\(\int \frac {32 x+16 x^2+2 x^3+e^{\frac {5}{-2+\log (5)}} (72 x-7 x^2-2 x^3)+(-144-40 x+7 x^2+2 x^3) \log (-9+2 x)+(e^{\frac {5}{-2+\log (5)}} (36 x+x^2-2 x^3)+(-144-40 x+7 x^2+2 x^3) \log (-9+2 x)) \log (\frac {e^{\frac {5}{-2+\log (5)}} x^2+(-4 x-x^2) \log (-9+2 x)}{4+x})}{e^{\frac {5}{-2+\log (5)}} (36 x+x^2-2 x^3)+(-144-40 x+7 x^2+2 x^3) \log (-9+2 x)} \, dx\) [3710]

Optimal result

Integrand size = 195, antiderivative size = 31 \[ \int \frac {32 x+16 x^2+2 x^3+e^{\frac {5}{-2+\log (5)}} \left (72 x-7 x^2-2 x^3\right )+\left (-144-40 x+7 x^2+2 x^3\right ) \log (-9+2 x)+\left (e^{\frac {5}{-2+\log (5)}} \left (36 x+x^2-2 x^3\right )+\left (-144-40 x+7 x^2+2 x^3\right ) \log (-9+2 x)\right ) \log \left (\frac {e^{\frac {5}{-2+\log (5)}} x^2+\left (-4 x-x^2\right ) \log (-9+2 x)}{4+x}\right )}{e^{\frac {5}{-2+\log (5)}} \left (36 x+x^2-2 x^3\right )+\left (-144-40 x+7 x^2+2 x^3\right ) \log (-9+2 x)} \, dx=x \log \left (x \left (\frac {e^{\frac {5}{-2+\log (5)}} x}{4+x}-\log (-9+2 x)\right )\right ) \]

[Out]

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

Rubi [F]

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

[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{align*} \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{align*}

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \frac {32 x+16 x^2+2 x^3+e^{\frac {5}{-2+\log (5)}} \left (72 x-7 x^2-2 x^3\right )+\left (-144-40 x+7 x^2+2 x^3\right ) \log (-9+2 x)+\left (e^{\frac {5}{-2+\log (5)}} \left (36 x+x^2-2 x^3\right )+\left (-144-40 x+7 x^2+2 x^3\right ) \log (-9+2 x)\right ) \log \left (\frac {e^{\frac {5}{-2+\log (5)}} x^2+\left (-4 x-x^2\right ) \log (-9+2 x)}{4+x}\right )}{e^{\frac {5}{-2+\log (5)}} \left (36 x+x^2-2 x^3\right )+\left (-144-40 x+7 x^2+2 x^3\right ) \log (-9+2 x)} \, dx=x \log \left (x \left (\frac {e^{\frac {5}{-2+\log (5)}} x}{4+x}-\log (-9+2 x)\right )\right ) \]

[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])]

Maple [A] (verified)

