time = 0.51, size = 20, normalized size = 0.67 \begin {gather*} -\frac {5 \, x}{x^{3} + x e^{x} + e^{x} + \log \left (x^{2}\right ) - 9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*log(exp((-x-1)*exp(x)-x^3+9)/x^2)+(5*x^2+10*x)*exp(x)+15*x^3+10)/log(exp((-x-1)*exp(x)-x^3+9)/x^2
)^2,x, algorithm="giac")

[Out]

-5*x/(x^3 + x*e^x + e^x + log(x^2) - 9)

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maple [C]  time = 0.18, size = 246, normalized size = 8.20

method	result	size
risch	\(-\frac {10 i x}{\pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+\pi \mathrm {csgn}\left (i x^{2}\right )^{3}-\pi \,\mathrm {csgn}\left (\frac {i}{x^{2}}\right ) \mathrm {csgn}\left (i {\mathrm e}^{9-{\mathrm e}^{x} x -x^{3}-{\mathrm e}^{x}}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{9-{\mathrm e}^{x} x -x^{3}-{\mathrm e}^{x}}}{x^{2}}\right )+\pi \,\mathrm {csgn}\left (\frac {i}{x^{2}}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{9-{\mathrm e}^{x} x -x^{3}-{\mathrm e}^{x}}}{x^{2}}\right )^{2}+\pi \,\mathrm {csgn}\left (i {\mathrm e}^{9-{\mathrm e}^{x} x -x^{3}-{\mathrm e}^{x}}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{9-{\mathrm e}^{x} x -x^{3}-{\mathrm e}^{x}}}{x^{2}}\right )^{2}-\pi \mathrm {csgn}\left (\frac {i {\mathrm e}^{9-{\mathrm e}^{x} x -x^{3}-{\mathrm e}^{x}}}{x^{2}}\right )^{3}+4 i \ln \relax (x )-2 i \ln \left ({\mathrm e}^{9-{\mathrm e}^{x} x -x^{3}-{\mathrm e}^{x}}\right )}\)	\(246\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*ln(exp((-x-1)*exp(x)-x^3+9)/x^2)+(5*x^2+10*x)*exp(x)+15*x^3+10)/ln(exp((-x-1)*exp(x)-x^3+9)/x^2)^2,x,me
thod=_RETURNVERBOSE)

[Out]

-10*I*x/(Pi*csgn(I*x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2+Pi*csgn(I*x^2)^3-Pi*csgn(I/x^2)*csgn(I*exp(9-
exp(x)*x-x^3-exp(x)))*csgn(I*exp(9-exp(x)*x-x^3-exp(x))/x^2)+Pi*csgn(I/x^2)*csgn(I*exp(9-exp(x)*x-x^3-exp(x))/
x^2)^2+Pi*csgn(I*exp(9-exp(x)*x-x^3-exp(x)))*csgn(I*exp(9-exp(x)*x-x^3-exp(x))/x^2)^2-Pi*csgn(I*exp(9-exp(x)*x
-x^3-exp(x))/x^2)^3+4*I*ln(x)-2*I*ln(exp(9-exp(x)*x-x^3-exp(x))))

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maxima [A]  time = 0.46, size = 20, normalized size = 0.67 \begin {gather*} -\frac {5 \, x}{x^{3} + {\left (x + 1\right )} e^{x} + 2 \, \log \relax (x) - 9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*log(exp((-x-1)*exp(x)-x^3+9)/x^2)+(5*x^2+10*x)*exp(x)+15*x^3+10)/log(exp((-x-1)*exp(x)-x^3+9)/x^2
)^2,x, algorithm="maxima")

[Out]

-5*x/(x^3 + (x + 1)*e^x + 2*log(x) - 9)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {5\,\ln \left (\frac {{\mathrm {e}}^{9-x^3-{\mathrm {e}}^x\,\left (x+1\right )}}{x^2}\right )+{\mathrm {e}}^x\,\left (5\,x^2+10\,x\right )+15\,x^3+10}{{\ln \left (\frac {{\mathrm {e}}^{9-x^3-{\mathrm {e}}^x\,\left (x+1\right )}}{x^2}\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*log(exp(9 - x^3 - exp(x)*(x + 1))/x^2) + exp(x)*(10*x + 5*x^2) + 15*x^3 + 10)/log(exp(9 - x^3 - exp(x)*
(x + 1))/x^2)^2,x)

[Out]

int((5*log(exp(9 - x^3 - exp(x)*(x + 1))/x^2) + exp(x)*(10*x + 5*x^2) + 15*x^3 + 10)/log(exp(9 - x^3 - exp(x)*
(x + 1))/x^2)^2, x)

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sympy [A]  time = 0.34, size = 22, normalized size = 0.73 \begin {gather*} \frac {5 x}{\log {\left (\frac {e^{- x^{3} + \left (- x - 1\right ) e^{x} + 9}}{x^{2}} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*ln(exp((-x-1)*exp(x)-x**3+9)/x**2)+(5*x**2+10*x)*exp(x)+15*x**3+10)/ln(exp((-x-1)*exp(x)-x**3+9)/
x**2)**2,x)

[Out]

5*x/log(exp(-x**3 + (-x - 1)*exp(x) + 9)/x**2)

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