Integrand size = 49, antiderivative size = 29 \[ \int \frac {e^{\frac {5-x \log (3)}{\log (3)}} \left (-15-15 x-10 x^2+\left (2+2 x-2 x^2\right ) \log (2)+(1+x) \log ^2(2)\right )}{x^2} \, dx=\frac {e^{-x+\frac {5}{\log (3)}} (3+2 x-\log (2)) (5+\log (2))}{x} \]
Time = 0.03 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {5-x \log (3)}{\log (3)}} \left (-15-15 x-10 x^2+\left (2+2 x-2 x^2\right ) \log (2)+(1+x) \log ^2(2)\right )}{x^2} \, dx=\frac {e^{-x+\frac {5}{\log (3)}} (3+2 x-\log (2)) (5+\log (2))}{x} \]
Integrate[(E^((5 - x*Log[3])/Log[3])*(-15 - 15*x - 10*x^2 + (2 + 2*x - 2*x ^2)*Log[2] + (1 + x)*Log[2]^2))/x^2,x]
Time = 0.33 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.97, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.041, Rules used = {2629, 2009}
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.
| \(\displaystyle \int \frac {e^{\frac {5-x \log (3)}{\log (3)}} \left (-10 x^2+\left (-2 x^2+2 x+2\right ) \log (2)-15 x+(x+1) \log ^2(2)-15\right )}{x^2} \, dx\) |
| \(\Big \downarrow \) 2629 |
| \(\displaystyle \int \left (\frac {\left (-15+\log ^2(2)+\log (4)\right ) e^{\frac {5-x \log (3)}{\log (3)}}}{x^2}+\frac {\left (-15+\log ^2(2)+\log (4)\right ) e^{\frac {5-x \log (3)}{\log (3)}}}{x}-2 (5+\log (2)) e^{\frac {5-x \log (3)}{\log (3)}}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {e^{\frac {5}{\log (3)}} \left (15-\log ^2(2)-\log (4)\right ) 3^{-\frac {x}{\log (3)}}}{x}+2 e^{\frac {5}{\log (3)}} (5+\log (2)) 3^{-\frac {x}{\log (3)}}\) |
Int[(E^((5 - x*Log[3])/Log[3])*(-15 - 15*x - 10*x^2 + (2 + 2*x - 2*x^2)*Lo g[2] + (1 + x)*Log[2]^2))/x^2,x]
(2*E^(5/Log[3])*(5 + Log[2]))/3^(x/Log[3]) + (E^(5/Log[3])*(15 - Log[2]^2 - Log[4]))/(3^(x/Log[3])*x)
3.9.47.3.1 Defintions of rubi rules used
Int[(F_)^(v_)*(Px_)*((d_.) + (e_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandInte grand[F^v, Px*(d + e*x)^m, x], x] /; FreeQ[{F, d, e, m}, x] && PolynomialQ[ Px, x] && LinearQ[v, x] && !TrueQ[$UseGamma]
Time = 2.30 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.03
| method | result | size |
| gosper | \(-\frac {{\mathrm e}^{-\frac {x \ln \left (3\right )-5}{\ln \left (3\right )}} \left (\ln \left (2\right )+5\right ) \left (-3-2 x +\ln \left (2\right )\right )}{x}\) | \(30\) |
| risch | \(-\frac {\left (\ln \left (2\right )^{2}-2 x \ln \left (2\right )+2 \ln \left (2\right )-10 x -15\right ) {\mathrm e}^{-\frac {x \ln \left (3\right )-5}{\ln \left (3\right )}}}{x}\) | \(37\) |
| norman | \(\frac {\left (-\ln \left (2\right )^{2}-2 \ln \left (2\right )+15\right ) {\mathrm e}^{\frac {-x \ln \left (3\right )+5}{\ln \left (3\right )}}+\left (2 \ln \left (2\right )+10\right ) x \,{\mathrm e}^{\frac {-x \ln \left (3\right )+5}{\ln \left (3\right )}}}{x}\) | \(53\) |
