Optimal. Leaf size=33 \[ \frac {e^x}{2 \left (\frac {1}{5} \left (3+e^2\right )+\frac {5}{2 x}+x\right ) (2+\log (x))} \]
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Rubi [F] time = 4.72, 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 {e^x \left (125+220 x-90 x^2+100 x^3+e^2 \left (-10 x+20 x^2\right )\right )+e^x \left (125+125 x-20 x^2+10 e^2 x^2+50 x^3\right ) \log (x)}{2500+1200 x+2144 x^2+16 e^4 x^2+480 x^3+400 x^4+e^2 \left (400 x+96 x^2+160 x^3\right )+\left (2500+1200 x+2144 x^2+16 e^4 x^2+480 x^3+400 x^4+e^2 \left (400 x+96 x^2+160 x^3\right )\right ) \log (x)+\left (625+300 x+536 x^2+4 e^4 x^2+120 x^3+100 x^4+e^2 \left (100 x+24 x^2+40 x^3\right )\right ) \log ^2(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 {e^x \left (125+220 x-90 x^2+100 x^3+e^2 \left (-10 x+20 x^2\right )\right )+e^x \left (125+125 x-20 x^2+10 e^2 x^2+50 x^3\right ) \log (x)}{2500+1200 x+\left (2144+16 e^4\right ) x^2+480 x^3+400 x^4+e^2 \left (400 x+96 x^2+160 x^3\right )+\left (2500+1200 x+2144 x^2+16 e^4 x^2+480 x^3+400 x^4+e^2 \left (400 x+96 x^2+160 x^3\right )\right ) \log (x)+\left (625+300 x+536 x^2+4 e^4 x^2+120 x^3+100 x^4+e^2 \left (100 x+24 x^2+40 x^3\right )\right ) \log ^2(x)} \, dx\\ &=\int \frac {5 e^x \left (25-2 \left (-22+e^2\right ) x+2 \left (-9+2 e^2\right ) x^2+20 x^3+\left (25+25 x+2 \left (-2+e^2\right ) x^2+10 x^3\right ) \log (x)\right )}{\left (25+2 \left (3+e^2\right ) x+10 x^2\right )^2 (2+\log (x))^2} \, dx\\ &=5 \int \frac {e^x \left (25-2 \left (-22+e^2\right ) x+2 \left (-9+2 e^2\right ) x^2+20 x^3+\left (25+25 x+2 \left (-2+e^2\right ) x^2+10 x^3\right ) \log (x)\right )}{\left (25+2 \left (3+e^2\right ) x+10 x^2\right )^2 (2+\log (x))^2} \, dx\\ &=5 \int \left (\frac {e^x}{\left (-25-2 \left (3+e^2\right ) x-10 x^2\right ) (2+\log (x))^2}+\frac {e^x \left (25+25 x-2 \left (2-e^2\right ) x^2+10 x^3\right )}{\left (25+2 \left (3+e^2\right ) x+10 x^2\right )^2 (2+\log (x))}\right ) \, dx\\ &=5 \int \frac {e^x}{\left (-25-2 \left (3+e^2\right ) x-10 x^2\right ) (2+\log (x))^2} \, dx+5 \int \frac {e^x \left (25+25 x-2 \left (2-e^2\right ) x^2+10 x^3\right )}{\left (25+2 \left (3+e^2\right ) x+10 x^2\right )^2 (2+\log (x))} \, dx\\ &=5 \int \left (-\frac {10 i e^x}{\sqrt {241-6 e^2-e^4} \left (-6-2 e^2+2 i \sqrt {241-6 e^2-e^4}-20 x\right ) (2+\log (x))^2}-\frac {10 i e^x}{\sqrt {241-6 e^2-e^4} \left (6+2 e^2+2 i \sqrt {241-6 e^2-e^4}+20 x\right ) (2+\log (x))^2}\right ) \, dx+5 \int \left (\frac {2 e^x \left (25+\left (3+e^2\right ) x\right )}{\left (25+2 \left (3+e^2\right ) x+10 x^2\right )^2 (2+\log (x))}+\frac {e^x (-1+x)}{\left (25+2 \left (3+e^2\right ) x+10 x^2\right ) (2+\log (x))}\right ) \, dx\\ &=5 \int \frac {e^x (-1+x)}{\left (25+2 \left (3+e^2\right ) x+10 x^2\right ) (2+\log (x))} \, dx+10 \int \frac {e^x \left (25+\left (3+e^2\right ) x\right )}{\left (25+2 \left (3+e^2\right ) x+10 x^2\right )^2 (2+\log (x))} \, dx-\frac {(50 i) \int \frac {e^x}{\left (-6-2 e^2+2 i \sqrt {241-6 e^2-e^4}-20 x\right ) (2+\log (x))^2} \, dx}{\sqrt {241-6 e^2-e^4}}-\frac {(50 i) \int \frac {e^x}{\left (6+2 e^2+2 i \sqrt {241-6 e^2-e^4}+20 x\right ) (2+\log (x))^2} \, dx}{\sqrt {241-6 e^2-e^4}}\\ &=5 \int \left (\frac {e^x}{\left (-25-2 \left (3+e^2\right ) x-10 x^2\right ) (2+\log (x))}+\frac {e^x x}{\left (25+2 \left (3+e^2\right ) x+10 x^2\right ) (2+\log (x))}\right ) \, dx+10 \int \left (\frac {25 e^x}{\left (25+2 \left (3+e^2\right ) x+10 x^2\right )^2 (2+\log (x))}+\frac {e^x \left (3+e^2\right ) x}{\left (25+2 \left (3+e^2\right ) x+10 x^2\right )^2 (2+\log (x))}\right ) \, dx-\frac {(50 i) \int \frac {e^x}{\left (-6-2 e^2+2 i \sqrt {241-6 e^2-e^4}-20 x\right ) (2+\log (x))^2} \, dx}{\sqrt {241-6 e^2-e^4}}-\frac {(50 i) \int \frac {e^x}{\left (6+2 e^2+2 i \sqrt {241-6 e^2-e^4}+20 x\right ) (2+\log (x))^2} \, dx}{\sqrt {241-6 e^2-e^4}}\\ &=5 \int \frac {e^x}{\left (-25-2 \left (3+e^2\right ) x-10 x^2\right ) (2+\log (x))} \, dx+5 \int \frac {e^x x}{\left (25+2 \left (3+e^2\right ) x+10 x^2\right ) (2+\log (x))} \, dx+250 \int \frac {e^x}{\left (25+2 \left (3+e^2\right ) x+10 x^2\right )^2 (2+\log (x))} \, dx+\left (10 \left (3+e^2\right )\right ) \int \frac {e^x x}{\left (25+2 \left (3+e^2\right ) x+10 x^2\right )^2 (2+\log (x))} \, dx-\frac {(50 i) \int \frac {e^x}{\left (-6-2 e^2+2 i \sqrt {241-6 e^2-e^4}-20 x\right ) (2+\log (x))^2} \, dx}{\sqrt {241-6 e^2-e^4}}-\frac {(50 i) \int \frac {e^x}{\left (6+2 e^2+2 i \sqrt {241-6 e^2-e^4}+20 x\right ) (2+\log (x))^2} \, dx}{\sqrt {241-6 e^2-e^4}}\\ &=5 \int \left (-\frac {10 i e^x}{\sqrt {241-6 e^2-e^4} \left (-6-2 e^2+2 i \sqrt {241-6 e^2-e^4}-20 x\right ) (2+\log (x))}-\frac {10 i e^x}{\sqrt {241-6 e^2-e^4} \left (6+2 e^2+2 i \sqrt {241-6 e^2-e^4}+20 x\right ) (2+\log (x))}\right ) \, dx+5 \int \left (\frac {e^x \left (1+\frac {i \left (3+e^2\right )}{\sqrt {241-6 e^2-e^4}}\right )}{\left (2 \left (3+e^2\right )-2 i \sqrt {241-6 e^2-e^4}+20 x\right ) (2+\log (x))}+\frac {e^x \left (1-\frac {i \left (3+e^2\right )}{\sqrt {241-6 e^2-e^4}}\right )}{\left (2 \left (3+e^2\right )+2 i \sqrt {241-6 e^2-e^4}+20 x\right ) (2+\log (x))}\right ) \, dx+250 \int \frac {e^x}{\left (25+2 \left (3+e^2\right ) x+10 x^2\right )^2 (2+\log (x))} \, dx+\left (10 \left (3+e^2\right )\right ) \int \frac {e^x x}{\left (25+2 \left (3+e^2\right ) x+10 x^2\right )^2 (2+\log (x))} \, dx-\frac {(50 i) \int \frac {e^x}{\left (-6-2 e^2+2 i \sqrt {241-6 e^2-e^4}-20 x\right ) (2+\log (x))^2} \, dx}{\sqrt {241-6 e^2-e^4}}-\frac {(50 i) \int \frac {e^x}{\left (6+2 e^2+2 i \sqrt {241-6 e^2-e^4}+20 x\right ) (2+\log (x))^2} \, dx}{\sqrt {241-6 e^2-e^4}}\\ &=250 \int \frac {e^x}{\left (25+2 \left (3+e^2\right ) x+10 x^2\right )^2 (2+\log (x))} \, dx+\left (10 \left (3+e^2\right )\right ) \int \frac {e^x x}{\left (25+2 \left (3+e^2\right ) x+10 x^2\right )^2 (2+\log (x))} \, dx-\frac {(50 i) \int \frac {e^x}{\left (-6-2 e^2+2 i \sqrt {241-6 e^2-e^4}-20 x\right ) (2+\log (x))^2} \, dx}{\sqrt {241-6 e^2-e^4}}-\frac {(50 i) \int \frac {e^x}{\left (6+2 e^2+2 i \sqrt {241-6 e^2-e^4}+20 x\right ) (2+\log (x))^2} \, dx}{\sqrt {241-6 e^2-e^4}}-\frac {(50 i) \int \frac {e^x}{\left (-6-2 e^2+2 i \sqrt {241-6 e^2-e^4}-20 x\right ) (2+\log (x))} \, dx}{\sqrt {241-6 e^2-e^4}}-\frac {(50 i) \int \frac {e^x}{\left (6+2 e^2+2 i \sqrt {241-6 e^2-e^4}+20 x\right ) (2+\log (x))} \, dx}{\sqrt {241-6 e^2-e^4}}+\left (5 \left (1-\frac {i \left (3+e^2\right )}{\sqrt {241-6 e^2-e^4}}\right )\right ) \int \frac {e^x}{\left (2 \left (3+e^2\right )+2 i \sqrt {241-6 e^2-e^4}+20 x\right ) (2+\log (x))} \, dx+\left (5 \left (1+\frac {i \left (3+e^2\right )}{\sqrt {241-6 e^2-e^4}}\right )\right ) \int \frac {e^x}{\left (2 \left (3+e^2\right )-2 i \sqrt {241-6 e^2-e^4}+20 x\right ) (2+\log (x))} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.98, size = 30, normalized size = 0.91 \begin {gather*} \frac {5 e^x x}{\left (25+6 x+2 e^2 x+10 x^2\right ) (2+\log (x))} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.53, size = 40, normalized size = 1.21 \begin {gather*} \frac {5 \, x e^{x}}{20 \, x^{2} + 4 \, x e^{2} + {\left (10 \, x^{2} + 2 \, x e^{2} + 6 \, x + 25\right )} \log \relax (x) + 12 \, x + 50} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.62, size = 45, normalized size = 1.36 \begin {gather*} \frac {5 \, x e^{x}}{10 \, x^{2} \log \relax (x) + 2 \, x e^{2} \log \relax (x) + 20 \, x^{2} + 4 \, x e^{2} + 6 \, x \log \relax (x) + 12 \, x + 25 \, \log \relax (x) + 50} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.06, size = 29, normalized size = 0.88
method | result | size |
risch | \(\frac {5 x \,{\mathrm e}^{x}}{\left (2 \,{\mathrm e}^{2} x +10 x^{2}+6 x +25\right ) \left (\ln \relax (x )+2\right )}\) | \(29\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.42, size = 38, normalized size = 1.15 \begin {gather*} \frac {5 \, x e^{x}}{20 \, x^{2} + 4 \, x {\left (e^{2} + 3\right )} + {\left (10 \, x^{2} + 2 \, x {\left (e^{2} + 3\right )} + 25\right )} \log \relax (x) + 50} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 4.42, size = 28, normalized size = 0.85 \begin {gather*} \frac {5\,x\,{\mathrm {e}}^x}{\left (\ln \relax (x)+2\right )\,\left (6\,x+2\,x\,{\mathrm {e}}^2+10\,x^2+25\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 0.51, size = 53, normalized size = 1.61 \begin {gather*} \frac {5 x e^{x}}{10 x^{2} \log {\relax (x )} + 20 x^{2} + 6 x \log {\relax (x )} + 2 x e^{2} \log {\relax (x )} + 12 x + 4 x e^{2} + 25 \log {\relax (x )} + 50} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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