Optimal. Leaf size=16 \[ \frac {4 x^2 \log (2 x)}{(1+8 x)^2} \]
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Rubi [A] time = 0.17, antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.147, Rules used = {6688, 12, 6742, 43, 2335} \begin {gather*} \frac {4 x^2 \log (2 x)}{(8 x+1)^2} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 43
Rule 2335
Rule 6688
Rule 6742
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {4 x (1+8 x+2 \log (2 x))}{(1+8 x)^3} \, dx\\ &=4 \int \frac {x (1+8 x+2 \log (2 x))}{(1+8 x)^3} \, dx\\ &=4 \int \left (\frac {x}{(1+8 x)^2}+\frac {2 x \log (2 x)}{(1+8 x)^3}\right ) \, dx\\ &=4 \int \frac {x}{(1+8 x)^2} \, dx+8 \int \frac {x \log (2 x)}{(1+8 x)^3} \, dx\\ &=\frac {4 x^2 \log (2 x)}{(1+8 x)^2}-4 \int \frac {x}{(1+8 x)^2} \, dx+4 \int \left (-\frac {1}{8 (1+8 x)^2}+\frac {1}{8 (1+8 x)}\right ) \, dx\\ &=\frac {1}{16 (1+8 x)}+\frac {4 x^2 \log (2 x)}{(1+8 x)^2}+\frac {1}{16} \log (1+8 x)-4 \int \left (-\frac {1}{8 (1+8 x)^2}+\frac {1}{8 (1+8 x)}\right ) \, dx\\ &=\frac {4 x^2 \log (2 x)}{(1+8 x)^2}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.06, size = 32, normalized size = 2.00 \begin {gather*} \frac {-x \log (65536)+\log (x)+64 x^2 \log (x)-\log (2 x)}{16 (1+8 x)^2} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.75, size = 21, normalized size = 1.31 \begin {gather*} \frac {4 \, x^{2} \log \left (2 \, x\right )}{64 \, x^{2} + 16 \, x + 1} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.20, size = 28, normalized size = 1.75 \begin {gather*} -\frac {{\left (16 \, x + 1\right )} \log \left (2 \, x\right )}{16 \, {\left (64 \, x^{2} + 16 \, x + 1\right )}} + \frac {1}{16} \, \log \relax (x) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.08, size = 17, normalized size = 1.06
method | result | size |
norman | \(\frac {4 x^{2} \ln \left (2 x \right )}{\left (8 x +1\right )^{2}}\) | \(17\) |
risch | \(-\frac {\left (16 x +1\right ) \ln \left (2 x \right )}{16 \left (64 x^{2}+16 x +1\right )}+\frac {\ln \relax (x )}{16}\) | \(29\) |
derivativedivides | \(\frac {\ln \left (2 x \right ) x}{8 x +1}-\frac {\ln \left (2 x \right ) x \left (4 x +1\right )}{\left (8 x +1\right )^{2}}\) | \(34\) |
default | \(\frac {\ln \left (2 x \right ) x}{8 x +1}-\frac {\ln \left (2 x \right ) x \left (4 x +1\right )}{\left (8 x +1\right )^{2}}\) | \(34\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.38, size = 75, normalized size = 4.69 \begin {gather*} -\frac {{\left (16 \, x + 1\right )} \log \left (2 \, x\right )}{16 \, {\left (64 \, x^{2} + 16 \, x + 1\right )}} + \frac {32 \, x + 3}{32 \, {\left (64 \, x^{2} + 16 \, x + 1\right )}} - \frac {16 \, x + 1}{32 \, {\left (64 \, x^{2} + 16 \, x + 1\right )}} - \frac {1}{16 \, {\left (8 \, x + 1\right )}} + \frac {1}{16} \, \log \relax (x) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 4.36, size = 16, normalized size = 1.00 \begin {gather*} \frac {4\,x^2\,\ln \left (2\,x\right )}{{\left (8\,x+1\right )}^2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.16, size = 26, normalized size = 1.62 \begin {gather*} \frac {\left (- 16 x - 1\right ) \log {\left (2 x \right )}}{1024 x^{2} + 256 x + 16} + \frac {\log {\relax (x )}}{16} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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