Optimal. Leaf size=22 \[ \frac {3 \left (e^x+x\right )^2 (x-\log (2))}{(5-x)^4} \]
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Rubi [C] time = 1.40, antiderivative size = 553, normalized size of antiderivative = 25.14, number of steps used = 39, number of rules used = 8, integrand size = 104, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {6688, 12, 6742, 37, 43, 2199, 2177, 2178} \begin {gather*} 12 e^{10} \text {Ei}(-2 (5-x))+6 e^5 \text {Ei}(x-5)-\frac {1}{8} e^5 (200-\log (1099511627776)) \text {Ei}(x-5)-\frac {1}{2} e^5 (10+\log (16)) \text {Ei}(x-5)-2 e^{10} (20-\log (16)) \text {Ei}(-2 (5-x))+\frac {3}{2} e^5 (16-\log (4)) \text {Ei}(x-5)+4 e^{10} (7-\log (4)) \text {Ei}(-2 (5-x))+\frac {3 x^4}{20 (5-x)^4}+\frac {6 e^x}{5-x}+\frac {6 e^{2 x}}{5-x}-\frac {3 e^{2 x}}{(5-x)^2}-\frac {e^x (200-\log (1099511627776))}{8 (5-x)}+\frac {e^x (200-\log (1099511627776))}{8 (5-x)^2}-\frac {e^x (200-\log (1099511627776))}{4 (5-x)^3}+\frac {3 e^x (200-\log (1099511627776))}{4 (5-x)^4}+\frac {\log (1024)}{(5-x)^3}-\frac {15 \log (1024)}{4 (5-x)^4}-\frac {e^x (10+\log (16))}{2 (5-x)}+\frac {e^x (10+\log (16))}{2 (5-x)^2}-\frac {e^x (10+\log (16))}{(5-x)^3}-\frac {e^{2 x} (20-\log (16))}{5-x}+\frac {e^{2 x} (20-\log (16))}{2 (5-x)^2}-\frac {e^{2 x} (20-\log (16))}{2 (5-x)^3}+\frac {3 e^{2 x} (20-\log (16))}{4 (5-x)^4}+\frac {3 e^x (16-\log (4))}{2 (5-x)}-\frac {3 e^x (16-\log (4))}{2 (5-x)^2}+\frac {3 (15-\log (4))}{2 (5-x)^2}-\frac {10 (15-\log (4))}{(5-x)^3}+\frac {75 (15-\log (4))}{4 (5-x)^4}+\frac {2 e^{2 x} (7-\log (4))}{5-x}-\frac {e^{2 x} (7-\log (4))}{(5-x)^2}+\frac {e^{2 x} (7-\log (4))}{(5-x)^3} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 37
Rule 43
Rule 2177
Rule 2178
Rule 2199
Rule 6688
Rule 6742
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {3 \left (e^x+x\right ) \left (x^2-x (-15+\log (4))-e^x \left (-5+2 x^2+14 \log (2)-x (13+\log (4))\right )-\log (1024)\right )}{(5-x)^5} \, dx\\ &=3 \int \frac {\left (e^x+x\right ) \left (x^2-x (-15+\log (4))-e^x \left (-5+2 x^2+14 \log (2)-x (13+\log (4))\right )-\log (1024)\right )}{(5-x)^5} \, dx\\ &=3 \int \left (-\frac {x^3}{(-5+x)^5}+\frac {x^2 (-15+\log (4))}{(-5+x)^5}+\frac {e^{2 x} \left (5-2 x^2-14 \log (2)+x (13+\log (4))\right )}{(5-x)^5}+\frac {x \log (1024)}{(-5+x)^5}+\frac {e^x \left (-2 x^3+x^2 (14+\log (4))-\log (1024)+x (20-\log (65536))\right )}{(5-x)^5}\right ) \, dx\\ &=-\left (3 \int \frac {x^3}{(-5+x)^5} \, dx\right )+3 \int \frac {e^{2 x} \left (5-2 x^2-14 \log (2)+x (13+\log (4))\right )}{(5-x)^5} \, dx+3 \int \frac {e^x \left (-2 x^3+x^2 (14+\log (4))-\log (1024)+x (20-\log (65536))\right )}{(5-x)^5} \, dx+(3 (-15+\log (4))) \int \frac {x^2}{(-5+x)^5} \, dx+(3 \log (1024)) \int \frac {x}{(-5+x)^5} \, dx\\ &=\frac {3 x^4}{20 (5-x)^4}+3 \int \left (\frac {2 e^{2 x}}{(-5+x)^3}+\frac {e^{2 x} (7-\log (4))}{(-5+x)^4}+\frac {e^{2 x} (-20+\log (16))}{(-5+x)^5}\right ) \, dx+3 \int \left (\frac {2 e^x}{(-5+x)^2}+\frac {e^x (16-\log (4))}{(-5+x)^3}+\frac {e^x (-10-\log (16))}{(-5+x)^4}+\frac {e^x (-200+\log (1099511627776))}{(-5+x)^5}\right ) \, dx+(3 (-15+\log (4))) \int \left (\frac {25}{(-5+x)^5}+\frac {10}{(-5+x)^4}+\frac {1}{(-5+x)^3}\right ) \, dx+(3 \log (1024)) \int \left (\frac {5}{(-5+x)^5}+\frac {1}{(-5+x)^4}\right ) \, dx\\ &=\frac {3 x^4}{20 (5-x)^4}+\frac {75 (15-\log (4))}{4 (5-x)^4}-\frac {10 (15-\log (4))}{(5-x)^3}+\frac {3 (15-\log (4))}{2 (5-x)^2}-\frac {15 \log (1024)}{4 (5-x)^4}+\frac {\log (1024)}{(5-x)^3}+6 \int \frac {e^{2 x}}{(-5+x)^3} \, dx+6 \int \frac {e^x}{(-5+x)^2} \, dx+(3 (7-\log (4))) \int \frac {e^{2 x}}{(-5+x)^4} \, dx+(3 (16-\log (4))) \int \frac {e^x}{(-5+x)^3} \, dx+(3 (-20+\log (16))) \int \frac {e^{2 x}}{(-5+x)^5} \, dx-(3 (10+\log (16))) \int \frac {e^x}{(-5+x)^4} \, dx-(3 (200-\log (1099511627776))) \int \frac {e^x}{(-5+x)^5} \, dx\\ &=-\frac {3 e^{2 x}}{(5-x)^2}+\frac {6 e^x}{5-x}+\frac {3 x^4}{20 (5-x)^4}+\frac {e^{2 x} (7-\log (4))}{(5-x)^3}+\frac {75 (15-\log (4))}{4 (5-x)^4}-\frac {10 (15-\log (4))}{(5-x)^3}+\frac {3 (15-\log (4))}{2 (5-x)^2}-\frac {3 e^x (16-\log (4))}{2 (5-x)^2}+\frac {3 e^{2 x} (20-\log (16))}{4 (5-x)^4}-\frac {e^x (10+\log (16))}{(5-x)^3}-\frac {15 \log (1024)}{4 (5-x)^4}+\frac {\log (1024)}{(5-x)^3}+\frac {3 e^x (200-\log (1099511627776))}{4 (5-x)^4}+6 \int \frac {e^{2 x}}{(-5+x)^2} \, dx+6 \int \frac {e^x}{-5+x} \, dx+(2 (7-\log (4))) \int \frac {e^{2 x}}{(-5+x)^3} \, dx+\frac {1}{2} (3 (16-\log (4))) \int \frac {e^x}{(-5+x)^2} \, dx+\frac {1}{2} (3 (-20+\log (16))) \int \frac {e^{2 x}}{(-5+x)^4} \, dx-(10+\log (16)) \int \frac {e^x}{(-5+x)^3} \, dx-\frac {1}{4} (3 (200-\log (1099511627776))) \int \frac {e^x}{(-5+x)^4} \, dx\\ &=-\frac {3 e^{2 x}}{(5-x)^2}+\frac {6 e^x}{5-x}+\frac {6 e^{2 x}}{5-x}+\frac {3 x^4}{20 (5-x)^4}+6 e^5 \text {Ei}(-5+x)+\frac {e^{2 x} (7-\log (4))}{(5-x)^3}-\frac {e^{2 x} (7-\log (4))}{(5-x)^2}+\frac {75 (15-\log (4))}{4 (5-x)^4}-\frac {10 (15-\log (4))}{(5-x)^3}+\frac {3 (15-\log (4))}{2 (5-x)^2}-\frac {3 e^x (16-\log (4))}{2 (5-x)^2}+\frac {3 e^x (16-\log (4))}{2 (5-x)}+\frac {3 e^{2 x} (20-\log (16))}{4 (5-x)^4}-\frac {e^{2 x} (20-\log (16))}{2 (5-x)^3}-\frac {e^x (10+\log (16))}{(5-x)^3}+\frac {e^x (10+\log (16))}{2 (5-x)^2}-\frac {15 \log (1024)}{4 (5-x)^4}+\frac {\log (1024)}{(5-x)^3}+\frac {3 e^x (200-\log (1099511627776))}{4 (5-x)^4}-\frac {e^x (200-\log (1099511627776))}{4 (5-x)^3}+12 \int \frac {e^{2 x}}{-5+x} \, dx+(2 (7-\log (4))) \int \frac {e^{2 x}}{(-5+x)^2} \, dx+\frac {1}{2} (3 (16-\log (4))) \int \frac {e^x}{-5+x} \, dx+(-20+\log (16)) \int \frac {e^{2 x}}{(-5+x)^3} \, dx-\frac {1}{2} (10+\log (16)) \int \frac {e^x}{(-5+x)^2} \, dx-\frac {1}{4} (200-\log (1099511627776)) \int \frac {e^x}{(-5+x)^3} \, dx\\ &=-\frac {3 e^{2 x}}{(5-x)^2}+\frac {6 e^x}{5-x}+\frac {6 e^{2 x}}{5-x}+\frac {3 x^4}{20 (5-x)^4}+12 e^{10} \text {Ei}(-2 (5-x))+6 e^5 \text {Ei}(-5+x)+\frac {e^{2 x} (7-\log (4))}{(5-x)^3}-\frac {e^{2 x} (7-\log (4))}{(5-x)^2}+\frac {2 e^{2 x} (7-\log (4))}{5-x}+\frac {75 (15-\log (4))}{4 (5-x)^4}-\frac {10 (15-\log (4))}{(5-x)^3}+\frac {3 (15-\log (4))}{2 (5-x)^2}-\frac {3 e^x (16-\log (4))}{2 (5-x)^2}+\frac {3 e^x (16-\log (4))}{2 (5-x)}+\frac {3}{2} e^5 \text {Ei}(-5+x) (16-\log (4))+\frac {3 e^{2 x} (20-\log (16))}{4 (5-x)^4}-\frac {e^{2 x} (20-\log (16))}{2 (5-x)^3}+\frac {e^{2 x} (20-\log (16))}{2 (5-x)^2}-\frac {e^x (10+\log (16))}{(5-x)^3}+\frac {e^x (10+\log (16))}{2 (5-x)^2}-\frac {e^x (10+\log (16))}{2 (5-x)}-\frac {15 \log (1024)}{4 (5-x)^4}+\frac {\log (1024)}{(5-x)^3}+\frac {3 e^x (200-\log (1099511627776))}{4 (5-x)^4}-\frac {e^x (200-\log (1099511627776))}{4 (5-x)^3}+\frac {e^x (200-\log (1099511627776))}{8 (5-x)^2}+(4 (7-\log (4))) \int \frac {e^{2 x}}{-5+x} \, dx+(-20+\log (16)) \int \frac {e^{2 x}}{(-5+x)^2} \, dx-\frac {1}{2} (10+\log (16)) \int \frac {e^x}{-5+x} \, dx-\frac {1}{8} (200-\log (1099511627776)) \int \frac {e^x}{(-5+x)^2} \, dx\\ &=-\frac {3 e^{2 x}}{(5-x)^2}+\frac {6 e^x}{5-x}+\frac {6 e^{2 x}}{5-x}+\frac {3 x^4}{20 (5-x)^4}+12 e^{10} \text {Ei}(-2 (5-x))+6 e^5 \text {Ei}(-5+x)+\frac {e^{2 x} (7-\log (4))}{(5-x)^3}-\frac {e^{2 x} (7-\log (4))}{(5-x)^2}+\frac {2 e^{2 x} (7-\log (4))}{5-x}+4 e^{10} \text {Ei}(-2 (5-x)) (7-\log (4))+\frac {75 (15-\log (4))}{4 (5-x)^4}-\frac {10 (15-\log (4))}{(5-x)^3}+\frac {3 (15-\log (4))}{2 (5-x)^2}-\frac {3 e^x (16-\log (4))}{2 (5-x)^2}+\frac {3 e^x (16-\log (4))}{2 (5-x)}+\frac {3}{2} e^5 \text {Ei}(-5+x) (16-\log (4))+\frac {3 e^{2 x} (20-\log (16))}{4 (5-x)^4}-\frac {e^{2 x} (20-\log (16))}{2 (5-x)^3}+\frac {e^{2 x} (20-\log (16))}{2 (5-x)^2}-\frac {e^{2 x} (20-\log (16))}{5-x}-\frac {e^x (10+\log (16))}{(5-x)^3}+\frac {e^x (10+\log (16))}{2 (5-x)^2}-\frac {e^x (10+\log (16))}{2 (5-x)}-\frac {1}{2} e^5 \text {Ei}(-5+x) (10+\log (16))-\frac {15 \log (1024)}{4 (5-x)^4}+\frac {\log (1024)}{(5-x)^3}+\frac {3 e^x (200-\log (1099511627776))}{4 (5-x)^4}-\frac {e^x (200-\log (1099511627776))}{4 (5-x)^3}+\frac {e^x (200-\log (1099511627776))}{8 (5-x)^2}-\frac {e^x (200-\log (1099511627776))}{8 (5-x)}+(2 (-20+\log (16))) \int \frac {e^{2 x}}{-5+x} \, dx-\frac {1}{8} (200-\log (1099511627776)) \int \frac {e^x}{-5+x} \, dx\\ &=-\frac {3 e^{2 x}}{(5-x)^2}+\frac {6 e^x}{5-x}+\frac {6 e^{2 x}}{5-x}+\frac {3 x^4}{20 (5-x)^4}+12 e^{10} \text {Ei}(-2 (5-x))+6 e^5 \text {Ei}(-5+x)+\frac {e^{2 x} (7-\log (4))}{(5-x)^3}-\frac {e^{2 x} (7-\log (4))}{(5-x)^2}+\frac {2 e^{2 x} (7-\log (4))}{5-x}+4 e^{10} \text {Ei}(-2 (5-x)) (7-\log (4))+\frac {75 (15-\log (4))}{4 (5-x)^4}-\frac {10 (15-\log (4))}{(5-x)^3}+\frac {3 (15-\log (4))}{2 (5-x)^2}-\frac {3 e^x (16-\log (4))}{2 (5-x)^2}+\frac {3 e^x (16-\log (4))}{2 (5-x)}+\frac {3}{2} e^5 \text {Ei}(-5+x) (16-\log (4))+\frac {3 e^{2 x} (20-\log (16))}{4 (5-x)^4}-\frac {e^{2 x} (20-\log (16))}{2 (5-x)^3}+\frac {e^{2 x} (20-\log (16))}{2 (5-x)^2}-\frac {e^{2 x} (20-\log (16))}{5-x}-2 e^{10} \text {Ei}(-2 (5-x)) (20-\log (16))-\frac {e^x (10+\log (16))}{(5-x)^3}+\frac {e^x (10+\log (16))}{2 (5-x)^2}-\frac {e^x (10+\log (16))}{2 (5-x)}-\frac {1}{2} e^5 \text {Ei}(-5+x) (10+\log (16))-\frac {15 \log (1024)}{4 (5-x)^4}+\frac {\log (1024)}{(5-x)^3}+\frac {3 e^x (200-\log (1099511627776))}{4 (5-x)^4}-\frac {e^x (200-\log (1099511627776))}{4 (5-x)^3}+\frac {e^x (200-\log (1099511627776))}{8 (5-x)^2}-\frac {e^x (200-\log (1099511627776))}{8 (5-x)}-\frac {1}{8} e^5 \text {Ei}(-5+x) (200-\log (1099511627776))\\ \end {aligned} \end {gather*}
