Integrand size = 64, antiderivative size = 35 \[ \int \frac {e^{-6-2 x} \left (120+105 x+18 x^2+4 e^{6+2 x} x^3+\left (-70-68 x-12 x^2\right ) \log (x)+\left (10+11 x+2 x^2\right ) \log ^2(x)\right )}{3 x^3} \, dx=\frac {x}{3}+\frac {1}{3} (5+x) \left (3-\frac {e^{-6-2 x} (3-\log (x))^2}{x^2}\right ) \]
Time = 5.06 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.34 \[ \int \frac {e^{-6-2 x} \left (120+105 x+18 x^2+4 e^{6+2 x} x^3+\left (-70-68 x-12 x^2\right ) \log (x)+\left (10+11 x+2 x^2\right ) \log ^2(x)\right )}{3 x^3} \, dx=\frac {e^{-2 (3+x)} \left (-45-9 x+4 e^{6+2 x} x^3+6 (5+x) \log (x)-(5+x) \log ^2(x)\right )}{3 x^2} \]
Integrate[(E^(-6 - 2*x)*(120 + 105*x + 18*x^2 + 4*E^(6 + 2*x)*x^3 + (-70 - 68*x - 12*x^2)*Log[x] + (10 + 11*x + 2*x^2)*Log[x]^2))/(3*x^3),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.
\(\displaystyle \int \frac {e^{-2 x-6} \left (4 e^{2 x+6} x^3+18 x^2+\left (2 x^2+11 x+10\right ) \log ^2(x)+\left (-12 x^2-68 x-70\right ) \log (x)+105 x+120\right )}{3 x^3} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{3} \int \frac {e^{-2 x-6} \left (4 e^{2 x+6} x^3+18 x^2+105 x+\left (2 x^2+11 x+10\right ) \log ^2(x)-2 \left (6 x^2+34 x+35\right ) \log (x)+120\right )}{x^3}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{3} \int \left (\frac {e^{-2 x-6} (\log (x)-3) \left (2 \log (x) x^2-6 x^2+11 \log (x) x-35 x+10 \log (x)-40\right )}{x^3}+4\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{3} \left (10 \int \frac {e^{-2 x-6} \log ^2(x)}{x^3}dx+11 \int \frac {e^{-2 x-6} \log ^2(x)}{x^2}dx+2 \int \frac {e^{-2 x-6} \log ^2(x)}{x}dx-\frac {32 x \, _3F_3(1,1,1;2,2,2;-2 x)}{e^6}+\frac {16 \log (x) (\operatorname {ExpIntegralE}(1,2 x)+\operatorname {ExpIntegralEi}(-2 x))}{e^6}-\frac {26 \operatorname {ExpIntegralEi}(-2 x)}{e^6}-\frac {16 \operatorname {ExpIntegralEi}(-2 x) \log (x)}{e^6}-\frac {85 e^{-2 x-6}}{2 x^2}+\frac {35 e^{-2 x-6} \log (x)}{x^2}+4 x-\frac {22 e^{-2 x-6}}{x}+\frac {8 \log ^2(2 x)}{e^6}-\frac {2 e^{-2 x-6} \log (x)}{x}+\frac {16 \gamma \log (x)}{e^6}\right )\) |
Int[(E^(-6 - 2*x)*(120 + 105*x + 18*x^2 + 4*E^(6 + 2*x)*x^3 + (-70 - 68*x - 12*x^2)*Log[x] + (10 + 11*x + 2*x^2)*Log[x]^2))/(3*x^3),x]
3.9.45.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Time = 1.58 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.43
method | result | size |
parallelrisch | \(\frac {\left (-45+4 x^{3} {\mathrm e}^{2 x +6}-x \ln \left (x \right )^{2}+6 x \ln \left (x \right )-5 \ln \left (x \right )^{2}-9 x +30 \ln \left (x \right )\right ) {\mathrm e}^{-2 x -6}}{3 x^{2}}\) | \(50\) |
risch | \(-\frac {\left (5+x \right ) {\mathrm e}^{-2 x -6} \ln \left (x \right )^{2}}{3 x^{2}}+\frac {2 \left (5+x \right ) {\mathrm e}^{-2 x -6} \ln \left (x \right )}{x^{2}}+\frac {\left (4 x^{3} {\mathrm e}^{2 x +6}-9 x -45\right ) {\mathrm e}^{-2 x -6}}{3 x^{2}}\) | \(63\) |
int(1/3*((2*x^2+11*x+10)*ln(x)^2+(-12*x^2-68*x-70)*ln(x)+4*x^3*exp(3+x)^2+ 18*x^2+105*x+120)/x^3/exp(3+x)^2,x,method=_RETURNVERBOSE)
Time = 0.26 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.23 \[ \int \frac {e^{-6-2 x} \left (120+105 x+18 x^2+4 e^{6+2 x} x^3+\left (-70-68 x-12 x^2\right ) \log (x)+\left (10+11 x+2 x^2\right ) \log ^2(x)\right )}{3 x^3} \, dx=\frac {{\left (4 \, x^{3} e^{\left (2 \, x + 6\right )} - {\left (x + 5\right )} \log \left (x\right )^{2} + 6 \, {\left (x + 5\right )} \log \left (x\right ) - 9 \, x - 45\right )} e^{\left (-2 \, x - 6\right )}}{3 \, x^{2}} \]
