Integrand size = 104, antiderivative size = 26 \[ \int \frac {-8 e^{x/2} x^2+e^{x/2} \left (8 x+2 x^2\right ) \log \left (4 e^{5+x}\right )}{-64 e^{3 x/4} x^3+48 e^{x/2} x^2 \log \left (4 e^{5+x}\right )-12 e^{x/4} x \log ^2\left (4 e^{5+x}\right )+\log ^3\left (4 e^{5+x}\right )} \, dx=\frac {4}{\left (4-\frac {e^{-x/4} \log \left (4 e^{5+x}\right )}{x}\right )^2} \] Output:
4/(4-ln(4*exp(5+x))/x/exp(1/4*x))^2
Time = 0.29 (sec) , antiderivative size = 52, normalized size of antiderivative = 2.00 \[ \int \frac {-8 e^{x/2} x^2+e^{x/2} \left (8 x+2 x^2\right ) \log \left (4 e^{5+x}\right )}{-64 e^{3 x/4} x^3+48 e^{x/2} x^2 \log \left (4 e^{5+x}\right )-12 e^{x/4} x \log ^2\left (4 e^{5+x}\right )+\log ^3\left (4 e^{5+x}\right )} \, dx=-\frac {\log \left (4 e^{5+x}\right ) \left (-8 e^{x/4} x+\log \left (4 e^{5+x}\right )\right )}{4 \left (-4 e^{x/4} x+\log \left (4 e^{5+x}\right )\right )^2} \] Input:
Integrate[(-8*E^(x/2)*x^2 + E^(x/2)*(8*x + 2*x^2)*Log[4*E^(5 + x)])/(-64*E ^((3*x)/4)*x^3 + 48*E^(x/2)*x^2*Log[4*E^(5 + x)] - 12*E^(x/4)*x*Log[4*E^(5 + x)]^2 + Log[4*E^(5 + x)]^3),x]
Output:
-1/4*(Log[4*E^(5 + x)]*(-8*E^(x/4)*x + Log[4*E^(5 + x)]))/(-4*E^(x/4)*x + Log[4*E^(5 + x)])^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.
| \(\displaystyle \int \frac {e^{x/2} \left (2 x^2+8 x\right ) \log \left (4 e^{x+5}\right )-8 e^{x/2} x^2}{-64 e^{3 x/4} x^3+48 e^{x/2} x^2 \log \left (4 e^{x+5}\right )+\log ^3\left (4 e^{x+5}\right )-12 e^{x/4} x \log ^2\left (4 e^{x+5}\right )} \, dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {2 e^{x/2} x \left (4 x-(x+4) \log \left (4 e^{x+5}\right )\right )}{\left (4 e^{x/4} x-\log \left (4 e^{x+5}\right )\right )^3}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle 2 \int \frac {e^{x/2} x \left (4 x-(x+4) \log \left (4 e^{x+5}\right )\right )}{\left (4 e^{x/4} x-\log \left (4 e^{x+5}\right )\right )^3}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle 2 \int \left (-\frac {e^{x/2} \log \left (4 e^{x+5}\right ) x^2}{\left (4 e^{x/4} x-\log \left (4 e^{x+5}\right )\right )^3}+\frac {4 e^{x/2} x^2}{\left (4 e^{x/4} x-\log \left (4 e^{x+5}\right )\right )^3}-\frac {4 e^{x/2} \log \left (4 e^{x+5}\right ) x}{\left (4 e^{x/4} x-\log \left (4 e^{x+5}\right )\right )^3}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle 2 \left (4 \int \frac {e^{x/2} x^2}{\left (4 e^{x/4} x-\log \left (4 e^{x+5}\right )\right )^3}dx-\int \frac {e^{x/2} x^2 \log \left (4 e^{x+5}\right )}{\left (4 e^{x/4} x-\log \left (4 e^{x+5}\right )\right )^3}dx-4 \int \frac {e^{x/2} x \log \left (4 e^{x+5}\right )}{\left (4 e^{x/4} x-\log \left (4 e^{x+5}\right )\right )^3}dx\right )\) |
Input:
