Integrand size = 109, antiderivative size = 32 \[ \int \frac {-e^{3 x}+e^{2 x} (4-2 x)+4 x^2+e^x \left (8 x-x^2\right )+\left (4 e^x x \log (5)+4 x^2 \log (5)\right ) \log \left (3 x^2\right )+\left (x^2 \log (5)+e^x \left (2 x-x^2\right ) \log (5)\right ) \log ^2\left (3 x^2\right )}{e^{2 x}+2 e^x x+x^2} \, dx=-e^x-x+x \left (5+\frac {x \log (5) \log ^2\left (3 x^2\right )}{e^x+x}\right ) \]
Time = 5.50 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.94 \[ \int \frac {-e^{3 x}+e^{2 x} (4-2 x)+4 x^2+e^x \left (8 x-x^2\right )+\left (4 e^x x \log (5)+4 x^2 \log (5)\right ) \log \left (3 x^2\right )+\left (x^2 \log (5)+e^x \left (2 x-x^2\right ) \log (5)\right ) \log ^2\left (3 x^2\right )}{e^{2 x}+2 e^x x+x^2} \, dx=-e^x+4 x+\frac {x^2 \log (5) \log ^2\left (3 x^2\right )}{e^x+x} \]
Integrate[(-E^(3*x) + E^(2*x)*(4 - 2*x) + 4*x^2 + E^x*(8*x - x^2) + (4*E^x *x*Log[5] + 4*x^2*Log[5])*Log[3*x^2] + (x^2*Log[5] + E^x*(2*x - x^2)*Log[5 ])*Log[3*x^2]^2)/(E^(2*x) + 2*E^x*x + 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.
\(\displaystyle \int \frac {4 x^2+e^x \left (8 x-x^2\right )+\left (x^2 \log (5)+e^x \left (2 x-x^2\right ) \log (5)\right ) \log ^2\left (3 x^2\right )+\left (4 x^2 \log (5)+4 e^x x \log (5)\right ) \log \left (3 x^2\right )-e^{3 x}+e^{2 x} (4-2 x)}{x^2+2 e^x x+e^{2 x}} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \left (-\frac {\left (e^x (x-2)-x\right ) x \log (5) \log ^2\left (3 x^2\right )}{\left (x+e^x\right )^2}+\frac {4 x \log (5) \log \left (3 x^2\right )}{x+e^x}-e^x+4\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\log (5) \int \frac {x^2 \log ^2\left (3 x^2\right )}{\left (x+e^x\right )^2}dx+2 \log (5) \int \frac {x \log ^2\left (3 x^2\right )}{x+e^x}dx-\log (5) \int \frac {x^2 \log ^2\left (3 x^2\right )}{x+e^x}dx+4 \log (5) \log \left (3 x^2\right ) \int \frac {x}{x+e^x}dx+\log (5) \int \frac {x^3 \log ^2\left (3 x^2\right )}{\left (x+e^x\right )^2}dx-8 \log (5) \int \frac {\int \frac {x}{x+e^x}dx}{x}dx+4 x-e^x\) |
Int[(-E^(3*x) + E^(2*x)*(4 - 2*x) + 4*x^2 + E^x*(8*x - x^2) + (4*E^x*x*Log [5] + 4*x^2*Log[5])*Log[3*x^2] + (x^2*Log[5] + E^x*(2*x - x^2)*Log[5])*Log [3*x^2]^2)/(E^(2*x) + 2*E^x*x + x^2),x]
3.30.69.3.1 Defintions of rubi rules used
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 0.68 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.28
method | result | size |
parallelrisch | \(\frac {2 \ln \left (5\right ) \ln \left (3 x^{2}\right )^{2} x^{2}+8 x^{2}+6 \,{\mathrm e}^{x} x -2 \,{\mathrm e}^{2 x}}{2 \,{\mathrm e}^{x}+2 x}\) | \(41\) |
