Integrand size = 108, antiderivative size = 27 \[ \int \frac {e^{-2+2 x+\frac {-10+2 e^{-2+2 x} x^3-2 \log (3)-5 \log (5)}{-25+5 e^{-2+2 x} x^3-5 \log (3)}} \left (3 x^2+2 x^3\right ) \log (5)}{25+e^{-4+4 x} x^6+10 \log (3)+\log ^2(3)+e^{-2+2 x} \left (-10 x^3-2 x^3 \log (3)\right )} \, dx=e^{\frac {2}{5}+\frac {\log (5)}{5-e^{-2+2 x} x^3+\log (3)}} \] Output:
exp(2/5+ln(5)/(ln(3)-x^3*exp(-1+x)^2+5))
Time = 0.31 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.22 \[ \int \frac {e^{-2+2 x+\frac {-10+2 e^{-2+2 x} x^3-2 \log (3)-5 \log (5)}{-25+5 e^{-2+2 x} x^3-5 \log (3)}} \left (3 x^2+2 x^3\right ) \log (5)}{25+e^{-4+4 x} x^6+10 \log (3)+\log ^2(3)+e^{-2+2 x} \left (-10 x^3-2 x^3 \log (3)\right )} \, dx=5^{\frac {e^2}{-e^{2 x} x^3+e^2 (5+\log (3))}} e^{2/5} \] Input:
Integrate[(E^(-2 + 2*x + (-10 + 2*E^(-2 + 2*x)*x^3 - 2*Log[3] - 5*Log[5])/ (-25 + 5*E^(-2 + 2*x)*x^3 - 5*Log[3]))*(3*x^2 + 2*x^3)*Log[5])/(25 + E^(-4 + 4*x)*x^6 + 10*Log[3] + Log[3]^2 + E^(-2 + 2*x)*(-10*x^3 - 2*x^3*Log[3]) ),x]
Output:
5^(E^2/(-(E^(2*x)*x^3) + E^2*(5 + Log[3])))*E^(2/5)
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.
| \(\displaystyle \int \frac {\left (2 x^3+3 x^2\right ) \log (5) \exp \left (\frac {2 e^{2 x-2} x^3-10-5 \log (5)-2 \log (3)}{5 e^{2 x-2} x^3-25-5 \log (3)}+2 x-2\right )}{e^{4 x-4} x^6+e^{2 x-2} \left (-10 x^3-2 x^3 \log (3)\right )+25+\log ^2(3)+10 \log (3)} \, dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \log (5) \int \frac {\exp \left (2 x+\frac {-2 e^{2 x-2} x^3+\log (28125)+10}{5 \left (-e^{2 x-2} x^3+\log (3)+5\right )}-2\right ) \left (2 x^3+3 x^2\right )}{e^{4 x-4} x^6-2 e^{2 x-2} \left (\log (3) x^3+5 x^3\right )+(5+\log (3))^2}dx\) |
| \(\Big \downarrow \) 2027 |
| \(\displaystyle \log (5) \int \frac {\exp \left (2 x+\frac {-2 e^{2 x-2} x^3+\log (28125)+10}{5 \left (-e^{2 x-2} x^3+\log (3)+5\right )}-2\right ) x^2 (2 x+3)}{e^{4 x-4} x^6-2 e^{2 x-2} \left (\log (3) x^3+5 x^3\right )+(5+\log (3))^2}dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \log (5) \int \frac {\exp \left (2 x+\frac {-2 e^{2 x-2} x^3+\log (28125)+10}{5 \left (-e^{2 x-2} x^3+\log (3)+5\right )}+2\right ) x^2 (2 x+3)}{\left (e^{2 x} x^3-5 e^2 \left (1+\frac {\log (3)}{5}\right )\right )^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \log (5) \int \left (\frac {2 \exp \left (2 x+\frac {-2 e^{2 x-2} x^3+\log (28125)+10}{5 \left (-e^{2 x-2} x^3+\log (3)+5\right )}+2\right ) x^3}{\left (e^{2 x} x^3-5 e^2 \left (1+\frac {\log (3)}{5}\right )\right )^2}+\frac {3 \exp \left (2 x+\frac {-2 e^{2 x-2} x^3+\log (28125)+10}{5 \left (-e^{2 x-2} x^3+\log (3)+5\right )}+2\right ) x^2}{\left (e^{2 x} x^3-5 e^2 \left (1+\frac {\log (3)}{5}\right )\right )^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \log (5) \left (2 \int \frac {\exp \left (2 x+\frac {-2 e^{2 x-2} x^3+\log (28125)+10}{5 \left (-e^{2 x-2} x^3+\log (3)+5\right )}+2\right ) x^3}{\left (e^{2 x} x^3-5 e^2 \left (1+\frac {\log (3)}{5}\right )\right )^2}dx+3 \int \frac {\exp \left (2 x+\frac {-2 e^{2 x-2} x^3+\log (28125)+10}{5 \left (-e^{2 x-2} x^3+\log (3)+5\right )}+2\right ) x^2}{\left (e^{2 x} x^3-5 e^2 \left (1+\frac {\log (3)}{5}\right )\right )^2}dx\right )\) |
Input:
Int[(E^(-2 + 2*x + (-10 + 2*E^(-2 + 2*x)*x^3 - 2*Log[3] - 5*Log[5])/(-25 + 5*E^(-2 + 2*x)*x^3 - 5*Log[3]))*(3*x^2 + 2*x^3)*Log[5])/(25 + E^(-4 + 4*x )*x^6 + 10*Log[3] + Log[3]^2 + E^(-2 + 2*x)*(-10*x^3 - 2*x^3*Log[3])),x]
Output:
$Aborted
Time = 174.08 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85
| method | result | size |
| risch | \(5^{\frac {1}{-x^{3} {\mathrm e}^{-2+2 x}+\ln \left (3\right )+5}} {\mathrm e}^{\frac {2}{5}}\) | \(23\) |
| parallelrisch | \({\mathrm e}^{\frac {2 x^{3} {\mathrm e}^{-2+2 x}-5 \ln \left (5\right )-2 \ln \left (3\right )-10}{5 x^{3} {\mathrm e}^{-2+2 x}-5 \ln \left (3\right )-25}}\) | \(43\) |
Input:
int((2*x^3+3*x^2)*ln(5)*exp(-1+x)^2*exp((2*x^3*exp(-1+x)^2-5*ln(5)-2*ln(3) -10)/(5*x^3*exp(-1+x)^2-5*ln(3)-25))/(x^6*exp(-1+x)^4+(-2*x^3*ln(3)-10*x^3 )*exp(-1+x)^2+ln(3)^2+10*ln(3)+25),x,method=_RETURNVERBOSE)
Output:
5^(1/(-x^3*exp(-2+2*x)+ln(3)+5))*exp(2/5)
Leaf count of result is larger than twice the leaf count of optimal. 63 vs. \(2 (25) = 50\).
