Integrand size = 116, antiderivative size = 29 \[ \int \frac {e^{1-x} \left (4+4 x+7 x^2-9 x^3-36 x^4+\left (-4 x^2+4 x^3+9 x^4-9 x^5\right ) \log (3)\right )-36 e^{1-x} x^3 \log \left (4-9 x^2\right )+e^{1-x} \left (-4 x^2+9 x^4\right ) \log ^2\left (4-9 x^2\right )}{-4 x^2+9 x^4} \, dx=e^{1-x} \left (4+\frac {1}{x}+x \log (3)-\log ^2\left (4-9 x^2\right )\right ) \]
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\[ \int \frac {e^{1-x} \left (4+4 x+7 x^2-9 x^3-36 x^4+\left (-4 x^2+4 x^3+9 x^4-9 x^5\right ) \log (3)\right )-36 e^{1-x} x^3 \log \left (4-9 x^2\right )+e^{1-x} \left (-4 x^2+9 x^4\right ) \log ^2\left (4-9 x^2\right )}{-4 x^2+9 x^4} \, dx=\int \frac {e^{1-x} \left (4+4 x+7 x^2-9 x^3-36 x^4+\left (-4 x^2+4 x^3+9 x^4-9 x^5\right ) \log (3)\right )-36 e^{1-x} x^3 \log \left (4-9 x^2\right )+e^{1-x} \left (-4 x^2+9 x^4\right ) \log ^2\left (4-9 x^2\right )}{-4 x^2+9 x^4} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{1-x} \left (4+4 x+7 x^2-9 x^3-36 x^4+\left (-4 x^2+4 x^3+9 x^4-9 x^5\right ) \log (3)\right )-36 e^{1-x} x^3 \log \left (4-9 x^2\right )+e^{1-x} \left (-4 x^2+9 x^4\right ) \log ^2\left (4-9 x^2\right )}{x^2 \left (-4+9 x^2\right )} \, dx \\ & = \int \frac {e^{1-x} \left (\left (-4+9 x^2\right ) \left (1+x-x^2 (-4+\log (3))+x^3 \log (3)\right )+36 x^3 \log \left (4-9 x^2\right )+\left (4 x^2-9 x^4\right ) \log ^2\left (4-9 x^2\right )\right )}{x^2 \left (4-9 x^2\right )} \, dx \\ & = \int \left (\frac {e^{1-x} \left (-1-x-x^2 (4-\log (3))-x^3 \log (3)\right )}{x^2}-\frac {36 e^{1-x} x \log \left (4-9 x^2\right )}{-4+9 x^2}+e^{1-x} \log ^2\left (4-9 x^2\right )\right ) \, dx \\ & = -\left (36 \int \frac {e^{1-x} x \log \left (4-9 x^2\right )}{-4+9 x^2} \, dx\right )+\int \frac {e^{1-x} \left (-1-x-x^2 (4-\log (3))-x^3 \log (3)\right )}{x^2} \, dx+\int e^{1-x} \log ^2\left (4-9 x^2\right ) \, dx \\ & = -2 e^{5/3} \operatorname {ExpIntegralEi}\left (\frac {1}{3} (-2-3 x)\right ) \log \left (4-9 x^2\right )-2 \sqrt [3]{e} \operatorname {ExpIntegralEi}\left (\frac {1}{3} (2-3 x)\right ) \log \left (4-9 x^2\right )+36 \int \frac {\sqrt [3]{e} x \left (-e^{4/3} \operatorname {ExpIntegralEi}\left (-\frac {2}{3}-x\right )-\operatorname {ExpIntegralEi}\left (\frac {2}{3}-x\right )\right )}{4-9 x^2} \, dx+\int \left (-\frac {e^{1-x}}{x^2}-\frac {e^{1-x}}{x}-4 e^{1-x} \left (1-\frac {\log (3)}{4}\right )-e^{1-x} x \log (3)\right ) \, dx+\int e^{1-x} \log ^2\left (4-9 x^2\right ) \, dx \\ & = -2 e^{5/3} \operatorname {ExpIntegralEi}\left (\frac {1}{3} (-2-3 x)\right ) \log \left (4-9 x^2\right )-2 \sqrt [3]{e} \operatorname {ExpIntegralEi}\left (\frac {1}{3} (2-3 x)\right ) \log \left (4-9 x^2\right )+\left (36 \sqrt [3]{e}\right ) \int \frac {x \left (-e^{4/3} \operatorname {ExpIntegralEi}\left (-\frac {2}{3}-x\right )-\operatorname {ExpIntegralEi}\left (\frac {2}{3}-x\right )\right )}{4-9 x^2} \, dx-(4-\log (3)) \int e^{1-x} \, dx-\log (3) \int e^{1-x} x \, dx-\int \frac {e^{1-x}}{x^2} \, dx-\int \frac {e^{1-x}}{x} \, dx+\int e^{1-x} \log ^2\left (4-9 x^2\right ) \, dx \\ & = \frac {e^{1-x}}{x}-e \operatorname {ExpIntegralEi}(-x)+e^{1-x} (4-\log (3))+e^{1-x} x \log (3)-2 e^{5/3} \operatorname {ExpIntegralEi}\left (\frac {1}{3} (-2-3 x)\right ) \log \left (4-9 x^2\right )-2 \sqrt [3]{e} \operatorname {ExpIntegralEi}\left (\frac {1}{3} (2-3 x)\right ) \log \left (4-9 x^2\right )+\left (36 \sqrt [3]{e}\right ) \int \left (\frac {e^{4/3} x \operatorname {ExpIntegralEi}\left (-\frac {2}{3}-x\right )}{-4+9 x^2}+\frac {x \operatorname {ExpIntegralEi}\left (\frac {2}{3}-x\right )}{-4+9 x^2}\right ) \, dx-\log (3) \int e^{1-x} \, dx+\int \frac {e^{1-x}}{x} \, dx+\int e^{1-x} \log ^2\left (4-9 x^2\right ) \, dx \\ & = \frac {e^{1-x}}{x}+e^{1-x} (4-\log (3))+e^{1-x} \log (3)+e^{1-x} x \log (3)-2 e^{5/3} \operatorname {ExpIntegralEi}\left (\frac {1}{3} (-2-3 x)\right ) \log \left (4-9 x^2\right )-2 \sqrt [3]{e} \operatorname {ExpIntegralEi}\left (\frac {1}{3} (2-3 x)\right ) \log \left (4-9 x^2\right )+\left (36 \sqrt [3]{e}\right ) \int \frac {x \operatorname {ExpIntegralEi}\left (\frac {2}{3}-x\right )}{-4+9 x^2} \, dx+\left (36 e^{5/3}\right ) \int \frac {x \operatorname {ExpIntegralEi}\left (-\frac {2}{3}-x\right )}{-4+9 x^2} \, dx+\int e^{1-x} \log ^2\left (4-9 x^2\right ) \, dx \\ & = \frac {e^{1-x}}{x}+e^{1-x} (4-\log (3))+e^{1-x} \log (3)+e^{1-x} x \log (3)-2 e^{5/3} \operatorname {ExpIntegralEi}\left (\frac {1}{3} (-2-3 x)\right ) \log \left (4-9 x^2\right )-2 \sqrt [3]{e} \operatorname {ExpIntegralEi}\left (\frac {1}{3} (2-3 x)\right ) \log \left (4-9 x^2\right )+\left (36 \sqrt [3]{e}\right ) \int \left (\frac {\operatorname {ExpIntegralEi}\left (\frac {2}{3}-x\right )}{6 (-2+3 x)}+\frac {\operatorname {ExpIntegralEi}\left (\frac {2}{3}-x\right )}{6 (2+3 x)}\right ) \, dx+\left (36 e^{5/3}\right ) \int \left (\frac {\operatorname {ExpIntegralEi}\left (-\frac {2}{3}-x\right )}{6 (-2+3 x)}+\frac {\operatorname {ExpIntegralEi}\left (-\frac {2}{3}-x\right )}{6 (2+3 x)}\right ) \, dx+\int e^{1-x} \log ^2\left (4-9 x^2\right ) \, dx \\ & = \frac {e^{1-x}}{x}+e^{1-x} (4-\log (3))+e^{1-x} \log (3)+e^{1-x} x \log (3)-2 e^{5/3} \operatorname {ExpIntegralEi}\left (\frac {1}{3} (-2-3 x)\right ) \log \left (4-9 x^2\right )-2 \sqrt [3]{e} \operatorname {ExpIntegralEi}\left (\frac {1}{3} (2-3 x)\right ) \log \left (4-9 x^2\right )+\left (6 \sqrt [3]{e}\right ) \int \frac {\operatorname {ExpIntegralEi}\left (\frac {2}{3}-x\right )}{-2+3 x} \, dx+\left (6 \sqrt [3]{e}\right ) \int \frac {\operatorname {ExpIntegralEi}\left (\frac {2}{3}-x\right )}{2+3 x} \, dx+\left (6 e^{5/3}\right ) \int \frac {\operatorname {ExpIntegralEi}\left (-\frac {2}{3}-x\right )}{-2+3 x} \, dx+\left (6 e^{5/3}\right ) \int \frac {\operatorname {ExpIntegralEi}\left (-\frac {2}{3}-x\right )}{2+3 x} \, dx+\int e^{1-x} \log ^2\left (4-9 x^2\right ) \, dx \\ \end{align*}
Time = 1.48 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.48 \[ \int \frac {e^{1-x} \left (4+4 x+7 x^2-9 x^3-36 x^4+\left (-4 x^2+4 x^3+9 x^4-9 x^5\right ) \log (3)\right )-36 e^{1-x} x^3 \log \left (4-9 x^2\right )+e^{1-x} \left (-4 x^2+9 x^4\right ) \log ^2\left (4-9 x^2\right )}{-4 x^2+9 x^4} \, dx=-e^{-x} \left (-4 e-\frac {e}{x}-e x \log (3)\right )-e^{1-x} \log ^2\left (4-9 x^2\right ) \]
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Time = 2.00 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.41
method | result | size |
