3.38.10 \(\int \frac {32 x+16 x^2+2 x^3+e^{\frac {5}{-2+\log (5)}} (72 x-7 x^2-2 x^3)+(-144-40 x+7 x^2+2 x^3) \log (-9+2 x)+(e^{\frac {5}{-2+\log (5)}} (36 x+x^2-2 x^3)+(-144-40 x+7 x^2+2 x^3) \log (-9+2 x)) \log (\frac {e^{\frac {5}{-2+\log (5)}} x^2+(-4 x-x^2) \log (-9+2 x)}{4+x})}{e^{\frac {5}{-2+\log (5)}} (36 x+x^2-2 x^3)+(-144-40 x+7 x^2+2 x^3) \log (-9+2 x)} \, dx\)

Optimal. Leaf size=31 \[ x \log \left (x \left (\frac {e^{\frac {5}{-2+\log (5)}} x}{4+x}-\log (-9+2 x)\right )\right ) \]

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Rubi [F]  time = 7.28, 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 {32 x+16 x^2+2 x^3+e^{\frac {5}{-2+\log (5)}} \left (72 x-7 x^2-2 x^3\right )+\left (-144-40 x+7 x^2+2 x^3\right ) \log (-9+2 x)+\left (e^{\frac {5}{-2+\log (5)}} \left (36 x+x^2-2 x^3\right )+\left (-144-40 x+7 x^2+2 x^3\right ) \log (-9+2 x)\right ) \log \left (\frac {e^{\frac {5}{-2+\log (5)}} x^2+\left (-4 x-x^2\right ) \log (-9+2 x)}{4+x}\right )}{e^{\frac {5}{-2+\log (5)}} \left (36 x+x^2-2 x^3\right )+\left (-144-40 x+7 x^2+2 x^3\right ) \log (-9+2 x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(32*x + 16*x^2 + 2*x^3 + E^(5/(-2 + Log[5]))*(72*x - 7*x^2 - 2*x^3) + (-144 - 40*x + 7*x^2 + 2*x^3)*Log[-9
+ 2*x] + (E^(5/(-2 + Log[5]))*(36*x + x^2 - 2*x^3) + (-144 - 40*x + 7*x^2 + 2*x^3)*Log[-9 + 2*x])*Log[(E^(5/(
-2 + Log[5]))*x^2 + (-4*x - x^2)*Log[-9 + 2*x])/(4 + x)])/(E^(5/(-2 + Log[5]))*(36*x + x^2 - 2*x^3) + (-144 -
40*x + 7*x^2 + 2*x^3)*Log[-9 + 2*x]),x]

[Out]

