Optimal. Leaf size=27 \[ e^{2 x} \left (e^3+e^x+x-\log \left (1+5 x+4 x^2\right )\right ) \]
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Rubi [A]
time = 1.28, antiderivative size = 38, normalized size of antiderivative = 1.41, number of steps
used = 36, number of rules used = 9, integrand size = 92, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.098, Rules used = {6860, 2225,
6820, 6874, 2300, 2209, 2302, 2207, 2634} \begin {gather*} -e^{2 x} \log \left (4 x^2+5 x+1\right )+e^{2 x} x+e^{3 x}+e^{2 x+3} \end {gather*}
Antiderivative was successfully verified.
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Rule 2207
Rule 2209
Rule 2225
Rule 2300
Rule 2302
Rule 2634
Rule 6820
Rule 6860
Rule 6874
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \left (3 e^{3 x}+\frac {e^{2 x} \left (-4 \left (1-\frac {e^3}{2}\right )-\left (1-10 e^3\right ) x+14 \left (1+\frac {4 e^3}{7}\right ) x^2+8 x^3-2 \log \left (1+5 x+4 x^2\right )-10 x \log \left (1+5 x+4 x^2\right )-8 x^2 \log \left (1+5 x+4 x^2\right )\right )}{(1+x) (1+4 x)}\right ) \, dx\\ &=3 \int e^{3 x} \, dx+\int \frac {e^{2 x} \left (-4 \left (1-\frac {e^3}{2}\right )-\left (1-10 e^3\right ) x+14 \left (1+\frac {4 e^3}{7}\right ) x^2+8 x^3-2 \log \left (1+5 x+4 x^2\right )-10 x \log \left (1+5 x+4 x^2\right )-8 x^2 \log \left (1+5 x+4 x^2\right )\right )}{(1+x) (1+4 x)} \, dx\\ &=e^{3 x}+\int \frac {e^{2 x} \left (-4-x+14 x^2+8 x^3+2 e^3 \left (1+5 x+4 x^2\right )-2 \left (1+5 x+4 x^2\right ) \log \left (1+5 x+4 x^2\right )\right )}{(1+x) (1+4 x)} \, dx\\ &=e^{3 x}+\int \left (2 e^{3+2 x}-\frac {4 e^{2 x}}{1+5 x+4 x^2}-\frac {e^{2 x} x}{1+5 x+4 x^2}+\frac {14 e^{2 x} x^2}{1+5 x+4 x^2}+\frac {8 e^{2 x} x^3}{1+5 x+4 x^2}-2 e^{2 x} \log \left (1+5 x+4 x^2\right )\right ) \, dx\\ &=e^{3 x}+2 \int e^{3+2 x} \, dx-2 \int e^{2 x} \log \left (1+5 x+4 x^2\right ) \, dx-4 \int \frac {e^{2 x}}{1+5 x+4 x^2} \, dx+8 \int \frac {e^{2 x} x^3}{1+5 x+4 x^2} \, dx+14 \int \frac {e^{2 x} x^2}{1+5 x+4 x^2} \, dx-\int \frac {e^{2 x} x}{1+5 x+4 x^2} \, dx\\ &=e^{3 x}+e^{3+2 x}-e^{2 x} \log \left (1+5 x+4 x^2\right )+2 \int \frac {e^{2 x} (5+8 x)}{2+10 x+8 x^2} \, dx-4 \int \left (-\frac {8 e^{2 x}}{3 (-2-8 x)}-\frac {8 e^{2 x}}{3 (8+8 x)}\right ) \, dx+8 \int \left (-\frac {5 e^{2 x}}{16}+\frac {1}{4} e^{2 x} x+\frac {e^{2 x} (5+21 x)}{16 \left (1+5 x+4 x^2\right )}\right ) \, dx+14 \int \left (\frac {e^{2 x}}{4}-\frac {e^{2 x} (1+5 x)}{4 \left (1+5 x+4 x^2\right )}\right ) \, dx-\int \left (-\frac {2 e^{2 x}}{3 (2+8 x)}+\frac {8 e^{2 x}}{3 (8+8 x)}\right ) \, dx\\ &=e^{3 x}+e^{3+2 x}-e^{2 x} \log \left (1+5 x+4 x^2\right )+\frac {1}{2} \int \frac {e^{2 x} (5+21 x)}{1+5 x+4 x^2} \, dx+\frac {2}{3} \int \frac {e^{2 x}}{2+8 x} \, dx+2 \int e^{2 x} x \, dx+2 \int \left (\frac {8 e^{2 x}}{4+16 x}+\frac {8 e^{2 x}}{16+16 x}\right ) \, dx-\frac {5}{2} \int e^{2 x} \, dx-\frac {8}{3} \int \frac {e^{2 x}}{8+8 x} \, dx+\frac {7}{2} \int e^{2 x} \, dx-\frac {7}{2} \int \frac {e^{2 x} (1+5 x)}{1+5 