Int[(-8*E^(x/2)*x^2 + E^(x/2)*(8*x + 2*x^2)*Log[4*E^(5 + x)])/(-64*E^((3*x )/4)*x^3 + 48*E^(x/2)*x^2*Log[4*E^(5 + x)] - 12*E^(x/4)*x*Log[4*E^(5 + x)] ^2 + Log[4*E^(5 + x)]^3),x]
Output:
$Aborted
Time = 3.71 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19
| method | result | size |
| default | \(\frac {4 x^{2} {\mathrm e}^{\frac {x}{2}}}{\left (4 \,{\mathrm e}^{\frac {x}{4}} x -\ln \left (4 \,{\mathrm e}^{5+x}\right )\right )^{2}}\) | \(31\) |
| parallelrisch | \(\frac {4 x^{2} {\mathrm e}^{\frac {x}{2}}}{16 x^{2} {\mathrm e}^{\frac {x}{2}}-8 x \,{\mathrm e}^{\frac {x}{4}} \ln \left (4 \,{\mathrm e}^{5+x}\right )+\ln \left (4 \,{\mathrm e}^{5+x}\right )^{2}}\) | \(49\) |
| risch | \(\frac {16 x^{2} {\mathrm e}^{\frac {x}{2}}}{{\left (i \pi \,\operatorname {csgn}\left (i {\mathrm e}^{\frac {x}{4}}\right ) \operatorname {csgn}\left (i {\mathrm e}^{\frac {x}{2}}\right ) \operatorname {csgn}\left (i {\mathrm e}^{\frac {3 x}{4}}\right )+i \pi \,\operatorname {csgn}\left (i {\mathrm e}^{\frac {x}{4}}\right ) \operatorname {csgn}\left (i {\mathrm e}^{\frac {3 x}{4}}\right ) \operatorname {csgn}\left (i {\mathrm e}^{x}\right )+i \pi \operatorname {csgn}\left (i {\mathrm e}^{\frac {3 x}{4}}\right )^{3}-i \pi \,\operatorname {csgn}\left (i {\mathrm e}^{\frac {x}{4}}\right ) \operatorname {csgn}\left (i {\mathrm e}^{\frac {3 x}{4}}\right )^{2}-i \pi \,\operatorname {csgn}\left (i {\mathrm e}^{\frac {x}{2}}\right ) \operatorname {csgn}\left (i {\mathrm e}^{\frac {3 x}{4}}\right )^{2}+i \pi \operatorname {csgn}\left (i {\mathrm e}^{\frac {x}{2}}\right )^{3}-i \pi \,\operatorname {csgn}\left (i {\mathrm e}^{\frac {x}{4}}\right ) \operatorname {csgn}\left (i {\mathrm e}^{x}\right )^{2}+i \pi \operatorname {csgn}\left (i {\mathrm e}^{x}\right )^{3}+i \pi \operatorname {csgn}\left (i {\mathrm e}^{\frac {x}{4}}\right )^{2} \operatorname {csgn}\left (i {\mathrm e}^{\frac {x}{2}}\right )-2 i \pi \,\operatorname {csgn}\left (i {\mathrm e}^{\frac {x}{4}}\right ) \operatorname {csgn}\left (i {\mathrm e}^{\frac {x}{2}}\right )^{2}-i \pi \,\operatorname {csgn}\left (i {\mathrm e}^{\frac {3 x}{4}}\right ) \operatorname {csgn}\left (i {\mathrm e}^{x}\right )^{2}-10+8 \,{\mathrm e}^{\frac {x}{4}} x -4 \ln \left (2\right )-8 \ln \left ({\mathrm e}^{\frac {x}{4}}\right )\right )}^{2}}\) | \(254\) |
Input:
int(((2*x^2+8*x)*exp(1/4*x)^2*ln(4*exp(5+x))-8*x^2*exp(1/4*x)^2)/(ln(4*exp (5+x))^3-12*x*exp(1/4*x)*ln(4*exp(5+x))^2+48*x^2*exp(1/4*x)^2*ln(4*exp(5+x ))-64*x^3*exp(1/4*x)^3),x,method=_RETURNVERBOSE)
Output:
4*x^2*exp(1/4*x)^2/(4*exp(1/4*x)*x-ln(4*exp(5+x)))^2
Leaf count of result is larger than twice the leaf count of optimal. 91 vs. \(2 (21) = 42\).
