Optimal. Leaf size=34 \[ \frac {4 x}{\log (x) \left (-\frac {1+4 x}{x}+\frac {x \log (x)}{-e^x+\log (3)}\right )} \]
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Rubi [F]
time = 50.20, 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^{2 x} \left (4 x+16 x^2\right )+e^x \left (-8 x-32 x^2\right ) \log (3)+\left (4 x+16 x^2\right ) \log ^2(3)+\left (e^{2 x} \left (-8 x-16 x^2\right )-8 x^3 \log (3)+\left (-8 x-16 x^2\right ) \log ^2(3)+e^x \left (8 x^3+\left (16 x+32 x^2\right ) \log (3)\right )\right ) \log (x)-4 e^x x^4 \log ^2(x)}{\left (e^{2 x} \left (1+8 x+16 x^2\right )+e^x \left (-2-16 x-32 x^2\right ) \log (3)+\left (1+8 x+16 x^2\right ) \log ^2(3)\right ) \log ^2(x)+\left (e^x \left (2 x^2+8 x^3\right )+\left (-2 x^2-8 x^3\right ) \log (3)\right ) \log ^3(x)+x^4 \log ^4(x)} \, 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 x \left (e^{2 x} (1+4 x)+(1+4 x) \log ^2(3)-e^x (8 x \log (3)+\log (9))-\left (e^{2 x} (2+4 x)+2 \log ^2(3)+4 x \log ^2(3)+x^2 \log (9)-2 e^x \left (x^2+\log (9)+x \log (81)\right )\right ) \log (x)-e^x x^3 \log ^2(x)\right )}{\log ^2(x) \left ((1+4 x) \left (e^x-\log (3)\right )+x^2 \log (x)\right )^2} \, dx\\ &=4 \int \frac {x \left (e^{2 x} (1+4 x)+(1+4 x) \log ^2(3)-e^x (8 x \log (3)+\log (9))-\left (e^{2 x} (2+4 x)+2 \log ^2(3)+4 x \log ^2(3)+x^2 \log (9)-2 e^x \left (x^2+\log (9)+x \log (81)\right )\right ) \log (x)-e^x x^3 \log ^2(x)\right )}{\log ^2(x) \left ((1+4 x) \left (e^x-\log (3)\right )+x^2 \log (x)\right )^2} \, dx\\ &=4 \int \left (-\frac {x (-1-4 x+2 \log (x)+4 x \log (x))}{(1+4 x)^2 \log ^2(x)}-\frac {x^3 \left (-4-7 x+4 x^2\right )}{(1+4 x)^2 \left (e^x+4 e^x x-\log (3)-4 x \log (3)+x^2 \log (x)\right )}+\frac {x \left (-4 \log ^2(3) \left (1-\frac {\log ^2(9)}{4 \log ^2(3)}\right )-40 x \log ^2(3) \left (1-\frac {\log (9) (16 \log (3)+\log (81))}{40 \log ^2(3)}\right )-x^3 \log (3) \log (x)-4 x^5 (1+\log (81)) \log (x)-x^4 (1+\log (6561)) \log (x)-2 x^4 \log ^2(x)-3 x^5 \log ^2(x)+4 x^6 \log ^2(x)\right )}{(1+4 x)^2 \log (x) \left (e^x+4 e^x x-\log (3)-4 x \log (3)+x^2 \log (x)\right )^2}\right ) \, dx\\ &=-\left (4 \int \frac {x (-1-4 x+2 \log (x)+4 x \log (x))}{(1+4 x)^2 \log ^2(x)} \, dx\right )-4 \int \frac {x^3 \left (-4-7 x+4 x^2\right )}{(1+4 x)^2 \left (e^x+4 e^x x-\log (3)-4 x \log (3)+x^2 \log (x)\right )} \, dx+4 \int \frac {x \left (-4 \log ^2(3) \left (1-\frac {\log ^2(9)}{4 \log ^2(3)}\right )-40 x \log ^2(3) \left (1-\frac {\log (9) (16 \log (3)+\log (81))}{40 \log ^2(3)}\right )-x^3 \log (3) \log (x)-4 x^5 (1+\log (81)) \log (x)-x^4 (1+\log (6561)) \log (x)-2 x^4 \log ^2(x)-3 x^5 \log ^2(x)+4 x^6 \log ^2(x)\right )}{(1+4 x)^2 \log (x) \left (e^x+4 e^x x-\log (3)-4 x \log (3)+x^2 \log (x)\right )^2} \, dx\\ &=-\left (4 \int \left (-\frac {x}{(1+4 x) \log ^2(x)}+\frac {2 x (1+2 x)}{(1+4 x)^2 \log (x)}\right ) \, dx\right )-4 \int \left (\frac {7}{256 \left (e^x+4 e^x x-\log (3)-4 x \log (3)+x^2 \log (x)\right )}+\frac {x}{64 \left (e^x+4 e^x x-\log (3)-4 x \log (3)+x^2 \log (x)\right )}-\frac {9 x^2}{16 \left (e^x+4 e^x x-\log (3)-4 x \log (3)+x^2 \log (x)\right )}+\frac {x^3}{4 \left (e^x+4 e^x x-\log (3)-4 x \log (3)+x^2 \log (x)\right )}+\frac {1}{32 (1+4 x)^2 \left (e^x+4 e^x x-\log (3)-4 x \log (3)+x^2 \log (x)\right )}-\frac {15}{256 (1+4 x) \left (e^x+4 e^x x-\log (3)-4 x \log (3)+x^2 \log (x)\right )}\right ) \, dx+4 \int \left (\frac {4 \log ^2(3) \left (1-\frac {\log ^2(9)}{4 \log ^2(3)}\right )+40 x \log ^2(3) \left (1-\frac {\log (9) (16 \log (3)+\log (81))}{40 \log ^2(3)}\right )+x^3 \log (3) \log (x)+4 x^5 (1+\log (81)) \log (x)+x^4 (1+\log (6561)) \log (x)+2 x^4 \log ^2(x)+3 x^5 \log ^2(x)-4 x^6 \log ^2(x)}{4 (1+4 x)^2 \log (x) \left (e^x+4 e^x x-\log (3)-4 x \log (3)+x^2 \log (x)\right )^2}+\frac {-4 \log ^2(3) \left (1-\frac {\log ^2(9)}{4 \log ^2(3)}\right )-40 x \log ^2(3) \left (1-\frac {\log (9) (16 \log (3)+\log (81))}{40 \log ^2(3)}\right )-x^3 \log (3) \log (x)-4 x^5 (1+\log (81)) \log (x)-x^4 (1+\log (6561)) \log (x)-2 x^4 \log ^2(x)-3 x^5 \log ^2(x)+4 x^6 \log ^2(x)}{4 (1+4 x) \log (x) \left (e^x+4 e^x x-\log (3)-4 x \log (3)+x^2 \log (x)\right )^2}\right ) \, dx\\ &=-\left (\frac {1}{16} \int \frac {x}{e^x+4 e^x x-\log (3)-4 x \log (3)+x^2 \log (x)} \, dx\right )-\frac {7}{64} \int \frac {1}{e^x+4 e^x x-\log (3)-4 x \log (3)+x^2 \log (x)} \, dx-\frac {1}{8} \int \frac {1}{(1+4 x)^2 \left (e^x+4 e^x x-\log (3)-4 x \log (3)+x^2 \log (x)\right )} \, dx+\frac {15}{64} \int \frac {1}{(1+4 x) \left (e^x+4 e^x x-\log (3)-4 x \log (3)+x^2 \log (x)\right )} \, dx+\frac {9}{4} \int \frac {x^2}{e^x+4 e^x x-\log (3)-4 x \log (3)+x^2 \log (x)} \, dx+4 \int \frac {x}{(1+4 x) \log ^2(x)} \, dx-8 \int \frac {x (1+2 x)}{(1+4 x)^2 \log (x)} \, dx-\int \frac {x^3}{e^x+4 e^x x-\log (3)-4 x \log (3)+x^2 \log (x)} \, dx+\int \frac {4 \log ^2(3) \left (1-\frac {\log ^2(9)}{4 \log ^2(3)}\right )+40 x \log ^2(3) \left (1-\frac {\log (9) (16 \log (3)+\log (81))}{40 \log ^2(3)}\right )+x^3 \log (3) \log (x)+4 x^5 (1+\log (81)) \log (x)+x^4 (1+\log (6561)) \log (x)+2 x^4 \log ^2(x)+3 x^5 \log ^2(x)-4 x^6 \log ^2(x)}{(1+4 x)^2 \log (x) \left (e^x+4 e^x x-\log (3)-4 x \log (3)+x^2 \log (x)\right )^2} \, dx+\int \frac {-4 \log ^2(3) \left (1-\frac {\log ^2(9)}{4 \log ^2(3)}\right )-40 x \log ^2(3) \left (1-\frac {\log (9) (16 \log (3)+\log (81))}{40 \log ^2(3)}\right )-x^3 \log (3) \log (x)-4 x^5 (1+\log (81)) \log (x)-x^4 (1+\log (6561)) \log (x)-2 x^4 \log ^2(x)-3 x^5 \log ^2(x)+4 x^6 \log ^2(x)}{(1+4 x) \log (x) \left (e^x+4 e^x x-\log (3)-4 x \log (3)+x^2 \log (x)\right )^2} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}
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Mathematica [F]
time = 3.14, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {e^{2 x} \left (4 x+16 x^2\right )+e^x \left (-8 x-32 x^2\right ) \log (3)+\left (4 x+16 x^2\right ) \log ^2(3)+\left (e^{2 x} \left (-8 x-16 x^2\right )-8 x^3 \log (3)+\left (-8 x-16 x^2\right ) \log ^2(3)+e^x \left (8 x^3+\left (16 x+32 x^2\right ) \log (3)\right )\right ) \log (x)-4 e^x x^4 \log ^2(x)}{\left (e^{2 x} \left (1+8 x+16 x^2\right )+e^x \left (-2-16 x-32 x^2\right ) \log (3)+\left (1+8 x+16 x^2\right ) \log ^2(3)\right ) \log ^2(x)+\left (e^x \left (2 x^2+8 x^3\right )+\left (-2 x^2-8 x^3\right ) \log (3)\right ) \log ^3(x)+x^4 \log ^4(x)} \, dx \end {gather*}
Verification is not applicable to the result.
