Integrand size = 81, antiderivative size = 26 \[ \int \frac {\left (e^{x^2} (-2+4 x)+(-2+4 x) \log (4)\right ) \log \left (x-x^2\right )+\left (e^{x^2} \left (-1+x-2 x^2+2 x^3\right )+(-1+x) \log (4)\right ) \log ^2\left (x-x^2\right )}{-10+10 x+(-1+x) \log (3)} \, dx=\frac {x \left (e^{x^2}+\log (4)\right ) \log ^2\left (x-x^2\right )}{10+\log (3)} \]
\[ \int \frac {\left (e^{x^2} (-2+4 x)+(-2+4 x) \log (4)\right ) \log \left (x-x^2\right )+\left (e^{x^2} \left (-1+x-2 x^2+2 x^3\right )+(-1+x) \log (4)\right ) \log ^2\left (x-x^2\right )}{-10+10 x+(-1+x) \log (3)} \, dx=\int \frac {\left (e^{x^2} (-2+4 x)+(-2+4 x) \log (4)\right ) \log \left (x-x^2\right )+\left (e^{x^2} \left (-1+x-2 x^2+2 x^3\right )+(-1+x) \log (4)\right ) \log ^2\left (x-x^2\right )}{-10+10 x+(-1+x) \log (3)} \, dx \]
Integrate[((E^x^2*(-2 + 4*x) + (-2 + 4*x)*Log[4])*Log[x - x^2] + (E^x^2*(- 1 + x - 2*x^2 + 2*x^3) + (-1 + x)*Log[4])*Log[x - x^2]^2)/(-10 + 10*x + (- 1 + x)*Log[3]),x]
Integrate[((E^x^2*(-2 + 4*x) + (-2 + 4*x)*Log[4])*Log[x - x^2] + (E^x^2*(- 1 + x - 2*x^2 + 2*x^3) + (-1 + x)*Log[4])*Log[x - x^2]^2)/(-10 + 10*x + (- 1 + x)*Log[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.
\(\displaystyle \int \frac {\left (e^{x^2} (4 x-2)+(4 x-2) \log (4)\right ) \log \left (x-x^2\right )+\left (e^{x^2} \left (2 x^3-2 x^2+x-1\right )+(x-1) \log (4)\right ) \log ^2\left (x-x^2\right )}{10 x+(x-1) \log (3)-10} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {-\left (e^{x^2} (4 x-2)+(4 x-2) \log (4)\right ) \log \left (x-x^2\right )-\left (\left (e^{x^2} \left (2 x^3-2 x^2+x-1\right )+(x-1) \log (4)\right ) \log ^2\left (x-x^2\right )\right )}{-(x (10+\log (3)))+10+\log (3)}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {e^{x^2} \log ((1-x) x) \left (-2 x^3 \log (-((x-1) x))+2 x^2 \log (-((x-1) x))-4 x-x \log (-((x-1) x))+\log (-((x-1) x))+2\right )}{(1-x) (10+\log (3))}+\frac {\log (4) \log ((1-x) x) (-4 x+x (-\log (-((x-1) x)))+\log (-((x-1) x))+2)}{(1-x) (10+\log (3))}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 \sqrt {\pi } \int \frac {\text {erfi}(x)}{x-1}dx}{10+\log (3)}+\frac {\int e^{x^2} \log ^2((1-x) x)dx}{10+\log (3)}+\frac {2 \int e^{x^2} x^2 \log ^2((1-x) x)dx}{10+\log (3)}+\frac {2 \log ((1-x) x) \int \frac {e^{x^2}}{x-1}dx}{10+\log (3)}-\frac {2 \int \frac {\int \frac {e^{x^2}}{x-1}dx}{x-1}dx}{10+\log (3)}-\frac {2 \int \frac {\int \frac {e^{x^2}}{x-1}dx}{x}dx}{10+\log (3)}-\frac {4 x \, _2F_2\left (\frac {1}{2},\frac {1}{2};\frac {3}{2},\frac {3}{2};x^2\right )}{10+\log (3)}+\frac {2 \sqrt {\pi } \text {erfi}(x) \log ((1-x) x)}{10+\log (3)}+\frac {\log (4) \log ^2(1-x)}{10+\log (3)}-\frac {\log (4) \log ^2(x)}{10+\log (3)}-\frac {(1-x) \log (4) \log ^2((1-x) x)}{10+\log (3)}-\frac {2 \log (4) (\log (1-x)+\log (x)-\log ((1-x) x)) \log (1-x)}{10+\log (3)}-\frac {4 (1-x) \log (4) \log (1-x)}{10+\log (3)}+\frac {4 x \log (4) \log (x)}{10+\log (3)}-\frac {4 \log (4) \log (x)}{10+\log (3)}-\frac {4 x \log (4) (\log (1-x)+\log (x)-\log ((1-x) x))}{10+\log (3)}+\frac {2 \log (4) \log (x) \log ((1-x) x)}{10+\log (3)}+\frac {4 (1-x) \log (4) \log ((1-x) x)}{10+\log (3)}\) |
Int[((E^x^2*(-2 + 4*x) + (-2 + 4*x)*Log[4])*Log[x - x^2] + (E^x^2*(-1 + x - 2*x^2 + 2*x^3) + (-1 + x)*Log[4])*Log[x - x^2]^2)/(-10 + 10*x + (-1 + x) *Log[3]),x]
3.16.11.3.1 Defintions of rubi rules used
Time = 1.04 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.54
method | result | size |
parallelrisch | \(\frac {2 \ln \left (2\right ) \ln \left (-x^{2}+x \right )^{2} x +{\mathrm e}^{x^{2}} \ln \left (-x^{2}+x \right )^{2} x}{10+\ln \left (3\right )}\) | \(40\) |
risch | \(\text {Expression too large to display}\) | \(1460\) |
int((((2*x^3-2*x^2+x-1)*exp(x^2)+2*(-1+x)*ln(2))*ln(-x^2+x)^2+((4*x-2)*exp (x^2)+2*(4*x-2)*ln(2))*ln(-x^2+x))/((-1+x)*ln(3)+10*x-10),x,method=_RETURN VERBOSE)
Time = 0.26 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12 \[ \int \frac {\left (e^{x^2} (-2+4 x)+(-2+4 x) \log (4)\right ) \log \left (x-x^2\right )+\left (e^{x^2} \left (-1+x-2 x^2+2 x^3\right )+(-1+x) \log (4)\right ) \log ^2\left (x-x^2\right )}{-10+10 x+(-1+x) \log (3)} \, dx=\frac {{\left (x e^{\left (x^{2}\right )} + 2 \, x \log \left (2\right )\right )} \log \left (-x^{2} + x\right )^{2}}{\log \left (3\right ) + 10} \]
integrate((((2*x^3-2*x^2+x-1)*exp(x^2)+2*(-1+x)*log(2))*log(-x^2+x)^2+((4* x-2)*exp(x^2)+2*(4*x-2)*log(2))*log(-x^2+x))/((-1+x)*log(3)+10*x-10),x, al gorithm=\
Timed out. \[ \int \frac {\left (e^{x^2} (-2+4 x)+(-2+4 x) \log (4)\right ) \log \left (x-x^2\right )+\left (e^{x^2} \left (-1+x-2 x^2+2 x^3\right )+(-1+x) \log (4)\right ) \log ^2\left (x-x^2\right )}{-10+10 x+(-1+x) \log (3)} \, dx=\text {Timed out} \]
integrate((((2*x**3-2*x**2+x-1)*exp(x**2)+2*(-1+x)*ln(2))*ln(-x**2+x)**2+( (4*x-2)*exp(x**2)+2*(4*x-2)*ln(2))*ln(-x**2+x))/((-1+x)*ln(3)+10*x-10),x)
Leaf count of result is larger than twice the leaf count of optimal. 72 vs. \(2 (27) = 54\).
