3.81.41 \(\int \frac {e^{\frac {-5+x}{x}} (2 x^2-32 x^3-8 x^5)+(16 x^2-x^3+8 x^4+x^6+e^{\frac {2 (-5+x)}{x}} (-160+10 x-80 x^2-10 x^4)) \log (-16+x-8 x^2-x^4) \log (\frac {1}{4} \log (-16+x-8 x^2-x^4))+(2 x^2-32 x^3-8 x^5+e^{\frac {-5+x}{x}} (-160+10 x-80 x^2-10 x^4) \log (-16+x-8 x^2-x^4) \log (\frac {1}{4} \log (-16+x-8 x^2-x^4))) \log (\log (\frac {1}{4} \log (-16+x-8 x^2-x^4)))}{(16 x^2-x^3+8 x^4+x^6) \log (-16+x-8 x^2-x^4) \log (\frac {1}{4} \log (-16+x-8 x^2-x^4))} \, dx\)

Optimal. Leaf size=35 \[ -1+x-\left (e^{1-\frac {5}{x}}+\log \left (\log \left (\frac {1}{4} \log \left (x-\left (4+x^2\right )^2\right )\right )\right )\right )^2 \]

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Rubi [F]  time = 8.16, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {e^{\frac {-5+x}{x}} \left (2 x^2-32 x^3-8 x^5\right )+\left (16 x^2-x^3+8 x^4+x^6+e^{\frac {2 (-5+x)}{x}} \left (-160+10 x-80 x^2-10 x^4\right )\right ) \log \left (-16+x-8 x^2-x^4\right ) \log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right )+\left (2 x^2-32 x^3-8 x^5+e^{\frac {-5+x}{x}} \left (-160+10 x-80 x^2-10 x^4\right ) \log \left (-16+x-8 x^2-x^4\right ) \log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right )\right ) \log \left (\log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right )\right )}{\left (16 x^2-x^3+8 x^4+x^6\right ) \log \left (-16+x-8 x^2-x^4\right ) \log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((-5 + x)/x)*(2*x^2 - 32*x^3 - 8*x^5) + (16*x^2 - x^3 + 8*x^4 + x^6 + E^((2*(-5 + x))/x)*(-160 + 10*x -
80*x^2 - 10*x^4))*Log[-16 + x - 8*x^2 - x^4]*Log[Log[-16 + x - 8*x^2 - x^4]/4] + (2*x^2 - 32*x^3 - 8*x^5 + E^
((-5 + x)/x)*(-160 + 10*x - 80*x^2 - 10*x^4)*Log[-16 + x - 8*x^2 - x^4]*Log[Log[-16 + x - 8*x^2 - x^4]/4])*Log
[Log[Log[-16 + x - 8*x^2 - x^4]/4]])/((16*x^2 - x^3 + 8*x^4 + x^6)*Log[-16 + x - 8*x^2 - x^4]*Log[Log[-16 + x
- 8*x^2 - x^4]/4]),x]

[Out]

