Integrand size = 80, antiderivative size = 22 \[ \int \frac {-16 e^x x-16 e^x \log (x)+\left (e^x \left (-8-4 x+4 x^2\right )+e^x (4+4 x) \log (x)\right ) \log \left (9 x^2\right )}{\left (x^3+3 x^2 \log (x)+3 x \log ^2(x)+\log ^3(x)\right ) \log ^3\left (9 x^2\right )} \, dx=-2+\frac {4 e^x x}{(x+\log (x))^2 \log ^2\left (9 x^2\right )} \] Output:
4*exp(x)*x/ln(9*x^2)^2/(x+ln(x))^2-2
Time = 0.21 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {-16 e^x x-16 e^x \log (x)+\left (e^x \left (-8-4 x+4 x^2\right )+e^x (4+4 x) \log (x)\right ) \log \left (9 x^2\right )}{\left (x^3+3 x^2 \log (x)+3 x \log ^2(x)+\log ^3(x)\right ) \log ^3\left (9 x^2\right )} \, dx=\frac {4 e^x x}{(x+\log (x))^2 \log ^2\left (9 x^2\right )} \] Input:
Integrate[(-16*E^x*x - 16*E^x*Log[x] + (E^x*(-8 - 4*x + 4*x^2) + E^x*(4 + 4*x)*Log[x])*Log[9*x^2])/((x^3 + 3*x^2*Log[x] + 3*x*Log[x]^2 + Log[x]^3)*L og[9*x^2]^3),x]
Output:
(4*E^x*x)/((x + Log[x])^2*Log[9*x^2]^2)
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\left (e^x \left (4 x^2-4 x-8\right )+e^x (4 x+4) \log (x)\right ) \log \left (9 x^2\right )-16 e^x x-16 e^x \log (x)}{\left (x^3+3 x^2 \log (x)+\log ^3(x)+3 x \log ^2(x)\right ) \log ^3\left (9 x^2\right )} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {\left (e^x \left (4 x^2-4 x-8\right )+e^x (4 x+4) \log (x)\right ) \log \left (9 x^2\right )-16 e^x x-16 e^x \log (x)}{(x+\log (x))^3 \log ^3\left (9 x^2\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (-\frac {16 e^x x}{(x+\log (x))^3 \log ^3\left (9 x^2\right )}-\frac {16 e^x \log (x)}{(x+\log (x))^3 \log ^3\left (9 x^2\right )}+\frac {4 e^x x^2}{(x+\log (x))^3 \log ^2\left (9 x^2\right )}-\frac {4 e^x x}{(x+\log (x))^3 \log ^2\left (9 x^2\right )}+\frac {4 e^x x \log (x)}{(x+\log (x))^3 \log ^2\left (9 x^2\right )}-\frac {8 e^x}{(x+\log (x))^3 \log ^2\left (9 x^2\right )}+\frac {4 e^x \log (x)}{(x+\log (x))^3 \log ^2\left (9 x^2\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -16 \int \frac {e^x x}{(x+\log (x))^3 \log ^3\left (9 x^2\right )}dx-16 \int \frac {e^x \log (x)}{(x+\log (x))^3 \log ^3\left (9 x^2\right )}dx-8 \int \frac {e^x}{(x+\log (x))^3 \log ^2\left (9 x^2\right )}dx-4 \int \frac {e^x x}{(x+\log (x))^3 \log ^2\left (9 x^2\right )}dx+4 \int \frac {e^x x^2}{(x+\log (x))^3 \log ^2\left (9 x^2\right )}dx+4 \int \frac {e^x \log (x)}{(x+\log (x))^3 \log ^2\left (9 x^2\right )}dx+4 \int \frac {e^x x \log (x)}{(x+\log (x))^3 \log ^2\left (9 x^2\right )}dx\) |
Input:
Int[(-16*E^x*x - 16*E^x*Log[x] + (E^x*(-8 - 4*x + 4*x^2) + E^x*(4 + 4*x)*L og[x])*Log[9*x^2])/((x^3 + 3*x^2*Log[x] + 3*x*Log[x]^2 + Log[x]^3)*Log[9*x ^2]^3),x]
Output:
$Aborted
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 51.11 (sec) , antiderivative size = 71, normalized size of antiderivative = 3.23
\[-\frac {16 x \,{\mathrm e}^{x}}{{\left (-\pi \operatorname {csgn}\left (i x^{2}\right )^{3}-\pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )+2 \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}-4 i \ln \left (3\right )-4 i \ln \left (x \right )\right )}^{2} \left (x +\ln \left (x \right )\right )^{2}}\]
Input:
int((((4+4*x)*exp(x)*ln(x)+(4*x^2-4*x-8)*exp(x))*ln(9*x^2)-16*exp(x)*ln(x) -16*exp(x)*x)/(ln(x)^3+3*x*ln(x)^2+3*x^2*ln(x)+x^3)/ln(9*x^2)^3,x)
Output:
-16*x*exp(x)/(-Pi*csgn(I*x^2)^3-Pi*csgn(I*x)^2*csgn(I*x^2)+2*Pi*csgn(I*x)* csgn(I*x^2)^2-4*I*ln(3)-4*I*ln(x))^2/(x+ln(x))^2
Leaf count of result is larger than twice the leaf count of optimal. 64 vs. \(2 (21) = 42\).
