Integrand size = 154, antiderivative size = 24 \[ \int \frac {8 x+4 e^{2 x} x+e^x \left (-4-4 x-4 x^2\right )}{4 e^{3 x} x-4 e^{2 x} x^2+\left (-4 e^{2 x} x+4 e^x x^2\right ) \log \left (x^2\right )+\left (e^x x-x^2\right ) \log ^2\left (x^2\right )+\left (-8 e^{2 x} x+8 e^x x^2+\left (4 e^x x-4 x^2\right ) \log \left (x^2\right )\right ) \log \left (-e^x+x\right )+\left (4 e^x x-4 x^2\right ) \log ^2\left (-e^x+x\right )} \, dx=\frac {1}{-e^x+\frac {\log \left (x^2\right )}{2}+\log \left (-e^x+x\right )} \] Output:
1/(1/2*ln(x^2)+ln(x-exp(x))-exp(x))
Time = 0.05 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {8 x+4 e^{2 x} x+e^x \left (-4-4 x-4 x^2\right )}{4 e^{3 x} x-4 e^{2 x} x^2+\left (-4 e^{2 x} x+4 e^x x^2\right ) \log \left (x^2\right )+\left (e^x x-x^2\right ) \log ^2\left (x^2\right )+\left (-8 e^{2 x} x+8 e^x x^2+\left (4 e^x x-4 x^2\right ) \log \left (x^2\right )\right ) \log \left (-e^x+x\right )+\left (4 e^x x-4 x^2\right ) \log ^2\left (-e^x+x\right )} \, dx=\frac {2}{-2 e^x+\log \left (x^2\right )+2 \log \left (-e^x+x\right )} \] Input:
Integrate[(8*x + 4*E^(2*x)*x + E^x*(-4 - 4*x - 4*x^2))/(4*E^(3*x)*x - 4*E^ (2*x)*x^2 + (-4*E^(2*x)*x + 4*E^x*x^2)*Log[x^2] + (E^x*x - x^2)*Log[x^2]^2 + (-8*E^(2*x)*x + 8*E^x*x^2 + (4*E^x*x - 4*x^2)*Log[x^2])*Log[-E^x + x] + (4*E^x*x - 4*x^2)*Log[-E^x + x]^2),x]
Output:
2/(-2*E^x + Log[x^2] + 2*Log[-E^x + x])
Time = 0.86 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.013, Rules used = {7239, 7237}
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.
\(\displaystyle \int \frac {e^x \left (-4 x^2-4 x-4\right )+4 e^{2 x} x+8 x}{-4 e^{2 x} x^2+\left (e^x x-x^2\right ) \log ^2\left (x^2\right )+\left (4 e^x x-4 x^2\right ) \log ^2\left (x-e^x\right )+\left (4 e^x x^2-4 e^{2 x} x\right ) \log \left (x^2\right )+\left (8 e^x x^2+\left (4 e^x x-4 x^2\right ) \log \left (x^2\right )-8 e^{2 x} x\right ) \log \left (x-e^x\right )+4 e^{3 x} x} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {-4 e^x \left (x^2+x+1\right )+4 e^{2 x} x+8 x}{\left (e^x-x\right ) x \left (-\log \left (x^2\right )+2 e^x-2 \log \left (x-e^x\right )\right )^2}dx\) |
\(\Big \downarrow \) 7237 |
\(\displaystyle -\frac {2}{-\log \left (x^2\right )+2 e^x-2 \log \left (x-e^x\right )}\) |
Input:
Int[(8*x + 4*E^(2*x)*x + E^x*(-4 - 4*x - 4*x^2))/(4*E^(3*x)*x - 4*E^(2*x)* x^2 + (-4*E^(2*x)*x + 4*E^x*x^2)*Log[x^2] + (E^x*x - x^2)*Log[x^2]^2 + (-8 *E^(2*x)*x + 8*E^x*x^2 + (4*E^x*x - 4*x^2)*Log[x^2])*Log[-E^x + x] + (4*E^ x*x - 4*x^2)*Log[-E^x + x]^2),x]
Output:
-2/(2*E^x - Log[x^2] - 2*Log[-E^x + x])
Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Si mp[q*(y^(m + 1)/(m + 1)), x] /; !FalseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 2.30 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04
method | result | size |
parallelrisch | \(-\frac {2}{2 \,{\mathrm e}^{x}-2 \ln \left (x -{\mathrm e}^{x}\right )-\ln \left (x^{2}\right )}\) | \(25\) |
