Integrand size = 151, antiderivative size = 22 \begin {dmath*} \int \frac {-4 x^2-4 e^{x^2} x^3+\left (8 x^2+e^{x^2} \left (2 x+4 x^3\right )+2 x \log (8)\right ) \log \left (e^{x^2}+2 x+\log (8)\right )+\left (-4 e^{x^2} x-8 x^2-4 x \log (8)\right ) \log ^2\left (e^{x^2}+2 x+\log (8)\right )+\left (2 e^{x^2} x+4 x^2+2 x \log (8)\right ) \log ^3\left (e^{x^2}+2 x+\log (8)\right )}{\left (e^{x^2}+2 x+\log (8)\right ) \log ^3\left (e^{x^2}+2 x+\log (8)\right )} \, dx=\left (-x+\frac {x}{\log \left (e^{x^2}+2 x+\log (8)\right )}\right )^2 \end {dmath*}
Leaf count is larger than twice the leaf count of optimal. \(50\) vs. \(2(22)=44\).
Time = 0.06 (sec) , antiderivative size = 50, normalized size of antiderivative = 2.27 \begin {dmath*} \int \frac {-4 x^2-4 e^{x^2} x^3+\left (8 x^2+e^{x^2} \left (2 x+4 x^3\right )+2 x \log (8)\right ) \log \left (e^{x^2}+2 x+\log (8)\right )+\left (-4 e^{x^2} x-8 x^2-4 x \log (8)\right ) \log ^2\left (e^{x^2}+2 x+\log (8)\right )+\left (2 e^{x^2} x+4 x^2+2 x \log (8)\right ) \log ^3\left (e^{x^2}+2 x+\log (8)\right )}{\left (e^{x^2}+2 x+\log (8)\right ) \log ^3\left (e^{x^2}+2 x+\log (8)\right )} \, dx=2 \left (\frac {x^2}{2}+\frac {x^2}{2 \log ^2\left (e^{x^2}+2 x+\log (8)\right )}-\frac {x^2}{\log \left (e^{x^2}+2 x+\log (8)\right )}\right ) \end {dmath*}
Integrate[(-4*x^2 - 4*E^x^2*x^3 + (8*x^2 + E^x^2*(2*x + 4*x^3) + 2*x*Log[8 ])*Log[E^x^2 + 2*x + Log[8]] + (-4*E^x^2*x - 8*x^2 - 4*x*Log[8])*Log[E^x^2 + 2*x + Log[8]]^2 + (2*E^x^2*x + 4*x^2 + 2*x*Log[8])*Log[E^x^2 + 2*x + Lo g[8]]^3)/((E^x^2 + 2*x + Log[8])*Log[E^x^2 + 2*x + Log[8]]^3),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.
