Integrand size = 120, antiderivative size = 27 \[ \int \frac {15+e^{2 x} \left (15+5 x^2\right )+e^x \left (-30-5 x^2-5 x^3\right )}{\left (3 x+x^2 \log (3)+e^x \left (-6 x+x^3-2 x^2 \log (3)\right )+e^{2 x} \left (3 x-x^3+x^2 \log (3)\right )\right ) \log ^2\left (\frac {-3-x \log (3)+e^x \left (3-x^2+x \log (3)\right )}{-x+e^x x}\right )} \, dx=\frac {5}{\log \left (\frac {3}{x}-x+\frac {x}{1-e^x}+\log (3)\right )} \] Output:
5/ln(ln(3)+x/(1-exp(x))+3/x-x)
Time = 0.07 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.41 \[ \int \frac {15+e^{2 x} \left (15+5 x^2\right )+e^x \left (-30-5 x^2-5 x^3\right )}{\left (3 x+x^2 \log (3)+e^x \left (-6 x+x^3-2 x^2 \log (3)\right )+e^{2 x} \left (3 x-x^3+x^2 \log (3)\right )\right ) \log ^2\left (\frac {-3-x \log (3)+e^x \left (3-x^2+x \log (3)\right )}{-x+e^x x}\right )} \, dx=\frac {5}{\log \left (\frac {-3-x \log (3)+e^x \left (3-x^2+x \log (3)\right )}{\left (-1+e^x\right ) x}\right )} \] Input:
Integrate[(15 + E^(2*x)*(15 + 5*x^2) + E^x*(-30 - 5*x^2 - 5*x^3))/((3*x + x^2*Log[3] + E^x*(-6*x + x^3 - 2*x^2*Log[3]) + E^(2*x)*(3*x - x^3 + x^2*Lo g[3]))*Log[(-3 - x*Log[3] + E^x*(3 - x^2 + x*Log[3]))/(-x + E^x*x)]^2),x]
Output:
5/Log[(-3 - x*Log[3] + E^x*(3 - x^2 + x*Log[3]))/((-1 + E^x)*x)]
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.
| \(\displaystyle \int \frac {e^{2 x} \left (5 x^2+15\right )+e^x \left (-5 x^3-5 x^2-30\right )+15}{\left (x^2 \log (3)+e^x \left (x^3-2 x^2 \log (3)-6 x\right )+e^{2 x} \left (-x^3+x^2 \log (3)+3 x\right )+3 x\right ) \log ^2\left (\frac {e^x \left (-x^2+x \log (3)+3\right )+x (-\log (3))-3}{e^x x-x}\right )} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {e^{2 x} \left (5 x^2+15\right )+e^x \left (-5 x^3-5 x^2-30\right )+15}{\left (1-e^x\right ) x \left (e^x x^2-3 e^x-e^x x \log (3)+x \log (3)+3\right ) \log ^2\left (\frac {e^x \left (-x^2+x \log (3)+3\right )+x (-\log (3))-3}{e^x x-x}\right )}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (-\frac {5 \left (x^2+3\right )}{x \left (x^2-x \log (3)-3\right ) \log ^2\left (\frac {e^x \left (-x^2+x \log (3)+3\right )+x (-\log (3))-3}{\left (e^x-1\right ) x}\right )}+\frac {5}{\left (e^x-1\right ) \log ^2\left (\frac {e^x \left (-x^2+x \log (3)+3\right )+x (-\log (3))-3}{\left (e^x-1\right ) x}\right )}+\frac {5 \left (x^3 (-\log (3))-x^2 \left (3-\log ^2(3)+\log (3)\right )-6 x (1-\log (3))+9\right )}{\left (-x^2+x \log (3)+3\right ) \left (e^x x^2-3 e^x-e^x x \log (3)+x \log (3)+3\right ) \log ^2\left (\frac {e^x \left (-x^2+x \log (3)+3\right )+x (-\log (3))-3}{\left (e^x-1\right ) x}\right )}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle 5 \int \frac {1}{\left (-1+e^x\right ) \log ^2\left (\frac {-\log (3) x+e^x \left (-x^2+\log (3) x+3\right )-3}{\left (-1+e^x\right ) x}\right )}dx+5 \int \frac {1}{x \log ^2\left (\frac {-\log (3) x+e^x \left (-x^2+\log (3) x+3\right )-3}{\left (-1+e^x\right ) x}\right )}dx+5 (3+\log (3)) \int \frac {1}{\left (e^x x^2-e^x \log (3) x+\log (3) x-3 e^x+3\right ) \log ^2\left (\frac {-\log (3) x+e^x \left (-x^2+\log (3) x+3\right )-3}{\left (-1+e^x\right ) x}\right )}dx+5 \log (3) \int \frac {x}{\left (e^x x^2-e^x \log (3) x+\log (3) x-3 e^x+3\right ) \log ^2\left (\frac {-\log (3) x+e^x \left (-x^2+\log (3) x+3\right )-3}{\left (-1+e^x\right ) x}\right )}dx-10 \left (1+\frac {\log (3)}{\sqrt {12+\log ^2(3)}}\right ) \int \frac {1}{\left (2 x-\sqrt {12+\log ^2(3)}-\log (3)\right ) \log ^2\left (\frac {-\log (3) x+e^x \left (-x^2+\log (3) x+3\right )-3}{\left (-1+e^x\right ) x}\right )}dx+5 \left (6+\log ^2(3)\right ) \left (1+\frac {\log (3)}{\sqrt {12+\log ^2(3)}}\right ) \int \frac {1}{\left (e^x x^2-e^x \log (3) x+\log (3) x-3 e^x+3\right ) \left (2 x-\sqrt {12+\log ^2(3)}-\log (3)\right ) \log ^2\left (\frac {-\log (3) x+e^x \left (-x^2+\log (3) x+3\right )-3}{\left (-1+e^x\right ) x}\right )}dx-10 \left (1-\frac {\log (3)}{\sqrt {12+\log ^2(3)}}\right ) \int \frac {1}{\left (2 x+\sqrt {12+\log ^2(3)}-\log (3)\right ) \log ^2\left (\frac {-\log (3) x+e^x \left (-x^2+\log (3) x+3\right )-3}{\left (-1+e^x\right ) x}\right )}dx+5 \left (6+\log ^2(3)\right ) \left (1-\frac {\log (3)}{\sqrt {12+\log ^2(3)}}\right ) \int \frac {1}{\left (e^x x^2-e^x \log (3) x+\log (3) x-3 e^x+3\right ) \left (2 x+\sqrt {12+\log ^2(3)}-\log (3)\right ) \log ^2\left (\frac {-\log (3) x+e^x \left (-x^2+\log (3) x+3\right )-3}{\left (-1+e^x\right ) x}\right )}dx-\frac {10 \log (3) \int \frac {1}{\left (-2 x+\sqrt {12+\log ^2(3)}+\log (3)\right ) \log ^2\left (\frac {-\log (3) x+e^x \left (-x^2+\log (3) x+3\right )-3}{\left (-1+e^x\right ) x}\right )}dx}{\sqrt {12+\log ^2(3)}}-\frac {10 \log (27) \int \frac {1}{\left (e^x x^2-e^x \log (3) x+\log (3) x-3 e^x+3\right ) \left (-2 x+\sqrt {12+\log ^2(3)}+\log (3)\right ) \log ^2\left (\frac {-\log (3) x+e^x \left (-x^2+\log (3) x+3\right )-3}{\left (-1+e^x\right ) x}\right )}dx}{\sqrt {12+\log ^2(3)}}-\frac {20 \log (3) \int \frac {1}{\left (4 x-\log (9)+2 \sqrt {12+\log ^2(3)}\right ) \log ^2\left (\frac {-\log (3) x+e^x \left (-x^2+\log (3) x+3\right )-3}{\left (-1+e^x\right ) x}\right )}dx}{\sqrt {12+\log ^2(3)}}-\frac {20 \log (27) \int \frac {1}{\left (e^x x^2-e^x \log (3) x+\log (3) x-3 e^x+3\right ) \left (4 x-\log (9)+2 \sqrt {12+\log ^2(3)}\right ) \log ^2\left (\frac {-\log (3) x+e^x \left (-x^2+\log (3) x+3\right )-3}{\left (-1+e^x\right ) x}\right )}dx}{\sqrt {12+\log ^2(3)}}\) |
Input:
Int[(15 + E^(2*x)*(15 + 5*x^2) + E^x*(-30 - 5*x^2 - 5*x^3))/((3*x + x^2*Lo g[3] + E^x*(-6*x + x^3 - 2*x^2*Log[3]) + E^(2*x)*(3*x - x^3 + x^2*Log[3])) *Log[(-3 - x*Log[3] + E^x*(3 - x^2 + x*Log[3]))/(-x + E^x*x)]^2),x]
