Optimal. Leaf size=31 \[ \frac {x^2 \left (\frac {x}{2}-\frac {1}{2} \log (-4+x)\right )^2}{4 e^8 (5+x)^2} \]
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Rubi [C] Result contains higher order function than in optimal. Order 4 vs. order 3 in
optimal.
time = 0.44, antiderivative size = 257, normalized size of antiderivative = 8.29, number of steps
used = 38, number of rules used = 22, integrand size = 77, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {12, 6874, 90,
2465, 2436, 2332, 2437, 2338, 2442, 46, 36, 31, 2441, 2440, 2438, 2463, 2445, 2458, 2389, 2379,
2351, 2444} \begin {gather*} -\frac {25 \text {PolyLog}\left (2,\frac {9}{4-x}\right )}{648 e^8}-\frac {25 \text {PolyLog}\left (2,\frac {4-x}{9}\right )}{648 e^8}+\frac {x^2}{16 e^8}-\frac {5 x}{8 e^8}-\frac {125}{4 e^8 (x+5)}+\frac {625}{16 e^8 (x+5)^2}-\frac {5 (4-x) \log ^2(x-4)}{72 e^8 (x+5)}+\frac {25 \log ^2(x-4)}{16 e^8 (x+5)^2}+\frac {\log ^2(x-4)}{81 e^8}+\frac {461 \log (4-x)}{648 e^8}+\frac {(4-x) \log (x-4)}{8 e^8}+\frac {25 \log \left (1-\frac {9}{4-x}\right ) \log (x-4)}{648 e^8}-\frac {25 (4-x) \log (x-4)}{648 e^8 (x+5)}-\frac {325 \log (x-4)}{36 e^8 (x+5)}+\frac {125 \log (x-4)}{8 e^8 (x+5)^2}-\frac {25 \log (x-4) \log \left (\frac {x+5}{9}\right )}{648 e^8} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 31
Rule 36
Rule 46
Rule 90
Rule 2332
Rule 2338
Rule 2351
Rule 2379
Rule 2389
Rule 2436
Rule 2437
Rule 2438
Rule 2440
Rule 2441
Rule 2442
Rule 2444
Rule 2445
Rule 2458
Rule 2463
Rule 2465
Rule 6874
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {-45 x^3+5 x^4+x^5+\left (65 x^2-10 x^3-x^4\right ) \log (-4+x)+\left (-20 x+5 x^2\right ) \log ^2(-4+x)}{-4000-1400 x+120 x^2+88 x^3+8 x^4} \, dx}{e^8}\\ &=\frac {\int \left (-\frac {45 x^3}{8 (-4+x) (5+x)^3}+\frac {5 x^4}{8 (-4+x) (5+x)^3}+\frac {x^5}{8 (-4+x) (5+x)^3}-\frac {x^2 \left (-65+10 x+x^2\right ) \log (-4+x)}{8 (-4+x) (5+x)^3}+\frac {5 x \log ^2(-4+x)}{8 (5+x)^3}\right ) \, dx}{e^8}\\ &=\frac {\int \frac {x^5}{(-4+x) (5+x)^3} \, dx}{8 e^8}-\frac {\int \frac {x^2 \left (-65+10 x+x^2\right ) \log (-4+x)}{(-4+x) (5+x)^3} \, dx}{8 e^8}+\frac {5 \int \frac {x^4}{(-4+x) (5+x)^3} \, dx}{8 e^8}+\frac {5 \int \frac {x \log ^2(-4+x)}{(5+x)^3} \, dx}{8 e^8}-\frac {45 \int \frac {x^3}{(-4+x) (5+x)^3} \, dx}{8 e^8}\\ &=\frac {\int \left (-11+\frac {1024}{729 (-4+x)}+x+\frac {3125}{9 (5+x)^3}-\frac {25000}{81 (5+x)^2}+\frac {76250}{729 (5+x)}\right ) \, dx}{8 e^8}-\frac {\int \left (\log (-4+x)-\frac {16 \log (-4+x)}{81 (-4+x)}+\frac {250 \log (-4+x)}{(5+x)^3}-\frac {650 \log (-4+x)}{9 (5+x)^2}-\frac {65 \log (-4+x)}{81 (5+x)}\right ) \, dx}{8 e^8}+\frac {5 \int \left (1+\frac {256}{729 (-4+x)}-\frac {625}{9 (5+x)^3}+\frac {3875}{81 (5+x)^2}-\frac {8275}{729 (5+x)}\right ) \, dx}{8 e^8}+\frac {5 \int \left (-\frac {5 \log ^2(-4+x)}{(5+x)^3}+\frac {\log ^2(-4+x)}{(5+x)^2}\right ) \, dx}{8 e^8}-\frac {45 \int \left (\frac {64}{729 (-4+x)}+\frac {125}{9 (5+x)^3}-\frac {550}{81 (5+x)^2}+\frac {665}{729 (5+x)}\right ) \, dx}{8 e^8}\\ &=-\frac {3 x}{4 e^8}+\frac {x^2}{16 e^8}+\frac {625}{16 e^8 (5+x)^2}-\frac {2125}{72 e^8 (5+x)}-\frac {8 \log (4-x)}{81 e^8}+\frac {275 \log (5+x)}{324 e^8}+\frac {2 \int \frac {\log (-4+x)}{-4+x} \, dx}{81 e^8}+\frac {65 \int \frac {\log (-4+x)}{5+x} \, dx}{648 e^8}-\frac {\int \log (-4+x) \, dx}{8 e^8}+\frac {5 \int \frac {\log ^2(-4+x)}{(5+x)^2} \, dx}{8 e^8}-\frac {25 \int \frac {\log ^2(-4+x)}{(5+x)^3} \, dx}{8 e^8}+\frac {325 \int \frac {\log (-4+x)}{(5+x)^2} \, dx}{36 e^8}-\frac {125 \int \frac {\log (-4+x)}{(5+x)^3} \, dx}{4 e^8}\\ &=-\frac {3 x}{4 e^8}+\frac {x^2}{16 e^8}+\frac {625}{16 e^8 (5+x)^2}-\frac {2125}{72 e^8 (5+x)}-\frac {8 \log (4-x)}{81 e^8}+\frac {125 \log (-4+x)}{8 e^8 (5+x)^2}-\frac {325 \log (-4+x)}{36 e^8 (5+x)}+\frac {25 \log ^2(-4+x)}{16 e^8 (5+x)^2}-\frac {5 (4-x) \log ^2(-4+x)}{72 e^8 (5+x)}+\frac {65 \log (-4+x) \log \left (\frac {5+x}{9}\right )}{648 e^8}+\frac {275 \log (5+x)}{324 e^8}+\frac {2 \text {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,-4+x\right )}{81 e^8}-\frac {65 \int \frac {\log \left (\frac {5+x}{9}\right )}{-4+x} \, dx}{648 e^8}-\frac {\text {Subst}(\int \log (x) \, dx,x,-4+x)}{8 e^8}-\frac {5 \int \frac {\log (-4+x)}{5+x} \, dx}{36 e^8}-\frac {25 \int \frac {\log (-4+x)}{(-4+x) (5+x)^2} \, dx}{8 e^8}+\frac {325 \int \frac {1}{(-4+x) (5+x)} \, dx}{36 e^8}-\frac {125 \int \frac {1}{(-4+x) (5+x)^2} \, dx}{8 e^8}\\ &=-\frac {5 x}{8 e^8}+\frac {x^2}{16 e^8}+\frac {625}{16 e^8 (5+x)^2}-\frac {2125}{72 e^8 (5+x)}-\frac {8 \log (4-x)}{81 e^8}+\frac {(4-x) \log (-4+x)}{8 e^8}+\frac {125 \log (-4+x)}{8 e^8 (5+x)^2}-\frac {325 \log (-4+x)}{36 e^8 (5+x)}+\frac {\log ^2(-4+x)}{81 e^8}+\frac {25 \log ^2(-4+x)}{16 e^8 (5+x)^2}-\frac {5 (4-x) \log ^2(-4+x)}{72 e^8 (5+x)}-\frac {25 \log (-4+x) \log \left (\frac {5+x}{9}\right )}{648 e^8}+\frac {275 \log (5+x)}{324 e^8}-\frac {65 \text {Subst}\left (\int \frac {\log \left (1+\frac {x}{9}\right )}{x} \, dx,x,-4+x\right )}{648 e^8}+\frac {5 \int \frac {\log \left (\frac {5+x}{9}\right )}{-4+x} \, dx}{36 e^8}+\frac {325 \int \frac {1}{-4+x} \, dx}{324 e^8}-\frac {325 \int \frac {1}{5+x} \, dx}{324 e^8}-\frac {25 \text {Subst}\left (\int \frac {\log (x)}{x (9+x)^2} \, dx,x,-4+x\right )}{8 e^8}-\frac {125 \int \left (\frac {1}{81 (-4+x)}-\frac {1}{9 (5+x)^2}-\frac {1}{81 (5+x)}\right ) \, dx}{8 e^8}\\ &=-\frac {5 x}{8 e^8}+\frac {x^2}{16 e^8}+\frac {625}{16 