Integrand size = 48, antiderivative size = 21 \[ \int \frac {-19200+3840 x+(-1-32 x) \log \left (1+64 x+1024 x^2\right )}{\left (-5-159 x+32 x^2\right ) \log \left (1+64 x+1024 x^2\right )} \, dx=-3-\log (-5+x)+60 \left (-5+\log \left (\log \left ((1+32 x)^2\right )\right )\right ) \]
[Out]
Time = 0.10 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {6860, 2437, 2339, 29} \[ \int \frac {-19200+3840 x+(-1-32 x) \log \left (1+64 x+1024 x^2\right )}{\left (-5-159 x+32 x^2\right ) \log \left (1+64 x+1024 x^2\right )} \, dx=60 \log \left (\log \left ((32 x+1)^2\right )\right )-\log (5-x) \]
[In]
[Out]
Rule 29
Rule 2339
Rule 2437
Rule 6860
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {1}{5-x}+\frac {3840}{(1+32 x) \log \left ((1+32 x)^2\right )}\right ) \, dx \\ & = -\log (5-x)+3840 \int \frac {1}{(1+32 x) \log \left ((1+32 x)^2\right )} \, dx \\ & = -\log (5-x)+120 \text {Subst}\left (\int \frac {1}{x \log \left (x^2\right )} \, dx,x,1+32 x\right ) \\ & = -\log (5-x)+60 \text {Subst}\left (\int \frac {1}{x} \, dx,x,\log \left ((1+32 x)^2\right )\right ) \\ & = -\log (5-x)+60 \log \left (\log \left ((1+32 x)^2\right )\right ) \\ \end{align*}
Time = 0.29 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int \frac {-19200+3840 x+(-1-32 x) \log \left (1+64 x+1024 x^2\right )}{\left (-5-159 x+32 x^2\right ) \log \left (1+64 x+1024 x^2\right )} \, dx=-\log (160-32 x)+60 \log \left (\log \left ((1+32 x)^2\right )\right ) \]
[In]
[Out]
Time = 2.78 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.05
method | result | size |
default | \(-\ln \left (-5+x \right )+60 \ln \left (\ln \left (1024 x^{2}+64 x +1\right )\right )\) | \(22\) |
norman | \(-\ln \left (-5+x \right )+60 \ln \left (\ln \left (1024 x^{2}+64 x +1\right )\right )\) | \(22\) |
risch | \(-\ln \left (-5+x \right )+60 \ln \left (\ln \left (1024 x^{2}+64 x +1\right )\right )\) | \(22\) |
parallelrisch | \(-\ln \left (-5+x \right )+60 \ln \left (\ln \left (1024 x^{2}+64 x +1\right )\right )\) | \(22\) |
parts | \(-\ln \left (-5+x \right )+60 \ln \left (\ln \left (1024 x^{2}+64 x +1\right )\right )\) | \(22\) |
[In]
[Out]
none
Time = 0.26 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {-19200+3840 x+(-1-32 x) \log \left (1+64 x+1024 x^2\right )}{\left (-5-159 x+32 x^2\right ) \log \left (1+64 x+1024 x^2\right )} \, dx=-\log \left (x - 5\right ) + 60 \, \log \left (\log \left (1024 \, x^{2} + 64 \, x + 1\right )\right ) \]
[In]
[Out]
Time = 0.06 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {-19200+3840 x+(-1-32 x) \log \left (1+64 x+1024 x^2\right )}{\left (-5-159 x+32 x^2\right ) \log \left (1+64 x+1024 x^2\right )} \, dx=- \log {\left (x - 5 \right )} + 60 \log {\left (\log {\left (1024 x^{2} + 64 x + 1 \right )} \right )} \]
[In]
[Out]
none
Time = 0.24 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.76 \[ \int \frac {-19200+3840 x+(-1-32 x) \log \left (1+64 x+1024 x^2\right )}{\left (-5-159 x+32 x^2\right ) \log \left (1+64 x+1024 x^2\right )} \, dx=-\log \left (x - 5\right ) + 60 \, \log \left (\log \left (32 \, x + 1\right )\right ) \]
[In]
[Out]
none
Time = 0.28 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {-19200+3840 x+(-1-32 x) \log \left (1+64 x+1024 x^2\right )}{\left (-5-159 x+32 x^2\right ) \log \left (1+64 x+1024 x^2\right )} \, dx=-\log \left (x - 5\right ) + 60 \, \log \left (\log \left (1024 \, x^{2} + 64 \, x + 1\right )\right ) \]
[In]
[Out]
Time = 10.32 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {-19200+3840 x+(-1-32 x) \log \left (1+64 x+1024 x^2\right )}{\left (-5-159 x+32 x^2\right ) \log \left (1+64 x+1024 x^2\right )} \, dx=60\,\ln \left (\ln \left (1024\,x^2+64\,x+1\right )\right )-\ln \left (x-5\right ) \]
[In]
[Out]