Integrand size = 80, antiderivative size = 34 \begin {dmath*} \int \frac {e^{\frac {20+29 x-20 x^2-5 x^3-5 \log (4)}{225 x+45 x^2}} \left (-100-40 x-129 x^2-50 x^3-5 x^4+(25+10 x) \log (4)\right )}{3375 x^2+1350 x^3+135 x^4} \, dx=-5+\frac {1}{3} e^{\frac {\frac {4}{5}+x-x^2-\frac {\log (4)}{5+x}}{9 x}} \end {dmath*}
Time = 2.88 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.24 \begin {dmath*} \int \frac {e^{\frac {20+29 x-20 x^2-5 x^3-5 \log (4)}{225 x+45 x^2}} \left (-100-40 x-129 x^2-50 x^3-5 x^4+(25+10 x) \log (4)\right )}{3375 x^2+1350 x^3+135 x^4} \, dx=\frac {2^{-1-\frac {2}{9 x (5+x)}} e^{\frac {1}{45} \left (5+\frac {4}{x}-5 x\right )} \log (4)}{3 \log (2)} \end {dmath*}
Integrate[(E^((20 + 29*x - 20*x^2 - 5*x^3 - 5*Log[4])/(225*x + 45*x^2))*(- 100 - 40*x - 129*x^2 - 50*x^3 - 5*x^4 + (25 + 10*x)*Log[4]))/(3375*x^2 + 1 350*x^3 + 135*x^4),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.
\(\displaystyle \int \frac {\left (-5 x^4-50 x^3-129 x^2-40 x+(10 x+25) \log (4)-100\right ) \exp \left (\frac {-5 x^3-20 x^2+29 x+20-5 \log (4)}{45 x^2+225 x}\right )}{135 x^4+1350 x^3+3375 x^2} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {\left (-5 x^4-50 x^3-129 x^2-40 x+(10 x+25) \log (4)-100\right ) \exp \left (\frac {-5 x^3-20 x^2+29 x+20-5 \log (4)}{45 x^2+225 x}\right )}{x^2 \left (135 x^2+1350 x+3375\right )}dx\) |
\(\Big \downarrow \) 2007 |
\(\displaystyle \int \frac {\left (-5 x^4-50 x^3-129 x^2-40 x+(10 x+25) \log (4)-100\right ) \exp \left (\frac {-5 x^3-20 x^2+29 x+20-5 \log (4)}{45 x^2+225 x}\right )}{x^2 \left (3 \sqrt {15} x+15 \sqrt {15}\right )^2}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {\left (-5 x^4-50 x^3-129 x^2-10 x (4-\log (4))-25 (4-\log (4))\right ) \exp \left (\frac {-5 x^3-20 x^2+29 x+5 (4-\log (4))}{x (45 x+225)}\right )}{x^2 \left (3 \sqrt {15} x+15 \sqrt {15}\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (-\frac {1}{27} \exp \left (\frac {-5 x^3-20 x^2+29 x+5 (4-\log (4))}{x (45 x+225)}\right )-\frac {\log (4) \exp \left (\frac {-5 x^3-20 x^2+29 x+5 (4-\log (4))}{x (45 x+225)}\right )}{135 (x+5)^2}+\frac {(\log (4)-4) \exp \left (\frac {-5 x^3-20 x^2+29 x+5 (4-\log (4))}{x (45 x+225)}\right )}{135 x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {1}{27} \int \exp \left (\frac {-5 x^3-20 x^2+29 x+5 (4-\log (4))}{x (45 x+225)}\right )dx-\frac {1}{135} (4-\log (4)) \int \frac {\exp \left (\frac {-5 x^3-20 x^2+29 x+5 (4-\log (4))}{x (45 x+225)}\right )}{x^2}dx-\frac {1}{135} \log (4) \int \frac {\exp \left (\frac {-5 x^3-20 x^2+29 x+5 (4-\log (4))}{x (45 x+225)}\right )}{(x+5)^2}dx\) |