x - (17*Defer[Int][(E^(5/(-2 + Log[5]))*x - 4*Log[-9 + 2*x] - x*Log[-9 + 2*x])^(-1), x])/2 + (4*Defer[Int][(E^
(5/(-2 + Log[5]))*x - 4*Log[-9 + 2*x] - x*Log[-9 + 2*x])^(-1), x])/E^(5/(2 - Log[5])) - Defer[Int][x/(E^(5/(-2
+ Log[5]))*x - 4*Log[-9 + 2*x] - x*Log[-9 + 2*x]), x] - (16*Defer[Int][1/((4 + x)*(E^(5/(-2 + Log[5]))*x - 4*
Log[-9 + 2*x] - x*Log[-9 + 2*x])), x])/E^(5/(2 - Log[5])) - (153*Defer[Int][1/((-9 + 2*x)*(E^(5/(-2 + Log[5]))
*x - 4*Log[-9 + 2*x] - x*Log[-9 + 2*x])), x])/2 + Defer[Int][Log[(E^(5/(-2 + Log[5]))*x^2)/(4 + x) - x*Log[-9
+ 2*x]], x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {32 x+16 x^2+2 x^3+e^{\frac {5}{-2+\log (5)}} \left (72 x-7 x^2-2 x^3\right )+\left (-144-40 x+7 x^2+2 x^3\right ) \log (-9+2 x)+\left (e^{\frac {5}{-2+\log (5)}} \left (36 x+x^2-2 x^3\right )+\left (-144-40 x+7 x^2+2 x^3\right ) \log (-9+2 x)\right ) \log \left (\frac {e^{\frac {5}{-2+\log (5)}} x^2+\left (-4 x-x^2\right ) \log (-9+2 x)}{4+x}\right )}{\left (36+x-2 x^2\right ) \left (e^{\frac {5}{-2+\log (5)}} x-4 \log (-9+2 x)-x \log (-9+2 x)\right )} \, dx\\ &=\int \left (\frac {e^{\frac {5}{-2+\log (5)}} x (8+x)}{(4+x) \left (e^{\frac {5}{-2+\log (5)}} x-4 \log (-9+2 x)-x \log (-9+2 x)\right )}-\frac {32 x}{(4+x) (-9+2 x) \left (e^{\frac {5}{-2+\log (5)}} x-4 \log (-9+2 x)-x \log (-9+2 x)\right )}-\frac {16 x^2}{(4+x) (-9+2 x) \left (e^{\frac {5}{-2+\log (5)}} x-4 \log (-9+2 x)-x \log (-9+2 x)\right )}-\frac {2 x^3}{(4+x) (-9+2 x) \left (e^{\frac {5}{-2+\log (5)}} x-4 \log (-9+2 x)-x \log (-9+2 x)\right )}+\frac {(4+x) \log (-9+2 x)}{-e^{\frac {5}{-2+\log (5)}} x+4 \log (-9+2 x)+x \log (-9+2 x)}+\log \left (\frac {e^{\frac {5}{-2+\log (5)}} x^2}{4+x}-x \log (-9+2 x)\right )\right ) \, dx\\ &=-\left (2 \int \frac {x^3}{(4+x) (-9+2 x) \left (e^{\frac {5}{-2+\log (5)}} x-4 \log (-9+2 x)-x \log (-9+2 x)\right )} \, dx\right )-16 \int \frac {x^2}{(4+x) (-9+2 x) \left (e^{\frac {5}{-2+\log (5)}} x-4 \log (-9+2 x)-x \log (-9+2 x)\right )} \, dx-32 \int \frac {x}{(4+x) (-9+2 x) \left (e^{\frac {5}{-2+\log (5)}} x-4 \log (-9+2 x)-x \log (-9+2 x)\right )} \, dx+e^{-\frac {5}{2-\log (5)}} \int \frac {x (8+x)}{(4+x) \left (e^{\frac {5}{-2+\log (5)}} x-4 \log (-9+2 x)-x \log (-9+2 x)\right )} \, dx+\int \frac {(4+x) \log (-9+2 x)}{-e^{\frac {5}{-2+\log (5)}} x+4 \log (-9+2 x)+x \log (-9+2 x)} \, dx+\int \log \left (\frac {e^{\frac {5}{-2+\log (5)}} x^2}{4+x}-x \log (-9+2 x)\right ) \, dx\\ &=-\left (2 \int \left (\frac {1}{4 \left (e^{\frac {5}{-2+\log (5)}} x-4 \log (-9+2 x)-x \log (-9+2 x)\right )}+\frac {x}{2 \left (e^{\frac {5}{-2+\log (5)}} x-4 \log (-9+2 x)-x \log (-9+2 x)\right )}+\frac {64}{17 (4+x) \left (e^{\frac {5}{-2+\log (5)}} x-4 \log (-9+2 x)-x \log (-9+2 x)\right )}+\frac {729}{68 (-9+2 x) \left (e^{\frac {5}{-2+\log (5)}} x-4 \log (-9+2 x)-x \log (-9+2 x)\right )}\right ) \, dx\right )-16 \int \left (\frac {1}{2 \left (e^{\frac {5}{-2+\log (5)}} x-4 \log (-9+2 x)-x \log (-9+2 x)\right )}-\frac {16}{17 (4+x) \left (e^{\frac {5}{-2+\log (5)}} x-4 \log (-9+2 x)-x \log (-9+2 x)\right )}+\frac {81}{34 (-9+2 x) \left (e^{\frac {5}{-2+\log (5)}} x-4 \log (-9+2 x)-x \log (-9+2 x)\right )}\right ) \, dx-32 \int \left (\frac {4}{17 (4+x) \left (e^{\frac {5}{-2+\log (5)}} x-4 \log (-9+2 x)-x \log (-9+2 x)\right )}+\frac {9}{17 (-9+2 x) \left (e^{\frac {5}{-2+\log (5)}} x-4 \log (-9+2 x)-x \log (-9+2 x)\right )}\right ) \, dx+e^{-\frac {5}{2-\log (5)}} \int \left (\frac {4}{e^{\frac {5}{-2+\log (5)}} x-4 \log (-9+2 x)-x \log (-9+2 x)}+\frac {x}{e^{\frac {5}{-2+\log (5)}} x-4 \log (-9+2 x)-x \log (-9+2 x)}-\frac {16}{(4+x) \left (e^{\frac {5}{-2+\log (5)}} x-4 \log (-9+2 x)-x \log (-9+2 x)\right )}\right ) \, dx+\int \left (1-\frac {e^{\frac {5}{-2+\log (5)}} x}{e^{\frac {5}{-2+\log (5)}} x-4 \log (-9+2 x)-x \log (-9+2 x)}\right ) \, dx+\int \log \left (\frac {e^{\frac {5}{-2+\log (5)}} x^2}{4+x}-x \log (-9+2 x)\right ) \, dx\\ &=x-\frac {1}{2} \int \frac {1}{e^{\frac {5}{-2+\log (5)}} x-4 \log (-9+2 x)-x \log (-9+2 x)} \, dx-2 \left (\frac {128}{17} \int \frac {1}{(4+x) \left (e^{\frac {5}{-2+\log (5)}} x-4 \log (-9+2 x)-x \log (-9+2 x)\right )} \, dx\right )-8 \int \frac {1}{e^{\frac {5}{-2+\log (5)}} x-4 \log (-9+2 x)-x \log (-9+2 x)} \, dx+\frac {256}{17} \int \frac {1}{(4+x) \left (e^{\frac {5}{-2+\log (5)}} x-4 \log (-9+2 x)-x \log (-9+2 x)\right )} \, dx-\frac {288}{17} \int \frac {1}{(-9+2 x) \left (e^{\frac {5}{-2+\log (5)}} x-4 \log (-9+2 x)-x \log (-9+2 x)\right )} \, dx-\frac {729}{34} \int \frac {1}{(-9+2 x) \left (e^{\frac {5}{-2+\log (5)}} x-4 \log (-9+2 x)-x \log (-9+2 x)\right )} \, dx-\frac {648}{17} \int \frac {1}{(-9+2 x) \left (e^{\frac {5}{-2+\log (5)}} x-4 \log (-9+2 x)-x \log (-9+2 x)\right )} \, dx+\left (4 e^{-\frac {5}{2-\log (5)}}\right ) \int \frac {1}{e^{\frac {5}{-2+\log (5)}} x-4 \log (-9+2 x)-x \log (-9+2 x)} \, dx-\left (16 e^{-\frac {5}{2-\log (5)}}\right ) \int \frac {1}{(4+x) \left (e^{\frac {5}{-2+\log (5)}} x-4 \log (-9+2 x)-x \log (-9+2 x)\right )} \, dx-\int \frac {x}{e^{\frac {5}{-2+\log (5)}} x-4 \log (-9+2 x)-x \log (-9+2 x)} \, dx+\int \log \left (\frac {e^{\frac {5}{-2+\log (5)}} x^2}{4+x}-x \log (-9+2 x)\right ) \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.23, size = 31, normalized size = 1.00 \begin {gather*} x \log \left (x \left (\frac {e^{\frac {5}{-2+\log (5)}} x}{4+x}-\log (-9+2 x)\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(32*x + 16*x^2 + 2*x^3 + E^(5/(-2 + Log[5]))*(72*x - 7*x^2 - 2*x^3) + (-144 - 40*x + 7*x^2 + 2*x^3)*
Log[-9 + 2*x] + (E^(5/(-2 + Log[5]))*(36*x + x^2 - 2*x^3) + (-144 - 40*x + 7*x^2 + 2*x^3)*Log[-9 + 2*x])*Log[(
E^(5/(-2 + Log[5]))*x^2 + (-4*x - x^2)*Log[-9 + 2*x])/(4 + x)])/(E^(5/(-2 + Log[5]))*(36*x + x^2 - 2*x^3) + (-
144 - 40*x + 7*x^2 + 2*x^3)*Log[-9 + 2*x]),x]