risch | \(-{\mathrm e}^{1-x} \ln \left (-9 x^{2}+4\right )^{2}+\frac {\left (x^{2} \ln \left (3\right )+4 x +1\right ) {\mathrm e}^{1-x}}{x}\) | \(41\) |
parallelrisch | \(\frac {81 \ln \left (3\right ) {\mathrm e}^{1-x} x^{2}-81 \ln \left (-9 x^{2}+4\right )^{2} x \,{\mathrm e}^{1-x}+324 x \,{\mathrm e}^{1-x}+81 \,{\mathrm e}^{1-x}}{81 x}\) | \(56\) |
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Time = 0.25 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.48 \[ \int \frac {e^{1-x} \left (4+4 x+7 x^2-9 x^3-36 x^4+\left (-4 x^2+4 x^3+9 x^4-9 x^5\right ) \log (3)\right )-36 e^{1-x} x^3 \log \left (4-9 x^2\right )+e^{1-x} \left (-4 x^2+9 x^4\right ) \log ^2\left (4-9 x^2\right )}{-4 x^2+9 x^4} \, dx=-\frac {x e^{\left (-x + 1\right )} \log \left (-9 \, x^{2} + 4\right )^{2} - {\left (x^{2} \log \left (3\right ) + 4 \, x + 1\right )} e^{\left (-x + 1\right )}}{x} \]
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Time = 0.13 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {e^{1-x} \left (4+4 x+7 x^2-9 x^3-36 x^4+\left (-4 x^2+4 x^3+9 x^4-9 x^5\right ) \log (3)\right )-36 e^{1-x} x^3 \log \left (4-9 x^2\right )+e^{1-x} \left (-4 x^2+9 x^4\right ) \log ^2\left (4-9 x^2\right )}{-4 x^2+9 x^4} \, dx=\frac {\left (x^{2} \log {\left (3 \right )} - x \log {\left (4 - 9 x^{2} \right )}^{2} + 4 x + 1\right ) e^{1 - x}}{x} \]
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Leaf count of result is larger than twice the leaf count of optimal. 67 vs. \(2 (28) = 56\).
Time = 0.35 (sec) , antiderivative size = 67, normalized size of antiderivative = 2.31 \[ \int \frac {e^{1-x} \left (4+4 x+7 x^2-9 x^3-36 x^4+\left (-4 x^2+4 x^3+9 x^4-9 x^5\right ) \log (3)\right )-36 e^{1-x} x^3 \log \left (4-9 x^2\right )+e^{1-x} \left (-4 x^2+9 x^4\right ) \log ^2\left (4-9 x^2\right )}{-4 x^2+9 x^4} \, dx=\frac {{\left (x^{2} e \log \left (3\right ) - x e \log \left (3 \, x + 2\right )^{2} - 2 \, x e \log \left (3 \, x + 2\right ) \log \left (-3 \, x + 2\right ) - x e \log \left (-3 \, x + 2\right )^{2} + 4 \, x e + e\right )} e^{\left (-x\right )}}{x} \]
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Time = 0.34 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.76 \[ \int \frac {e^{1-x} \left (4+4 x+7 x^2-9 x^3-36 x^4+\left (-4 x^2+4 x^3+9 x^4-9 x^5\right ) \log (3)\right )-36 e^{1-x} x^3 \log \left (4-9 x^2\right )+e^{1-x} \left (-4 x^2+9 x^4\right ) \log ^2\left (4-9 x^2\right )}{-4 x^2+9 x^4} \, dx=\frac {x^{2} e^{\left (-x + 1\right )} \log \left (3\right ) - x e^{\left (-x + 1\right )} \log \left (-9 \, x^{2} + 4\right )^{2} + 4 \, x e^{\left (-x + 1\right )} + e^{\left (-x + 1\right )}}{x} \]
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Time = 11.60 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.38 \[ \int \frac {e^{1-x} \left (4+4 x+7 x^2-9 x^3-36 x^4+\left (-4 x^2+4 x^3+9 x^4-9 x^5\right ) \log (3)\right )-36 e^{1-x} x^3 \log \left (4-9 x^2\right )+e^{1-x} \left (-4 x^2+9 x^4\right ) \log ^2\left (4-9 x^2\right )}{-4 x^2+9 x^4} \, dx=\frac {{\mathrm {e}}^{1-x}\,\left (\ln \left (3\right )\,x^2+4\,x+1\right )}{x}-{\ln \left (4-9\,x^2\right )}^2\,{\mathrm {e}}^{1-x} \]
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