Optimal. Leaf size=20 \[ -x+x^3 \log ^2(1-x+\log (4+x)) \]
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Rubi [F] time = 1.69, 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 {-4+3 x+x^2+(-4-x) \log (4+x)+\left (-6 x^3-2 x^4\right ) \log (1-x+\log (4+x))+\left (12 x^2-9 x^3-3 x^4+\left (12 x^2+3 x^3\right ) \log (4+x)\right ) \log ^2(1-x+\log (4+x))}{4-3 x-x^2+(4+x) \log (4+x)} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-4+3 x+x^2+(-4-x) \log (4+x)+\left (-6 x^3-2 x^4\right ) \log (1-x+\log (4+x))+\left (12 x^2-9 x^3-3 x^4+\left (12 x^2+3 x^3\right ) \log (4+x)\right ) \log ^2(1-x+\log (4+x))}{(4+x) (1-x+\log (4+x))} \, dx\\ &=\int \left (\frac {4}{(4+x) (-1+x-\log (4+x))}-\frac {3 x}{(4+x) (-1+x-\log (4+x))}-\frac {x^2}{(4+x) (-1+x-\log (4+x))}+\frac {\log (4+x)}{-1+x-\log (4+x)}+\frac {2 x^3 (3+x) \log (1-x+\log (4+x))}{(4+x) (-1+x-\log (4+x))}+3 x^2 \log ^2(1-x+\log (4+x))\right ) \, dx\\ &=2 \int \frac {x^3 (3+x) \log (1-x+\log (4+x))}{(4+x) (-1+x-\log (4+x))} \, dx-3 \int \frac {x}{(4+x) (-1+x-\log (4+x))} \, dx+3 \int x^2 \log ^2(1-x+\log (4+x)) \, dx+4 \int \frac {1}{(4+x) (-1+x-\log (4+x))} \, dx-\int \frac {x^2}{(4+x) (-1+x-\log (4+x))} \, dx+\int \frac {\log (4+x)}{-1+x-\log (4+x)} \, dx\\ &=2 \int \left (-\frac {16 \log (1-x+\log (4+x))}{-1+x-\log (4+x)}+\frac {4 x \log (1-x+\log (4+x))}{-1+x-\log (4+x)}-\frac {x^2 \log (1-x+\log (4+x))}{-1+x-\log (4+x)}+\frac {x^3 \log (1-x+\log (4+x))}{-1+x-\log (4+x)}+\frac {64 \log (1-x+\log (4+x))}{(4+x) (-1+x-\log (4+x))}\right ) \, dx-3 \int \left (\frac {1}{-1+x-\log (4+x)}-\frac {4}{(4+x) (-1+x-\log (4+x))}\right ) \, dx+3 \int x^2 \log ^2(1-x+\log (4+x)) \, dx+4 \int \frac {1}{(4+x) (-1+x-\log (4+x))} \, dx+\int \left (-1+\frac {-1+x}{-1+x-\log (4+x)}\right ) \, dx-\int \left (-\frac {4}{-1+x-\log (4+x)}+\frac {x}{-1+x-\log (4+x)}+\frac {16}{(4+x) (-1+x-\log (4+x))}\right ) \, dx\\ &=-x-2 \int \frac {x^2 \log (1-x+\log (4+x))}{-1+x-\log (4+x)} \, dx+2 \int \frac {x^3 \log (1-x+\log (4+x))}{-1+x-\log (4+x)} \, dx-3 \int \frac {1}{-1+x-\log (4+x)} \, dx+3 \int x^2 \log ^2(1-x+\log (4+x)) \, dx+4 \int \frac {1}{-1+x-\log (4+x)} \, dx+4 \int \frac {1}{(4+x) (-1+x-\log (4+x))} \, dx+8 \int \frac {x \log (1-x+\log (4+x))}{-1+x-\log (4+x)} \, dx+12 \int \frac {1}{(4+x) (-1+x-\log (4+x))} \, dx-16 \int \frac {1}{(4+x) (-1+x-\log (4+x))} \, dx-32 \int \frac {\log (1-x+\log (4+x))}{-1+x-\log (4+x)} \, dx+128 \int \frac {\log (1-x+\log (4+x))}{(4+x) (-1+x-\log (4+x))} \, dx+\int \frac {-1+x}{-1+x-\log (4+x)} \, dx-\int \frac {x}{-1+x-\log (4+x)} \, dx\\ &=-x-2 \int \frac {x^2 \log (1-x+\log (4+x))}{-1+x-\log (4+x)} \, dx+2 \int \frac {x^3 \log (1-x+\log (4+x))}{-1+x-\log (4+x)} \, dx-3 \int \frac {1}{-1+x-\log (4+x)} \, dx+3 \int x^2 \log ^2(1-x+\log (4+x)) \, dx+4 \int \frac {1}{-1+x-\log (4+x)} \, dx+4 \int \frac {1}{(4+x) (-1+x-\log (4+x))} \, dx+8 \int \frac {x \log (1-x+\log (4+x))}{-1+x-\log (4+x)} \, dx+12 \int \frac {1}{(4+x) (-1+x-\log (4+x))} \, dx-16 \int \frac {1}{(4+x) (-1+x-\log (4+x))} \, dx-32 \int \frac {\log (1-x+\log (4+x))}{-1+x-\log (4+x)} \, dx+128 \int \frac {\log (1-x+\log (4+x))}{(4+x) (-1+x-\log (4+x))} \, dx-\int \frac {x}{-1+x-\log (4+x)} \, dx+\int \left (\frac {x}{-1+x-\log (4+x)}+\frac {1}{1-x+\log (4+x)}\right ) \, dx\\ &=-x-2 \int \frac {x^2 \log (1-x+\log (4+x))}{-1+x-\log (4+x)} \, dx+2 \int \frac {x^3 \log (1-x+\log (4+x))}{-1+x-\log (4+x)} \, dx-3 \int \frac {1}{-1+x-\log (4+x)} \, dx+3 \int x^2 \log ^2(1-x+\log (4+x)) \, dx+4 \int \frac {1}{-1+x-\log (4+x)} \, dx+4 \int \frac {1}{(4+x) (-1+x-\log (4+x))} \, dx+8 \int \frac {x \log (1-x+\log (4+x))}{-1+x-\log (4+x)} \, dx+12 \int \frac {1}{(4+x) (-1+x-\log (4+x))} \, dx-16 \int \frac {1}{(4+x) (-1+x-\log (4+x))} \, dx-32 \int \frac {\log (1-x+\log (4+x))}{-1+x-\log (4+x)} \, dx+128 \int \frac {\log (1-x+\log (4+x))}{(4+x) (-1+x-\log (4+x))} \, dx+\int \frac {1}{1-x+\log (4+x)} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [F] time = 0.57, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {-4+3 x+x^2+(-4-x) \log (4+x)+\left (-6 x^3-2 x^4\right ) \log (1-x+\log (4+x))+\left (12 x^2-9 x^3-3 x^4+\left (12 x^2+3 x^3\right ) \log (4+x)\right ) \log ^2(1-x+\log (4+x))}{4-3 x-x^2+(4+x) \log (4+x)} \, dx \end {gather*}
Verification is not applicable to the result.
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fricas [A] time = 0.68, size = 20, normalized size = 1.00 \begin {gather*} x^{3} \log \left (-x + \log \left (x + 4\right ) + 1\right )^{2} - x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.27, size = 20, normalized size = 1.00 \begin {gather*} x^{3} \log \left (-x + \log \left (x + 4\right ) + 1\right )^{2} - x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.04, size = 21, normalized size = 1.05
method | result | size |
risch | \(x^{3} \ln \left (\ln \left (4+x \right )-x +1\right )^{2}-x\) | \(21\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.40, size = 20, normalized size = 1.00 \begin {gather*} x^{3} \log \left (-x + \log \left (x + 4\right ) + 1\right )^{2} - x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.26, size = 20, normalized size = 1.00 \begin {gather*} x^3\,{\ln \left (\ln \left (x+4\right )-x+1\right )}^2-x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.65, size = 15, normalized size = 0.75 \begin {gather*} x^{3} \log {\left (- x + \log {\left (x + 4 \right )} + 1 \right )}^{2} - x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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