Integrand size = 136, antiderivative size = 23 \[ \int \frac {e^{24} \left (-16+(-8 x-8 \log (3)) \log (x)+\left (5-x-x^2-2 x \log (3)-\log ^2(3)\right ) \log ^2(x)\right ) \left (-32+(-8 x-8 \log (3)) \log (x)+8 x \log ^2(x)+\left (x+2 x^2+2 x \log (3)\right ) \log ^3(x)\right )}{\log ^2(x) \left (16 x \log (x)+\left (8 x^2+8 x \log (3)\right ) \log ^2(x)+\left (-5 x+x^2+x^3+2 x^2 \log (3)+x \log ^2(3)\right ) \log ^3(x)\right )} \, dx=e^{24} \left (5-x-\left (x+\log (3)+\frac {4}{\log (x)}\right )^2\right ) \]
[Out]
Leaf count is larger than twice the leaf count of optimal. \(49\) vs. \(2(23)=46\).
Time = 0.27 (sec) , antiderivative size = 49, normalized size of antiderivative = 2.13, number of steps used = 13, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {12, 6820, 6874, 2339, 30, 2395, 2334, 2335} \[ \int \frac {e^{24} \left (-16+(-8 x-8 \log (3)) \log (x)+\left (5-x-x^2-2 x \log (3)-\log ^2(3)\right ) \log ^2(x)\right ) \left (-32+(-8 x-8 \log (3)) \log (x)+8 x \log ^2(x)+\left (x+2 x^2+2 x \log (3)\right ) \log ^3(x)\right )}{\log ^2(x) \left (16 x \log (x)+\left (8 x^2+8 x \log (3)\right ) \log ^2(x)+\left (-5 x+x^2+x^3+2 x^2 \log (3)+x \log ^2(3)\right ) \log ^3(x)\right )} \, dx=-e^{24} x^2-\frac {16 e^{24}}{\log ^2(x)}-\frac {8 e^{24} x}{\log (x)}-e^{24} x (1+\log (9))-\frac {8 e^{24} \log (3)}{\log (x)} \]
[In]
[Out]
Rule 12
Rule 30
Rule 2334
Rule 2335
Rule 2339
Rule 2395
Rule 6820
Rule 6874
Rubi steps \begin{align*} \text {integral}& = e^{24} \int \frac {\left (-16+(-8 x-8 \log (3)) \log (x)+\left (5-x-x^2-2 x \log (3)-\log ^2(3)\right ) \log ^2(x)\right ) \left (-32+(-8 x-8 \log (3)) \log (x)+8 x \log ^2(x)+\left (x+2 x^2+2 x \log (3)\right ) \log ^3(x)\right )}{\log ^2(x) \left (16 x \log (x)+\left (8 x^2+8 x \log (3)\right ) \log ^2(x)+\left (-5 x+x^2+x^3+2 x^2 \log (3)+x \log ^2(3)\right ) \log ^3(x)\right )} \, dx \\ & = e^{24} \int \frac {32+8 (x+\log (3)) \log (x)-8 x \log ^2(x)-x (1+2 x+\log (9)) \log ^3(x)}{x \log ^3(x)} \, dx \\ & = e^{24} \int \left (-1-2 x-\log (9)+\frac {32}{x \log ^3(x)}+\frac {8 (x+\log (3))}{x \log ^2(x)}-\frac {8}{\log (x)}\right ) \, dx \\ & = -e^{24} x^2-e^{24} x (1+\log (9))+\left (8 e^{24}\right ) \int \frac {x+\log (3)}{x \log ^2(x)} \, dx-\left (8 e^{24}\right ) \int \frac {1}{\log (x)} \, dx+\left (32 e^{24}\right ) \int \frac {1}{x \log ^3(x)} \, dx \\ & = -e^{24} x^2-e^{24} x (1+\log (9))-8 e^{24} \operatorname {LogIntegral}(x)+\left (8 e^{24}\right ) \int \left (\frac {1}{\log ^2(x)}+\frac {\log (3)}{x \log ^2(x)}\right ) \, dx+\left (32 e^{24}\right ) \text {Subst}\left (\int \frac {1}{x^3} \, dx,x,\log (x)\right ) \\ & = -e^{24} x^2-e^{24} x (1+\log (9))-\frac {16 e^{24}}{\log ^2(x)}-8 e^{24} \operatorname {LogIntegral}(x)+\left (8 e^{24}\right ) \int \frac {1}{\log ^2(x)} \, dx+\left (8 e^{24} \log (3)\right ) \int \frac {1}{x \log ^2(x)} \, dx \\ & = -e^{24} x^2-e^{24} x (1+\log (9))-\frac {16 e^{24}}{\log ^2(x)}-\frac {8 e^{24} x}{\log (x)}-8 e^{24} \operatorname {LogIntegral}(x)+\left (8 e^{24}\right ) \int \frac {1}{\log (x)} \, dx+\left (8 e^{24} \log (3)\right ) \text {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log (x)\right ) \\ & = -e^{24} x^2-e^{24} x (1+\log (9))-\frac {16 e^{24}}{\log ^2(x)}-\frac {8 e^{24} x}{\log (x)}-\frac {8 e^{24} \log (3)}{\log (x)} \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(53\) vs. \(2(23)=46\).