risch | \(\frac {4 x^{2} \ln \left (5\right ) \ln \left (x \right )^{2}}{{\mathrm e}^{x}+x}+\frac {2 \left (-i \pi \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x \right )^{2}+2 i \pi \operatorname {csgn}\left (i x^{2}\right )^{2} \operatorname {csgn}\left (i x \right )-i \pi \operatorname {csgn}\left (i x^{2}\right )^{3}+2 \ln \left (3\right )\right ) x^{2} \ln \left (5\right ) \ln \left (x \right )}{{\mathrm e}^{x}+x}-\frac {-12 \,{\mathrm e}^{x} x -16 x^{2}+4 \,{\mathrm e}^{2 x}+\pi ^{2} \ln \left (5\right ) x^{2} \operatorname {csgn}\left (i x^{2}\right )^{6}+\pi ^{2} \ln \left (5\right ) x^{2} \operatorname {csgn}\left (i x \right )^{4} \operatorname {csgn}\left (i x^{2}\right )^{2}-4 \pi ^{2} \ln \left (5\right ) x^{2} \operatorname {csgn}\left (i x \right )^{3} \operatorname {csgn}\left (i x^{2}\right )^{3}+6 \pi ^{2} \ln \left (5\right ) x^{2} \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )^{4}-4 \pi ^{2} \ln \left (5\right ) x^{2} \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{5}-4 \ln \left (3\right )^{2} \ln \left (5\right ) x^{2}+4 i \pi \ln \left (3\right ) \ln \left (5\right ) x^{2} \operatorname {csgn}\left (i x^{2}\right )^{3}-8 i \pi \ln \left (3\right ) \ln \left (5\right ) x^{2} \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}+4 i \pi \ln \left (3\right ) \ln \left (5\right ) x^{2} \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )}{4 \left ({\mathrm e}^{x}+x \right )}\) | \(313\) |
int((((-x^2+2*x)*ln(5)*exp(x)+x^2*ln(5))*ln(3*x^2)^2+(4*x*exp(x)*ln(5)+4*x ^2*ln(5))*ln(3*x^2)-exp(x)^3+(4-2*x)*exp(x)^2+(-x^2+8*x)*exp(x)+4*x^2)/(ex p(x)^2+2*exp(x)*x+x^2),x,method=_RETURNVERBOSE)
Time = 0.24 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.19 \[ \int \frac {-e^{3 x}+e^{2 x} (4-2 x)+4 x^2+e^x \left (8 x-x^2\right )+\left (4 e^x x \log (5)+4 x^2 \log (5)\right ) \log \left (3 x^2\right )+\left (x^2 \log (5)+e^x \left (2 x-x^2\right ) \log (5)\right ) \log ^2\left (3 x^2\right )}{e^{2 x}+2 e^x x+x^2} \, dx=\frac {x^{2} \log \left (5\right ) \log \left (3 \, x^{2}\right )^{2} + 4 \, x^{2} + 3 \, x e^{x} - e^{\left (2 \, x\right )}}{x + e^{x}} \]
integrate((((-x^2+2*x)*log(5)*exp(x)+x^2*log(5))*log(3*x^2)^2+(4*x*exp(x)* log(5)+4*x^2*log(5))*log(3*x^2)-exp(x)^3+(4-2*x)*exp(x)^2+(-x^2+8*x)*exp(x )+4*x^2)/(exp(x)^2+2*exp(x)*x+x^2),x, algorithm=\
Time = 0.10 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81 \[ \int \frac {-e^{3 x}+e^{2 x} (4-2 x)+4 x^2+e^x \left (8 x-x^2\right )+\left (4 e^x x \log (5)+4 x^2 \log (5)\right ) \log \left (3 x^2\right )+\left (x^2 \log (5)+e^x \left (2 x-x^2\right ) \log (5)\right ) \log ^2\left (3 x^2\right )}{e^{2 x}+2 e^x x+x^2} \, dx=\frac {x^{2} \log {\left (5 \right )} \log {\left (3 x^{2} \right )}^{2}}{x + e^{x}} + 4 x - e^{x} \]