\(\displaystyle \int \frac {-4 x^2+\left (4 x^2+2 e^{x^2} x+2 x \log (8)\right ) \log ^3\left (e^{x^2}+2 x+\log (8)\right )+\left (-8 x^2-4 e^{x^2} x-4 x \log (8)\right ) \log ^2\left (e^{x^2}+2 x+\log (8)\right )-4 e^{x^2} x^3+\left (8 x^2+e^{x^2} \left (4 x^3+2 x\right )+2 x \log (8)\right ) \log \left (e^{x^2}+2 x+\log (8)\right )}{\left (e^{x^2}+2 x+\log (8)\right ) \log ^3\left (e^{x^2}+2 x+\log (8)\right )} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {2 x \left (\log \left (e^{x^2}+2 x+\log (8)\right )-1\right ) \left (2 x^2+\log ^2\left (e^{x^2}+2 x+\log (8)\right )-\log \left (e^{x^2}+2 x+\log (8)\right )\right )}{\log ^3\left (e^{x^2}+2 x+\log (8)\right )}-\frac {4 x^2 \left (2 x^2+x \log (8)-1\right ) \left (\log \left (e^{x^2}+2 x+\log (8)\right )-1\right )}{\left (e^{x^2}+2 x+\log (8)\right ) \log ^3\left (e^{x^2}+2 x+\log (8)\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -4 \int \frac {x^2}{\left (2 x+e^{x^2}+\log (8)\right ) \log ^3\left (2 x+e^{x^2}+\log (8)\right )}dx+2 \int \frac {x}{\log ^2\left (2 x+e^{x^2}+\log (8)\right )}dx+4 \int \frac {x^2}{\left (2 x+e^{x^2}+\log (8)\right ) \log ^2\left (2 x+e^{x^2}+\log (8)\right )}dx-4 \int \frac {x}{\log \left (2 x+e^{x^2}+\log (8)\right )}dx+8 \int \frac {x^4}{\left (2 x+e^{x^2}+\log (8)\right ) \log ^3\left (2 x+e^{x^2}+\log (8)\right )}dx-8 \int \frac {x^4}{\left (2 x+e^{x^2}+\log (8)\right ) \log ^2\left (2 x+e^{x^2}+\log (8)\right )}dx-4 \int \frac {x^3}{\log ^3\left (2 x+e^{x^2}+\log (8)\right )}dx+4 \log (8) \int \frac {x^3}{\left (2 x+e^{x^2}+\log (8)\right ) \log ^3\left (2 x+e^{x^2}+\log (8)\right )}dx+4 \int \frac {x^3}{\log ^2\left (2 x+e^{x^2}+\log (8)\right )}dx-4 \log (8) \int \frac {x^3}{\left (2 x+e^{x^2}+\log (8)\right ) \log ^2\left (2 x+e^{x^2}+\log (8)\right )}dx+x^2\) |
Int[(-4*x^2 - 4*E^x^2*x^3 + (8*x^2 + E^x^2*(2*x + 4*x^3) + 2*x*Log[8])*Log [E^x^2 + 2*x + Log[8]] + (-4*E^x^2*x - 8*x^2 - 4*x*Log[8])*Log[E^x^2 + 2*x + Log[8]]^2 + (2*E^x^2*x + 4*x^2 + 2*x*Log[8])*Log[E^x^2 + 2*x + Log[8]]^ 3)/((E^x^2 + 2*x + Log[8])*Log[E^x^2 + 2*x + Log[8]]^3),x]
3.12.72.3.1 Defintions of rubi rules used
Time = 0.95 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.91
method | result | size |
risch | \(x^{2}-\frac {\left (2 \ln \left ({\mathrm e}^{x^{2}}+3 \ln \left (2\right )+2 x \right )-1\right ) x^{2}}{\ln \left ({\mathrm e}^{x^{2}}+3 \ln \left (2\right )+2 x \right )^{2}}\) | \(42\) |
parallelrisch | \(\frac {\ln \left ({\mathrm e}^{x^{2}}+3 \ln \left (2\right )+2 x \right )^{2} x^{2}-2 \ln \left ({\mathrm e}^{x^{2}}+3 \ln \left (2\right )+2 x \right ) x^{2}+x^{2}}{\ln \left ({\mathrm e}^{x^{2}}+3 \ln \left (2\right )+2 x \right )^{2}}\) | \(58\) |
int(((2*exp(x^2)*x+6*x*ln(2)+4*x^2)*ln(exp(x^2)+3*ln(2)+2*x)^3+(-4*exp(x^2 )*x-12*x*ln(2)-8*x^2)*ln(exp(x^2)+3*ln(2)+2*x)^2+((4*x^3+2*x)*exp(x^2)+6*x *ln(2)+8*x^2)*ln(exp(x^2)+3*ln(2)+2*x)-4*x^3*exp(x^2)-4*x^2)/(exp(x^2)+3*l n(2)+2*x)/ln(exp(x^2)+3*ln(2)+2*x)^3,x,method=_RETURNVERBOSE)
Leaf count of result is larger than twice the leaf count of optimal. 57 vs. \(2 (22) = 44\).
