Optimal. Leaf size=37 \[ x+\frac {1}{20} e^{2 x} x \left (-x+\log \left (\frac {(3-x) \left (-e^x+5 x\right )}{x}\right )\right )^2 \]
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Rubi [F] time = 53.69, 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 {300 x-100 x^2+e^x (-60+20 x)+e^{3 x} \left (-6 x-3 x^2-5 x^3+2 x^4\right )+e^{2 x} \left (55 x^3+15 x^4-10 x^5\right )+\left (e^{3 x} \left (6+6 x+10 x^2-4 x^3\right )+e^{2 x} \left (-70 x^2-40 x^3+20 x^4\right )\right ) \log \left (\frac {e^x (-3+x)+15 x-5 x^2}{x}\right )+\left (e^{3 x} \left (-3-5 x+2 x^2\right )+e^{2 x} \left (15 x+25 x^2-10 x^3\right )\right ) \log ^2\left (\frac {e^x (-3+x)+15 x-5 x^2}{x}\right )}{300 x-100 x^2+e^x (-60+20 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 {-300 x+100 x^2-e^x (-60+20 x)-e^{3 x} \left (-6 x-3 x^2-5 x^3+2 x^4\right )-e^{2 x} \left (55 x^3+15 x^4-10 x^5\right )-\left (e^{3 x} \left (6+6 x+10 x^2-4 x^3\right )+e^{2 x} \left (-70 x^2-40 x^3+20 x^4\right )\right ) \log \left (\frac {e^x (-3+x)+15 x-5 x^2}{x}\right )-\left (e^{3 x} \left (-3-5 x+2 x^2\right )+e^{2 x} \left (15 x+25 x^2-10 x^3\right )\right ) \log ^2\left (\frac {e^x (-3+x)+15 x-5 x^2}{x}\right )}{20 \left (e^x-5 x\right ) (3-x)} \, dx\\ &=\frac {1}{20} \int \frac {-300 x+100 x^2-e^x (-60+20 x)-e^{3 x} \left (-6 x-3 x^2-5 x^3+2 x^4\right )-e^{2 x} \left (55 x^3+15 x^4-10 x^5\right )-\left (e^{3 x} \left (6+6 x+10 x^2-4 x^3\right )+e^{2 x} \left (-70 x^2-40 x^3+20 x^4\right )\right ) \log \left (\frac {e^x (-3+x)+15 x-5 x^2}{x}\right )-\left (e^{3 x} \left (-3-5 x+2 x^2\right )+e^{2 x} \left (15 x+25 x^2-10 x^3\right )\right ) \log ^2\left (\frac {e^x (-3+x)+15 x-5 x^2}{x}\right )}{\left (e^x-5 x\right ) (3-x)} \, dx\\ &=\frac {1}{20} \int \left (-10 e^x (-1+x) x \left (x-\log \left (\frac {\left (e^x-5 x\right ) (-3+x)}{x}\right )\right )-\frac {250 (-1+x) x^3 \left (x-\log \left (\frac {\left (e^x-5 x\right ) (-3+x)}{x}\right )\right )}{e^x-5 x}-10 \left (-2-5 x^3+5 x^4+5 x^2 \log \left (\frac {\left (e^x-5 x\right ) (-3+x)}{x}\right )-5 x^3 \log \left (\frac {\left (e^x-5 x\right ) (-3+x)}{x}\right )\right )+\frac {e^{2 x} \left (-6 x-3 x^2-5 x^3+2 x^4+6 \log \left (\frac {\left (e^x-5 x\right ) (-3+x)}{x}\right )+6 x \log \left (\frac {\left (e^x-5 x\right ) (-3+x)}{x}\right )+10 x^2 \log \left (\frac {\left (e^x-5 x\right ) (-3+x)}{x}\right )-4 x^3 \log \left (\frac {\left (e^x-5 x\right ) (-3+x)}{x}\right )-3 \log ^2\left (\frac {\left (e^x-5 x\right ) (-3+x)}{x}\right )-5 x \log ^2\left (\frac {\left (e^x-5 x\right ) (-3+x)}{x}\right )+2 x^2 \log ^2\left (\frac {\left (e^x-5 x\right ) (-3+x)}{x}\right )\right )}{-3+x}\right ) \, dx\\ &=\frac {1}{20} \int \frac {e^{2 x} \left (-6 x-3 x^2-5 x^3+2 x^4+6 \log \left (\frac {\left (e^x-5 x\right ) (-3+x)}{x}\right )+6 x \log \left (\frac {\left (e^x-5 x\right ) (-3+x)}{x}\right )+10 x^2 \log \left (\frac {\left (e^x-5 x\right ) (-3+x)}{x}\right )-4 x^3 \log \left (\frac {\left (e^x-5 x\right ) (-3+x)}{x}\right )-3 \log ^2\left (\frac {\left (e^x-5 x\right ) (-3+x)}{x}\right )-5 x \log ^2\left (\frac {\left (e^x-5 x\right ) (-3+x)}{x}\right )+2 x^2 \log ^2\left (\frac {\left (e^x-5 x\right ) (-3+x)}{x}\right )\right )}{-3+x} \, dx-\frac {1}{2} \int e^x (-1+x) x \left (x-\log \left (\frac {\left (e^x-5 x\right ) (-3+x)}{x}\right )\right ) \, dx-\frac {1}{2} \int \left (-2-5 x^3+5 x^4+5 x^2 \log \left (\frac {\left (e^x-5 x\right ) (-3+x)}{x}\right )-5 x^3 \log \left (\frac {\left (e^x-5 x\right ) (-3+x)}{x}\right )\right ) \, dx-\frac {25}{2} \int \frac {(-1+x) x^3 \left (x-\log \left (\frac {\left (e^x-5 x\right ) (-3+x)}{x}\right )\right )}{e^x-5 x} \, dx\\ &=x+\frac {5 x^4}{8}-\frac {x^5}{2}+\frac {1}{20} \int \frac {e^{2 x} \left (-x \left (-6-3 x-5 x^2+2 x^3\right )-\left (6+6 x+10 x^2-4 x^3\right ) \log \left (\frac {\left (e^x-5 x\right ) (-3+x)}{x}\right )-\left (-3-5 x+2 x^2\right ) \log ^2\left (\frac {\left (e^x-5 x\right ) (-3+x)}{x}\right )\right )}{3-x} \, dx-\frac {1}{2} \int \left (e^x (-1+x) x^2-e^x (-1+x) x \log \left (\frac {\left (e^x-5 x\right ) (-3+x)}{x}\right )\right ) \, dx-\frac {5}{2} \int x^2 \log \left (\frac {\left (e^x-5 x\right ) (-3+x)}{x}\right ) \, dx+\frac {5}{2} \int x^3 \log \left (\frac {\left (e^x-5 x\right ) (-3+x)}{x}\right ) \, dx-\frac {25}{2} \int \left (-\frac {x^3 \left (x-\log \left (\frac {\left (e^x-5 x\right ) (-3+x)}{x}\right )\right )}{e^x-5 x}+\frac {x^4 \left (x-\log \left (\frac {\left (e^x-5 x\right ) (-3+x)}{x}\right )\right )}{e^x-5 x}\right ) \, dx\\ &=x+\frac {5 x^4}{8}-\frac {x^5}{2}-\frac {5}{6} x^3 \log \left (-\frac {\left (e^x-5 x\right ) (3-x)}{x}\right )+\frac {5}{8} x^4 \log \left (-\frac {\left (e^x-5 x\right ) (3-x)}{x}\right )+\frac {1}{20} \int \left (\frac {e^{2 x} x \left (-6-3 x-5 x^2+2 x^3\right )}{-3+x}-\frac {2 e^{2 x} \left (-3-3 x-5 x^2+2 x^3\right ) \log \left (\frac {\left (e^x-5 x\right ) (-3+x)}{x}\right )}{-3+x}+e^{2 x} (1+2 x) \log ^2\left (\frac {\left (e^x-5 x\right ) (-3+x)}{x}\right )\right ) \, dx-\frac {1}{2} \int e^x (-1+x) x^2 \, dx+\frac {1}{2} \int e^x (-1+x) x \log \left (\frac {\left (e^x-5 x\right ) (-3+x)}{x}\right ) \, dx-\frac {5}{8} \int \frac {5 x^5-e^x x^3 \left (3-3 x+x^2\right )}{\left (e^x-5 x\right ) (3-x)} \, dx+\frac {5}{6} \int \frac {5 x^4-e^x x^2 \left (3-3 x+x^2\right )}{\left (e^x-5 x\right ) (3-x)} \, dx+\frac {25}{2} \int \frac {x^3 \left (x-\log \left (\frac {\left (e^x-5 x\right ) (-3+x)}{x}\right )\right )}{e^x-5 x} \, dx-\frac {25}{2} \int \frac {x^4 \left (x-\log \left (\frac {\left (e^x-5 x\right ) (-3+x)}{x}\right )\right )}{e^x-5 x} \, dx\\ &=x+\frac {5 x^4}{8}-\frac {x^5}{2}+\frac {3}{2} e^x \log \left (-\frac {\left (e^x-5 x\right ) (3-x)}{x}\right )-\frac {3}{2} e^x x \log \left (-\frac {\left (e^x-5 x\right ) (3-x)}{x}\right )+\frac {1}{2} e^x x^2 \log \left (-\frac {\left (e^x-5 x\right ) (3-x)}{x}\right )-\frac {5}{6} x^3 \log \left (-\frac {\left (e^x-5 x\right ) (3-x)}{x}\right )+\frac {5}{8} x^4 \log \left (-\frac {\left (e^x-5 x\right ) (3-x)}{x}\right )+\frac {1}{20} \int \frac {e^{2 x} x \left (-6-3 x-5 x^2+2 x^3\right )}{-3+x} \, dx+\frac {1}{20} \int e^{2 x} (1+2 x) \log ^2\left (\frac {\left (e^x-5 x\right ) (-3+x)}{x}\right ) \, dx-\frac {1}{10} \int \frac {e^{2 x} \left (-3-3 x-5 x^2+2 x^3\right ) \log \left (\frac {\left (e^x-5 x\right ) (-3+x)}{x}\right )}{-3+x} \, dx-\frac {1}{2} \int \left (-e^x x^2+e^x x^3\right ) \, dx-\frac {1}{2} \int \frac {e^x \left (3-3 x+x^2\right ) \left (5 x^2-e^x \left (3-3 x+x^2\right )\right )}{\left (e^x-5 x\right ) (3-x) x} \, dx-\frac {5}{8} \int \left (\frac {5 (-1+x) x^4}{e^x-5 x}+\frac {x^3 \left (3-3 x+x^2\right )}{-3+x}\right ) \, dx+\frac {5}{6} \int \left (\frac {5 (-1+x) x^3}{e^x-5 x}+\frac {x^2 \left (3-3 x+x^2\right )}{-3+x}\right ) \, dx+\frac {25}{2} \int \left (\frac {x^4}{e^x-5 x}-\frac {x^3 \log \left (\frac {\left (e^x-5 x\right ) (-3+x)}{x}\right )}{e^x-5 x}\right ) \, dx-\frac {25}{2} \int \left (\frac {x^5}{e^x-5 x}-\frac {x^4 \log \left (\frac {\left (e^x-5 x\right ) (-3+x)}{x}\right )}{e^x-5 x}\right ) \, dx\\ &=x+\frac {5 x^4}{8}-\frac {x^5}{2}+\frac {3}{2} e^x \log \left (-\frac {\left (e^x-5 x\right ) (3-x)}{x}\right )-\frac {1}{40} e^{2 x} \log \left (-\frac {\left (e^x-5 x\right ) (3-x)}{x}\right )-\frac {3}{2} e^x x \log \left (-\frac {\left (e^x-5 x\right ) (3-x)}{x}\right )+\frac {1}{20} e^{2 x} x \log \left (-\frac {\left (e^x-5 x\right ) (3-x)}{x}\right )+\frac {1}{2} e^x x^2 \log \left (-\frac {\left (e^x-5 x\right ) (3-x)}{x}\right )-\frac {1}{10} e^{2 x} x^2 \log \left (-\frac {\left (e^x-5 x\right ) (3-x)}{x}\right )-\frac {5}{6} x^3 \log \left (-\frac {\left (e^x-5 x\right ) (3-x)}{x}\right )+\frac {5}{8} x^4 \log \left (-\frac {\left (e^x-5 x\right ) (3-x)}{x}\right )+\frac {3}{10} e^6 \text {Ei}(-2 (3-x)) \log \left (-\frac {\left (e^x-5 x\right ) (3-x)}{x}\right )+\frac {1}{20} \int \left (-6 e^{2 x}-\frac {18 e^{2 x}}{-3+x}+e^{2 x} x^2+2 e^{2 x} x^3\right ) \, dx+\frac {1}{20} \int \left (e^{2 x} \log ^2\left (\frac {\left (e^x-5 x\right ) (-3+x)}{x}\right )+2 e^{2 x} x \log ^2\left (\frac {\left (e^x-5 x\right ) (-3+x)}{x}\right )\right ) \, dx+\frac {1}{10} \int \frac {\left (5 x^2-e^x \left (3-3 x+x^2\right )\right ) \left (e^{2 x} \left (1-2 x+4 x^2\right )-12 e^6 \text {Ei}(-6+2 x)\right )}{4 \left (e^x-5 x\right ) (3-x) x} \, dx+\frac {1}{2} \int e^x x^2 \, dx-\frac {1}{2} \int e^x x^3 \, dx-\frac {1}{2} \int \left (\frac {e^x \left (3-3 x+x^2\right )^2}{(-3+x) x}+\frac {5 e^x \left (-3+6 x-4 x^2+x^3\right )}{e^x-5 x}\right ) \, dx-\frac {5}{8} \int \frac {x^3 \left (3-3 x+x^2\right )}{-3+x} \, dx+\frac {5}{6} \int \frac {x^2 \left (3-3 x+x^2\right )}{-3+x} \, dx-\frac {25}{8} \int \frac {(-1+x) x^4}{e^x-5 x} \, dx+\frac {25}{6} \int \frac {(-1+x) x^3}{e^x-5 x} \, dx+\frac {25}{2} \int \frac {x^4}{e^x-5 x} \, dx-\frac {25}{2} \int \frac {x^5}{e^x-5 x} \, dx-\frac {25}{2} \int \frac {x^3 \log \left (\frac {\left (e^x-5 x\right ) (-3+x)}{x}\right )}{e^x-5 x} \, dx+\frac {25}{2} \int \frac {x^4 \log \left (\frac {\left (e^x-5 x\right ) (-3+x)}{x}\right )}{e^x-5 x} \, 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 = 62, normalized size = 1.68 \begin {gather*} \frac {1}{20} x \left (20+e^{2 x} x^2-2 e^{2 x} x \log \left (\frac {\left (e^x-5 x\right ) (-3+x)}{x}\right )+e^{2 x} \log ^2\left (\frac {\left (e^x-5 x\right ) (-3+x)}{x}\right )\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.88, size = 73, normalized size = 1.97 \begin {gather*} \frac {1}{20} \, x^{3} e^{\left (2 \, x\right )} - \frac {1}{10} \, x^{2} e^{\left (2 \, x\right )} \log \left (-\frac {5 \, x^{2} - {\left (x - 3\right )} e^{x} - 15 \, x}{x}\right ) + \frac {1}{20} \, x e^{\left (2 \, x\right )} \log \left (-\frac {5 \, x^{2} - {\left (x - 3\right )} e^{x} - 15 \, x}{x}\right )^{2} + x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 1.18, size = 77, normalized size = 2.08 \begin {gather*} \frac {1}{20} \, x^{3} e^{\left (2 \, x\right )} - \frac {1}{10} \, x^{2} e^{\left (2 \, x\right )} \log \left (-\frac {5 \, x^{2} - x e^{x} - 15 \, x + 3 \, e^{x}}{x}\right ) + \frac {1}{20} \, x e^{\left (2 \, x\right )} \log \left (-\frac {5 \, x^{2} - x e^{x} - 15 \, x + 3 \, e^{x}}{x}\right )^{2} + x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.40, size = 2157, normalized size = 58.30
method | result | size |
risch | \(\text {Expression too large to display}\) | \(2157\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.75, size = 106, normalized size = 2.86 \begin {gather*} \frac {1}{20} \, x e^{\left (2 \, x\right )} \log \left (x - 3\right )^{2} + \frac {1}{20} \, x e^{\left (2 \, x\right )} \log \left (-5 \, x + e^{x}\right )^{2} - \frac {1}{10} \, {\left (x^{2} + x \log \relax (x)\right )} e^{\left (2 \, x\right )} \log \left (x - 3\right ) + \frac {1}{20} \, {\left (x^{3} + 2 \, x^{2} \log \relax (x) + x \log \relax (x)^{2}\right )} e^{\left (2 \, x\right )} + \frac {1}{10} \, {\left (x e^{\left (2 \, x\right )} \log \left (x - 3\right ) - {\left (x^{2} + x \log \relax (x)\right )} e^{\left (2 \, x\right )}\right )} \log \left (-5 \, x + e^{x}\right ) + x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 4.45, size = 69, normalized size = 1.86 \begin {gather*} x+\frac {x^3\,{\mathrm {e}}^{2\,x}}{20}+\frac {x\,{\ln \left (\frac {15\,x+{\mathrm {e}}^x\,\left (x-3\right )-5\,x^2}{x}\right )}^2\,{\mathrm {e}}^{2\,x}}{20}-\frac {x^2\,\ln \left (\frac {15\,x+{\mathrm {e}}^x\,\left (x-3\right )-5\,x^2}{x}\right )\,{\mathrm {e}}^{2\,x}}{10} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ShapeError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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