risch | \(\frac {4 i}{\pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )-2 \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}+\pi \operatorname {csgn}\left (i x^{2}\right )^{3}-4 i {\mathrm e}^{x}+4 i \ln \left (x \right )+4 i \ln \left (x -{\mathrm e}^{x}\right )}\) | \(71\) |
Input:
int((4*x*exp(x)^2+(-4*x^2-4*x-4)*exp(x)+8*x)/((4*exp(x)*x-4*x^2)*ln(x-exp( x))^2+((4*exp(x)*x-4*x^2)*ln(x^2)-8*x*exp(x)^2+8*exp(x)*x^2)*ln(x-exp(x))+ (exp(x)*x-x^2)*ln(x^2)^2+(-4*x*exp(x)^2+4*exp(x)*x^2)*ln(x^2)+4*x*exp(x)^3 -4*exp(x)^2*x^2),x,method=_RETURNVERBOSE)
Output:
-2/(2*exp(x)-2*ln(x-exp(x))-ln(x^2))
Time = 0.09 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {8 x+4 e^{2 x} x+e^x \left (-4-4 x-4 x^2\right )}{4 e^{3 x} x-4 e^{2 x} x^2+\left (-4 e^{2 x} x+4 e^x x^2\right ) \log \left (x^2\right )+\left (e^x x-x^2\right ) \log ^2\left (x^2\right )+\left (-8 e^{2 x} x+8 e^x x^2+\left (4 e^x x-4 x^2\right ) \log \left (x^2\right )\right ) \log \left (-e^x+x\right )+\left (4 e^x x-4 x^2\right ) \log ^2\left (-e^x+x\right )} \, dx=-\frac {2}{2 \, e^{x} - \log \left (x^{2}\right ) - 2 \, \log \left (x - e^{x}\right )} \] Input:
integrate((4*x*exp(x)^2+(-4*x^2-4*x-4)*exp(x)+8*x)/((4*exp(x)*x-4*x^2)*log (x-exp(x))^2+((4*exp(x)*x-4*x^2)*log(x^2)-8*x*exp(x)^2+8*exp(x)*x^2)*log(x -exp(x))+(exp(x)*x-x^2)*log(x^2)^2+(-4*x*exp(x)^2+4*exp(x)*x^2)*log(x^2)+4 *x*exp(x)^3-4*exp(x)^2*x^2),x, algorithm="fricas")
Output:
-2/(2*e^x - log(x^2) - 2*log(x - e^x))
Time = 0.20 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79 \[ \int \frac {8 x+4 e^{2 x} x+e^x \left (-4-4 x-4 x^2\right )}{4 e^{3 x} x-4 e^{2 x} x^2+\left (-4 e^{2 x} x+4 e^x x^2\right ) \log \left (x^2\right )+\left (e^x x-x^2\right ) \log ^2\left (x^2\right )+\left (-8 e^{2 x} x+8 e^x x^2+\left (4 e^x x-4 x^2\right ) \log \left (x^2\right )\right ) \log \left (-e^x+x\right )+\left (4 e^x x-4 x^2\right ) \log ^2\left (-e^x+x\right )} \, dx=\frac {2}{- 2 e^{x} + \log {\left (x^{2} \right )} + 2 \log {\left (x - e^{x} \right )}} \] Input:
integrate((4*x*exp(x)**2+(-4*x**2-4*x-4)*exp(x)+8*x)/((4*exp(x)*x-4*x**2)* ln(x-exp(x))**2+((4*exp(x)*x-4*x**2)*ln(x**2)-8*x*exp(x)**2+8*exp(x)*x**2) *ln(x-exp(x))+(exp(x)*x-x**2)*ln(x**2)**2+(-4*x*exp(x)**2+4*exp(x)*x**2)*l n(x**2)+4*x*exp(x)**3-4*exp(x)**2*x**2),x)
Output:
2/(-2*exp(x) + log(x**2) + 2*log(x - exp(x)))
Time = 0.17 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \frac {8 x+4 e^{2 x} x+e^x \left (-4-4 x-4 x^2\right )}{4 e^{3 x} x-4 e^{2 x} x^2+\left (-4 e^{2 x} x+4 e^x x^2\right ) \log \left (x^2\right )+\left (e^x x-x^2\right ) \log ^2\left (x^2\right )+\left (-8 e^{2 x} x+8 e^x x^2+\left (4 e^x x-4 x^2\right ) \log \left (x^2\right )\right ) \log \left (-e^x+x\right )+\left (4 e^x x-4 x^2\right ) \log ^2\left (-e^x+x\right )} \, dx=-\frac {1}{e^{x} - \log \left (x - e^{x}\right ) - \log \left (x\right )} \] Input:
integrate((4*x*exp(x)^2+(-4*x^2-4*x-4)*exp(x)+8*x)/((4*exp(x)*x-4*x^2)*log (x-exp(x))^2+((4*exp(x)*x-4*x^2)*log(x^2)-8*x*exp(x)^2+8*exp(x)*x^2)*log(x -exp(x))+(exp(x)*x-x^2)*log(x^2)^2+(-4*x*exp(x)^2+4*exp(x)*x^2)*log(x^2)+4 *x*exp(x)^3-4*exp(x)^2*x^2),x, algorithm="maxima")
Output:
-1/(e^x - log(x - e^x) - log(x))
Time = 0.24 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {8 x+4 e^{2 x} x+e^x \left (-4-4 x-4 x^2\right )}{4 e^{3 x} x-4 e^{2 x} x^2+\left (-4 e^{2 x} x+4 e^x x^2\right ) \log \left (x^2\right )+\left (e^x x-x^2\right ) \log ^2\left (x^2\right )+\left (-8 e^{2 x} x+8 e^x x^2+\left (4 e^x x-4 x^2\right ) \log \left (x^2\right )\right ) \log \left (-e^x+x\right )+\left (4 e^x x-4 x^2\right ) \log ^2\left (-e^x+x\right )} \, dx=-\frac {2}{2 \, e^{x} - \log \left (x^{2}\right ) - 2 \, \log \left (x - e^{x}\right )} \] Input:
integrate((4*x*exp(x)^2+(-4*x^2-4*x-4)*exp(x)+8*x)/((4*exp(x)*x-4*x^2)*log (x-exp(x))^2+((4*exp(x)*x-4*x^2)*log(x^2)-8*x*exp(x)^2+8*exp(x)*x^2)*log(x -exp(x))+(exp(x)*x-x^2)*log(x^2)^2+(-4*x*exp(x)^2+4*exp(x)*x^2)*log(x^2)+4 *x*exp(x)^3-4*exp(x)^2*x^2),x, algorithm="giac")
Output:
-2/(2*e^x - log(x^2) - 2*log(x - e^x))
Time = 3.51 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {8 x+4 e^{2 x} x+e^x \left (-4-4 x-4 x^2\right )}{4 e^{3 x} x-4 e^{2 x} x^2+\left (-4 e^{2 x} x+4 e^x x^2\right ) \log \left (x^2\right )+\left (e^x x-x^2\right ) \log ^2\left (x^2\right )+\left (-8 e^{2 x} x+8 e^x x^2+\left (4 e^x x-4 x^2\right ) \log \left (x^2\right )\right ) \log \left (-e^x+x\right )+\left (4 e^x x-4 x^2\right ) \log ^2\left (-e^x+x\right )} \, dx=\frac {2}{\ln \left (x^2\right )+2\,\ln \left (x-{\mathrm {e}}^x\right )-2\,{\mathrm {e}}^x} \] Input:
int((8*x + 4*x*exp(2*x) - exp(x)*(4*x + 4*x^2 + 4))/(4*x*exp(3*x) - log(x^ 2)*(4*x*exp(2*x) - 4*x^2*exp(x)) + log(x^2)^2*(x*exp(x) - x^2) + log(x - e xp(x))*(8*x^2*exp(x) - 8*x*exp(2*x) + log(x^2)*(4*x*exp(x) - 4*x^2)) - 4*x ^2*exp(2*x) + log(x - exp(x))^2*(4*x*exp(x) - 4*x^2)),x)
Output:
2/(log(x^2) + 2*log(x - exp(x)) - 2*exp(x))
Time = 0.52 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {8 x+4 e^{2 x} x+e^x \left (-4-4 x-4 x^2\right )}{4 e^{3 x} x-4 e^{2 x} x^2+\left (-4 e^{2 x} x+4 e^x x^2\right ) \log \left (x^2\right )+\left (e^x x-x^2\right ) \log ^2\left (x^2\right )+\left (-8 e^{2 x} x+8 e^x x^2+\left (4 e^x x-4 x^2\right ) \log \left (x^2\right )\right ) \log \left (-e^x+x\right )+\left (4 e^x x-4 x^2\right ) \log ^2\left (-e^x+x\right )} \, dx=-\frac {2}{2 e^{x}-\mathrm {log}\left (x^{2}\right )-2 \,\mathrm {log}\left (-e^{x}+x \right )} \] Input:
int((4*x*exp(x)^2+(-4*x^2-4*x-4)*exp(x)+8*x)/((4*exp(x)*x-4*x^2)*log(x-exp (x))^2+((4*exp(x)*x-4*x^2)*log(x^2)-8*x*exp(x)^2+8*exp(x)*x^2)*log(x-exp(x ))+(exp(x)*x-x^2)*log(x^2)^2+(-4*x*exp(x)^2+4*exp(x)*x^2)*log(x^2)+4*x*exp (x)^3-4*exp(x)^2*x^2),x)
Output:
( - 2)/(2*e**x - log(x**2) - 2*log( - e**x + x))