Integrand size = 135, antiderivative size = 29 \[ \int \frac {e^5 \left (-10-2 x+10 x^2+2 x^3\right )+e^{10} \left (10-8 x-2 x^2\right ) \log ^2(5+x)+\log \left (\frac {3 e^x}{x}\right ) \left (e^5 \left (-10 x-2 x^2\right )+4 e^{10} x \log (5+x)\right )}{5 x+11 x^2+7 x^3+x^4+e^5 \left (-10 x-12 x^2-2 x^3\right ) \log ^2(5+x)+e^{10} \left (5 x+x^2\right ) \log ^4(5+x)} \, dx=\frac {2 \log \left (\frac {3 e^x}{x}\right )}{\frac {1+x}{e^5}-\log ^2(5+x)} \] Output:
2*ln(exp(ln(3)+x)/x)/((1+x)/exp(5)-ln(5+x)^2)
Time = 0.27 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.07 \[ \int \frac {e^5 \left (-10-2 x+10 x^2+2 x^3\right )+e^{10} \left (10-8 x-2 x^2\right ) \log ^2(5+x)+\log \left (\frac {3 e^x}{x}\right ) \left (e^5 \left (-10 x-2 x^2\right )+4 e^{10} x \log (5+x)\right )}{5 x+11 x^2+7 x^3+x^4+e^5 \left (-10 x-12 x^2-2 x^3\right ) \log ^2(5+x)+e^{10} \left (5 x+x^2\right ) \log ^4(5+x)} \, dx=-\frac {2 e^5 \log \left (\frac {3 e^x}{x}\right )}{-1-x+e^5 \log ^2(5+x)} \] Input:
Integrate[(E^5*(-10 - 2*x + 10*x^2 + 2*x^3) + E^10*(10 - 8*x - 2*x^2)*Log[ 5 + x]^2 + Log[(3*E^x)/x]*(E^5*(-10*x - 2*x^2) + 4*E^10*x*Log[5 + x]))/(5* x + 11*x^2 + 7*x^3 + x^4 + E^5*(-10*x - 12*x^2 - 2*x^3)*Log[5 + x]^2 + E^1 0*(5*x + x^2)*Log[5 + x]^4),x]
Output:
(-2*E^5*Log[(3*E^x)/x])/(-1 - x + E^5*Log[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.
| \(\displaystyle \int \frac {e^{10} \left (-2 x^2-8 x+10\right ) \log ^2(x+5)+\log \left (\frac {3 e^x}{x}\right ) \left (e^5 \left (-2 x^2-10 x\right )+4 e^{10} x \log (x+5)\right )+e^5 \left (2 x^3+10 x^2-2 x-10\right )}{x^4+7 x^3+11 x^2+e^{10} \left (x^2+5 x\right ) \log ^4(x+5)+e^5 \left (-2 x^3-12 x^2-10 x\right ) \log ^2(x+5)+5 x} \, dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {2 e^5 \left (\left (x^2+4 x-5\right ) \left (x-e^5 \log ^2(x+5)+1\right )-x \log \left (\frac {3 e^x}{x}\right ) \left (x-2 e^5 \log (x+5)+5\right )\right )}{x (x+5) \left (x-e^5 \log ^2(x+5)+1\right )^2}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle 2 e^5 \int -\frac {x \log \left (\frac {3 e^x}{x}\right ) \left (x-2 e^5 \log (x+5)+5\right )+\left (-x^2-4 x+5\right ) \left (-e^5 \log ^2(x+5)+x+1\right )}{x (x+5) \left (-e^5 \log ^2(x+5)+x+1\right )^2}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -2 e^5 \int \frac {x \log \left (\frac {3 e^x}{x}\right ) \left (x-2 e^5 \log (x+5)+5\right )+\left (-x^2-4 x+5\right ) \left (-e^5 \log ^2(x+5)+x+1\right )}{x (x+5) \left (-e^5 \log ^2(x+5)+x+1\right )^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle -2 e^5 \int \left (\frac {\log \left (\frac {3 e^x}{x}\right ) \left (x-2 e^5 \log (x+5)+5\right )}{(x+5) \left (-e^5 \log ^2(x+5)+x+1\right )^2}-\frac {x-1}{x \left (-e^5 \log ^2(x+5)+x+1\right )}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -2 e^5 \left (\int \frac {\log \left (\frac {3 e^x}{x}\right )}{\left (-e^5 \log ^2(x+5)+x+1\right )^2}dx-2 e^5 \int \frac {\log \left (\frac {3 e^x}{x}\right ) \log (x+5)}{(x+5) \left (-e^5 \log ^2(x+5)+x+1\right )^2}dx-\int \frac {1}{-e^5 \log ^2(x+5)+x+1}dx+\int \frac {1}{x \left (-e^5 \log ^2(x+5)+x+1\right )}dx\right )\) |
Input:
Int[(E^5*(-10 - 2*x + 10*x^2 + 2*x^3) + E^10*(10 - 8*x - 2*x^2)*Log[5 + x] ^2 + Log[(3*E^x)/x]*(E^5*(-10*x - 2*x^2) + 4*E^10*x*Log[5 + x]))/(5*x + 11 *x^2 + 7*x^3 + x^4 + E^5*(-10*x - 12*x^2 - 2*x^3)*Log[5 + x]^2 + E^10*(5*x + x^2)*Log[5 + x]^4),x]
Output:
$Aborted
Time = 53.03 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.07
| method | result | size |
| parallelrisch | \(-\frac {2 \ln \left (\frac {{\mathrm e}^{\ln \left (3\right )+x}}{x}\right ) {\mathrm e}^{5}}{{\mathrm e}^{5} \ln \left (5+x \right )^{2}-x -1}\) | \(31\) |
| risch | \(-\frac {2 \,{\mathrm e}^{5} \ln \left (3 \,{\mathrm e}^{x}\right )}{{\mathrm e}^{5} \ln \left (5+x \right )^{2}-x -1}+\frac {{\mathrm e}^{5} \left (i \pi \,\operatorname {csgn}\left (\frac {i}{x}\right ) \operatorname {csgn}\left (i {\mathrm e}^{x}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{x}}{x}\right )-i \pi \,\operatorname {csgn}\left (\frac {i}{x}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{x}}{x}\right )^{2}-i \pi \,\operatorname {csgn}\left (i {\mathrm e}^{x}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{x}}{x}\right )^{2}+i \pi \operatorname {csgn}\left (\frac {i {\mathrm e}^{x}}{x}\right )^{3}+2 \ln \left (x \right )\right )}{{\mathrm e}^{5} \ln \left (5+x \right )^{2}-x -1}\) | \(135\) |
Input:
int(((4*x*exp(5)^2*ln(5+x)+(-2*x^2-10*x)*exp(5))*ln(exp(ln(3)+x)/x)+(-2*x^ 2-8*x+10)*exp(5)^2*ln(5+x)^2+(2*x^3+10*x^2-2*x-10)*exp(5))/((x^2+5*x)*exp( 5)^2*ln(5+x)^4+(-2*x^3-12*x^2-10*x)*exp(5)*ln(5+x)^2+x^4+7*x^3+11*x^2+5*x) ,x,method=_RETURNVERBOSE)
Output:
-2*ln(exp(ln(3)+x)/x)*exp(5)/(exp(5)*ln(5+x)^2-x-1)
Time = 0.10 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.03 \[ \int \frac {e^5 \left (-10-2 x+10 x^2+2 x^3\right )+e^{10} \left (10-8 x-2 x^2\right ) \log ^2(5+x)+\log \left (\frac {3 e^x}{x}\right ) \left (e^5 \left (-10 x-2 x^2\right )+4 e^{10} x \log (5+x)\right )}{5 x+11 x^2+7 x^3+x^4+e^5 \left (-10 x-12 x^2-2 x^3\right ) \log ^2(5+x)+e^{10} \left (5 x+x^2\right ) \log ^4(5+x)} \, dx=-\frac {2 \, e^{5} \log \left (\frac {e^{\left (x + \log \left (3\right )\right )}}{x}\right )}{e^{5} \log \left (x + 5\right )^{2} - x - 1} \] Input:
integrate(((4*x*exp(5)^2*log(5+x)+(-2*x^2-10*x)*exp(5))*log(exp(log(3)+x)/ x)+(-2*x^2-8*x+10)*exp(5)^2*log(5+x)^2+(2*x^3+10*x^2-2*x-10)*exp(5))/((x^2 +5*x)*exp(5)^2*log(5+x)^4+(-2*x^3-12*x^2-10*x)*exp(5)*log(5+x)^2+x^4+7*x^3 +11*x^2+5*x),x, algorithm="fricas")
Output:
-2*e^5*log(e^(x + log(3))/x)/(e^5*log(x + 5)^2 - x - 1)
Exception generated. \[ \int \frac {e^5 \left (-10-2 x+10 x^2+2 x^3\right )+e^{10} \left (10-8 x-2 x^2\right ) \log ^2(5+x)+\log \left (\frac {3 e^x}{x}\right ) \left (e^5 \left (-10 x-2 x^2\right )+4 e^{10} x \log (5+x)\right )}{5 x+11 x^2+7 x^3+x^4+e^5 \left (-10 x-12 x^2-2 x^3\right ) \log ^2(5+x)+e^{10} \left (5 x+x^2\right ) \log ^4(5+x)} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(((4*x*exp(5)**2*ln(5+x)+(-2*x**2-10*x)*exp(5))*ln(exp(ln(3)+x)/x )+(-2*x**2-8*x+10)*exp(5)**2*ln(5+x)**2+(2*x**3+10*x**2-2*x-10)*exp(5))/(( x**2+5*x)*exp(5)**2*ln(5+x)**4+(-2*x**3-12*x**2-10*x)*exp(5)*ln(5+x)**2+x* *4+7*x**3+11*x**2+5*x),x)
Output:
Exception raised: TypeError >> '>' not supported between instances of 'Pol y' and 'int'
Time = 0.17 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.17 \[ \int \frac {e^5 \left (-10-2 x+10 x^2+2 x^3\right )+e^{10} \left (10-8 x-2 x^2\right ) \log ^2(5+x)+\log \left (\frac {3 e^x}{x}\right ) \left (e^5 \left (-10 x-2 x^2\right )+4 e^{10} x \log (5+x)\right )}{5 x+11 x^2+7 x^3+x^4+e^5 \left (-10 x-12 x^2-2 x^3\right ) \log ^2(5+x)+e^{10} \left (5 x+x^2\right ) \log ^4(5+x)} \, dx=-\frac {2 \, {\left (x e^{5} + e^{5} \log \left (3\right ) - e^{5} \log \left (x\right )\right )}}{e^{5} \log \left (x + 5\right )^{2} - x - 1} \] Input:
integrate(((4*x*exp(5)^2*log(5+x)+(-2*x^2-10*x)*exp(5))*log(exp(log(3)+x)/ x)+(-2*x^2-8*x+10)*exp(5)^2*log(5+x)^2+(2*x^3+10*x^2-2*x-10)*exp(5))/((x^2 +5*x)*exp(5)^2*log(5+x)^4+(-2*x^3-12*x^2-10*x)*exp(5)*log(5+x)^2+x^4+7*x^3 +11*x^2+5*x),x, algorithm="maxima")
Output:
-2*(x*e^5 + e^5*log(3) - e^5*log(x))/(e^5*log(x + 5)^2 - x - 1)
Time = 0.31 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.17 \[ \int \frac {e^5 \left (-10-2 x+10 x^2+2 x^3\right )+e^{10} \left (10-8 x-2 x^2\right ) \log ^2(5+x)+\log \left (\frac {3 e^x}{x}\right ) \left (e^5 \left (-10 x-2 x^2\right )+4 e^{10} x \log (5+x)\right )}{5 x+11 x^2+7 x^3+x^4+e^5 \left (-10 x-12 x^2-2 x^3\right ) \log ^2(5+x)+e^{10} \left (5 x+x^2\right ) \log ^4(5+x)} \, dx=-\frac {4 \, {\left (x e^{5} + e^{5} \log \left (3\right ) - e^{5} \log \left (x\right )\right )}}{e^{5} \log \left (x + 5\right )^{2} - x - 1} \] Input:
integrate(((4*x*exp(5)^2*log(5+x)+(-2*x^2-10*x)*exp(5))*log(exp(log(3)+x)/ x)+(-2*x^2-8*x+10)*exp(5)^2*log(5+x)^2+(2*x^3+10*x^2-2*x-10)*exp(5))/((x^2 +5*x)*exp(5)^2*log(5+x)^4+(-2*x^3-12*x^2-10*x)*exp(5)*log(5+x)^2+x^4+7*x^3 +11*x^2+5*x),x, algorithm="giac")
Output:
-4*(x*e^5 + e^5*log(3) - e^5*log(x))/(e^5*log(x + 5)^2 - x - 1)
Time = 3.67 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.93 \[ \int \frac {e^5 \left (-10-2 x+10 x^2+2 x^3\right )+e^{10} \left (10-8 x-2 x^2\right ) \log ^2(5+x)+\log \left (\frac {3 e^x}{x}\right ) \left (e^5 \left (-10 x-2 x^2\right )+4 e^{10} x \log (5+x)\right )}{5 x+11 x^2+7 x^3+x^4+e^5 \left (-10 x-12 x^2-2 x^3\right ) \log ^2(5+x)+e^{10} \left (5 x+x^2\right ) \log ^4(5+x)} \, dx=\frac {2\,{\mathrm {e}}^5\,\ln \left (\frac {3\,{\mathrm {e}}^x}{x}\right )}{-{\mathrm {e}}^5\,{\ln \left (x+5\right )}^2+x+1} \] Input:
int(-(exp(5)*(2*x - 10*x^2 - 2*x^3 + 10) + log(exp(x + log(3))/x)*(exp(5)* (10*x + 2*x^2) - 4*x*log(x + 5)*exp(10)) + log(x + 5)^2*exp(10)*(8*x + 2*x ^2 - 10))/(5*x + 11*x^2 + 7*x^3 + x^4 - log(x + 5)^2*exp(5)*(10*x + 12*x^2 + 2*x^3) + log(x + 5)^4*exp(10)*(5*x + x^2)),x)
Output:
(2*exp(5)*log((3*exp(x))/x))/(x - log(x + 5)^2*exp(5) + 1)
Time = 0.21 (sec) , antiderivative size = 101, normalized size of antiderivative = 3.48 \[ \int \frac {e^5 \left (-10-2 x+10 x^2+2 x^3\right )+e^{10} \left (10-8 x-2 x^2\right ) \log ^2(5+x)+\log \left (\frac {3 e^x}{x}\right ) \left (e^5 \left (-10 x-2 x^2\right )+4 e^{10} x \log (5+x)\right )}{5 x+11 x^2+7 x^3+x^4+e^5 \left (-10 x-12 x^2-2 x^3\right ) \log ^2(5+x)+e^{10} \left (5 x+x^2\right ) \log ^4(5+x)} \, dx=\frac {2 e^{5} \left (-\mathrm {log}\left (x +5\right )^{2} \mathrm {log}\left (\frac {3 e^{x}}{x}\right ) e^{5}-\mathrm {log}\left (x +5\right )^{2} \mathrm {log}\left (x \right ) e^{5}+\mathrm {log}\left (x +5\right )^{2} e^{5} x -\mathrm {log}\left (x +5\right )^{2} e^{5}+\mathrm {log}\left (\frac {3 e^{x}}{x}\right ) x +\mathrm {log}\left (x \right ) x +\mathrm {log}\left (x \right )-x^{2}+1\right )}{\mathrm {log}\left (x +5\right )^{2} e^{5}-x -1} \] Input:
int(((4*x*exp(5)^2*log(5+x)+(-2*x^2-10*x)*exp(5))*log(exp(log(3)+x)/x)+(-2 *x^2-8*x+10)*exp(5)^2*log(5+x)^2+(2*x^3+10*x^2-2*x-10)*exp(5))/((x^2+5*x)* exp(5)^2*log(5+x)^4+(-2*x^3-12*x^2-10*x)*exp(5)*log(5+x)^2+x^4+7*x^3+11*x^ 2+5*x),x)
Output:
(2*e**5*( - log(x + 5)**2*log((3*e**x)/x)*e**5 - log(x + 5)**2*log(x)*e**5 + log(x + 5)**2*e**5*x - log(x + 5)**2*e**5 + log((3*e**x)/x)*x + log(x)* x + log(x) - x**2 + 1))/(log(x + 5)**2*e**5 - x - 1)