| parallelrisch | \(-\frac {\ln \left (2\right )^{2} {\mathrm e}^{-\frac {x \ln \left (3\right )-5}{\ln \left (3\right )}}-2 \ln \left (2\right ) {\mathrm e}^{-\frac {x \ln \left (3\right )-5}{\ln \left (3\right )}} x +2 \ln \left (2\right ) {\mathrm e}^{-\frac {x \ln \left (3\right )-5}{\ln \left (3\right )}}-10 x \,{\mathrm e}^{-\frac {x \ln \left (3\right )-5}{\ln \left (3\right )}}-15 \,{\mathrm e}^{-\frac {x \ln \left (3\right )-5}{\ln \left (3\right )}}}{x}\) | \(91\) |
| meijerg | \(-\left (\ln \left (2\right )^{2}+2 \ln \left (2\right )-15\right ) {\mathrm e}^{\frac {5}{\ln \left (3\right )}} \operatorname {Ei}_{1}\left (x \right )+\left (-2 \ln \left (2\right )-10\right ) {\mathrm e}^{\frac {5}{\ln \left (3\right )}} \left (1-{\mathrm e}^{-x}\right )+\ln \left (2\right )^{2} {\mathrm e}^{\frac {5}{\ln \left (3\right )}} \left (-\frac {1}{x}+1+\frac {2-2 x}{2 x}-\frac {{\mathrm e}^{-x}}{x}+\operatorname {Ei}_{1}\left (x \right )\right )+2 \ln \left (2\right ) {\mathrm e}^{\frac {5}{\ln \left (3\right )}} \left (-\frac {1}{x}+1+\frac {2-2 x}{2 x}-\frac {{\mathrm e}^{-x}}{x}+\operatorname {Ei}_{1}\left (x \right )\right )-15 \,{\mathrm e}^{\frac {5}{\ln \left (3\right )}} \left (-\frac {1}{x}+1+\frac {2-2 x}{2 x}-\frac {{\mathrm e}^{-x}}{x}+\operatorname {Ei}_{1}\left (x \right )\right )\) | \(165\) |
| derivativedivides | \(-\ln \left (2\right )^{2} \left (\frac {{\mathrm e}^{\frac {5}{\ln \left (3\right )}-x}}{x}-{\mathrm e}^{\frac {5}{\ln \left (3\right )}} \operatorname {Ei}_{1}\left (x \right )\right )+\frac {15 \,{\mathrm e}^{\frac {5}{\ln \left (3\right )}-x}}{x}+10 \,{\mathrm e}^{\frac {5}{\ln \left (3\right )}-x}+\frac {\frac {500 \,{\mathrm e}^{\frac {5}{\ln \left (3\right )}-x}}{x}-500 \,{\mathrm e}^{\frac {5}{\ln \left (3\right )}} \operatorname {Ei}_{1}\left (x \right )}{\ln \left (3\right )^{2}}-2 \ln \left (2\right ) \left (\frac {{\mathrm e}^{\frac {5}{\ln \left (3\right )}-x}}{x}-{\mathrm e}^{\frac {5}{\ln \left (3\right )}} \operatorname {Ei}_{1}\left (x \right )\right )+2 \ln \left (2\right ) \left (-{\mathrm e}^{\frac {5}{\ln \left (3\right )}} \operatorname {Ei}_{1}\left (x \right )+\frac {\frac {5 \,{\mathrm e}^{\frac {5}{\ln \left (3\right )}-x}}{x}-5 \,{\mathrm e}^{\frac {5}{\ln \left (3\right )}} \operatorname {Ei}_{1}\left (x \right )}{\ln \left (3\right )}\right )+2 \ln \left (2\right ) \left ({\mathrm e}^{\frac {5}{\ln \left (3\right )}-x}-\frac {10 \,{\mathrm e}^{\frac {5}{\ln \left (3\right )}} \operatorname {Ei}_{1}\left (x \right )}{\ln \left (3\right )}+\frac {\frac {25 \,{\mathrm e}^{\frac {5}{\ln \left (3\right )}-x}}{x}-25 \,{\mathrm e}^{\frac {5}{\ln \left (3\right )}} \operatorname {Ei}_{1}\left (x \right )}{\ln \left (3\right )^{2}}\right )+\frac {50 \ln \left (2\right ) \left (\frac {{\mathrm e}^{\frac {5}{\ln \left (3\right )}-x}}{x}-{\mathrm