Time = 0.10 (sec) , antiderivative size = 64, normalized size of antiderivative = 2.91 \[ \int \frac {-16 e^x x-16 e^x \log (x)+\left (e^x \left (-8-4 x+4 x^2\right )+e^x (4+4 x) \log (x)\right ) \log \left (9 x^2\right )}{\left (x^3+3 x^2 \log (x)+3 x \log ^2(x)+\log ^3(x)\right ) \log ^3\left (9 x^2\right )} \, dx=\frac {x e^{x}}{x^{2} \log \left (3\right )^{2} + 2 \, {\left (x + \log \left (3\right )\right )} \log \left (x\right )^{3} + \log \left (x\right )^{4} + {\left (x^{2} + 4 \, x \log \left (3\right ) + \log \left (3\right )^{2}\right )} \log \left (x\right )^{2} + 2 \, {\left (x^{2} \log \left (3\right ) + x \log \left (3\right )^{2}\right )} \log \left (x\right )} \] Input:
integrate((((4+4*x)*exp(x)*log(x)+(4*x^2-4*x-8)*exp(x))*log(9*x^2)-16*exp( x)*log(x)-16*exp(x)*x)/(log(x)^3+3*x*log(x)^2+3*x^2*log(x)+x^3)/log(9*x^2) ^3,x, algorithm="fricas")
Output:
x*e^x/(x^2*log(3)^2 + 2*(x + log(3))*log(x)^3 + log(x)^4 + (x^2 + 4*x*log( 3) + log(3)^2)*log(x)^2 + 2*(x^2*log(3) + x*log(3)^2)*log(x))
Leaf count of result is larger than twice the leaf count of optimal. 90 vs. \(2 (22) = 44\).
Time = 0.20 (sec) , antiderivative size = 90, normalized size of antiderivative = 4.09 \[ \int \frac {-16 e^x x-16 e^x \log (x)+\left (e^x \left (-8-4 x+4 x^2\right )+e^x (4+4 x) \log (x)\right ) \log \left (9 x^2\right )}{\left (x^3+3 x^2 \log (x)+3 x \log ^2(x)+\log ^3(x)\right ) \log ^3\left (9 x^2\right )} \, dx=\frac {x e^{x}}{x^{2} \log {\left (x \right )}^{2} + 2 x^{2} \log {\left (3 \right )} \log {\left (x \right )} + x^{2} \log {\left (3 \right )}^{2} + 2 x \log {\left (x \right )}^{3} + 4 x \log {\left (3 \right )} \log {\left (x \right )}^{2} + 2 x \log {\left (3 \right )}^{2} \log {\left (x \right )} + \log {\left (x \right )}^{4} + 2 \log {\left (3 \right )} \log {\left (x \right )}^{3} + \log {\left (3 \right )}^{2} \log {\left (x \right )}^{2}} \] Input:
integrate((((4+4*x)*exp(x)*ln(x)+(4*x**2-4*x-8)*exp(x))*ln(9*x**2)-16*exp( x)*ln(x)-16*exp(x)*x)/(ln(x)**3+3*x*ln(x)**2+3*x**2*ln(x)+x**3)/ln(9*x**2) **3,x)
Output:
x*exp(x)/(x**2*log(x)**2 + 2*x**2*log(3)*log(x) + x**2*log(3)**2 + 2*x*log (x)**3 + 4*x*log(3)*log(x)**2 + 2*x*log(3)**2*log(x) + log(x)**4 + 2*log(3 )*log(x)**3 + log(3)**2*log(x)**2)
Leaf count of result is larger than twice the leaf count of optimal. 64 vs. \(2 (21) = 42\).
Time = 0.19 (sec) , antiderivative size = 64, normalized size of antiderivative = 2.91 \[ \int \frac {-16 e^x x-16 e^x \log (x)+\left (e^x \left (-8-4 x+4 x^2\right )+e^x (4+4 x) \log (x)\right ) \log \left (9 x^2\right )}{\left (x^3+3 x^2 \log (x)+3 x \log ^2(x)+\log ^3(x)\right ) \log ^3\left (9 x^2\right )} \, dx=\frac {x e^{x}}{x^{2} \log \left (3\right )^{2} + 2 \, {\left (x + \log \left (3\right )\right )} \log \left (x\right )^{3} + \log \left (x\right )^{4} + {\left (x^{2} + 4 \, x \log \left (3\right ) + \log \left (3\right )^{2}\right )} \log \left (x\right )^{2} + 2 \, {\left (x^{2} \log \left (3\right ) + x \log \left (3\right )^{2}\right )} \log \left (x\right )} \] Input:
integrate((((4+4*x)*exp(x)*log(x)+(4*x^2-4*x-8)*exp(x))*log(9*x^2)-16*exp( x)*log(x)-16*exp(x)*x)/(log(x)^3+3*x*log(x)^2+3*x^2*log(x)+x^3)/log(9*x^2) ^3,x, algorithm="maxima")
Output:
x*e^x/(x^2*log(3)^2 + 2*(x + log(3))*log(x)^3 + log(x)^4 + (x^2 + 4*x*log( 3) + log(3)^2)*log(x)^2 + 2*(x^2*log(3) + x*log(3)^2)*log(x))
Leaf count of result is larger than twice the leaf count of optimal. 78 vs. \(2 (21) = 42\).
