Integrand size = 122, antiderivative size = 35 \[ \int \frac {16+8 x^2+x^4+e^{5+x} \left (-16 x-8 x^3-x^5\right )+e^5 \left (-4 x+x^3\right ) \log (2)+\left (e^x \left (-16 x-8 x^3-x^5\right )+\left (-4 x+x^3\right ) \log (2)\right ) \log \left (\frac {x}{4}\right )}{e^5 \left (256 x+128 x^3+16 x^5\right )+\left (256 x+128 x^3+16 x^5\right ) \log \left (\frac {x}{4}\right )} \, dx=\frac {1}{16} \left (2-e^x-\frac {\log (2)}{\frac {4}{x}+x}+\log \left (e^5+\log \left (\frac {x}{4}\right )\right )\right ) \]
[Out]
\[ \int \frac {16+8 x^2+x^4+e^{5+x} \left (-16 x-8 x^3-x^5\right )+e^5 \left (-4 x+x^3\right ) \log (2)+\left (e^x \left (-16 x-8 x^3-x^5\right )+\left (-4 x+x^3\right ) \log (2)\right ) \log \left (\frac {x}{4}\right )}{e^5 \left (256 x+128 x^3+16 x^5\right )+\left (256 x+128 x^3+16 x^5\right ) \log \left (\frac {x}{4}\right )} \, dx=\int \frac {16+8 x^2+x^4+e^{5+x} \left (-16 x-8 x^3-x^5\right )+e^5 \left (-4 x+x^3\right ) \log (2)+\left (e^x \left (-16 x-8 x^3-x^5\right )+\left (-4 x+x^3\right ) \log (2)\right ) \log \left (\frac {x}{4}\right )}{e^5 \left (256 x+128 x^3+16 x^5\right )+\left (256 x+128 x^3+16 x^5\right ) \log \left (\frac {x}{4}\right )} \, dx \]
[In]
[Out]
Rubi steps \begin{align*} \text {integral}& = \int \frac {\left (4+x^2\right )^2-e^{5+x} x \left (4+x^2\right )^2+e^5 x \left (-4+x^2\right ) \log (2)-x \left (e^x \left (4+x^2\right )^2-x^2 \log (2)+\log (16)\right ) \log \left (\frac {x}{4}\right )}{16 x \left (4+x^2\right )^2 \left (e^5+\log \left (\frac {x}{4}\right )\right )} \, dx \\ & = \frac {1}{16} \int \frac {\left (4+x^2\right )^2-e^{5+x} x \left (4+x^2\right )^2+e^5 x \left (-4+x^2\right ) \log (2)-x \left (e^x \left (4+x^2\right )^2-x^2 \log (2)+\log (16)\right ) \log \left (\frac {x}{4}\right )}{x \left (4+x^2\right )^2 \left (e^5+\log \left (\frac {x}{4}\right )\right )} \, dx \\ & = \frac {1}{16} \int \left (-e^x+\frac {1}{x \left (e^5+\log \left (\frac {x}{4}\right )\right )}+\frac {e^5 (-2+x) (2+x) \log (2)}{\left (4+x^2\right )^2 \left (e^5+\log \left (\frac {x}{4}\right )\right )}+\frac {x^2 \log (2) \log \left (\frac {x}{4}\right )}{\left (4+x^2\right )^2 \left (e^5+\log \left (\frac {x}{4}\right )\right )}-\frac {\log (16) \log \left (\frac {x}{4}\right )}{\left (4+x^2\right )^2 \left (e^5+\log \left (\frac {x}{4}\right )\right )}\right ) \, dx \\ & = -\frac {\int e^x \, dx}{16}+\frac {1}{16} \int \frac {1}{x \left (e^5+\log \left (\frac {x}{4}\right )\right )} \, dx+\frac {1}{16} \log (2) \int \frac {x^2 \log \left (\frac {x}{4}\right )}{\left (4+x^2\right )^2 \left (e^5+\log \left (\frac {x}{4}\right )\right )} \, dx+\frac {1}{16} \left (e^5 \log (2)\right ) \int \frac {(-2+x) (2+x)}{\left (4+x^2\right )^2 \left (e^5+\log \left (\frac {x}{4}\right )\right )} \, dx-\frac {1}{16} \log (16) \int \frac {\log \left (\frac {x}{4}\right )}{\left (4+x^2\right )^2 \left (e^5+\log \left (\frac {x}{4}\right )\right )} \, dx \\ & = -\frac {e^x}{16}+\frac {1}{16} \text {Subst}\left (\int \frac {1}{x} \, dx,x,e^5+\log \left (\frac {x}{4}\right )\right )+\frac {1}{16} \log (2) \int \left (\frac {x^2}{\left (4+x^2\right )^2}-\frac {e^5 x^2}{\left (4+x^2\right )^2 \left (e^5+\log \left (\frac {x}{4}\right )\right )}\right ) \, dx+\frac {1}{16} \left (e^5 \log (2)\right ) \int \frac {-4+x^2}{\left (4+x^2\right )^2 \left (e^5+\log \left (\frac {x}{4}\right )\right )} \, dx-\frac {1}{4} \log (16) \text {Subst}\left (\int \frac {\log (x)}{\left (4+16 x^2\right )^2 \left (e^5+\log (x)\right )} \, dx,x,\frac {x}{4}\right ) \\ & = -\frac {e^x}{16}+\frac {1}{16} \log \left (e^5+\log \left (\frac {x}{4}\right )\right )+\frac {1}{16} \log (2) \int \frac {x^2}{\left (4+x^2\right )^2} \, dx-\frac {1}{16} \left (e^5 \log (2)\right ) \int \frac {x^2}{\left (4+x^2\right )^2 \left (e^5+\log \left (\frac {x}{4}\right )\right )} \, dx+\frac {1}{16} \left (e^5 \log (2)\right ) \int \frac {-4+x^2}{\left (4+x^2\right )^2 \left (e^5+\log \left (\frac {x}{4}\right )\right )} \, dx-\frac {1}{4} \log (16) \text {Subst}\left (\int \left (\frac {1}{16 \left (1+4 x^2\right )^2}-\frac {e^5}{16 \left (1+4 x^2\right )^2 \left (e^5+\log (x)\right )}\right ) \, dx,x,\frac {x}{4}\right ) \\ & = -\frac {e^x}{16}-\frac {x \log (2)}{32 \left (4+x^2\right )}+\frac {1}{16} \log \left (e^5+\log \left (\frac {x}{4}\right )\right )+\frac {1}{32} \log (2) \int \frac {1}{4+x^2} \, dx-\frac {1}{16} \left (e^5 \log (2)\right ) \int \frac {x^2}{\left (4+x^2\right )^2 \left (e^5+\log \left (\frac {x}{4}\right )\right )} \, dx+\frac {1}{16} \left (e^5 \log (2)\right ) \int \frac {-4+x^2}{\left (4+x^2\right )^2 \left (e^5+\log \left (\frac {x}{4}\right )\right )} \, dx-\frac {1}{64} \log (16) \text {Subst}\left (\int \frac {1}{\left (1+4 x^2\right )^2} \, dx,x,\frac {x}{4}\right )+\frac {1}{64} \left (e^5 \log (16)\right ) \text {Subst}\left (\int \frac {1}{\left (1+4 x^2\right )^2 \left (e^5+\log (x)\right )} \, dx,x,\frac {x}{4}\right ) \\ & = -\frac {e^x}{16}-\frac {x \log (2)}{32 \left (4+x^2\right )}+\frac {1}{64} \arctan \left (\frac {x}{2}\right ) \log (2)-\frac {x \log (16)}{128 \left (4+x^2\right )}+\frac {1}{16} \log \left (e^5+\log \left (\frac {x}{4}\right )\right )-\frac {1}{16} \left (e^5 \log (2)\right ) \int \frac {x^2}{\left (4+x^2\right )^2 \left (e^5+\log \left (\frac {x}{4}\right )\right )} \, dx+\frac {1}{16} \left (e^5 \log (2)\right ) \int \frac {-4+x^2}{\left (4+x^2\right )^2 \left (e^5+\log \left (\frac {x}{4}\right )\right )} \, dx-\frac {1}{128} \log (16) \text {Subst}\left (\int \frac {1}{1+4 x^2} \, dx,x,\frac {x}{4}\right )+\frac {1}{64} \left (e^5 \log (16)\right ) \text {Subst}\left (\int \frac {1}{\left (1+4 x^2\right )^2 \left (e^5+\log (x)\right )} \, dx,x,\frac {x}{4}\right ) \\ & = -\frac {e^x}{16}-\frac {x \log (2)}{32 \left (4+x^2\right )}+\frac {1}{64} \arctan \left (\frac {x}{2}\right ) \log (2)-\frac {x \log (16)}{128 \left (4+x^2\right )}-\frac {1}{256} \arctan \left (\frac {x}{2}\right ) \log (16)+\frac {1}{16} \log \left (e^5+\log \left (\frac {x}{4}\right )\right )-\frac {1}{16} \left (e^5 \log (2)\right ) \int \frac {x^2}{\left (4+x^2\right )^2 \left (e^5+\log \left (\frac {x}{4}\right )\right )} \, dx+\frac {1}{16} \left (e^5 \log (2)\right ) \int \frac {-4+x^2}{\left (4+x^2\right )^2 \left (e^5+\log \left (\frac {x}{4}\right )\right )} \, dx+\frac {1}{64} \left (e^5 \log (16)\right ) \text {Subst}\left (\int \frac {1}{\left (1+4 x^2\right )^2 \left (e^5+\log (x)\right )} \, dx,x,\frac {x}{4}\right ) \\ \end{align*}
Time = 11.17 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.03 \[ \int \frac {16+8 x^2+x^4+e^{5+x} \left (-16 x-8 x^3-x^5\right )+e^5 \left (-4 x+x^3\right ) \log (2)+\left (e^x \left (-16 x-8 x^3-x^5\right )+\left (-4 x+x^3\right ) \log (2)\right ) \log \left (\frac {x}{4}\right )}{e^5 \left (256 x+128 x^3+16 x^5\right )+\left (256 x+128 x^3+16 x^5\right ) \log \left (\frac {x}{4}\right )} \, dx=\frac {1}{16} \left (1-e^x-\frac {x \log (256)}{8 \left (4+x^2\right )}+\log \left (e^5+\log \left (\frac {x}{4}\right )\right )\right ) \]
[In]
[Out]
Time = 15.53 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.03
method | result | size |
risch | \(-\frac {{\mathrm e}^{x} x^{2}+x \ln \left (2\right )+4 \,{\mathrm e}^{x}}{16 \left (x^{2}+4\right )}+\frac {\ln \left ({\mathrm e}^{5}+\ln \left (\frac {x}{4}\right )\right )}{16}\) | \(36\) |
parallelrisch | \(-\frac {{\mathrm e}^{x} x^{2}-\ln \left ({\mathrm e}^{5}+\ln \left (\frac {x}{4}\right )\right ) x^{2}+x \ln \left (2\right )+4 \,{\mathrm e}^{x}-4 \ln \left ({\mathrm e}^{5}+\ln \left (\frac {x}{4}\right )\right )}{16 \left (x^{2}+4\right )}\) | \(48\) |
[In]
[Out]
none
Time = 0.26 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.29 \[ \int \frac {16+8 x^2+x^4+e^{5+x} \left (-16 x-8 x^3-x^5\right )+e^5 \left (-4 x+x^3\right ) \log (2)+\left (e^x \left (-16 x-8 x^3-x^5\right )+\left (-4 x+x^3\right ) \log (2)\right ) \log \left (\frac {x}{4}\right )}{e^5 \left (256 x+128 x^3+16 x^5\right )+\left (256 x+128 x^3+16 x^5\right ) \log \left (\frac {x}{4}\right )} \, dx=-\frac {{\left (x e^{5} \log \left (2\right ) - {\left (x^{2} + 4\right )} e^{5} \log \left (e^{5} + \log \left (\frac {1}{4} \, x\right )\right ) + {\left (x^{2} + 4\right )} e^{\left (x + 5\right )}\right )} e^{\left (-5\right )}}{16 \, {\left (x^{2} + 4\right )}} \]
