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

Optimal. Leaf size=26 \[ \frac {x \left (e^{x^2}+\log (4)\right ) \log ^2\left (x-x^2\right )}{10+\log (3)} \]

[Out]

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

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Rubi [F]
time = 2.48, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \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 \end {gather*}

Verification is not applicable to the result.

[In]

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]

[Out]

(-4*x*HypergeometricPFQ[{1/2, 1/2}, {3/2, 3/2}, x^2])/(10 + Log[3]) - (4*(1 - x)*Log[4]*Log[1 - x])/(10 + Log[
3]) + (Log[4]*Log[1 - x]^2)/(10 + Log[3]) - (4*Log[4]*Log[x])/(10 + Log[3]) + (4*x*Log[4]*Log[x])/(10 + Log[3]
) - (Log[4]*Log[x]^2)/(10 + Log[3]) - (4*x*Log[4]*(Log[1 - x] + Log[x] - Log[(1 - x)*x]))/(10 + Log[3]) - (2*L
og[4]*Log[1 - x]*(Log[1 - x] + Log[x] - Log[(1 - x)*x]))/(10 + Log[3]) + (2*Sqrt[Pi]*Erfi[x]*Log[(1 - x)*x])/(
10 + Log[3]) + (4*(1 - x)*Log[4]*Log[(1 - x)*x])/(10 + Log[3]) + (2*Log[4]*Log[x]*Log[(1 - x)*x])/(10 + Log[3]
) - ((1 - x)*Log[4]*Log[(1 - x)*x]^2)/(10 + Log[3]) + (2*Log[(1 - x)*x]*Defer[Int][E^x^2/(-1 + x), x])/(10 + L
og[3]) - (2*Sqrt[Pi]*Defer[Int][Erfi[x]/(-1 + x), x])/(10 + Log[3]) + Defer[Int][E^x^2*Log[(1 - x)*x]^2, x]/(1
0 + Log[3]) + (2*Defer[Int][E^x^2*x^2*Log[(1 - x)*x]^2, x])/(10 + Log[3]) - (2*Defer[Int][Defer[Int][E^x^2/(-1
+ x), x]/(-1 + x), x])/(10 + Log[3]) - (2*Defer[Int][Defer[Int][E^x^2/(-1 + x), x]/x, x])/(10 + Log[3])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-\left (\left (e^{x^2} (-2+4 x)+(-2+4 x) \log (4)\right ) \log \left (x-x^2\right )\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+\log (3)-x (10+\log (3))} \, dx\\ &=\int \left (\frac {\log (4) \log ((1-x) x) (2-4 x+\log (-((-1+x) x))-x \log (-((-1+x) x)))}{(1-x) (10+\log (3))}+\frac {e^{x^2} \log ((1-x) x) \left (2-4 x+\log (-((-1+x) x))-x \log (-((-1+x) x))+2 x^2 \log (-((-1+x) x))-2 x^3 \log (-((-1+x) x))\right )}{(1-x) (10+\log (3))}\right ) \, dx\\ &=\frac {\int \frac {e^{x^2} \log ((1-x) x) \left (2-4 x+\log (-((-1+x) x))-x \log (-((-1+x) x))+2 x^2 \log (-((-1+x) x))-2 x^3 \log (-((-1+x) x))\right )}{1-x} \, dx}{10+\log (3)}+\frac {\log (4) \int \frac {\log ((1-x) x) (2-4 x+\log (-((-1+x) x))-x \log (-((-1+x) x)))}{1-x} \, dx}{10+\log (3)}\\ &=\frac {\int \frac {e^{x^2} \log ((1-x) x) \left (2-4 x-\left (-1+x-2 x^2+2 x^3\right ) \log (-((-1+x) x))\right )}{1-x} \, dx}{10+\log (3)}+\frac {\log (4) \int \left (\frac {2 (1-2 x) \log ((1-x) x)}{1-x}+\log ^2((1-x) x)\right ) \, dx}{10+\log (3)}\\ &=\frac {\int \left (\frac {2 e^{x^2} (1-2 x) \log ((1-x) x)}{1-x}+e^{x^2} \left (1+2 x^2\right ) \log ^2((1-x) x)\right ) \, dx}{10+\log (3)}+\frac {\log (4) \int \log ^2((1-x) x) \, dx}{10+\log (3)}+\frac {(2 \log (4)) \int \frac {(1-2 x) \log ((1-x) x)}{1-x} \, dx}{10+\log (3)}\\ &=-\frac {(1-x) \log (4) \log ^2((1-x) x)}{10+\log (3)}+\frac {\int e^{x^2} \left (1+2 x^2\right ) \log ^2((1-x) x) \, dx}{10+\log (3)}+\frac {2 \int \frac {e^{x^2} (1-2 x) \log ((1-x) x)}{1-x} \, dx}{10+\log (3)}+\frac {(2 \log (4)) \int \frac {(1-2 x) \log (1-x)}{1-x} \, dx}{10+\log (3)}+\frac {(2 \log (4)) \int \frac {(1-2 x) \log (x)}{1-x} \, dx}{10+\log (3)}+\frac {(2 \log (4)) \int \frac {\log ((1-x) x)}{x} \, dx}{10+\log (3)}-\frac {(4 \log (4)) \int \log ((1-x) x) \, dx}{10+\log (3)}-\frac {(2 \log (4) (\log (1-x)+\log (x)-\log ((1-x) x))) \int \frac {1-2 x}{1-x} \, dx}{10+\log (3)}\\ &=\frac {2 \sqrt {\pi } \text {erfi}(x) \log ((1-x) x)}{10+\log (3)}+\frac {4 (1-x) \log (4) \log ((1-x) x)}{10+\log (3)}+\frac {2 \log (4) \log (x) \log ((1-x) x)}{10+\log (3)}-\frac {(1-x) \log (4) \log ^2((1-x) x)}{10+\log (3)}+\frac {\int \left (e^{x^2} \log ^2((1-x) x)+2 e^{x^2} x^2 \log ^2((1-x) x)\right ) \, dx}{10+\log (3)}-\frac {2 \int \frac {(1-2 x) \left (\sqrt {\pi } \text {erfi}(x)+\int \frac {e^{x^2}}{-1+x} \, dx\right )}{(1-x) x} \, dx}{10+\log (3)}+\frac {(2 \log (4)) \int \frac {\log (x)}{1-x} \, dx}{10+\log (3)}-\frac {(2 \log (4)) \int \frac {\log (x)}{x} \, dx}{10+\log (3)}+\frac {(2 \log (4)) \int \left (2 \log (x)+\frac {\log (x)}{-1+x}\right ) \, dx}{10+\log (3)}-\frac {(2 \log (4)) \text {Subst}\left (\int \frac {(-1+2 x) \log (x)}{x} \, dx,x,1-x\right )}{10+\log (3)}-\frac {(4 \log (4)) \int \frac {1}{x} \, dx}{10+\log (3)}+\frac {(8 \log (4)) \int 1 \, dx}{10+\log (3)}-\frac {(2 \log (4) (\log (1-x)+\log (x)-\log ((1-x) x))) \int \left (2+\frac {1}{-1+x}\right ) \, dx}{10+\log (3)}+\frac {(2 \log ((1-x) x)) \int \frac {e^{x^2}}{-1+x} \, dx}{10+\log (3)}\\ &=\frac {8 x \log (4)}{10+\log (3)}-\frac {4 \log (4) \log (x)}{10+\log (3)}-\frac {\log (4) \log ^2(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 (1-x) (\log (1-x)+\log (x)-\log ((1-x) x))}{10+\log (3)}+\frac {2 \sqrt {\pi } \text {erfi}(x) \log ((1-x) x)}{10+\log (3)}+\frac {4 (1-x) \log (4) \log ((1-x) x)}{10+\log (3)}+\frac {2 \log (4) \log (x) \log ((1-x) x)}{10+\log (3)}-\frac {(1-x) \log (4) \log ^2((1-x) x)}{10+\log (3)}+\frac {2 \log (4) \text {Li}_2(1-x)}{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 \int \left (\frac {\sqrt {\pi } (-1+2 x) \text {erfi}(x)}{(-1+x) x}+\frac {(-1+2 x) \int \frac {e^{x^2}}{-1+x} \, dx}{(-1+x) x}\right ) \, dx}{10+\log (3)}+\frac {(2 \log (4)) \int \frac {\log (x)}{-1+x} \, dx}{10+\log (3)}+\frac {(2 \log (4)) \text {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,1-x\right )}{10+\log (3)}+\frac {(4 \log (4)) \int \log (x) \, dx}{10+\log (3)}-\frac {(4 \log (4)) \text {Subst}(\int \log (x) \, dx,x,1-x)}{10+\log (3)}+\frac {(2 \log ((1-x) x)) \int \frac {e^{x^2}}{-1+x} \, dx}{10+\log (3)}\\ &=-\frac {4 (1-x) \log (4) \log (1-x)}{10+\log (3)}+\frac {\log (4) \log ^2(1-x)}{10+\log (3)}-\frac {4 \log (4) \log (x)}{10+\log (3)}+\frac {4 x \log (4) \log (x)}{10+\log (3)}-\frac {\log (4) \log ^2(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 (1-x) (\log (1-x)+\log (x)-\log ((1-x) x))}{10+\log (3)}+\frac {2 \sqrt {\pi } \text {erfi}(x) \log ((1-x) x)}{10+\log (3)}+\frac {4 (1-x) \log (4) \log ((1-x) x)}{10+\log (3)}+\frac {2 \log (4) \log (x) \log ((1-x) x)}{10+\log (3)}-\frac {(1-x) \log (4) \log ^2((1-x) x)}{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 \int \frac {(-1+2 x) \int \frac {e^{x^2}}{-1+x} \, dx}{(-1+x) x} \, dx}{10+\log (3)}-\frac {\left (2 \sqrt {\pi }\right ) \int \frac {(-1+2 x) \text {erfi}(x)}{(-1+x) x} \, dx}{10+\log (3)}+\frac {(2 \log ((1-x) x)) \int \frac {e^{x^2}}{-1+x} \, dx}{10+\log (3)}\\ &=-\frac {4 (1-x) \log (4) \log (1-x)}{10+\log (3)}+\frac {\log (4) \log ^2(1-x)}{10+\log (3)}-\frac {4 \log (4) \log (x)}{10+\log (3)}+\frac {4 x \log (4) \log (x)}{10+\log (3)}-\frac {\log (4) \log ^2(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 (1-x) (\log (1-x)+\log (x)-\log ((1-x) x))}{10+\log (3)}+\frac {2 \sqrt {\pi } \text {erfi}(x) \log ((1-x) x)}{10+\log (3)}+\frac {4 (1-x) \log (4) \log ((1-x) x)}{10+\log (3)}+\frac {2 \log (4) \log (x) \log ((1-x) x)}{10+\log (3)}-\frac {(1-x) \log (4) \log ^2((1-x) x)}{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 \int \left (\frac {\int \frac {e^{x^2}}{-1+x} \, dx}{-1+x}+\frac {\int \frac {e^{x^2}}{-1+x} \, dx}{x}\right ) \, dx}{10+\log (3)}-\frac {\left (2 \sqrt {\pi }\right ) \int \left (\frac {\text {erfi}(x)}{-1+x}+\frac {\text {erfi}(x)}{x}\right ) \, dx}{10+\log (3)}+\frac {(2 \log ((1-x) x)) \int \frac {e^{x^2}}{-1+x} \, dx}{10+\log (3)}\\ &=-\frac {4 (1-x) \log (4) \log (1-x)}{10+\log (3)}+\frac {\log (4) \log ^2(1-x)}{10+\log (3)}-\frac {4 \log (4) \log (x)}{10+\log (3)}+\frac {4 x \log (4) \log (x)}{10+\log (3)}-\frac {\log (4) \log ^2(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 (1-x) (\log (1-x)+\log (x)-\log ((1-x) x))}{10+\log (3)}+\frac {2 \sqrt {\pi } \text {erfi}(x) \log ((1-x) x)}{10+\log (3)}+\frac {4 (1-x) \log (4) \log ((1-x) x)}{10+\log (3)}+\frac {2 \log (4) \log (x) \log ((1-x) x)}{10+\log (3)}-\frac {(1-x) \log (4) \log ^2((1-x) x)}{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 \int \frac {\int \frac {e^{x^2}}{-1+x} \, dx}{-1+x} \, dx}{10+\log (3)}-\frac {2 \int \frac {\int \frac {e^{x^2}}{-1+x} \, dx}{x} \, dx}{10+\log (3)}-\frac {\left (2 \sqrt {\pi }\right ) \int \frac {\text {erfi}(x)}{-1+x} \, dx}{10+\log (3)}-\frac {\left (2 \sqrt {\pi }\right ) \int \frac {\text {erfi}(x)}{x} \, dx}{10+\log (3)}+\frac {(2 \log ((1-x) x)) \int \frac {e^{x^2}}{-1+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 {4 (1-x) \log (4) \log (1-x)}{10+\log (3)}+\frac {\log (4) \log ^2(1-x)}{10+\log (3)}-\frac {4 \log (4) \log (x)}{10+\log (3)}+\frac {4 x \log (4) \log (x)}{10+\log (3)}-\frac {\log (4) \log ^2(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 (1-x) (\log (1-x)+\log (x)-\log ((1-x) x))}{10+\log (3)}+\frac {2 \sqrt {\pi } \text {erfi}(x) \log ((1-x) x)}{10+\log (3)}+\frac {4 (1-x) \log (4) \log ((1-x) x)}{10+\log (3)}+\frac {2 \log (4) \log (x) \log ((1-x) x)}{10+\log (3)}-\frac {(1-x) \log (4) \log ^2((1-x) x)}{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 \int \frac {\int \frac {e^{x^2}}{-1+x} \, dx}{-1+x} \, dx}{10+\log (3)}-\frac {2 \int \frac {\int \frac {e^{x^2}}{-1+x} \, dx}{x} \, dx}{10+\log (3)}-\frac {\left (2 \sqrt {\pi }\right ) \int \frac {\text {erfi}(x)}{-1+x} \, dx}{10+\log (3)}+\frac {(2 \log ((1-x) x)) \int \frac {e^{x^2}}{-1+x} \, dx}{10+\log (3)}\\ \end {aligned} \end {gather*}

