Optimal. Leaf size=25 \[ 2 x+\frac {\left (\frac {1}{e^5}+3 (2+x+\log (-4+x))\right )^2 \log (x)}{x^3} \]
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Rubi [F] time = 17.46, 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+x+e^5 \left (-48-12 x+6 x^2\right )+e^{10} \left (-144-108 x+9 x^3-8 x^4+2 x^5\right )+\left (e^5 (-24+6 x)+e^{10} \left (-144-36 x+18 x^2\right )\right ) \log (-4+x)+e^{10} (-36+9 x) \log ^2(-4+x)+\left (12-3 x+e^5 \left (144+18 x-12 x^2\right )+e^{10} \left (432+216 x-18 x^2-9 x^3\right )+\left (e^5 (72-18 x)+e^{10} \left (432+54 x-36 x^2\right )\right ) \log (-4+x)+e^{10} (108-27 x) \log ^2(-4+x)\right ) \log (x)}{e^{10} \left (-4 x^4+x^5\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {-4+x+e^5 \left (-48-12 x+6 x^2\right )+e^{10} \left (-144-108 x+9 x^3-8 x^4+2 x^5\right )+\left (e^5 (-24+6 x)+e^{10} \left (-144-36 x+18 x^2\right )\right ) \log (-4+x)+e^{10} (-36+9 x) \log ^2(-4+x)+\left (12-3 x+e^5 \left (144+18 x-12 x^2\right )+e^{10} \left (432+216 x-18 x^2-9 x^3\right )+\left (e^5 (72-18 x)+e^{10} \left (432+54 x-36 x^2\right )\right ) \log (-4+x)+e^{10} (108-27 x) \log ^2(-4+x)\right ) \log (x)}{-4 x^4+x^5} \, dx}{e^{10}}\\ &=\frac {\int \frac {-4+x+e^5 \left (-48-12 x+6 x^2\right )+e^{10} \left (-144-108 x+9 x^3-8 x^4+2 x^5\right )+\left (e^5 (-24+6 x)+e^{10} \left (-144-36 x+18 x^2\right )\right ) \log (-4+x)+e^{10} (-36+9 x) \log ^2(-4+x)+\left (12-3 x+e^5 \left (144+18 x-12 x^2\right )+e^{10} \left (432+216 x-18 x^2-9 x^3\right )+\left (e^5 (72-18 x)+e^{10} \left (432+54 x-36 x^2\right )\right ) \log (-4+x)+e^{10} (108-27 x) \log ^2(-4+x)\right ) \log (x)}{(-4+x) x^4} \, dx}{e^{10}}\\ &=\frac {\int \left (\frac {1+12 e^5 \left (1+3 e^5\right )+6 e^5 \left (1+6 e^5\right ) x+9 e^{10} x^2+2 e^{10} x^4+6 e^5 \left (1+6 e^5\right ) \log (-4+x)+18 e^{10} x \log (-4+x)+9 e^{10} \log ^2(-4+x)}{x^4}+\frac {3 \left (1+6 e^5+3 e^5 x+3 e^5 \log (-4+x)\right ) \left (-4 \left (1+6 e^5\right )+x+e^5 x^2-12 e^5 \log (-4+x)+3 e^5 x \log (-4+x)\right ) \log (x)}{(4-x) x^4}\right ) \, dx}{e^{10}}\\ &=\frac {\int \frac {1+12 e^5 \left (1+3 e^5\right )+6 e^5 \left (1+6 e^5\right ) x+9 e^{10} x^2+2 e^{10} x^4+6 e^5 \left (1+6 e^5\right ) \log (-4+x)+18 e^{10} x \log (-4+x)+9 e^{10} \log ^2(-4+x)}{x^4} \, dx}{e^{10}}+\frac {3 \int \frac {\left (1+6 e^5+3 e^5 x+3 e^5 \log (-4+x)\right ) \left (-4 \left (1+6 e^5\right )+x+e^5 x^2-12 e^5 \log (-4+x)+3 e^5 x \log (-4+x)\right ) \log (x)}{(4-x) x^4} \, dx}{e^{10}}\\ &=\frac {\int \left (\frac {\left (1+6 e^5\right )^2+6 e^5 \left (1+6 e^5\right ) x+9 e^{10} x^2+2 e^{10} x^4}{x^4}+\frac {6 e^5 \left (1+6 e^5+3 e^5 x\right ) \log (-4+x)}{x^4}+\frac {9 e^{10} \log ^2(-4+x)}{x^4}\right ) \, dx}{e^{10}}+\frac {3 \int \frac {\left (1+3 e^5 (2+x)+3 e^5 \log (-4+x)\right ) \left (-4+x+e^5 \left (-24+x^2\right )+3 e^5 (-4+x) \log (-4+x)\right ) \log (x)}{(4-x) x^4} \, dx}{e^{10}}\\ &=9 \int \frac {\log ^2(-4+x)}{x^4} \, dx+\frac {\int \frac {\left (1+6 e^5\right )^2+6 e^5 \left (1+6 e^5\right ) x+9 e^{10} x^2+2 e^{10} x^4}{x^4} \, dx}{e^{10}}+\frac {3 \int \left (\frac {\left (1+6 e^5+3 e^5 x+3 e^5 \log (-4+x)\right ) \left (-4 \left (1+6 e^5\right )+x+e^5 x^2-12 e^5 \log (-4+x)+3 e^5 x \log (-4+x)\right ) \log (x)}{256 (4-x)}+\frac {\left (1+6 e^5+3 e^5 x+3 e^5 \log (-4+x)\right ) \left (-4 \left (1+6 e^5\right )+x+e^5 x^2-12 e^5 \log (-4+x)+3 e^5 x \log (-4+x)\right ) \log (x)}{4 x^4}+\frac {\left (1+6 e^5+3 e^5 x+3 e^5 \log (-4+x)\right ) \left (-4 \left (1+6 e^5\right )+x+e^5 x^2-12 e^5 \log (-4+x)+3 e^5 x \log (-4+x)\right ) \log (x)}{16 x^3}+\frac {\left (1+6 e^5+3 e^5 x+3 e^5 \log (-4+x)\right ) \left (-4 \left (1+6 e^5\right )+x+e^5 x^2-12 e^5 \log (-4+x)+3 e^5 x \log (-4+x)\right ) \log (x)}{64 x^2}+\frac {\left (1+6 e^5+3 e^5 x+3 e^5 \log (-4+x)\right ) \left (-4 \left (1+6 e^5\right )+x+e^5 x^2-12 e^5 \log (-4+x)+3 e^5 x \log (-4+x)\right ) \log (x)}{256 x}\right ) \, dx}{e^{10}}+\frac {6 \int \frac {\left (1+6 e^5+3 e^5 x\right ) \log (-4+x)}{x^4} \, dx}{e^5}\\ &=-\frac {2 \left (6+\frac {1}{e^5}\right ) \log (-4+x)}{x^3}-\frac {9 \log (-4+x)}{x^2}-\frac {3 \log ^2(-4+x)}{x^3}+6 \int \frac {\log (-4+x)}{(-4+x) x^3} \, dx+\frac {3 \int \frac {\left (1+6 e^5+3 e^5 x+3 e^5 \log (-4+x)\right ) \left (-4 \left (1+6 e^5\right )+x+e^5 x^2-12 e^5 \log (-4+x)+3 e^5 x \log (-4+x)\right ) \log (x)}{4-x} \, dx}{256 e^{10}}+\frac {3 \int \frac {\left (1+6 e^5+3 e^5 x+3 e^5 \log (-4+x)\right ) \left (-4 \left (1+6 e^5\right )+x+e^5 x^2-12 e^5 \log (-4+x)+3 e^5 x \log (-4+x)\right ) \log (x)}{x} \, dx}{256 e^{10}}+\frac {3 \int \frac {\left (1+6 e^5+3 e^5 x+3 e^5 \log (-4+x)\right ) \left (-4 \left (1+6 e^5\right )+x+e^5 x^2-12 e^5 \log (-4+x)+3 e^5 x \log (-4+x)\right ) \log (x)}{x^2} \, dx}{64 e^{10}}+\frac {3 \int \frac {\left (1+6 e^5+3 e^5 x+3 e^5 \log (-4+x)\right ) \left (-4 \left (1+6 e^5\right )+x+e^5 x^2-12 e^5 \log (-4+x)+3 e^5 x \log (-4+x)\right ) \log (x)}{x^3} \, dx}{16 e^{10}}+\frac {3 \int \frac {\left (1+6 e^5+3 e^5 x+3 e^5 \log (-4+x)\right ) \left (-4 \left (1+6 e^5\right )+x+e^5 x^2-12 e^5 \log (-4+x)+3 e^5 x \log (-4+x)\right ) \log (x)}{x^4} \, dx}{4 e^{10}}+\frac {\int \left (2 e^{10}+\frac {\left (1+6 e^5\right )^2}{x^4}+\frac {6 e^5 \left (1+6 e^5\right )}{x^3}+\frac {9 e^{10}}{x^2}\right ) \, dx}{e^{10}}-\frac {6 \int \frac {2 \left (1+6 e^5\right )+9 e^5 x}{6 (4-x) x^3} \, dx}{e^5}\\ &=-\frac {\left (1+6 e^5\right )^2}{3 