Integrand size = 137, antiderivative size = 24 \[ \int \frac {\left (-12 x^4-x^5+x^6+e^x \left (6 x^4+2 x^5\right )\right ) \log ^2(3+x)+\log \left (e^{2 x}+e^x (-5+x)\right ) \left (\left (-10 x^4+2 e^x x^4+2 x^5\right ) \log (3+x)+\left (-60 x^3-8 x^4+4 x^5+e^x \left (12 x^3+4 x^4\right )\right ) \log ^2(3+x)\right )}{-75-10 x+5 x^2+e^x (15+5 x)} \, dx=\frac {1}{5} x^4 \log ^2(3+x) \log \left (e^x \left (-5+e^x+x\right )\right ) \] Output:
1/5*ln(3+x)^2*x^4*ln(exp(x)*(exp(x)+x-5))
Time = 0.52 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.79 \[ \int \frac {\left (-12 x^4-x^5+x^6+e^x \left (6 x^4+2 x^5\right )\right ) \log ^2(3+x)+\log \left (e^{2 x}+e^x (-5+x)\right ) \left (\left (-10 x^4+2 e^x x^4+2 x^5\right ) \log (3+x)+\left (-60 x^3-8 x^4+4 x^5+e^x \left (12 x^3+4 x^4\right )\right ) \log ^2(3+x)\right )}{-75-10 x+5 x^2+e^x (15+5 x)} \, dx=\frac {1}{5} \left (-162 (3+x)-162 \log \left (-5+e^x+x\right )+\left (162+x^4 \log ^2(3+x)\right ) \log \left (e^x \left (-5+e^x+x\right )\right )\right ) \] Input:
Integrate[((-12*x^4 - x^5 + x^6 + E^x*(6*x^4 + 2*x^5))*Log[3 + x]^2 + Log[ E^(2*x) + E^x*(-5 + x)]*((-10*x^4 + 2*E^x*x^4 + 2*x^5)*Log[3 + x] + (-60*x ^3 - 8*x^4 + 4*x^5 + E^x*(12*x^3 + 4*x^4))*Log[3 + x]^2))/(-75 - 10*x + 5* x^2 + E^x*(15 + 5*x)),x]
Output:
(-162*(3 + x) - 162*Log[-5 + E^x + x] + (162 + x^4*Log[3 + x]^2)*Log[E^x*( -5 + E^x + x)])/5
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.
| \(\displaystyle \int \frac {\left (x^6-x^5-12 x^4+e^x \left (2 x^5+6 x^4\right )\right ) \log ^2(x+3)+\log \left (e^x (x-5)+e^{2 x}\right ) \left (\left (2 x^5+2 e^x x^4-10 x^4\right ) \log (x+3)+\left (4 x^5-8 x^4-60 x^3+e^x \left (4 x^4+12 x^3\right )\right ) \log ^2(x+3)\right )}{5 x^2-10 x+e^x (5 x+15)-75} \, dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {1}{5} x^3 \log (x+3) \left (\frac {2 x \log \left (e^x \left (x+e^x-5\right )\right )}{x+3}+\log (x+3) \left (\frac {x \left (x+2 e^x-4\right )}{x+e^x-5}+4 \log \left (e^x \left (x+e^x-5\right )\right )\right )\right )dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \int x^3 \log (x+3) \left (\frac {2 x \log \left (-e^x \left (-x-e^x+5\right )\right )}{x+3}+\left (\frac {\left (-x-2 e^x+4\right ) x}{-x-e^x+5}+4 \log \left (-e^x \left (-x-e^x+5\right )\right )\right ) \log (x+3)\right )dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \frac {1}{5} \int \left (\frac {2 x^3 \log (x+3) \left (\log (x+3) x^2+3 \log (x+3) x+2 \log (x+3) \log \left (e^x \left (x+e^x-5\right )\right ) x+\log \left (e^x \left (x+e^x-5\right )\right ) x+6 \log (x+3) \log \left (e^x \left (x+e^x-5\right )\right )\right )}{x+3}-\frac {(x-6) x^4 \log ^2(x+3)}{x+e^x-5}\right )dx\) |
