Integrand size = 140, antiderivative size = 26 \[ \int \frac {16 x^2+e^x \left (8 x+8 x^2\right )+\left (6 x+e^{2 x} x+2 x^3+e^x \left (3+3 x^2\right )\right ) \log \left (3+e^x x+x^2\right )+\left (64 x+e^x (32+32 x)+\left (24+8 e^x x+8 x^2\right ) \log \left (3+e^x x+x^2\right )\right ) \log \left ((-1+\log (9)) \log \left (3+e^x x+x^2\right )\right )}{\left (3+e^x x+x^2\right ) \log \left (3+e^x x+x^2\right )} \, dx=e^x+\left (x+4 \log \left ((-1+\log (9)) \log \left (3+x \left (e^x+x\right )\right )\right )\right )^2 \]
Time = 0.11 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.85 \[ \int \frac {16 x^2+e^x \left (8 x+8 x^2\right )+\left (6 x+e^{2 x} x+2 x^3+e^x \left (3+3 x^2\right )\right ) \log \left (3+e^x x+x^2\right )+\left (64 x+e^x (32+32 x)+\left (24+8 e^x x+8 x^2\right ) \log \left (3+e^x x+x^2\right )\right ) \log \left ((-1+\log (9)) \log \left (3+e^x x+x^2\right )\right )}{\left (3+e^x x+x^2\right ) \log \left (3+e^x x+x^2\right )} \, dx=e^x+x^2+8 x \log \left ((-1+\log (9)) \log \left (3+e^x x+x^2\right )\right )+16 \log ^2\left ((-1+\log (9)) \log \left (3+e^x x+x^2\right )\right ) \]
Integrate[(16*x^2 + E^x*(8*x + 8*x^2) + (6*x + E^(2*x)*x + 2*x^3 + E^x*(3 + 3*x^2))*Log[3 + E^x*x + x^2] + (64*x + E^x*(32 + 32*x) + (24 + 8*E^x*x + 8*x^2)*Log[3 + E^x*x + x^2])*Log[(-1 + Log[9])*Log[3 + E^x*x + x^2]])/((3 + E^x*x + x^2)*Log[3 + E^x*x + x^2]),x]
E^x + x^2 + 8*x*Log[(-1 + Log[9])*Log[3 + E^x*x + x^2]] + 16*Log[(-1 + Log [9])*Log[3 + E^x*x + 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.
\(\displaystyle \int \frac {16 x^2+e^x \left (8 x^2+8 x\right )+\left (\left (8 x^2+8 e^x x+24\right ) \log \left (x^2+e^x x+3\right )+64 x+e^x (32 x+32)\right ) \log \left ((\log (9)-1) \log \left (x^2+e^x x+3\right )\right )+\left (2 x^3+e^x \left (3 x^2+3\right )+e^{2 x} x+6 x\right ) \log \left (x^2+e^x x+3\right )}{\left (x^2+e^x x+3\right ) \log \left (x^2+e^x x+3\right )} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {2 \left (x \log \left (x^2+e^x x+3\right )+4 x+4\right ) \left (4 \log \left ((\log (9)-1) \log \left (x^2+e^x x+3\right )\right )+x\right )}{x \log \left (x^2+e^x x+3\right )}-\frac {8 \left (x^3-x^2+3 x+3\right ) \left (4 \log \left ((\log (9)-1) \log \left (x^2+e^x x+3\right )\right )+x\right )}{x \left (x^2+e^x x+3\right ) \log \left (x^2+e^x x+3\right )}+e^x\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 32 \int \frac {\log \left ((-1+\log (9)) \log \left (x^2+e^x x+3\right )\right )}{\log \left (x^2+e^x x+3\right )}dx+32 \int \frac {\log \left ((-1+\log (9)) \log \left (x^2+e^x x+3\right )\right )}{x \log \left (x^2+e^x x+3\right )}dx-96 \int \frac {\log \left ((-1+\log (9)) \log \left (x^2+e^x x+3\right )\right )}{\left (x^2+e^x x+3\right ) \log \left (x^2+e^x x+3\right )}dx-96 \int \frac {\log \left ((-1+\log (9)) \log \left (x^2+e^x x+3\right )\right )}{x \left (x^2+e^x x+3\right ) \log \left (x^2+e^x x+3\right )}dx+32 \int \frac {x \log \left ((-1+\log (9)) \log \left (x^2+e^x x+3\right )\right )}{\left (x^2+e^x x+3\right ) \log \left (x^2+e^x x+3\right )}dx-32 \int \frac {x^2 \log \left ((-1+\log (9)) \log \left (x^2+e^x x+3\right )\right )}{\left (x^2+e^x x+3\right ) \log \left (x^2+e^x x+3\right )}dx+x^2+8 x \log \left (-\left ((1-\log (9)) \log \left (x^2+e^x x+3\right )\right )\right )+e^x\) |
Int[(16*x^2 + E^x*(8*x + 8*x^2) + (6*x + E^(2*x)*x + 2*x^3 + E^x*(3 + 3*x^ 2))*Log[3 + E^x*x + x^2] + (64*x + E^x*(32 + 32*x) + (24 + 8*E^x*x + 8*x^2 )*Log[3 + E^x*x + x^2])*Log[(-1 + Log[9])*Log[3 + E^x*x + x^2]])/((3 + E^x *x + x^2)*Log[3 + E^x*x + x^2]),x]
3.12.94.3.1 Defintions of rubi rules used
Time = 0.60 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.92
method | result | size |
risch | \(x^{2}+8 \ln \left (\left (2 \ln \left (3\right )-1\right ) \ln \left ({\mathrm e}^{x} x +x^{2}+3\right )\right ) x +16 {\ln \left (\left (2 \ln \left (3\right )-1\right ) \ln \left ({\mathrm e}^{x} x +x^{2}+3\right )\right )}^{2}+{\mathrm e}^{x}\) | \(50\) |
parallelrisch | \(-3+x^{2}+8 \ln \left (\left (2 \ln \left (3\right )-1\right ) \ln \left ({\mathrm e}^{x} x +x^{2}+3\right )\right ) x +16 {\ln \left (\left (2 \ln \left (3\right )-1\right ) \ln \left ({\mathrm e}^{x} x +x^{2}+3\right )\right )}^{2}+{\mathrm e}^{x}\) | \(51\) |
int((((8*exp(x)*x+8*x^2+24)*ln(exp(x)*x+x^2+3)+(32*x+32)*exp(x)+64*x)*ln(( 2*ln(3)-1)*ln(exp(x)*x+x^2+3))+(x*exp(x)^2+(3*x^2+3)*exp(x)+2*x^3+6*x)*ln( exp(x)*x+x^2+3)+(8*x^2+8*x)*exp(x)+16*x^2)/(exp(x)*x+x^2+3)/ln(exp(x)*x+x^ 2+3),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.88 \[ \int \frac {16 x^2+e^x \left (8 x+8 x^2\right )+\left (6 x+e^{2 x} x+2 x^3+e^x \left (3+3 x^2\right )\right ) \log \left (3+e^x x+x^2\right )+\left (64 x+e^x (32+32 x)+\left (24+8 e^x x+8 x^2\right ) \log \left (3+e^x x+x^2\right )\right ) \log \left ((-1+\log (9)) \log \left (3+e^x x+x^2\right )\right )}{\left (3+e^x x+x^2\right ) \log \left (3+e^x x+x^2\right )} \, dx=x^{2} + 8 \, x \log \left ({\left (2 \, \log \left (3\right ) - 1\right )} \log \left (x^{2} + x e^{x} + 3\right )\right ) + 16 \, \log \left ({\left (2 \, \log \left (3\right ) - 1\right )} \log \left (x^{2} + x e^{x} + 3\right )\right )^{2} + e^{x} \]
integrate((((8*exp(x)*x+8*x^2+24)*log(exp(x)*x+x^2+3)+(32*x+32)*exp(x)+64* x)*log((2*log(3)-1)*log(exp(x)*x+x^2+3))+(x*exp(x)^2+(3*x^2+3)*exp(x)+2*x^ 3+6*x)*log(exp(x)*x+x^2+3)+(8*x^2+8*x)*exp(x)+16*x^2)/(exp(x)*x+x^2+3)/log (exp(x)*x+x^2+3),x, algorithm=\
x^2 + 8*x*log((2*log(3) - 1)*log(x^2 + x*e^x + 3)) + 16*log((2*log(3) - 1) *log(x^2 + x*e^x + 3))^2 + e^x
Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (26) = 52\).