[Out]

x*Log[x*((E^(5/(-2 + Log[5]))*x)/(4 + x) - Log[-9 + 2*x])]

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fricas [A]  time = 0.57, size = 38, normalized size = 1.23 \begin {gather*} x \log \left (\frac {x^{2} e^{\left (\frac {5}{\log \relax (5) - 2}\right )} - {\left (x^{2} + 4 \, x\right )} \log \left (2 \, x - 9\right )}{x + 4}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^3+7*x^2-40*x-144)*log(2*x-9)+(-2*x^3+x^2+36*x)*exp(5/(log(5)-2)))*log(((-x^2-4*x)*log(2*x-9)+
x^2*exp(5/(log(5)-2)))/(4+x))+(2*x^3+7*x^2-40*x-144)*log(2*x-9)+(-2*x^3-7*x^2+72*x)*exp(5/(log(5)-2))+2*x^3+16
*x^2+32*x)/((2*x^3+7*x^2-40*x-144)*log(2*x-9)+(-2*x^3+x^2+36*x)*exp(5/(log(5)-2))),x, algorithm="fricas")

[Out]

x*log((x^2*e^(5/(log(5) - 2)) - (x^2 + 4*x)*log(2*x - 9))/(x + 4))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {sage}_{0} x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^3+7*x^2-40*x-144)*log(2*x-9)+(-2*x^3+x^2+36*x)*exp(5/(log(5)-2)))*log(((-x^2-4*x)*log(2*x-9)+
x^2*exp(5/(log(5)-2)))/(4+x))+(2*x^3+7*x^2-40*x-144)*log(2*x-9)+(-2*x^3-7*x^2+72*x)*exp(5/(log(5)-2))+2*x^3+16
*x^2+32*x)/((2*x^3+7*x^2-40*x-144)*log(2*x-9)+(-2*x^3+x^2+36*x)*exp(5/(log(5)-2))),x, algorithm="giac")

[Out]

sage0*x

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maple [C]  time = 0.38, size = 575, normalized size = 18.55

method	result	size
risch	\(x \ln \left (-\ln \left (2 x -9\right ) x +x \,{\mathrm e}^{\frac {5}{\ln \relax (5)-2}}-4 \ln \left (2 x -9\right )\right )-x \ln \left (4+x \right )+x \ln \relax (x )-\frac {i \pi x \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i \left (\ln \left (2 x -9\right ) x -x \,{\mathrm e}^{\frac {5}{\ln \relax (5)-2}}+4 \ln \left (2 x -9\right )\right )}{4+x}\right ) \mathrm {csgn}\left (\frac {i x \left (\ln \left (2 x -9\right ) x -x \,{\mathrm e}^{\frac {5}{\ln \relax (5)-2}}+4 \ln \left (2 x -9\right )\right )}{4+x}\right )}{2}+\frac {i \pi x \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i x \left (\ln \left (2 x -9\right ) x -x \,{\mathrm e}^{\frac {5}{\ln \relax (5)-2}}+4 \ln \left (2 x -9\right )\right )}{4+x}\right )^{2}}{2}-\frac {i \pi x \,\mathrm {csgn}\left (\frac {i}{4+x}\right ) \mathrm {csgn}\left (i \left (\ln \left (2 x -9\right ) x -x \,{\mathrm e}^{\frac {5}{\ln \relax (5)-2}}+4 \ln \left (2 x -9\right )\right )\right ) \mathrm {csgn}\left (\frac {i \left (\ln \left (2 x -9\right ) x -x \,{\mathrm e}^{\frac {5}{\ln \relax (5)-2}}+4 \ln \left (2 x -9\right )\right )}{4+x}\right )}{2}+\frac {i \pi x \,\mathrm {csgn}\left (\frac {i}{4+x}\right ) \mathrm {csgn}\left (\frac {i \left (\ln \left (2 x -9\right ) x -x \,{\mathrm e}^{\frac {5}{\ln \relax (5)-2}}+4 \ln \left (2 x -9\right )\right )}{4+x}\right )^{2}}{2}-\frac {i \pi x \,\mathrm {csgn}\left (i \left (\ln \left (2 x -9\right ) x -x \,{\mathrm e}^{\frac {5}{\ln \relax (5)-2}}+4 \ln \left (2 x -9\right )\right )\right ) \mathrm {csgn}\left (\frac {i \left (\ln \left (2 x -9\right ) x -x \,{\mathrm e}^{\frac {5}{\ln \relax (5)-2}}+4 \ln \left (2 x -9\right )\right )}{4+x}\right )^{2}}{2}+\frac {i \pi x \mathrm {csgn}\left (\frac {i \left (\ln \left (2 x -9\right ) x -x \,{\mathrm e}^{\frac {5}{\ln \relax (5)-2}}+4 \ln \left (2 x -9\right )\right )}{4+x}\right )^{3}}{2}-\frac {i \pi x \,\mathrm {csgn}\left (\frac {i \left (\ln \left (2 x -9\right ) x -x \,{\mathrm e}^{\frac {5}{\ln \relax (5)-2}}+4 \ln \left (2 x -9\right )\right )}{4+x}\right ) \mathrm {csgn}\left (\frac {i x \left (\ln \left (2 x -9\right ) x -x \,{\mathrm e}^{\frac {5}{\ln \relax (5)-2}}+4 \ln \left (2 x -9\right )\right )}{4+x}\right )^{2}}{2}+\frac {i \pi x \mathrm {csgn}\left (\frac {i x \left (\ln \left (2 x -9\right ) x -x \,{\mathrm e}^{\frac {5}{\ln \relax (5)-2}}+4 \ln \left (2 x -9\right )\right )}{4+x}\right )^{3}}{2}\)	\(575\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((2*x^3+7*x^2-40*x-144)*ln(2*x-9)+(-2*x^3+x^2+36*x)*exp(5/(ln(5)-2)))*ln(((-x^2-4*x)*ln(2*x-9)+x^2*exp(5/
(ln(5)-2)))/(4+x))+(2*x^3+7*x^2-40*x-144)*ln(2*x-9)+(-2*x^3-7*x^2+72*x)*exp(5/(ln(5)-2))+2*x^3+16*x^2+32*x)/((
2*x^3+7*x^2-40*x-144)*ln(2*x-9)+(-2*x^3+x^2+36*x)*exp(5/(ln(5)-2))),x,method=_RETURNVERBOSE)