| \(\Big \downarrow \) 7299 |
| \(\displaystyle \frac {1}{5} \int \left (\frac {2 x^3 \log (x+3) \left (\log (x+3) x^2+3 \log (x+3) x+2 \log (x+3) \log \left (e^x \left (x+e^x-5\right )\right ) x+\log \left (e^x \left (x+e^x-5\right )\right ) x+6 \log (x+3) \log \left (e^x \left (x+e^x-5\right )\right )\right )}{x+3}-\frac {(x-6) x^4 \log ^2(x+3)}{x+e^x-5}\right )dx\) |
Input:
Int[((-12*x^4 - x^5 + x^6 + E^x*(6*x^4 + 2*x^5))*Log[3 + x]^2 + Log[E^(2*x ) + E^x*(-5 + x)]*((-10*x^4 + 2*E^x*x^4 + 2*x^5)*Log[3 + x] + (-60*x^3 - 8 *x^4 + 4*x^5 + E^x*(12*x^3 + 4*x^4))*Log[3 + x]^2))/(-75 - 10*x + 5*x^2 + E^x*(15 + 5*x)),x]
Output:
$Aborted
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.04 (sec) , antiderivative size = 165, normalized size of antiderivative = 6.88
\[\frac {x^{4} \ln \left (3+x \right )^{2} \ln \left ({\mathrm e}^{x}\right )}{5}+\frac {x^{4} \ln \left (3+x \right )^{2} \ln \left ({\mathrm e}^{x}+x -5\right )}{5}-\frac {i \pi \,x^{4} \operatorname {csgn}\left (i \left ({\mathrm e}^{x}+x -5\right )\right ) \operatorname {csgn}\left (i {\mathrm e}^{x}\right ) \operatorname {csgn}\left (i {\mathrm e}^{x} \left ({\mathrm e}^{x}+x -5\right )\right ) \ln \left (3+x \right )^{2}}{10}+\frac {i \pi \,x^{4} \operatorname {csgn}\left (i \left ({\mathrm e}^{x}+x -5\right )\right ) \operatorname {csgn}\left (i {\mathrm e}^{x} \left ({\mathrm e}^{x}+x -5\right )\right )^{2} \ln \left (3+x \right )^{2}}{10}+\frac {i \pi \,x^{4} \operatorname {csgn}\left (i {\mathrm e}^{x}\right ) \operatorname {csgn}\left (i {\mathrm e}^{x} \left ({\mathrm e}^{x}+x -5\right )\right )^{2} \ln \left (3+x \right )^{2}}{10}-\frac {i \pi \,x^{4} \operatorname {csgn}\left (i {\mathrm e}^{x} \left ({\mathrm e}^{x}+x -5\right )\right )^{3} \ln \left (3+x \right )^{2}}{10}\]
Input:
int(((((4*x^4+12*x^3)*exp(x)+4*x^5-8*x^4-60*x^3)*ln(3+x)^2+(2*exp(x)*x^4+2 *x^5-10*x^4)*ln(3+x))*ln(exp(x)^2+(-5+x)*exp(x))+((2*x^5+6*x^4)*exp(x)+x^6 -x^5-12*x^4)*ln(3+x)^2)/((5*x+15)*exp(x)+5*x^2-10*x-75),x)
Output:
1/5*x^4*ln(3+x)^2*ln(exp(x))+1/5*x^4*ln(3+x)^2*ln(exp(x)+x-5)-1/10*I*Pi*x^ 4*csgn(I*(exp(x)+x-5))*csgn(I*exp(x))*csgn(I*exp(x)*(exp(x)+x-5))*ln(3+x)^ 2+1/10*I*Pi*x^4*csgn(I*(exp(x)+x-5))*csgn(I*exp(x)*(exp(x)+x-5))^2*ln(3+x) ^2+1/10*I*Pi*x^4*csgn(I*exp(x))*csgn(I*exp(x)*(exp(x)+x-5))^2*ln(3+x)^2-1/ 10*I*Pi*x^4*csgn(I*exp(x)*(exp(x)+x-5))^3*ln(3+x)^2