x+4 x^2} \, dx+\frac {32}{3} \int \frac {e^{2 x}}{-2-8 x} \, dx+\frac {32}{3} \int \frac {e^{2 x}}{8+8 x} \, dx\\ &=\frac {e^{2 x}}{2}+e^{3 x}+e^{3+2 x}+e^{2 x} x+\frac {\text {Ei}(2 (1+x))}{e^2}-\frac {5 \text {Ei}\left (\frac {1}{2} (1+4 x)\right )}{4 \sqrt {e}}-e^{2 x} \log \left (1+5 x+4 x^2\right )+\frac {1}{2} \int \left (-\frac {2 e^{2 x}}{3 (2+8 x)}+\frac {128 e^{2 x}}{3 (8+8 x)}\right ) \, dx-\frac {7}{2} \int \left (-\frac {2 e^{2 x}}{3 (2+8 x)}+\frac {32 e^{2 x}}{3 (8+8 x)}\right ) \, dx+16 \int \frac {e^{2 x}}{4+16 x} \, dx+16 \int \frac {e^{2 x}}{16+16 x} \, dx-\int e^{2 x} \, dx\\ &=e^{3 x}+e^{3+2 x}+e^{2 x} x+\frac {2 \text {Ei}(2 (1+x))}{e^2}-\frac {\text {Ei}\left (\frac {1}{2} (1+4 x)\right )}{4 \sqrt {e}}-e^{2 x} \log \left (1+5 x+4 x^2\right )-\frac {1}{3} \int \frac {e^{2 x}}{2+8 x} \, dx+\frac {7}{3} \int \frac {e^{2 x}}{2+8 x} \, dx+\frac {64}{3} \int \frac {e^{2 x}}{8+8 x} \, dx-\frac {112}{3} \int \frac {e^{2 x}}{8+8 x} \, dx\\ &=e^{3 x}+e^{3+2 x}+e^{2 x} x-e^{2 x} \log \left (1+5 x+4 x^2\right )\\ \end {aligned} \end {gather*}
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Mathematica [A]
time = 0.04, size = 27, normalized size = 1.00 \begin {gather*} e^{2 x} \left (e^3+e^x+x-\log \left (1+5 x+4 x^2\right )\right ) \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 3.18, size = 36, normalized size = 1.33
method | result | size |
risch | \({\mathrm e}^{2 x +3}+x \,{\mathrm e}^{2 x}+{\mathrm e}^{3 x}-{\mathrm e}^{2 x} \ln \left (4 x^{2}+5 x +1\right )\) | \(35\) |
default | \({\mathrm e}^{3} {\mathrm e}^{2 x}+x \,{\mathrm e}^{2 x}-{\mathrm e}^{2 x} \ln \left (4 x^{2}+5 x +1\right )+{\mathrm e}^{3 x}\) | \(36\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.33, size = 36, normalized size = 1.33 \begin {gather*} {\left (x + e^{3}\right )} e^{\left (2 \, x\right )} - e^{\left (2 \, x\right )} \log \left (4 \, x + 1\right ) - e^{\left (2 \, x\right )} \log \left (x + 1\right ) + e^{\left (3 \, x\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.41, size = 31, normalized size = 1.15 \begin {gather*} {\left (x + e^{3}\right )} e^{\left (2 \, x\right )} - e^{\left (2 \, x\right )} \log \left (4 \, x^{2} + 5 \, x + 1\right ) + e^{\left (3 \, x\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 0.23, size = 26, normalized size = 0.96 \begin {gather*} \left (x - \log {\left (4 x^{2} + 5 x + 1 \right )} + e^{3}\right ) e^{2 x} + e^{3 x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.42, size = 34, normalized size = 1.26 \begin {gather*} x e^{\left (2 \, x\right )} - e^{\left (2 \, x\right )} \log \left (4 \, x^{2} + 5 \, x + 1\right ) + e^{\left (3 \, x\right )} + e^{\left (2 \, x + 3\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 0.38, size = 24, normalized size = 0.89 \begin {gather*} {\mathrm {e}}^{2\,x}\,\left (x+{\mathrm {e}}^3-\ln \left (4\,x^2+5\,x+1\right )+{\mathrm {e}}^x\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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