Optimal. Leaf size=23 \[ \frac {7+x}{-e^9+25 (x+5 x \log (x))^2} \]
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Rubi [F] time = 2.07, 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^9-2100 x-275 x^2+\left (-12250 x-1500 x^2\right ) \log (x)+\left (-8750 x-625 x^2\right ) \log ^2(x)}{e^{18}-50 e^9 x^2+625 x^4+\left (-500 e^9 x^2+12500 x^4\right ) \log (x)+\left (-1250 e^9 x^2+93750 x^4\right ) \log ^2(x)+312500 x^4 \log ^3(x)+390625 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 {-e^9-25 x (84+11 x)-250 x (49+6 x) \log (x)-625 x (14+x) \log ^2(x)}{\left (e^9-25 x^2-250 x^2 \log (x)-625 x^2 \log ^2(x)\right )^2} \, dx\\ &=\int \left (\frac {14+x}{x \left (e^9-25 x^2-250 x^2 \log (x)-625 x^2 \log ^2(x)\right )}-\frac {2 (7+x) \left (e^9+125 x^2+625 x^2 \log (x)\right )}{x \left (-e^9+25 x^2+250 x^2 \log (x)+625 x^2 \log ^2(x)\right )^2}\right ) \, dx\\ &=-\left (2 \int \frac {(7+x) \left (e^9+125 x^2+625 x^2 \log (x)\right )}{x \left (-e^9+25 x^2+250 x^2 \log (x)+625 x^2 \log ^2(x)\right )^2} \, dx\right )+\int \frac {14+x}{x \left (e^9-25 x^2-250 x^2 \log (x)-625 x^2 \log ^2(x)\right )} \, dx\\ &=-\left (2 \int \left (\frac {e^9+125 x^2+625 x^2 \log (x)}{\left (e^9-25 x^2-250 x^2 \log (x)-625 x^2 \log ^2(x)\right )^2}+\frac {7 \left (e^9+125 x^2+625 x^2 \log (x)\right )}{x \left (-e^9+25 x^2+250 x^2 \log (x)+625 x^2 \log ^2(x)\right )^2}\right ) \, dx\right )+\int \left (\frac {1}{e^9-25 x^2-250 x^2 \log (x)-625 x^2 \log ^2(x)}+\frac {14}{x \left (e^9-25 x^2-250 x^2 \log (x)-625 x^2 \log ^2(x)\right )}\right ) \, dx\\ &=-\left (2 \int \frac {e^9+125 x^2+625 x^2 \log (x)}{\left (e^9-25 x^2-250 x^2 \log (x)-625 x^2 \log ^2(x)\right )^2} \, dx\right )+14 \int \frac {1}{x \left (e^9-25 x^2-250 x^2 \log (x)-625 x^2 \log ^2(x)\right )} \, dx-14 \int \frac {e^9+125 x^2+625 x^2 \log (x)}{x \left (-e^9+25 x^2+250 x^2 \log (x)+625 x^2 \log ^2(x)\right )^2} \, dx+\int \frac {1}{e^9-25 x^2-250 x^2 \log (x)-625 x^2 \log ^2(x)} \, dx\\ &=-\left (2 \int \left (\frac {e^9}{\left (e^9-25 x^2-250 x^2 \log (x)-625 x^2 \log ^2(x)\right )^2}+\frac {125 x^2}{\left (e^9-25 x^2-250 x^2 \log (x)-625 x^2 \log ^2(x)\right )^2}+\frac {625 x^2 \log (x)}{\left (e^9-25 x^2-250 x^2 \log (x)-625 x^2 \log ^2(x)\right )^2}\right ) \, dx\right )+14 \int \frac {1}{x \left (e^9-25 x^2-250 x^2 \log (x)-625 x^2 \log ^2(x)\right )} \, dx-14 \int \left (\frac {125 x}{\left (e^9-25 x^2-250 x^2 \log (x)-625 x^2 \log ^2(x)\right )^2}+\frac {625 x \log (x)}{\left (e^9-25 x^2-250 x^2 \log (x)-625 x^2 \log ^2(x)\right )^2}+\frac {e^9}{x \left (-e^9+25 x^2+250 x^2 \log (x)+625 x^2 \log ^2(x)\right )^2}\right ) \, dx+\int \frac {1}{e^9-25 x^2-250 x^2 \log (x)-625 x^2 \log ^2(x)} \, dx\\ &=14 \int \frac {1}{x \left (e^9-25 x^2-250 x^2 \log (x)-625 x^2 \log ^2(x)\right )} \, dx-250 \int \frac {x^2}{\left (e^9-25 x^2-250 x^2 \log (x)-625 x^2 \log ^2(x)\right )^2} \, dx-1250 \int \frac {x^2 \log (x)}{\left (e^9-25 x^2-250 x^2 \log (x)-625 x^2 \log ^2(x)\right )^2} \, dx-1750 \int \frac {x}{\left (e^9-25 x^2-250 x^2 \log (x)-625 x^2 \log ^2(x)\right )^2} \, dx-8750 \int \frac {x \log (x)}{\left (e^9-25 x^2-250 x^2 \log (x)-625 x^2 \log ^2(x)\right )^2} \, dx-\left (2 e^9\right ) \int \frac {1}{\left (e^9-25 x^2-250 x^2 \log (x)-625 x^2 \log ^2(x)\right )^2} \, dx-\left (14 e^9\right ) \int \frac {1}{x \left (-e^9+25 x^2+250 x^2 \log (x)+625 x^2 \log ^2(x)\right )^2} \, dx+\int \frac {1}{e^9-25 x^2-250 x^2 \log (x)-625 x^2 \log ^2(x)} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.50, size = 32, normalized size = 1.39 \begin {gather*} -\frac {7+x}{e^9-25 x^2-250 x^2 \log (x)-625 x^2 \log ^2(x)} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.59, size = 32, normalized size = 1.39 \begin {gather*} \frac {x + 7}{625 \, x^{2} \log \relax (x)^{2} + 250 \, x^{2} \log \relax (x) + 25 \, x^{2} - e^{9}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.36, size = 33, normalized size = 1.43 \begin {gather*} \frac {2 \, {\left (x + 7\right )}}{625 \, x^{2} \log \relax (x)^{2} + 250 \, x^{2} \log \relax (x) + 25 \, x^{2} - e^{9}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.19, size = 32, normalized size = 1.39
method | result | size |
risch | \(-\frac {x +7}{-625 x^{2} \ln \relax (x )^{2}-250 x^{2} \ln \relax (x )-25 x^{2}+{\mathrm e}^{9}}\) | \(32\) |
norman | \(\frac {-x -7}{-625 x^{2} \ln \relax (x )^{2}-250 x^{2} \ln \relax (x )-25 x^{2}+{\mathrm e}^{9}}\) | \(33\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.41, size = 32, normalized size = 1.39 \begin {gather*} \frac {x + 7}{625 \, x^{2} \log \relax (x)^{2} + 250 \, x^{2} \log \relax (x) + 25 \, x^{2} - e^{9}} \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 {2100\,x+{\mathrm {e}}^9+{\ln \relax (x)}^2\,\left (625\,x^2+8750\,x\right )+\ln \relax (x)\,\left (1500\,x^2+12250\,x\right )+275\,x^2}{{\mathrm {e}}^{18}-\ln \relax (x)\,\left (500\,x^2\,{\mathrm {e}}^9-12500\,x^4\right )+312500\,x^4\,{\ln \relax (x)}^3+390625\,x^4\,{\ln \relax (x)}^4-50\,x^2\,{\mathrm {e}}^9-{\ln \relax (x)}^2\,\left (1250\,x^2\,{\mathrm {e}}^9-93750\,x^4\right )+625\,x^4} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.24, size = 29, normalized size = 1.26 \begin {gather*} \frac {x + 7}{625 x^{2} \log {\relax (x )}^{2} + 250 x^{2} \log {\relax (x )} + 25 x^{2} - e^{9}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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