Integrand size = 89, antiderivative size = 30 \[ \int \frac {4 x-2 x^2+e^x \left (2 x-4 x^2-2 x^3\right )+e^{2 x} \left (-2 x^2-2 x^3\right )+\left (-2+2 x+e^x \left (2 x+2 x^2\right )\right ) \log (x)-16 x^2 \log \left (x^2\right )-8 x^2 \log ^2\left (x^2\right )}{x} \, dx=2 x-\left (x+e^x x-\log (x)\right )^2-4 x^2 \log ^2\left (x^2\right ) \]
[Out]
Time = 0.11 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.83, number of steps used = 20, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.112, Rules used = {14, 2227, 2207, 2225, 2326, 2388, 2338, 2332, 2341, 2342} \[ \int \frac {4 x-2 x^2+e^x \left (2 x-4 x^2-2 x^3\right )+e^{2 x} \left (-2 x^2-2 x^3\right )+\left (-2+2 x+e^x \left (2 x+2 x^2\right )\right ) \log (x)-16 x^2 \log \left (x^2\right )-8 x^2 \log ^2\left (x^2\right )}{x} \, dx=-e^{2 x} x^2-x^2-4 x^2 \log ^2\left (x^2\right )-2 e^x \left (x^2-x \log (x)\right )+2 x-\log ^2(x)+2 x \log (x) \]
[In]
[Out]
Rule 14
Rule 2207
Rule 2225
Rule 2227
Rule 2326
Rule 2332
Rule 2338
Rule 2341
Rule 2342
Rule 2388
Rubi steps \begin{align*} \text {integral}& = \int \left (-2 e^{2 x} x (1+x)-2 e^x \left (-1+2 x+x^2-\log (x)-x \log (x)\right )-\frac {2 \left (-2 x+x^2+\log (x)-x \log (x)+8 x^2 \log \left (x^2\right )+4 x^2 \log ^2\left (x^2\right )\right )}{x}\right ) \, dx \\ & = -\left (2 \int e^{2 x} x (1+x) \, dx\right )-2 \int e^x \left (-1+2 x+x^2-\log (x)-x \log (x)\right ) \, dx-2 \int \frac {-2 x+x^2+\log (x)-x \log (x)+8 x^2 \log \left (x^2\right )+4 x^2 \log ^2\left (x^2\right )}{x} \, dx \\ & = -2 e^x \left (x^2-x \log (x)\right )-2 \int \left (e^{2 x} x+e^{2 x} x^2\right ) \, dx-2 \int \left (\frac {-2 x+x^2+\log (x)-x \log (x)}{x}+8 x \log \left (x^2\right )+4 x \log ^2\left (x^2\right )\right ) \, dx \\ & = -2 e^x \left (x^2-x \log (x)\right )-2 \int e^{2 x} x \, dx-2 \int e^{2 x} x^2 \, dx-2 \int \frac {-2 x+x^2+\log (x)-x \log (x)}{x} \, dx-8 \int x \log ^2\left (x^2\right ) \, dx-16 \int x \log \left (x^2\right ) \, dx \\ & = -e^{2 x} x+8 x^2-e^{2 x} x^2-2 e^x \left (x^2-x \log (x)\right )-8 x^2 \log \left (x^2\right )-4 x^2 \log ^2\left (x^2\right )+2 \int e^{2 x} x \, dx-2 \int \left (-2+x-\frac {(-1+x) \log (x)}{x}\right ) \, dx+16 \int x \log \left (x^2\right ) \, dx+\int e^{2 x} \, dx \\ & = \frac {e^{2 x}}{2}+4 x-x^2-e^{2 x} x^2-2 e^x \left (x^2-x \log (x)\right )-4 x^2 \log ^2\left (x^2\right )+2 \int \frac {(-1+x) \log (x)}{x} \, dx-\int e^{2 x} \, dx \\ & = 4 x-x^2-e^{2 x} x^2-2 e^x \left (x^2-x \log (x)\right )-4 x^2 \log ^2\left (x^2\right )+2 \int \log (x) \, dx-2 \int \frac {\log (x)}{x} \, dx \\ & = 2 x-x^2-e^{2 x} x^2+2 x \log (x)-\log ^2(x)-2 e^x \left (x^2-x \log (x)\right )-4 x^2 \log ^2\left (x^2\right ) \\ \end{align*}
Time = 0.20 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.33 \[ \int \frac {4 x-2 x^2+e^x \left (2 x-4 x^2-2 x^3\right )+e^{2 x} \left (-2 x^2-2 x^3\right )+\left (-2+2 x+e^x \left (2 x+2 x^2\right )\right ) \log (x)-16 x^2 \log \left (x^2\right )-8 x^2 \log ^2\left (x^2\right )}{x} \, dx=2 \left (1+e^x\right ) x \log (x)-\log ^2(x)-x \left (-2+\left (1+e^x\right )^2 x+4 x \log ^2\left (x^2\right )\right ) \]
[In]
[Out]
Time = 0.18 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.83
method | result | size |
default | \(2 x -2 \,{\mathrm e}^{x} x^{2}+2 x \,{\mathrm e}^{x} \ln \left (x \right )-x^{2}-4 x^{2} \ln \left (x^{2}\right )^{2}+2 x \ln \left (x \right )-\ln \left (x \right )^{2}-{\mathrm e}^{2 x} x^{2}\) | \(55\) |
parallelrisch | \(2 x -2 \,{\mathrm e}^{x} x^{2}+2 x \,{\mathrm e}^{x} \ln \left (x \right )-x^{2}-4 x^{2} \ln \left (x^{2}\right )^{2}+2 x \ln \left (x \right )-\ln \left (x \right )^{2}-{\mathrm e}^{2 x} x^{2}\) | \(55\) |
