Optimal. Leaf size=29 \[ 4+\frac {x}{x^2-5 (1-x) \log \left (-2 x+\frac {5+x}{5}\right )} \]
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Rubi [F] time = 2.13, 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 {45 x-40 x^2-9 x^3+(25-45 x) \log \left (\frac {1}{5} (5-9 x)\right )}{-5 x^4+9 x^5+\left (50 x^2-140 x^3+90 x^4\right ) \log \left (\frac {1}{5} (5-9 x)\right )+\left (-125+475 x-575 x^2+225 x^3\right ) \log ^2\left (\frac {1}{5} (5-9 x)\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 \left (45-40 x-9 x^2\right )+5 (-5+9 x) \log \left (1-\frac {9 x}{5}\right )}{(5-9 x) \left (x^2+5 (-1+x) \log \left (1-\frac {9 x}{5}\right )\right )^2} \, dx\\ &=\int \left (-\frac {x \left (45-80 x+22 x^2+9 x^3\right )}{(-1+x) (-5+9 x) \left (x^2-5 \log \left (1-\frac {9 x}{5}\right )+5 x \log \left (1-\frac {9 x}{5}\right )\right )^2}-\frac {1}{(-1+x) \left (x^2-5 \log \left (1-\frac {9 x}{5}\right )+5 x \log \left (1-\frac {9 x}{5}\right )\right )}\right ) \, dx\\ &=-\int \frac {x \left (45-80 x+22 x^2+9 x^3\right )}{(-1+x) (-5+9 x) \left (x^2-5 \log \left (1-\frac {9 x}{5}\right )+5 x \log \left (1-\frac {9 x}{5}\right )\right )^2} \, dx-\int \frac {1}{(-1+x) \left (x^2-5 \log \left (1-\frac {9 x}{5}\right )+5 x \log \left (1-\frac {9 x}{5}\right )\right )} \, dx\\ &=-\int \frac {1}{(-1+x) \left (x^2+5 (-1+x) \log \left (1-\frac {9 x}{5}\right )\right )} \, dx-\int \left (-\frac {29}{9 \left (x^2-5 \log \left (1-\frac {9 x}{5}\right )+5 x \log \left (1-\frac {9 x}{5}\right )\right )^2}-\frac {1}{(-1+x) \left (x^2-5 \log \left (1-\frac {9 x}{5}\right )+5 x \log \left (1-\frac {9 x}{5}\right )\right )^2}+\frac {4 x}{\left (x^2-5 \log \left (1-\frac {9 x}{5}\right )+5 x \log \left (1-\frac {9 x}{5}\right )\right )^2}+\frac {x^2}{\left (x^2-5 \log \left (1-\frac {9 x}{5}\right )+5 x \log \left (1-\frac {9 x}{5}\right )\right )^2}-\frac {100}{9 (-5+9 x) \left (x^2-5 \log \left (1-\frac {9 x}{5}\right )+5 x \log \left (1-\frac {9 x}{5}\right )\right )^2}\right ) \, dx\\ &=\frac {29}{9} \int \frac {1}{\left (x^2-5 \log \left (1-\frac {9 x}{5}\right )+5 x \log \left (1-\frac {9 x}{5}\right )\right )^2} \, dx-4 \int \frac {x}{\left (x^2-5 \log \left (1-\frac {9 x}{5}\right )+5 x \log \left (1-\frac {9 x}{5}\right )\right )^2} \, dx+\frac {100}{9} \int \frac {1}{(-5+9 x) \left (x^2-5 \log \left (1-\frac {9 x}{5}\right )+5 x \log \left (1-\frac {9 x}{5}\right )\right )^2} \, dx-\int \frac {1}{(-1+x) \left (x^2+5 (-1+x) \log \left (1-\frac {9 x}{5}\right )\right )} \, dx+\int \frac {1}{(-1+x) \left (x^2-5 \log \left (1-\frac {9 x}{5}\right )+5 x \log \left (1-\frac {9 x}{5}\right )\right )^2} \, dx-\int \frac {x^2}{\left (x^2-5 \log \left (1-\frac {9 x}{5}\right )+5 x \log \left (1-\frac {9 x}{5}\right )\right )^2} \, dx\\ &=\frac {29}{9} \int \frac {1}{\left (x^2+5 (-1+x) \log \left (1-\frac {9 x}{5}\right )\right )^2} \, dx-4 \int \frac {x}{\left (x^2+5 (-1+x) \log \left (1-\frac {9 x}{5}\right )\right )^2} \, dx+\frac {100}{9} \int \frac {1}{(-5+9 x) \left (x^2+5 (-1+x) \log \left (1-\frac {9 x}{5}\right )\right )^2} \, dx+\int \frac {1}{(-1+x) \left (x^2+5 (-1+x) \log \left (1-\frac {9 x}{5}\right )\right )^2} \, dx-\int \frac {x^2}{\left (x^2+5 (-1+x) \log \left (1-\frac {9 x}{5}\right )\right )^2} \, dx-\int \frac {1}{(-1+x) \left (x^2+5 (-1+x) \log \left (1-\frac {9 x}{5}\right )\right )} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.73, size = 21, normalized size = 0.72 \begin {gather*} \frac {x}{x^2+5 (-1+x) \log \left (1-\frac {9 x}{5}\right )} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.75, size = 19, normalized size = 0.66 \begin {gather*} \frac {x}{x^{2} + 5 \, {\left (x - 1\right )} \log \left (-\frac {9}{5} \, x + 1\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.23, size = 25, normalized size = 0.86 \begin {gather*} \frac {x}{x^{2} + 5 \, x \log \left (-\frac {9}{5} \, x + 1\right ) - 5 \, \log \left (-\frac {9}{5} \, x + 1\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.12, size = 26, normalized size = 0.90
method | result | size |
norman | \(\frac {x}{5 \ln \left (-\frac {9 x}{5}+1\right ) x +x^{2}-5 \ln \left (-\frac {9 x}{5}+1\right )}\) | \(26\) |
risch | \(\frac {x}{5 \ln \left (-\frac {9 x}{5}+1\right ) x +x^{2}-5 \ln \left (-\frac {9 x}{5}+1\right )}\) | \(26\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.48, size = 28, normalized size = 0.97 \begin {gather*} \frac {x}{x^{2} - 5 \, x \log \relax (5) + 5 \, {\left (x - 1\right )} \log \left (-9 \, x + 5\right ) + 5 \, \log \relax (5)} \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 (1-\frac {9\,x}{5}\right )\,\left (45\,x-25\right )-45\,x+40\,x^2+9\,x^3}{\ln \left (1-\frac {9\,x}{5}\right )\,\left (90\,x^4-140\,x^3+50\,x^2\right )+{\ln \left (1-\frac {9\,x}{5}\right )}^2\,\left (225\,x^3-575\,x^2+475\,x-125\right )-5\,x^4+9\,x^5} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.18, size = 17, normalized size = 0.59 \begin {gather*} \frac {x}{x^{2} + \left (5 x - 5\right ) \log {\left (1 - \frac {9 x}{5} \right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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