-E^(2 - 10/x) + x - (2*E^(1 - 5/x)*(16*Log[-16 + x - 8*x^2 - x^4]*Log[Log[-16 + x - 8*x^2 - x^4]/4]*Log[Log[Lo
g[-16 + x - 8*x^2 - x^4]/4]] - x*Log[-16 + x - 8*x^2 - x^4]*Log[Log[-16 + x - 8*x^2 - x^4]/4]*Log[Log[Log[-16
+ x - 8*x^2 - x^4]/4]] + 8*x^2*Log[-16 + x - 8*x^2 - x^4]*Log[Log[-16 + x - 8*x^2 - x^4]/4]*Log[Log[Log[-16 +
x - 8*x^2 - x^4]/4]] + x^4*Log[-16 + x - 8*x^2 - x^4]*Log[Log[-16 + x - 8*x^2 - x^4]/4]*Log[Log[Log[-16 + x -
8*x^2 - x^4]/4]]))/((16 - x + 8*x^2 + x^4)*Log[-16 + x - 8*x^2 - x^4]*Log[Log[-16 + x - 8*x^2 - x^4]/4]) + 2*D
efer[Int][Log[Log[Log[-16 + x - 8*x^2 - x^4]/4]]/((16 - x + 8*x^2 + x^4)*Log[-16 + x - 8*x^2 - x^4]*Log[Log[-1
6 + x - 8*x^2 - x^4]/4]), x] - 32*Defer[Int][(x*Log[Log[Log[-16 + x - 8*x^2 - x^4]/4]])/((16 - x + 8*x^2 + x^4
)*Log[-16 + x - 8*x^2 - x^4]*Log[Log[-16 + x - 8*x^2 - x^4]/4]), x] - 8*Defer[Int][(x^3*Log[Log[Log[-16 + x -
8*x^2 - x^4]/4]])/((16 - x + 8*x^2 + x^4)*Log[-16 + x - 8*x^2 - x^4]*Log[Log[-16 + x - 8*x^2 - x^4]/4]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{\frac {-5+x}{x}} \left (2 x^2-32 x^3-8 x^5\right )+\left (16 x^2-x^3+8 x^4+x^6+e^{\frac {2 (-5+x)}{x}} \left (-160+10 x-80 x^2-10 x^4\right )\right ) \log \left (-16+x-8 x^2-x^4\right ) \log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right )+\left (2 x^2-32 x^3-8 x^5+e^{\frac {-5+x}{x}} \left (-160+10 x-80 x^2-10 x^4\right ) \log \left (-16+x-8 x^2-x^4\right ) \log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right )\right ) \log \left (\log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right )\right )}{x^2 \left (16-x+8 x^2+x^4\right ) \log \left (-16+x-8 x^2-x^4\right ) \log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right )} \, dx\\ &=\int \left (-\frac {10 e^{2-\frac {10}{x}}}{x^2}+\frac {16}{16-x+8 x^2+x^4}-\frac {x}{16-x+8 x^2+x^4}+\frac {8 x^2}{16-x+8 x^2+x^4}+\frac {x^4}{16-x+8 x^2+x^4}+\frac {2 \log \left (\log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right )\right )}{\left (16-x+8 x^2+x^4\right ) \log \left (-16+x-8 x^2-x^4\right ) \log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right )}-\frac {32 x \log \left (\log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right )\right )}{\left (16-x+8 x^2+x^4\right ) \log \left (-16+x-8 x^2-x^4\right ) \log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right )}-\frac {8 x^3 \log \left (\log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right )\right )}{\left (16-x+8 x^2+x^4\right ) \log \left (-16+x-8 x^2-x^4\right ) \log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right )}-\frac {2 e^{1-\frac {5}{x}} \left (-x^2+16 x^3+4 x^5+80 \log \left (-16+x-8 x^2-x^4\right ) \log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right ) \log \left (\log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right )\right )-5 x \log \left (-16+x-8 x^2-x^4\right ) \log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right ) \log \left (\log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right )\right )+40 x^2 \log \left (-16+x-8 x^2-x^4\right ) \log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right ) \log \left (\log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right )\right )+5 x^4 \log \left (-16+x-8 x^2-x^4\right ) \log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right ) \log \left (\log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right )\right )\right )}{x^2 \left (16-x+8 x^2+x^4\right ) \log \left (-16+x-8 x^2-x^4\right ) \log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right )}\right ) \, dx\\ &=2 \int \frac {\log \left (\log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right )\right )}{\left (16-x+8 x^2+x^4\right ) \log \left (-16+x-8 x^2-x^4\right ) \log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right )} \, dx-2 \int \frac {e^{1-\frac {5}{x}} \left (-x^2+16 x^3+4 x^5+80 \log \left (-16+x-8 x^2-x^4\right ) \log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right ) \log \left (\log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right )\right )-5 x \log \left (-16+x-8 x^2-x^4\right ) \log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right ) \log \left (\log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right )\right )+40 x^2 \log \left (-16+x-8 x^2-x^4\right ) \log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right ) \log \left (\log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right )\right )+5 x^4 \log \left (-16+x-8 x^2-x^4\right ) \log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right ) \log \left (\log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right )\right )\right )}{x^2 \left (16-x+8 x^2+x^4\right ) \log \left (-16+x-8 x^2-x^4\right ) \log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right )} \, dx+8 \int \frac {x^2}{16-x+8 x^2+x^4} \, dx-8 \int \frac {x^3 \log \left (\log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right )\right )}{\left (16-x+8 x^2+x^4\right ) \log \left (-16+x-8 x^2-x^4\right ) \log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right )} \, dx-10 \int \frac {e^{2-\frac {10}{x}}}{x^2} \, dx+16 \int \frac {1}{16-x+8 x^2+x^4} \, dx-32 \int \frac {x \log \left (\log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right )\right )}{\left (16-x+8 x^2+x^4\right ) \log \left (-16+x-8 x^2-x^4\right ) \log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right )} \, dx-\int \frac {x}{16-x+8 x^2+x^4} \, dx+\int \frac {x^4}{16-x+8 x^2+x^4} \, dx\\ &=-e^{2-\frac {10}{x}}-\frac {2 e^{1-\frac {5}{x}} \left (16 \log \left (-16+x-8 x^2-x^4\right ) \log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right ) \log \left (\log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right )\right )-x \log \left (-16+x-8 x^2-x^4\right ) \log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right ) \log \left (\log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right )\right )+8 x^2 \log \left (-16+x-8 x^2-x^4\right ) \log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right ) \log \left (\log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right )\right )+x^4 \log \left (-16+x-8 x^2-x^4\right ) \log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right ) \log \left (\log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right )\right )\right )}{\left (16-x+8 x^2+x^4\right ) \log \left (-16+x-8 x^2-x^4\right ) \log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right )}+2 \int \frac {\log \left (\log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right )\right )}{\left (16-x+8 x^2+x^4\right ) \log \left (-16+x-8 x^2-x^4\right ) \log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right )} \, dx+8 \int \frac {x^2}{16-x+8 x^2+x^4} \, dx-8 \int \frac {x^3 \log \left (\log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right )\right )}{\left (16-x+8 x^2+x^4\right ) \log \left (-16+x-8 x^2-x^4\right ) \log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right )} \, dx+16 \int \frac {1}{16-x+8 x^2+x^4} \, dx-32 \int \frac {x \log \left (\log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right )\right )}{\left (16-x+8 x^2+x^4\right ) \log \left (-16+x-8 x^2-x^4\right ) \log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right )} \, dx-\int \frac {x}{16-x+8 x^2+x^4} \, dx+\int \left (1-\frac {16-x+8 x^2}{16-x+8 x^2+x^4}\right ) \, dx\\ &=-e^{2-\frac {10}{x}}+x-\frac {2 e^{1-\frac {5}{x}} \left (16 \log \left (-16+x-8 x^2-x^4\right ) \log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right ) \log \left (\log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right )\right )-x \log \left (-16+x-8 x^2-x^4\right ) \log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right ) \log \left (\log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right )\right )+8 x^2 \log \left (-16+x-8 x^2-x^4\right ) \log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right ) \log \left (\log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right )\right )+x^4 \log \left (-16+x-8 x^2-x^4\right ) \log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right ) \log \left (\log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right )\right )\right )}{\left (16-x+8 x^2+x^4\right ) \log \left (-16+x-8 x^2-x^4\right ) \log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right )}+2 \int \frac {\log \left (\log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right )\right )}{\left (16-x+8 x^2+x^4\right ) \log \left (-16+x-8 x^2-x^4\right ) \log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right )} \, dx+8 \int \frac {x^2}{16-x+8 x^2+x^4} \, dx-8 \int \frac {x^3 \log \left (\log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right )\right )}{\left (16-x+8 x^2+x^4\right ) \log \left (-16+x-8 x^2-x^4\right ) \log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right )} \, dx+16 \int \frac {1}{16-x+8 x^2+x^4} \, dx-32 \int \frac {x \log \left (\log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right )\right )}{\left (16-x+8 x^2+x^4\right ) \log \left (-16+x-8 x^2-x^4\right ) \log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right )} \, dx-\int \frac {x}{16-x+8 x^2+x^4} \, dx-\int \frac {16-x+8 x^2}{16-x+8 x^2+x^4} \, dx\\ &=-e^{2-\frac {10}{x}}+x-\frac {2 e^{1-\frac {5}{x}} \left (16 \log \left (-16+x-8 x^2-x^4\right ) \log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right ) \log \left (\log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right )\right )-x \log \left (-16+x-8 x^2-x^4\right ) \log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right ) \log \left (\log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right )\right )+8 x^2 \log \left (-16+x-8 x^2-x^4\right ) \log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right ) \log \left (\log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right )\right )+x^4 \log \left (-16+x-8 x^2-x^4\right ) \log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right ) \log \left (\log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right )\right )\right )}{\left (16-x+8 x^2+x^4\right ) \log \left (-16+x-8 x^2-x^4\right ) \log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right )}+2 \int \frac {\log \left (\log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right )\right )}{\left (16-x+8 x^2+x^4\right ) \log \left (-16+x-8 x^2-x^4\right ) \log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right )} \, dx+8 \int \frac {x^2}{16-x+8 x^2+x^4} \, dx-8 \int \frac {x^3 \log \left (\log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right )\right )}{\left (16-x+8 x^2+x^4\right ) \log \left (-16+x-8 x^2-x^4\right ) \log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right )} \, dx+16 \int \frac {1}{16-x+8 x^2+x^4} \, dx-32 \int \frac {x \log \left (\log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right )\right )}{\left (16-x+8 x^2+x^4\right ) \log \left (-16+x-8 x^2-x^4\right ) \log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right )} \, dx-\int \frac {x}{16-x+8 x^2+x^4} \, dx-\int \left (\frac {16}{16-x+8 x^2+x^4}-\frac {x}{16-x+8 x^2+x^4}+\frac {8 x^2}{16-x+8 x^2+x^4}\right ) \, dx\\ &=-e^{2-\frac {10}{x}}+x-\frac {2 e^{1-\frac {5}{x}} \left (16 \log \left (-16+x-8 x^2-x^4\right ) \log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right ) \log \left (\log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right )\right )-x \log \left (-16+x-8 x^2-x^4\right ) \log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right ) \log \left (\log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right )\right )+8 x^2 \log \left (-16+x-8 x^2-x^4\right ) \log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right ) \log \left (\log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right )\right )+x^4 \log \left (-16+x-8 x^2-x^4\right ) \log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right ) \log \left (\log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right )\right )\right )}{\left (16-x+8 x^2+x^4\right ) \log \left (-16+x-8 x^2-x^4\right ) \log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right )}+2 \int \frac {\log \left (\log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right )\right )}{\left (16-x+8 x^2+x^4\right ) \log \left (-16+x-8 x^2-x^4\right ) \log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right )} \, dx-8 \int \frac {x^3 \log \left (\log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right )\right )}{\left (16-x+8 x^2+x^4\right ) \log \left (-16+x-8 x^2-x^4\right ) \log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right )} \, dx-32 \int \frac {x \log \left (\log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right )\right )}{\left (16-x+8 x^2+x^4\right ) \log \left (-16+x-8 x^2-x^4\right ) \log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.24, size = 68, normalized size = 1.94 \begin {gather*} -e^{2-\frac {10}{x}}+x-2 e^{1-\frac {5}{x}} \log \left (\log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right )\right )-\log ^2\left (\log \left (\frac {1}{4} \log \left (-16+x-8 x^2-x^4\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((-5 + x)/x)*(2*x^2 - 32*x^3 - 8*x^5) + (16*x^2 - x^3 + 8*x^4 + x^6 + E^((2*(-5 + x))/x)*(-160 +
10*x - 80*x^2 - 10*x^4))*Log[-16 + x - 8*x^2 - x^4]*Log[Log[-16 + x - 8*x^2 - x^4]/4] + (2*x^2 - 32*x^3 - 8*x^
5 + E^((-5 + x)/x)*(-160 + 10*x - 80*x^2 - 10*x^4)*Log[-16 + x - 8*x^2 - x^4]*Log[Log[-16 + x - 8*x^2 - x^4]/4
])*Log[Log[Log[-16 + x - 8*x^2 - x^4]/4]])/((16*x^2 - x^3 + 8*x^4 + x^6)*Log[-16 + x - 8*x^2 - x^4]*Log[Log[-1
6 + x - 8*x^2 - x^4]/4]),x]