e}^{\frac {5}{\ln \left (3\right )}} \operatorname {Ei}_{1}\left (x \right )\right )}{\ln \left (3\right )^{2}}-\frac {10 \ln \left (2\right ) \left (\frac {{\mathrm e}^{\frac {5}{\ln \left (3\right )}-x}}{x}-{\mathrm e}^{\frac {5}{\ln \left (3\right )}} \operatorname {Ei}_{1}\left (x \right )\right )}{\ln \left (3\right )}+\ln \left (2\right )^{2} \left (-{\mathrm e}^{\frac {5}{\ln \left (3\right )}} \operatorname {Ei}_{1}\left (x \right )+\frac {\frac {5 \,{\mathrm e}^{\frac {5}{\ln \left (3\right )}-x}}{x}-5 \,{\mathrm e}^{\frac {5}{\ln \left (3\right )}} \operatorname {Ei}_{1}\left (x \right )}{\ln \left (3\right )}\right )-\frac {5 \ln \left (2\right )^{2} \left (\frac {{\mathrm e}^{\frac {5}{\ln \left (3\right )}-x}}{x}-{\mathrm e}^{\frac {5}{\ln \left (3\right )}} \operatorname {Ei}_{1}\left (x \right )\right )}{\ln \left (3\right )}-\frac {500 \,{\mathrm e}^{\frac {5}{\ln \left (3\right )}-x}}{\ln \left (3\right )^{2} x}+\frac {500 \,{\mathrm e}^{\frac {5}{\ln \left (3\right )}} \operatorname {Ei}_{1}\left (x \right )}{\ln \left (3\right )^{2}}-\frac {20 \ln \left (2\right ) \left (-{\mathrm e}^{\frac {5}{\ln \left (3\right )}} \operatorname {Ei}_{1}\left (x \right )+\frac {\frac {5 \,{\mathrm e}^{\frac {5}{\ln \left (3\right )}-x}}{x}-5 \,{\mathrm e}^{\frac {5}{\ln \left (3\right )}} \operatorname {Ei}_{1}\left (x \right )}{\ln \left (3\right )}\right )}{\ln \left (3\right )}\) | \(501\) |
| default | \(-\ln \left (2\right )^{2} \left (\frac {{\mathrm e}^{\frac {5}{\ln \left (3\right )}-x}}{x}-{\mathrm e}^{\frac {5}{\ln \left (3\right )}} \operatorname {Ei}_{1}\left (x \right )\right )+\frac {15 \,{\mathrm e}^{\frac {5}{\ln \left (3\right )}-x}}{x}+10 \,{\mathrm e}^{\frac {5}{\ln \left (3\right )}-x}+\frac {\frac {500 \,{\mathrm e}^{\frac {5}{\ln \left (3\right )}-x}}{x}-500 \,{\mathrm e}^{\frac {5}{\ln \left (3\right )}} \operatorname {Ei}_{1}\left (x \right )}{\ln \left (3\right )^{2}}-2 \ln \left (2\right ) \left (\frac {{\mathrm e}^{\frac {5}{\ln \left (3\right )}-x}}{x}-{\mathrm e}^{\frac {5}{\ln \left (3\right )}} \operatorname {Ei}_{1}\left (x \right )\right )+2 \ln \left (2\right ) \left (-{\mathrm e}^{\frac {5}{\ln \left (3\right )}} \operatorname {Ei}_{1}\left (x \right )+\frac {\frac {5 \,{\mathrm e}^{\frac {5}{\ln \left (3\right )}-x}}{x}-5 \,{\mathrm e}^{\frac {5}{\ln \left (3\right )}} \operatorname {Ei}_{1}\left (x \right )}{\ln \left (3\right )}\right )+2 \ln \left (2\right ) \left ({\mathrm e}^{\frac {5}{\ln \left (3\right )}-x}-\frac {10 \,{\mathrm e}^{\frac {5}{\ln \left (3\right )}} \operatorname {Ei}_{1}\left (x \right )}{\ln \left (3\right )}+\frac {\frac {25 \,{\mathrm e}^{\frac {5}{\ln \left (3\right )}-x}}{x}-25 \,{\mathrm e}^{\frac {5}{\ln \left (3\right )}} \operatorname {Ei}_{1}\left (x \right )}{\ln \left (3\right )^{2}}\right )+\frac {50 \ln \left (2\right ) \left (\frac {{\mathrm e}^{\frac {5}{\ln \left (3\right )}-x}}{x}-{\mathrm e}^{\frac {5}{\ln \left (3\right )}} \operatorname {Ei}_{1}\left (x \right )\right )}{\ln \left (3\right )^{2}}-\frac {10 \ln \left (2\right ) \left (\frac {{\mathrm e}^{\frac {5}{\ln \left (3\right )}-x}}{x}-{\mathrm e}^{\frac {5}{\ln \left (3\right )}} \operatorname {Ei}_{1}\left (x \right )\right )}{\ln \left (3\right )}+\ln \left (2\right )^{2} \left (-{\mathrm e}^{\frac {5}{\ln \left (3\right )}} \operatorname {Ei}_{1}\left (x \right )+\frac {\frac {5 \,{\mathrm e}^{\frac {5}{\ln \left (3\right )}-x}}{x}-5 \,{\mathrm e}^{\frac {5}{\ln \left (3\right )}} \operatorname {Ei}_{1}\left (x \right )}{\ln \left (3\right )}\right )-\frac {5 \ln \left (2\right )^{2} \left (\frac {{\mathrm e}^{\frac {5}{\ln \left (3\right )}-x}}{x}-{\mathrm e}^{\frac {5}{\ln \left (3\right )}} \operatorname {Ei}_{1}\left (x \right )\right )}{\ln \left (3\right )}-\frac {500 \,{\mathrm e}^{\frac {5}{\ln \left (3\right )}-x}}{\ln \left (3\right )^{2} x}+\frac {500 \,{\mathrm e}^{\frac {5}{\ln \left (3\right )}} \operatorname {Ei}_{1}\left (x \right )}{\ln \left (3\right )^{2}}-\frac {20 \ln \left (2\right ) \left (-{\mathrm e}^{\frac {5}{\ln \left (3\right )}} \operatorname {Ei}_{1}\left (x \right )+\frac {\frac {5 \,{\mathrm e}^{\frac {5}{\ln \left (3\right )}-x}}{x}-5 \,{\mathrm e}^{\frac {5}{\ln \left (3\right )}} \operatorname {Ei}_{1}\left (x \right )}{\ln \left (3\right )}\right )}{\ln \left (3\right )}\) | \(501\) |
int(((1+x)*ln(2)^2+(-2*x^2+2*x+2)*ln(2)-10*x^2-15*x-15)*exp((-x*ln(3)+5)/l n(3))/x^2,x,method=_RETURNVERBOSE)
Time = 0.24 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.21 \[ \int \frac {e^{\frac {5-x \log (3)}{\log (3)}} \left (-15-15 x-10 x^2+\left (2+2 x-2 x^2\right ) \log (2)+(1+x) \log ^2(2)\right )}{x^2} \, dx=\frac {{\left (2 \, {\left (x - 1\right )} \log \left (2\right ) - \log \left (2\right )^{2} + 10 \, x + 15\right )} e^{\left (-\frac {x \log \left (3\right ) - 5}{\log \left (3\right )}\right )}}{x} \]
integrate(((1+x)*log(2)^2+(-2*x^2+2*x+2)*log(2)-10*x^2-15*x-15)*exp((-x*lo g(3)+5)/log(3))/x^2,x, algorithm=\
Time = 0.08 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.17 \[ \int \frac {e^{\frac {5-x \log (3)}{\log (3)}} \left (-15-15 x-10 x^2+\left (2+2 x-2 x^2\right ) \log (2)+(1+x) \log ^2(2)\right )}{x^2} \, dx=\frac {\left (2 x \log {\left (2 \right )} + 10 x - 2 \log {\left (2 \right )} - \log {\left (2 \right )}^{2} + 15\right ) e^{\frac {- x \log {\left (3 \right )} + 5}{\log {\left (3 \right )}}}}{x} \]
integrate(((1+x)*ln(2)**2+(-2*x**2+2*x+2)*ln(2)-10*x**2-15*x-15)*exp((-x*l n(3)+5)/ln(3))/x**2,x)
Result contains higher order function than in optimal. Order 4 vs. order 3.