e^{10} x^3}-\frac {3 \left (6+\frac {1}{e^5}\right )}{x^2}-\frac {9}{x}+2 x-\frac {2 \left (6+\frac {1}{e^5}\right ) \log (-4+x)}{x^3}-\frac {9 \log (-4+x)}{x^2}-\frac {3 \log ^2(-4+x)}{x^3}+6 \operatorname {Subst}\left (\int \frac {\log (x)}{x (4+x)^3} \, dx,x,-4+x\right )+\frac {3 \int \frac {\left (1+3 e^5 (2+x)+3 e^5 \log (-4+x)\right ) \left (-4+x+e^5 \left (-24+x^2\right )+3 e^5 (-4+x) \log (-4+x)\right ) \log (x)}{4-x} \, dx}{256 e^{10}}+\frac {3 \int \frac {\left (1+3 e^5 (2+x)+3 e^5 \log (-4+x)\right ) \left (-4+x+e^5 \left (-24+x^2\right )+3 e^5 (-4+x) \log (-4+x)\right ) \log (x)}{x} \, dx}{256 e^{10}}+\frac {3 \int \frac {\left (1+3 e^5 (2+x)+3 e^5 \log (-4+x)\right ) \left (-4+x+e^5 \left (-24+x^2\right )+3 e^5 (-4+x) \log (-4+x)\right ) \log (x)}{x^2} \, dx}{64 e^{10}}+\frac {3 \int \frac {\left (1+3 e^5 (2+x)+3 e^5 \log (-4+x)\right ) \left (-4+x+e^5 \left (-24+x^2\right )+3 e^5 (-4+x) \log (-4+x)\right ) \log (x)}{x^3} \, dx}{16 e^{10}}+\frac {3 \int \frac {\left (1+3 e^5 (2+x)+3 e^5 \log (-4+x)\right ) \left (-4+x+e^5 \left (-24+x^2\right )+3 e^5 (-4+x) \log (-4+x)\right ) \log (x)}{x^4} \, dx}{4 e^{10}}-\frac {\int \frac {2 \left (1+6 e^5\right )+9 e^5 x}{(4-x) x^3} \, dx}{e^5}\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}
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Mathematica [A] time = 0.18, size = 40, normalized size = 1.60 \begin {gather*} -\frac {-2 e^{10} x-\frac {\left (1+3 e^5 (2+x)+3 e^5 \log (-4+x)\right )^2 \log (x)}{x^3}}{e^{10}} \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.70, size = 64, normalized size = 2.56 \begin {gather*} \frac {{\left (2 \, x^{4} e^{10} + {\left (9 \, e^{10} \log \left (x - 4\right )^{2} + 9 \, {\left (x^{2} + 4 \, x + 4\right )} e^{10} + 6 \, {\left (x + 2\right )} e^{5} + 6 \, {\left (3 \, {\left (x + 2\right )} e^{10} + e^{5}\right )} \log \left (x - 4\right ) + 1\right )} \log \relax (x)\right )} e^{\left (-10\right )}}{x^{3}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.20, size = 94, normalized size = 3.76 \begin {gather*} \frac {{\left (2 \, x^{4} e^{10} + 9 \, x^{2} e^{10} \log \relax (x) + 18 \, x e^{10} \log \left (x - 4\right ) \log \relax (x) + 9 \, e^{10} \log \left (x - 4\right )^{2} \log \relax (x) + 36 \, x e^{10} \log \relax (x) + 6 \, x e^{5} \log \relax (x) + 36 \, e^{10} \log \left (x - 4\right ) \log \relax (x) + 6 \, e^{5} \log \left (x - 4\right ) \log \relax (x) + 36 \, e^{10} \log \relax (x) + 12 \, e^{5} \log \relax (x) + \log \relax (x)\right )} e^{\left (-10\right )}}{x^{3}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.33, size = 90, normalized size = 3.60
method | result | size |
