Integrand size = 120, antiderivative size = 23 \[ \int \frac {-16 x^3+\left (32 x^2-32 x^3\right ) \log (e (-1+x))+e^{6+2 x+2 \log ^2(x)} \left (-16 x+\log (e (-1+x)) \left (32 x-32 x^2+(64-64 x) \log (x)\right )\right )+e^{3+x+\log ^2(x)} \left (-32 x^2+\log (e (-1+x)) \left (32 x-32 x^3+\left (64 x-64 x^2\right ) \log (x)\right )\right )}{-x+x^2} \, dx=5-16 \left (e^{3+x+\log ^2(x)}+x\right )^2 \log (e (-1+x)) \] Output:
5-4*ln((-1+x)*exp(1))*(x+exp(ln(x)^2+3+x))*(4*x+4*exp(ln(x)^2+3+x))
Time = 3.95 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {-16 x^3+\left (32 x^2-32 x^3\right ) \log (e (-1+x))+e^{6+2 x+2 \log ^2(x)} \left (-16 x+\log (e (-1+x)) \left (32 x-32 x^2+(64-64 x) \log (x)\right )\right )+e^{3+x+\log ^2(x)} \left (-32 x^2+\log (e (-1+x)) \left (32 x-32 x^3+\left (64 x-64 x^2\right ) \log (x)\right )\right )}{-x+x^2} \, dx=-16 \left (e^{3+x+\log ^2(x)}+x\right )^2 (1+\log (-1+x)) \] Input:
Integrate[(-16*x^3 + (32*x^2 - 32*x^3)*Log[E*(-1 + x)] + E^(6 + 2*x + 2*Lo g[x]^2)*(-16*x + Log[E*(-1 + x)]*(32*x - 32*x^2 + (64 - 64*x)*Log[x])) + E ^(3 + x + Log[x]^2)*(-32*x^2 + Log[E*(-1 + x)]*(32*x - 32*x^3 + (64*x - 64 *x^2)*Log[x])))/(-x + x^2),x]
Output:
-16*(E^(3 + x + Log[x]^2) + x)^2*(1 + Log[-1 + 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.
| \(\displaystyle \int \frac {-16 x^3+e^{2 x+2 \log ^2(x)+6} \left (\log (e (x-1)) \left (-32 x^2+32 x+(64-64 x) \log (x)\right )-16 x\right )+e^{x+\log ^2(x)+3} \left (\log (e (x-1)) \left (-32 x^3+\left (64 x-64 x^2\right ) \log (x)+32 x\right )-32 x^2\right )+\left (32 x^2-32 x^3\right ) \log (e (x-1))}{x^2-x} \, dx\) |
| \(\Big \downarrow \) 2026 |
| \(\displaystyle \int \frac {-16 x^3+e^{2 x+2 \log ^2(x)+6} \left (\log (e (x-1)) \left (-32 x^2+32 x+(64-64 x) \log (x)\right )-16 x\right )+e^{x+\log ^2(x)+3} \left (\log (e (x-1)) \left (-32 x^3+\left (64 x-64 x^2\right ) \log (x)+32 x\right )-32 x^2\right )+\left (32 x^2-32 x^3\right ) \log (e (x-1))}{(x-1) x}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (-\frac {32 e^{x+\log ^2(x)+3} \left (x^2+x^2 \log (x-1)+x+2 x \log (x-1) \log (x)+2 x \log (x)-\log (x-1)-2 \log (x-1) \log (x)-2 \log (x)-1\right )}{x-1}+\frac {16 e^{2 \left (x+\log ^2(x)+3\right )} \left (2 x^2+2 x^2 \log (x-1)-x-2 x \log (x-1)+4 x \log (x-1) \log (x)+4 