Optimal. Leaf size=29 \[ \frac {4}{2 x-\log (5)+\frac {8 x}{x+e^4 \log ^2(-3+x)}} \]
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Rubi [F] time = 49.70, 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 {24 x^2-8 x^3+64 e^4 x \log (-3+x)+e^4 \left (96+16 x-16 x^2\right ) \log ^2(-3+x)+e^8 (24-8 x) \log ^4(-3+x)}{-192 x^2-32 x^3+20 x^4+4 x^5+\left (48 x^2-4 x^3-4 x^4\right ) \log (5)+\left (-3 x^2+x^3\right ) \log ^2(5)+\left (e^4 \left (-96 x^2+8 x^3+8 x^4\right )+e^4 \left (48 x+8 x^2-8 x^3\right ) \log (5)+e^4 \left (-6 x+2 x^2\right ) \log ^2(5)\right ) \log ^2(-3+x)+\left (e^8 \left (-12 x^2+4 x^3\right )+e^8 \left (12 x-4 x^2\right ) \log (5)+e^8 (-3+x) \log ^2(5)\right ) \log ^4(-3+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 {8 \left ((-3+x) x^2-8 e^4 x \log (-3+x)+2 e^4 \left (-6-x+x^2\right ) \log ^2(-3+x)+e^8 (-3+x) \log ^4(-3+x)\right )}{(3-x) \left (x (8+2 x-\log (5))+e^4 (2 x-\log (5)) \log ^2(-3+x)\right )^2} \, dx\\ &=8 \int \frac {(-3+x) x^2-8 e^4 x \log (-3+x)+2 e^4 \left (-6-x+x^2\right ) \log ^2(-3+x)+e^8 (-3+x) \log ^4(-3+x)}{(3-x) \left (x (8+2 x-\log (5))+e^4 (2 x-\log (5)) \log ^2(-3+x)\right )^2} \, dx\\ &=8 \int \left (-\frac {1}{(2 x-\log (5))^2}+\frac {4 x \left (-4 x^3+12 x^2 \left (1+\frac {\log (5)}{3}\right )-24 \left (1-\frac {\log (5)}{8}\right ) \log (5)-4 x \left (1+\frac {\log (5)}{4}\right ) \log (5)-8 e^4 x^2 \log (-3+x)+8 e^4 x \log (5) \log (-3+x)-2 e^4 \log ^2(5) \log (-3+x)\right )}{(3-x) (2 x-\log (5))^2 \left (2 x^2+8 x \left (1-\frac {\log (5)}{8}\right )+2 e^4 x \log ^2(-3+x)-e^4 \log (5) \log ^2(-3+x)\right )^2}+\frac {8 x+\log (625)}{(2 x-\log (5))^2 \left (2 x^2+8 x \left (1-\frac {\log (5)}{8}\right )+2 e^4 x \log ^2(-3+x)-e^4 \log (5) \log ^2(-3+x)\right )}\right ) \, dx\\ &=\frac {4}{2 x-\log (5)}+8 \int \frac {8 x+\log (625)}{(2 x-\log (5))^2 \left (2 x^2+8 x \left (1-\frac {\log (5)}{8}\right )+2 e^4 x \log ^2(-3+x)-e^4 \log (5) \log ^2(-3+x)\right )} \, dx+32 \int \frac {x \left (-4 x^3+12 x^2 \left (1+\frac {\log (5)}{3}\right )-24 \left (1-\frac {\log (5)}{8}\right ) \log (5)-4 x \left (1+\frac {\log (5)}{4}\right ) \log (5)-8 e^4 x^2 \log (-3+x)+8 e^4 x \log (5) \log (-3+x)-2 e^4 \log ^2(5) \log (-3+x)\right )}{(3-x) (2 x-\log (5))^2 \left (2 x^2+8 x \left (1-\frac {\log (5)}{8}\right )+2 e^4 x \log ^2(-3+x)-e^4 \log (5) \log ^2(-3+x)\right )^2} \, dx\\ &=\frac {4}{2 x-\log (5)}+8 \int \frac {8 x+\log (625)}{(2 x-\log (5))^2 \left (x (8+2 x-\log (5))+e^4 (2 x-\log (5)) \log ^2(-3+x)\right )} \, dx+32 \int \frac {x \left (-\left ((-3+x) \left (4 