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Mathematica [F]
time = 0.14, size = 0, normalized size = 0.00 \begin {gather*} \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 \end {gather*}

Verification is not applicable to the result.

[In]

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)*Lo
g[4])*Log[x - x^2]^2)/(-10 + 10*x + (-1 + x)*Log[3]),x]

[Out]

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)*Lo
g[4])*Log[x - x^2]^2)/(-10 + 10*x + (-1 + x)*Log[3]), x]

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.52, size = 1460, normalized size = 56.15

method	result	size
risch	\(\text {Expression too large to display}\)	\(1460\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((2*x^3-2*x^2+x-1)*exp(x^2)+2*(x-1)*ln(2))*ln(-x^2+x)^2+((4*x-2)*exp(x^2)+2*(4*x-2)*ln(2))*ln(-x^2+x))/((
x-1)*ln(3)+10*x-10),x,method=_RETURNVERBOSE)

[Out]

x*(2*ln(2)+exp(x^2))/(10+ln(3))*ln(x-1)^2+x*(I*Pi*csgn(I*(x-1))*csgn(I*x*(x-1))^2*exp(x^2)+4*I*Pi*ln(2)+2*I*Pi
*exp(x^2)+2*I*Pi*ln(2)*csgn(I*x*(x-1))^2*csgn(I*x)-2*I*Pi*csgn(I*x*(x-1))^2*exp(x^2)+I*Pi*csgn(I*x*(x-1))^2*cs
gn(I*x)*exp(x^2)-2*I*Pi*ln(2)*csgn(I*(x-1))*csgn(I*x*(x-1))*csgn(I*x)-4*I*Pi*ln(2)*csgn(I*x*(x-1))^2+2*I*Pi*ln
(2)*csgn(I*(x-1))*csgn(I*x*(x-1))^2+I*Pi*csgn(I*x*(x-1))^3*exp(x^2)-I*Pi*csgn(I*(x-1))*csgn(I*x*(x-1))*csgn(I*
x)*exp(x^2)+2*I*Pi*ln(2)*csgn(I*x*(x-1))^3+4*ln(2)*ln(x)+2*exp(x^2)*ln(x))/(10+ln(3))*ln(x-1)+1/4*x*(8*ln(2)*l
n(x)^2-8*Pi^2*ln(2)*csgn(I*(x-1))*csgn(I*x*(x-1))^2-8*Pi^2*ln(2)*csgn(I*x*(x-1))^2*csgn(I*x)+4*Pi^2*csgn(I*x*(
x-1))^4*csgn(I*x)*exp(x^2)-2*Pi^2*ln(2)*csgn(I*(x-1))^2*csgn(I*x*(x-1))^4-4*Pi^2*ln(2)*csgn(I*(x-1))*csgn(I*x*
(x-1))^5-4*Pi^2*ln(2)*csgn(I*x*(x-1))^5*csgn(I*x)-Pi^2*csgn(I*(x-1))^2*csgn(I*x*(x-1))^4*exp(x^2)-2*Pi^2*csgn(
I*(x-1))*csgn(I*x*(x-1))^5*exp(x^2)-2*Pi^2*csgn(I*x*(x-1))^5*csgn(I*x)*exp(x^2)-Pi^2*csgn(I*x*(x-1))^4*csgn(I*
x)^2*exp(x^2)-4*Pi^2*csgn(I*(x-1))*csgn(I*x*(x-1))^2*exp(x^2)-4*Pi^2*csgn(I*x*(x-1))^2*csgn(I*x)*exp(x^2)-2*Pi
^2*ln(2)*csgn(I*x*(x-1))^4*csgn(I*x)^2+8*I*Pi*ln(2)*csgn(I*x*(x-1))^2*csgn(I*x)*ln(x)+4*I*Pi*csgn(I*x*(x-1))^2
*csgn(I*x)*exp(x^2)*ln(x)+8*I*Pi*ln(2)*csgn(I*x*(x-1))^3*ln(x)+4*I*Pi*csgn(I*x*(x-1))^3*exp(x^2)*ln(x)+8*I*Pi*
ln(2)*csgn(I*(x-1))*csgn(I*x*(x-1))^2*ln(x)+4*I*Pi*csgn(I*(x-1))*csgn(I*x*(x-1))^2*exp(x^2)*ln(x)+4*Pi^2*ln(2)
*csgn(I*(x-1))^2*csgn(I*x*(x-1))^3*csgn(I*x)-Pi^2*csgn(I*(x-1))^2*csgn(I*x*(x-1))^2*csgn(I*x)^2*exp(x^2)+2*Pi^
