Integrand size = 112, antiderivative size = 28 \[ \int \frac {e^{\left .-\frac {3}{2}\right /x} \left (e^{\left .\frac {3}{2}\right /x} \left (480 x^3-320 x^4\right )+e^{e^{\left .-\frac {3}{2}\right /x} \left (-3+e^{\left .\frac {3}{2}\right /x} x\right )} \left (-81+216 x-144 x^2+e^{\left .\frac {3}{2}\right /x} \left (18 x^2-48 x^3+32 x^4\right )\right )\right )}{18 x^2-48 x^3+32 x^4} \, dx=e^{-3 e^{\left .-\frac {3}{2}\right /x}+x}-\frac {10 x^2}{-\frac {3}{4}+x} \]
Time = 0.14 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07 \[ \int \frac {e^{\left .-\frac {3}{2}\right /x} \left (e^{\left .\frac {3}{2}\right /x} \left (480 x^3-320 x^4\right )+e^{e^{\left .-\frac {3}{2}\right /x} \left (-3+e^{\left .\frac {3}{2}\right /x} x\right )} \left (-81+216 x-144 x^2+e^{\left .\frac {3}{2}\right /x} \left (18 x^2-48 x^3+32 x^4\right )\right )\right )}{18 x^2-48 x^3+32 x^4} \, dx=e^{-3 e^{\left .-\frac {3}{2}\right /x}+x}-10 x-\frac {45}{2 (-3+4 x)} \]
Integrate[(E^(3/(2*x))*(480*x^3 - 320*x^4) + E^((-3 + E^(3/(2*x))*x)/E^(3/ (2*x)))*(-81 + 216*x - 144*x^2 + E^(3/(2*x))*(18*x^2 - 48*x^3 + 32*x^4)))/ (E^(3/(2*x))*(18*x^2 - 48*x^3 + 32*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.
| \(\displaystyle \int \frac {e^{\left .-\frac {3}{2}\right /x} \left (e^{\left .\frac {3}{2}\right /x} \left (480 x^3-320 x^4\right )+e^{e^{\left .-\frac {3}{2}\right /x} \left (e^{\left .\frac {3}{2}\right /x} x-3\right )} \left (-144 x^2+e^{\left .\frac {3}{2}\right /x} \left (32 x^4-48 x^3+18 x^2\right )+216 x-81\right )\right )}{32 x^4-48 x^3+18 x^2} \, dx\) |
| \(\Big \downarrow \) 2026 |
| \(\displaystyle \int \frac {e^{\left .-\frac {3}{2}\right /x} \left (e^{\left .\frac {3}{2}\right /x} \left (480 x^3-320 x^4\right )+e^{e^{\left .-\frac {3}{2}\right /x} \left (e^{\left .\frac {3}{2}\right /x} x-3\right )} \left (-144 x^2+e^{\left .\frac {3}{2}\right /x} \left (32 x^4-48 x^3+18 x^2\right )+216 x-81\right )\right )}{x^2 \left (32 x^2-48 x+18\right )}dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \left (-\frac {9 e^{x-3 e^{\left .-\frac {3}{2}\right /x}-\frac {3}{2 x}}}{2 x^2}-\frac {80 x (2 x-3)}{(3-4 x)^2}+e^{x-3 e^{\left .-\frac {3}{2}\right /x}}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {9}{2} \int \frac {e^{x-3 e^{\left .-\frac {3}{2}\right /x}-\frac {3}{2 x}}}{x^2}dx+\int e^{x-3 e^{\left .-\frac {3}{2}\right /x}}dx+\frac {10 (3-2 x)^2}{3-4 x}\) |
Int[(E^(3/(2*x))*(480*x^3 - 320*x^4) + E^((-3 + E^(3/(2*x))*x)/E^(3/(2*x)) )*(-81 + 216*x - 144*x^2 + E^(3/(2*x))*(18*x^2 - 48*x^3 + 32*x^4)))/(E^(3/ (2*x))*(18*x^2 - 48*x^3 + 32*x^4)),x]
3.1.42.3.1 Defintions of rubi rules used
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])
