Integrand size = 91, antiderivative size = 28 \[ \int \frac {e^{-x+\frac {1+e^{4 x} (-2-2 x)+x+e^{8 x} (1+x)}{x^3 \log ^2(5)}} \left (-9-6 x+e^{4 x} \left (18-12 x-24 x^2\right )+e^{8 x} \left (-9+18 x+24 x^2\right )-3 x^4 \log ^2(5)\right )}{x^4 \log ^2(5)} \, dx=3 e^{-x+\frac {\left (-1+e^{4 x}\right )^2 (1+x)}{x^3 \log ^2(5)}} \] Output:
3*exp((exp(4*x)-1)^2/x^3/ln(5)^2*(1+x))/exp(x)
Time = 0.12 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.54 \[ \int \frac {e^{-x+\frac {1+e^{4 x} (-2-2 x)+x+e^{8 x} (1+x)}{x^3 \log ^2(5)}} \left (-9-6 x+e^{4 x} \left (18-12 x-24 x^2\right )+e^{8 x} \left (-9+18 x+24 x^2\right )-3 x^4 \log ^2(5)\right )}{x^4 \log ^2(5)} \, dx=3 e^{\frac {1+x-2 e^{4 x} (1+x)+e^{8 x} (1+x)-x^4 \log ^2(5)}{x^3 \log ^2(5)}} \] Input:
Integrate[(E^(-x + (1 + E^(4*x)*(-2 - 2*x) + x + E^(8*x)*(1 + x))/(x^3*Log [5]^2))*(-9 - 6*x + E^(4*x)*(18 - 12*x - 24*x^2) + E^(8*x)*(-9 + 18*x + 24 *x^2) - 3*x^4*Log[5]^2))/(x^4*Log[5]^2),x]
Output:
3*E^((1 + x - 2*E^(4*x)*(1 + x) + E^(8*x)*(1 + x) - x^4*Log[5]^2)/(x^3*Log [5]^2))
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {\left (-3 x^4 \log ^2(5)+e^{4 x} \left (-24 x^2-12 x+18\right )+e^{8 x} \left (24 x^2+18 x-9\right )-6 x-9\right ) \exp \left (\frac {e^{4 x} (-2 x-2)+x+e^{8 x} (x+1)+1}{x^3 \log ^2(5)}-x\right )}{x^4 \log ^2(5)} \, dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int -\frac {3 \exp \left (\frac {x-2 e^{4 x} (x+1)+e^{8 x} (x+1)+1}{x^3 \log ^2(5)}-x\right ) \left (\log ^2(5) x^4+2 x+e^{8 x} \left (-8 x^2-6 x+3\right )-2 e^{4 x} \left (-4 x^2-2 x+3\right )+3\right )}{x^4}dx}{\log ^2(5)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {3 \int \frac {\exp \left (\frac {x-2 e^{4 x} (x+1)+e^{8 x} (x+1)+1}{x^3 \log ^2(5)}-x\right ) \left (\log ^2(5) x^4+2 x+e^{8 x} \left (-8 x^2-6 x+3\right )-2 e^{4 x} \left (-4 x^2-2 x+3\right )+3\right )}{x^4}dx}{\log ^2(5)}\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle -\frac {3 \int \left (\frac {2 \exp \left (3 x+\frac {x-2 e^{4 x} (x+1)+e^{8 x} (x+1)+1}{\log ^2(5) x^3}\right ) \left (4 x^2+2 x-3\right )}{x^4}-\frac {\exp \left (7 x+\frac {x-2 e^{4 x} (x+1)+e^{8 x} (x+1)+1}{\log ^2(5) x^3}\right ) \left (8 x^2+6 x-3\right )}{x^4}+\frac {\exp \left (\frac {x-2 e^{4 x} (x+1)+e^{8 x} (x+1)+1}{x^3 \log ^2(5)}-x\right ) \left (\log ^2(5) x^4+2 x+3\right )}{x^4}\right )dx}{\log ^2(5)}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {3 \left (\log ^2(5) \int \exp \left (\frac {x-2 e^{4 