Optimal. Leaf size=32 \[ \log \left (-2+\frac {x^4}{\left (1-x-\frac {3+e^{-5 x} x}{x}\right ) \log (x)}\right ) \]
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Rubi [F] time = 17.83, 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^{5 x} x^5+e^{10 x} \left (-3 x^4+x^5-x^6\right )+\left (e^{10 x} \left (15 x^4-4 x^5+3 x^6\right )+e^{5 x} \left (4 x^5+5 x^6\right )\right ) \log (x)}{\left (e^{5 x} x^6+e^{10 x} \left (3 x^5-x^6+x^7\right )\right ) \log (x)+\left (2 x^2+e^{5 x} \left (12 x-4 x^2+4 x^3\right )+e^{10 x} \left (18-12 x+14 x^2-4 x^3+2 x^4\right )\right ) \log ^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 {e^{5 x} x^4 \left (-x+e^{5 x} \left (-3+x-x^2\right )+\left (x (4+5 x)+e^{5 x} \left (15-4 x+3 x^2\right )\right ) \log (x)\right )}{\left (x+e^{5 x} \left (3-x+x^2\right )\right ) \log (x) \left (e^{5 x} x^5+2 \left (x+e^{5 x} \left (3-x+x^2\right )\right ) \log (x)\right )} \, dx\\ &=\int \left (-\frac {e^{5 x} \left (-3+15 x-4 x^2+5 x^3\right )}{x \left (3 e^{5 x}+x-e^{5 x} x+e^{5 x} x^2\right )}+\frac {e^{5 x} \left (-x^5+4 x^5 \log (x)+5 x^6 \log (x)-6 \log ^2(x)+30 x \log ^2(x)-8 x^2 \log ^2(x)+10 x^3 \log ^2(x)\right )}{x \log (x) \left (e^{5 x} x^5+6 e^{5 x} \log (x)+2 x \log (x)-2 e^{5 x} x \log (x)+2 e^{5 x} x^2 \log (x)\right )}\right ) \, dx\\ &=-\int \frac {e^{5 x} \left (-3+15 x-4 x^2+5 x^3\right )}{x \left (3 e^{5 x}+x-e^{5 x} x+e^{5 x} x^2\right )} \, dx+\int \frac {e^{5 x} \left (-x^5+4 x^5 \log (x)+5 x^6 \log (x)-6 \log ^2(x)+30 x \log ^2(x)-8 x^2 \log ^2(x)+10 x^3 \log ^2(x)\right )}{x \log (x) \left (e^{5 x} x^5+6 e^{5 x} \log (x)+2 x \log (x)-2 e^{5 x} x \log (x)+2 e^{5 x} x^2 \log (x)\right )} \, dx\\ &=-\int \frac {e^{5 x} \left (-3+15 x-4 x^2+5 x^3\right )}{x \left (x+e^{5 x} \left (3-x+x^2\right )\right )} \, dx+\int \frac {e^{5 x} \left (-x^5+x^5 (4+5 x) \log (x)+2 \left (-3+15 x-4 x^2+5 x^3\right ) \log ^2(x)\right )}{x \log (x) \left (e^{5 x} x^5+2 \left (x+e^{5 x} \left (3-x+x^2\right )\right ) \log (x)\right )} \, dx\\ &=\int \left (\frac {4 e^{5 x} x^4}{e^{5 x} x^5+6 e^{5 x} \log (x)+2 x \log (x)-2 e^{5 x} x \log (x)+2 e^{5 x} x^2 \log (x)}+\frac {5 e^{5 x} x^5}{e^{5 x} x^5+6 e^{5 x} \log (x)+2 x \log (x)-2 e^{5 x} x \log (x)+2 e^{5 x} x^2 \log (x)}-\frac {e^{5 x} x^4}{\log (x) \left (e^{5 x} x^5+6 e^{5 x} \log (x)+2 x \log (x)-2 e^{5 x} x \log (x)+2 e^{5 x} x^2 \log (x)\right )}+\frac {30 e^{5 x} \log (x)}{e^{5 x} x^5+6 e^{5 x} \log (x)+2 x \log (x)-2 e^{5 x} x \log (x)+2 e^{5 x} x^2 \log (x)}-\frac {6 e^{5 x} \log (x)}{x \left (e^{5 x} x^5+6 e^{5 x} \log (x)+2 x \log (x)-2 e^{5 x} x \log (x)+2 e^{5 x} x^2 \log (x)\right )}-\frac {8 e^{5 x} x \log (x)}{e^{5 x} x^5+6 e^{5 x} \log (x)+2 x \log (x)-2 e^{5 x} x \log (x)+2 e^{5 x} x^2 \log (x)}+\frac {10 e^{5 x} x^2 \log (x)}{e^{5 x} x^5+6 e^{5 x} \log (x)+2 x \log (x)-2 e^{5 x} x \log (x)+2 e^{5 x} x^2 \log (x)}\right ) \, dx-\operatorname {Subst}\left (\int \frac {1}{1+x} \, dx,x,\frac {e^{5 x} \left (3-x+x^2\right )}{x}\right )\\ &=-\log \left (1+\frac {e^{5 x} \left (3-x+x^2\right )}{x}\right )+4 \int \frac {e^{5 x} x^4}{e^{5 x} x^5+6 e^{5 x} \log (x)+2 x \log (x)-2 e^{5 x} x \log (x)+2 e^{5 x} x^2 \log (x)} \, dx+5 \int \frac {e^{5 x} x^5}{e^{5 x} x^5+6 e^{5 x} \log (x)+2 x \log (x)-2 e^{5 x} x \log (x)+2 e^{5 x} x^2 \log (x)} \, dx-6 \int \frac {e^{5 x} \log (x)}{x \left (e^{5 x} x^5+6 e^{5 x} \log (x)+2 x \log (x)-2 e^{5 x} x \log (x)+2 e^{5 x} x^2 \log (x)\right )} \, dx-8 \int \frac {e^{5 x} x \log (x)}{e^{5 x} x^5+6 e^{5 x} \log (x)+2 x \log (x)-2 e^{5 x} x \log (x)+2 e^{5 x} x^2 \log (x)} \, dx+10 \int \frac {e^{5 x} x^2 \log (x)}{e^{5 x} x^5+6 