Optimal. Leaf size=25 \[ e^{4 x^2}+\frac {x}{\log \left (\frac {\left (\frac {5}{8}+x\right )^2}{\log ^2(12)}\right )} \]
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Rubi [B] time = 0.44, antiderivative size = 57, normalized size of antiderivative = 2.28, number of steps used = 14, number of rules used = 10, integrand size = 96, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.104, Rules used = {6688, 2209, 2411, 2353, 2297, 2300, 2178, 2302, 30, 2389} \begin {gather*} e^{4 x^2}+\frac {8 x+5}{8 \log \left (\frac {(8 x+5)^2}{64 \log ^2(12)}\right )}-\frac {5}{8 \log \left (\frac {(8 x+5)^2}{64 \log ^2(12)}\right )} \end {gather*}
Antiderivative was successfully verified.
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Rule 30
Rule 2178
Rule 2209
Rule 2297
Rule 2300
Rule 2302
Rule 2353
Rule 2389
Rule 2411
Rule 6688
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \left (8 e^{4 x^2} x-\frac {16 x}{(5+8 x) \log ^2\left (\frac {(5+8 x)^2}{64 \log ^2(12)}\right )}+\frac {1}{\log \left (\frac {(5+8 x)^2}{64 \log ^2(12)}\right )}\right ) \, dx\\ &=8 \int e^{4 x^2} x \, dx-16 \int \frac {x}{(5+8 x) \log ^2\left (\frac {(5+8 x)^2}{64 \log ^2(12)}\right )} \, dx+\int \frac {1}{\log \left (\frac {(5+8 x)^2}{64 \log ^2(12)}\right )} \, dx\\ &=e^{4 x^2}+\frac {1}{8} \operatorname {Subst}\left (\int \frac {1}{\log \left (\frac {x^2}{64 \log ^2(12)}\right )} \, dx,x,5+8 x\right )-2 \operatorname {Subst}\left (\int \frac {-\frac {5}{8}+\frac {x}{8}}{x \log ^2\left (\frac {x^2}{64 \log ^2(12)}\right )} \, dx,x,5+8 x\right )\\ &=e^{4 x^2}-2 \operatorname {Subst}\left (\int \left (\frac {1}{8 \log ^2\left (\frac {x^2}{64 \log ^2(12)}\right )}-\frac {5}{8 x \log ^2\left (\frac {x^2}{64 \log ^2(12)}\right )}\right ) \, dx,x,5+8 x\right )+\frac {((5+8 x) \log (12)) \operatorname {Subst}\left (\int \frac {e^{x/2}}{x} \, dx,x,\log \left (\frac {(5+8 x)^2}{64 \log ^2(12)}\right )\right )}{2 \sqrt {(5+8 x)^2}}\\ &=e^{4 x^2}+\frac {(5+8 x) \text {Ei}\left (\frac {1}{2} \log \left (\frac {(5+8 x)^2}{64 \log ^2(12)}\right )\right ) \log (12)}{2 \sqrt {(5+8 x)^2}}-\frac {1}{4} \operatorname {Subst}\left (\int \frac {1}{\log ^2\left (\frac {x^2}{64 \log ^2(12)}\right )} \, dx,x,5+8 x\right )+\frac {5}{4} \operatorname {Subst}\left (\int \frac {1}{x \log ^2\left (\frac {x^2}{64 \log ^2(12)}\right )} \, dx,x,5+8 x\right )\\ &=e^{4 x^2}+\frac {(5+8 x) \text {Ei}\left (\frac {1}{2} \log \left (\frac {(5+8 x)^2}{64 \log ^2(12)}\right )\right ) \log (12)}{2 \sqrt {(5+8 x)^2}}+\frac {5+8 x}{8 \log \left (\frac {(5+8 x)^2}{64 \log ^2(12)}\right )}-\frac {1}{8} \operatorname {Subst}\left (\int \frac {1}{\log \left (\frac {x^2}{64 \log ^2(12)}\right )} \, dx,x,5+8 x\right )+\frac {5}{8} \operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log \left (\frac {(5+8 x)^2}{64 \log ^2(12)}\right )\right )\\ &=e^{4 x^2}+\frac {(5+8 x) \text {Ei}\left (\frac {1}{2} \log \left (\frac {(5+8 x)^2}{64 \log ^2(12)}\right )\right ) \log (12)}{2 \sqrt {(5+8 x)^2}}-\frac {5}{8 \log \left (\frac {(5+8 x)^2}{64 \log ^2(12)}\right )}+\frac {5+8 x}{8 \log \left (\frac {(5+8 x)^2}{64 \log ^2(12)}\right )}-\frac {((5+8 x) \log (12)) \operatorname {Subst}\left (\int \frac {e^{x/2}}{x} \, dx,x,\log \left (\frac {(5+8 x)^2}{64 \log ^2(12)}\right )\right )}{2 \sqrt {(5+8 x)^2}}\\ &=e^{4 x^2}-\frac {5}{8 \log \left (\frac {(5+8 x)^2}{64 \log ^2(12)}\right )}+\frac {5+8 x}{8 \log \left (\frac {(5+8 x)^2}{64 \log ^2(12)}\right )}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.13, size = 28, normalized size = 1.12 \begin {gather*} e^{4 x^2}+\frac {x}{\log \left (\frac {(5+8 x)^2}{64 \log ^2(12)}\right )} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.60, size = 46, normalized size = 1.84 \begin {gather*} \frac {e^{\left (4 \, x^{2}\right )} \log \left (\frac {64 \, x^{2} + 80 \, x + 25}{64 \, \log \left (12\right )^{2}}\right ) + x}{\log \left (\frac {64 \, x^{2} + 80 \, x + 25}{64 \, \log \left (12\right )^{2}}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.38, size = 70, normalized size = 2.80 \begin {gather*} \frac {6 \, e^{\left (4 \, x^{2}\right )} \log \relax (2) - e^{\left (4 \, x^{2}\right )} \log \left (64 \, x^{2} + 80 \, x + 25\right ) + 2 \, e^{\left (4 \, x^{2}\right )} \log \left (\log \left (12\right )\right ) - x}{6 \, \log \relax (2) - \log \left (64 \, x^{2} + 80 \, x + 25\right ) + 2 \, \log \left (\log \left (12\right )\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.34, size = 38, normalized size = 1.52
method | result | size |
default | \(-\frac {x}{6 \ln \relax (2)+2 \ln \left (\ln \relax (3)+2 \ln \relax (2)\right )-\ln \left (\left (8 x +5\right )^{2}\right )}+{\mathrm e}^{4 x^{2}}\) | \(38\) |
norman | \(\frac {x +{\mathrm e}^{4 x^{2}} \ln \left (\frac {64 x^{2}+80 x +25}{64 \ln \left (12\right )^{2}}\right )}{\ln \left (\frac {64 x^{2}+80 x +25}{64 \ln \left (12\right )^{2}}\right )}\) | \(47\) |
risch | \({\mathrm e}^{4 x^{2}}+\frac {2 i x}{\pi \mathrm {csgn}\left (i \left (\frac {5}{8}+x \right )\right )^{2} \mathrm {csgn}\left (i \left (\frac {5}{8}+x \right )^{2}\right )-2 \pi \,\mathrm {csgn}\left (i \left (\frac {5}{8}+x \right )\right ) \mathrm {csgn}\left (i \left (\frac {5}{8}+x \right )^{2}\right )^{2}+\pi \mathrm {csgn}\left (i \left (\frac {5}{8}+x \right )^{2}\right )^{3}-4 i \ln \left (\ln \relax (3)+2 \ln \relax (2)\right )+4 i \ln \left (\frac {5}{8}+x \right )}\) | \(87\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.51, size = 64, normalized size = 2.56 \begin {gather*} \frac {2 \, {\left (3 \, \log \relax (2) + \log \left (\log \relax (3) + 2 \, \log \relax (2)\right )\right )} e^{\left (4 \, x^{2}\right )} - 2 \, e^{\left (4 \, x^{2}\right )} \log \left (8 \, x + 5\right ) - x}{2 \, {\left (3 \, \log \relax (2) - \log \left (8 \, x + 5\right ) + \log \left (\log \relax (3) + 2 \, \log \relax (2)\right )\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 3.85, size = 25, normalized size = 1.00 \begin {gather*} {\mathrm {e}}^{4\,x^2}+\frac {x}{\ln \left (\frac {x^2+\frac {5\,x}{4}+\frac {25}{64}}{{\ln \left (12\right )}^2}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.38, size = 26, normalized size = 1.04 \begin {gather*} \frac {x}{\log {\left (\frac {x^{2} + \frac {5 x}{4} + \frac {25}{64}}{\log {\left (12 \right )}^{2}} \right )}} + e^{4 x^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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