Optimal. Leaf size=28 \[ \frac {5}{x \left (-x+(-3+x) x \left (2+e^x \log (2+2 x)\right )\right )} \]
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Rubi [F] time = 75.13, 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 {70+40 x-30 x^2+e^x \left (15 x-5 x^2\right )+e^x \left (30+30 x-5 x^2-5 x^3\right ) \log (2+2 x)}{49 x^3+21 x^4-24 x^5+4 x^6+e^x \left (42 x^3+16 x^4-22 x^5+4 x^6\right ) \log (2+2 x)+e^{2 x} \left (9 x^3+3 x^4-5 x^5+x^6\right ) \log ^2(2+2 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 {70+5 \left (8+3 e^x\right ) x-5 \left (6+e^x\right ) x^2-5 e^x \left (-6-6 x+x^2+x^3\right ) \log (2 (1+x))}{x^3 (1+x) \left (7-2 x-e^x (-3+x) \log (2 (1+x))\right )^2} \, dx\\ &=\int \left (\frac {5 \left (-21+13 x-2 x^2-20 \log (2 (1+x))-7 x \log (2 (1+x))+11 x^2 \log (2 (1+x))-2 x^3 \log (2 (1+x))\right )}{(3-x) x^2 (1+x) \left (7-2 x+3 e^x \log (2 (1+x))-e^x x \log (2 (1+x))\right )^2 \log (2+2 x)}+\frac {5 \left (3 x-x^2+6 \log (2 (1+x))+6 x \log (2 (1+x))-x^2 \log (2 (1+x))-x^3 \log (2 (1+x))\right )}{(3-x) x^3 (1+x) \left (7-2 x+3 e^x \log (2 (1+x))-e^x x \log (2 (1+x))\right ) \log (2+2 x)}\right ) \, dx\\ &=5 \int \frac {-21+13 x-2 x^2-20 \log (2 (1+x))-7 x \log (2 (1+x))+11 x^2 \log (2 (1+x))-2 x^3 \log (2 (1+x))}{(3-x) x^2 (1+x) \left (7-2 x+3 e^x \log (2 (1+x))-e^x x \log (2 (1+x))\right )^2 \log (2+2 x)} \, dx+5 \int \frac {3 x-x^2+6 \log (2 (1+x))+6 x \log (2 (1+x))-x^2 \log (2 (1+x))-x^3 \log (2 (1+x))}{(3-x) x^3 (1+x) \left (7-2 x+3 e^x \log (2 (1+x))-e^x x \log (2 (1+x))\right ) \log (2+2 x)} \, dx\\ &=5 \int \frac {-((-3+x) x)-\left (-6-6 x+x^2+x^3\right ) \log (2 (1+x))}{(3-x) x^3 (1+x) \left (7-2 x-e^x (-3+x) \log (2 (1+x))\right ) \log (2+2 x)} \, dx+5 \int \frac {-21+13 x-2 x^2-\left (20+7 x-11 x^2+2 x^3\right ) \log (2 (1+x))}{(3-x) x^2 (1+x) \left (7-2 x-e^x (-3+x) \log (2 (1+x))\right )^2 \log (2+2 x)} \, dx\\ &=5 \int \left (\frac {3 x-x^2+6 \log (2 (1+x))+6 x \log (2 (1+x))-x^2 \log (2 (1+x))-x^3 \log (2 (1+x))}{108 (3-x) \left (7-2 x+3 e^x \log (2 (1+x))-e^x x \log (2 (1+x))\right ) \log (2+2 x)}+\frac {3 x-x^2+6 \log (2 (1+x))+6 x \log (2 (1+x))-x^2 \log (2 (1+x))-x^3 \log (2 (1+x))}{3 x^3 \left (7-2 x+3 e^x \log (2 (1+x))-e^x x \log (2 (1+x))\right ) \log (2+2 x)}+\frac {7 \left (3 x-x^2+6 \log (2 (1+x))+6 x \log (2 (1+x))-x^2 \log (2 (1+x))-x^3 \log (2 (1+x))\right )}{27 x \left (7-2 x+3 e^x \log (2 (1+x))-e^x x \log (2 (1+x))\right ) \log (2+2 x)}+\frac {2 \left (-3 x+x^2-6 \log (2 (1+x))-6 x \log (2 (1+x))+x^2 \log (2 (1+x))+x^3 \log (2 (1+x))\right )}{9 x^2 \left (7-2 x+3 e^x \log (2 (1+x))-e^x x \log (2 (1+x))\right ) \log (2+2 x)}+\frac {-3 x+x^2-6 \log (2 (1+x))-6 x \log (2 (1+x))+x^2 \log (2 (1+x))+x^3 \log (2 (1+x))}{4 (1+x) \left (7-2 x+3 e^x \log (2 (1+x))-e^x x \log (2 (1+x))\right ) \log (2+2 x)}\right ) \, dx+5 \int \left (\frac {-21+13 x-2 x^2-20 \log (2 (1+x))-7 x \log (2 (1+x))+11 x^2 \log (2 (1+x))-2 x^3 \log (2 (1+x))}{36 (3-x) \left (7-2 x+3 e^x \log (2 (1+x))-e^x x \log (2 (1+x))\right )^2 \log (2+2 x)}+\frac {-21+13 x-2 x^2-20 \log (2 (1+x))-7 x \log (2 (1+x))+11 x^2 \log (2 (1+x))-2 x^3 \log (2 (1+x))}{3 x^2 \left (7-2 x+3 e^x \log (2 (1+x))-e^x x \log (2 (1+x))\right )^2 \log (2+2 x)}+\frac {-21+13 x-2 x^2-20 \log (2 (1+x))-7 x \log (2 (1+x))+11 x^2 \log (2 (1+x))-2 x^3 \log (2 (1+x))}{4 (1+x) \left (7-2 x+3 e^x \log (2 (1+x))-e^x x \log (2 (1+x))\right )^2 \log (2+2 x)}+\frac {2 \left (21-13 x+2 x^2+20 \log (2 (1+x))+7 x \log (2 (1+x))-11 x^2 \log (2 (1+x))+2 x^3 \log (2 (1+x))\right )}{9 x \left (7-2 x+3 e^x \log (2 (1+x))-e^x x \log (2 (1+x))\right )^2 \log (2+2 x)}\right ) \, dx\\ &=\frac {5}{108} \int \frac {3 x-x^2+6 \log (2 (1+x))+6 x \log (2 (1+x))-x^2 \log (2 (1+x))-x^3 \log (2 (1+x))}{(3-x) \left (7-2 x+3 e^x \log (2 (1+x))-e^x x \log (2 (1+x))\right ) \log (2+2 x)} \, dx+\frac {5}{36} \int \frac {-21+13 x-2 x^2-20 \log (2 (1+x))-7 x \log (2 (1+x))+11 x^2 \log (2 (1+x))-2 x^3 \log (2 (1+x))}{(3-x) \left (7-2 x+3 e^x \log (2 (1+x))-e^x x \log (2 (1+x))\right )^2 \log (2+2 x)} \, dx+\frac {10}{9} \int \frac {-3 x+x^2-6 \log (2 (1+x))-6 x \log (2 (1+x))+x^2 \log (2 (1+x))+x^3 \log (2 (1+x))}{x^2 \left (7-2 x+3 e^x \log (2 (1+x))-e^x x \log (2 (1+x))\right ) \log (2+2 x)} \, dx+\frac {10}{9} \int \frac {21-13 x+2 x^2+20 \log (2 (1+x))+7 x \log (2 (1+x))-11 x^2 \log (2 (1+x))+2 x^3 \log (2 (1+x))}{x \left (7-2 x+3 e^x \log (2 (1+x))-e^x x \log (2 (1+x))\right )^2 \log (2+2 x)} \, dx+\frac {5}{4} \int \frac {-21+13 x-2 x^2-20 \log (2 (1+x))-7 x \log (2 (1+x))+11 x^2 \log (2 (1+x))-2 x^3 \log (2 (1+x))}{(1+x) \left (7-2 x+3 e^x \log (2 (1+x))-e^x x \log (2 (1+x))\right )^2 \log (2+2 x)} \, dx+\frac {5}{4} \int \frac {-3 x+x^2-6 \log (2 (1+x))-6 x \log (2 (1+x))+x^2 \log (2 (1+x))+x^3 \log (2 (1+x))}{(1+x) \left (7-2 x+3 e^x \log (2 (1+x))-e^x x \log (2 (1+x))\right ) \log (2+2 x)} \, dx+\frac {35}{27} \int \frac {3 x-x^2+6 \log (2 (1+x))+6 x \log (2 (1+x))-x^2 \log (2 (1+x))-x^3 \log (2 (1+x))}{x \left (7-2 x+3 e^x \log (2 (1+x))-e^x x \log (2 (1+x))\right ) \log (2+2 x)} \, dx+\frac {5}{3} \int \frac {-21+13 x-2 x^2-20 \log (2 (1+x))-7 x \log (2 (1+x))+11 x^2 \log (2 (1+x))-2 x^3 \log (2 (1+x))}{x^2 \left (7-2 x+3 e^x \log (2 (1+x))-e^x x \log (2 (1+x))\right )^2 \log (2+2 x)} \, dx+\frac {5}{3} \int \frac {3 x-x^2+6 \log (2 (1+x))+6 x \log (2 (1+x))-x^2 \log (2 (1+x))-x^3 \log (2 (1+x))}{x^3 \left (7-2 x+3 e^x \log (2 (1+x))-e^x x \log (2 (1+x))\right ) \log (2+2 x)} \, dx\\ &=\frac {5}{108} \int \frac {-((-3+x) x)-\left (-6-6 x+x^2+x^3\right ) \log (2 (1+x))}{(3-x) \left (7-2 x-e^x (-3+x) \log (2 (1+x))\right ) \log (2+2 x)} \, dx+\frac {5}{36} \int \frac {-21+13 x-2 x^2-\left (20+7 x-11 x^2+2 x^3\right ) \log (2 (1+x))}{(3-x) \left (7-2 x-e^x (-3+x) \log (2 (1+x))\right )^2 \log (2+2 x)} \, dx+\frac {10}{9} \int \frac {(-3+x) x+\left (-6-6 x+x^2+x^3\right ) \log (2 (1+x))}{x^2 \left (7-2 x-e^x (-3+x) \log (2 (1+x))\right ) \log (2+2 x)} \, dx+\frac {10}{9} \int \frac {21-13 x+2 x^2+\left (20+7 x-11 