integrate(1/3*((2*x^2+11*x+10)*log(x)^2+(-12*x^2-68*x-70)*log(x)+4*x^3*exp (3+x)^2+18*x^2+105*x+120)/x^3/exp(3+x)^2,x, algorithm=\
Time = 0.13 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.37 \[ \int \frac {e^{-6-2 x} \left (120+105 x+18 x^2+4 e^{6+2 x} x^3+\left (-70-68 x-12 x^2\right ) \log (x)+\left (10+11 x+2 x^2\right ) \log ^2(x)\right )}{3 x^3} \, dx=\frac {4 x}{3} + \frac {\left (- x \log {\left (x \right )}^{2} + 6 x \log {\left (x \right )} - 9 x - 5 \log {\left (x \right )}^{2} + 30 \log {\left (x \right )} - 45\right ) e^{- 2 x - 6}}{3 x^{2}} \]
integrate(1/3*((2*x**2+11*x+10)*ln(x)**2+(-12*x**2-68*x-70)*ln(x)+4*x**3*e xp(3+x)**2+18*x**2+105*x+120)/x**3/exp(3+x)**2,x)
4*x/3 + (-x*log(x)**2 + 6*x*log(x) - 9*x - 5*log(x)**2 + 30*log(x) - 45)*e xp(-2*x - 6)/(3*x**2)
\[ \int \frac {e^{-6-2 x} \left (120+105 x+18 x^2+4 e^{6+2 x} x^3+\left (-70-68 x-12 x^2\right ) \log (x)+\left (10+11 x+2 x^2\right ) \log ^2(x)\right )}{3 x^3} \, dx=\int { \frac {{\left (4 \, x^{3} e^{\left (2 \, x + 6\right )} + {\left (2 \, x^{2} + 11 \, x + 10\right )} \log \left (x\right )^{2} + 18 \, x^{2} - 2 \, {\left (6 \, x^{2} + 34 \, x + 35\right )} \log \left (x\right ) + 105 \, x + 120\right )} e^{\left (-2 \, x - 6\right )}}{3 \, x^{3}} \,d x } \]
integrate(1/3*((2*x^2+11*x+10)*log(x)^2+(-12*x^2-68*x-70)*log(x)+4*x^3*exp (3+x)^2+18*x^2+105*x+120)/x^3/exp(3+x)^2,x, algorithm=\
6*Ei(-2*x)*e^(-6) - 70*e^(-6)*gamma(-1, 2*x) - 160*e^(-6)*gamma(-2, 2*x) + 4/3*x - 1/3*((x + 5)*log(x)^2 - 6*(x + 5)*log(x))*e^(-2*x - 6)/x^2 - 1/3* integrate(6*(x + 5)*e^(-2*x - 6)/x^3, x)
Leaf count of result is larger than twice the leaf count of optimal. 66 vs. \(2 (27) = 54\).
Time = 0.25 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.89 \[ \int \frac {e^{-6-2 x} \left (120+105 x+18 x^2+4 e^{6+2 x} x^3+\left (-70-68 x-12 x^2\right ) \log (x)+\left (10+11 x+2 x^2\right ) \log ^2(x)\right )}{3 x^3} \, dx=\frac {{\left (4 \, x^{3} e^{6} - x e^{\left (-2 \, x\right )} \log \left (x\right )^{2} + 6 \, x e^{\left (-2 \, x\right )} \log \left (x\right ) - 5 \, e^{\left (-2 \, x\right )} \log \left (x\right )^{2} - 9 \, x e^{\left (-2 \, x\right )} + 30 \, e^{\left (-2 \, x\right )} \log \left (x\right ) - 45 \, e^{\left (-2 \, x\right )}\right )} e^{\left (-6\right )}}{3 \, x^{2}} \]
integrate(1/3*((2*x^2+11*x+10)*log(x)^2+(-12*x^2-68*x-70)*log(x)+4*x^3*exp (3+x)^2+18*x^2+105*x+120)/x^3/exp(3+x)^2,x, algorithm=\
1/3*(4*x^3*e^6 - x*e^(-2*x)*log(x)^2 + 6*x*e^(-2*x)*log(x) - 5*e^(-2*x)*lo g(x)^2 - 9*x*e^(-2*x) + 30*e^(-2*x)*log(x) - 45*e^(-2*x))*e^(-6)/x^2
Time = 9.31 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.40 \[ \int \frac {e^{-6-2 x} \left (120+105 x+18 x^2+4 e^{6+2 x} x^3+\left (-70-68 x-12 x^2\right ) \log (x)+\left (10+11 x+2 x^2\right ) \log ^2(x)\right )}{3 x^3} \, dx=\frac {4\,x}{3}-\frac {\frac {{\mathrm {e}}^{-2\,x-6}\,\left (5\,{\ln \left (x\right )}^2-30\,\ln \left (x\right )+45\right )}{3}+\frac {x\,{\mathrm {e}}^{-2\,x-6}\,\left ({\ln \left (x\right )}^2-6\,\ln \left (x\right )+9\right )}{3}}{x^2} \]
int((exp(- 2*x - 6)*(35*x + (log(x)^2*(11*x + 2*x^2 + 10))/3 - (log(x)*(68 *x + 12*x^2 + 70))/3 + (4*x^3*exp(2*x + 6))/3 + 6*x^2 + 40))/x^3,x)