Integrand size = 116, antiderivative size = 28 \[ \int \frac {(-16+4 x+(4+4 x) \log (5+5 x)) \log (\log (3))}{e^5 \left (-100-75 x+25 x^2\right ) \log (5+5 x)+e^5 \left (-40-30 x+10 x^2\right ) \log (5+5 x) \log \left (\frac {2}{(-4+x) \log (5+5 x)}\right )+e^5 \left (-4-3 x+x^2\right ) \log (5+5 x) \log ^2\left (\frac {2}{(-4+x) \log (5+5 x)}\right )} \, dx=\frac {4 \log (\log (3))}{e^5 \left (5+\log \left (\frac {2}{(-4+x) \log (5 (1+x))}\right )\right )} \]
[Out]
Time = 0.37 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.29, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.034, Rules used = {12, 6820, 6843, 32} \[ \int \frac {(-16+4 x+(4+4 x) \log (5+5 x)) \log (\log (3))}{e^5 \left (-100-75 x+25 x^2\right ) \log (5+5 x)+e^5 \left (-40-30 x+10 x^2\right ) \log (5+5 x) \log \left (\frac {2}{(-4+x) \log (5+5 x)}\right )+e^5 \left (-4-3 x+x^2\right ) \log (5+5 x) \log ^2\left (\frac {2}{(-4+x) \log (5+5 x)}\right )} \, dx=-\frac {4 \log (\log (3))}{5 e^5 \left (\frac {5}{\log \left (-\frac {2}{(4-x) \log (5 (x+1))}\right )}+1\right )} \]
[In]
[Out]
Rule 12
Rule 32
Rule 6820
Rule 6843
Rubi steps \begin{align*} \text {integral}& = \log (\log (3)) \int \frac {-16+4 x+(4+4 x) \log (5+5 x)}{e^5 \left (-100-75 x+25 x^2\right ) \log (5+5 x)+e^5 \left (-40-30 x+10 x^2\right ) \log (5+5 x) \log \left (\frac {2}{(-4+x) \log (5+5 x)}\right )+e^5 \left (-4-3 x+x^2\right ) \log (5+5 x) \log ^2\left (\frac {2}{(-4+x) \log (5+5 x)}\right )} \, dx \\ & = \log (\log (3)) \int \frac {4 (4-x-(1+x) \log (5 (1+x)))}{e^5 \left (4+3 x-x^2\right ) \log (5+5 x) \left (5+\log \left (\frac {2}{(-4+x) \log (5 (1+x))}\right )\right )^2} \, dx \\ & = \frac {(4 \log (\log (3))) \int \frac {4-x-(1+x) \log (5 (1+x))}{\left (4+3 x-x^2\right ) \log (5+5 x) \left (5+\log \left (\frac {2}{(-4+x) \log (5 (1+x))}\right )\right )^2} \, dx}{e^5} \\ & = \frac {(4 \log (\log (3))) \text {Subst}\left (\int \frac {1}{(1+x)^2} \, dx,x,\frac {5}{\log \left (\frac {2}{(-4+x) \log (5 (1+x))}\right )}\right )}{5 e^5} \\ & = -\frac {4 \log (\log (3))}{5 e^5 \left (1+\frac {5}{\log \left (-\frac {2}{(4-x) \log (5 (1+x))}\right )}\right )} \\ \end{align*}
Time = 0.23 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {(-16+4 x+(4+4 x) \log (5+5 x)) \log (\log (3))}{e^5 \left (-100-75 x+25 x^2\right ) \log (5+5 x)+e^5 \left (-40-30 x+10 x^2\right ) \log (5+5 x) \log \left (\frac {2}{(-4+x) \log (5+5 x)}\right )+e^5 \left (-4-3 x+x^2\right ) \log (5+5 x) \log ^2\left (\frac {2}{(-4+x) \log (5+5 x)}\right )} \, dx=\frac {4 \log (\log (3))}{e^5 \left (5+\log \left (\frac {2}{(-4+x) \log (5 (1+x))}\right )\right )} \]
[In]
[Out]
Time = 2.80 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07
method | result | size |
parallelrisch | \(\frac {4 \,{\mathrm e}^{-5} \ln \left (\ln \left (3\right )\right )}{\ln \left (\frac {2}{\left (x -4\right ) \ln \left (5 x +5\right )}\right )+5}\) | \(30\) |
default | \(\frac {4 \ln \left (\ln \left (3\right )\right ) {\mathrm e}^{-5}}{\ln \left (2\right )+\ln \left (\frac {1}{\left (x -4\right ) \left (\ln \left (5\right )+\ln \left (1+x \right )\right )}\right )+5}\) | \(32\) |