parts | \(2 x -2 \,{\mathrm e}^{x} x^{2}+2 x \,{\mathrm e}^{x} \ln \left (x \right )-x^{2}-4 x^{2} \ln \left (x^{2}\right )^{2}+2 x \ln \left (x \right )-\ln \left (x \right )^{2}-{\mathrm e}^{2 x} x^{2}\) | \(55\) |
risch | \(\left (-16 x^{2}-1\right ) \ln \left (x \right )^{2}+\left (8 i x^{2} \pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )-16 i x^{2} \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}+8 i x^{2} \pi \operatorname {csgn}\left (i x^{2}\right )^{3}+2 \,{\mathrm e}^{x} x +2 x \right ) \ln \left (x \right )+x^{2} \pi ^{2} \operatorname {csgn}\left (i x \right )^{4} \operatorname {csgn}\left (i x^{2}\right )^{2}-4 x^{2} \pi ^{2} \operatorname {csgn}\left (i x \right )^{3} \operatorname {csgn}\left (i x^{2}\right )^{3}+6 x^{2} \pi ^{2} \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )^{4}-4 x^{2} \pi ^{2} \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{5}+x^{2} \pi ^{2} \operatorname {csgn}\left (i x^{2}\right )^{6}-{\mathrm e}^{2 x} x^{2}-2 \,{\mathrm e}^{x} x^{2}-x^{2}+2 x\) | \(217\) |
[In]
[Out]
none
Time = 0.26 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.60 \[ \int \frac {4 x-2 x^2+e^x \left (2 x-4 x^2-2 x^3\right )+e^{2 x} \left (-2 x^2-2 x^3\right )+\left (-2+2 x+e^x \left (2 x+2 x^2\right )\right ) \log (x)-16 x^2 \log \left (x^2\right )-8 x^2 \log ^2\left (x^2\right )}{x} \, dx=-x^{2} e^{\left (2 \, x\right )} - 2 \, x^{2} e^{x} - {\left (16 \, x^{2} + 1\right )} \log \left (x\right )^{2} - x^{2} + 2 \, {\left (x e^{x} + x\right )} \log \left (x\right ) + 2 \, x \]
[In]
[Out]
Time = 0.16 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.63 \[ \int \frac {4 x-2 x^2+e^x \left (2 x-4 x^2-2 x^3\right )+e^{2 x} \left (-2 x^2-2 x^3\right )+\left (-2+2 x+e^x \left (2 x+2 x^2\right )\right ) \log (x)-16 x^2 \log \left (x^2\right )-8 x^2 \log ^2\left (x^2\right )}{x} \, dx=- x^{2} e^{2 x} - x^{2} + 2 x \log {\left (x \right )} + 2 x + \left (- 16 x^{2} - 1\right ) \log {\left (x \right )}^{2} + \left (- 2 x^{2} + 2 x \log {\left (x \right )}\right ) e^{x} \]
[In]
[Out]
\[ \int \frac {4 x-2 x^2+e^x \left (2 x-4 x^2-2 x^3\right )+e^{2 x} \left (-2 x^2-2 x^3\right )+\left (-2+2 x+e^x \left (2 x+2 x^2\right )\right ) \log (x)-16 x^2 \log \left (x^2\right )-8 x^2 \log ^2\left (x^2\right )}{x} \, dx=\int { -\frac {2 \, {\left (4 \, x^{2} \log \left (x^{2}\right )^{2} + 8 \, x^{2} \log \left (x^{2}\right ) + x^{2} + {\left (x^{3} + x^{2}\right )} e^{\left (2 \, x\right )} + {\left (x^{3} + 2 \, x^{2} - x\right )} e^{x} - {\left ({\left (x^{2} + x\right )} e^{x} + x - 1\right )} \log \left (x\right ) - 2 \, x\right )}}{x} \,d x } \]
[In]
[Out]
none
Time = 0.27 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.73 \[ \int \frac {4 x-2 x^2+e^x \left (2 x-4 x^2-2 x^3\right )+e^{2 x} \left (-2 x^2-2 x^3\right )+\left (-2+2 x+e^x \left (2 x+2 x^2\right )\right ) \log (x)-16 x^2 \log \left (x^2\right )-8 x^2 \log ^2\left (x^2\right )}{x} \, dx=-16 \, x^{2} \log \left (x\right )^{2} - x^{2} e^{\left (2 \, x\right )} - 2 \, x^{2} e^{x} + 2 \, x e^{x} \log \left (x\right ) - x^{2} + 2 \, x \log \left (x\right ) - \log \left (x\right )^{2} + 2 \, x \]
[In]
[Out]
Time = 12.74 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.80 \[ \int \frac {4 x-2 x^2+e^x \left (2 x-4 x^2-2 x^3\right )+e^{2 x} \left (-2 x^2-2 x^3\right )+\left (-2+2 x+e^x \left (2 x+2 x^2\right )\right ) \log (x)-16 x^2 \log \left (x^2\right )-8 x^2 \log ^2\left (x^2\right )}{x} \, dx=2\,x-2\,x^2\,{\mathrm {e}}^x-{\ln \left (x\right )}^2-x^2\,{\mathrm {e}}^{2\,x}+2\,x\,\ln \left (x\right )-x^2-4\,x^2\,{\ln \left (x^2\right )}^2+2\,x\,{\mathrm {e}}^x\,\ln \left (x\right ) \]
[In]
[Out]