Optimal. Leaf size=27 \[ \log (2)+\frac {1}{3} \left (1-x+\frac {x}{-25+x-\log (4+3 x)}\right ) \]
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Rubi [F] time = 0.80, 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 {-2600-1747 x+146 x^2-3 x^3+\left (-204-145 x+6 x^2\right ) \log (4+3 x)+(-4-3 x) \log ^2(4+3 x)}{7500+5025 x-438 x^2+9 x^3+\left (600+426 x-18 x^2\right ) \log (4+3 x)+(12+9 x) \log ^2(4+3 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 {-2600-1747 x+146 x^2-3 x^3+\left (-204-145 x+6 x^2\right ) \log (4+3 x)+(-4-3 x) \log ^2(4+3 x)}{3 (4+3 x) (25-x+\log (4+3 x))^2} \, dx\\ &=\frac {1}{3} \int \frac {-2600-1747 x+146 x^2-3 x^3+\left (-204-145 x+6 x^2\right ) \log (4+3 x)+(-4-3 x) \log ^2(4+3 x)}{(4+3 x) (25-x+\log (4+3 x))^2} \, dx\\ &=\frac {1}{3} \int \left (-1-\frac {x (1+3 x)}{(4+3 x) (-25+x-\log (4+3 x))^2}+\frac {1}{-25+x-\log (4+3 x)}\right ) \, dx\\ &=-\frac {x}{3}-\frac {1}{3} \int \frac {x (1+3 x)}{(4+3 x) (-25+x-\log (4+3 x))^2} \, dx+\frac {1}{3} \int \frac {1}{-25+x-\log (4+3 x)} \, dx\\ &=-\frac {x}{3}-\frac {1}{3} \int \left (-\frac {1}{(-25+x-\log (4+3 x))^2}+\frac {x}{(-25+x-\log (4+3 x))^2}+\frac {4}{(4+3 x) (-25+x-\log (4+3 x))^2}\right ) \, dx+\frac {1}{3} \int \frac {1}{-25+x-\log (4+3 x)} \, dx\\ &=-\frac {x}{3}+\frac {1}{3} \int \frac {1}{(-25+x-\log (4+3 x))^2} \, dx-\frac {1}{3} \int \frac {x}{(-25+x-\log (4+3 x))^2} \, dx+\frac {1}{3} \int \frac {1}{-25+x-\log (4+3 x)} \, dx-\frac {4}{3} \int \frac {1}{(4+3 x) (-25+x-\log (4+3 x))^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.42, size = 20, normalized size = 0.74 \begin {gather*} -\frac {1}{3} x \left (1+\frac {1}{25-x+\log (4+3 x)}\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 1.01, size = 31, normalized size = 1.15 \begin {gather*} -\frac {x^{2} - x \log \left (3 \, x + 4\right ) - 26 \, x}{3 \, {\left (x - \log \left (3 \, x + 4\right ) - 25\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.25, size = 20, normalized size = 0.74 \begin {gather*} -\frac {1}{3} \, x + \frac {x}{3 \, {\left (x - \log \left (3 \, x + 4\right ) - 25\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.07, size = 21, normalized size = 0.78
method | result | size |
risch | \(-\frac {x}{3}+\frac {x}{3 x -3 \ln \left (4+3 x \right )-75}\) | \(21\) |
norman | \(\frac {\frac {\ln \left (4+3 x \right )^{2}}{3}+17 x -\frac {x^{2}}{3}-\frac {625}{3}}{x -\ln \left (4+3 x \right )-25}+\frac {\ln \left (4+3 x \right )}{3}\) | \(44\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.52, size = 31, normalized size = 1.15 \begin {gather*} -\frac {x^{2} - x \log \left (3 \, x + 4\right ) - 26 \, x}{3 \, {\left (x - \log \left (3 \, x + 4\right ) - 25\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.25, size = 41, normalized size = 1.52 \begin {gather*} -\frac {25\,x+\ln \left (3\,x+4\right )+x\,\ln \left (3\,x+4\right )-x^2+25}{3\,\left (\ln \left (3\,x+4\right )-x+25\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.15, size = 19, normalized size = 0.70 \begin {gather*} - \frac {x}{3} - \frac {x}{- 3 x + 3 \log {\left (3 x + 4 \right )} + 75} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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