Time = 0.67 (sec) , antiderivative size = 53, normalized size of antiderivative = 2.04 \[ \int \frac {16 x^2+e^x \left (8 x+8 x^2\right )+\left (6 x+e^{2 x} x+2 x^3+e^x \left (3+3 x^2\right )\right ) \log \left (3+e^x x+x^2\right )+\left (64 x+e^x (32+32 x)+\left (24+8 e^x x+8 x^2\right ) \log \left (3+e^x x+x^2\right )\right ) \log \left ((-1+\log (9)) \log \left (3+e^x x+x^2\right )\right )}{\left (3+e^x x+x^2\right ) \log \left (3+e^x x+x^2\right )} \, dx=x^{2} + 8 x \log {\left (\left (-1 + 2 \log {\left (3 \right )}\right ) \log {\left (x^{2} + x e^{x} + 3 \right )} \right )} + e^{x} + 16 \log {\left (\left (-1 + 2 \log {\left (3 \right )}\right ) \log {\left (x^{2} + x e^{x} + 3 \right )} \right )}^{2} \]
integrate((((8*exp(x)*x+8*x**2+24)*ln(exp(x)*x+x**2+3)+(32*x+32)*exp(x)+64 *x)*ln((2*ln(3)-1)*ln(exp(x)*x+x**2+3))+(x*exp(x)**2+(3*x**2+3)*exp(x)+2*x **3+6*x)*ln(exp(x)*x+x**2+3)+(8*x**2+8*x)*exp(x)+16*x**2)/(exp(x)*x+x**2+3 )/ln(exp(x)*x+x**2+3),x)
x**2 + 8*x*log((-1 + 2*log(3))*log(x**2 + x*exp(x) + 3)) + exp(x) + 16*log ((-1 + 2*log(3))*log(x**2 + x*exp(x) + 3))**2
Leaf count of result is larger than twice the leaf count of optimal. 55 vs. \(2 (26) = 52\).
Time = 0.33 (sec) , antiderivative size = 55, normalized size of antiderivative = 2.12 \[ \int \frac {16 x^2+e^x \left (8 x+8 x^2\right )+\left (6 x+e^{2 x} x+2 x^3+e^x \left (3+3 x^2\right )\right ) \log \left (3+e^x x+x^2\right )+\left (64 x+e^x (32+32 x)+\left (24+8 e^x x+8 x^2\right ) \log \left (3+e^x x+x^2\right )\right ) \log \left ((-1+\log (9)) \log \left (3+e^x x+x^2\right )\right )}{\left (3+e^x x+x^2\right ) \log \left (3+e^x x+x^2\right )} \, dx=x^{2} + 8 \, x \log \left (2 \, \log \left (3\right ) - 1\right ) + 8 \, {\left (x + 4 \, \log \left (2 \, \log \left (3\right ) - 1\right )\right )} \log \left (\log \left (x^{2} + x e^{x} + 3\right )\right ) + 16 \, \log \left (\log \left (x^{2} + x e^{x} + 3\right )\right )^{2} + e^{x} \]
integrate((((8*exp(x)*x+8*x^2+24)*log(exp(x)*x+x^2+3)+(32*x+32)*exp(x)+64* x)*log((2*log(3)-1)*log(exp(x)*x+x^2+3))+(x*exp(x)^2+(3*x^2+3)*exp(x)+2*x^ 3+6*x)*log(exp(x)*x+x^2+3)+(8*x^2+8*x)*exp(x)+16*x^2)/(exp(x)*x+x^2+3)/log (exp(x)*x+x^2+3),x, algorithm=\
x^2 + 8*x*log(2*log(3) - 1) + 8*(x + 4*log(2*log(3) - 1))*log(log(x^2 + x* e^x + 3)) + 16*log(log(x^2 + x*e^x + 3))^2 + e^x
Leaf count of result is larger than twice the leaf count of optimal. 69 vs. \(2 (26) = 52\).