Time = 19.27 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.29

method	result	size
parallelrisch	\(x \ln \left (\frac {\left (-x^{2}-4 x \right ) \ln \left (2 x -9\right )+x^{2} {\mathrm e}^{\frac {5}{\ln \left (5\right )-2}}}{4+x}\right )\)	\(40\)
risch	\(x \ln \left (x \,{\mathrm e}^{\frac {5}{\ln \left (5\right )-2}}-\ln \left (2 x -9\right ) x -4 \ln \left (2 x -9\right )\right )-\ln \left (4+x \right ) x +x \ln \left (x \right )-\frac {i \pi x \,\operatorname {csgn}\left (\frac {i}{4+x}\right ) \operatorname {csgn}\left (i \left (\ln \left (2 x -9\right ) x -x \,{\mathrm e}^{\frac {5}{\ln \left (5\right )-2}}+4 \ln \left (2 x -9\right )\right )\right ) \operatorname {csgn}\left (\frac {i \left (\ln \left (2 x -9\right ) x -x \,{\mathrm e}^{\frac {5}{\ln \left (5\right )-2}}+4 \ln \left (2 x -9\right )\right )}{4+x}\right )}{2}+\frac {i \pi x \,\operatorname {csgn}\left (\frac {i}{4+x}\right ) {\operatorname {csgn}\left (\frac {i \left (\ln \left (2 x -9\right ) x -x \,{\mathrm e}^{\frac {5}{\ln \left (5\right )-2}}+4 \ln \left (2 x -9\right )\right )}{4+x}\right )}^{2}}{2}-\frac {i \pi x \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (\frac {i \left (\ln \left (2 x -9\right ) x -x \,{\mathrm e}^{\frac {5}{\ln \left (5\right )-2}}+4 \ln \left (2 x -9\right )\right )}{4+x}\right ) \operatorname {csgn}\left (\frac {i x \left (\ln \left (2 x -9\right ) x -x \,{\mathrm e}^{\frac {5}{\ln \left (5\right )-2}}+4 \ln \left (2 x -9\right )\right )}{4+x}\right )}{2}+\frac {i \pi x \,\operatorname {csgn}\left (i x \right ) {\operatorname {csgn}\left (\frac {i x \left (\ln \left (2 x -9\right ) x -x \,{\mathrm e}^{\frac {5}{\ln \left (5\right )-2}}+4 \ln \left (2 x -9\right )\right )}{4+x}\right )}^{2}}{2}-\frac {i \pi x \,\operatorname {csgn}\left (i \left (\ln \left (2 x -9\right ) x -x \,{\mathrm e}^{\frac {5}{\ln \left (5\right )-2}}+4 \ln \left (2 x -9\right )\right )\right ) {\operatorname {csgn}\left (\frac {i \left (\ln \left (2 x -9\right ) x -x \,{\mathrm e}^{\frac {5}{\ln \left (5\right )-2}}+4 \ln \left (2 x -9\right )\right )}{4+x}\right )}^{2}}{2}+\frac {i \pi x {\operatorname {csgn}\left (\frac {i \left (\ln \left (2 x -9\right ) x -x \,{\mathrm e}^{\frac {5}{\ln \left (5\right )-2}}+4 \ln \left (2 x -9\right )\right )}{4+x}\right )}^{3}}{2}-\frac {i \pi x \,\operatorname {csgn}\left (\frac {i \left (\ln \left (2 x -9\right ) x -x \,{\mathrm e}^{\frac {5}{\ln \left (5\right )-2}}+4 \ln \left (2 x -9\right )\right )}{4+x}\right ) {\operatorname {csgn}\left (\frac {i x \left (\ln \left (2 x -9\right ) x -x \,{\mathrm e}^{\frac {5}{\ln \left (5\right )-2}}+4 \ln \left (2 x -9\right )\right )}{4+x}\right )}^{2}}{2}+\frac {i \pi x {\operatorname {csgn}\left (\frac {i x \left (\ln \left (2 x -9\right ) x -x \,{\mathrm e}^{\frac {5}{\ln \left (5\right )-2}}+4 \ln \left (2 x -9\right )\right )}{4+x}\right )}^{3}}{2}\)	\(575\)

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.23 \[ \int \frac {32 x+16 x^2+2 x^3+e^{\frac {5}{-2+\log (5)}} \left (72 x-7 x^2-2 x^3\right )+\left (-144-40 x+7 x^2+2 x^3\right ) \log (-9+2 x)+\left (e^{\frac {5}{-2+\log (5)}} \left (36 x+x^2-2 x^3\right )+\left (-144-40 x+7 x^2+2 x^3\right ) \log (-9+2 x)\right ) \log \left (\frac {e^{\frac {5}{-2+\log (5)}} x^2+\left (-4 x-x^2\right ) \log (-9+2 x)}{4+x}\right )}{e^{\frac {5}{-2+\log (5)}} \left (36 x+x^2-2 x^3\right )+\left (-144-40 x+7 x^2+2 x^3\right ) \log (-9+2 x)} \, dx=x \log \left (\frac {x^{2} e^{\left (\frac {5}{\log \left (5\right ) - 2}\right )} - {\left (x^{2} + 4 \, x\right )} \log \left (2 \, x - 9\right )}{x + 4}\right ) \]

[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))

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 75 vs. \(2 (26) = 52\).

Time = 1.39 (sec) , antiderivative size = 75, normalized size of antiderivative = 2.42 \[ \int \frac {32 x+16 x^2+2 x^3+e^{\frac {5}{-2+\log (5)}} \left (72 x-7 x^2-2 x^3\right )+\left (-144-40 x+7 x^2+2 x^3\right ) \log (-9+2 x)+\left (e^{\frac {5}{-2+\log (5)}} \left (36 x+x^2-2 x^3\right )+\left (-144-40 x+7 x^2+2 x^3\right ) \log (-9+2 x)\right ) \log \left (\frac {e^{\frac {5}{-2+\log (5)}} x^2+\left (-4 x-x^2\right ) \log (-9+2 x)}{4+x}\right )}{e^{\frac {5}{-2+\log (5)}} \left (36 x+x^2-2 x^3\right )+\left (-144-40 x+7 x^2+2 x^3\right ) \log (-9+2 x)} \, dx=\left (x - \frac {1}{24}\right ) \log {\left (\frac {\frac {x^{2}}{e^{- \frac {5}{-2 + \log {\left (5 \right )}}}} + \left (- x^{2} - 4 x\right ) \log {\left (2 x - 9 \right )}}{x + 4} \right )} + \frac {\log {\left (x \right )}}{24} + \frac {\log {\left (- \frac {x}{x e^{\frac {5}{2 - \log {\left (5 \right )}}} + 4 e^{\frac {5}{2 - \log {\left (5 \right )}}}} + \log {\left (2 x - 9 \right )} \right )}}{24} \]