Time = 0.08 (sec) , antiderivative size = 91, normalized size of antiderivative = 3.50 \[ \int \frac {-8 e^{x/2} x^2+e^{x/2} \left (8 x+2 x^2\right ) \log \left (4 e^{5+x}\right )}{-64 e^{3 x/4} x^3+48 e^{x/2} x^2 \log \left (4 e^{5+x}\right )-12 e^{x/4} x \log ^2\left (4 e^{5+x}\right )+\log ^3\left (4 e^{5+x}\right )} \, dx=-\frac {x^{2} - 8 \, {\left (x^{2} + 2 \, x \log \left (2\right ) + 5 \, x\right )} e^{\left (\frac {1}{4} \, x\right )} + 4 \, {\left (x + 5\right )} \log \left (2\right ) + 4 \, \log \left (2\right )^{2} + 10 \, x + 25}{4 \, {\left (16 \, x^{2} e^{\left (\frac {1}{2} \, x\right )} + x^{2} - 8 \, {\left (x^{2} + 2 \, x \log \left (2\right ) + 5 \, x\right )} e^{\left (\frac {1}{4} \, x\right )} + 4 \, {\left (x + 5\right )} \log \left (2\right ) + 4 \, \log \left (2\right )^{2} + 10 \, x + 25\right )}} \] Input:
integrate(((2*x^2+8*x)*exp(1/4*x)^2*log(4*exp(5+x))-8*x^2*exp(1/4*x)^2)/(l og(4*exp(5+x))^3-12*x*exp(1/4*x)*log(4*exp(5+x))^2+48*x^2*exp(1/4*x)^2*log (4*exp(5+x))-64*x^3*exp(1/4*x)^3),x, algorithm="fricas")
Output:
-1/4*(x^2 - 8*(x^2 + 2*x*log(2) + 5*x)*e^(1/4*x) + 4*(x + 5)*log(2) + 4*lo g(2)^2 + 10*x + 25)/(16*x^2*e^(1/2*x) + x^2 - 8*(x^2 + 2*x*log(2) + 5*x)*e ^(1/4*x) + 4*(x + 5)*log(2) + 4*log(2)^2 + 10*x + 25)
Leaf count of result is larger than twice the leaf count of optimal. 107 vs. \(2 (19) = 38\).
Time = 0.25 (sec) , antiderivative size = 107, normalized size of antiderivative = 4.12 \[ \int \frac {-8 e^{x/2} x^2+e^{x/2} \left (8 x+2 x^2\right ) \log \left (4 e^{5+x}\right )}{-64 e^{3 x/4} x^3+48 e^{x/2} x^2 \log \left (4 e^{5+x}\right )-12 e^{x/4} x \log ^2\left (4 e^{5+x}\right )+\log ^3\left (4 e^{5+x}\right )} \, dx=\frac {- \frac {x^{2}}{4} - \frac {5 x}{2} - x \log {\left (2 \right )} + \left (2 x^{2} + 4 x \log {\left (2 \right )} + 10 x\right ) e^{\frac {x}{4}} - \frac {25}{4} - 5 \log {\left (2 \right )} - \log {\left (2 \right )}^{2}}{16 x^{2} e^{\frac {x}{2}} + x^{2} + 4 x \log {\left (2 \right )} + 10 x + \left (- 8 x^{2} - 40 x - 16 x \log {\left (2 \right )}\right ) e^{\frac {x}{4}} + 4 \log {\left (2 \right )}^{2} + 20 \log {\left (2 \right )} + 25} \] Input:
integrate(((2*x**2+8*x)*exp(1/4*x)**2*ln(4*exp(5+x))-8*x**2*exp(1/4*x)**2) /(ln(4*exp(5+x))**3-12*x*exp(1/4*x)*ln(4*exp(5+x))**2+48*x**2*exp(1/4*x)** 2*ln(4*exp(5+x))-64*x**3*exp(1/4*x)**3),x)
Output:
(-x**2/4 - 5*x/2 - x*log(2) + (2*x**2 + 4*x*log(2) + 10*x)*exp(x/4) - 25/4 - 5*log(2) - log(2)**2)/(16*x**2*exp(x/2) + x**2 + 4*x*log(2) + 10*x + (- 8*x**2 - 40*x - 16*x*log(2))*exp(x/4) + 4*log(2)**2 + 20*log(2) + 25)
Leaf count of result is larger than twice the leaf count of optimal. 97 vs. \(2 (21) = 42\).
