Optimal. Leaf size=20 \[ 2 e^{1+x} \left (1-x+\log \left (-5+4 x^2\right )\right ) \]
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Rubi [A] time = 2.49, antiderivative size = 31, normalized size of antiderivative = 1.55, number of steps used = 36, number of rules used = 8, integrand size = 47, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.170, Rules used = {6725, 2271, 2178, 2176, 2194, 2269, 2554, 12} \begin {gather*} 2 e^{x+1} \log \left (4 x^2-5\right )-2 e^{x+1} x+2 e^{x+1} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 2176
Rule 2178
Rule 2194
Rule 2269
Rule 2271
Rule 2554
Rule 6725
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {26 e^{1+x} x}{-5+4 x^2}-\frac {8 e^{1+x} x^3}{-5+4 x^2}-\frac {10 e^{1+x} \log \left (-5+4 x^2\right )}{-5+4 x^2}+\frac {8 e^{1+x} x^2 \log \left (-5+4 x^2\right )}{-5+4 x^2}\right ) \, dx\\ &=-\left (8 \int \frac {e^{1+x} x^3}{-5+4 x^2} \, dx\right )+8 \int \frac {e^{1+x} x^2 \log \left (-5+4 x^2\right )}{-5+4 x^2} \, dx-10 \int \frac {e^{1+x} \log \left (-5+4 x^2\right )}{-5+4 x^2} \, dx+26 \int \frac {e^{1+x} x}{-5+4 x^2} \, dx\\ &=2 e^{1+x} \log \left (-5+4 x^2\right )-8 \int \left (\frac {1}{4} e^{1+x} x+\frac {5 e^{1+x} x}{4 \left (-5+4 x^2\right )}\right ) \, dx-8 \int \frac {e x \left (-4 e^x-\sqrt {5} e^{\frac {\sqrt {5}}{2}} \text {Ei}\left (-\frac {\sqrt {5}}{2}+x\right )+\sqrt {5} e^{-\frac {\sqrt {5}}{2}} \text {Ei}\left (\frac {\sqrt {5}}{2}+x\right )\right )}{10-8 x^2} \, dx+10 \int \frac {2 e^{1-\frac {\sqrt {5}}{2}} x \left (-e^{\sqrt {5}} \text {Ei}\left (-\frac {\sqrt {5}}{2}+x\right )+\text {Ei}\left (\frac {\sqrt {5}}{2}+x\right )\right )}{\sqrt {5} \left (5-4 x^2\right )} \, dx+26 \int \left (-\frac {e^{1+x}}{4 \left (\sqrt {5}-2 x\right )}+\frac {e^{1+x}}{4 \left (\sqrt {5}+2 x\right )}\right ) \, dx\\ &=2 e^{1+x} \log \left (-5+4 x^2\right )-2 \int e^{1+x} x \, dx-\frac {13}{2} \int \frac {e^{1+x}}{\sqrt {5}-2 x} \, dx+\frac {13}{2} \int \frac {e^{1+x}}{\sqrt {5}+2 x} \, dx-10 \int \frac {e^{1+x} x}{-5+4 x^2} \, dx-(8 e) \int \frac {x \left (-4 e^x-\sqrt {5} e^{\frac {\sqrt {5}}{2}} \text {Ei}\left (-\frac {\sqrt {5}}{2}+x\right )+\sqrt {5} e^{-\frac {\sqrt {5}}{2}} \text {Ei}\left (\frac {\sqrt {5}}{2}+x\right )\right )}{10-8 x^2} \, dx+\left (4 \sqrt {5} e^{1-\frac {\sqrt {5}}{2}}\right ) \int \frac {x \left (-e^{\sqrt {5}} \text {Ei}\left (-\frac {\sqrt {5}}{2}+x\right )+\text {Ei}\left (\frac {\sqrt {5}}{2}+x\right )\right )}{5-4 x^2} \, dx\\ &=-2 e^{1+x} x+\frac {13}{4} e^{1+\frac {\sqrt {5}}{2}} \text {Ei}\left (\frac {1}{2} \left (-\sqrt {5}+2 x\right )\right )+\frac {13}{4} e^{1-\frac {\sqrt {5}}{2}} \text {Ei}\left (\frac {1}{2} \left (\sqrt {5}+2 x\right )\right )+2 e^{1+x} \log \left (-5+4 x^2\right )+2 \int e^{1+x} \, dx-10 \int \left (-\frac {e^{1+x}}{4 \left (\sqrt {5}-2 x\right )}+\frac {e^{1+x}}{4 \left (\sqrt {5}+2 x\right )}\right ) \, dx-(8 e) \int \left (\frac {2 e^x x}{-5+4 x^2}+\frac {\sqrt {5} e^{-\frac {\sqrt {5}}{2}} x \left (e^{\sqrt {5}} \text {Ei}\left (-\frac {\sqrt {5}}{2}+x\right )-\text {Ei}\left (\frac {\sqrt {5}}{2}+x\right )\right )}{2 \left (-5+4 x^2\right )}\right ) \, dx+\left (4 \sqrt {5} e^{1-\frac {\sqrt {5}}{2}}\right ) \int \left (\frac {e^{\sqrt {5}} x \text {Ei}\left (-\frac {\sqrt {5}}{2}+x\right )}{-5+4 x^2}-\frac {x \text {Ei}\left (\frac {\sqrt {5}}{2}+x\right )}{-5+4 x^2}\right ) \, dx\\ &=2 e^{1+x}-2 e^{1+x} x+\frac {13}{4} e^{1+\frac {\sqrt {5}}{2}} \text {Ei}\left (\frac {1}{2} \left (-\sqrt {5}+2 x\right )\right )+\frac {13}{4} e^{1-\frac {\sqrt {5}}{2}} \text {Ei}\left (\frac {1}{2} \left (\sqrt {5}+2 x\right )\right )+2 e^{1+x} \log \left (-5+4 x^2\right )+\frac {5}{2} \int \frac {e^{1+x}}{\sqrt {5}-2 x} \, dx-\frac {5}{2} \int \frac {e^{1+x}}{\sqrt {5}+2 x} \, dx-(16 e) \int \frac {e^x x}{-5+4 x^2} \, dx-\left (4 \sqrt {5} e^{1-\frac {\sqrt {5}}{2}}\right ) \int \frac {x \left (e^{\sqrt {5}} \text {Ei}\left (-\frac {\sqrt {5}}{2}+x\right )-\text {Ei}\left (\frac {\sqrt {5}}{2}+x\right )\right )}{-5+4 x^2} \, dx-\left (4 \sqrt {5} e^{1-\frac {\sqrt {5}}{2}}\right ) \int \frac {x \text {Ei}\left (\frac {\sqrt {5}}{2}+x\right )}{-5+4 x^2} \, dx+\left (4 \sqrt {5} e^{1+\frac {\sqrt {5}}{2}}\right ) \int \frac {x \text {Ei}\left (-\frac {\sqrt {5}}{2}+x\right )}{-5+4 x^2} \, dx\\ &=2 e^{1+x}-2 e^{1+x} x+2 e^{1+\frac {\sqrt {5}}{2}} \text {Ei}\left (\frac {1}{2} \left (-\sqrt {5}+2 x\right )\right )+2 e^{1-\frac {\sqrt {5}}{2}} \text {Ei}\left (\frac {1}{2} \left (\sqrt {5}+2 x\right )\right )+2 e^{1+x} \log \left (-5+4 x^2\right )-(16 e) \int \left (-\frac {e^x}{4 \left (\sqrt {5}-2 x\right )}+\frac {e^x}{4 \left (\sqrt {5}+2 x\right )}\right ) \, dx-\left (4 \sqrt {5} e^{1-\frac {\sqrt {5}}{2}}\right ) \int \left (-\frac {\text {Ei}\left (\frac {\sqrt {5}}{2}+x\right )}{4 \left (\sqrt {5}-2 x\right )}+\frac {\text {Ei}\left (\frac {\sqrt {5}}{2}+x\right )}{4 \left (\sqrt {5}+2 x\right )}\right ) \, dx-\left (4 \sqrt {5} e^{1-\frac {\sqrt {5}}{2}}\right ) \int \left (\frac {e^{\sqrt {5}} x \text {Ei}\left (-\frac {\sqrt {5}}{2}+x\right )}{-5+4 x^2}-\frac {x \text {Ei}\left (\frac {\sqrt {5}}{2}+x\right )}{-5+4 x^2}\right ) \, dx+\left (4 \sqrt {5} e^{1+\frac {\sqrt {5}}{2}}\right ) \int \left (-\frac {\text {Ei}\left (-\frac {\sqrt {5}}{2}+x\right )}{4 \left (\sqrt {5}-2 x\right )}+\frac {\text {Ei}\left (-\frac {\sqrt {5}}{2}+x\right )}{4 \left (\sqrt {5}+2 x\right )}\right ) \, dx\\ &=2 e^{1+x}-2 e^{1+x} x+2 e^{1+\frac {\sqrt {5}}{2}} \text {Ei}\left (\frac {1}{2} \left (-\sqrt {5}+2 x\right )\right )+2 e^{1-\frac {\sqrt {5}}{2}} \text {Ei}\left (\frac {1}{2} \left (\sqrt {5}+2 x\right )\right )+2 e^{1+x} \log \left (-5+4 x^2\right )+(4 e) \int \frac {e^x}{\sqrt {5}-2 x} \, dx-(4 e) \int \frac {e^x}{\sqrt {5}+2 x} \, dx+\left (\sqrt {5} e^{1-\frac {\sqrt {5}}{2}}\right ) \int \frac {\text {Ei}\left (\frac {\sqrt {5}}{2}+x\right )}{\sqrt {5}-2 x} \, dx-\left (\sqrt {5} e^{1-\frac {\sqrt {5}}{2}}\right ) \int \frac {\text {Ei}\left (\frac {\sqrt {5}}{2}+x\right )}{\sqrt {5}+2 x} \, dx+\left (4 \sqrt {5} e^{1-\frac {\sqrt {5}}{2}}\right ) \int \frac {x \text {Ei}\left (\frac {\sqrt {5}}{2}+x\right )}{-5+4 x^2} \, dx-\left (\sqrt {5} e^{1+\frac {\sqrt {5}}{2}}\right ) \int \frac {\text {Ei}\left (-\frac {\sqrt {5}}{2}+x\right )}{\sqrt {5}-2 x} \, dx+\left (\sqrt {5} e^{1+\frac {\sqrt {5}}{2}}\right ) \int \frac {\text {Ei}\left (-\frac {\sqrt {5}}{2}+x\right )}{\sqrt {5}+2 x} \, dx-\left (4 \sqrt {5} e^{1+\frac {\sqrt {5}}{2}}\right ) \int \frac {x \text {Ei}\left (-\frac {\sqrt {5}}{2}+x\right )}{-5+4 x^2} \, dx\\ &=2 e^{1+x}-2 e^{1+x} x+2 e^{1+x} \log \left (-5+4 x^2\right )+\left (\sqrt {5} e^{1-\frac {\sqrt {5}}{2}}\right ) \int \frac {\text {Ei}\left (\frac {\sqrt {5}}{2}+x\right )}{\sqrt {5}-2 x} \, dx-\left (\sqrt {5} e^{1-\frac {\sqrt {5}}{2}}\right ) \int \frac {\text {Ei}\left (\frac {\sqrt {5}}{2}+x\right )}{\sqrt {5}+2 x} \, dx+\left (4 \sqrt {5} e^{1-\frac {\sqrt {5}}{2}}\right ) \int \left (-\frac {\text {Ei}\left (\frac {\sqrt {5}}{2}+x\right )}{4 \left (\sqrt {5}-2 x\right )}+\frac {\text {Ei}\left (\frac {\sqrt {5}}{2}+x\right )}{4 \left (\sqrt {5}+2 x\right )}\right ) \, dx-\left (\sqrt {5} e^{1+\frac {\sqrt {5}}{2}}\right ) \int \frac {\text {Ei}\left (-\frac {\sqrt {5}}{2}+x\right )}{\sqrt {5}-2 x} \, dx+\left (\sqrt {5} e^{1+\frac {\sqrt {5}}{2}}\right ) \int \frac {\text {Ei}\left (-\frac {\sqrt {5}}{2}+x\right )}{\sqrt {5}+2 x} \, dx-\left (4 \sqrt {5} e^{1+\frac {\sqrt {5}}{2}}\right ) \int \left (-\frac {\text {Ei}\left (-\frac {\sqrt {5}}{2}+x\right )}{4 \left (\sqrt {5}-2 x\right )}+\frac {\text {Ei}\left (-\frac {\sqrt {5}}{2}+x\right )}{4 \left (\sqrt {5}+2 x\right )}\right ) \, dx\\ &=2 e^{1+x}-2 e^{1+x} x+2 e^{1+x} \log \left (-5+4 x^2\right )\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.10, size = 20, normalized size = 1.00 \begin {gather*} -2 e^{1+x} \left (-1+x-\log \left (-5+4 x^2\right )\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.57, size = 24, normalized size = 1.20 \begin {gather*} -2 \, {\left (x - 1\right )} e^{\left (x + 1\right )} + 2 \, e^{\left (x + 1\right )} \log \left (4 \, x^{2} - 5\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.37, size = 28, normalized size = 1.40 \begin {gather*} -2 \, x e^{\left (x + 1\right )} + 2 \, e^{\left (x + 1\right )} \log \left (4 \, x^{2} - 5\right ) + 2 \, e^{\left (x + 1\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.47, size = 25, normalized size = 1.25
method | result | size |
risch | \(2 \ln \left (4 x^{2}-5\right ) {\mathrm e}^{x +1}-2 \left (x -1\right ) {\mathrm e}^{x +1}\) | \(25\) |
norman | \(2 \,{\mathrm e} \,{\mathrm e}^{x}-2 x \,{\mathrm e} \,{\mathrm e}^{x}+2 \,{\mathrm e} \,{\mathrm e}^{x} \ln \left (4 x^{2}-5\right )\) | \(29\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.45, size = 28, normalized size = 1.40 \begin {gather*} -2 \, {\left (x e - e\right )} e^{x} + 2 \, e^{\left (x + 1\right )} \log \left (4 \, x^{2} - 5\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.27, size = 28, normalized size = 1.40 \begin {gather*} 2\,\mathrm {e}\,{\mathrm {e}}^x-2\,x\,\mathrm {e}\,{\mathrm {e}}^x+2\,\ln \left (4\,x^2-5\right )\,\mathrm {e}\,{\mathrm {e}}^x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.52, size = 27, normalized size = 1.35 \begin {gather*} \left (- 2 e x + 2 e \log {\left (4 x^{2} - 5 \right )} + 2 e\right ) e^{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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