[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

Maxima [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.42 \[ \int \frac {32 x+16 x^2+2 x^3+e^{\frac {5}{-2+\log (5)}} \left (72 x-7 x^2-2 x^3\right )+\left (-144-40 x+7 x^2+2 x^3\right ) \log (-9+2 x)+\left (e^{\frac {5}{-2+\log (5)}} \left (36 x+x^2-2 x^3\right )+\left (-144-40 x+7 x^2+2 x^3\right ) \log (-9+2 x)\right ) \log \left (\frac {e^{\frac {5}{-2+\log (5)}} x^2+\left (-4 x-x^2\right ) \log (-9+2 x)}{4+x}\right )}{e^{\frac {5}{-2+\log (5)}} \left (36 x+x^2-2 x^3\right )+\left (-144-40 x+7 x^2+2 x^3\right ) \log (-9+2 x)} \, dx=x \log \left (x {\left (e^{\left (\frac {5}{\log \left (5\right ) - 2}\right )} - \log \left (2 \, x - 9\right )\right )} - 4 \, \log \left (2 \, x - 9\right )\right ) - x \log \left (x + 4\right ) + x \log \left (x\right ) \]

[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)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 70 vs. \(2 (30) = 60\).

Time = 0.65 (sec) , antiderivative size = 70, normalized size of antiderivative = 2.26 \[ \int \frac {32 x+16 x^2+2 x^3+e^{\frac {5}{-2+\log (5)}} \left (72 x-7 x^2-2 x^3\right )+\left (-144-40 x+7 x^2+2 x^3\right ) \log (-9+2 x)+\left (e^{\frac {5}{-2+\log (5)}} \left (36 x+x^2-2 x^3\right )+\left (-144-40 x+7 x^2+2 x^3\right ) \log (-9+2 x)\right ) \log \left (\frac {e^{\frac {5}{-2+\log (5)}} x^2+\left (-4 x-x^2\right ) \log (-9+2 x)}{4+x}\right )}{e^{\frac {5}{-2+\log (5)}} \left (36 x+x^2-2 x^3\right )+\left (-144-40 x+7 x^2+2 x^3\right ) \log (-9+2 x)} \, dx=x \log \left (-\frac {1}{5} \cdot 125^{\frac {3}{4}} x^{2} \log \left (2 \, x - 9\right ) + x^{2} e^{\left (\frac {5 \, {\left (\log \left (5\right )^{2} - 2 \, \log \left (5\right ) + 4\right )}}{4 \, {\left (\log \left (5\right ) - 2\right )}}\right )} - 20 \cdot 5^{\frac {1}{4}} x \log \left (2 \, x - 9\right )\right ) - x \log \left (5 \cdot 5^{\frac {1}{4}} x + 20 \cdot 5^{\frac {1}{4}}\right ) \]

[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]

x*log(-1/5*125^(3/4)*x^2*log(2*x - 9) + x^2*e^(5/4*(log(5)^2 - 2*log(5) + 4)/(log(5) - 2)) - 20*5^(1/4)*x*log(
2*x - 9)) - x*log(5*5^(1/4)*x + 20*5^(1/4))

Mupad [B] (verification not implemented)

Time = 9.32 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.26 \[ \int \frac {32 x+16 x^2+2 x^3+e^{\frac {5}{-2+\log (5)}} \left (72 x-7 x^2-2 x^3\right )+\left (-144-40 x+7 x^2+2 x^3\right ) \log (-9+2 x)+\left (e^{\frac {5}{-2+\log (5)}} \left (36 x+x^2-2 x^3\right )+\left (-144-40 x+7 x^2+2 x^3\right ) \log (-9+2 x)\right ) \log \left (\frac {e^{\frac {5}{-2+\log (5)}} x^2+\left (-4 x-x^2\right ) \log (-9+2 x)}{4+x}\right )}{e^{\frac {5}{-2+\log (5)}} \left (36 x+x^2-2 x^3\right )+\left (-144-40 x+7 x^2+2 x^3\right ) \log (-9+2 x)} \, dx=x\,\ln \left (-\frac {\ln \left (2\,x-9\right )\,\left (x^2+4\,x\right )-x^2\,{\mathrm {e}}^{\frac {5}{\ln \left (5\right )-2}}}{x+4}\right ) \]

[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))