Optimal. Leaf size=31 \[ 4+x-e^{3+2 x} \left (5-x+\frac {x^2}{\log ^2\left (x-4 x^2\right )}\right ) \]
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Rubi [F] time = 1.39, 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^{3+2 x} \left (-2 x+16 x^2\right )+e^{3+2 x} \left (2 x-6 x^2-8 x^3\right ) \log \left (x-4 x^2\right )+\left (-1+4 x+e^{3+2 x} \left (9-38 x+8 x^2\right )\right ) \log ^3\left (x-4 x^2\right )}{(-1+4 x) \log ^3\left (x-4 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 \left (1+e^{3+2 x} (-9+2 x)+\frac {2 e^{3+2 x} x (-1+8 x)}{(-1+4 x) \log ^3((1-4 x) x)}-\frac {2 e^{3+2 x} x (1+x)}{\log ^2((1-4 x) x)}\right ) \, dx\\ &=x+2 \int \frac {e^{3+2 x} x (-1+8 x)}{(-1+4 x) \log ^3((1-4 x) x)} \, dx-2 \int \frac {e^{3+2 x} x (1+x)}{\log ^2((1-4 x) x)} \, dx+\int e^{3+2 x} (-9+2 x) \, dx\\ &=-\frac {1}{2} e^{3+2 x} (9-2 x)+x+2 \int \left (\frac {e^{3+2 x}}{4 \log ^3((1-4 x) x)}+\frac {2 e^{3+2 x} x}{\log ^3((1-4 x) x)}+\frac {e^{3+2 x}}{4 (-1+4 x) \log ^3((1-4 x) x)}\right ) \, dx-2 \int \left (\frac {e^{3+2 x} x}{\log ^2((1-4 x) x)}+\frac {e^{3+2 x} x^2}{\log ^2((1-4 x) x)}\right ) \, dx-\int e^{3+2 x} \, dx\\ &=-\frac {1}{2} e^{3+2 x}-\frac {1}{2} e^{3+2 x} (9-2 x)+x+\frac {1}{2} \int \frac {e^{3+2 x}}{\log ^3((1-4 x) x)} \, dx+\frac {1}{2} \int \frac {e^{3+2 x}}{(-1+4 x) \log ^3((1-4 x) x)} \, dx-2 \int \frac {e^{3+2 x} x}{\log ^2((1-4 x) x)} \, dx-2 \int \frac {e^{3+2 x} x^2}{\log ^2((1-4 x) x)} \, dx+4 \int \frac {e^{3+2 x} x}{\log ^3((1-4 x) x)} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.35, size = 35, normalized size = 1.13 \begin {gather*} e^{3+2 x} (-5+x)+x-\frac {e^{3+2 x} x^2}{\log ^2\left (x-4 x^2\right )} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.95, size = 47, normalized size = 1.52 \begin {gather*} -\frac {x^{2} e^{\left (2 \, x + 3\right )} - {\left ({\left (x - 5\right )} e^{\left (2 \, x + 3\right )} + x\right )} \log \left (-4 \, x^{2} + x\right )^{2}}{\log \left (-4 \, x^{2} + x\right )^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 1.22, size = 71, normalized size = 2.29 \begin {gather*} \frac {x e^{\left (2 \, x + 3\right )} \log \left (-4 \, x^{2} + x\right )^{2} - x^{2} e^{\left (2 \, x + 3\right )} + x \log \left (-4 \, x^{2} + x\right )^{2} - 5 \, e^{\left (2 \, x + 3\right )} \log \left (-4 \, x^{2} + x\right )^{2}}{\log \left (-4 \, x^{2} + x\right )^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.66, size = 134, normalized size = 4.32
method | result | size |
risch | \({\mathrm e}^{2 x +3} x +x -5 \,{\mathrm e}^{2 x +3}+\frac {4 x^{2} {\mathrm e}^{2 x +3}}{\left (\pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i \left (x -\frac {1}{4}\right )\right ) \mathrm {csgn}\left (i x \left (x -\frac {1}{4}\right )\right )-\pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \left (x -\frac {1}{4}\right )\right )^{2}+2 \pi \mathrm {csgn}\left (i x \left (x -\frac {1}{4}\right )\right )^{2}-\pi \,\mathrm {csgn}\left (i \left (x -\frac {1}{4}\right )\right ) \mathrm {csgn}\left (i x \left (x -\frac {1}{4}\right )\right )^{2}-\pi \mathrm {csgn}\left (i x \left (x -\frac {1}{4}\right )\right )^{3}-2 \pi +2 i \ln \relax (x )+2 i \ln \left (x -\frac {1}{4}\right )\right )^{2}}\) | \(134\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.72, size = 115, normalized size = 3.71 \begin {gather*} \frac {x \log \relax (x)^{2} + {\left ({\left (x e^{3} - 5 \, e^{3}\right )} e^{\left (2 \, x\right )} + x\right )} \log \left (-4 \, x + 1\right )^{2} - {\left (x^{2} e^{3} - {\left (x e^{3} - 5 \, e^{3}\right )} \log \relax (x)^{2}\right )} e^{\left (2 \, x\right )} + 2 \, {\left ({\left (x e^{3} - 5 \, e^{3}\right )} e^{\left (2 \, x\right )} \log \relax (x) + x \log \relax (x)\right )} \log \left (-4 \, x + 1\right )}{\log \relax (x)^{2} + 2 \, \log \relax (x) \log \left (-4 \, x + 1\right ) + \log \left (-4 \, x + 1\right )^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.65, size = 216, normalized size = 6.97 \begin {gather*} x+\frac {\frac {\left (x-4\,x^2\right )\,\left (x\,{\mathrm {e}}^{2\,x+3}+x^2\,{\mathrm {e}}^{2\,x+3}\right )}{8\,x-1}-\frac {{\mathrm {e}}^{2\,x+3}\,\ln \left (x-4\,x^2\right )\,\left (x-4\,x^2\right )\,\left (64\,x^5+136\,x^4+10\,x^3-15\,x^2+2\,x\right )}{{\left (8\,x-1\right )}^3}}{\ln \left (x-4\,x^2\right )}-\frac {x^2\,{\mathrm {e}}^{2\,x+3}+\frac {x\,{\mathrm {e}}^{2\,x+3}\,\ln \left (x-4\,x^2\right )\,\left (x-4\,x^2\right )\,\left (x+1\right )}{8\,x-1}}{{\ln \left (x-4\,x^2\right )}^2}-\frac {{\mathrm {e}}^{2\,x+3}\,\left (\frac {x^7}{2}+\frac {15\,x^6}{16}-\frac {3\,x^5}{16}-\frac {291\,x^4}{256}+\frac {2775\,x^3}{512}-\frac {493\,x^2}{256}+\frac {121\,x}{512}-\frac {5}{512}\right )}{x^3-\frac {3\,x^2}{8}+\frac {3\,x}{64}-\frac {1}{512}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.41, size = 44, normalized size = 1.42 \begin {gather*} x + \frac {\left (- x^{2} + x \log {\left (- 4 x^{2} + x \right )}^{2} - 5 \log {\left (- 4 x^{2} + x \right )}^{2}\right ) e^{2 x + 3}}{\log {\left (- 4 x^{2} + x \right )}^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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