Optimal. Leaf size=22 \[ 2+\frac {x}{x+\log (5)+2 \log (16)-\log \left ((25+x)^2\right )} \]
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Rubi [F]
time = 0.64, 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 {2 x+(25+x) \log (5)+(50+2 x) \log (16)+(-25-x) \log \left (625+50 x+x^2\right )}{25 x^2+x^3+\left (50 x+2 x^2\right ) \log (5)+(25+x) \log ^2(5)+\left (100 x+4 x^2+(100+4 x) \log (5)\right ) \log (16)+(100+4 x) \log ^2(16)+\left (-50 x-2 x^2+(-50-2 x) \log (5)+(-100-4 x) \log (16)\right ) \log \left (625+50 x+x^2\right )+(25+x) \log ^2\left (625+50 x+x^2\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 {25 \log (1280)+x (2+\log (1280))-(25+x) \log \left ((25+x)^2\right )}{(25+x) \left (x+\log (1280)-\log \left ((25+x)^2\right )\right )^2} \, dx\\ &=\int \left (-\frac {x (23+x)}{(25+x) \left (x+\log (1280)-\log \left ((25+x)^2\right )\right )^2}+\frac {1}{x+\log (1280)-\log \left ((25+x)^2\right )}\right ) \, dx\\ &=-\int \frac {x (23+x)}{(25+x) \left (x+\log (1280)-\log \left ((25+x)^2\right )\right )^2} \, dx+\int \frac {1}{x+\log (1280)-\log \left ((25+x)^2\right )} \, dx\\ &=-\int \left (-\frac {2}{\left (x+\log (1280)-\log \left ((25+x)^2\right )\right )^2}+\frac {x}{\left (x+\log (1280)-\log \left ((25+x)^2\right )\right )^2}+\frac {50}{(25+x) \left (x+\log (1280)-\log \left ((25+x)^2\right )\right )^2}\right ) \, dx+\int \frac {1}{x+\log (1280)-\log \left ((25+x)^2\right )} \, dx\\ &=2 \int \frac {1}{\left (x+\log (1280)-\log \left ((25+x)^2\right )\right )^2} \, dx-50 \int \frac {1}{(25+x) \left (x+\log (1280)-\log \left ((25+x)^2\right )\right )^2} \, dx-\int \frac {x}{\left (x+\log (1280)-\log \left ((25+x)^2\right )\right )^2} \, dx+\int \frac {1}{x+\log (1280)-\log \left ((25+x)^2\right )} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A]
time = 0.17, size = 16, normalized size = 0.73 \begin {gather*} \frac {x}{x+\log (1280)-\log \left ((25+x)^2\right )} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 4.15, size = 24, normalized size = 1.09
method | result | size |
risch | \(\frac {x}{8 \ln \left (2\right )+\ln \left (5\right )-\ln \left (x^{2}+50 x +625\right )+x}\) | \(24\) |
norman | \(\frac {\ln \left (x^{2}+50 x +625\right )-8 \ln \left (2\right )-\ln \left (5\right )}{8 \ln \left (2\right )+\ln \left (5\right )-\ln \left (x^{2}+50 x +625\right )+x}\) | \(41\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.51, size = 18, normalized size = 0.82 \begin {gather*} \frac {x}{x + \log \left (5\right ) + 8 \, \log \left (2\right ) - 2 \, \log \left (x + 25\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.35, size = 23, normalized size = 1.05 \begin {gather*} \frac {x}{x + \log \left (5\right ) + 8 \, \log \left (2\right ) - \log \left (x^{2} + 50 \, x + 625\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 0.08, size = 22, normalized size = 1.00 \begin {gather*} - \frac {x}{- x + \log {\left (x^{2} + 50 x + 625 \right )} - 8 \log {\left (2 \right )} - \log {\left (5 \right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.44, size = 23, normalized size = 1.05 \begin {gather*} \frac {x}{x + \log \left (5\right ) + 8 \, \log \left (2\right ) - \log \left (x^{2} + 50 \, x + 625\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 = -0.05 \begin {gather*} \int \frac {2\,x+4\,\ln \left (2\right )\,\left (2\,x+50\right )+\ln \left (5\right )\,\left (x+25\right )-\ln \left (x^2+50\,x+625\right )\,\left (x+25\right )}{{\ln \left (5\right )}^2\,\left (x+25\right )+{\ln \left (x^2+50\,x+625\right )}^2\,\left (x+25\right )+\ln \left (5\right )\,\left (2\,x^2+50\,x\right )+4\,\ln \left (2\right )\,\left (100\,x+\ln \left (5\right )\,\left (4\,x+100\right )+4\,x^2\right )-\ln \left (x^2+50\,x+625\right )\,\left (50\,x+\ln \left (5\right )\,\left (2\,x+50\right )+4\,\ln \left (2\right )\,\left (4\,x+100\right )+2\,x^2\right )+16\,{\ln \left (2\right )}^2\,\left (4\,x+100\right )+25\,x^2+x^3} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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