e^8 (5+x)^2}-\frac {125}{4 e^8 (5+x)}+\frac {461 \log (4-x)}{648 e^8}+\frac {(4-x) \log (-4+x)}{8 e^8}+\frac {125 \log (-4+x)}{8 e^8 (5+x)^2}-\frac {325 \log (-4+x)}{36 e^8 (5+x)}+\frac {\log ^2(-4+x)}{81 e^8}+\frac {25 \log ^2(-4+x)}{16 e^8 (5+x)^2}-\frac {5 (4-x) \log ^2(-4+x)}{72 e^8 (5+x)}-\frac {25 \log (-4+x) \log \left (\frac {5+x}{9}\right )}{648 e^8}+\frac {25 \log (5+x)}{648 e^8}+\frac {65 \text {Li}_2\left (\frac {4-x}{9}\right )}{648 e^8}+\frac {5 \text {Subst}\left (\int \frac {\log \left (1+\frac {x}{9}\right )}{x} \, dx,x,-4+x\right )}{36 e^8}+\frac {25 \text {Subst}\left (\int \frac {\log (x)}{(9+x)^2} \, dx,x,-4+x\right )}{72 e^8}-\frac {25 \text {Subst}\left (\int \frac {\log (x)}{x (9+x)} \, dx,x,-4+x\right )}{72 e^8}\\ &=-\frac {5 x}{8 e^8}+\frac {x^2}{16 e^8}+\frac {625}{16 e^8 (5+x)^2}-\frac {125}{4 e^8 (5+x)}+\frac {461 \log (4-x)}{648 e^8}+\frac {(4-x) \log (-4+x)}{8 e^8}+\frac {125 \log (-4+x)}{8 e^8 (5+x)^2}-\frac {325 \log (-4+x)}{36 e^8 (5+x)}-\frac {25 (4-x) \log (-4+x)}{648 e^8 (5+x)}+\frac {\log ^2(-4+x)}{81 e^8}+\frac {25 \log ^2(-4+x)}{16 e^8 (5+x)^2}-\frac {5 (4-x) \log ^2(-4+x)}{72 e^8 (5+x)}-\frac {25 \log (-4+x) \log \left (\frac {5+x}{9}\right )}{648 e^8}+\frac {25 \log (5+x)}{648 e^8}-\frac {25 \text {Li}_2\left (\frac {4-x}{9}\right )}{648 e^8}-\frac {25 \text {Subst}\left (\int \frac {1}{9+x} \, dx,x,-4+x\right )}{648 e^8}-\frac {25 \text {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,-4+x\right )}{648 e^8}+\frac {25 \text {Subst}\left (\int \frac {\log (x)}{9+x} \, dx,x,-4+x\right )}{648 e^8}\\ &=-\frac {5 x}{8 e^8}+\frac {x^2}{16 e^8}+\frac {625}{16 e^8 (5+x)^2}-\frac {125}{4 e^8 (5+x)}+\frac {461 \log (4-x)}{648 e^8}+\frac {(4-x) \log (-4+x)}{8 e^8}+\frac {125 \log (-4+x)}{8 e^8 (5+x)^2}-\frac {325 \log (-4+x)}{36 e^8 (5+x)}-\frac {25 (4-x) \log (-4+x)}{648 e^8 (5+x)}-\frac {\log ^2(-4+x)}{144 e^8}+\frac {25 \log ^2(-4+x)}{16 e^8 (5+x)^2}-\frac {5 (4-x) \log ^2(-4+x)}{72 e^8 (5+x)}-\frac {25 \text {Li}_2\left (\frac {4-x}{9}\right )}{648 e^8}-\frac {25 \text {Subst}\left (\int \frac {\log \left (1+\frac {x}{9}\right )}{x} \, dx,x,-4+x\right )}{648 e^8}\\ &=-\frac {5 x}{8 e^8}+\frac {x^2}{16 e^8}+\frac {625}{16 e^8 (5+x)^2}-\frac {125}{4 e^8 (5+x)}+\frac {461 \log (4-x)}{648 e^8}+\frac {(4-x) \log (-4+x)}{8 e^8}+\frac {125 \log (-4+x)}{8 e^8 (5+x)^2}-\frac {325 \log (-4+x)}{36 e^8 (5+x)}-\frac {25 (4-x) \log (-4+x)}{648 e^8 (5+x)}-\frac {\log ^2(-4+x)}{144 e^8}+\frac {25 \log ^2(-4+x)}{16 e^8 (5+x)^2}-\frac {5 (4-x) \log ^2(-4+x)}{72 e^8 (5+x)}\\ \end {aligned} \end {gather*}
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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(67\) vs. \(2(31)=62\).