Output:
$Aborted
Time = 9.87 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.37
| method | result | size |
| parallelrisch | \(\frac {5}{\ln \left (\frac {\left (x \ln \left (3\right )-x^{2}+3\right ) {\mathrm e}^{x}-x \ln \left (3\right )-3}{x \left ({\mathrm e}^{x}-1\right )}\right )}\) | \(37\) |
| risch | \(\frac {10 i}{\pi \,\operatorname {csgn}\left (\frac {i}{x}\right ) \operatorname {csgn}\left (\frac {i \left (-{\mathrm e}^{x} x^{2}+\ln \left (3\right ) \left ({\mathrm e}^{x}-1\right ) x +3 \,{\mathrm e}^{x}-3\right )}{{\mathrm e}^{x}-1}\right ) \operatorname {csgn}\left (\frac {i \left (-{\mathrm e}^{x} x^{2}+\ln \left (3\right ) \left ({\mathrm e}^{x}-1\right ) x +3 \,{\mathrm e}^{x}-3\right )}{x \left ({\mathrm e}^{x}-1\right )}\right )-\pi \,\operatorname {csgn}\left (\frac {i}{x}\right ) {\operatorname {csgn}\left (\frac {i \left (-{\mathrm e}^{x} x^{2}+\ln \left (3\right ) \left ({\mathrm e}^{x}-1\right ) x +3 \,{\mathrm e}^{x}-3\right )}{x \left ({\mathrm e}^{x}-1\right )}\right )}^{2}+\pi \,\operatorname {csgn}\left (\frac {i}{{\mathrm e}^{x}-1}\right ) \operatorname {csgn}\left (i \left (-{\mathrm e}^{x} x^{2}+\ln \left (3\right ) \left ({\mathrm e}^{x}-1\right ) x +3 \,{\mathrm e}^{x}-3\right )\right ) \operatorname {csgn}\left (\frac {i \left (-{\mathrm e}^{x} x^{2}+\ln \left (3\right ) \left ({\mathrm e}^{x}-1\right ) x +3 \,{\mathrm e}^{x}-3\right )}{{\mathrm e}^{x}-1}\right )-\pi \,\operatorname {csgn}\left (\frac {i}{{\mathrm e}^{x}-1}\right ) {\operatorname {csgn}\left (\frac {i \left (-{\mathrm e}^{x} x^{2}+\ln \left (3\right ) \left ({\mathrm e}^{x}-1\right ) x +3 \,{\mathrm e}^{x}-3\right )}{{\mathrm e}^{x}-1}\right )}^{2}-\pi \,\operatorname {csgn}\left (i \left (-{\mathrm e}^{x} x^{2}+\ln \left (3\right ) \left ({\mathrm e}^{x}-1\right ) x +3 \,{\mathrm e}^{x}-3\right )\right ) {\operatorname {csgn}\left (\frac {i \left (-{\mathrm e}^{x} x^{2}+\ln \left (3\right ) \left ({\mathrm e}^{x}-1\right ) x +3 \,{\mathrm e}^{x}-3\right )}{{\mathrm e}^{x}-1}\right )}^{2}+\pi {\operatorname {csgn}\left (\frac {i \left (-{\mathrm e}^{x} x^{2}+\ln \left (3\right ) \left ({\mathrm e}^{x}-1\right ) x +3 \,{\mathrm e}^{x}-3\right )}{{\mathrm e}^{x}-1}\right )}^{3}-\pi \,\operatorname {csgn}\left (\frac {i \left (-{\mathrm e}^{x} x^{2}+\ln \left (3\right ) \left ({\mathrm e}^{x}-1\right ) x +3 \,{\mathrm e}^{x}-3\right )}{{\mathrm e}^{x}-1}\right ) {\operatorname {csgn}\left (\frac {i \left (-{\mathrm e}^{x} x^{2}+\ln \left (3\right ) \left ({\mathrm e}^{x}-1\right ) x +3 \,{\mathrm e}^{x}-3\right )}{x \left ({\mathrm e}^{x}-1\right )}\right )}^{2}+\pi {\operatorname {csgn}\left (\frac {i \left (-{\mathrm e}^{x} x^{2}+\ln \left (3\right ) \left ({\mathrm e}^{x}-1\right ) x +3 \,{\mathrm e}^{x}-3\right )}{x \left ({\mathrm e}^{x}-1\right )}\right )}^{3}-2 i \ln \left (x \right )-2 i \ln \left ({\mathrm e}^{x}-1\right )+2 i \ln \left (-{\mathrm e}^{x} x^{2}+\ln \left (3\right ) \left ({\mathrm e}^{x}-1\right ) x +3 \,{\mathrm e}^{x}-3\right )}\) | \(483\) |