risch | \(\frac {9 \ln \relax (x ) \ln \left (x -4\right )^{2}}{x^{3}}+\frac {6 \,{\mathrm e}^{-5} \left (3 x \,{\mathrm e}^{5}+6 \,{\mathrm e}^{5}+1\right ) \ln \relax (x ) \ln \left (x -4\right )}{x^{3}}+\frac {{\mathrm e}^{-10} \left (2 x^{4} {\mathrm e}^{10}+9 x^{2} {\mathrm e}^{10} \ln \relax (x )+36 \,{\mathrm e}^{10} \ln \relax (x ) x +36 \,{\mathrm e}^{10} \ln \relax (x )+6 x \,{\mathrm e}^{5} \ln \relax (x )+12 \,{\mathrm e}^{5} \ln \relax (x )+\ln \relax (x )\right )}{x^{3}}\) | \(90\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.52, size = 234, normalized size = 9.36 \begin {gather*} -\frac {1}{192} \, {\left (36 \, {\left (\frac {4 \, {\left (3 \, x^{2} + 6 \, x + 16\right )}}{x^{3}} + 3 \, \log \left (x - 4\right ) - 3 \, \log \relax (x)\right )} e^{10} + 12 \, {\left (\frac {4 \, {\left (3 \, x^{2} + 6 \, x + 16\right )}}{x^{3}} + 3 \, \log \left (x - 4\right ) - 3 \, \log \relax (x)\right )} e^{5} - \frac {384 \, x^{4} e^{10} + 1728 \, e^{10} \log \left (x - 4\right )^{2} \log \relax (x) + 12 \, x^{2} {\left (36 \, e^{10} + 12 \, e^{5} + 1\right )} + 24 \, x {\left (36 \, e^{10} + 12 \, e^{5} + 1\right )} + 3 \, {\left (x^{3} {\left (36 \, e^{10} + 12 \, e^{5} + 1\right )} + 384 \, {\left (3 \, x e^{10} + 6 \, e^{10} + e^{5}\right )} \log \relax (x)\right )} \log \left (x - 4\right ) - 3 \, {\left (x^{3} {\left (36 \, e^{10} + 12 \, e^{5} + 1\right )} - 576 \, x^{2} e^{10} - 384 \, x {\left (6 \, e^{10} + e^{5}\right )} - 2304 \, e^{10} - 768 \, e^{5} - 64\right )} \log \relax (x) + 2304 \, e^{10} + 768 \, e^{5} + 64}{x^{3}} + \frac {4 \, {\left (3 \, x^{2} + 6 \, x + 16\right )}}{x^{3}} + 3 \, \log \left (x - 4\right ) - 3 \, \log \relax (x)\right )} e^{\left (-10\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 5.28, size = 74, normalized size = 2.96 \begin {gather*} 2\,x+\frac {\ln \relax (x)\,\left (9\,x^2+6\,{\mathrm {e}}^{-5}\,\left (6\,{\mathrm {e}}^5+1\right )\,x+{\mathrm {e}}^{-10}\,{\left (6\,{\mathrm {e}}^5+1\right )}^2\right )}{x^3}+\frac {9\,{\ln \left (x-4\right )}^2\,\ln \relax (x)}{x^3}+\frac {\ln \left (x-4\right )\,{\mathrm {e}}^{-5}\,\ln \relax (x)\,\left (36\,{\mathrm {e}}^5+18\,x\,{\mathrm {e}}^5+6\right )}{x^3} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 1.33, size = 97, normalized size = 3.88 \begin {gather*} 2 x + \frac {\left (18 x e^{5} \log {\relax (x )} + 6 \log {\relax (x )} + 36 e^{5} \log {\relax (x )}\right ) \log {\left (x - 4 \right )}}{x^{3} e^{5}} + \frac {\left (9 x^{2} e^{10} + 6 x e^{5} + 36 x e^{10} + 1 + 12 e^{5} + 36 e^{10}\right ) \log {\relax (x )}}{x^{3} e^{10}} + \frac {9 \log {\relax (x )} \log {\left (x - 4 \right )}^{2}}{x^{3}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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