Time = 0.26 (sec) , antiderivative size = 57, normalized size of antiderivative = 2.59 \begin {dmath*} \int \frac {-4 x^2-4 e^{x^2} x^3+\left (8 x^2+e^{x^2} \left (2 x+4 x^3\right )+2 x \log (8)\right ) \log \left (e^{x^2}+2 x+\log (8)\right )+\left (-4 e^{x^2} x-8 x^2-4 x \log (8)\right ) \log ^2\left (e^{x^2}+2 x+\log (8)\right )+\left (2 e^{x^2} x+4 x^2+2 x \log (8)\right ) \log ^3\left (e^{x^2}+2 x+\log (8)\right )}{\left (e^{x^2}+2 x+\log (8)\right ) \log ^3\left (e^{x^2}+2 x+\log (8)\right )} \, dx=\frac {x^{2} \log \left (2 \, x + e^{\left (x^{2}\right )} + 3 \, \log \left (2\right )\right )^{2} - 2 \, x^{2} \log \left (2 \, x + e^{\left (x^{2}\right )} + 3 \, \log \left (2\right )\right ) + x^{2}}{\log \left (2 \, x + e^{\left (x^{2}\right )} + 3 \, \log \left (2\right )\right )^{2}} \end {dmath*}
integrate(((2*exp(x^2)*x+6*x*log(2)+4*x^2)*log(exp(x^2)+3*log(2)+2*x)^3+(- 4*exp(x^2)*x-12*x*log(2)-8*x^2)*log(exp(x^2)+3*log(2)+2*x)^2+((4*x^3+2*x)* exp(x^2)+6*x*log(2)+8*x^2)*log(exp(x^2)+3*log(2)+2*x)-4*x^3*exp(x^2)-4*x^2 )/(exp(x^2)+3*log(2)+2*x)/log(exp(x^2)+3*log(2)+2*x)^3,x, algorithm=\
(x^2*log(2*x + e^(x^2) + 3*log(2))^2 - 2*x^2*log(2*x + e^(x^2) + 3*log(2)) + x^2)/log(2*x + e^(x^2) + 3*log(2))^2
Leaf count of result is larger than twice the leaf count of optimal. 42 vs. \(2 (19) = 38\).
Time = 0.16 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.91 \begin {dmath*} \int \frac {-4 x^2-4 e^{x^2} x^3+\left (8 x^2+e^{x^2} \left (2 x+4 x^3\right )+2 x \log (8)\right ) \log \left (e^{x^2}+2 x+\log (8)\right )+\left (-4 e^{x^2} x-8 x^2-4 x \log (8)\right ) \log ^2\left (e^{x^2}+2 x+\log (8)\right )+\left (2 e^{x^2} x+4 x^2+2 x \log (8)\right ) \log ^3\left (e^{x^2}+2 x+\log (8)\right )}{\left (e^{x^2}+2 x+\log (8)\right ) \log ^3\left (e^{x^2}+2 x+\log (8)\right )} \, dx=x^{2} + \frac {- 2 x^{2} \log {\left (2 x + e^{x^{2}} + 3 \log {\left (2 \right )} \right )} + x^{2}}{\log {\left (2 x + e^{x^{2}} + 3 \log {\left (2 \right )} \right )}^{2}} \end {dmath*}
integrate(((2*exp(x**2)*x+6*x*ln(2)+4*x**2)*ln(exp(x**2)+3*ln(2)+2*x)**3+( -4*exp(x**2)*x-12*x*ln(2)-8*x**2)*ln(exp(x**2)+3*ln(2)+2*x)**2+((4*x**3+2* x)*exp(x**2)+6*x*ln(2)+8*x**2)*ln(exp(x**2)+3*ln(2)+2*x)-4*x**3*exp(x**2)- 4*x**2)/(exp(x**2)+3*ln(2)+2*x)/ln(exp(x**2)+3*ln(2)+2*x)**3,x)
Leaf count of result is larger than twice the leaf count of optimal. 57 vs. \(2 (22) = 44\).