Time = 0.07 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {\left (-12 x^4-x^5+x^6+e^x \left (6 x^4+2 x^5\right )\right ) \log ^2(3+x)+\log \left (e^{2 x}+e^x (-5+x)\right ) \left (\left (-10 x^4+2 e^x x^4+2 x^5\right ) \log (3+x)+\left (-60 x^3-8 x^4+4 x^5+e^x \left (12 x^3+4 x^4\right )\right ) \log ^2(3+x)\right )}{-75-10 x+5 x^2+e^x (15+5 x)} \, dx=\frac {1}{5} \, x^{4} \log \left ({\left (x - 5\right )} e^{x} + e^{\left (2 \, x\right )}\right ) \log \left (x + 3\right )^{2} \] Input:
integrate(((((4*x^4+12*x^3)*exp(x)+4*x^5-8*x^4-60*x^3)*log(3+x)^2+(2*exp(x )*x^4+2*x^5-10*x^4)*log(3+x))*log(exp(x)^2+(-5+x)*exp(x))+((2*x^5+6*x^4)*e xp(x)+x^6-x^5-12*x^4)*log(3+x)^2)/((5*x+15)*exp(x)+5*x^2-10*x-75),x, algor ithm="fricas")
Output:
1/5*x^4*log((x - 5)*e^x + e^(2*x))*log(x + 3)^2
Time = 0.51 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {\left (-12 x^4-x^5+x^6+e^x \left (6 x^4+2 x^5\right )\right ) \log ^2(3+x)+\log \left (e^{2 x}+e^x (-5+x)\right ) \left (\left (-10 x^4+2 e^x x^4+2 x^5\right ) \log (3+x)+\left (-60 x^3-8 x^4+4 x^5+e^x \left (12 x^3+4 x^4\right )\right ) \log ^2(3+x)\right )}{-75-10 x+5 x^2+e^x (15+5 x)} \, dx=\frac {x^{4} \log {\left (x + 3 \right )}^{2} \log {\left (\left (x - 5\right ) e^{x} + e^{2 x} \right )}}{5} \] Input:
integrate(((((4*x**4+12*x**3)*exp(x)+4*x**5-8*x**4-60*x**3)*ln(3+x)**2+(2* exp(x)*x**4+2*x**5-10*x**4)*ln(3+x))*ln(exp(x)**2+(-5+x)*exp(x))+((2*x**5+ 6*x**4)*exp(x)+x**6-x**5-12*x**4)*ln(3+x)**2)/((5*x+15)*exp(x)+5*x**2-10*x -75),x)
Output:
x**4*log(x + 3)**2*log((x - 5)*exp(x) + exp(2*x))/5
Time = 0.09 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21 \[ \int \frac {\left (-12 x^4-x^5+x^6+e^x \left (6 x^4+2 x^5\right )\right ) \log ^2(3+x)+\log \left (e^{2 x}+e^x (-5+x)\right ) \left (\left (-10 x^4+2 e^x x^4+2 x^5\right ) \log (3+x)+\left (-60 x^3-8 x^4+4 x^5+e^x \left (12 x^3+4 x^4\right )\right ) \log ^2(3+x)\right )}{-75-10 x+5 x^2+e^x (15+5 x)} \, dx=\frac {1}{5} \, x^{5} \log \left (x + 3\right )^{2} + \frac {1}{5} \, x^{4} \log \left (x + e^{x} - 5\right ) \log \left (x + 3\right )^{2} \] Input:
integrate(((((4*x^4+12*x^3)*exp(x)+4*x^5-8*x^4-60*x^3)*log(3+x)^2+(2*exp(x )*x^4+2*x^5-10*x^4)*log(3+x))*log(exp(x)^2+(-5+x)*exp(x))+((2*x^5+6*x^4)*e xp(x)+x^6-x^5-12*x^4)*log(3+x)^2)/((5*x+15)*exp(x)+5*x^2-10*x-75),x, algor ithm="maxima")
Output:
1/5*x^5*log(x + 3)^2 + 1/5*x^4*log(x + e^x - 5)*log(x + 3)^2
Time = 0.22 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21 \[ \int \frac {\left (-12 x^4-x^5+x^6+e^x \left (6 x^4+2 x^5\right )\right ) \log ^2(3+x)+\log \left (e^{2 x}+e^x (-5+x)\right ) \left (\left (-10 x^4+2 e^x x^4+2 x^5\right ) \log (3+x)+\left (-60 x^3-8 x^4+4 x^5+e^x \left (12 x^3+4 x^4\right )\right ) \log ^2(3+x)\right )}{-75-10 x+5 x^2+e^x (15+5 x)} \, dx=\frac {1}{5} \, x^{5} \log \left (x + 3\right )^{2} + \frac {1}{5} \, x^{4} \log \left (x + e^{x} - 5\right ) \log \left (x + 3\right )^{2} \] Input:
integrate(((((4*x^4+12*x^3)*exp(x)+4*x^5-8*x^4-60*x^3)*log(3+x)^2+(2*exp(x )*x^4+2*x^5-10*x^4)*log(3+x))*log(exp(x)^2+(-5+x)*exp(x))+((2*x^5+6*x^4)*e xp(x)+x^6-x^5-12*x^4)*log(3+x)^2)/((5*x+15)*exp(x)+5*x^2-10*x-75),x, algor ithm="giac")
Output:
1/5*x^5*log(x + 3)^2 + 1/5*x^4*log(x + e^x - 5)*log(x + 3)^2
Time = 0.25 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {\left (-12 x^4-x^5+x^6+e^x \left (6 x^4+2 x^5\right )\right ) \log ^2(3+x)+\log \left (e^{2 x}+e^x (-5+x)\right ) \left (\left (-10 x^4+2 e^x x^4+2 x^5\right ) \log (3+x)+\left (-60 x^3-8 x^4+4 x^5+e^x \left (12 x^3+4 x^4\right )\right ) \log ^2(3+x)\right )}{-75-10 x+5 x^2+e^x (15+5 x)} \, dx=\frac {x^4\,\ln \left ({\mathrm {e}}^{2\,x}+{\mathrm {e}}^x\,\left (x-5\right )\right )\,{\ln \left (x+3\right )}^2}{5} \] Input:
int(-(log(exp(2*x) + exp(x)*(x - 5))*(log(x + 3)^2*(exp(x)*(12*x^3 + 4*x^4 ) - 60*x^3 - 8*x^4 + 4*x^5) + log(x + 3)*(2*x^4*exp(x) - 10*x^4 + 2*x^5)) + log(x + 3)^2*(exp(x)*(6*x^4 + 2*x^5) - 12*x^4 - x^5 + x^6))/(10*x - exp( x)*(5*x + 15) - 5*x^2 + 75),x)
Output:
(x^4*log(exp(2*x) + exp(x)*(x - 5))*log(x + 3)^2)/5
Time = 0.23 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17 \[ \int \frac {\left (-12 x^4-x^5+x^6+e^x \left (6 x^4+2 x^5\right )\right ) \log ^2(3+x)+\log \left (e^{2 x}+e^x (-5+x)\right ) \left (\left (-10 x^4+2 e^x x^4+2 x^5\right ) \log (3+x)+\left (-60 x^3-8 x^4+4 x^5+e^x \left (12 x^3+4 x^4\right )\right ) \log ^2(3+x)\right )}{-75-10 x+5 x^2+e^x (15+5 x)} \, dx=\frac {\mathrm {log}\left (e^{2 x}+e^{x} x -5 e^{x}\right ) \mathrm {log}\left (x +3\right )^{2} x^{4}}{5} \] Input:
int(((((4*x^4+12*x^3)*exp(x)+4*x^5-8*x^4-60*x^3)*log(3+x)^2+(2*exp(x)*x^4+ 2*x^5-10*x^4)*log(3+x))*log(exp(x)^2+(-5+x)*exp(x))+((2*x^5+6*x^4)*exp(x)+ x^6-x^5-12*x^4)*log(3+x)^2)/((5*x+15)*exp(x)+5*x^2-10*x-75),x)
Output:
(log(e**(2*x) + e**x*x - 5*e**x)*log(x + 3)**2*x**4)/5