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 0.31 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07
| method | result | size |
| risch | \(-10 x -\frac {45}{8 \left (x -\frac {3}{4}\right )}+{\mathrm e}^{\left (x \,{\mathrm e}^{\frac {3}{2 x}}-3\right ) {\mathrm e}^{-\frac {3}{2 x}}}\) | \(30\) |
| parts | \({\mathrm e}^{\left (x \,{\mathrm e}^{\frac {3}{2 x}}-3\right ) {\mathrm e}^{-\frac {3}{2 x}}}-10 x -\frac {45}{2 \left (-3+4 x \right )}\) | \(34\) |
| parallelrisch | \(\frac {-2880 x^{3}+288 \,{\mathrm e}^{\left (x \,{\mathrm e}^{\frac {3}{2 x}}-3\right ) {\mathrm e}^{-\frac {3}{2 x}}} x^{2}-216 \,{\mathrm e}^{\left (x \,{\mathrm e}^{\frac {3}{2 x}}-3\right ) {\mathrm e}^{-\frac {3}{2 x}}} x}{72 x \left (-3+4 x \right )}\) | \(67\) |
| norman | \(\frac {\left (-40 \,{\mathrm e}^{\frac {3}{2 x}} x^{3}+4 x^{2} {\mathrm e}^{\left (x \,{\mathrm e}^{\frac {3}{2 x}}-3\right ) {\mathrm e}^{-\frac {3}{2 x}}} {\mathrm e}^{\frac {3}{2 x}}-3 \,{\mathrm e}^{\frac {3}{2 x}} {\mathrm e}^{\left (x \,{\mathrm e}^{\frac {3}{2 x}}-3\right ) {\mathrm e}^{-\frac {3}{2 x}}} x \right ) {\mathrm e}^{-\frac {3}{2 x}}}{x \left (-3+4 x \right )}\) | \(92\) |
int((((32*x^4-48*x^3+18*x^2)*exp(3/2/x)-144*x^2+216*x-81)*exp((x*exp(3/2/x )-3)/exp(3/2/x))+(-320*x^4+480*x^3)*exp(3/2/x))/(32*x^4-48*x^3+18*x^2)/exp (3/2/x),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.57 \[ \int \frac {e^{\left .-\frac {3}{2}\right /x} \left (e^{\left .\frac {3}{2}\right /x} \left (480 x^3-320 x^4\right )+e^{e^{\left .-\frac {3}{2}\right /x} \left (-3+e^{\left .\frac {3}{2}\right /x} x\right )} \left (-81+216 x-144 x^2+e^{\left .\frac {3}{2}\right /x} \left (18 x^2-48 x^3+32 x^4\right )\right )\right )}{18 x^2-48 x^3+32 x^4} \, dx=-\frac {80 \, x^{2} - 2 \, {\left (4 \, x - 3\right )} e^{\left ({\left (x e^{\left (\frac {3}{2 \, x}\right )} - 3\right )} e^{\left (-\frac {3}{2 \, x}\right )}\right )} - 60 \, x + 45}{2 \, {\left (4 \, x - 3\right )}} \]
integrate((((32*x^4-48*x^3+18*x^2)*exp(3/2/x)-144*x^2+216*x-81)*exp((x*exp (3/2/x)-3)/exp(3/2/x))+(-320*x^4+480*x^3)*exp(3/2/x))/(32*x^4-48*x^3+18*x^ 2)/exp(3/2/x),x, algorithm=\
Time = 0.19 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96 \[ \int \frac {e^{\left .-\frac {3}{2}\right /x} \left (e^{\left .\frac {3}{2}\right /x} \left (480 x^3-320 x^4\right )+e^{e^{\left .-\frac {3}{2}\right /x} \left (-3+e^{\left .\frac {3}{2}\right /x} x\right )} \left (-81+216 x-144 x^2+e^{\left .\frac {3}{2}\right /x} \left (18 x^2-48 x^3+32 x^4\right )\right )\right )}{18 x^2-48 x^3+32 x^4} \, dx=- 10 x + e^{\left (x e^{\frac {3}{2 x}} - 3\right ) e^{- \frac {3}{2 x}}} - \frac {45}{8 x - 6} \]