time = 0.06, size = 67, normalized size = 2.16 \begin {gather*} \frac {-1875-750 x-75 x^2+x^4+20 (5+x)^2 \log (4-x)-2 \left (250+100 x+10 x^2+x^3\right ) \log (-4+x)+x^2 \log ^2(-4+x)}{16 e^8 (5+x)^2} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.73, size = 49, normalized size = 1.58
method | result | size |
norman | \(\frac {\left (\frac {x^{4} {\mathrm e}^{-4}}{16}+\frac {x^{2} {\mathrm e}^{-4} \ln \left (x -4\right )^{2}}{16}-\frac {{\mathrm e}^{-4} x^{3} \ln \left (x -4\right )}{8}\right ) {\mathrm e}^{-4}}{\left (5+x \right )^{2}}\) | \(49\) |
risch | \(\frac {{\mathrm e}^{-8} x^{2} \ln \left (x -4\right )^{2}}{16 x^{2}+160 x +400}-\frac {{\mathrm e}^{-8} \left (x^{3}+10 x^{2}+100 x +250\right ) \ln \left (x -4\right )}{8 \left (x^{2}+10 x +25\right )}+\frac {{\mathrm e}^{-8} \left (x^{4}+20 x^{2} \ln \left (x -4\right )+200 x \ln \left (x -4\right )-75 x^{2}+500 \ln \left (x -4\right )-750 x -1875\right )}{16 x^{2}+160 x +400}\) | \(105\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 46 vs.
\(2 (22) = 44\).
time = 0.47, size = 46, normalized size = 1.48 \begin {gather*} \frac {{\left (x^{4} - 2 \, x^{3} \log \left (x - 4\right ) + x^{2} \log \left (x - 4\right )^{2} - 75 \, x^{2} - 750 \, x - 1875\right )} e^{\left (-8\right )}}{16 \, {\left (x^{2} + 10 \, x + 25\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 46 vs.
\(2 (22) = 44\).
time = 0.36, size = 46, normalized size = 1.48 \begin {gather*} \frac {{\left (x^{4} - 2 \, x^{3} \log \left (x - 4\right ) + x^{2} \log \left (x - 4\right )^{2} - 75 \, x^{2} - 750 \, x - 1875\right )} e^{\left (-8\right )}}{16 \, {\left (x^{2} + 10 \, x + 25\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 126 vs.
\(2 (24) = 48\).
time = 0.37, size = 126, normalized size = 4.06 \begin {gather*} \frac {x^{2}}{16 e^{8}} + \frac {x^{2} \log {\left (x - 4 \right )}^{2}}{16 x^{2} e^{8} + 160 x e^{8} + 400 e^{8}} - \frac {5 x}{8 e^{8}} + \frac {- 500 x - 1875}{16 x^{2} e^{8} + 160 x e^{8} + 400 e^{8}} + \frac {5 \log {\left (x - 4 \right )}}{4 e^{8}} + \frac {\left (- x^{3} - 10 x^{2} - 100 x - 250\right ) \log {\left (x - 4 \right )}}{8 x^{2} e^{8} + 80 x e^{8} + 200 e^{8}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 0.37, size = 22, normalized size = 0.71 \begin {gather*} \frac {x^2\,{\mathrm {e}}^{-8}\,{\left (x-\ln \left (x-4\right )\right )}^2}{16\,{\left (x+5\right )}^2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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