x \log (x)-4 \log (x-1) \log (x)-4 \log (x)\right )}{(1-x) x}-\frac {16 x (3 x+2 x \log (x-1)-2 \log (x-1)-2)}{x-1}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -64 \int e^{\log ^2(x)+x+3}dx-32 \int e^{2 \left (\log ^2(x)+x+3\right )}dx+16 \int \frac {e^{2 \left (\log ^2(x)+x+3\right )}}{1-x}dx-32 \int \frac {e^{\log ^2(x)+x+3}}{x-1}dx-32 \int e^{\log ^2(x)+x+3} xdx-32 \int e^{\log ^2(x)+x+3} \log (x-1)dx-32 \int e^{2 \left (\log ^2(x)+x+3\right )} \log (x-1)dx-32 \int e^{\log ^2(x)+x+3} x \log (x-1)dx-64 \int e^{\log ^2(x)+x+3} \log (x)dx-64 \int \frac {e^{2 \left (\log ^2(x)+x+3\right )} \log (x)}{x}dx-64 \int e^{\log ^2(x)+x+3} \log (x-1) \log (x)dx-64 \int \frac {e^{2 \left (\log ^2(x)+x+3\right )} \log (x-1) \log (x)}{x}dx-16 x^2-16 x^2 \log (x-1)\) |
Input:
Int[(-16*x^3 + (32*x^2 - 32*x^3)*Log[E*(-1 + x)] + E^(6 + 2*x + 2*Log[x]^2 )*(-16*x + Log[E*(-1 + x)]*(32*x - 32*x^2 + (64 - 64*x)*Log[x])) + E^(3 + x + Log[x]^2)*(-32*x^2 + Log[E*(-1 + x)]*(32*x - 32*x^3 + (64*x - 64*x^2)* Log[x])))/(-x + x^2),x]
Output:
$Aborted
Time = 1.29 (sec) , antiderivative size = 53, normalized size of antiderivative = 2.30
| method | result | size |
| risch | \(-16 x^{2} \ln \left (\left (-1+x \right ) {\mathrm e}\right )-32 \ln \left (\left (-1+x \right ) {\mathrm e}\right ) {\mathrm e}^{\ln \left (x \right )^{2}+3+x} x -16 \ln \left (\left (-1+x \right ) {\mathrm e}\right ) {\mathrm e}^{2 \ln \left (x \right )^{2}+6+2 x}\) | \(53\) |
| parallelrisch | \(-16 x^{2} \ln \left (\left (-1+x \right ) {\mathrm e}\right )-32 \ln \left (\left (-1+x \right ) {\mathrm e}\right ) {\mathrm e}^{\ln \left (x \right )^{2}+3+x} x -16 \ln \left (\left (-1+x \right ) {\mathrm e}\right ) {\mathrm e}^{2 \ln \left (x \right )^{2}+6+2 x}-16 \ln \left (-1+x \right )+16 \ln \left (\left (-1+x \right ) {\mathrm e}\right )\) | \(66\) |
Input:
int(((((-64*x+64)*ln(x)-32*x^2+32*x)*ln((-1+x)*exp(1))-16*x)*exp(ln(x)^2+3 +x)^2+(((-64*x^2+64*x)*ln(x)-32*x^3+32*x)*ln((-1+x)*exp(1))-32*x^2)*exp(ln (x)^2+3+x)+(-32*x^3+32*x^2)*ln((-1+x)*exp(1))-16*x^3)/(x^2-x),x,method=_RE TURNVERBOSE)
Output:
-16*x^2*ln((-1+x)*exp(1))-32*ln((-1+x)*exp(1))*exp(ln(x)^2+3+x)*x-16*ln((- 1+x)*exp(1))*exp(2*ln(x)^2+6+2*x)
Leaf count of result is larger than twice the leaf count of optimal. 52 vs. \(2 (23) = 46\).
