Integrand size = 82, antiderivative size = 23 \[ \int \frac {-36+e^{x+x^2 \log (3 x)} \left (-16+16 x+16 x^2+32 x^2 \log (3 x)\right )}{81+16 e^{2 x+2 x^2 \log (3 x)}+e^{x+x^2 \log (3 x)} (72-32 x)-72 x+16 x^2} \, dx=\frac {x}{-\frac {9}{4}-e^{x+x^2 \log (3 x)}+x} \] Output:
x/(x-9/4-exp(x^2*ln(3*x)+x))
\[ \int \frac {-36+e^{x+x^2 \log (3 x)} \left (-16+16 x+16 x^2+32 x^2 \log (3 x)\right )}{81+16 e^{2 x+2 x^2 \log (3 x)}+e^{x+x^2 \log (3 x)} (72-32 x)-72 x+16 x^2} \, dx=\int \frac {-36+e^{x+x^2 \log (3 x)} \left (-16+16 x+16 x^2+32 x^2 \log (3 x)\right )}{81+16 e^{2 x+2 x^2 \log (3 x)}+e^{x+x^2 \log (3 x)} (72-32 x)-72 x+16 x^2} \, dx \] Input:
Integrate[(-36 + E^(x + x^2*Log[3*x])*(-16 + 16*x + 16*x^2 + 32*x^2*Log[3* x]))/(81 + 16*E^(2*x + 2*x^2*Log[3*x]) + E^(x + x^2*Log[3*x])*(72 - 32*x) - 72*x + 16*x^2),x]
Output:
Integrate[(-36 + E^(x + x^2*Log[3*x])*(-16 + 16*x + 16*x^2 + 32*x^2*Log[3* x]))/(81 + 16*E^(2*x + 2*x^2*Log[3*x]) + E^(x + x^2*Log[3*x])*(72 - 32*x) - 72*x + 16*x^2), x]
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.
\(\displaystyle \int \frac {e^{x^2 \log (3 x)+x} \left (16 x^2+32 x^2 \log (3 x)+16 x-16\right )-36}{16 x^2+16 e^{2 x^2 \log (3 x)+2 x}+(72-32 x) e^{x^2 \log (3 x)+x}-72 x+81} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{x^2 \log (3 x)+x} \left (16 x^2+32 x^2 \log (3 x)+16 x-16\right )-36}{\left (4\ 3^{x^2} e^x x^{x^2}-4 x+9\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (-\frac {36}{\left (4\ 3^{x^2} e^x x^{x^2}-4 x+9\right )^2}+\frac {16 x e^{x^2 \log (3 x)+x}}{\left (4\ 3^{x^2} e^x x^{x^2}-4 x+9\right )^2}-\frac {16 e^{x^2 \log (3 x)+x}}{\left (4\ 3^{x^2} e^x x^{x^2}-4 x+9\right )^2}+\frac {16\ 3^{x^2} e^x x^{x^2+2}}{\left (4\ 3^{x^2} e^x x^{x^2}-4 x+9\right )^2}+\frac {32\ 3^{x^2} e^x x^{x^2+2} \log (3 x)}{\left (4\ 3^{x^2} e^x x^{x^2}-4 x+9\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -36 \int \frac {1}{\left (4\ 3^{x^2} e^x x^{x^2}-4 x+9\right )^2}dx-16 \int \frac {e^{\log (3 x) x^2+x}}{\left (4\ 3^{x^2} e^x x^{x^2}-4 x+9\right )^2}dx+16 \int \frac {e^{\log (3 x) x^2+x} x}{\left (4\ 3^{x^2} e^x x^{x^2}-4 x+9\right )^2}dx-32 \int \frac {\int \frac {3^{x^2} e^x x^{x^2+2}}{\left (4\ 3^{x^2} e^x x^{x^2}-4 x+9\right )^2}dx}{x}dx+32 \log (3 x) \int \frac {e^{\log (3) x^2+x} x^{x^2+2}}{\left (4\ 3^{x^2} e^x x^{x^2}-4 x+9\right )^2}dx+16 \int \frac {e^{\log (3) x^2+x} x^{x^2+2}}{\left (4\ 3^{x^2} e^x x^{x^2}-4 x+9\right )^2}dx\) |