Time = 0.32 (sec) , antiderivative size = 57, normalized size of antiderivative = 2.59 \begin {dmath*} \int \frac {-4 x^2-4 e^{x^2} x^3+\left (8 x^2+e^{x^2} \left (2 x+4 x^3\right )+2 x \log (8)\right ) \log \left (e^{x^2}+2 x+\log (8)\right )+\left (-4 e^{x^2} x-8 x^2-4 x \log (8)\right ) \log ^2\left (e^{x^2}+2 x+\log (8)\right )+\left (2 e^{x^2} x+4 x^2+2 x \log (8)\right ) \log ^3\left (e^{x^2}+2 x+\log (8)\right )}{\left (e^{x^2}+2 x+\log (8)\right ) \log ^3\left (e^{x^2}+2 x+\log (8)\right )} \, dx=\frac {x^{2} \log \left (2 \, x + e^{\left (x^{2}\right )} + 3 \, \log \left (2\right )\right )^{2} - 2 \, x^{2} \log \left (2 \, x + e^{\left (x^{2}\right )} + 3 \, \log \left (2\right )\right ) + x^{2}}{\log \left (2 \, x + e^{\left (x^{2}\right )} + 3 \, \log \left (2\right )\right )^{2}} \end {dmath*}
integrate(((2*exp(x^2)*x+6*x*log(2)+4*x^2)*log(exp(x^2)+3*log(2)+2*x)^3+(- 4*exp(x^2)*x-12*x*log(2)-8*x^2)*log(exp(x^2)+3*log(2)+2*x)^2+((4*x^3+2*x)* exp(x^2)+6*x*log(2)+8*x^2)*log(exp(x^2)+3*log(2)+2*x)-4*x^3*exp(x^2)-4*x^2 )/(exp(x^2)+3*log(2)+2*x)/log(exp(x^2)+3*log(2)+2*x)^3,x, algorithm=\
(x^2*log(2*x + e^(x^2) + 3*log(2))^2 - 2*x^2*log(2*x + e^(x^2) + 3*log(2)) + x^2)/log(2*x + e^(x^2) + 3*log(2))^2
Leaf count of result is larger than twice the leaf count of optimal. 57 vs. \(2 (22) = 44\).
Time = 0.71 (sec) , antiderivative size = 57, normalized size of antiderivative = 2.59 \begin {dmath*} \int \frac {-4 x^2-4 e^{x^2} x^3+\left (8 x^2+e^{x^2} \left (2 x+4 x^3\right )+2 x \log (8)\right ) \log \left (e^{x^2}+2 x+\log (8)\right )+\left (-4 e^{x^2} x-8 x^2-4 x \log (8)\right ) \log ^2\left (e^{x^2}+2 x+\log (8)\right )+\left (2 e^{x^2} x+4 x^2+2 x \log (8)\right ) \log ^3\left (e^{x^2}+2 x+\log (8)\right )}{\left (e^{x^2}+2 x+\log (8)\right ) \log ^3\left (e^{x^2}+2 x+\log (8)\right )} \, dx=\frac {x^{2} \log \left (2 \, x + e^{\left (x^{2}\right )} + 3 \, \log \left (2\right )\right )^{2} - 2 \, x^{2} \log \left (2 \, x + e^{\left (x^{2}\right )} + 3 \, \log \left (2\right )\right ) + x^{2}}{\log \left (2 \, x + e^{\left (x^{2}\right )} + 3 \, \log \left (2\right )\right )^{2}} \end {dmath*}
