Optimal. Leaf size=26 \[ \frac {x \left (e^{x^2}+\log (4)\right ) \log ^2\left (x-x^2\right )}{10+\log (3)} \]
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Rubi [F] time = 3.92, 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.
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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)) \operatorname {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)) \operatorname {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)) \operatorname {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 = 0.55, 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.
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fricas [A] time = 1.29, size = 29, normalized size = 1.12 \begin {gather*} \frac {{\left (x e^{\left (x^{2}\right )} + 2 \, x \log \relax (2)\right )} \log \left (-x^{2} + x\right )^{2}}{\log \relax (3) + 10} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.27, 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 \relax (2) \log \left (-x^{2} + x\right )^{2}}{\log \relax (3) + 10} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.57, size = 1460, normalized size = 56.15
method | result | size |
risch | \(\frac {x \left (2 \ln \relax (2)+{\mathrm e}^{x^{2}}\right ) \ln \left (x -1\right )^{2}}{10+\ln \relax (3)}+\frac {x \left (4 i \pi \ln \relax (2)+i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \left (x -1\right )\right )^{2} {\mathrm e}^{x^{2}}-i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i \left (x -1\right )\right ) \mathrm {csgn}\left (i x \left (x -1\right )\right ) {\mathrm e}^{x^{2}}+2 i \pi \ln \relax (2) \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \left (x -1\right )\right )^{2}+2 i \pi \ln \relax (2) \mathrm {csgn}\left (i x \left (x -1\right )\right )^{3}+i \pi \mathrm {csgn}\left (i x \left (x -1\right )\right )^{3} {\mathrm e}^{x^{2}}+2 i \pi \ln \relax (2) \mathrm {csgn}\left (i \left (x -1\right )\right ) \mathrm {csgn}\left (i x \left (x -1\right )\right )^{2}+2 i \pi \,{\mathrm e}^{x^{2}}-2 i \pi \mathrm {csgn}\left (i x \left (x -1\right )\right )^{2} {\mathrm e}^{x^{2}}-2 i \pi \ln \relax (2) \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i \left (x -1\right )\right ) \mathrm {csgn}\left (i x \left (x -1\right )\right )+i \pi \,\mathrm {csgn}\left (i \left (x -1\right )\right ) \mathrm {csgn}\left (i x \left (x -1\right )\right )^{2} {\mathrm e}^{x^{2}}-4 i \pi \ln \relax (2) \mathrm {csgn}\left (i x \left (x -1\right )\right )^{2}+4 \ln \relax (2) \ln \relax (x )+2 \,{\mathrm e}^{x^{2}} \ln \relax (x )\right ) \ln \left (x -1\right )}{10+\ln \relax (3)}+\frac {x \left (8 \ln \relax (2) \ln \relax (x )^{2}+8 \pi ^{2} \ln \relax (2) \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \left (x -1\right )\right )^{4}+4 \pi ^{2} \mathrm {csgn}\left (i \left (x -1\right )\right ) \mathrm {csgn}\left (i x \left (x -1\right )\right )^{4} {\mathrm e}^{x^{2}}+4 \pi ^{2} \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \left (x -1\right )\right )^{4} {\mathrm e}^{x^{2}}+16 i \pi \ln \relax (2) \ln \relax (x )+8 i \pi \,{\mathrm e}^{x^{2}} \ln \relax (x )-\pi ^{2} \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x \left (x -1\right )\right )^{4} {\mathrm e}^{x^{2}}-2 \pi ^{2} \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \left (x -1\right )\right )^{5} {\mathrm e}^{x^{2}}-\pi ^{2} \mathrm {csgn}\left (i \left (x -1\right )\right )^{2} \mathrm {csgn}\left (i x \left (x -1\right )\right )^{4} {\mathrm e}^{x^{2}}-2 \pi ^{2} \mathrm {csgn}\left (i \left (x -1\right )\right ) \mathrm {csgn}\left (i x \left (x -1\right )\right )^{5} {\mathrm e}^{x^{2}}-4 \pi ^{2} \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \left (x -1\right )\right )^{2} {\mathrm e}^{x^{2}}-8 \pi ^{2} \ln \relax (2) \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \left (x -1\right )\right )^{2}-8 \pi ^{2} \ln \relax (2) \mathrm {csgn}\left (i \left (x -1\right )\right ) \mathrm {csgn}\left (i x \left (x -1\right )\right )^{2}-4 \pi ^{2} \mathrm {csgn}\left (i \left (x -1\right )\right ) \mathrm {csgn}\left (i x \left (x -1\right )\right )^{2} {\mathrm e}^{x^{2}}-2 \pi ^{2} \ln \relax (2) \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x \left (x -1\right )\right )^{4}-4 \pi ^{2} \ln \relax (2) \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \left (x -1\right )\right )^{5}-2 \pi ^{2} \ln \relax (2) \mathrm {csgn}\left (i \left (x -1\right )\right )^{2} \mathrm {csgn}\left (i x \left (x -1\right )\right )^{4}-4 \pi ^{2} \ln \relax (2) \mathrm {csgn}\left (i \left (x -1\right )\right ) \mathrm {csgn}\left (i x \left (x -1\right )\right )^{5}+8 \pi ^{2} \ln \relax (2) \mathrm {csgn}\left (i \left (x -1\right )\right ) \mathrm {csgn}\left (i x \left (x -1\right )\right )^{4}-4 i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i \left (x -1\right )\right ) \mathrm {csgn}\left (i x \left (x -1\right )\right ) {\mathrm e}^{x^{2}} \ln \relax (x )-8 \pi ^{2} \ln \relax (2)+4 \,{\mathrm e}^{x^{2}} \ln \relax (x )^{2}-4 \pi ^{2} {\mathrm e}^{x^{2}}+8 i \pi \ln \relax (2) \mathrm {csgn}\left (i \left (x -1\right )\right ) \mathrm {csgn}\left (i x \left (x -1\right )\right )^{2} \ln \relax (x )-8 \pi ^{2} \ln \relax (2) \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i \left (x -1\right )\right ) \mathrm {csgn}\left (i x \left (x -1\right )\right )^{3}-4 \pi ^{2} \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i \left (x -1\right )\right ) \mathrm {csgn}\left (i x \left (x -1\right )\right )^{3} {\mathrm e}^{x^{2}}+8 i \pi \ln \relax (2) \mathrm {csgn}\left (i x \left (x -1\right )\right )^{3} \ln \relax (x )+4 i \pi \mathrm {csgn}\left (i x \left (x -1\right )\right )^{3} {\mathrm e}^{x^{2}} \ln \relax (x )+8 \pi ^{2} \ln \relax (2) \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i \left (x -1\right )\right ) \mathrm {csgn}\left (i x \left (x -1\right )\right )-2 \pi ^{2} \ln \relax (2) \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i \left (x -1\right )\right )^{2} \mathrm {csgn}\left (i x \left (x -1\right )\right )^{2}+4 \pi ^{2} \ln \relax (2) \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i \left (x -1\right )\right ) \mathrm {csgn}\left (i x \left (x -1\right )\right )^{3}+4 \pi ^{2} \ln \relax (2) \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i \left (x -1\right )\right )^{2} \mathrm {csgn}\left (i x \left (x -1\right )\right )^{3}+4 \pi ^{2} \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i \left (x -1\right )\right ) \mathrm {csgn}\left (i x \left (x -1\right )\right ) {\mathrm e}^{x^{2}}-\pi ^{2} \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i \left (x -1\right )\right )^{2} \mathrm {csgn}\left (i x \left (x -1\right )\right )^{2} {\mathrm e}^{x^{2}}+2 \pi ^{2} \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i \left (x -1\right )\right ) \mathrm {csgn}\left (i x \left (x -1\right )\right )^{3} {\mathrm e}^{x^{2}}+2 \pi ^{2} \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i \left (x -1\right )\right )^{2} \mathrm {csgn}\left (i x \left (x -1\right )\right )^{3} {\mathrm e}^{x^{2}}-16 i \pi \ln \relax (2) \mathrm {csgn}\left (i x \left (x -1\right )\right )^{2} \ln \relax (x )-8 i \pi \mathrm {csgn}\left (i x \left (x -1\right )\right )^{2} {\mathrm e}^{x^{2}} \ln \relax (x )+16 \pi ^{2} \ln \relax (2) \mathrm {csgn}\left (i x \left (x -1\right )\right )^{2}+8 \pi ^{2} \mathrm {csgn}\left (i x \left (x -1\right )\right )^{2} {\mathrm e}^{x^{2}}-8 \pi ^{2} \ln \relax (2) \mathrm {csgn}\left (i x \left (x -1\right )\right )^{3}-2 \pi ^{2} \ln \relax (2) \mathrm {csgn}\left (i x \left (x -1\right )\right )^{6}-8 \pi ^{2} \ln \relax (2) \mathrm {csgn}\left (i x \left (x -1\right )\right )^{4}+8 \pi ^{2} \ln \relax (2) \mathrm {csgn}\left (i x \left (x -1\right )\right )^{5}-4 \pi ^{2} \mathrm {csgn}\left (i x \left (x -1\right )\right )^{3} {\mathrm e}^{x^{2}}-\pi ^{2} \mathrm {csgn}\left (i x \left (x -1\right )\right )^{6} {\mathrm e}^{x^{2}}-4 \pi ^{2} \mathrm {csgn}\left (i x \left (x -1\right )\right )^{4} {\mathrm e}^{x^{2}}+4 \pi ^{2} \mathrm {csgn}\left (i x \left (x -1\right )\right )^{5} {\mathrm e}^{x^{2}}-8 i \pi \ln \relax (2) \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i \left (x -1\right )\right ) \mathrm {csgn}\left (i x \left (x -1\right )\right ) \ln \relax (x )+4 i \pi \,\mathrm {csgn}\left (i \left (x -1\right )\right ) \mathrm {csgn}\left (i x \left (x -1\right )\right )^{2} {\mathrm e}^{x^{2}} \ln \relax (x )+4 i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \left (x -1\right )\right )^{2} {\mathrm e}^{x^{2}} \ln \relax (x )+8 i \pi \ln \relax (2) \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \left (x -1\right )\right )^{2} \ln \relax (x )\right )}{40+4 \ln \relax (3)}\) | \(1460\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.58, size = 72, normalized size = 2.77 \begin {gather*} \frac {x e^{\left (x^{2}\right )} \log \relax (x)^{2} + 2 \, x \log \relax (2) \log \relax (x)^{2} + {\left (x e^{\left (x^{2}\right )} + 2 \, x \log \relax (2)\right )} \log \left (-x + 1\right )^{2} + 2 \, {\left (x e^{\left (x^{2}\right )} \log \relax (x) + 2 \, x \log \relax (2) \log \relax (x)\right )} \log \left (-x + 1\right )}{\log \relax (3) + 10} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {\left (2\,\ln \relax (2)\,\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 \relax (2)\,\left (4\,x-2\right )+{\mathrm {e}}^{x^2}\,\left (4\,x-2\right )\right )\,\ln \left (x-x^2\right )}{10\,x+\ln \relax (3)\,\left (x-1\right )-10} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] 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.
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