[Out]

-E^(2 - 10/x) + x - 2*E^(1 - 5/x)*Log[Log[Log[-16 + x - 8*x^2 - x^4]/4]] - Log[Log[Log[-16 + x - 8*x^2 - x^4]/
4]]^2

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fricas [A]  time = 2.53, size = 63, normalized size = 1.80 \begin {gather*} -2 \, e^{\left (\frac {x - 5}{x}\right )} \log \left (\log \left (\frac {1}{4} \, \log \left (-x^{4} - 8 \, x^{2} + x - 16\right )\right )\right ) - \log \left (\log \left (\frac {1}{4} \, \log \left (-x^{4} - 8 \, x^{2} + x - 16\right )\right )\right )^{2} + x - e^{\left (\frac {2 \, {\left (x - 5\right )}}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-10*x^4-80*x^2+10*x-160)*exp((x-5)/x)*log(-x^4-8*x^2+x-16)*log(1/4*log(-x^4-8*x^2+x-16))-8*x^5-32
*x^3+2*x^2)*log(log(1/4*log(-x^4-8*x^2+x-16)))+((-10*x^4-80*x^2+10*x-160)*exp((x-5)/x)^2+x^6+8*x^4-x^3+16*x^2)
*log(-x^4-8*x^2+x-16)*log(1/4*log(-x^4-8*x^2+x-16))+(-8*x^5-32*x^3+2*x^2)*exp((x-5)/x))/(x^6+8*x^4-x^3+16*x^2)
/log(-x^4-8*x^2+x-16)/log(1/4*log(-x^4-8*x^2+x-16)),x, algorithm="fricas")

[Out]

-2*e^((x - 5)/x)*log(log(1/4*log(-x^4 - 8*x^2 + x - 16))) - log(log(1/4*log(-x^4 - 8*x^2 + x - 16)))^2 + x - e
^(2*(x - 5)/x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (x^{6} + 8 \, x^{4} - x^{3} + 16 \, x^{2} - 10 \, {\left (x^{4} + 8 \, x^{2} - x + 16\right )} e^{\left (\frac {2 \, {\left (x - 5\right )}}{x}\right )}\right )} \log \left (-x^{4} - 8 \, x^{2} + x - 16\right ) \log \left (\frac {1}{4} \, \log \left (-x^{4} - 8 \, x^{2} + x - 16\right )\right ) - 2 \, {\left (4 \, x^{5} + 16 \, x^{3} - x^{2}\right )} e^{\left (\frac {x - 5}{x}\right )} - 2 \, {\left (4 \, x^{5} + 5 \, {\left (x^{4} + 8 \, x^{2} - x + 16\right )} e^{\left (\frac {x - 5}{x}\right )} \log \left (-x^{4} - 8 \, x^{2} + x - 16\right ) \log \left (\frac {1}{4} \, \log \left (-x^{4} - 8 \, x^{2} + x - 16\right )\right ) + 16 \, x^{3} - x^{2}\right )} \log \left (\log \left (\frac {1}{4} \, \log \left (-x^{4} - 8 \, x^{2} + x - 16\right )\right )\right )}{{\left (x^{6} + 8 \, x^{4} - x^{3} + 16 \, x^{2}\right )} \log \left (-x^{4} - 8 \, x^{2} + x - 16\right ) \log \left (\frac {1}{4} \, \log \left (-x^{4} - 8 \, x^{2} + x - 16\right )\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-10*x^4-80*x^2+10*x-160)*exp((x-5)/x)*log(-x^4-8*x^2+x-16)*log(1/4*log(-x^4-8*x^2+x-16))-8*x^5-32
*x^3+2*x^2)*log(log(1/4*log(-x^4-8*x^2+x-16)))+((-10*x^4-80*x^2+10*x-160)*exp((x-5)/x)^2+x^6+8*x^4-x^3+16*x^2)
*log(-x^4-8*x^2+x-16)*log(1/4*log(-x^4-8*x^2+x-16))+(-8*x^5-32*x^3+2*x^2)*exp((x-5)/x))/(x^6+8*x^4-x^3+16*x^2)
/log(-x^4-8*x^2+x-16)/log(1/4*log(-x^4-8*x^2+x-16)),x, algorithm="giac")

