Optimal. Leaf size=30 \[ \frac {5}{(4+5 x)^2}+\log \left (-1-\log (2)+\frac {\log (x)}{\log \left (1+\frac {x}{5}\right )}\right ) \]
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Rubi [F] time = 8.11, 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 {\left (320+1264 x+1740 x^2+925 x^3+125 x^4\right ) \log \left (\frac {5+x}{5}\right )+\left (250 x+50 x^2+\left (250 x+50 x^2\right ) \log (2)\right ) \log ^2\left (\frac {5+x}{5}\right )+\log (x) \left (-64 x-240 x^2-300 x^3-125 x^4+\left (-250 x-50 x^2\right ) \log \left (\frac {5+x}{5}\right )\right )}{\left (320 x+1264 x^2+1740 x^3+925 x^4+125 x^5\right ) \log (x) \log \left (\frac {5+x}{5}\right )+\left (-320 x-1264 x^2-1740 x^3-925 x^4-125 x^5+\left (-320 x-1264 x^2-1740 x^3-925 x^4-125 x^5\right ) \log (2)\right ) \log ^2\left (\frac {5+x}{5}\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-x \log (x) \left ((4+5 x)^3+50 (5+x) \log \left (\frac {5+x}{5}\right )\right )+(5+x) \log \left (\frac {5+x}{5}\right ) \left ((4+5 x)^3+50 x (1+\log (2)) \log \left (\frac {5+x}{5}\right )\right )}{x (5+x) (4+5 x)^3 \log \left (1+\frac {x}{5}\right ) \left (\log (x)-(1+\log (2)) \log \left (\frac {5+x}{5}\right )\right )} \, dx\\ &=\int \left (-\frac {50}{(4+5 x)^3}+\frac {1}{x (-1-\log (2)) \log \left (1+\frac {x}{5}\right )}+\frac {(5-x \log (2)) \log (x)}{x (5+x) (1+\log (2)) \log \left (1+\frac {x}{5}\right ) \left (\log (x)-(1+\log (2)) \log \left (\frac {5+x}{5}\right )\right )}\right ) \, dx\\ &=\frac {5}{(4+5 x)^2}-\frac {\int \frac {1}{x \log \left (1+\frac {x}{5}\right )} \, dx}{1+\log (2)}+\frac {\int \frac {(5-x \log (2)) \log (x)}{x (5+x) \log \left (1+\frac {x}{5}\right ) \left (\log (x)-(1+\log (2)) \log \left (\frac {5+x}{5}\right )\right )} \, dx}{1+\log (2)}\\ &=\frac {5}{(4+5 x)^2}-\frac {\int \frac {1}{x \log \left (1+\frac {x}{5}\right )} \, dx}{1+\log (2)}+\frac {\int \left (\frac {\log (x)}{x \log \left (1+\frac {x}{5}\right ) \left (\log (x)-(1+\log (2)) \log \left (\frac {5+x}{5}\right )\right )}+\frac {(-1-\log (2)) \log (x)}{(5+x) \log \left (1+\frac {x}{5}\right ) \left (\log (x)-(1+\log (2)) \log \left (\frac {5+x}{5}\right )\right )}\right ) \, dx}{1+\log (2)}\\ &=\frac {5}{(4+5 x)^2}-\frac {\int \frac {1}{x \log \left (1+\frac {x}{5}\right )} \, dx}{1+\log (2)}+\frac {\int \frac {\log (x)}{x \log \left (1+\frac {x}{5}\right ) \left (\log (x)-(1+\log (2)) \log \left (\frac {5+x}{5}\right )\right )} \, dx}{1+\log (2)}-\int \frac {\log (x)}{(5+x) \log \left (1+\frac {x}{5}\right ) \left (\log (x)-(1+\log (2)) \log \left (\frac {5+x}{5}\right )\right )} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.08, size = 40, normalized size = 1.33 \begin {gather*} \frac {5}{(4+5 x)^2}-\log \left (\log \left (\frac {5+x}{5}\right )\right )+\log \left (-\log (x)+(1+\log (2)) \log \left (\frac {5+x}{5}\right )\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.66, size = 61, normalized size = 2.03 \begin {gather*} \frac {{\left (25 \, x^{2} + 40 \, x + 16\right )} \log \left (-{\left (\log \relax (2) + 1\right )} \log \left (\frac {1}{5} \, x + 1\right ) + \log \relax (x)\right ) - {\left (25 \, x^{2} + 40 \, x + 16\right )} \log \left (\log \left (\frac {1}{5} \, x + 1\right )\right ) + 5}{25 \, x^{2} + 40 \, x + 16} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 1.03, size = 52, normalized size = 1.73 \begin {gather*} \frac {5}{25 \, x^{2} + 40 \, x + 16} + \log \left (\log \relax (5) \log \relax (2) - \log \relax (2) \log \left (x + 5\right ) + \log \relax (5) - \log \left (x + 5\right ) + \log \relax (x)\right ) - \log \left (\log \relax (5) - \log \left (x + 5\right )\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.06, size = 43, normalized size = 1.43
method | result | size |
risch | \(\frac {5}{25 x^{2}+40 x +16}+\ln \left (\ln \left (1+\frac {x}{5}\right )-\frac {\ln \relax (x )}{1+\ln \relax (2)}\right )-\ln \left (\ln \left (1+\frac {x}{5}\right )\right )\) | \(43\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.74, size = 56, normalized size = 1.87 \begin {gather*} \frac {5}{25 \, x^{2} + 40 \, x + 16} + \log \left (-\frac {\log \relax (5) \log \relax (2) - {\left (\log \relax (2) + 1\right )} \log \left (x + 5\right ) + \log \relax (5) + \log \relax (x)}{\log \relax (2) + 1}\right ) - \log \left (-\log \relax (5) + \log \left (x + 5\right )\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 {{\ln \left (\frac {x}{5}+1\right )}^2\,\left (250\,x+\ln \relax (2)\,\left (50\,x^2+250\,x\right )+50\,x^2\right )-\ln \relax (x)\,\left (64\,x+\ln \left (\frac {x}{5}+1\right )\,\left (50\,x^2+250\,x\right )+240\,x^2+300\,x^3+125\,x^4\right )+\ln \left (\frac {x}{5}+1\right )\,\left (125\,x^4+925\,x^3+1740\,x^2+1264\,x+320\right )}{{\ln \left (\frac {x}{5}+1\right )}^2\,\left (320\,x+1264\,x^2+1740\,x^3+925\,x^4+125\,x^5+\ln \relax (2)\,\left (125\,x^5+925\,x^4+1740\,x^3+1264\,x^2+320\,x\right )\right )-\ln \left (\frac {x}{5}+1\right )\,\ln \relax (x)\,\left (125\,x^5+925\,x^4+1740\,x^3+1264\,x^2+320\,x\right )} \,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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