Input:
int(((5*x^2+15)*exp(x)^2+(-5*x^3-5*x^2-30)*exp(x)+15)/((x^2*ln(3)-x^3+3*x) *exp(x)^2+(-2*x^2*ln(3)+x^3-6*x)*exp(x)+x^2*ln(3)+3*x)/ln(((x*ln(3)-x^2+3) *exp(x)-x*ln(3)-3)/(exp(x)*x-x))^2,x,method=_RETURNVERBOSE)
Output:
5/ln(((x*ln(3)-x^2+3)*exp(x)-x*ln(3)-3)/x/(exp(x)-1))
Time = 0.09 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.33 \[ \int \frac {15+e^{2 x} \left (15+5 x^2\right )+e^x \left (-30-5 x^2-5 x^3\right )}{\left (3 x+x^2 \log (3)+e^x \left (-6 x+x^3-2 x^2 \log (3)\right )+e^{2 x} \left (3 x-x^3+x^2 \log (3)\right )\right ) \log ^2\left (\frac {-3-x \log (3)+e^x \left (3-x^2+x \log (3)\right )}{-x+e^x x}\right )} \, dx=\frac {5}{\log \left (-\frac {{\left (x^{2} - x \log \left (3\right ) - 3\right )} e^{x} + x \log \left (3\right ) + 3}{x e^{x} - x}\right )} \] Input:
integrate(((5*x^2+15)*exp(x)^2+(-5*x^3-5*x^2-30)*exp(x)+15)/((x^2*log(3)-x ^3+3*x)*exp(x)^2+(-2*x^2*log(3)+x^3-6*x)*exp(x)+x^2*log(3)+3*x)/log(((x*lo g(3)-x^2+3)*exp(x)-x*log(3)-3)/(exp(x)*x-x))^2,x, algorithm="fricas")
Output:
5/log(-((x^2 - x*log(3) - 3)*e^x + x*log(3) + 3)/(x*e^x - x))
Time = 0.31 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07 \[ \int \frac {15+e^{2 x} \left (15+5 x^2\right )+e^x \left (-30-5 x^2-5 x^3\right )}{\left (3 x+x^2 \log (3)+e^x \left (-6 x+x^3-2 x^2 \log (3)\right )+e^{2 x} \left (3 x-x^3+x^2 \log (3)\right )\right ) \log ^2\left (\frac {-3-x \log (3)+e^x \left (3-x^2+x \log (3)\right )}{-x+e^x x}\right )} \, dx=\frac {5}{\log {\left (\frac {- x \log {\left (3 \right )} + \left (- x^{2} + x \log {\left (3 \right )} + 3\right ) e^{x} - 3}{x e^{x} - x} \right )}} \] Input:
integrate(((5*x**2+15)*exp(x)**2+(-5*x**3-5*x**2-30)*exp(x)+15)/((x**2*ln( 3)-x**3+3*x)*exp(x)**2+(-2*x**2*ln(3)+x**3-6*x)*exp(x)+x**2*ln(3)+3*x)/ln( ((x*ln(3)-x**2+3)*exp(x)-x*ln(3)-3)/(exp(x)*x-x))**2,x)
Output:
5/log((-x*log(3) + (-x**2 + x*log(3) + 3)*exp(x) - 3)/(x*exp(x) - x))
Time = 1.50 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.41 \[ \int \frac {15+e^{2 x} \left (15+5 x^2\right )+e^x \left (-30-5 x^2-5 x^3\right )}{\left (3 x+x^2 \log (3)+e^x \left (-6 x+x^3-2 x^2 \log (3)\right )+e^{2 x} \left (3 x-x^3+x^2 \log (3)\right )\right ) \log ^2\left (\frac {-3-x \log (3)+e^x \left (3-x^2+x \log (3)\right )}{-x+e^x x}\right )} \, dx=\frac {5}{\log \left (-{\left (x^{2} - x \log \left (3\right ) - 3\right )} e^{x} - x \log \left (3\right ) - 3\right ) - \log \left (x\right ) - \log \left (e^{x} - 1\right )} \] Input:
integrate(((5*x^2+15)*exp(x)^2+(-5*x^3-5*x^2-30)*exp(x)+15)/((x^2*log(3)-x ^3+3*x)*exp(x)^2+(-2*x^2*log(3)+x^3-6*x)*exp(x)+x^2*log(3)+3*x)/log(((x*lo g(3)-x^2+3)*exp(x)-x*log(3)-3)/(exp(x)*x-x))^2,x, algorithm="maxima")
Output:
5/(log(-(x^2 - x*log(3) - 3)*e^x - x*log(3) - 3) - log(x) - log(e^x - 1))
Time = 0.45 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.52 \[ \int \frac {15+e^{2 x} \left (15+5 x^2\right )+e^x \left (-30-5 x^2-5 x^3\right )}{\left (3 x+x^2 \log (3)+e^x \left (-6 x+x^3-2 x^2 \log (3)\right )+e^{2 x} \left (3 x-x^3+x^2 \log (3)\right )\right ) \log ^2\left (\frac {-3-x \log (3)+e^x \left (3-x^2+x \log (3)\right )}{-x+e^x x}\right )} \, dx=\frac {5}{\log \left (-x^{2} e^{x} + x e^{x} \log \left (3\right ) - x \log \left (3\right ) + 3 \, e^{x} - 3\right ) - \log \left (x e^{x} - x\right )} \] Input:
integrate(((5*x^2+15)*exp(x)^2+(-5*x^3-5*x^2-30)*exp(x)+15)/((x^2*log(3)-x ^3+3*x)*exp(x)^2+(-2*x^2*log(3)+x^3-6*x)*exp(x)+x^2*log(3)+3*x)/log(((x*lo g(3)-x^2+3)*exp(x)-x*log(3)-3)/(exp(x)*x-x))^2,x, algorithm="giac")
Output:
5/(log(-x^2*e^x + x*e^x*log(3) - x*log(3) + 3*e^x - 3) - log(x*e^x - x))
Time = 2.45 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.33 \[ \int \frac {15+e^{2 x} \left (15+5 x^2\right )+e^x \left (-30-5 x^2-5 x^3\right )}{\left (3 x+x^2 \log (3)+e^x \left (-6 x+x^3-2 x^2 \log (3)\right )+e^{2 x} \left (3 x-x^3+x^2 \log (3)\right )\right ) \log ^2\left (\frac {-3-x \log (3)+e^x \left (3-x^2+x \log (3)\right )}{-x+e^x x}\right )} \, dx=\frac {5}{\ln \left (\frac {x\,\ln \left (3\right )-{\mathrm {e}}^x\,\left (-x^2+\ln \left (3\right )\,x+3\right )+3}{x-x\,{\mathrm {e}}^x}\right )} \] Input:
int((exp(2*x)*(5*x^2 + 15) - exp(x)*(5*x^2 + 5*x^3 + 30) + 15)/(log((x*log (3) - exp(x)*(x*log(3) - x^2 + 3) + 3)/(x - x*exp(x)))^2*(3*x - exp(x)*(6* x + 2*x^2*log(3) - x^3) + x^2*log(3) + exp(2*x)*(3*x + x^2*log(3) - x^3))) ,x)
Output:
5/log((x*log(3) - exp(x)*(x*log(3) - x^2 + 3) + 3)/(x - x*exp(x)))
Time = 0.23 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.63 \[ \int \frac {15+e^{2 x} \left (15+5 x^2\right )+e^x \left (-30-5 x^2-5 x^3\right )}{\left (3 x+x^2 \log (3)+e^x \left (-6 x+x^3-2 x^2 \log (3)\right )+e^{2 x} \left (3 x-x^3+x^2 \log (3)\right )\right ) \log ^2\left (\frac {-3-x \log (3)+e^x \left (3-x^2+x \log (3)\right )}{-x+e^x x}\right )} \, dx=\frac {5}{\mathrm {log}\left (\frac {e^{x} \mathrm {log}\left (3\right ) x -e^{x} x^{2}+3 e^{x}-\mathrm {log}\left (3\right ) x -3}{e^{x} x -x}\right )} \] Input:
int(((5*x^2+15)*exp(x)^2+(-5*x^3-5*x^2-30)*exp(x)+15)/((x^2*log(3)-x^3+3*x )*exp(x)^2+(-2*x^2*log(3)+x^3-6*x)*exp(x)+x^2*log(3)+3*x)/log(((x*log(3)-x ^2+3)*exp(x)-x*log(3)-3)/(exp(x)*x-x))^2,x)
Output:
5/log((e**x*log(3)*x - e**x*x**2 + 3*e**x - log(3)*x - 3)/(e**x*x - x))