x} (x+1)+e^{8 x} (x+1)+1}{x^3 \log ^2(5)}-x\right )dx+2 \int \frac {\exp \left (\frac {x-2 e^{4 x} (x+1)+e^{8 x} (x+1)+1}{x^3 \log ^2(5)}-x\right )}{x^3}dx+4 \int \frac {\exp \left (3 x+\frac {x-2 e^{4 x} (x+1)+e^{8 x} (x+1)+1}{\log ^2(5) x^3}\right )}{x^3}dx-6 \int \frac {\exp \left (7 x+\frac {x-2 e^{4 x} (x+1)+e^{8 x} (x+1)+1}{\log ^2(5) x^3}\right )}{x^3}dx+3 \int \frac {\exp \left (\frac {x-2 e^{4 x} (x+1)+e^{8 x} (x+1)+1}{x^3 \log ^2(5)}-x\right )}{x^4}dx-6 \int \frac {\exp \left (3 x+\frac {x-2 e^{4 x} (x+1)+e^{8 x} (x+1)+1}{\log ^2(5) x^3}\right )}{x^4}dx+3 \int \frac {\exp \left (7 x+\frac {x-2 e^{4 x} (x+1)+e^{8 x} (x+1)+1}{\log ^2(5) x^3}\right )}{x^4}dx+8 \int \frac {\exp \left (3 x+\frac {x-2 e^{4 x} (x+1)+e^{8 x} (x+1)+1}{\log ^2(5) x^3}\right )}{x^2}dx-8 \int \frac {\exp \left (7 x+\frac {x-2 e^{4 x} (x+1)+e^{8 x} (x+1)+1}{\log ^2(5) x^3}\right )}{x^2}dx\right )}{\log ^2(5)}\) |
Input:
Int[(E^(-x + (1 + E^(4*x)*(-2 - 2*x) + x + E^(8*x)*(1 + x))/(x^3*Log[5]^2) )*(-9 - 6*x + E^(4*x)*(18 - 12*x - 24*x^2) + E^(8*x)*(-9 + 18*x + 24*x^2) - 3*x^4*Log[5]^2))/(x^4*Log[5]^2),x]
Output:
$Aborted
Time = 20.22 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.39
| method | result | size |
| parallelrisch | \(3 \,{\mathrm e}^{\frac {\left (1+x \right ) {\mathrm e}^{8 x}+\left (-2-2 x \right ) {\mathrm e}^{4 x}+x +1}{x^{3} \ln \left (5\right )^{2}}} {\mathrm e}^{-x}\) | \(39\) |
| risch | \(3 \,{\mathrm e}^{-\frac {x^{4} \ln \left (5\right )^{2}+2 x \,{\mathrm e}^{4 x}-x \,{\mathrm e}^{8 x}+2 \,{\mathrm e}^{4 x}-{\mathrm e}^{8 x}-x -1}{x^{3} \ln \left (5\right )^{2}}}\) | \(52\) |
Input:
int(((24*x^2+18*x-9)*exp(4*x)^2+(-24*x^2-12*x+18)*exp(4*x)-3*x^4*ln(5)^2-6 *x-9)*exp(((1+x)*exp(4*x)^2+(-2-2*x)*exp(4*x)+x+1)/x^3/ln(5)^2)/x^4/ln(5)^ 2/exp(x),x,method=_RETURNVERBOSE)
Output:
3*exp(((1+x)*exp(4*x)^2+(-2-2*x)*exp(4*x)+x+1)/x^3/ln(5)^2)/exp(x)
Time = 0.09 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.54 \[ \int \frac {e^{-x+\frac {1+e^{4 x} (-2-2 x)+x+e^{8 x} (1+x)}{x^3 \log ^2(5)}} \left (-9-6 x+e^{4 x} \left (18-12 x-24 x^2\right )+e^{8 x} \left (-9+18 x+24 x^2\right )-3 x^4 \log ^2(5)\right )}{x^4 \log ^2(5)} \, dx=3 \, e^{\left (-\frac {x^{4} \log \left (5\right )^{2} - {\left (x + 1\right )} e^{\left (8 \, x\right )} + 2 \, {\left (x + 1\right )} e^{\left (4 \, x\right )} - x - 1}{x^{3} \log \left (5\right )^{2}}\right )} \] Input:
integrate(((24*x^2+18*x-9)*exp(4*x)^2+(-24*x^2-12*x+18)*exp(4*x)-3*x^4*log (5)^2-6*x-9)*exp(((1+x)*exp(4*x)^2+(-2-2*x)*exp(4*x)+x+1)/x^3/log(5)^2)/x^ 4/log(5)^2/exp(x),x, algorithm="fricas")
Output:
3*e^(-(x^4*log(5)^2 - (x + 1)*e^(8*x) + 2*(x + 1)*e^(4*x) - x - 1)/(x^3*lo g(5)^2))
Time = 12.64 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.32 \[ \int \frac {e^{-x+\frac {1+e^{4 x} (-2-2 x)+x+e^{8 x} (1+x)}{x^3 \log ^2(5)}} \left (-9-6 x+e^{4 x} \left (18-12 x-24 x^2\right )+e^{8 x} \left (-9+18 x+24 x^2\right )-3 x^4 \log ^2(5)\right )}{x^4 \log ^2(5)} \, dx=3 e^{- x} e^{\frac {x + \left (- 2 x - 2\right ) e^{4 x} + \left (x + 1\right ) e^{8 x} + 1}{x^{3} \log {\left (5 \right )}^{2}}} \] Input:
integrate(((24*x**2+18*x-9)*exp(4*x)**2+(-24*x**2-12*x+18)*exp(4*x)-3*x**4 *ln(5)**2-6*x-9)*exp(((1+x)*exp(4*x)**2+(-2-2*x)*exp(4*x)+x+1)/x**3/ln(5)* *2)/x**4/ln(5)**2/exp(x),x)
Output:
3*exp(-x)*exp((x + (-2*x - 2)*exp(4*x) + (x + 1)*exp(8*x) + 1)/(x**3*log(5 )**2))
Leaf count of result is larger than twice the leaf count of optimal. 73 vs. \(2 (26) = 52\).
Time = 0.28 (sec) , antiderivative size = 73, normalized size of antiderivative = 2.61 \[ \int \frac {e^{-x+\frac {1+e^{4 x} (-2-2 x)+x+e^{8 x} (1+x)}{x^3 \log ^2(5)}} \left (-9-6 x+e^{4 x} \left (18-12 x-24 x^2\right )+e^{8 x} \left (-9+18 x+24 x^2\right )-3 x^4 \log ^2(5)\right )}{x^4 \log ^2(5)} \, dx=3 \, e^{\left (-x + \frac {e^{\left (8 \, x\right )}}{x^{2} \log \left (5\right )^{2}} - \frac {2 \, e^{\left (4 \, x\right )}}{x^{2} \log \left (5\right )^{2}} + \frac {1}{x^{2} \log \left (5\right )^{2}} + \frac {e^{\left (8 \, x\right )}}{x^{3} \log \left (5\right )^{2}} - \frac {2 \, e^{\left (4 \, x\right )}}{x^{3} \log \left (5\right )^{2}} + \frac {1}{x^{3} \log \left (5\right )^{2}}\right )} \] Input:
integrate(((24*x^2+18*x-9)*exp(4*x)^2+(-24*x^2-12*x+18)*exp(4*x)-3*x^4*log (5)^2-6*x-9)*exp(((1+x)*exp(4*x)^2+(-2-2*x)*exp(4*x)+x+1)/x^3/log(5)^2)/x^ 4/log(5)^2/exp(x),x, algorithm="maxima")
Output:
3*e^(-x + e^(8*x)/(x^2*log(5)^2) - 2*e^(4*x)/(x^2*log(5)^2) + 1/(x^2*log(5 )^2) + e^(8*x)/(x^3*log(5)^2) - 2*e^(4*x)/(x^3*log(5)^2) + 1/(x^3*log(5)^2 ))
\[ \int \frac {e^{-x+\frac {1+e^{4 x} (-2-2 x)+x+e^{8 x} (1+x)}{x^3 \log ^2(5)}} \left (-9-6 x+e^{4 x} \left (18-12 x-24 x^2\right )+e^{8 x} \left (-9+18 x+24 x^2\right )-3 x^4 \log ^2(5)\right )}{x^4 \log ^2(5)} \, dx=\int { -\frac {3 \, {\left (x^{4} \log \left (5\right )^{2} - {\left (8 \, x^{2} + 6 \, x - 3\right )} e^{\left (8 \, x\right )} + 2 \, {\left (4 \, x^{2} + 2 \, x - 3\right )} e^{\left (4 \, x\right )} + 2 \, x + 3\right )} e^{\left (-x + \frac {{\left (x + 1\right )} e^{\left (8 \, x\right )} - 2 \, {\left (x + 1\right )} e^{\left (4 \, x\right )} + x + 1}{x^{3} \log \left (5\right )^{2}}\right )}}{x^{4} \log \left (5\right )^{2}} \,d x } \] Input:
integrate(((24*x^2+18*x-9)*exp(4*x)^2+(-24*x^2-12*x+18)*exp(4*x)-3*x^4*log (5)^2-6*x-9)*exp(((1+x)*exp(4*x)^2+(-2-2*x)*exp(4*x)+x+1)/x^3/log(5)^2)/x^ 4/log(5)^2/exp(x),x, algorithm="giac")
Output:
undef
Time = 8.14 (sec) , antiderivative size = 78, normalized size of antiderivative = 2.79 \[ \int \frac {e^{-x+\frac {1+e^{4 x} (-2-2 x)+x+e^{8 x} (1+x)}{x^3 \log ^2(5)}} \left (-9-6 x+e^{4 x} \left (18-12 x-24 x^2\right )+e^{8 x} \left (-9+18 x+24 x^2\right )-3 x^4 \log ^2(5)\right )}{x^4 \log ^2(5)} \, dx=3\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^{\frac {1}{x^2\,{\ln \left (5\right )}^2}}\,{\mathrm {e}}^{\frac {1}{x^3\,{\ln \left (5\right )}^2}}\,{\mathrm {e}}^{-\frac {2\,{\mathrm {e}}^{4\,x}}{x^2\,{\ln \left (5\right )}^2}}\,{\mathrm {e}}^{-\frac {2\,{\mathrm {e}}^{4\,x}}{x^3\,{\ln \left (5\right )}^2}}\,{\mathrm {e}}^{\frac {{\mathrm {e}}^{8\,x}}{x^2\,{\ln \left (5\right )}^2}}\,{\mathrm {e}}^{\frac {{\mathrm {e}}^{8\,x}}{x^3\,{\ln \left (5\right )}^2}} \] Input:
int(-(exp(-x)*exp((x + exp(8*x)*(x + 1) - exp(4*x)*(2*x + 2) + 1)/(x^3*log (5)^2))*(6*x + 3*x^4*log(5)^2 + exp(4*x)*(12*x + 24*x^2 - 18) - exp(8*x)*( 18*x + 24*x^2 - 9) + 9))/(x^4*log(5)^2),x)
Output:
3*exp(-x)*exp(1/(x^2*log(5)^2))*exp(1/(x^3*log(5)^2))*exp(-(2*exp(4*x))/(x ^2*log(5)^2))*exp(-(2*exp(4*x))/(x^3*log(5)^2))*exp(exp(8*x)/(x^2*log(5)^2 ))*exp(exp(8*x)/(x^3*log(5)^2))
Time = 3.43 (sec) , antiderivative size = 63, normalized size of antiderivative = 2.25 \[ \int \frac {e^{-x+\frac {1+e^{4 x} (-2-2 x)+x+e^{8 x} (1+x)}{x^3 \log ^2(5)}} \left (-9-6 x+e^{4 x} \left (18-12 x-24 x^2\right )+e^{8 x} \left (-9+18 x+24 x^2\right )-3 x^4 \log ^2(5)\right )}{x^4 \log ^2(5)} \, dx=\frac {3 e^{\frac {e^{8 x} x +e^{8 x}+x +1}{\mathrm {log}\left (5\right )^{2} x^{3}}}}{e^{\frac {2 e^{4 x} x +2 e^{4 x}+\mathrm {log}\left (5\right )^{2} x^{4}}{\mathrm {log}\left (5\right )^{2} x^{3}}}} \] Input:
int(((24*x^2+18*x-9)*exp(4*x)^2+(-24*x^2-12*x+18)*exp(4*x)-3*x^4*log(5)^2- 6*x-9)*exp(((1+x)*exp(4*x)^2+(-2-2*x)*exp(4*x)+x+1)/x^3/log(5)^2)/x^4/log( 5)^2/exp(x),x)
Output:
(3*e**((e**(8*x)*x + e**(8*x) + x + 1)/(log(5)**2*x**3)))/e**((2*e**(4*x)* x + 2*e**(4*x) + log(5)**2*x**4)/(log(5)**2*x**3))