\(\int \frac {(e^{x^2} (-2+4 x)+(-2+4 x) \log (4)) \log (x-x^2)+(e^{x^2} (-1+x-2 x^2+2 x^3)+(-1+x) \log (4)) \log ^2(x-x^2)}{-10+10 x+(-1+x) \log (3)} \, dx\) [1511]

Optimal result

Integrand size = 81, antiderivative size = 26 \[ \int \frac {\left (e^{x^2} (-2+4 x)+(-2+4 x) \log (4)\right ) \log \left (x-x^2\right )+\left (e^{x^2} \left (-1+x-2 x^2+2 x^3\right )+(-1+x) \log (4)\right ) \log ^2\left (x-x^2\right )}{-10+10 x+(-1+x) \log (3)} \, dx=\frac {x \left (e^{x^2}+\log (4)\right ) \log ^2\left (x-x^2\right )}{10+\log (3)} \] Output:

x*(2*ln(2)+exp(x^2))*ln(-x^2+x)^2/(10+ln(3))
                                                                                    
                                                                                    
 

Mathematica [F]

\[ \int \frac {\left (e^{x^2} (-2+4 x)+(-2+4 x) \log (4)\right ) \log \left (x-x^2\right )+\left (e^{x^2} \left (-1+x-2 x^2+2 x^3\right )+(-1+x) \log (4)\right ) \log ^2\left (x-x^2\right )}{-10+10 x+(-1+x) \log (3)} \, dx=\int \frac {\left (e^{x^2} (-2+4 x)+(-2+4 x) \log (4)\right ) \log \left (x-x^2\right )+\left (e^{x^2} \left (-1+x-2 x^2+2 x^3\right )+(-1+x) \log (4)\right ) \log ^2\left (x-x^2\right )}{-10+10 x+(-1+x) \log (3)} \, dx \] Input:

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]
 

Output:

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]
 

Rubi [F]

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

\(\displaystyle -\frac {2 \sqrt {\pi } \int \frac {\text {erfi}(x)}{x-1}dx}{10+\log (3)}+\frac {\int e^{x^2} \log ^2((1-x) x)dx}{10+\log (3)}+\frac {2 \int e^{x^2} x^2 \log ^2((1-x) x)dx}{10+\log (3)}+\frac {2 \log ((1-x) x) \int \frac {e^{x^2}}{x-1}dx}{10+\log (3)}-\frac {2 \int \frac {\int \frac {e^{x^2}}{x-1}dx}{x-1}dx}{10+\log (3)}-\frac {2 \int \frac {\int \frac {e^{x^2}}{x-1}dx}{x}dx}{10+\log (3)}-\frac {4 x \, _2F_2\left (\frac {1}{2},\frac {1}{2};\frac {3}{2},\frac {3}{2};x^2\right )}{10+\log (3)}+\frac {2 \sqrt {\pi } \text {erfi}(x) \log ((1-x) x)}{10+\log (3)}+\frac {\log (4) \log ^2(1-x)}{10+\log (3)}-\frac {\log (4) \log ^2(x)}{10+\log (3)}-\frac {(1-x) \log (4) \log ^2((1-x) x)}{10+\log (3)}-\frac {2 \log (4) (\log (1-x)+\log (x)-\log ((1-x) x)) \log (1-x)}{10+\log (3)}-\frac {4 (1-x) \log (4) \log (1-x)}{10+\log (3)}+\frac {4 x \log (4) \log (x)}{10+\log (3)}-\frac {4 \log (4) \log (x)}{10+\log (3)}-\frac {4 x \log (4) (\log (1-x)+\log (x)-\log ((1-x) x))}{10+\log (3)}+\frac {2 \log (4) \log (x) \log ((1-x) x)}{10+\log (3)}+\frac {4 (1-x) \log (4) \log ((1-x) x)}{10+\log (3)}\)

Input:

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]
 

Output:

$Aborted
 

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.12 (sec) , antiderivative size = 1460, normalized size of antiderivative = 56.15

\[\text {Expression too large to display}\]

Input:

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)
 