integrate((((-x**2+2*x)*ln(5)*exp(x)+x**2*ln(5))*ln(3*x**2)**2+(4*x*exp(x) *ln(5)+4*x**2*ln(5))*ln(3*x**2)-exp(x)**3+(4-2*x)*exp(x)**2+(-x**2+8*x)*ex p(x)+4*x**2)/(exp(x)**2+2*exp(x)*x+x**2),x)
Time = 0.32 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.69 \[ \int \frac {-e^{3 x}+e^{2 x} (4-2 x)+4 x^2+e^x \left (8 x-x^2\right )+\left (4 e^x x \log (5)+4 x^2 \log (5)\right ) \log \left (3 x^2\right )+\left (x^2 \log (5)+e^x \left (2 x-x^2\right ) \log (5)\right ) \log ^2\left (3 x^2\right )}{e^{2 x}+2 e^x x+x^2} \, dx=\frac {4 \, x^{2} \log \left (5\right ) \log \left (3\right ) \log \left (x\right ) + 4 \, x^{2} \log \left (5\right ) \log \left (x\right )^{2} + {\left (\log \left (5\right ) \log \left (3\right )^{2} + 4\right )} x^{2} + 3 \, x e^{x} - e^{\left (2 \, x\right )}}{x + e^{x}} \]
integrate((((-x^2+2*x)*log(5)*exp(x)+x^2*log(5))*log(3*x^2)^2+(4*x*exp(x)* log(5)+4*x^2*log(5))*log(3*x^2)-exp(x)^3+(4-2*x)*exp(x)^2+(-x^2+8*x)*exp(x )+4*x^2)/(exp(x)^2+2*exp(x)*x+x^2),x, algorithm=\
(4*x^2*log(5)*log(3)*log(x) + 4*x^2*log(5)*log(x)^2 + (log(5)*log(3)^2 + 4 )*x^2 + 3*x*e^x - e^(2*x))/(x + e^x)
Time = 0.29 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.19 \[ \int \frac {-e^{3 x}+e^{2 x} (4-2 x)+4 x^2+e^x \left (8 x-x^2\right )+\left (4 e^x x \log (5)+4 x^2 \log (5)\right ) \log \left (3 x^2\right )+\left (x^2 \log (5)+e^x \left (2 x-x^2\right ) \log (5)\right ) \log ^2\left (3 x^2\right )}{e^{2 x}+2 e^x x+x^2} \, dx=\frac {x^{2} \log \left (5\right ) \log \left (3 \, x^{2}\right )^{2} + 4 \, x^{2} + 3 \, x e^{x} - e^{\left (2 \, x\right )}}{x + e^{x}} \]
integrate((((-x^2+2*x)*log(5)*exp(x)+x^2*log(5))*log(3*x^2)^2+(4*x*exp(x)* log(5)+4*x^2*log(5))*log(3*x^2)-exp(x)^3+(4-2*x)*exp(x)^2+(-x^2+8*x)*exp(x )+4*x^2)/(exp(x)^2+2*exp(x)*x+x^2),x, algorithm=\
Timed out. \[ \int \frac {-e^{3 x}+e^{2 x} (4-2 x)+4 x^2+e^x \left (8 x-x^2\right )+\left (4 e^x x \log (5)+4 x^2 \log (5)\right ) \log \left (3 x^2\right )+\left (x^2 \log (5)+e^x \left (2 x-x^2\right ) \log (5)\right ) \log ^2\left (3 x^2\right )}{e^{2 x}+2 e^x x+x^2} \, dx=\int \frac {\ln \left (3\,x^2\right )\,\left (4\,x^2\,\ln \left (5\right )+4\,x\,{\mathrm {e}}^x\,\ln \left (5\right )\right )-{\mathrm {e}}^{3\,x}+{\ln \left (3\,x^2\right )}^2\,\left (x^2\,\ln \left (5\right )+{\mathrm {e}}^x\,\ln \left (5\right )\,\left (2\,x-x^2\right )\right )+{\mathrm {e}}^x\,\left (8\,x-x^2\right )-{\mathrm {e}}^{2\,x}\,\left (2\,x-4\right )+4\,x^2}{{\mathrm {e}}^{2\,x}+2\,x\,{\mathrm {e}}^x+x^2} \,d x \]
int((log(3*x^2)*(4*x^2*log(5) + 4*x*exp(x)*log(5)) - exp(3*x) + log(3*x^2) ^2*(x^2*log(5) + exp(x)*log(5)*(2*x - x^2)) + exp(x)*(8*x - x^2) - exp(2*x )*(2*x - 4) + 4*x^2)/(exp(2*x) + 2*x*exp(x) + x^2),x)