Time = 0.32 (sec) , antiderivative size = 72, normalized size of antiderivative = 2.77 \[ \int \frac {\left (e^{x^2} (-2+4 x)+(-2+4 x) \log (4)\right ) \log \left (x-x^2\right )+\left (e^{x^2} \left (-1+x-2 x^2+2 x^3\right )+(-1+x) \log (4)\right ) \log ^2\left (x-x^2\right )}{-10+10 x+(-1+x) \log (3)} \, dx=\frac {x e^{\left (x^{2}\right )} \log \left (x\right )^{2} + 2 \, x \log \left (2\right ) \log \left (x\right )^{2} + {\left (x e^{\left (x^{2}\right )} + 2 \, x \log \left (2\right )\right )} \log \left (-x + 1\right )^{2} + 2 \, {\left (x e^{\left (x^{2}\right )} \log \left (x\right ) + 2 \, x \log \left (2\right ) \log \left (x\right )\right )} \log \left (-x + 1\right )}{\log \left (3\right ) + 10} \]
integrate((((2*x^3-2*x^2+x-1)*exp(x^2)+2*(-1+x)*log(2))*log(-x^2+x)^2+((4* x-2)*exp(x^2)+2*(4*x-2)*log(2))*log(-x^2+x))/((-1+x)*log(3)+10*x-10),x, al gorithm=\
(x*e^(x^2)*log(x)^2 + 2*x*log(2)*log(x)^2 + (x*e^(x^2) + 2*x*log(2))*log(- x + 1)^2 + 2*(x*e^(x^2)*log(x) + 2*x*log(2)*log(x))*log(-x + 1))/(log(3) + 10)
Time = 0.31 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.50 \[ \int \frac {\left (e^{x^2} (-2+4 x)+(-2+4 x) \log (4)\right ) \log \left (x-x^2\right )+\left (e^{x^2} \left (-1+x-2 x^2+2 x^3\right )+(-1+x) \log (4)\right ) \log ^2\left (x-x^2\right )}{-10+10 x+(-1+x) \log (3)} \, dx=\frac {x e^{\left (x^{2}\right )} \log \left (-x^{2} + x\right )^{2} + 2 \, x \log \left (2\right ) \log \left (-x^{2} + x\right )^{2}}{\log \left (3\right ) + 10} \]
integrate((((2*x^3-2*x^2+x-1)*exp(x^2)+2*(-1+x)*log(2))*log(-x^2+x)^2+((4* x-2)*exp(x^2)+2*(4*x-2)*log(2))*log(-x^2+x))/((-1+x)*log(3)+10*x-10),x, al gorithm=\
Timed out. \[ \int \frac {\left (e^{x^2} (-2+4 x)+(-2+4 x) \log (4)\right ) \log \left (x-x^2\right )+\left (e^{x^2} \left (-1+x-2 x^2+2 x^3\right )+(-1+x) \log (4)\right ) \log ^2\left (x-x^2\right )}{-10+10 x+(-1+x) \log (3)} \, dx=\int \frac {\left (2\,\ln \left (2\right )\,\left (x-1\right )+{\mathrm {e}}^{x^2}\,\left (2\,x^3-2\,x^2+x-1\right )\right )\,{\ln \left (x-x^2\right )}^2+\left (2\,\ln \left (2\right )\,\left (4\,x-2\right )+{\mathrm {e}}^{x^2}\,\left (4\,x-2\right )\right )\,\ln \left (x-x^2\right )}{10\,x+\ln \left (3\right )\,\left (x-1\right )-10} \,d x \]
int((log(x - x^2)*(2*log(2)*(4*x - 2) + exp(x^2)*(4*x - 2)) + log(x - x^2) ^2*(2*log(2)*(x - 1) + exp(x^2)*(x - 2*x^2 + 2*x^3 - 1)))/(10*x + log(3)*( x - 1) - 10),x)