Time = 0.19 (sec) , antiderivative size = 97, normalized size of antiderivative = 3.73 \[ \int \frac {-8 e^{x/2} x^2+e^{x/2} \left (8 x+2 x^2\right ) \log \left (4 e^{5+x}\right )}{-64 e^{3 x/4} x^3+48 e^{x/2} x^2 \log \left (4 e^{5+x}\right )-12 e^{x/4} x \log ^2\left (4 e^{5+x}\right )+\log ^3\left (4 e^{5+x}\right )} \, dx=-\frac {x^{2} + 2 \, x {\left (2 \, \log \left (2\right ) + 5\right )} - 8 \, {\left (x^{2} + x {\left (2 \, \log \left (2\right ) + 5\right )}\right )} e^{\left (\frac {1}{4} \, x\right )} + 4 \, \log \left (2\right )^{2} + 20 \, \log \left (2\right ) + 25}{4 \, {\left (16 \, x^{2} e^{\left (\frac {1}{2} \, x\right )} + x^{2} + 2 \, x {\left (2 \, \log \left (2\right ) + 5\right )} - 8 \, {\left (x^{2} + x {\left (2 \, \log \left (2\right ) + 5\right )}\right )} e^{\left (\frac {1}{4} \, x\right )} + 4 \, \log \left (2\right )^{2} + 20 \, \log \left (2\right ) + 25\right )}} \] Input:
integrate(((2*x^2+8*x)*exp(1/4*x)^2*log(4*exp(5+x))-8*x^2*exp(1/4*x)^2)/(l og(4*exp(5+x))^3-12*x*exp(1/4*x)*log(4*exp(5+x))^2+48*x^2*exp(1/4*x)^2*log (4*exp(5+x))-64*x^3*exp(1/4*x)^3),x, algorithm="maxima")
Output:
-1/4*(x^2 + 2*x*(2*log(2) + 5) - 8*(x^2 + x*(2*log(2) + 5))*e^(1/4*x) + 4* log(2)^2 + 20*log(2) + 25)/(16*x^2*e^(1/2*x) + x^2 + 2*x*(2*log(2) + 5) - 8*(x^2 + x*(2*log(2) + 5))*e^(1/4*x) + 4*log(2)^2 + 20*log(2) + 25)
Leaf count of result is larger than twice the leaf count of optimal. 148 vs. \(2 (21) = 42\).
Time = 0.16 (sec) , antiderivative size = 148, normalized size of antiderivative = 5.69 \[ \int \frac {-8 e^{x/2} x^2+e^{x/2} \left (8 x+2 x^2\right ) \log \left (4 e^{5+x}\right )}{-64 e^{3 x/4} x^3+48 e^{x/2} x^2 \log \left (4 e^{5+x}\right )-12 e^{x/4} x \log ^2\left (4 e^{5+x}\right )+\log ^3\left (4 e^{5+x}\right )} \, dx=\frac {8 \, {\left (x + 5\right )}^{2} e^{\left (\frac {1}{4} \, x\right )} + 16 \, {\left (x + 5\right )} e^{\left (\frac {1}{4} \, x\right )} \log \left (2\right ) - {\left (x + 5\right )}^{2} - 40 \, {\left (x + 5\right )} e^{\left (\frac {1}{4} \, x\right )} - 4 \, {\left (x + 5\right )} \log \left (2\right ) - 80 \, e^{\left (\frac {1}{4} \, x\right )} \log \left (2\right ) - 4 \, \log \left (2\right )^{2}}{4 \, {\left (16 \, {\left (x + 5\right )}^{2} e^{\left (\frac {1}{2} \, x\right )} - 8 \, {\left (x + 5\right )}^{2} e^{\left (\frac {1}{4} \, x\right )} - 16 \, {\left (x + 5\right )} e^{\left (\frac {1}{4} \, x\right )} \log \left (2\right ) + {\left (x + 5\right )}^{2} - 160 \, {\left (x + 5\right )} e^{\left (\frac {1}{2} \, x\right )} + 40 \, {\left (x + 5\right )} e^{\left (\frac {1}{4} \, x\right )} + 4 \, {\left (x + 5\right )} \log \left (2\right ) + 80 \, e^{\left (\frac {1}{4} \, x\right )} \log \left (2\right ) + 4 \, \log \left (2\right )^{2} + 400 \, e^{\left (\frac {1}{2} \, x\right )}\right )}} \] Input:
integrate(((2*x^2+8*x)*exp(1/4*x)^2*log(4*exp(5+x))-8*x^2*exp(1/4*x)^2)/(l og(4*exp(5+x))^3-12*x*exp(1/4*x)*log(4*exp(5+x))^2+48*x^2*exp(1/4*x)^2*log (4*exp(5+x))-64*x^3*exp(1/4*x)^3),x, algorithm="giac")
Output:
1/4*(8*(x + 5)^2*e^(1/4*x) + 16*(x + 5)*e^(1/4*x)*log(2) - (x + 5)^2 - 40* (x + 5)*e^(1/4*x) - 4*(x + 5)*log(2) - 80*e^(1/4*x)*log(2) - 4*log(2)^2)/( 16*(x + 5)^2*e^(1/2*x) - 8*(x + 5)^2*e^(1/4*x) - 16*(x + 5)*e^(1/4*x)*log( 2) + (x + 5)^2 - 160*(x + 5)*e^(1/2*x) + 40*(x + 5)*e^(1/4*x) + 4*(x + 5)* log(2) + 80*e^(1/4*x)*log(2) + 4*log(2)^2 + 400*e^(1/2*x))
Timed out. \[ \int \frac {-8 e^{x/2} x^2+e^{x/2} \left (8 x+2 x^2\right ) \log \left (4 e^{5+x}\right )}{-64 e^{3 x/4} x^3+48 e^{x/2} x^2 \log \left (4 e^{5+x}\right )-12 e^{x/4} x \log ^2\left (4 e^{5+x}\right )+\log ^3\left (4 e^{5+x}\right )} \, dx=\int -\frac {8\,x^2\,{\mathrm {e}}^{x/2}-\ln \left (4\,{\mathrm {e}}^{x+5}\right )\,{\mathrm {e}}^{x/2}\,\left (2\,x^2+8\,x\right )}{{\ln \left (4\,{\mathrm {e}}^{x+5}\right )}^3-64\,x^3\,{\mathrm {e}}^{\frac {3\,x}{4}}+48\,x^2\,\ln \left (4\,{\mathrm {e}}^{x+5}\right )\,{\mathrm {e}}^{x/2}-12\,x\,{\ln \left (4\,{\mathrm {e}}^{x+5}\right )}^2\,{\mathrm {e}}^{x/4}} \,d x \] Input:
int(-(8*x^2*exp(x/2) - log(4*exp(x + 5))*exp(x/2)*(8*x + 2*x^2))/(log(4*ex p(x + 5))^3 - 64*x^3*exp((3*x)/4) + 48*x^2*log(4*exp(x + 5))*exp(x/2) - 12 *x*log(4*exp(x + 5))^2*exp(x/4)),x)
Output:
int(-(8*x^2*exp(x/2) - log(4*exp(x + 5))*exp(x/2)*(8*x + 2*x^2))/(log(4*ex p(x + 5))^3 - 64*x^3*exp((3*x)/4) + 48*x^2*log(4*exp(x + 5))*exp(x/2) - 12 *x*log(4*exp(x + 5))^2*exp(x/4)), x)
Time = 0.21 (sec) , antiderivative size = 53, normalized size of antiderivative = 2.04 \[ \int \frac {-8 e^{x/2} x^2+e^{x/2} \left (8 x+2 x^2\right ) \log \left (4 e^{5+x}\right )}{-64 e^{3 x/4} x^3+48 e^{x/2} x^2 \log \left (4 e^{5+x}\right )-12 e^{x/4} x \log ^2\left (4 e^{5+x}\right )+\log ^3\left (4 e^{5+x}\right )} \, dx=-\frac {4 e^{\frac {x}{2}} x^{2}}{8 e^{\frac {x}{4}} \mathrm {log}\left (4 e^{x} e^{5}\right ) x -16 e^{\frac {x}{2}} x^{2}-\mathrm {log}\left (4 e^{x} e^{5}\right )^{2}} \] Input:
int(((2*x^2+8*x)*exp(1/4*x)^2*log(4*exp(5+x))-8*x^2*exp(1/4*x)^2)/(log(4*e xp(5+x))^3-12*x*exp(1/4*x)*log(4*exp(5+x))^2+48*x^2*exp(1/4*x)^2*log(4*exp (5+x))-64*x^3*exp(1/4*x)^3),x)
Output:
( - 4*e**(x/2)*x**2)/(8*e**(x/4)*log(4*e**x*e**5)*x - 16*e**(x/2)*x**2 - l og(4*e**x*e**5)**2)