e^{5 x} \log (x)+2 x \log (x)-2 e^{5 x} x \log (x)+2 e^{5 x} x^2 \log (x)} \, dx+30 \int \frac {e^{5 x} \log (x)}{e^{5 x} x^5+6 e^{5 x} \log (x)+2 x \log (x)-2 e^{5 x} x \log (x)+2 e^{5 x} x^2 \log (x)} \, dx-\int \frac {e^{5 x} x^4}{\log (x) \left (e^{5 x} x^5+6 e^{5 x} \log (x)+2 x \log (x)-2 e^{5 x} x \log (x)+2 e^{5 x} x^2 \log (x)\right )} \, dx\\ &=-\log \left (1+\frac {e^{5 x} \left (3-x+x^2\right )}{x}\right )+4 \int \frac {e^{5 x} x^4}{e^{5 x} x^5+2 \left (x+e^{5 x} \left (3-x+x^2\right )\right ) \log (x)} \, dx+5 \int \frac {e^{5 x} x^5}{e^{5 x} x^5+2 \left (x+e^{5 x} \left (3-x+x^2\right )\right ) \log (x)} \, dx-6 \int \frac {e^{5 x} \log (x)}{e^{5 x} x^6+2 x \left (x+e^{5 x} \left (3-x+x^2\right )\right ) \log (x)} \, dx-8 \int \frac {e^{5 x} x \log (x)}{e^{5 x} x^5+2 \left (x+e^{5 x} \left (3-x+x^2\right )\right ) \log (x)} \, dx+10 \int \frac {e^{5 x} x^2 \log (x)}{e^{5 x} x^5+2 \left (x+e^{5 x} \left (3-x+x^2\right )\right ) \log (x)} \, dx+30 \int \frac {e^{5 x} \log (x)}{e^{5 x} x^5+2 \left (x+e^{5 x} \left (3-x+x^2\right )\right ) \log (x)} \, dx-\int \frac {e^{5 x} x^4}{\log (x) \left (e^{5 x} x^5+2 \left (x+e^{5 x} \left (3-x+x^2\right )\right ) \log (x)\right )} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [B] time = 1.01, size = 82, normalized size = 2.56 \begin {gather*} -\log \left (3 e^{5 x}+x-e^{5 x} x+e^{5 x} x^2\right )-\log (\log (x))+\log \left (e^{5 x} x^5+6 e^{5 x} \log (x)+2 x \log (x)-2 e^{5 x} x \log (x)+2 e^{5 x} x^2 \log (x)\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.92, size = 53, normalized size = 1.66 \begin {gather*} \log \left (\frac {x^{5} e^{\left (5 \, x\right )} + 2 \, {\left ({\left (x^{2} - x + 3\right )} e^{\left (5 \, x\right )} + x\right )} \log \relax (x)}{{\left (x^{2} - x + 3\right )} e^{\left (5 \, x\right )} + x}\right ) - \log \left (\log \relax (x)\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [F(-1)] 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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maple [A] time = 0.05, size = 45, normalized size = 1.41
method | result | size |
risch | \(-\ln \left (\ln \relax (x )\right )+\ln \left (\ln \relax (x )+\frac {x^{5} {\mathrm e}^{5 x}}{2 x^{2} {\mathrm e}^{5 x}-2 x \,{\mathrm e}^{5 x}+2 x +6 \,{\mathrm e}^{5 x}}\right )\) | \(45\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.44, size = 111, normalized size = 3.47 \begin {gather*} \log \left (\frac {x^{5} + 2 \, {\left (x^{2} - x + 3\right )} \log \relax (x)}{2 \, {\left (x^{2} - x + 3\right )}}\right ) + \log \left (\frac {{\left (x^{5} + 2 \, {\left (x^{2} - x + 3\right )} \log \relax (x)\right )} e^{\left (5 \, x\right )} + 2 \, x \log \relax (x)}{x^{5} + 2 \, {\left (x^{2} - x + 3\right )} \log \relax (x)}\right ) - \log \left (\frac {{\left (x^{2} - x + 3\right )} e^{\left (5 \, x\right )} + x}{x^{2} - x + 3}\right ) - \log \left (\log \relax (x)\right ) \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 {{\mathrm {e}}^{10\,x}\,\left (x^6-x^5+3\,x^4\right )+x^5\,{\mathrm {e}}^{5\,x}-\ln \relax (x)\,\left ({\mathrm {e}}^{5\,x}\,\left (5\,x^6+4\,x^5\right )+{\mathrm {e}}^{10\,x}\,\left (3\,x^6-4\,x^5+15\,x^4\right )\right )}{\left ({\mathrm {e}}^{5\,x}\,\left (4\,x^3-4\,x^2+12\,x\right )+{\mathrm {e}}^{10\,x}\,\left (2\,x^4-4\,x^3+14\,x^2-12\,x+18\right )+2\,x^2\right )\,{\ln \relax (x)}^2+\left ({\mathrm {e}}^{10\,x}\,\left (x^7-x^6+3\,x^5\right )+x^6\,{\mathrm {e}}^{5\,x}\right )\,\ln \relax (x)} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: PolynomialError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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