Time = 0.05 (sec) , antiderivative size = 53, normalized size of antiderivative = 2.30 \[ \int \frac {e^{24} \left (-16+(-8 x-8 \log (3)) \log (x)+\left (5-x-x^2-2 x \log (3)-\log ^2(3)\right ) \log ^2(x)\right ) \left (-32+(-8 x-8 \log (3)) \log (x)+8 x \log ^2(x)+\left (x+2 x^2+2 x \log (3)\right ) \log ^3(x)\right )}{\log ^2(x) \left (16 x \log (x)+\left (8 x^2+8 x \log (3)\right ) \log ^2(x)+\left (-5 x+x^2+x^3+2 x^2 \log (3)+x \log ^2(3)\right ) \log ^3(x)\right )} \, dx=-e^{24} x-e^{24} x^2-e^{24} x \log (9)-\frac {16 e^{24}}{\log ^2(x)}-\frac {8 e^{24} x}{\log (x)}-\frac {8 e^{24} \log (3)}{\log (x)} \]
[In]
[Out]
Time = 142.32 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.70
method | result | size |
default | \({\mathrm e}^{24} \left (-2 x \ln \left (3\right )-x^{2}-x -\frac {8 \ln \left (3\right )}{\ln \left (x \right )}-\frac {8 x}{\ln \left (x \right )}-\frac {16}{\ln \left (x \right )^{2}}\right )\) | \(39\) |
parallelrisch | \(\text {Expression too large to display}\) | \(1635\) |
[In]
[Out]
none
Time = 0.27 (sec) , antiderivative size = 46, normalized size of antiderivative = 2.00 \[ \int \frac {e^{24} \left (-16+(-8 x-8 \log (3)) \log (x)+\left (5-x-x^2-2 x \log (3)-\log ^2(3)\right ) \log ^2(x)\right ) \left (-32+(-8 x-8 \log (3)) \log (x)+8 x \log ^2(x)+\left (x+2 x^2+2 x \log (3)\right ) \log ^3(x)\right )}{\log ^2(x) \left (16 x \log (x)+\left (8 x^2+8 x \log (3)\right ) \log ^2(x)+\left (-5 x+x^2+x^3+2 x^2 \log (3)+x \log ^2(3)\right ) \log ^3(x)\right )} \, dx=-\frac {{\left (2 \, x e^{24} \log \left (3\right ) + {\left (x^{2} + x\right )} e^{24}\right )} \log \left (x\right )^{2} + 8 \, {\left (x e^{24} + e^{24} \log \left (3\right )\right )} \log \left (x\right ) + 16 \, e^{24}}{\log \left (x\right )^{2}} \]
[In]
[Out]
Leaf count of result is larger than twice the leaf count of optimal. 51 vs. \(2 (17) = 34\).