2*csgn(I*(x-1))*csgn(I*x*(x-1))^3*csgn(I*x)^2*exp(x^2)-8*Pi^2*ln(2)*csgn(I*(x-1))*csgn(I*x*(x-1))^3*csgn(I*x)-
4*Pi^2*csgn(I*(x-1))*csgn(I*x*(x-1))^3*csgn(I*x)*exp(x^2)-8*I*Pi*ln(2)*csgn(I*(x-1))*csgn(I*x*(x-1))*csgn(I*x)
*ln(x)-4*I*Pi*csgn(I*(x-1))*csgn(I*x*(x-1))*csgn(I*x)*exp(x^2)*ln(x)+4*Pi^2*csgn(I*(x-1))*csgn(I*x*(x-1))^4*ex
p(x^2)+8*Pi^2*ln(2)*csgn(I*x*(x-1))^4*csgn(I*x)+8*Pi^2*ln(2)*csgn(I*(x-1))*csgn(I*x*(x-1))^4+8*Pi^2*ln(2)*csgn
(I*(x-1))*csgn(I*x*(x-1))*csgn(I*x)-16*I*Pi*ln(2)*csgn(I*x*(x-1))^2*ln(x)-8*I*Pi*csgn(I*x*(x-1))^2*exp(x^2)*ln
(x)-2*Pi^2*ln(2)*csgn(I*x*(x-1))^6-Pi^2*csgn(I*x*(x-1))^6*exp(x^2)-4*Pi^2*csgn(I*x*(x-1))^4*exp(x^2)+4*Pi^2*cs
gn(I*x*(x-1))^5*exp(x^2)-4*Pi^2*csgn(I*x*(x-1))^3*exp(x^2)-8*Pi^2*ln(2)*csgn(I*x*(x-1))^4+8*Pi^2*ln(2)*csgn(I*
x*(x-1))^5-8*Pi^2*ln(2)*csgn(I*x*(x-1))^3+16*I*Pi*ln(2)*ln(x)+8*I*Pi*exp(x^2)*ln(x)-8*Pi^2*ln(2)+4*exp(x^2)*ln
(x)^2-4*Pi^2*exp(x^2)+8*Pi^2*csgn(I*x*(x-1))^2*exp(x^2)+16*Pi^2*ln(2)*csgn(I*x*(x-1))^2+4*Pi^2*csgn(I*(x-1))*c
sgn(I*x*(x-1))*csgn(I*x)*exp(x^2)+2*Pi^2*csgn(I*(x-1))^2*csgn(I*x*(x-1))^3*csgn(I*x)*exp(x^2)-2*Pi^2*ln(2)*csg
n(I*(x-1))^2*csgn(I*x*(x-1))^2*csgn(I*x)^2+4*Pi^2*ln(2)*csgn(I*(x-1))*csgn(I*x*(x-1))^3*csgn(I*x)^2)/(10+ln(3)
)

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 72 vs. \(2 (27) = 54\).
time = 0.50, size = 72, normalized size = 2.77 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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, algorithm="maxima")

[Out]

(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)

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Fricas [A]
time = 0.36, size = 29, normalized size = 1.12 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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, algorithm="fricas")

[Out]

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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))*l
n(-x**2+x))/((-1+x)*ln(3)+10*x-10),x)

[Out]

Timed out

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Giac [A]
time = 0.42, size = 39, normalized size = 1.50 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \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 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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)

[Out]

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)

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