Optimal. Leaf size=33 \[ \frac {2 \left (2 x+\frac {5}{5-\log \left (5-e^{x^2}\right )}\right )}{2-3 x^2} \]
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Rubi [F] time = 3.45, 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 {-1000-1500 x-1500 x^2+e^{x^2} \left (200+340 x+300 x^2-60 x^3\right )+\left (400+300 x+600 x^2+e^{x^2} \left (-80-60 x-120 x^2\right )\right ) \log \left (5-e^{x^2}\right )+\left (-40-60 x^2+e^{x^2} \left (8+12 x^2\right )\right ) \log ^2\left (5-e^{x^2}\right )}{-500+1500 x^2-1125 x^4+e^{x^2} \left (100-300 x^2+225 x^4\right )+\left (200-600 x^2+450 x^4+e^{x^2} \left (-40+120 x^2-90 x^4\right )\right ) \log \left (5-e^{x^2}\right )+\left (-20+60 x^2-45 x^4+e^{x^2} \left (4-12 x^2+9 x^4\right )\right ) \log ^2\left (5-e^{x^2}\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {4 \left (125 \left (2+3 x+3 x^2\right )+5 e^{x^2} \left (-10-17 x-15 x^2+3 x^3\right )+5 \left (-5+e^{x^2}\right ) \left (4+3 x+6 x^2\right ) \log \left (5-e^{x^2}\right )-\left (-5+e^{x^2}\right ) \left (2+3 x^2\right ) \log ^2\left (5-e^{x^2}\right )\right )}{\left (5-e^{x^2}\right ) \left (2-3 x^2\right )^2 \left (5-\log \left (5-e^{x^2}\right )\right )^2} \, dx\\ &=4 \int \frac {125 \left (2+3 x+3 x^2\right )+5 e^{x^2} \left (-10-17 x-15 x^2+3 x^3\right )+5 \left (-5+e^{x^2}\right ) \left (4+3 x+6 x^2\right ) \log \left (5-e^{x^2}\right )-\left (-5+e^{x^2}\right ) \left (2+3 x^2\right ) \log ^2\left (5-e^{x^2}\right )}{\left (5-e^{x^2}\right ) \left (2-3 x^2\right )^2 \left (5-\log \left (5-e^{x^2}\right )\right )^2} \, dx\\ &=4 \int \left (-\frac {25 x}{\left (-5+e^{x^2}\right ) \left (-2+3 x^2\right ) \left (-5+\log \left (5-e^{x^2}\right )\right )^2}+\frac {50+85 x+75 x^2-15 x^3-20 \log \left (5-e^{x^2}\right )-15 x \log \left (5-e^{x^2}\right )-30 x^2 \log \left (5-e^{x^2}\right )+2 \log ^2\left (5-e^{x^2}\right )+3 x^2 \log ^2\left (5-e^{x^2}\right )}{\left (-2+3 x^2\right )^2 \left (-5+\log \left (5-e^{x^2}\right )\right )^2}\right ) \, dx\\ &=4 \int \frac {50+85 x+75 x^2-15 x^3-20 \log \left (5-e^{x^2}\right )-15 x \log \left (5-e^{x^2}\right )-30 x^2 \log \left (5-e^{x^2}\right )+2 \log ^2\left (5-e^{x^2}\right )+3 x^2 \log ^2\left (5-e^{x^2}\right )}{\left (-2+3 x^2\right )^2 \left (-5+\log \left (5-e^{x^2}\right )\right )^2} \, dx-100 \int \frac {x}{\left (-5+e^{x^2}\right ) \left (-2+3 x^2\right ) \left (-5+\log \left (5-e^{x^2}\right )\right )^2} \, dx\\ &=4 \int \left (\frac {2+3 x^2}{\left (-2+3 x^2\right )^2}-\frac {5 x}{\left (-2+3 x^2\right ) \left (-5+\log \left (5-e^{x^2}\right )\right )^2}-\frac {15 x}{\left (-2+3 x^2\right )^2 \left (-5+\log \left (5-e^{x^2}\right )\right )}\right ) \, dx-50 \operatorname {Subst}\left (\int \frac {1}{\left (-5+e^x\right ) (-2+3 x) \left (-5+\log \left (5-e^x\right )\right )^2} \, dx,x,x^2\right )\\ &=4 \int \frac {2+3 x^2}{\left (-2+3 x^2\right )^2} \, dx-20 \int \frac {x}{\left (-2+3 x^2\right ) \left (-5+\log \left (5-e^{x^2}\right )\right )^2} \, dx-50 \operatorname {Subst}\left (\int \frac {1}{\left (-5+e^x\right ) (-2+3 x) \left (-5+\log \left (5-e^x\right )\right )^2} \, dx,x,x^2\right )-60 \int \frac {x}{\left (-2+3 x^2\right )^2 \left (-5+\log \left (5-e^{x^2}\right )\right )} \, dx\\ &=\frac {4 x}{2-3 x^2}-10 \operatorname {Subst}\left (\int \frac {1}{(-2+3 x) \left (-5+\log \left (5-e^x\right )\right )^2} \, dx,x,x^2\right )-30 \operatorname {Subst}\left (\int \frac {1}{(-2+3 x)^2 \left (-5+\log \left (5-e^x\right )\right )} \, dx,x,x^2\right )-50 \operatorname {Subst}\left (\int \frac {1}{\left (-5+e^x\right ) (-2+3 x) \left (-5+\log \left (5-e^x\right )\right )^2} \, dx,x,x^2\right )\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.12, size = 31, normalized size = 0.94 \begin {gather*} -\frac {2 \left (2 x-\frac {5}{-5+\log \left (5-e^{x^2}\right )}\right )}{-2+3 x^2} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.72, size = 46, normalized size = 1.39 \begin {gather*} \frac {2 \, {\left (2 \, x \log \left (-e^{\left (x^{2}\right )} + 5\right ) - 10 \, x - 5\right )}}{15 \, x^{2} - {\left (3 \, x^{2} - 2\right )} \log \left (-e^{\left (x^{2}\right )} + 5\right ) - 10} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.62, size = 53, normalized size = 1.61 \begin {gather*} -\frac {2 \, {\left (2 \, x \log \left (-e^{\left (x^{2}\right )} + 5\right ) - 10 \, x - 5\right )}}{3 \, x^{2} \log \left (-e^{\left (x^{2}\right )} + 5\right ) - 15 \, x^{2} - 2 \, \log \left (-e^{\left (x^{2}\right )} + 5\right ) + 10} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.04, size = 38, normalized size = 1.15
method | result | size |
risch | \(-\frac {4 x}{3 x^{2}-2}+\frac {10}{\left (3 x^{2}-2\right ) \left (\ln \left (5-{\mathrm e}^{x^{2}}\right )-5\right )}\) | \(38\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.44, size = 46, normalized size = 1.39 \begin {gather*} \frac {2 \, {\left (2 \, x \log \left (-e^{\left (x^{2}\right )} + 5\right ) - 10 \, x - 5\right )}}{15 \, x^{2} - {\left (3 \, x^{2} - 2\right )} \log \left (-e^{\left (x^{2}\right )} + 5\right ) - 10} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 2.82, size = 41, normalized size = 1.24 \begin {gather*} \frac {2\,\left (10\,x-2\,x\,\ln \left (5-{\mathrm {e}}^{x^2}\right )+5\right )}{\left (3\,x^2-2\right )\,\left (\ln \left (5-{\mathrm {e}}^{x^2}\right )-5\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.40, size = 32, normalized size = 0.97 \begin {gather*} - \frac {4 x}{3 x^{2} - 2} + \frac {10}{- 15 x^{2} + \left (3 x^{2} - 2\right ) \log {\left (5 - e^{x^{2}} \right )} + 10} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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