Integrand size = 58, antiderivative size = 25 \[ \int \frac {4+x-2 x \log \left (\frac {-5 x-5 e^5 x+\left (2 x+2 e^5 x\right ) \log (2)}{2 \log (2)}\right )}{64 x+48 x^2+12 x^3+x^4} \, dx=\frac {\log \left (\frac {1}{2} \left (1+e^5\right ) x \left (2-\frac {5}{\log (2)}\right )\right )}{(4+x)^2} \]
[Out]
Time = 0.15 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {6820, 6874, 46, 2356} \[ \int \frac {4+x-2 x \log \left (\frac {-5 x-5 e^5 x+\left (2 x+2 e^5 x\right ) \log (2)}{2 \log (2)}\right )}{64 x+48 x^2+12 x^3+x^4} \, dx=\frac {\log \left (-\frac {\left (1+e^5\right ) x (5-\log (4))}{\log (4)}\right )}{(x+4)^2} \]
[In]
[Out]
Rule 46
Rule 2356
Rule 6820
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {4+x-2 x \log \left (\frac {\left (1+e^5\right ) x (-5+\log (4))}{\log (4)}\right )}{x (4+x)^3} \, dx \\ & = \int \left (\frac {1}{x (4+x)^2}-\frac {2 \log \left (\frac {\left (1+e^5\right ) x (-5+\log (4))}{\log (4)}\right )}{(4+x)^3}\right ) \, dx \\ & = -\left (2 \int \frac {\log \left (\frac {\left (1+e^5\right ) x (-5+\log (4))}{\log (4)}\right )}{(4+x)^3} \, dx\right )+\int \frac {1}{x (4+x)^2} \, dx \\ & = \frac {\log \left (-\frac {\left (1+e^5\right ) x (5-\log (4))}{\log (4)}\right )}{(4+x)^2}-\int \frac {1}{x (4+x)^2} \, dx+\int \left (\frac {1}{16 x}-\frac {1}{4 (4+x)^2}-\frac {1}{16 (4+x)}\right ) \, dx \\ & = \frac {1}{4 (4+x)}+\frac {\log (x)}{16}-\frac {1}{16} \log (4+x)+\frac {\log \left (-\frac {\left (1+e^5\right ) x (5-\log (4))}{\log (4)}\right )}{(4+x)^2}-\int \left (\frac {1}{16 x}-\frac {1}{4 (4+x)^2}-\frac {1}{16 (4+x)}\right ) \, dx \\ & = \frac {\log \left (-\frac {\left (1+e^5\right ) x (5-\log (4))}{\log (4)}\right )}{(4+x)^2} \\ \end{align*}
Time = 0.06 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {4+x-2 x \log \left (\frac {-5 x-5 e^5 x+\left (2 x+2 e^5 x\right ) \log (2)}{2 \log (2)}\right )}{64 x+48 x^2+12 x^3+x^4} \, dx=\frac {\log \left (\frac {\left (1+e^5\right ) x (-5+\log (4))}{\log (4)}\right )}{(4+x)^2} \]
[In]
[Out]
Time = 0.37 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.40
method | result | size |
norman | \(\frac {\ln \left (\frac {\left (2 x \,{\mathrm e}^{5}+2 x \right ) \ln \left (2\right )-5 x \,{\mathrm e}^{5}-5 x}{2 \ln \left (2\right )}\right )}{\left (4+x \right )^{2}}\) | \(35\) |
parallelrisch | \(\frac {\ln \left (\frac {x \left (2 \,{\mathrm e}^{5} \ln \left (2\right )-5 \,{\mathrm e}^{5}+2 \ln \left (2\right )-5\right )}{2 \ln \left (2\right )}\right )}{x^{2}+8 x +16}\) | \(36\) |
risch | \(\frac {\ln \left (\frac {\left (2 x \,{\mathrm e}^{5}+2 x \right ) \ln \left (2\right )-5 x \,{\mathrm e}^{5}-5 x}{2 \ln \left (2\right )}\right )}{x^{2}+8 x +16}\) | \(40\) |
parts | \(\text {Expression too large to display}\) | \(6143\) |
derivativedivides | \(\text {Expression too large to display}\) | \(6305\) |
default | \(\text {Expression too large to display}\) | \(6305\) |
[In]
[Out]
none
Time = 0.25 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.48 \[ \int \frac {4+x-2 x \log \left (\frac {-5 x-5 e^5 x+\left (2 x+2 e^5 x\right ) \log (2)}{2 \log (2)}\right )}{64 x+48 x^2+12 x^3+x^4} \, dx=\frac {\log \left (-\frac {5 \, x e^{5} - 2 \, {\left (x e^{5} + x\right )} \log \left (2\right ) + 5 \, x}{2 \, \log \left (2\right )}\right )}{x^{2} + 8 \, x + 16} \]
[In]
[Out]
Time = 0.09 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.64 \[ \int \frac {4+x-2 x \log \left (\frac {-5 x-5 e^5 x+\left (2 x+2 e^5 x\right ) \log (2)}{2 \log (2)}\right )}{64 x+48 x^2+12 x^3+x^4} \, dx=\frac {\log {\left (\frac {- \frac {5 x e^{5}}{2} - \frac {5 x}{2} + \frac {\left (2 x + 2 x e^{5}\right ) \log {\left (2 \right )}}{2}}{\log {\left (2 \right )}} \right )}}{x^{2} + 8 x + 16} \]
[In]
[Out]
Leaf count of result is larger than twice the leaf count of optimal. 176 vs. \(2 (22) = 44\).