Time = 0.10 (sec) , antiderivative size = 63, normalized size of antiderivative = 2.33 \[ \int \frac {e^{-2+2 x+\frac {-10+2 e^{-2+2 x} x^3-2 \log (3)-5 \log (5)}{-25+5 e^{-2+2 x} x^3-5 \log (3)}} \left (3 x^2+2 x^3\right ) \log (5)}{25+e^{-4+4 x} x^6+10 \log (3)+\log ^2(3)+e^{-2+2 x} \left (-10 x^3-2 x^3 \log (3)\right )} \, dx=e^{\left (-2 \, x + \frac {2 \, {\left (5 \, x^{4} - 4 \, x^{3}\right )} e^{\left (2 \, x - 2\right )} - 2 \, {\left (5 \, x - 4\right )} \log \left (3\right ) - 50 \, x - 5 \, \log \left (5\right ) + 40}{5 \, {\left (x^{3} e^{\left (2 \, x - 2\right )} - \log \left (3\right ) - 5\right )}} + 2\right )} \] Input:
integrate((2*x^3+3*x^2)*log(5)*exp(-1+x)^2*exp((2*x^3*exp(-1+x)^2-5*log(5) -2*log(3)-10)/(5*x^3*exp(-1+x)^2-5*log(3)-25))/(x^6*exp(-1+x)^4+(-2*x^3*lo g(3)-10*x^3)*exp(-1+x)^2+log(3)^2+10*log(3)+25),x, algorithm="fricas")
Output:
e^(-2*x + 1/5*(2*(5*x^4 - 4*x^3)*e^(2*x - 2) - 2*(5*x - 4)*log(3) - 50*x - 5*log(5) + 40)/(x^3*e^(2*x - 2) - log(3) - 5) + 2)
Time = 0.28 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.56 \[ \int \frac {e^{-2+2 x+\frac {-10+2 e^{-2+2 x} x^3-2 \log (3)-5 \log (5)}{-25+5 e^{-2+2 x} x^3-5 \log (3)}} \left (3 x^2+2 x^3\right ) \log (5)}{25+e^{-4+4 x} x^6+10 \log (3)+\log ^2(3)+e^{-2+2 x} \left (-10 x^3-2 x^3 \log (3)\right )} \, dx=e^{\frac {2 x^{3} e^{2 x - 2} - 10 - 5 \log {\left (5 \right )} - 2 \log {\left (3 \right )}}{5 x^{3} e^{2 x - 2} - 25 - 5 \log {\left (3 \right )}}} \] Input:
integrate((2*x**3+3*x**2)*ln(5)*exp(-1+x)**2*exp((2*x**3*exp(-1+x)**2-5*ln (5)-2*ln(3)-10)/(5*x**3*exp(-1+x)**2-5*ln(3)-25))/(x**6*exp(-1+x)**4+(-2*x **3*ln(3)-10*x**3)*exp(-1+x)**2+ln(3)**2+10*ln(3)+25),x)
Output:
exp((2*x**3*exp(2*x - 2) - 10 - 5*log(5) - 2*log(3))/(5*x**3*exp(2*x - 2) - 25 - 5*log(3)))
Time = 0.37 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.11 \[ \int \frac {e^{-2+2 x+\frac {-10+2 e^{-2+2 x} x^3-2 \log (3)-5 \log (5)}{-25+5 e^{-2+2 x} x^3-5 \log (3)}} \left (3 x^2+2 x^3\right ) \log (5)}{25+e^{-4+4 x} x^6+10 \log (3)+\log ^2(3)+e^{-2+2 x} \left (-10 x^3-2 x^3 \log (3)\right )} \, dx=e^{\left (-\frac {e^{2} \log \left (5\right )}{x^{3} e^{\left (2 \, x\right )} - e^{2} \log \left (3\right ) - 5 \, e^{2}} + \frac {2}{5}\right )} \] Input:
integrate((2*x^3+3*x^2)*log(5)*exp(-1+x)^2*exp((2*x^3*exp(-1+x)^2-5*log(5) -2*log(3)-10)/(5*x^3*exp(-1+x)^2-5*log(3)-25))/(x^6*exp(-1+x)^4+(-2*x^3*lo g(3)-10*x^3)*exp(-1+x)^2+log(3)^2+10*log(3)+25),x, algorithm="maxima")
Output:
e^(-e^2*log(5)/(x^3*e^(2*x) - e^2*log(3) - 5*e^2) + 2/5)
\[ \int \frac {e^{-2+2 x+\frac {-10+2 e^{-2+2 x} x^3-2 \log (3)-5 \log (5)}{-25+5 e^{-2+2 x} x^3-5 \log (3)}} \left (3 x^2+2 x^3\right ) \log (5)}{25+e^{-4+4 x} x^6+10 \log (3)+\log ^2(3)+e^{-2+2 x} \left (-10 x^3-2 x^3 \log (3)\right )} \, dx=\int { \frac {{\left (2 \, x^{3} + 3 \, x^{2}\right )} e^{\left (2 \, x + \frac {2 \, x^{3} e^{\left (2 \, x - 2\right )} - 5 \, \log \left (5\right ) - 2 \, \log \left (3\right ) - 10}{5 \, {\left (x^{3} e^{\left (2 \, x - 2\right )} - \log \left (3\right ) - 5\right )}} - 2\right )} \log \left (5\right )}{x^{6} e^{\left (4 \, x - 4\right )} - 2 \, {\left (x^{3} \log \left (3\right ) + 5 \, x^{3}\right )} e^{\left (2 \, x - 2\right )} + \log \left (3\right )^{2} + 10 \, \log \left (3\right ) + 25} \,d x } \] Input:
integrate((2*x^3+3*x^2)*log(5)*exp(-1+x)^2*exp((2*x^3*exp(-1+x)^2-5*log(5) -2*log(3)-10)/(5*x^3*exp(-1+x)^2-5*log(3)-25))/(x^6*exp(-1+x)^4+(-2*x^3*lo g(3)-10*x^3)*exp(-1+x)^2+log(3)^2+10*log(3)+25),x, algorithm="giac")
Output:
integrate((2*x^3 + 3*x^2)*e^(2*x + 1/5*(2*x^3*e^(2*x - 2) - 5*log(5) - 2*l og(3) - 10)/(x^3*e^(2*x - 2) - log(3) - 5) - 2)*log(5)/(x^6*e^(4*x - 4) - 2*(x^3*log(3) + 5*x^3)*e^(2*x - 2) + log(3)^2 + 10*log(3) + 25), x)
Timed out. \[ \int \frac {e^{-2+2 x+\frac {-10+2 e^{-2+2 x} x^3-2 \log (3)-5 \log (5)}{-25+5 e^{-2+2 x} x^3-5 \log (3)}} \left (3 x^2+2 x^3\right ) \log (5)}{25+e^{-4+4 x} x^6+10 \log (3)+\log ^2(3)+e^{-2+2 x} \left (-10 x^3-2 x^3 \log (3)\right )} \, dx=\int \frac {{\mathrm {e}}^{2\,x-2}\,{\mathrm {e}}^{\frac {2\,\ln \left (3\right )+5\,\ln \left (5\right )-2\,x^3\,{\mathrm {e}}^{2\,x-2}+10}{5\,\ln \left (3\right )-5\,x^3\,{\mathrm {e}}^{2\,x-2}+25}}\,\ln \left (5\right )\,\left (2\,x^3+3\,x^2\right )}{10\,\ln \left (3\right )-{\mathrm {e}}^{2\,x-2}\,\left (2\,x^3\,\ln \left (3\right )+10\,x^3\right )+{\ln \left (3\right )}^2+x^6\,{\mathrm {e}}^{4\,x-4}+25} \,d x \] Input:
int((exp(2*x - 2)*exp((2*log(3) + 5*log(5) - 2*x^3*exp(2*x - 2) + 10)/(5*l og(3) - 5*x^3*exp(2*x - 2) + 25))*log(5)*(3*x^2 + 2*x^3))/(10*log(3) - exp (2*x - 2)*(2*x^3*log(3) + 10*x^3) + log(3)^2 + x^6*exp(4*x - 4) + 25),x)