Int[(E^((20 + 29*x - 20*x^2 - 5*x^3 - 5*Log[4])/(225*x + 45*x^2))*(-100 - 40*x - 129*x^2 - 50*x^3 - 5*x^4 + (25 + 10*x)*Log[4]))/(3375*x^2 + 1350*x^ 3 + 135*x^4),x]
3.6.55.3.1 Defintions of rubi rules used
Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{a = Rt[Coeff[Px, x, 0], Expon[Px, x]], b = Rt[Coeff[Px, x, Expon[Px, x]], Expon[Px, x]]}, Int[u*(a + b*x)^(Ex pon[Px, x]*p), x] /; EqQ[Px, (a + b*x)^Expon[Px, x]]] /; IntegerQ[p] && Pol yQ[Px, x] && GtQ[Expon[Px, x], 1] && NeQ[Coeff[Px, x, 0], 0]
Int[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p *r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ erQ[p] && !MonomialQ[Px, x] && (ILtQ[p, 0] || !PolyQ[u, x])
Time = 0.70 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.91
method | result | size |
risch | \(\frac {2^{-\frac {2}{9 \left (5+x \right ) x}} {\mathrm e}^{-\frac {5 x^{2}-5 x -4}{45 x}}}{3}\) | \(31\) |
gosper | \(\frac {{\mathrm e}^{-\frac {5 x^{3}+20 x^{2}+10 \ln \left (2\right )-29 x -20}{45 \left (5+x \right ) x}}}{3}\) | \(33\) |
parallelrisch | \(\frac {{\mathrm e}^{-\frac {5 x^{3}+20 x^{2}+10 \ln \left (2\right )-29 x -20}{45 \left (5+x \right ) x}}}{3}\) | \(33\) |
norman | \(\frac {\frac {5 x \,{\mathrm e}^{\frac {-10 \ln \left (2\right )-5 x^{3}-20 x^{2}+29 x +20}{45 x^{2}+225 x}}}{3}+\frac {x^{2} {\mathrm e}^{\frac {-10 \ln \left (2\right )-5 x^{3}-20 x^{2}+29 x +20}{45 x^{2}+225 x}}}{3}}{\left (5+x \right ) x}\) | \(83\) |
int((2*(10*x+25)*ln(2)-5*x^4-50*x^3-129*x^2-40*x-100)*exp((-10*ln(2)-5*x^3 -20*x^2+29*x+20)/(45*x^2+225*x))/(135*x^4+1350*x^3+3375*x^2),x,method=_RET URNVERBOSE)
Time = 0.24 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.97 \begin {dmath*} \int \frac {e^{\frac {20+29 x-20 x^2-5 x^3-5 \log (4)}{225 x+45 x^2}} \left (-100-40 x-129 x^2-50 x^3-5 x^4+(25+10 x) \log (4)\right )}{3375 x^2+1350 x^3+135 x^4} \, dx=\frac {1}{3} \, e^{\left (-\frac {5 \, x^{3} + 20 \, x^{2} - 29 \, x + 10 \, \log \left (2\right ) - 20}{45 \, {\left (x^{2} + 5 \, x\right )}}\right )} \end {dmath*}
integrate((2*(10*x+25)*log(2)-5*x^4-50*x^3-129*x^2-40*x-100)*exp((-10*log( 2)-5*x^3-20*x^2+29*x+20)/(45*x^2+225*x))/(135*x^4+1350*x^3+3375*x^2),x, al gorithm=\
Time = 0.34 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.91 \begin {dmath*} \int \frac {e^{\frac {20+29 x-20 x^2-5 x^3-5 \log (4)}{225 x+45 x^2}} \left (-100-40 x-129 x^2-50 x^3-5 x^4+(25+10 x) \log (4)\right )}{3375 x^2+1350 x^3+135 x^4} \, dx=\frac {e^{\frac {- 5 x^{3} - 20 x^{2} + 29 x - 10 \log {\left (2 \right )} + 20}{45 x^{2} + 225 x}}}{3} \end {dmath*}
