Optimal. Leaf size=28 \[ \frac {1-\frac {\log (2)}{\frac {2 x}{5}+\frac {3}{\log \left (49 x^4\right )}}}{x} \]
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Rubi [F] time = 0.69, 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 {-225-300 \log (2)+(-60 x+75 \log (2)) \log \left (49 x^4\right )+\left (-4 x^2+20 x \log (2)\right ) \log ^2\left (49 x^4\right )}{225 x^2+60 x^3 \log \left (49 x^4\right )+4 x^4 \log ^2\left (49 x^4\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 {-225 \left (1+\frac {4 \log (2)}{3}\right )+(-60 x+75 \log (2)) \log \left (49 x^4\right )+\left (-4 x^2+20 x \log (2)\right ) \log ^2\left (49 x^4\right )}{x^2 \left (15+2 x \log \left (49 x^4\right )\right )^2} \, dx\\ &=\int \left (\frac {-x+\log (32)}{x^3}-\frac {75 (-15+8 x) \log (2)}{2 x^3 \left (15+2 x \log \left (49 x^4\right )\right )^2}-\frac {225 \log (2)}{2 x^3 \left (15+2 x \log \left (49 x^4\right )\right )}\right ) \, dx\\ &=-\left (\frac {1}{2} (75 \log (2)) \int \frac {-15+8 x}{x^3 \left (15+2 x \log \left (49 x^4\right )\right )^2} \, dx\right )-\frac {1}{2} (225 \log (2)) \int \frac {1}{x^3 \left (15+2 x \log \left (49 x^4\right )\right )} \, dx+\int \frac {-x+\log (32)}{x^3} \, dx\\ &=-\frac {(x-\log (32))^2}{2 x^2 \log (32)}-\frac {1}{2} (75 \log (2)) \int \left (-\frac {15}{x^3 \left (15+2 x \log \left (49 x^4\right )\right )^2}+\frac {8}{x^2 \left (15+2 x \log \left (49 x^4\right )\right )^2}\right ) \, dx-\frac {1}{2} (225 \log (2)) \int \frac {1}{x^3 \left (15+2 x \log \left (49 x^4\right )\right )} \, dx\\ &=-\frac {(x-\log (32))^2}{2 x^2 \log (32)}-\frac {1}{2} (225 \log (2)) \int \frac {1}{x^3 \left (15+2 x \log \left (49 x^4\right )\right )} \, dx-(300 \log (2)) \int \frac {1}{x^2 \left (15+2 x \log \left (49 x^4\right )\right )^2} \, dx+\frac {1}{2} (1125 \log (2)) \int \frac {1}{x^3 \left (15+2 x \log \left (49 x^4\right )\right )^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.22, size = 34, normalized size = 1.21 \begin {gather*} \frac {15+(2 x-5 \log (2)) \log \left (49 x^4\right )}{x \left (15+2 x \log \left (49 x^4\right )\right )} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.60, size = 35, normalized size = 1.25 \begin {gather*} \frac {{\left (2 \, x - 5 \, \log \relax (2)\right )} \log \left (49 \, x^{4}\right ) + 15}{2 \, x^{2} \log \left (49 \, x^{4}\right ) + 15 \, x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.33, size = 37, normalized size = 1.32 \begin {gather*} \frac {75 \, \log \relax (2)}{2 \, {\left (2 \, x^{3} \log \left (49 \, x^{4}\right ) + 15 \, x^{2}\right )}} + \frac {2 \, x - 5 \, \log \relax (2)}{2 \, x^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.14, size = 35, normalized size = 1.25
method | result | size |
risch | \(-\frac {5 \ln \relax (2)-2 x}{2 x^{2}}+\frac {75 \ln \relax (2)}{2 x^{2} \left (2 \ln \left (49 x^{4}\right ) x +15\right )}\) | \(35\) |
norman | \(\frac {15+2 \ln \left (49 x^{4}\right ) x -5 \ln \left (49 x^{4}\right ) \ln \relax (2)}{x \left (2 \ln \left (49 x^{4}\right ) x +15\right )}\) | \(39\) |
default | \(\frac {1}{x}+\frac {-10 \ln \relax (2) \ln \relax (7)-5 \ln \relax (2) \ln \left (x^{4}\right )}{x \left (4 x \ln \relax (7)+2 x \ln \left (x^{4}\right )+15\right )}\) | \(40\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.47, size = 46, normalized size = 1.64 \begin {gather*} \frac {4 \, x \log \relax (7) - 10 \, \log \relax (7) \log \relax (2) + 4 \, {\left (2 \, x - 5 \, \log \relax (2)\right )} \log \relax (x) + 15}{4 \, x^{2} \log \relax (7) + 8 \, x^{2} \log \relax (x) + 15 \, x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 7.62, size = 38, normalized size = 1.36 \begin {gather*} \frac {2\,x\,\ln \left (49\,x^4\right )-\ln \left (32\right )\,\ln \left (49\,x^4\right )+15}{x\,\left (2\,x\,\ln \left (49\,x^4\right )+15\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.19, size = 34, normalized size = 1.21 \begin {gather*} \frac {75 \log {\relax (2 )}}{4 x^{3} \log {\left (49 x^{4} \right )} + 30 x^{2}} - \frac {- 2 x + 5 \log {\relax (2 )}}{2 x^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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