integrate(((2*exp(x^2)*x+6*x*log(2)+4*x^2)*log(exp(x^2)+3*log(2)+2*x)^3+(- 4*exp(x^2)*x-12*x*log(2)-8*x^2)*log(exp(x^2)+3*log(2)+2*x)^2+((4*x^3+2*x)* exp(x^2)+6*x*log(2)+8*x^2)*log(exp(x^2)+3*log(2)+2*x)-4*x^3*exp(x^2)-4*x^2 )/(exp(x^2)+3*log(2)+2*x)/log(exp(x^2)+3*log(2)+2*x)^3,x, algorithm=\
(x^2*log(2*x + e^(x^2) + 3*log(2))^2 - 2*x^2*log(2*x + e^(x^2) + 3*log(2)) + x^2)/log(2*x + e^(x^2) + 3*log(2))^2
Timed out. \begin {dmath*} \int \frac {-4 x^2-4 e^{x^2} x^3+\left (8 x^2+e^{x^2} \left (2 x+4 x^3\right )+2 x \log (8)\right ) \log \left (e^{x^2}+2 x+\log (8)\right )+\left (-4 e^{x^2} x-8 x^2-4 x \log (8)\right ) \log ^2\left (e^{x^2}+2 x+\log (8)\right )+\left (2 e^{x^2} x+4 x^2+2 x \log (8)\right ) \log ^3\left (e^{x^2}+2 x+\log (8)\right )}{\left (e^{x^2}+2 x+\log (8)\right ) \log ^3\left (e^{x^2}+2 x+\log (8)\right )} \, dx=-\int \frac {{\ln \left (2\,x+{\mathrm {e}}^{x^2}+3\,\ln \left (2\right )\right )}^2\,\left (4\,x\,{\mathrm {e}}^{x^2}+12\,x\,\ln \left (2\right )+8\,x^2\right )-{\ln \left (2\,x+{\mathrm {e}}^{x^2}+3\,\ln \left (2\right )\right )}^3\,\left (2\,x\,{\mathrm {e}}^{x^2}+6\,x\,\ln \left (2\right )+4\,x^2\right )+4\,x^3\,{\mathrm {e}}^{x^2}+4\,x^2-\ln \left (2\,x+{\mathrm {e}}^{x^2}+3\,\ln \left (2\right )\right )\,\left ({\mathrm {e}}^{x^2}\,\left (4\,x^3+2\,x\right )+6\,x\,\ln \left (2\right )+8\,x^2\right )}{{\ln \left (2\,x+{\mathrm {e}}^{x^2}+3\,\ln \left (2\right )\right )}^3\,\left (2\,x+{\mathrm {e}}^{x^2}+3\,\ln \left (2\right )\right )} \,d x \end {dmath*}
int(-(log(2*x + exp(x^2) + 3*log(2))^2*(4*x*exp(x^2) + 12*x*log(2) + 8*x^2 ) - log(2*x + exp(x^2) + 3*log(2))^3*(2*x*exp(x^2) + 6*x*log(2) + 4*x^2) + 4*x^3*exp(x^2) + 4*x^2 - log(2*x + exp(x^2) + 3*log(2))*(exp(x^2)*(2*x + 4*x^3) + 6*x*log(2) + 8*x^2))/(log(2*x + exp(x^2) + 3*log(2))^3*(2*x + exp (x^2) + 3*log(2))),x)
-int((log(2*x + exp(x^2) + 3*log(2))^2*(4*x*exp(x^2) + 12*x*log(2) + 8*x^2 ) - log(2*x + exp(x^2) + 3*log(2))^3*(2*x*exp(x^2) + 6*x*log(2) + 4*x^2) + 4*x^3*exp(x^2) + 4*x^2 - log(2*x + exp(x^2) + 3*log(2))*(exp(x^2)*(2*x + 4*x^3) + 6*x*log(2) + 8*x^2))/(log(2*x + exp(x^2) + 3*log(2))^3*(2*x + exp (x^2) + 3*log(2))), x)