[Out]

x*ln(-ln(2*x-9)*x+x*exp(5/(ln(5)-2))-4*ln(2*x-9))-x*ln(4+x)+x*ln(x)-1/2*I*Pi*x*csgn(I*x)*csgn(I*(ln(2*x-9)*x-x
*exp(5/(ln(5)-2))+4*ln(2*x-9))/(4+x))*csgn(I*x*(ln(2*x-9)*x-x*exp(5/(ln(5)-2))+4*ln(2*x-9))/(4+x))+1/2*I*Pi*x*
csgn(I*x)*csgn(I*x*(ln(2*x-9)*x-x*exp(5/(ln(5)-2))+4*ln(2*x-9))/(4+x))^2-1/2*I*Pi*x*csgn(I/(4+x))*csgn(I*(ln(2
*x-9)*x-x*exp(5/(ln(5)-2))+4*ln(2*x-9)))*csgn(I*(ln(2*x-9)*x-x*exp(5/(ln(5)-2))+4*ln(2*x-9))/(4+x))+1/2*I*Pi*x
*csgn(I/(4+x))*csgn(I*(ln(2*x-9)*x-x*exp(5/(ln(5)-2))+4*ln(2*x-9))/(4+x))^2-1/2*I*Pi*x*csgn(I*(ln(2*x-9)*x-x*e
xp(5/(ln(5)-2))+4*ln(2*x-9)))*csgn(I*(ln(2*x-9)*x-x*exp(5/(ln(5)-2))+4*ln(2*x-9))/(4+x))^2+1/2*I*Pi*x*csgn(I*(
ln(2*x-9)*x-x*exp(5/(ln(5)-2))+4*ln(2*x-9))/(4+x))^3-1/2*I*Pi*x*csgn(I*(ln(2*x-9)*x-x*exp(5/(ln(5)-2))+4*ln(2*
x-9))/(4+x))*csgn(I*x*(ln(2*x-9)*x-x*exp(5/(ln(5)-2))+4*ln(2*x-9))/(4+x))^2+1/2*I*Pi*x*csgn(I*x*(ln(2*x-9)*x-x
*exp(5/(ln(5)-2))+4*ln(2*x-9))/(4+x))^3