[Out]

integrate(((x^6 + 8*x^4 - x^3 + 16*x^2 - 10*(x^4 + 8*x^2 - x + 16)*e^(2*(x - 5)/x))*log(-x^4 - 8*x^2 + x - 16)
*log(1/4*log(-x^4 - 8*x^2 + x - 16)) - 2*(4*x^5 + 16*x^3 - x^2)*e^((x - 5)/x) - 2*(4*x^5 + 5*(x^4 + 8*x^2 - x
+ 16)*e^((x - 5)/x)*log(-x^4 - 8*x^2 + x - 16)*log(1/4*log(-x^4 - 8*x^2 + x - 16)) + 16*x^3 - x^2)*log(log(1/4
*log(-x^4 - 8*x^2 + x - 16))))/((x^6 + 8*x^4 - x^3 + 16*x^2)*log(-x^4 - 8*x^2 + x - 16)*log(1/4*log(-x^4 - 8*x
^2 + x - 16))), x)

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maple [A]  time = 0.12, size = 64, normalized size = 1.83

method	result	size
risch	\(-{\mathrm e}^{\frac {2 x -10}{x}}-2 \,{\mathrm e}^{\frac {x -5}{x}} \ln \left (\ln \left (\frac {\ln \left (-x^{4}-8 x^{2}+x -16\right )}{4}\right )\right )-\ln \left (\ln \left (\frac {\ln \left (-x^{4}-8 x^{2}+x -16\right )}{4}\right )\right )^{2}+x\)	\(64\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-10*x^4-80*x^2+10*x-160)*exp((x-5)/x)*ln(-x^4-8*x^2+x-16)*ln(1/4*ln(-x^4-8*x^2+x-16))-8*x^5-32*x^3+2*x^
2)*ln(ln(1/4*ln(-x^4-8*x^2+x-16)))+((-10*x^4-80*x^2+10*x-160)*exp((x-5)/x)^2+x^6+8*x^4-x^3+16*x^2)*ln(-x^4-8*x
^2+x-16)*ln(1/4*ln(-x^4-8*x^2+x-16))+(-8*x^5-32*x^3+2*x^2)*exp((x-5)/x))/(x^6+8*x^4-x^3+16*x^2)/ln(-x^4-8*x^2+
x-16)/ln(1/4*ln(-x^4-8*x^2+x-16)),x,method=_RETURNVERBOSE)

[Out]

-exp(2*(x-5)/x)-2*exp((x-5)/x)*ln(ln(1/4*ln(-x^4-8*x^2+x-16)))-ln(ln(1/4*ln(-x^4-8*x^2+x-16)))^2+x

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maxima [B]  time = 0.54, size = 81, normalized size = 2.31 \begin {gather*} -{\left (e^{\frac {10}{x}} \log \left (-2 \, \log \relax (2) + \log \left (\log \left (-x^{4} - 8 \, x^{2} + x - 16\right )\right )\right )^{2} - x e^{\frac {10}{x}} + 2 \, e^{\left (\frac {5}{x} + 1\right )} \log \left (-2 \, \log \relax (2) + \log \left (\log \left (-x^{4} - 8 \, x^{2} + x - 16\right )\right )\right ) + e^{2}\right )} e^{\left (-\frac {10}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-10*x^4-80*x^2+10*x-160)*exp((x-5)/x)*log(-x^4-8*x^2+x-16)*log(1/4*log(-x^4-8*x^2+x-16))-8*x^5-32
*x^3+2*x^2)*log(log(1/4*log(-x^4-8*x^2+x-16)))+((-10*x^4-80*x^2+10*x-160)*exp((x-5)/x)^2+x^6+8*x^4-x^3+16*x^2)
*log(-x^4-8*x^2+x-16)*log(1/4*log(-x^4-8*x^2+x-16))+(-8*x^5-32*x^3+2*x^2)*exp((x-5)/x))/(x^6+8*x^4-x^3+16*x^2)
/log(-x^4-8*x^2+x-16)/log(1/4*log(-x^4-8*x^2+x-16)),x, algorithm="maxima")

[Out]