risch | \(\frac {8 \ln \left (\ln \left (3\right )\right ) {\mathrm e}^{-5}}{10-i \pi \,\operatorname {csgn}\left (\frac {i}{x -4}\right ) \operatorname {csgn}\left (\frac {i}{\ln \left (5 x +5\right )}\right ) \operatorname {csgn}\left (\frac {i}{\ln \left (5 x +5\right ) \left (x -4\right )}\right )+i \pi \,\operatorname {csgn}\left (\frac {i}{x -4}\right ) \operatorname {csgn}\left (\frac {i}{\ln \left (5 x +5\right ) \left (x -4\right )}\right )^{2}+i \pi \,\operatorname {csgn}\left (\frac {i}{\ln \left (5 x +5\right )}\right ) \operatorname {csgn}\left (\frac {i}{\ln \left (5 x +5\right ) \left (x -4\right )}\right )^{2}-i \pi \operatorname {csgn}\left (\frac {i}{\ln \left (5 x +5\right ) \left (x -4\right )}\right )^{3}+2 \ln \left (2\right )-2 \ln \left (x -4\right )-2 \ln \left (\ln \left (5 x +5\right )\right )}\) | \(163\) |
[In]
[Out]
none
Time = 0.24 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.11 \[ \int \frac {(-16+4 x+(4+4 x) \log (5+5 x)) \log (\log (3))}{e^5 \left (-100-75 x+25 x^2\right ) \log (5+5 x)+e^5 \left (-40-30 x+10 x^2\right ) \log (5+5 x) \log \left (\frac {2}{(-4+x) \log (5+5 x)}\right )+e^5 \left (-4-3 x+x^2\right ) \log (5+5 x) \log ^2\left (\frac {2}{(-4+x) \log (5+5 x)}\right )} \, dx=\frac {4 \, \log \left (\log \left (3\right )\right )}{e^{5} \log \left (\frac {2}{{\left (x - 4\right )} \log \left (5 \, x + 5\right )}\right ) + 5 \, e^{5}} \]
[In]
[Out]
Time = 0.16 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96 \[ \int \frac {(-16+4 x+(4+4 x) \log (5+5 x)) \log (\log (3))}{e^5 \left (-100-75 x+25 x^2\right ) \log (5+5 x)+e^5 \left (-40-30 x+10 x^2\right ) \log (5+5 x) \log \left (\frac {2}{(-4+x) \log (5+5 x)}\right )+e^5 \left (-4-3 x+x^2\right ) \log (5+5 x) \log ^2\left (\frac {2}{(-4+x) \log (5+5 x)}\right )} \, dx=\frac {4 \log {\left (\log {\left (3 \right )} \right )}}{e^{5} \log {\left (\frac {2}{\left (x - 4\right ) \log {\left (5 x + 5 \right )}} \right )} + 5 e^{5}} \]
[In]
[Out]
none
Time = 0.32 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.25 \[ \int \frac {(-16+4 x+(4+4 x) \log (5+5 x)) \log (\log (3))}{e^5 \left (-100-75 x+25 x^2\right ) \log (5+5 x)+e^5 \left (-40-30 x+10 x^2\right ) \log (5+5 x) \log \left (\frac {2}{(-4+x) \log (5+5 x)}\right )+e^5 \left (-4-3 x+x^2\right ) \log (5+5 x) \log ^2\left (\frac {2}{(-4+x) \log (5+5 x)}\right )} \, dx=\frac {4 \, \log \left (\log \left (3\right )\right )}{{\left (\log \left (2\right ) + 5\right )} e^{5} - e^{5} \log \left (x - 4\right ) - e^{5} \log \left (\log \left (5\right ) + \log \left (x + 1\right )\right )} \]
[In]
[Out]
Leaf count of result is larger than twice the leaf count of optimal. 735 vs. \(2 (27) = 54\).
Time = 1.09 (sec) , antiderivative size = 735, normalized size of antiderivative = 26.25 \[ \int \frac {(-16+4 x+(4+4 x) \log (5+5 x)) \log (\log (3))}{e^5 \left (-100-75 x+25 x^2\right ) \log (5+5 x)+e^5 \left (-40-30 x+10 x^2\right ) \log (5+5 x) \log \left (\frac {2}{(-4+x) \log (5+5 x)}\right )+e^5 \left (-4-3 x+x^2\right ) \log (5+5 x) \log ^2\left (\frac {2}{(-4+x) \log (5+5 x)}\right )} \, dx=\text {Too large to display} \]
[In]
[Out]
Time = 9.81 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96 \[ \int \frac {(-16+4 x+(4+4 x) \log (5+5 x)) \log (\log (3))}{e^5 \left (-100-75 x+25 x^2\right ) \log (5+5 x)+e^5 \left (-40-30 x+10 x^2\right ) \log (5+5 x) \log \left (\frac {2}{(-4+x) \log (5+5 x)}\right )+e^5 \left (-4-3 x+x^2\right ) \log (5+5 x) \log ^2\left (\frac {2}{(-4+x) \log (5+5 x)}\right )} \, dx=\frac {4\,{\mathrm {e}}^{-5}\,\ln \left (\ln \left (3\right )\right )}{\ln \left (\frac {2}{\ln \left (5\,x+5\right )\,\left (x-4\right )}\right )+5} \]
[In]
[Out]