Time = 0.39 (sec) , antiderivative size = 69, normalized size of antiderivative = 2.65 \[ \int \frac {16 x^2+e^x \left (8 x+8 x^2\right )+\left (6 x+e^{2 x} x+2 x^3+e^x \left (3+3 x^2\right )\right ) \log \left (3+e^x x+x^2\right )+\left (64 x+e^x (32+32 x)+\left (24+8 e^x x+8 x^2\right ) \log \left (3+e^x x+x^2\right )\right ) \log \left ((-1+\log (9)) \log \left (3+e^x x+x^2\right )\right )}{\left (3+e^x x+x^2\right ) \log \left (3+e^x x+x^2\right )} \, dx=x^{2} + 8 \, x \log \left (2 \, \log \left (3\right ) \log \left (x^{2} + x e^{x} + 3\right ) - \log \left (x^{2} + x e^{x} + 3\right )\right ) + 16 \, \log \left (2 \, \log \left (3\right ) \log \left (x^{2} + x e^{x} + 3\right ) - \log \left (x^{2} + x e^{x} + 3\right )\right )^{2} + e^{x} \]
integrate((((8*exp(x)*x+8*x^2+24)*log(exp(x)*x+x^2+3)+(32*x+32)*exp(x)+64* x)*log((2*log(3)-1)*log(exp(x)*x+x^2+3))+(x*exp(x)^2+(3*x^2+3)*exp(x)+2*x^ 3+6*x)*log(exp(x)*x+x^2+3)+(8*x^2+8*x)*exp(x)+16*x^2)/(exp(x)*x+x^2+3)/log (exp(x)*x+x^2+3),x, algorithm=\
x^2 + 8*x*log(2*log(3)*log(x^2 + x*e^x + 3) - log(x^2 + x*e^x + 3)) + 16*l og(2*log(3)*log(x^2 + x*e^x + 3) - log(x^2 + x*e^x + 3))^2 + e^x
Time = 13.39 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.88 \[ \int \frac {16 x^2+e^x \left (8 x+8 x^2\right )+\left (6 x+e^{2 x} x+2 x^3+e^x \left (3+3 x^2\right )\right ) \log \left (3+e^x x+x^2\right )+\left (64 x+e^x (32+32 x)+\left (24+8 e^x x+8 x^2\right ) \log \left (3+e^x x+x^2\right )\right ) \log \left ((-1+\log (9)) \log \left (3+e^x x+x^2\right )\right )}{\left (3+e^x x+x^2\right ) \log \left (3+e^x x+x^2\right )} \, dx={\mathrm {e}}^x+8\,x\,\ln \left (\ln \left (x\,{\mathrm {e}}^x+x^2+3\right )\,\left (2\,\ln \left (3\right )-1\right )\right )+16\,{\ln \left (\ln \left (x\,{\mathrm {e}}^x+x^2+3\right )\,\left (2\,\ln \left (3\right )-1\right )\right )}^2+x^2 \]
int((log(x*exp(x) + x^2 + 3)*(6*x + x*exp(2*x) + exp(x)*(3*x^2 + 3) + 2*x^ 3) + log(log(x*exp(x) + x^2 + 3)*(2*log(3) - 1))*(64*x + log(x*exp(x) + x^ 2 + 3)*(8*x*exp(x) + 8*x^2 + 24) + exp(x)*(32*x + 32)) + exp(x)*(8*x + 8*x ^2) + 16*x^2)/(log(x*exp(x) + x^2 + 3)*(x*exp(x) + x^2 + 3)),x)