Time = 0.21 (sec) , antiderivative size = 176, normalized size of antiderivative = 7.04 \[ \int \frac {4+x-2 x \log \left (\frac {-5 x-5 e^5 x+\left (2 x+2 e^5 x\right ) \log (2)}{2 \log (2)}\right )}{64 x+48 x^2+12 x^3+x^4} \, dx=-\frac {1}{16} \, {\left (\frac {\log \left (2\right ) \log \left (x + 4\right )}{2 \, {\left (e^{5} + 1\right )} \log \left (2\right ) - 5 \, e^{5} - 5} - \frac {\log \left (2\right ) \log \left (x\right )}{2 \, {\left (e^{5} + 1\right )} \log \left (2\right ) - 5 \, e^{5} - 5} - \frac {4 \, \log \left (2\right )}{{\left (2 \, {\left (e^{5} + 1\right )} \log \left (2\right ) - 5 \, e^{5} - 5\right )} x + 8 \, {\left (e^{5} + 1\right )} \log \left (2\right ) - 20 \, e^{5} - 20}\right )} {\left (\frac {5 \, e^{5}}{\log \left (2\right )} + \frac {5}{\log \left (2\right )} - 2 \, e^{5} - 2\right )} + \frac {x + 6}{4 \, {\left (x^{2} + 8 \, x + 16\right )}} + \frac {\log \left (x e^{5} + x - \frac {5 \, x e^{5}}{2 \, \log \left (2\right )} - \frac {5 \, x}{2 \, \log \left (2\right )}\right )}{x^{2} + 8 \, x + 16} - \frac {1}{2 \, {\left (x^{2} + 8 \, x + 16\right )}} - \frac {1}{16} \, \log \left (x + 4\right ) + \frac {1}{16} \, \log \left (x\right ) \]
[In]
[Out]
none
Time = 0.29 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.68 \[ \int \frac {4+x-2 x \log \left (\frac {-5 x-5 e^5 x+\left (2 x+2 e^5 x\right ) \log (2)}{2 \log (2)}\right )}{64 x+48 x^2+12 x^3+x^4} \, dx=-\frac {\log \left (2\right ) - \log \left (2 \, x e^{5} \log \left (2\right ) - 5 \, x e^{5} + 2 \, x \log \left (2\right ) - 5 \, x\right ) + \log \left (\log \left (2\right )\right )}{x^{2} + 8 \, x + 16} \]
[In]
[Out]
Time = 14.11 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.96 \[ \int \frac {4+x-2 x \log \left (\frac {-5 x-5 e^5 x+\left (2 x+2 e^5 x\right ) \log (2)}{2 \log (2)}\right )}{64 x+48 x^2+12 x^3+x^4} \, dx=\frac {x^2\,\left (\ln \left (\frac {\ln \left (2\right )\,\left (2\,x+2\,x\,{\mathrm {e}}^5\right )}{2}-\frac {5\,x\,{\mathrm {e}}^5}{2}-\frac {5\,x}{2}\right )-\ln \left (\ln \left (2\right )\right )\right )}{x^4+8\,x^3+16\,x^2} \]
[In]
[Out]