integrate((2*(10*x+25)*ln(2)-5*x**4-50*x**3-129*x**2-40*x-100)*exp((-10*ln (2)-5*x**3-20*x**2+29*x+20)/(45*x**2+225*x))/(135*x**4+1350*x**3+3375*x**2 ),x)
Time = 0.51 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.85 \begin {dmath*} \int \frac {e^{\frac {20+29 x-20 x^2-5 x^3-5 \log (4)}{225 x+45 x^2}} \left (-100-40 x-129 x^2-50 x^3-5 x^4+(25+10 x) \log (4)\right )}{3375 x^2+1350 x^3+135 x^4} \, dx=\frac {1}{3} \, e^{\left (-\frac {1}{9} \, x + \frac {2 \, \log \left (2\right )}{45 \, {\left (x + 5\right )}} - \frac {2 \, \log \left (2\right )}{45 \, x} + \frac {4}{45 \, x} + \frac {1}{9}\right )} \end {dmath*}
integrate((2*(10*x+25)*log(2)-5*x^4-50*x^3-129*x^2-40*x-100)*exp((-10*log( 2)-5*x^3-20*x^2+29*x+20)/(45*x^2+225*x))/(135*x^4+1350*x^3+3375*x^2),x, al gorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 68 vs. \(2 (29) = 58\).
Time = 0.28 (sec) , antiderivative size = 68, normalized size of antiderivative = 2.00 \begin {dmath*} \int \frac {e^{\frac {20+29 x-20 x^2-5 x^3-5 \log (4)}{225 x+45 x^2}} \left (-100-40 x-129 x^2-50 x^3-5 x^4+(25+10 x) \log (4)\right )}{3375 x^2+1350 x^3+135 x^4} \, dx=\frac {1}{3} \, e^{\left (-\frac {x^{3}}{9 \, {\left (x^{2} + 5 \, x\right )}} - \frac {4 \, x^{2}}{9 \, {\left (x^{2} + 5 \, x\right )}} + \frac {29 \, x}{45 \, {\left (x^{2} + 5 \, x\right )}} - \frac {2 \, \log \left (2\right )}{9 \, {\left (x^{2} + 5 \, x\right )}} + \frac {4}{9 \, {\left (x^{2} + 5 \, x\right )}}\right )} \end {dmath*}
integrate((2*(10*x+25)*log(2)-5*x^4-50*x^3-129*x^2-40*x-100)*exp((-10*log( 2)-5*x^3-20*x^2+29*x+20)/(45*x^2+225*x))/(135*x^4+1350*x^3+3375*x^2),x, al gorithm=\
1/3*e^(-1/9*x^3/(x^2 + 5*x) - 4/9*x^2/(x^2 + 5*x) + 29/45*x/(x^2 + 5*x) - 2/9*log(2)/(x^2 + 5*x) + 4/9/(x^2 + 5*x))
Time = 14.01 (sec) , antiderivative size = 82, normalized size of antiderivative = 2.41 \begin {dmath*} \int \frac {e^{\frac {20+29 x-20 x^2-5 x^3-5 \log (4)}{225 x+45 x^2}} \left (-100-40 x-129 x^2-50 x^3-5 x^4+(25+10 x) \log (4)\right )}{3375 x^2+1350 x^3+135 x^4} \, dx=\frac {{\mathrm {e}}^{\frac {29\,x}{45\,x^2+225\,x}}\,{\mathrm {e}}^{-\frac {5\,x^3}{45\,x^2+225\,x}}\,{\mathrm {e}}^{-\frac {20\,x^2}{45\,x^2+225\,x}}\,{\mathrm {e}}^{\frac {20}{45\,x^2+225\,x}}}{3\,2^{\frac {10}{45\,x^2+225\,x}}} \end {dmath*}
int(-(exp(-(10*log(2) - 29*x + 20*x^2 + 5*x^3 - 20)/(225*x + 45*x^2))*(40* x - 2*log(2)*(10*x + 25) + 129*x^2 + 50*x^3 + 5*x^4 + 100))/(3375*x^2 + 13 50*x^3 + 135*x^4),x)