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maxima [A]  time = 0.52, size = 44, normalized size = 1.42 \begin {gather*} x \log \left (x {\left (e^{\left (\frac {5}{\log \relax (5) - 2}\right )} - \log \left (2 \, x - 9\right )\right )} - 4 \, \log \left (2 \, x - 9\right )\right ) - x \log \left (x + 4\right ) + x \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^3+7*x^2-40*x-144)*log(2*x-9)+(-2*x^3+x^2+36*x)*exp(5/(log(5)-2)))*log(((-x^2-4*x)*log(2*x-9)+
x^2*exp(5/(log(5)-2)))/(4+x))+(2*x^3+7*x^2-40*x-144)*log(2*x-9)+(-2*x^3-7*x^2+72*x)*exp(5/(log(5)-2))+2*x^3+16
*x^2+32*x)/((2*x^3+7*x^2-40*x-144)*log(2*x-9)+(-2*x^3+x^2+36*x)*exp(5/(log(5)-2))),x, algorithm="maxima")

[Out]

x*log(x*(e^(5/(log(5) - 2)) - log(2*x - 9)) - 4*log(2*x - 9)) - x*log(x + 4) + x*log(x)

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mupad [B]  time = 2.88, size = 39, normalized size = 1.26 \begin {gather*} x\,\ln \left (-\frac {\ln \left (2\,x-9\right )\,\left (x^2+4\,x\right )-x^2\,{\mathrm {e}}^{\frac {5}{\ln \relax (5)-2}}}{x+4}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((32*x + log(-(log(2*x - 9)*(4*x + x^2) - x^2*exp(5/(log(5) - 2)))/(x + 4))*(exp(5/(log(5) - 2))*(36*x + x^
2 - 2*x^3) - log(2*x - 9)*(40*x - 7*x^2 - 2*x^3 + 144)) - log(2*x - 9)*(40*x - 7*x^2 - 2*x^3 + 144) + 16*x^2 +
2*x^3 - exp(5/(log(5) - 2))*(7*x^2 - 72*x + 2*x^3))/(exp(5/(log(5) - 2))*(36*x + x^2 - 2*x^3) - log(2*x - 9)*
(40*x - 7*x^2 - 2*x^3 + 144)),x)

[Out]

x*log(-(log(2*x - 9)*(4*x + x^2) - x^2*exp(5/(log(5) - 2)))/(x + 4))

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sympy [B]  time = 2.41, size = 78, normalized size = 2.52 \begin {gather*} \left (x - \frac {1}{24}\right ) \log {\left (\frac {\frac {x^{2}}{e^{- \frac {5}{-2 + \log {\relax (5 )}}}} + \left (- x^{2} - 4 x\right ) \log {\left (2 x - 9 \right )}}{x + 4} \right )} + \frac {\log {\relax (x )}}{24} + \frac {\log {\left (- \frac {x}{x e^{- \frac {5}{-2 + \log {\relax (5 )}}} + 4 e^{- \frac {5}{-2 + \log {\relax (5 )}}}} + \log {\left (2 x - 9 \right )} \right )}}{24} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x**3+7*x**2-40*x-144)*ln(2*x-9)+(-2*x**3+x**2+36*x)*exp(5/(ln(5)-2)))*ln(((-x**2-4*x)*ln(2*x-9)
+x**2*exp(5/(ln(5)-2)))/(4+x))+(2*x**3+7*x**2-40*x-144)*ln(2*x-9)+(-2*x**3-7*x**2+72*x)*exp(5/(ln(5)-2))+2*x**
3+16*x**2+32*x)/((2*x**3+7*x**2-40*x-144)*ln(2*x-9)+(-2*x**3+x**2+36*x)*exp(5/(ln(5)-2))),x)

[Out]

(x - 1/24)*log((x**2*exp(5/(-2 + log(5))) + (-x**2 - 4*x)*log(2*x - 9))/(x + 4)) + log(x)/24 + log(-x/(x*exp(-
5/(-2 + log(5))) + 4*exp(-5/(-2 + log(5)))) + log(2*x - 9))/24

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