Time = 0.09 (sec) , antiderivative size = 52, normalized size of antiderivative = 2.26 \[ \int \frac {-16 x^3+\left (32 x^2-32 x^3\right ) \log (e (-1+x))+e^{6+2 x+2 \log ^2(x)} \left (-16 x+\log (e (-1+x)) \left (32 x-32 x^2+(64-64 x) \log (x)\right )\right )+e^{3+x+\log ^2(x)} \left (-32 x^2+\log (e (-1+x)) \left (32 x-32 x^3+\left (64 x-64 x^2\right ) \log (x)\right )\right )}{-x+x^2} \, dx=-16 \, x^{2} \log \left ({\left (x - 1\right )} e\right ) - 32 \, x e^{\left (\log \left (x\right )^{2} + x + 3\right )} \log \left ({\left (x - 1\right )} e\right ) - 16 \, e^{\left (2 \, \log \left (x\right )^{2} + 2 \, x + 6\right )} \log \left ({\left (x - 1\right )} e\right ) \] Input:
integrate(((((-64*x+64)*log(x)-32*x^2+32*x)*log((-1+x)*exp(1))-16*x)*exp(l og(x)^2+3+x)^2+(((-64*x^2+64*x)*log(x)-32*x^3+32*x)*log((-1+x)*exp(1))-32* x^2)*exp(log(x)^2+3+x)+(-32*x^3+32*x^2)*log((-1+x)*exp(1))-16*x^3)/(x^2-x) ,x, algorithm="fricas")
Output:
-16*x^2*log((x - 1)*e) - 32*x*e^(log(x)^2 + x + 3)*log((x - 1)*e) - 16*e^( 2*log(x)^2 + 2*x + 6)*log((x - 1)*e)
Time = 3.95 (sec) , antiderivative size = 71, normalized size of antiderivative = 3.09 \[ \int \frac {-16 x^3+\left (32 x^2-32 x^3\right ) \log (e (-1+x))+e^{6+2 x+2 \log ^2(x)} \left (-16 x+\log (e (-1+x)) \left (32 x-32 x^2+(64-64 x) \log (x)\right )\right )+e^{3+x+\log ^2(x)} \left (-32 x^2+\log (e (-1+x)) \left (32 x-32 x^3+\left (64 x-64 x^2\right ) \log (x)\right )\right )}{-x+x^2} \, dx=- 32 x e^{x + \log {\left (x \right )}^{2} + 3} \log {\left (e \left (x - 1\right ) \right )} + \left (\frac {16}{3} - 16 x^{2}\right ) \log {\left (e \left (x - 1\right ) \right )} - 16 e^{2 x + 2 \log {\left (x \right )}^{2} + 6} \log {\left (e \left (x - 1\right ) \right )} - \frac {16 \log {\left (3 x - 3 \right )}}{3} \] Input:
integrate(((((-64*x+64)*ln(x)-32*x**2+32*x)*ln((-1+x)*exp(1))-16*x)*exp(ln (x)**2+3+x)**2+(((-64*x**2+64*x)*ln(x)-32*x**3+32*x)*ln((-1+x)*exp(1))-32* x**2)*exp(ln(x)**2+3+x)+(-32*x**3+32*x**2)*ln((-1+x)*exp(1))-16*x**3)/(x** 2-x),x)
Output:
-32*x*exp(x + log(x)**2 + 3)*log(E*(x - 1)) + (16/3 - 16*x**2)*log(E*(x - 1)) - 16*exp(2*x + 2*log(x)**2 + 6)*log(E*(x - 1)) - 16*log(3*x - 3)/3
Leaf count of result is larger than twice the leaf count of optimal. 74 vs. \(2 (23) = 46\).