x^2+2 x^3\right ) \log (2 (1+x))}{x \left (7-2 x-e^x (-3+x) \log (2 (1+x))\right )^2 \log (2+2 x)} \, dx+\frac {5}{4} \int \frac {(-3+x) x+\left (-6-6 x+x^2+x^3\right ) \log (2 (1+x))}{(1+x) \left (7-2 x-e^x (-3+x) \log (2 (1+x))\right ) \log (2+2 x)} \, dx+\frac {5}{4} \int \frac {-21+13 x-2 x^2-\left (20+7 x-11 x^2+2 x^3\right ) \log (2 (1+x))}{(1+x) \left (7-2 x-e^x (-3+x) \log (2 (1+x))\right )^2 \log (2+2 x)} \, dx+\frac {35}{27} \int \frac {-((-3+x) x)-\left (-6-6 x+x^2+x^3\right ) \log (2 (1+x))}{x \left (7-2 x-e^x (-3+x) \log (2 (1+x))\right ) \log (2+2 x)} \, dx+\frac {5}{3} \int \frac {-((-3+x) x)-\left (-6-6 x+x^2+x^3\right ) \log (2 (1+x))}{x^3 \left (7-2 x-e^x (-3+x) \log (2 (1+x))\right ) \log (2+2 x)} \, dx+\frac {5}{3} \int \frac {-21+13 x-2 x^2-\left (20+7 x-11 x^2+2 x^3\right ) \log (2 (1+x))}{x^2 \left (7-2 x-e^x (-3+x) \log (2 (1+x))\right )^2 \log (2+2 x)} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}
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Mathematica [A] time = 5.15, size = 25, normalized size = 0.89 \begin {gather*} \frac {5}{x^2 \left (-7+2 x+e^x (-3+x) \log (2 (1+x))\right )} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.60, size = 33, normalized size = 1.18 \begin {gather*} \frac {5}{2 \, x^{3} + {\left (x^{3} - 3 \, x^{2}\right )} e^{x} \log \left (2 \, x + 2\right ) - 7 \, x^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 3.31, size = 53, normalized size = 1.89 \begin {gather*} \frac {5}{x^{3} e^{x} \log \relax (2) + x^{3} e^{x} \log \left (x + 1\right ) - 3 \, x^{2} e^{x} \log \relax (2) - 3 \, x^{2} e^{x} \log \left (x + 1\right ) + 2 \, x^{3} - 7 \, x^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.05, size = 33, normalized size = 1.18
method | result | size |
risch | \(\frac {5}{x^{2} \left (x \,{\mathrm e}^{x} \ln \left (2 x +2\right )-3 \ln \left (2 x +2\right ) {\mathrm e}^{x}+2 x -7\right )}\) | \(33\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.73, size = 48, normalized size = 1.71 \begin {gather*} \frac {5}{2 \, x^{3} + {\left (x^{3} - 3 \, x^{2}\right )} e^{x} \log \left (x + 1\right ) - 7 \, x^{2} + {\left (x^{3} \log \relax (2) - 3 \, x^{2} \log \relax (2)\right )} e^{x}} \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 = -0.04 \begin {gather*} \int \frac {40\,x+{\mathrm {e}}^x\,\left (15\,x-5\,x^2\right )-30\,x^2+{\mathrm {e}}^x\,\ln \left (2\,x+2\right )\,\left (-5\,x^3-5\,x^2+30\,x+30\right )+70}{49\,x^3+21\,x^4-24\,x^5+4\,x^6+{\mathrm {e}}^x\,\ln \left (2\,x+2\right )\,\left (4\,x^6-22\,x^5+16\,x^4+42\,x^3\right )+{\mathrm {e}}^{2\,x}\,{\ln \left (2\,x+2\right )}^2\,\left (x^6-5\,x^5+3\,x^4+9\,x^3\right )} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.49, size = 36, normalized size = 1.29 \begin {gather*} \frac {5}{2 x^{3} - 7 x^{2} + \left (x^{3} \log {\left (2 x + 2 \right )} - 3 x^{2} \log {\left (2 x + 2 \right )}\right ) e^{x}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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