Output:

x*(2*ln(2)+exp(x^2))/(10+ln(3))*ln(-1+x)^2+x*(-I*Pi*csgn(I*(-1+x))*csgn(I* 
x*(-1+x))*csgn(I*x)*exp(x^2)+I*Pi*csgn(I*x*(-1+x))^2*csgn(I*x)*exp(x^2)+2* 
I*Pi*ln(2)*csgn(I*(-1+x))*csgn(I*x*(-1+x))^2+2*I*Pi*ln(2)*csgn(I*x*(-1+x)) 
^2*csgn(I*x)+I*Pi*csgn(I*(-1+x))*csgn(I*x*(-1+x))^2*exp(x^2)+2*I*Pi*exp(x^ 
2)+2*I*Pi*ln(2)*csgn(I*x*(-1+x))^3-4*I*Pi*ln(2)*csgn(I*x*(-1+x))^2-2*I*Pi* 
ln(2)*csgn(I*(-1+x))*csgn(I*x*(-1+x))*csgn(I*x)-2*I*Pi*csgn(I*x*(-1+x))^2* 
exp(x^2)+4*I*Pi*ln(2)+I*Pi*csgn(I*x*(-1+x))^3*exp(x^2)+4*ln(2)*ln(x)+2*exp 
(x^2)*ln(x))/(10+ln(3))*ln(-1+x)+1/4*x*(-8*Pi^2*ln(2)*csgn(I*(-1+x))*csgn( 
I*x*(-1+x))^3*csgn(I*x)-4*Pi^2*csgn(I*(-1+x))*csgn(I*x*(-1+x))^3*csgn(I*x) 
*exp(x^2)-4*Pi^2*csgn(I*x*(-1+x))^3*exp(x^2)-4*Pi^2*csgn(I*x*(-1+x))^4*exp 
(x^2)+4*Pi^2*csgn(I*x*(-1+x))^5*exp(x^2)-2*Pi^2*ln(2)*csgn(I*(-1+x))^2*csg 
n(I*x*(-1+x))^2*csgn(I*x)^2+4*Pi^2*ln(2)*csgn(I*(-1+x))*csgn(I*x*(-1+x))^3 
*csgn(I*x)^2-Pi^2*csgn(I*(-1+x))^2*csgn(I*x*(-1+x))^2*csgn(I*x)^2*exp(x^2) 
+2*Pi^2*csgn(I*(-1+x))*csgn(I*x*(-1+x))^3*csgn(I*x)^2*exp(x^2)+4*Pi^2*csgn 
(I*(-1+x))*csgn(I*x*(-1+x))*csgn(I*x)*exp(x^2)+2*Pi^2*csgn(I*(-1+x))^2*csg 
n(I*x*(-1+x))^3*csgn(I*x)*exp(x^2)+8*Pi^2*ln(2)*csgn(I*(-1+x))*csgn(I*x*(- 
1+x))*csgn(I*x)+4*Pi^2*ln(2)*csgn(I*(-1+x))^2*csgn(I*x*(-1+x))^3*csgn(I*x) 
-8*Pi^2*ln(2)*csgn(I*x*(-1+x))^3-4*Pi^2*csgn(I*(-1+x))*csgn(I*x*(-1+x))^2* 
exp(x^2)-4*Pi^2*csgn(I*x*(-1+x))^2*csgn(I*x)*exp(x^2)-2*Pi^2*ln(2)*csgn(I* 
x*(-1+x))^4*csgn(I*x)^2-2*Pi^2*ln(2)*csgn(I*(-1+x))^2*csgn(I*x*(-1+x))^...
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12 \[ \int \frac {\left (e^{x^2} (-2+4 x)+(-2+4 x) \log (4)\right ) \log \left (x-x^2\right )+\left (e^{x^2} \left (-1+x-2 x^2+2 x^3\right )+(-1+x) \log (4)\right ) \log ^2\left (x-x^2\right )}{-10+10 x+(-1+x) \log (3)} \, dx=\frac {{\left (x e^{\left (x^{2}\right )} + 2 \, x \log \left (2\right )\right )} \log \left (-x^{2} + x\right )^{2}}{\log \left (3\right ) + 10} \] Input:

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="fricas")
 

Output:

(x*e^(x^2) + 2*x*log(2))*log(-x^2 + x)^2/(log(3) + 10)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (e^{x^2} (-2+4 x)+(-2+4 x) \log (4)\right ) \log \left (x-x^2\right )+\left (e^{x^2} \left (-1+x-2 x^2+2 x^3\right )+(-1+x) \log (4)\right ) \log ^2\left (x-x^2\right )}{-10+10 x+(-1+x) \log (3)} \, dx=\text {Timed out} \] Input:

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)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 72 vs. \(2 (27) = 54\).