Time = 0.10 (sec) , antiderivative size = 74, normalized size of antiderivative = 3.22 \[ \int \frac {-16 x^3+\left (32 x^2-32 x^3\right ) \log (e (-1+x))+e^{6+2 x+2 \log ^2(x)} \left (-16 x+\log (e (-1+x)) \left (32 x-32 x^2+(64-64 x) \log (x)\right )\right )+e^{3+x+\log ^2(x)} \left (-32 x^2+\log (e (-1+x)) \left (32 x-32 x^3+\left (64 x-64 x^2\right ) \log (x)\right )\right )}{-x+x^2} \, dx=-16 \, x^{2} - 16 \, {\left (e^{\left (2 \, x + 6\right )} \log \left (x - 1\right ) + e^{\left (2 \, x + 6\right )}\right )} e^{\left (2 \, \log \left (x\right )^{2}\right )} - 32 \, {\left (x e^{\left (x + 3\right )} \log \left (x - 1\right ) + x e^{\left (x + 3\right )}\right )} e^{\left (\log \left (x\right )^{2}\right )} - 16 \, {\left (x^{2} - 1\right )} \log \left (x - 1\right ) - 16 \, \log \left (x - 1\right ) \] Input:
integrate(((((-64*x+64)*log(x)-32*x^2+32*x)*log((-1+x)*exp(1))-16*x)*exp(l og(x)^2+3+x)^2+(((-64*x^2+64*x)*log(x)-32*x^3+32*x)*log((-1+x)*exp(1))-32* x^2)*exp(log(x)^2+3+x)+(-32*x^3+32*x^2)*log((-1+x)*exp(1))-16*x^3)/(x^2-x) ,x, algorithm="maxima")
Output:
-16*x^2 - 16*(e^(2*x + 6)*log(x - 1) + e^(2*x + 6))*e^(2*log(x)^2) - 32*(x *e^(x + 3)*log(x - 1) + x*e^(x + 3))*e^(log(x)^2) - 16*(x^2 - 1)*log(x - 1 ) - 16*log(x - 1)
Leaf count of result is larger than twice the leaf count of optimal. 73 vs. \(2 (23) = 46\).
Time = 0.16 (sec) , antiderivative size = 73, normalized size of antiderivative = 3.17 \[ \int \frac {-16 x^3+\left (32 x^2-32 x^3\right ) \log (e (-1+x))+e^{6+2 x+2 \log ^2(x)} \left (-16 x+\log (e (-1+x)) \left (32 x-32 x^2+(64-64 x) \log (x)\right )\right )+e^{3+x+\log ^2(x)} \left (-32 x^2+\log (e (-1+x)) \left (32 x-32 x^3+\left (64 x-64 x^2\right ) \log (x)\right )\right )}{-x+x^2} \, dx=-16 \, x^{2} \log \left (x - 1\right ) - 32 \, x e^{\left (\log \left (x\right )^{2} + x + 3\right )} \log \left (x - 1\right ) - 16 \, x^{2} - 32 \, x e^{\left (\log \left (x\right )^{2} + x + 3\right )} - 16 \, e^{\left (2 \, \log \left (x\right )^{2} + 2 \, x + 6\right )} \log \left (x - 1\right ) - 16 \, e^{\left (2 \, \log \left (x\right )^{2} + 2 \, x + 6\right )} \] Input:
integrate(((((-64*x+64)*log(x)-32*x^2+32*x)*log((-1+x)*exp(1))-16*x)*exp(l og(x)^2+3+x)^2+(((-64*x^2+64*x)*log(x)-32*x^3+32*x)*log((-1+x)*exp(1))-32* x^2)*exp(log(x)^2+3+x)+(-32*x^3+32*x^2)*log((-1+x)*exp(1))-16*x^3)/(x^2-x) ,x, algorithm="giac")
Output:
-16*x^2*log(x - 1) - 32*x*e^(log(x)^2 + x + 3)*log(x - 1) - 16*x^2 - 32*x* e^(log(x)^2 + x + 3) - 16*e^(2*log(x)^2 + 2*x + 6)*log(x - 1) - 16*e^(2*lo g(x)^2 + 2*x + 6)
