Optimal. Leaf size=31 \[ \frac {3}{2 \left (5 x+\frac {16}{x (i \pi +\log (3)) \log (5 x)}\right )} \]
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Rubi [F] time = 1.21, 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 {48 (i \pi +\log (3))+48 (i \pi +\log (3)) \log (5 x)-15 x^2 (i \pi +\log (3))^2 \log ^2(5 x)}{512+320 x^2 (i \pi +\log (3)) \log (5 x)+50 x^4 (i \pi +\log (3))^2 \log ^2(5 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 {3 (\pi -i \log (3)) \left (-16 i-16 i \log (5 x)-5 x^2 (\pi -i \log (3)) \log ^2(5 x)\right )}{2 \left (16 i-5 x^2 (\pi -i \log (3)) \log (5 x)\right )^2} \, dx\\ &=\frac {1}{2} (3 (\pi -i \log (3))) \int \frac {-16 i-16 i \log (5 x)-5 x^2 (\pi -i \log (3)) \log ^2(5 x)}{\left (16 i-5 x^2 (\pi -i \log (3)) \log (5 x)\right )^2} \, dx\\ &=\frac {1}{2} (3 (\pi -i \log (3))) \int \left (-\frac {1}{5 x^2 (\pi -i \log (3))}+\frac {16 \left (32-5 x^2 (i \pi +\log (3))\right )}{5 x^2 (\pi -i \log (3)) \left (16 i-5 \pi x^2 \left (1-\frac {i \log (3)}{\pi }\right ) \log (5 x)\right )^2}-\frac {48 i}{5 x^2 (-\pi +i \log (3)) \left (16 i-5 \pi x^2 \left (1-\frac {i \log (3)}{\pi }\right ) \log (5 x)\right )}\right ) \, dx\\ &=\frac {3}{10 x}+\frac {72}{5} i \int \frac {1}{x^2 \left (16 i-5 \pi x^2 \left (1-\frac {i \log (3)}{\pi }\right ) \log (5 x)\right )} \, dx+\frac {24}{5} \int \frac {32-5 x^2 (i \pi +\log (3))}{x^2 \left (16 i-5 \pi x^2 \left (1-\frac {i \log (3)}{\pi }\right ) \log (5 x)\right )^2} \, dx\\ &=\frac {3}{10 x}+\frac {72}{5} i \int \frac {1}{x^2 \left (16 i-5 \pi x^2 \left (1-\frac {i \log (3)}{\pi }\right ) \log (5 x)\right )} \, dx+\frac {24}{5} \int \left (\frac {32}{x^2 \left (16 i-5 \pi x^2 \left (1-\frac {i \log (3)}{\pi }\right ) \log (5 x)\right )^2}-\frac {5 i (\pi -i \log (3))}{\left (16 i-5 \pi x^2 \left (1-\frac {i \log (3)}{\pi }\right ) \log (5 x)\right )^2}\right ) \, dx\\ &=\frac {3}{10 x}+\frac {72}{5} i \int \frac {1}{x^2 \left (16 i-5 \pi x^2 \left (1-\frac {i \log (3)}{\pi }\right ) \log (5 x)\right )} \, dx+\frac {768}{5} \int \frac {1}{x^2 \left (16 i-5 \pi x^2 \left (1-\frac {i \log (3)}{\pi }\right ) \log (5 x)\right )^2} \, dx-(24 (i \pi +\log (3))) \int \frac {1}{\left (16 i-5 \pi x^2 \left (1-\frac {i \log (3)}{\pi }\right ) \log (5 x)\right )^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.25, size = 40, normalized size = 1.29 \begin {gather*} \frac {3 x (\pi -i \log (3)) \log (5 x)}{2 \left (-16 i+5 x^2 (\pi -i \log (3)) \log (5 x)\right )} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.71, size = 40, normalized size = 1.29 \begin {gather*} \frac {3 \, {\left (-i \, \pi x - x \log \relax (3)\right )} \log \left (5 \, x\right )}{2 \, {\left (5 \, {\left (-i \, \pi x^{2} - x^{2} \log \relax (3)\right )} \log \left (5 \, x\right ) - 16\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.24, size = 35, normalized size = 1.13 \begin {gather*} -\frac {24}{25 i \, \pi x^{3} \log \left (5 \, x\right ) + 25 \, x^{3} \log \relax (3) \log \left (5 \, x\right ) + 80 \, x} + \frac {3}{10 \, x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.25, size = 40, normalized size = 1.29
method | result | size |
risch | \(\frac {3}{10 x}+\frac {24 i}{5 x \left (-5 i \ln \relax (3) x^{2} \ln \left (5 x \right )+5 \pi \ln \left (5 x \right ) x^{2}-16 i\right )}\) | \(40\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.60, size = 59, normalized size = 1.90 \begin {gather*} \frac {3 \, {\left ({\left (\pi - i \, \log \relax (3)\right )} x \log \relax (x) + {\left (\pi \log \relax (5) - i \, \log \relax (5) \log \relax (3)\right )} x\right )}}{2 \, {\left (5 \, {\left (\pi - i \, \log \relax (3)\right )} x^{2} \log \relax (x) + 5 \, {\left (\pi \log \relax (5) - i \, \log \relax (5) \log \relax (3)\right )} x^{2} - 16 i\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 {\Pi \,48{}\mathrm {i}+48\,\ln \relax (3)+48\,\ln \left (5\,x\right )\,\left (\ln \relax (3)+\Pi \,1{}\mathrm {i}\right )-15\,x^2\,{\ln \left (5\,x\right )}^2\,{\left (\ln \relax (3)+\Pi \,1{}\mathrm {i}\right )}^2}{50\,x^4\,{\ln \left (5\,x\right )}^2\,{\left (\ln \relax (3)+\Pi \,1{}\mathrm {i}\right )}^2+320\,x^2\,\ln \left (5\,x\right )\,\left (\ln \relax (3)+\Pi \,1{}\mathrm {i}\right )+512} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 40.22, size = 54, normalized size = 1.74 \begin {gather*} \frac {24}{- 25 x^{3} \log {\relax (3 )} \log {\relax (5 )} - 25 i \pi x^{3} \log {\relax (5 )} - 80 x + \left (- 25 x^{3} \log {\relax (3 )} - 25 i \pi x^{3}\right ) \log {\relax (x )}} + \frac {3}{10 x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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