-(e^(10/x)*log(-2*log(2) + log(log(-x^4 - 8*x^2 + x - 16)))^2 - x*e^(10/x) + 2*e^(5/x + 1)*log(-2*log(2) + log
(log(-x^4 - 8*x^2 + x - 16))) + e^2)*e^(-10/x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int -\frac {{\mathrm {e}}^{\frac {x-5}{x}}\,\left (8\,x^5+32\,x^3-2\,x^2\right )+\ln \left (\ln \left (\frac {\ln \left (-x^4-8\,x^2+x-16\right )}{4}\right )\right )\,\left (32\,x^3-2\,x^2+8\,x^5+{\mathrm {e}}^{\frac {x-5}{x}}\,\ln \left (\frac {\ln \left (-x^4-8\,x^2+x-16\right )}{4}\right )\,\ln \left (-x^4-8\,x^2+x-16\right )\,\left (10\,x^4+80\,x^2-10\,x+160\right )\right )-\ln \left (\frac {\ln \left (-x^4-8\,x^2+x-16\right )}{4}\right )\,\ln \left (-x^4-8\,x^2+x-16\right )\,\left (16\,x^2-{\mathrm {e}}^{\frac {2\,\left (x-5\right )}{x}}\,\left (10\,x^4+80\,x^2-10\,x+160\right )-x^3+8\,x^4+x^6\right )}{\ln \left (\frac {\ln \left (-x^4-8\,x^2+x-16\right )}{4}\right )\,\ln \left (-x^4-8\,x^2+x-16\right )\,\left (x^6+8\,x^4-x^3+16\,x^2\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((x - 5)/x)*(32*x^3 - 2*x^2 + 8*x^5) + log(log(log(x - 8*x^2 - x^4 - 16)/4))*(32*x^3 - 2*x^2 + 8*x^5
+ exp((x - 5)/x)*log(log(x - 8*x^2 - x^4 - 16)/4)*log(x - 8*x^2 - x^4 - 16)*(80*x^2 - 10*x + 10*x^4 + 160)) -
log(log(x - 8*x^2 - x^4 - 16)/4)*log(x - 8*x^2 - x^4 - 16)*(16*x^2 - exp((2*(x - 5))/x)*(80*x^2 - 10*x + 10*x^
4 + 160) - x^3 + 8*x^4 + x^6))/(log(log(x - 8*x^2 - x^4 - 16)/4)*log(x - 8*x^2 - x^4 - 16)*(16*x^2 - x^3 + 8*x
^4 + x^6)),x)

[Out]

int(-(exp((x - 5)/x)*(32*x^3 - 2*x^2 + 8*x^5) + log(log(log(x - 8*x^2 - x^4 - 16)/4))*(32*x^3 - 2*x^2 + 8*x^5
+ exp((x - 5)/x)*log(log(x - 8*x^2 - x^4 - 16)/4)*log(x - 8*x^2 - x^4 - 16)*(80*x^2 - 10*x + 10*x^4 + 160)) -
log(log(x - 8*x^2 - x^4 - 16)/4)*log(x - 8*x^2 - x^4 - 16)*(16*x^2 - exp((2*(x - 5))/x)*(80*x^2 - 10*x + 10*x^
4 + 160) - x^3 + 8*x^4 + x^6))/(log(log(x - 8*x^2 - x^4 - 16)/4)*log(x - 8*x^2 - x^4 - 16)*(16*x^2 - x^3 + 8*x
^4 + x^6)), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-10*x**4-80*x**2+10*x-160)*exp((x-5)/x)*ln(-x**4-8*x**2+x-16)*ln(1/4*ln(-x**4-8*x**2+x-16))-8*x**
5-32*x**3+2*x**2)*ln(ln(1/4*ln(-x**4-8*x**2+x-16)))+((-10*x**4-80*x**2+10*x-160)*exp((x-5)/x)**2+x**6+8*x**4-x
**3+16*x**2)*ln(-x**4-8*x**2+x-16)*ln(1/4*ln(-x**4-8*x**2+x-16))+(-8*x**5-32*x**3+2*x**2)*exp((x-5)/x))/(x**6+
8*x**4-x**3+16*x**2)/ln(-x**4-8*x**2+x-16)/ln(1/4*ln(-x**4-8*x**2+x-16)),x)

[Out]

Timed out

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