Time = 0.16 (sec) , antiderivative size = 78, normalized size of antiderivative = 3.55 \[ \int \frac {-16 e^x x-16 e^x \log (x)+\left (e^x \left (-8-4 x+4 x^2\right )+e^x (4+4 x) \log (x)\right ) \log \left (9 x^2\right )}{\left (x^3+3 x^2 \log (x)+3 x \log ^2(x)+\log ^3(x)\right ) \log ^3\left (9 x^2\right )} \, dx=\frac {x e^{x}}{x^{2} \log \left (3\right )^{2} + 2 \, x^{2} \log \left (3\right ) \log \left (x\right ) + 2 \, x \log \left (3\right )^{2} \log \left (x\right ) + x^{2} \log \left (x\right )^{2} + 4 \, x \log \left (3\right ) \log \left (x\right )^{2} + \log \left (3\right )^{2} \log \left (x\right )^{2} + 2 \, x \log \left (x\right )^{3} + 2 \, \log \left (3\right ) \log \left (x\right )^{3} + \log \left (x\right )^{4}} \] Input:
integrate((((4+4*x)*exp(x)*log(x)+(4*x^2-4*x-8)*exp(x))*log(9*x^2)-16*exp( x)*log(x)-16*exp(x)*x)/(log(x)^3+3*x*log(x)^2+3*x^2*log(x)+x^3)/log(9*x^2) ^3,x, algorithm="giac")
Output:
x*e^x/(x^2*log(3)^2 + 2*x^2*log(3)*log(x) + 2*x*log(3)^2*log(x) + x^2*log( x)^2 + 4*x*log(3)*log(x)^2 + log(3)^2*log(x)^2 + 2*x*log(x)^3 + 2*log(3)*l og(x)^3 + log(x)^4)
Timed out. \[ \int \frac {-16 e^x x-16 e^x \log (x)+\left (e^x \left (-8-4 x+4 x^2\right )+e^x (4+4 x) \log (x)\right ) \log \left (9 x^2\right )}{\left (x^3+3 x^2 \log (x)+3 x \log ^2(x)+\log ^3(x)\right ) \log ^3\left (9 x^2\right )} \, dx=\int -\frac {16\,{\mathrm {e}}^x\,\ln \left (x\right )+\ln \left (9\,x^2\right )\,\left ({\mathrm {e}}^x\,\left (-4\,x^2+4\,x+8\right )-{\mathrm {e}}^x\,\ln \left (x\right )\,\left (4\,x+4\right )\right )+16\,x\,{\mathrm {e}}^x}{{\ln \left (9\,x^2\right )}^3\,\left (x^3+3\,x^2\,\ln \left (x\right )+3\,x\,{\ln \left (x\right )}^2+{\ln \left (x\right )}^3\right )} \,d x \] Input:
int(-(16*exp(x)*log(x) + log(9*x^2)*(exp(x)*(4*x - 4*x^2 + 8) - exp(x)*log (x)*(4*x + 4)) + 16*x*exp(x))/(log(9*x^2)^3*(3*x*log(x)^2 + 3*x^2*log(x) + log(x)^3 + x^3)),x)
Output:
int(-(16*exp(x)*log(x) + log(9*x^2)*(exp(x)*(4*x - 4*x^2 + 8) - exp(x)*log (x)*(4*x + 4)) + 16*x*exp(x))/(log(9*x^2)^3*(3*x*log(x)^2 + 3*x^2*log(x) + log(x)^3 + x^3)), x)
Time = 0.22 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.32 \[ \int \frac {-16 e^x x-16 e^x \log (x)+\left (e^x \left (-8-4 x+4 x^2\right )+e^x (4+4 x) \log (x)\right ) \log \left (9 x^2\right )}{\left (x^3+3 x^2 \log (x)+3 x \log ^2(x)+\log ^3(x)\right ) \log ^3\left (9 x^2\right )} \, dx=\frac {4 e^{x} x}{\mathrm {log}\left (9 x^{2}\right )^{2} \left (\mathrm {log}\left (x \right )^{2}+2 \,\mathrm {log}\left (x \right ) x +x^{2}\right )} \] Input:
int((((4+4*x)*exp(x)*log(x)+(4*x^2-4*x-8)*exp(x))*log(9*x^2)-16*exp(x)*log (x)-16*exp(x)*x)/(log(x)^3+3*x*log(x)^2+3*x^2*log(x)+x^3)/log(9*x^2)^3,x)
Output:
(4*e**x*x)/(log(9*x**2)**2*(log(x)**2 + 2*log(x)*x + x**2))