Time = 0.18 (sec) , antiderivative size = 72, normalized size of antiderivative = 2.77 \[ \int \frac {\left (e^{x^2} (-2+4 x)+(-2+4 x) \log (4)\right ) \log \left (x-x^2\right )+\left (e^{x^2} \left (-1+x-2 x^2+2 x^3\right )+(-1+x) \log (4)\right ) \log ^2\left (x-x^2\right )}{-10+10 x+(-1+x) \log (3)} \, dx=\frac {x e^{\left (x^{2}\right )} \log \left (x\right )^{2} + 2 \, x \log \left (2\right ) \log \left (x\right )^{2} + {\left (x e^{\left (x^{2}\right )} + 2 \, x \log \left (2\right )\right )} \log \left (-x + 1\right )^{2} + 2 \, {\left (x e^{\left (x^{2}\right )} \log \left (x\right ) + 2 \, x \log \left (2\right ) \log \left (x\right )\right )} \log \left (-x + 1\right )}{\log \left (3\right ) + 10} \] Input:

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="maxima")
 

Output:

(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)
 

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.50 \[ \int \frac {\left (e^{x^2} (-2+4 x)+(-2+4 x) \log (4)\right ) \log \left (x-x^2\right )+\left (e^{x^2} \left (-1+x-2 x^2+2 x^3\right )+(-1+x) \log (4)\right ) \log ^2\left (x-x^2\right )}{-10+10 x+(-1+x) \log (3)} \, dx=\frac {x e^{\left (x^{2}\right )} \log \left (-x^{2} + x\right )^{2} + 2 \, x \log \left (2\right ) \log \left (-x^{2} + x\right )^{2}}{\log \left (3\right ) + 10} \] Input:

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="giac")
 

Output:

(x*e^(x^2)*log(-x^2 + x)^2 + 2*x*log(2)*log(-x^2 + x)^2)/(log(3) + 10)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (e^{x^2} (-2+4 x)+(-2+4 x) \log (4)\right ) \log \left (x-x^2\right )+\left (e^{x^2} \left (-1+x-2 x^2+2 x^3\right )+(-1+x) \log (4)\right ) \log ^2\left (x-x^2\right )}{-10+10 x+(-1+x) \log (3)} \, dx=\int \frac {\left (2\,\ln \left (2\right )\,\left (x-1\right )+{\mathrm {e}}^{x^2}\,\left (2\,x^3-2\,x^2+x-1\right )\right )\,{\ln \left (x-x^2\right )}^2+\left (2\,\ln \left (2\right )\,\left (4\,x-2\right )+{\mathrm {e}}^{x^2}\,\left (4\,x-2\right )\right )\,\ln \left (x-x^2\right )}{10\,x+\ln \left (3\right )\,\left (x-1\right )-10} \,d x \] Input:

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)
 

Output:

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)
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08 \[ \int \frac {\left (e^{x^2} (-2+4 x)+(-2+4 x) \log (4)\right ) \log \left (x-x^2\right )+\left (e^{x^2} \left (-1+x-2 x^2+2 x^3\right )+(-1+x) \log (4)\right ) \log ^2\left (x-x^2\right )}{-10+10 x+(-1+x) \log (3)} \, dx=\frac {\mathrm {log}\left (-x^{2}+x \right )^{2} x \left (e^{x^{2}}+2 \,\mathrm {log}\left (2\right )\right )}{\mathrm {log}\left (3\right )+10} \] Input:

int((((2*x^3-2*x^2+x-1)*exp(x^2)+2*(-1+x)*log(2))*log(-x^2+x)^2+((4*x-2)*e 
xp(x^2)+2*(4*x-2)*log(2))*log(-x^2+x))/((-1+x)*log(3)+10*x-10),x)
 

Output:

(log( - x**2 + x)**2*x*(e**(x**2) + 2*log(2)))/(log(3) + 10)