Output:
int((exp(2*x - 2)*exp((2*log(3) + 5*log(5) - 2*x^3*exp(2*x - 2) + 10)/(5*l og(3) - 5*x^3*exp(2*x - 2) + 25))*log(5)*(3*x^2 + 2*x^3))/(10*log(3) - exp (2*x - 2)*(2*x^3*log(3) + 10*x^3) + log(3)^2 + x^6*exp(4*x - 4) + 25), x)
\[ \int \frac {e^{-2+2 x+\frac {-10+2 e^{-2+2 x} x^3-2 \log (3)-5 \log (5)}{-25+5 e^{-2+2 x} x^3-5 \log (3)}} \left (3 x^2+2 x^3\right ) \log (5)}{25+e^{-4+4 x} x^6+10 \log (3)+\log ^2(3)+e^{-2+2 x} \left (-10 x^3-2 x^3 \log (3)\right )} \, dx =\text {Too large to display} \] Input:
int((2*x^3+3*x^2)*log(5)*exp(-1+x)^2*exp((2*x^3*exp(-1+x)^2-5*log(5)-2*log (3)-10)/(5*x^3*exp(-1+x)^2-5*log(3)-25))/(x^6*exp(-1+x)^4+(-2*x^3*log(3)-1 0*x^3)*exp(-1+x)^2+log(3)^2+10*log(3)+25),x)
Output:
log(5)*e**2*(2*int((e**((10*e**(2*x)*x**4 + 2*e**(2*x)*x**3 - 10*log(3)*e* *2*x - 50*e**2*x)/(5*e**(2*x)*x**3 - 5*log(3)*e**2 - 25*e**2))*x**3)/(e**( (20*e**(2*x)*x**4 + 5*log(5)*e**2 - 20*log(3)*e**2*x + 2*log(3)*e**2 - 100 *e**2*x + 10*e**2)/(5*e**(2*x)*x**3 - 5*log(3)*e**2 - 25*e**2))*x**6 - 2*e **((10*e**(2*x)*x**4 + 5*log(5)*e**2 - 10*log(3)*e**2*x + 2*log(3)*e**2 - 50*e**2*x + 10*e**2)/(5*e**(2*x)*x**3 - 5*log(3)*e**2 - 25*e**2))*log(3)*e **2*x**3 - 10*e**((10*e**(2*x)*x**4 + 5*log(5)*e**2 - 10*log(3)*e**2*x + 2 *log(3)*e**2 - 50*e**2*x + 10*e**2)/(5*e**(2*x)*x**3 - 5*log(3)*e**2 - 25* e**2))*e**2*x**3 + e**((5*log(5)*e**2 + 2*log(3)*e**2 + 10*e**2)/(5*e**(2* x)*x**3 - 5*log(3)*e**2 - 25*e**2))*log(3)**2*e**4 + 10*e**((5*log(5)*e**2 + 2*log(3)*e**2 + 10*e**2)/(5*e**(2*x)*x**3 - 5*log(3)*e**2 - 25*e**2))*l og(3)*e**4 + 25*e**((5*log(5)*e**2 + 2*log(3)*e**2 + 10*e**2)/(5*e**(2*x)* x**3 - 5*log(3)*e**2 - 25*e**2))*e**4),x) + 3*int((e**((10*e**(2*x)*x**4 + 2*e**(2*x)*x**3 - 10*log(3)*e**2*x - 50*e**2*x)/(5*e**(2*x)*x**3 - 5*log( 3)*e**2 - 25*e**2))*x**2)/(e**((20*e**(2*x)*x**4 + 5*log(5)*e**2 - 20*log( 3)*e**2*x + 2*log(3)*e**2 - 100*e**2*x + 10*e**2)/(5*e**(2*x)*x**3 - 5*log (3)*e**2 - 25*e**2))*x**6 - 2*e**((10*e**(2*x)*x**4 + 5*log(5)*e**2 - 10*l og(3)*e**2*x + 2*log(3)*e**2 - 50*e**2*x + 10*e**2)/(5*e**(2*x)*x**3 - 5*l og(3)*e**2 - 25*e**2))*log(3)*e**2*x**3 - 10*e**((10*e**(2*x)*x**4 + 5*log (5)*e**2 - 10*log(3)*e**2*x + 2*log(3)*e**2 - 50*e**2*x + 10*e**2)/(5*e...