Input:
Int[(-36 + E^(x + x^2*Log[3*x])*(-16 + 16*x + 16*x^2 + 32*x^2*Log[3*x]))/( 81 + 16*E^(2*x + 2*x^2*Log[3*x]) + E^(x + x^2*Log[3*x])*(72 - 32*x) - 72*x + 16*x^2),x]
Output:
$Aborted
Time = 0.54 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96
method | result | size |
risch | \(\frac {4 x}{4 x -4 \left (3 x \right )^{x^{2}} {\mathrm e}^{x}-9}\) | \(22\) |
parallelrisch | \(\frac {4 x}{4 x -4 \,{\mathrm e}^{x^{2} \ln \left (3 x \right )+x}-9}\) | \(24\) |
norman | \(\frac {4 \,{\mathrm e}^{x^{2} \ln \left (3 x \right )+x}+9}{4 x -4 \,{\mathrm e}^{x^{2} \ln \left (3 x \right )+x}-9}\) | \(37\) |
Input:
int(((32*x^2*ln(3*x)+16*x^2+16*x-16)*exp(x^2*ln(3*x)+x)-36)/(16*exp(x^2*ln (3*x)+x)^2+(-32*x+72)*exp(x^2*ln(3*x)+x)+16*x^2-72*x+81),x,method=_RETURNV ERBOSE)
Output:
4*x/(4*x-4*(3*x)^(x^2)*exp(x)-9)
Time = 0.10 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {-36+e^{x+x^2 \log (3 x)} \left (-16+16 x+16 x^2+32 x^2 \log (3 x)\right )}{81+16 e^{2 x+2 x^2 \log (3 x)}+e^{x+x^2 \log (3 x)} (72-32 x)-72 x+16 x^2} \, dx=\frac {4 \, x}{4 \, x - 4 \, e^{\left (x^{2} \log \left (3 \, x\right ) + x\right )} - 9} \] Input:
integrate(((32*x^2*log(3*x)+16*x^2+16*x-16)*exp(x^2*log(3*x)+x)-36)/(16*ex p(x^2*log(3*x)+x)^2+(-32*x+72)*exp(x^2*log(3*x)+x)+16*x^2-72*x+81),x, algo rithm="fricas")
Output:
4*x/(4*x - 4*e^(x^2*log(3*x) + x) - 9)
Time = 0.12 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {-36+e^{x+x^2 \log (3 x)} \left (-16+16 x+16 x^2+32 x^2 \log (3 x)\right )}{81+16 e^{2 x+2 x^2 \log (3 x)}+e^{x+x^2 \log (3 x)} (72-32 x)-72 x+16 x^2} \, dx=- \frac {x}{- x + e^{x^{2} \log {\left (3 x \right )} + x} + \frac {9}{4}} \] Input:
integrate(((32*x**2*ln(3*x)+16*x**2+16*x-16)*exp(x**2*ln(3*x)+x)-36)/(16*e xp(x**2*ln(3*x)+x)**2+(-32*x+72)*exp(x**2*ln(3*x)+x)+16*x**2-72*x+81),x)
Output:
-x/(-x + exp(x**2*log(3*x) + x) + 9/4)
Time = 0.17 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.17 \[ \int \frac {-36+e^{x+x^2 \log (3 x)} \left (-16+16 x+16 x^2+32 x^2 \log (3 x)\right )}{81+16 e^{2 x+2 x^2 \log (3 x)}+e^{x+x^2 \log (3 x)} (72-32 x)-72 x+16 x^2} \, dx=\frac {4 \, x}{4 \, x - 4 \, e^{\left (x^{2} \log \left (3\right ) + x^{2} \log \left (x\right ) + x\right )} - 9} \] Input:
integrate(((32*x^2*log(3*x)+16*x^2+16*x-16)*exp(x^2*log(3*x)+x)-36)/(16*ex p(x^2*log(3*x)+x)^2+(-32*x+72)*exp(x^2*log(3*x)+x)+16*x^2-72*x+81),x, algo rithm="maxima")
Output:
4*x/(4*x - 4*e^(x^2*log(3) + x^2*log(x) + x) - 9)
Time = 0.18 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {-36+e^{x+x^2 \log (3 x)} \left (-16+16 x+16 x^2+32 x^2 \log (3 x)\right )}{81+16 e^{2 x+2 x^2 \log (3 x)}+e^{x+x^2 \log (3 x)} (72-32 x)-72 x+16 x^2} \, dx=\frac {4 \, x}{4 \, x - 4 \, e^{\left (x^{2} \log \left (3 \, x\right ) + x\right )} - 9} \] Input:
integrate(((32*x^2*log(3*x)+16*x^2+16*x-16)*exp(x^2*log(3*x)+x)-36)/(16*ex p(x^2*log(3*x)+x)^2+(-32*x+72)*exp(x^2*log(3*x)+x)+16*x^2-72*x+81),x, algo rithm="giac")
Output:
4*x/(4*x - 4*e^(x^2*log(3*x) + x) - 9)
Time = 3.54 (sec) , antiderivative size = 80, normalized size of antiderivative = 3.48 \[ \int \frac {-36+e^{x+x^2 \log (3 x)} \left (-16+16 x+16 x^2+32 x^2 \log (3 x)\right )}{81+16 e^{2 x+2 x^2 \log (3 x)}+e^{x+x^2 \log (3 x)} (72-32 x)-72 x+16 x^2} \, dx=-\frac {52\,x+72\,x^2\,\ln \left (3\,x\right )-32\,x^3\,\ln \left (3\,x\right )+20\,x^2-16\,x^3}{\left (4\,{\mathrm {e}}^x\,{\left (3\,x\right )}^{x^2}-4\,x+9\right )\,\left (5\,x+18\,x\,\ln \left (3\,x\right )-8\,x^2\,\ln \left (3\,x\right )-4\,x^2+13\right )} \] Input:
int((exp(x + x^2*log(3*x))*(16*x + 32*x^2*log(3*x) + 16*x^2 - 16) - 36)/(1 6*exp(2*x + 2*x^2*log(3*x)) - 72*x - exp(x + x^2*log(3*x))*(32*x - 72) + 1 6*x^2 + 81),x)
Output:
-(52*x + 72*x^2*log(3*x) - 32*x^3*log(3*x) + 20*x^2 - 16*x^3)/((4*exp(x)*( 3*x)^(x^2) - 4*x + 9)*(5*x + 18*x*log(3*x) - 8*x^2*log(3*x) - 4*x^2 + 13))
\[ \int \frac {-36+e^{x+x^2 \log (3 x)} \left (-16+16 x+16 x^2+32 x^2 \log (3 x)\right )}{81+16 e^{2 x+2 x^2 \log (3 x)}+e^{x+x^2 \log (3 x)} (72-32 x)-72 x+16 x^2} \, dx=\int \frac {\left (32 x^{2} \mathrm {log}\left (3 x \right )+16 x^{2}+16 x -16\right ) {\mathrm e}^{x^{2} \mathrm {log}\left (3 x \right )+x}-36}{16 \left ({\mathrm e}^{x^{2} \mathrm {log}\left (3 x \right )+x}\right )^{2}+\left (-32 x +72\right ) {\mathrm e}^{x^{2} \mathrm {log}\left (3 x \right )+x}+16 x^{2}-72 x +81}d x \] Input:
int(((32*x^2*log(3*x)+16*x^2+16*x-16)*exp(x^2*log(3*x)+x)-36)/(16*exp(x^2* log(3*x)+x)^2+(-32*x+72)*exp(x^2*log(3*x)+x)+16*x^2-72*x+81),x)
Output:
int(((32*x^2*log(3*x)+16*x^2+16*x-16)*exp(x^2*log(3*x)+x)-36)/(16*exp(x^2* log(3*x)+x)^2+(-32*x+72)*exp(x^2*log(3*x)+x)+16*x^2-72*x+81),x)