Time = 0.32 (sec) , antiderivative size = 115, normalized size of antiderivative = 3.97 \[ \int \frac {e^{\frac {5-x \log (3)}{\log (3)}} \left (-15-15 x-10 x^2+\left (2+2 x-2 x^2\right ) \log (2)+(1+x) \log ^2(2)\right )}{x^2} \, dx={\rm Ei}\left (-x\right ) e^{\frac {5}{\log \left (3\right )}} \log \left (2\right )^{2} - e^{\frac {5}{\log \left (3\right )}} \Gamma \left (-1, x\right ) \log \left (2\right )^{2} + 2 \, {\rm Ei}\left (-x\right ) e^{\frac {5}{\log \left (3\right )}} \log \left (2\right ) - 2 \, e^{\frac {5}{\log \left (3\right )}} \Gamma \left (-1, x\right ) \log \left (2\right ) - 15 \, {\rm Ei}\left (-x\right ) e^{\frac {5}{\log \left (3\right )}} + 15 \, e^{\frac {5}{\log \left (3\right )}} \Gamma \left (-1, x\right ) + 2 \, e^{\left (-x + \frac {5}{\log \left (3\right )}\right )} \log \left (2\right ) + 10 \, e^{\left (-x + \frac {5}{\log \left (3\right )}\right )} \]
integrate(((1+x)*log(2)^2+(-2*x^2+2*x+2)*log(2)-10*x^2-15*x-15)*exp((-x*lo g(3)+5)/log(3))/x^2,x, algorithm=\
Ei(-x)*e^(5/log(3))*log(2)^2 - e^(5/log(3))*gamma(-1, x)*log(2)^2 + 2*Ei(- x)*e^(5/log(3))*log(2) - 2*e^(5/log(3))*gamma(-1, x)*log(2) - 15*Ei(-x)*e^ (5/log(3)) + 15*e^(5/log(3))*gamma(-1, x) + 2*e^(-x + 5/log(3))*log(2) + 1 0*e^(-x + 5/log(3))
Leaf count of result is larger than twice the leaf count of optimal. 80 vs. \(2 (28) = 56\).
Time = 0.26 (sec) , antiderivative size = 80, normalized size of antiderivative = 2.76 \[ \int \frac {e^{\frac {5-x \log (3)}{\log (3)}} \left (-15-15 x-10 x^2+\left (2+2 x-2 x^2\right ) \log (2)+(1+x) \log ^2(2)\right )}{x^2} \, dx=\frac {2 \, x e^{\left (-x + \frac {5}{\log \left (3\right )}\right )} \log \left (2\right ) - e^{\left (-x + \frac {5}{\log \left (3\right )}\right )} \log \left (2\right )^{2} + 10 \, x e^{\left (-x + \frac {5}{\log \left (3\right )}\right )} - 2 \, e^{\left (-x + \frac {5}{\log \left (3\right )}\right )} \log \left (2\right ) + 15 \, e^{\left (-x + \frac {5}{\log \left (3\right )}\right )}}{x} \]
integrate(((1+x)*log(2)^2+(-2*x^2+2*x+2)*log(2)-10*x^2-15*x-15)*exp((-x*lo g(3)+5)/log(3))/x^2,x, algorithm=\
(2*x*e^(-x + 5/log(3))*log(2) - e^(-x + 5/log(3))*log(2)^2 + 10*x*e^(-x + 5/log(3)) - 2*e^(-x + 5/log(3))*log(2) + 15*e^(-x + 5/log(3)))/x
Time = 0.11 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.17 \[ \int \frac {e^{\frac {5-x \log (3)}{\log (3)}} \left (-15-15 x-10 x^2+\left (2+2 x-2 x^2\right ) \log (2)+(1+x) \log ^2(2)\right )}{x^2} \, dx=\frac {{\mathrm {e}}^{\frac {5}{\ln \left (3\right )}-x}\,\left (10\,x-\ln \left (4\right )+x\,\ln \left (4\right )-{\ln \left (2\right )}^2+15\right )}{x} \]