x^2-4 x \log (5)+(-8+\log (5)) \log (5)\right )\right )-2 e^4 (-2 x+\log (5))^2 \log (-3+x)\right )}{(3-x) (2 x-\log (5))^2 \left (x (8+2 x-\log (5))+e^4 (2 x-\log (5)) \log ^2(-3+x)\right )^2} \, dx\\ &=\frac {4}{2 x-\log (5)}+8 \int \left (\frac {4}{(2 x-\log (5)) \left (2 x^2+8 x \left (1-\frac {\log (5)}{8}\right )+2 e^4 x \log ^2(-3+x)-e^4 \log (5) \log ^2(-3+x)\right )}+\frac {2 \log (625)}{(2 x-\log (5))^2 \left (2 x^2+8 x \left (1-\frac {\log (5)}{8}\right )+2 e^4 x \log ^2(-3+x)-e^4 \log (5) \log ^2(-3+x)\right )}\right ) \, dx+32 \int \left (\frac {3 \left (-4 x^3+12 x^2 \left (1+\frac {\log (5)}{3}\right )-24 \left (1-\frac {\log (5)}{8}\right ) \log (5)-4 x \left (1+\frac {\log (5)}{4}\right ) \log (5)-8 e^4 x^2 \log (-3+x)+8 e^4 x \log (5) \log (-3+x)-2 e^4 \log ^2(5) \log (-3+x)\right )}{(3-x) (6-\log (5))^2 \left (2 x^2+8 x \left (1-\frac {\log (5)}{8}\right )+2 e^4 x \log ^2(-3+x)-e^4 \log (5) \log ^2(-3+x)\right )^2}+\frac {6 \left (-4 x^3+12 x^2 \left (1+\frac {\log (5)}{3}\right )-24 \left (1-\frac {\log (5)}{8}\right ) \log (5)-4 x \left (1+\frac {\log (5)}{4}\right ) \log (5)-8 e^4 x^2 \log (-3+x)+8 e^4 x \log (5) \log (-3+x)-2 e^4 \log ^2(5) \log (-3+x)\right )}{(6-\log (5))^2 (2 x-\log (5)) \left (2 x^2+8 x \left (1-\frac {\log (5)}{8}\right )+2 e^4 x \log ^2(-3+x)-e^4 \log (5) \log ^2(-3+x)\right )^2}+\frac {\log (5) \left (-4 x^3+12 x^2 \left (1+\frac {\log (5)}{3}\right )-24 \left (1-\frac {\log (5)}{8}\right ) \log (5)-4 x \left (1+\frac {\log (5)}{4}\right ) \log (5)-8 e^4 x^2 \log (-3+x)+8 e^4 x \log (5) \log (-3+x)-2 e^4 \log ^2(5) \log (-3+x)\right )}{(6-\log (5)) (2 x-\log (5))^2 \left (2 x^2+8 x \left (1-\frac {\log (5)}{8}\right )+2 e^4 x \log ^2(-3+x)-e^4 \log (5) \log ^2(-3+x)\right )^2}\right ) \, dx\\ &=\frac {4}{2 x-\log (5)}+32 \int \frac {1}{(2 x-\log (5)) \left (2 x^2+8 x \left (1-\frac {\log (5)}{8}\right )+2 e^4 x \log ^2(-3+x)-e^4 \log (5) \log ^2(-3+x)\right )} \, dx+\frac {96 \int \frac {-4 x^3+12 x^2 \left (1+\frac {\log (5)}{3}\right )-24 \left (1-\frac {\log (5)}{8}\right ) \log (5)-4 x \left (1+\frac {\log (5)}{4}\right ) \log (5)-8 e^4 x^2 \log (-3+x)+8 e^4 x \log (5) \log (-3+x)-2 e^4 \log ^2(5) \log (-3+x)}{(3-x) \left (2 x^2+8 x \left (1-\frac {\log (5)}{8}\right )+2 e^4 x \log ^2(-3+x)-e^4 \log (5) \log ^2(-3+x)\right )^2} \, dx}{(6-\log (5))^2}+\frac {192 \int \frac {-4 x^3+12 x^2 \left (1+\frac {\log (5)}{3}\right )-24 \left (1-\frac {\log (5)}{8}\right ) \log (5)-4 x \left (1+\frac {\log (5)}{4}\right ) \log (5)-8 e^4 x^2 \log (-3+x)+8 e^4 x \log (5) \log (-3+x)-2 e^4 \log ^2(5) \log (-3+x)}{(2 x-\log (5)) \left (2 x^2+8 x \left (1-\frac {\log (5)}{8}\right )+2 e^4 x \log ^2(-3+x)-e^4 \log (5) \log ^2(-3+x)\right )^2} \, dx}{(6-\log (5))^2}+\frac {(32 \log (5)) \int \frac {-4 x^3+12 x^2 \left (1+\frac {\log (5)}{3}\right )-24 \left (1-\frac {\log (5)}{8}\right ) \log (5)-4 x \left (1+\frac {\log (5)}{4}\right ) \log (5)-8 e^4 x^2 \log (-3+x)+8 e^4 x \log (5) \log (-3+x)-2 e^4 \log ^2(5) \log (-3+x)}{(2 x-\log (5))^2 \left (2 x^2+8 x \left (1-\frac {\log (5)}{8}\right )+2 e^4 x \log ^2(-3+x)-e^4 \log (5) \log ^2(-3+x)\right )^2} \, dx}{6-\log (5)}+(16 \log (625)) \int \frac {1}{(2 x-\log (5))^2 \left (2 x^2+8 x \left (1-\frac {\log (5)}{8}\right )+2 e^4 x \log ^2(-3+x)-e^4 \log (5) \log ^2(-3+x)\right )} \, dx\\ &=\frac {4}{2 x-\log (5)}+32 \int \frac {1}{(2 x-\log (5)) \left (x (8+2 x-\log (5))+e^4 (2 x-\log (5)) \log ^2(-3+x)\right )} \, dx+\frac {96 \int \frac {-\left ((-3+x) \left (4 x^2-4 x \log (5)+(-8+\log (5)) \log (5)\right )\right )-2 e^4 (-2 x+\log (5))^2 \log (-3+x)}{(3-x) \left (x (8+2 x-\log (5))+e^4 (2 x-\log (5)) \log ^2(-3+x)\right )^2} \, dx}{(6-\log (5))^2}+\frac {192 \int \frac {-\left ((-3+x) \left (4 x^2-4 x \log (5)+(-8+\log (5)) \log (5)\right )\right )-2 e^4 (-2 x+\log (5))^2 \log (-3+x)}{(2 x-\log (5)) \left (x (8+2 x-\log (5))+e^4 (2 x-\log (5)) \log ^2(-3+x)\right )^2} \, dx}{(6-\log (5))^2}+\frac {(32 \log (5)) \int \frac {-\left ((-3+x) \left (4 x^2-4 x \log (5)+(-8+\log (5)) \log (5)\right )\right )-2 e^4 (-2 x+\log (5))^2 \log (-3+x)}{(2 x-\log (5))^2 \left (x (8+2 x-\log (5))+e^4 (2 x-\log (5)) \log ^2(-3+x)\right )^2} \, dx}{6-\log (5)}+(16 \log (625)) \int \frac {1}{(2 x-\log (5))^2 \left (x (8+2 x-\log (5))+e^4 (2 x-\log (5)) \log ^2(-3+x)\right )} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}
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Mathematica [A] time = 0.12, size = 48, normalized size = 1.66 \begin {gather*} \frac {8 \left (x+e^4 \log ^2(-3+x)\right )}{2 x (8+2 x-\log (5))+2 e^4 (2 x-\log (5)) \log ^2(-3+x)} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.70, size = 48, normalized size = 1.66 \begin {gather*} \frac {4 \, {\left (e^{4} \log \left (x - 3\right )^{2} + x\right )}}{{\left (2 \, x e^{4} - e^{4} \log \relax (5)\right )} \log \left (x - 3\right )^{2} + 2 \, x^{2} - x \log \relax (5) + 8 \, x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 164.41, size = 109, normalized size = 3.76 \begin {gather*} \frac {4 \, {\left (2 \, x e^{4} \log \left (x - 3\right )^{2} - e^{4} \log \relax (5) \log \left (x - 3\right )^{2} + 2 \, x^{2} - x \log \relax (5) - 8 \, x\right )}}{4 \, x^{2} e^{4} \log \left (x - 3\right )^{2} - 4 \, x e^{4} \log \relax (5) \log \left (x - 3\right )^{2} + e^{4} \log \relax (5)^{2} \log \left (x - 3\right )^{2} + 4 \, x^{3} - 4 \, x^{2} \log \relax (5) + x \log \relax (5)^{2} + 16 \, x^{2} - 8 \, x \log \relax (5)} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.08, size = 60, normalized size = 2.07
method | result | size |
risch | \(-\frac {4}{\ln \relax (5)-2 x}-\frac {32 x}{\left (\ln \relax (5)-2 x \right ) \left (\ln \left (x -3\right )^{2} \ln \relax (5) {\mathrm e}^{4}-2 \,{\mathrm e}^{4} \ln \left (x -3\right )^{2} x +x \ln \relax (5)-2 x^{2}-8 x \right )}\) | \(60\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.86, size = 89, normalized size = 3.07 \begin {gather*} \frac {4 \, {\left ({\left (2 \, x e^{4} - e^{4} \log \relax (5)\right )} \log \left (x - 3\right )^{2} + 2 \, x^{2} - x \log \relax (5)\right )}}{4 \, x^{3} - 4 \, x^{2} {\left (\log \relax (5) - 4\right )} + {\left (4 \, x^{2} e^{4} - 4 \, x e^{4} \log \relax (5) + e^{4} \log \relax (5)^{2}\right )} \log \left (x - 3\right )^{2} + {\left (\log \relax (5)^{2} - 8 \, \log \relax (5)\right )} x} \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 = -0.03 \begin {gather*} \int -\frac {24\,x^2-8\,x^3+64\,x\,\ln \left (x-3\right )\,{\mathrm {e}}^4-{\ln \left (x-3\right )}^4\,{\mathrm {e}}^8\,\left (8\,x-24\right )+{\ln \left (x-3\right )}^2\,{\mathrm {e}}^4\,\left (-16\,x^2+16\,x+96\right )}{\ln \relax (5)\,\left (4\,x^4+4\,x^3-48\,x^2\right )+192\,x^2+32\,x^3-20\,x^4-4\,x^5+{\ln \relax (5)}^2\,\left (3\,x^2-x^3\right )-{\ln \left (x-3\right )}^4\,\left ({\mathrm {e}}^8\,{\ln \relax (5)}^2\,\left (x-3\right )-{\mathrm {e}}^8\,\left (12\,x^2-4\,x^3\right )+{\mathrm {e}}^8\,\ln \relax (5)\,\left (12\,x-4\,x^2\right )\right )-{\ln \left (x-3\right )}^2\,\left ({\mathrm {e}}^4\,\left (8\,x^4+8\,x^3-96\,x^2\right )+{\mathrm {e}}^4\,\ln \relax (5)\,\left (-8\,x^3+8\,x^2+48\,x\right )-{\mathrm {e}}^4\,{\ln \relax (5)}^2\,\left (6\,x-2\,x^2\right )\right )} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 0.59, size = 78, normalized size = 2.69 \begin {gather*} - \frac {32 x}{4 x^{3} - 4 x^{2} \log {\relax (5 )} + 16 x^{2} - 8 x \log {\relax (5 )} + x \log {\relax (5 )}^{2} + \left (4 x^{2} e^{4} - 4 x e^{4} \log {\relax (5 )} + e^{4} \log {\relax (5 )}^{2}\right ) \log {\left (x - 3 \right )}^{2}} + \frac {8}{4 x - 2 \log {\relax (5 )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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