Timed out. \[ \int \frac {-16 x^3+\left (32 x^2-32 x^3\right ) \log (e (-1+x))+e^{6+2 x+2 \log ^2(x)} \left (-16 x+\log (e (-1+x)) \left (32 x-32 x^2+(64-64 x) \log (x)\right )\right )+e^{3+x+\log ^2(x)} \left (-32 x^2+\log (e (-1+x)) \left (32 x-32 x^3+\left (64 x-64 x^2\right ) \log (x)\right )\right )}{-x+x^2} \, dx=\int \frac {{\mathrm {e}}^{2\,{\ln \left (x\right )}^2+2\,x+6}\,\left (16\,x+\ln \left (\mathrm {e}\,\left (x-1\right )\right )\,\left (\ln \left (x\right )\,\left (64\,x-64\right )-32\,x+32\,x^2\right )\right )-\ln \left (\mathrm {e}\,\left (x-1\right )\right )\,\left (32\,x^2-32\,x^3\right )+16\,x^3-{\mathrm {e}}^{{\ln \left (x\right )}^2+x+3}\,\left (\ln \left (\mathrm {e}\,\left (x-1\right )\right )\,\left (32\,x+\ln \left (x\right )\,\left (64\,x-64\,x^2\right )-32\,x^3\right )-32\,x^2\right )}{x-x^2} \,d x \] Input:
int((exp(2*x + 2*log(x)^2 + 6)*(16*x + log(exp(1)*(x - 1))*(log(x)*(64*x - 64) - 32*x + 32*x^2)) - log(exp(1)*(x - 1))*(32*x^2 - 32*x^3) + 16*x^3 - exp(x + log(x)^2 + 3)*(log(exp(1)*(x - 1))*(32*x + log(x)*(64*x - 64*x^2) - 32*x^3) - 32*x^2))/(x - x^2),x)
Output:
int((exp(2*x + 2*log(x)^2 + 6)*(16*x + log(exp(1)*(x - 1))*(log(x)*(64*x - 64) - 32*x + 32*x^2)) - log(exp(1)*(x - 1))*(32*x^2 - 32*x^3) + 16*x^3 - exp(x + log(x)^2 + 3)*(log(exp(1)*(x - 1))*(32*x + log(x)*(64*x - 64*x^2) - 32*x^3) - 32*x^2))/(x - x^2), x)
Time = 0.18 (sec) , antiderivative size = 77, normalized size of antiderivative = 3.35 \[ \int \frac {-16 x^3+\left (32 x^2-32 x^3\right ) \log (e (-1+x))+e^{6+2 x+2 \log ^2(x)} \left (-16 x+\log (e (-1+x)) \left (32 x-32 x^2+(64-64 x) \log (x)\right )\right )+e^{3+x+\log ^2(x)} \left (-32 x^2+\log (e (-1+x)) \left (32 x-32 x^3+\left (64 x-64 x^2\right ) \log (x)\right )\right )}{-x+x^2} \, dx=-16 e^{2 \mathrm {log}\left (x \right )^{2}+2 x} \mathrm {log}\left (e x -e \right ) e^{6}-32 e^{\mathrm {log}\left (x \right )^{2}+x} \mathrm {log}\left (e x -e \right ) e^{3} x -16 \,\mathrm {log}\left (e x -e \right ) x^{2}+16 \,\mathrm {log}\left (e x -e \right )-16 \,\mathrm {log}\left (x -1\right ) \] Input:
int(((((-64*x+64)*log(x)-32*x^2+32*x)*log((-1+x)*exp(1))-16*x)*exp(log(x)^ 2+3+x)^2+(((-64*x^2+64*x)*log(x)-32*x^3+32*x)*log((-1+x)*exp(1))-32*x^2)*e xp(log(x)^2+3+x)+(-32*x^3+32*x^2)*log((-1+x)*exp(1))-16*x^3)/(x^2-x),x)
Output:
16*( - e**(2*log(x)**2 + 2*x)*log(e*x - e)*e**6 - 2*e**(log(x)**2 + x)*log (e*x - e)*e**3*x - log(e*x - e)*x**2 + log(e*x - e) - log(x - 1))