[In]
[Out]
Time = 0.22 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.77 \[ \int \frac {16+8 x^2+x^4+e^{5+x} \left (-16 x-8 x^3-x^5\right )+e^5 \left (-4 x+x^3\right ) \log (2)+\left (e^x \left (-16 x-8 x^3-x^5\right )+\left (-4 x+x^3\right ) \log (2)\right ) \log \left (\frac {x}{4}\right )}{e^5 \left (256 x+128 x^3+16 x^5\right )+\left (256 x+128 x^3+16 x^5\right ) \log \left (\frac {x}{4}\right )} \, dx=- \frac {x \log {\left (2 \right )}}{16 x^{2} + 64} - \frac {e^{x}}{16} + \frac {\log {\left (\log {\left (\frac {x}{4} \right )} + e^{5} \right )}}{16} \]
[In]
[Out]
none
Time = 0.31 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00 \[ \int \frac {16+8 x^2+x^4+e^{5+x} \left (-16 x-8 x^3-x^5\right )+e^5 \left (-4 x+x^3\right ) \log (2)+\left (e^x \left (-16 x-8 x^3-x^5\right )+\left (-4 x+x^3\right ) \log (2)\right ) \log \left (\frac {x}{4}\right )}{e^5 \left (256 x+128 x^3+16 x^5\right )+\left (256 x+128 x^3+16 x^5\right ) \log \left (\frac {x}{4}\right )} \, dx=-\frac {{\left (x^{2} + 4\right )} e^{x} + x \log \left (2\right )}{16 \, {\left (x^{2} + 4\right )}} + \frac {1}{16} \, \log \left (e^{5} - 2 \, \log \left (2\right ) + \log \left (x\right )\right ) \]
[In]
[Out]
none
Time = 0.30 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.34 \[ \int \frac {16+8 x^2+x^4+e^{5+x} \left (-16 x-8 x^3-x^5\right )+e^5 \left (-4 x+x^3\right ) \log (2)+\left (e^x \left (-16 x-8 x^3-x^5\right )+\left (-4 x+x^3\right ) \log (2)\right ) \log \left (\frac {x}{4}\right )}{e^5 \left (256 x+128 x^3+16 x^5\right )+\left (256 x+128 x^3+16 x^5\right ) \log \left (\frac {x}{4}\right )} \, dx=-\frac {x^{2} e^{x} - x^{2} \log \left (e^{5} + \log \left (\frac {1}{4} \, x\right )\right ) + x \log \left (2\right ) + 4 \, e^{x} - 4 \, \log \left (e^{5} + \log \left (\frac {1}{4} \, x\right )\right )}{16 \, {\left (x^{2} + 4\right )}} \]
[In]
[Out]
Time = 13.65 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.83 \[ \int \frac {16+8 x^2+x^4+e^{5+x} \left (-16 x-8 x^3-x^5\right )+e^5 \left (-4 x+x^3\right ) \log (2)+\left (e^x \left (-16 x-8 x^3-x^5\right )+\left (-4 x+x^3\right ) \log (2)\right ) \log \left (\frac {x}{4}\right )}{e^5 \left (256 x+128 x^3+16 x^5\right )+\left (256 x+128 x^3+16 x^5\right ) \log \left (\frac {x}{4}\right )} \, dx=\frac {\ln \left (\ln \left (\frac {x}{4}\right )+{\mathrm {e}}^5\right )}{16}-\frac {{\mathrm {e}}^x}{16}-\frac {x\,\ln \left (2\right )}{16\,x^2+64} \]
[In]
[Out]