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Mathematica [B] time = 0.70, size = 51, normalized size = 2.32 \begin {gather*} \frac {8 e^x x (6 x-\log (64))+2 x^2 (12 x-\log (4096))+e^{2 x} (24 x-\log (16777216))}{8 (-5+x)^4} \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.98, size = 57, normalized size = 2.59 \begin {gather*} \frac {3 \, {\left (x^{3} - x^{2} \log \relax (2) + {\left (x - \log \relax (2)\right )} e^{\left (2 \, x\right )} + 2 \, {\left (x^{2} - x \log \relax (2)\right )} e^{x}\right )}}{x^{4} - 20 \, x^{3} + 150 \, x^{2} - 500 \, x + 625} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.16, size = 61, normalized size = 2.77 \begin {gather*} \frac {3 \, {\left (x^{3} + 2 \, x^{2} e^{x} - x^{2} \log \relax (2) - 2 \, x e^{x} \log \relax (2) + x e^{\left (2 \, x\right )} - e^{\left (2 \, x\right )} \log \relax (2)\right )}}{x^{4} - 20 \, x^{3} + 150 \, x^{2} - 500 \, x + 625} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.13, size = 49, normalized size = 2.23
method | result | size |
norman | \(\frac {3 x^{3}-3 x^{2} \ln \relax (2)+3 x \,{\mathrm e}^{2 x}+6 \,{\mathrm e}^{x} x^{2}-3 \ln \relax (2) {\mathrm e}^{2 x}-6 x \ln \relax (2) {\mathrm e}^{x}}{\left (x -5\right )^{4}}\) | \(49\) |
risch | \(\frac {-3 x^{2} \ln \relax (2)+3 x^{3}}{x^{4}-20 x^{3}+150 x^{2}-500 x +625}-\frac {3 \left (\ln \relax (2)-x \right ) {\mathrm e}^{2 x}}{\left (x -5\right )^{4}}-\frac {6 x \left (\ln \relax (2)-x \right ) {\mathrm e}^{x}}{\left (x -5\right )^{4}}\) | \(69\) |
default | \(\frac {375}{\left (x -5\right )^{4}}+\frac {3}{x -5}+\frac {15 \,{\mathrm e}^{2 x}}{\left (x -5\right )^{4}}+\frac {3 \,{\mathrm e}^{2 x}}{\left (x -5\right )^{3}}+\frac {60 \,{\mathrm e}^{x}}{\left (x -5\right )^{3}}+\frac {6 \,{\mathrm e}^{x}}{\left (x -5\right )^{2}}+\frac {150 \,{\mathrm e}^{x}}{\left (x -5\right )^{4}}-\frac {3 \ln \relax (2)}{\left (x -5\right )^{2}}-\frac {75 \ln \relax (2)}{\left (x -5\right )^{4}}-\frac {30 \ln \relax (2)}{\left (x -5\right )^{3}}-\frac {3 \ln \relax (2) {\mathrm e}^{2 x}}{\left (x -5\right )^{4}}+\frac {45}{\left (x -5\right )^{2}}+\frac {225}{\left (x -5\right )^{3}}-\frac {30 \ln \relax (2) {\mathrm e}^{x}}{\left (x -5\right )^{4}}-\frac {6 \ln \relax (2) {\mathrm e}^{x}}{\left (x -5\right )^{3}}\) | \(141\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -30 \, \int \frac {e^{x}}{x^{5} - 25 \, x^{4} + 250 \, x^{3} - 1250 \, x^{2} + 3125 \, x - 3125}\,{d x} \log \relax (2) - \frac {{\left (6 \, x^{2} - 20 \, x + 25\right )} \log \relax (2)}{2 \, {\left (x^{4} - 20 \, x^{3} + 150 \, x^{2} - 500 \, x + 625\right )}} - \frac {5 \, {\left (4 \, x - 5\right )} \log \relax (2)}{2 \, {\left (x^{4} - 20 \, x^{3} + 150 \, x^{2} - 500 \, x + 625\right )}} + \frac {3 \, {\left (4 \, x^{3} - 30 \, x^{2} + 100 \, x - 125\right )}}{4 \, {\left (x^{4} - 20 \, x^{3} + 150 \, x^{2} - 500 \, x + 625\right )}} + \frac {15 \, {\left (6 \, x^{2} - 20 \, x + 25\right )}}{4 \, {\left (x^{4} - 20 \, x^{3} + 150 \, x^{2} - 500 \, x + 625\right )}} + \frac {3 \, {\left ({\left (x - \log \relax (2)\right )} e^{\left (2 \, x\right )} + 2 \, {\left (x^{2} - x \log \relax (2)\right )} e^{x}\right )}}{x^{4} - 20 \, x^{3} + 150 \, x^{2} - 500 \, x + 625} - \frac {30 \, e^{5} E_{5}\left (-x + 5\right ) \log \relax (2)}{{\left (x - 5\right )}^{4}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.24, size = 62, normalized size = 2.82 \begin {gather*} -\frac {{\mathrm {e}}^{2\,x}\,\ln \relax (8)+x^2\,\left (\ln \relax (8)-6\,{\mathrm {e}}^x\right )-x\,\left (3\,{\mathrm {e}}^{2\,x}-{\mathrm {e}}^x\,\ln \left (64\right )\right )-3\,x^3}{x^4-20\,x^3+150\,x^2-500\,x+625} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 0.80, size = 206, normalized size = 9.36 \begin {gather*} - \frac {- 3 x^{3} + 3 x^{2} \log {\relax (2 )}}{x^{4} - 20 x^{3} + 150 x^{2} - 500 x + 625} + \frac {\left (3 x^{5} - 60 x^{4} - 3 x^{4} \log {\relax (2 )} + 60 x^{3} \log {\relax (2 )} + 450 x^{3} - 1500 x^{2} - 450 x^{2} \log {\relax (2 )} + 1500 x \log {\relax (2 )} + 1875 x - 1875 \log {\relax (2 )}\right ) e^{2 x} + \left (6 x^{6} - 120 x^{5} - 6 x^{5} \log {\relax (2 )} + 120 x^{4} \log {\relax (2 )} + 900 x^{4} - 3000 x^{3} - 900 x^{3} \log {\relax (2 )} + 3000 x^{2} \log {\relax (2 )} + 3750 x^{2} - 3750 x \log {\relax (2 )}\right ) e^{x}}{x^{8} - 40 x^{7} + 700 x^{6} - 7000 x^{5} + 43750 x^{4} - 175000 x^{3} + 437500 x^{2} - 625000 x + 390625} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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