Time = 0.16 (sec) , antiderivative size = 51, normalized size of antiderivative = 2.22 \[ \int \frac {e^{24} \left (-16+(-8 x-8 \log (3)) \log (x)+\left (5-x-x^2-2 x \log (3)-\log ^2(3)\right ) \log ^2(x)\right ) \left (-32+(-8 x-8 \log (3)) \log (x)+8 x \log ^2(x)+\left (x+2 x^2+2 x \log (3)\right ) \log ^3(x)\right )}{\log ^2(x) \left (16 x \log (x)+\left (8 x^2+8 x \log (3)\right ) \log ^2(x)+\left (-5 x+x^2+x^3+2 x^2 \log (3)+x \log ^2(3)\right ) \log ^3(x)\right )} \, dx=- x^{2} e^{24} + x \left (- 2 e^{24} \log {\left (3 \right )} - e^{24}\right ) + \frac {\left (- 8 x e^{24} - 8 e^{24} \log {\left (3 \right )}\right ) \log {\left (x \right )} - 16 e^{24}}{\log {\left (x \right )}^{2}} \]
[In]
[Out]
none
Time = 0.32 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.52 \[ \int \frac {e^{24} \left (-16+(-8 x-8 \log (3)) \log (x)+\left (5-x-x^2-2 x \log (3)-\log ^2(3)\right ) \log ^2(x)\right ) \left (-32+(-8 x-8 \log (3)) \log (x)+8 x \log ^2(x)+\left (x+2 x^2+2 x \log (3)\right ) \log ^3(x)\right )}{\log ^2(x) \left (16 x \log (x)+\left (8 x^2+8 x \log (3)\right ) \log ^2(x)+\left (-5 x+x^2+x^3+2 x^2 \log (3)+x \log ^2(3)\right ) \log ^3(x)\right )} \, dx=-\frac {{\left ({\left (x^{2} + x {\left (2 \, \log \left (3\right ) + 1\right )}\right )} \log \left (x\right )^{2} + 8 \, {\left (x + \log \left (3\right )\right )} \log \left (x\right ) + 16\right )} e^{24}}{\log \left (x\right )^{2}} \]
[In]
[Out]
Leaf count of result is larger than twice the leaf count of optimal. 55 vs. \(2 (23) = 46\).
Time = 0.27 (sec) , antiderivative size = 55, normalized size of antiderivative = 2.39 \[ \int \frac {e^{24} \left (-16+(-8 x-8 \log (3)) \log (x)+\left (5-x-x^2-2 x \log (3)-\log ^2(3)\right ) \log ^2(x)\right ) \left (-32+(-8 x-8 \log (3)) \log (x)+8 x \log ^2(x)+\left (x+2 x^2+2 x \log (3)\right ) \log ^3(x)\right )}{\log ^2(x) \left (16 x \log (x)+\left (8 x^2+8 x \log (3)\right ) \log ^2(x)+\left (-5 x+x^2+x^3+2 x^2 \log (3)+x \log ^2(3)\right ) \log ^3(x)\right )} \, dx=-\frac {x^{2} e^{24} \log \left (x\right )^{2} + 2 \, x e^{24} \log \left (3\right ) \log \left (x\right )^{2} + x e^{24} \log \left (x\right )^{2} + 8 \, x e^{24} \log \left (x\right ) + 8 \, e^{24} \log \left (3\right ) \log \left (x\right ) + 16 \, e^{24}}{\log \left (x\right )^{2}} \]
[In]
[Out]
Time = 8.68 (sec) , antiderivative size = 54, normalized size of antiderivative = 2.35 \[ \int \frac {e^{24} \left (-16+(-8 x-8 \log (3)) \log (x)+\left (5-x-x^2-2 x \log (3)-\log ^2(3)\right ) \log ^2(x)\right ) \left (-32+(-8 x-8 \log (3)) \log (x)+8 x \log ^2(x)+\left (x+2 x^2+2 x \log (3)\right ) \log ^3(x)\right )}{\log ^2(x) \left (16 x \log (x)+\left (8 x^2+8 x \log (3)\right ) \log ^2(x)+\left (-5 x+x^2+x^3+2 x^2 \log (3)+x \log ^2(3)\right ) \log ^3(x)\right )} \, dx=-x^2\,{\mathrm {e}}^{24}-\frac {16\,{\mathrm {e}}^{24}+8\,{\mathrm {e}}^{24}\,\ln \left (3\right )\,\ln \left (x\right )}{{\ln \left (x\right )}^2}-\frac {x\,\left ({\mathrm {e}}^{24}\,\left (2\,\ln \left (3\right )+1\right )\,{\ln \left (x\right )}^2+8\,{\mathrm {e}}^{24}\,\ln \left (x\right )\right )}{{\ln \left (x\right )}^2} \]
[In]
[Out]