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Maple [A]
time = 0.56, size = 43, normalized size = 1.26
method | result | size |
risch | \(-\frac {4 \left (\ln \left (3\right )-{\mathrm e}^{x}\right ) x^{2}}{\ln \left (x \right ) \left (-x^{2} \ln \left (x \right )+4 x \ln \left (3\right )-4 \,{\mathrm e}^{x} x +\ln \left (3\right )-{\mathrm e}^{x}\right )}\) | \(43\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.70, size = 49, normalized size = 1.44 \begin {gather*} -\frac {4 \, {\left (x^{2} e^{x} - x^{2} \log \left (3\right )\right )}}{x^{2} \log \left (x\right )^{2} + {\left (4 \, x + 1\right )} e^{x} \log \left (x\right ) - {\left (4 \, x \log \left (3\right ) + \log \left (3\right )\right )} \log \left (x\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.36, size = 48, normalized size = 1.41 \begin {gather*} -\frac {4 \, {\left (x^{2} e^{x} - x^{2} \log \left (3\right )\right )}}{x^{2} \log \left (x\right )^{2} + {\left ({\left (4 \, x + 1\right )} e^{x} - {\left (4 \, x + 1\right )} \log \left (3\right )\right )} \log \left (x\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 65 vs.
\(2 (24) = 48\).
time = 0.25, size = 65, normalized size = 1.91 \begin {gather*} \frac {4 x^{4}}{4 x^{3} \log {\left (x \right )} + x^{2} \log {\left (x \right )} - 16 x^{2} \log {\left (3 \right )} - 8 x \log {\left (3 \right )} + \left (16 x^{2} + 8 x + 1\right ) e^{x} - \log {\left (3 \right )}} - \frac {4 x^{2}}{\left (4 x + 1\right ) \log {\left (x \right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int -\frac {\ln \left (x\right )\,\left ({\mathrm {e}}^{2\,x}\,\left (16\,x^2+8\,x\right )-{\mathrm {e}}^x\,\left (\ln \left (3\right )\,\left (32\,x^2+16\,x\right )+8\,x^3\right )+{\ln \left (3\right )}^2\,\left (16\,x^2+8\,x\right )+8\,x^3\,\ln \left (3\right )\right )-{\ln \left (3\right )}^2\,\left (16\,x^2+4\,x\right )-{\mathrm {e}}^{2\,x}\,\left (16\,x^2+4\,x\right )+{\mathrm {e}}^x\,\ln \left (3\right )\,\left (32\,x^2+8\,x\right )+4\,x^4\,{\mathrm {e}}^x\,{\ln \left (x\right )}^2}{x^4\,{\ln \left (x\right )}^4+{\ln \left (x\right )}^3\,\left ({\mathrm {e}}^x\,\left (8\,x^3+2\,x^2\right )-\ln \left (3\right )\,\left (8\,x^3+2\,x^2\right )\right )+{\ln \left (x\right )}^2\,\left ({\mathrm {e}}^{2\,x}\,\left (16\,x^2+8\,x+1\right )+{\ln \left (3\right )}^2\,\left (16\,x^2+8\,x+1\right )-{\mathrm {e}}^x\,\ln \left (3\right )\,\left (32\,x^2+16\,x+2\right )\right )} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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