integrate((((32*x**4-48*x**3+18*x**2)*exp(3/2/x)-144*x**2+216*x-81)*exp((x *exp(3/2/x)-3)/exp(3/2/x))+(-320*x**4+480*x**3)*exp(3/2/x))/(32*x**4-48*x* *3+18*x**2)/exp(3/2/x),x)
\[ \int \frac {e^{\left .-\frac {3}{2}\right /x} \left (e^{\left .\frac {3}{2}\right /x} \left (480 x^3-320 x^4\right )+e^{e^{\left .-\frac {3}{2}\right /x} \left (-3+e^{\left .\frac {3}{2}\right /x} x\right )} \left (-81+216 x-144 x^2+e^{\left .\frac {3}{2}\right /x} \left (18 x^2-48 x^3+32 x^4\right )\right )\right )}{18 x^2-48 x^3+32 x^4} \, dx=\int { -\frac {{\left ({\left (144 \, x^{2} - 2 \, {\left (16 \, x^{4} - 24 \, x^{3} + 9 \, x^{2}\right )} e^{\left (\frac {3}{2 \, x}\right )} - 216 \, x + 81\right )} e^{\left ({\left (x e^{\left (\frac {3}{2 \, x}\right )} - 3\right )} e^{\left (-\frac {3}{2 \, x}\right )}\right )} + 160 \, {\left (2 \, x^{4} - 3 \, x^{3}\right )} e^{\left (\frac {3}{2 \, x}\right )}\right )} e^{\left (-\frac {3}{2 \, x}\right )}}{2 \, {\left (16 \, x^{4} - 24 \, x^{3} + 9 \, x^{2}\right )}} \,d x } \]
integrate((((32*x^4-48*x^3+18*x^2)*exp(3/2/x)-144*x^2+216*x-81)*exp((x*exp (3/2/x)-3)/exp(3/2/x))+(-320*x^4+480*x^3)*exp(3/2/x))/(32*x^4-48*x^3+18*x^ 2)/exp(3/2/x),x, algorithm=\
-5/2*(16*x^2 - 12*x - 9)/(4*x - 3) - 45/(4*x - 3) + 1/2*integrate((2*x^2*e ^(x + 3/2/x) - 9*e^x)*e^(-3/2/x - 3*e^(-3/2/x))/x^2, x)
\[ \int \frac {e^{\left .-\frac {3}{2}\right /x} \left (e^{\left .\frac {3}{2}\right /x} \left (480 x^3-320 x^4\right )+e^{e^{\left .-\frac {3}{2}\right /x} \left (-3+e^{\left .\frac {3}{2}\right /x} x\right )} \left (-81+216 x-144 x^2+e^{\left .\frac {3}{2}\right /x} \left (18 x^2-48 x^3+32 x^4\right )\right )\right )}{18 x^2-48 x^3+32 x^4} \, dx=\int { -\frac {{\left ({\left (144 \, x^{2} - 2 \, {\left (16 \, x^{4} - 24 \, x^{3} + 9 \, x^{2}\right )} e^{\left (\frac {3}{2 \, x}\right )} - 216 \, x + 81\right )} e^{\left ({\left (x e^{\left (\frac {3}{2 \, x}\right )} - 3\right )} e^{\left (-\frac {3}{2 \, x}\right )}\right )} + 160 \, {\left (2 \, x^{4} - 3 \, x^{3}\right )} e^{\left (\frac {3}{2 \, x}\right )}\right )} e^{\left (-\frac {3}{2 \, x}\right )}}{2 \, {\left (16 \, x^{4} - 24 \, x^{3} + 9 \, x^{2}\right )}} \,d x } \]
integrate((((32*x^4-48*x^3+18*x^2)*exp(3/2/x)-144*x^2+216*x-81)*exp((x*exp (3/2/x)-3)/exp(3/2/x))+(-320*x^4+480*x^3)*exp(3/2/x))/(32*x^4-48*x^3+18*x^ 2)/exp(3/2/x),x, algorithm=\
integrate(-1/2*((144*x^2 - 2*(16*x^4 - 24*x^3 + 9*x^2)*e^(3/2/x) - 216*x + 81)*e^((x*e^(3/2/x) - 3)*e^(-3/2/x)) + 160*(2*x^4 - 3*x^3)*e^(3/2/x))*e^( -3/2/x)/(16*x^4 - 24*x^3 + 9*x^2), x)
Time = 9.99 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89 \[ \int \frac {e^{\left .-\frac {3}{2}\right /x} \left (e^{\left .\frac {3}{2}\right /x} \left (480 x^3-320 x^4\right )+e^{e^{\left .-\frac {3}{2}\right /x} \left (-3+e^{\left .\frac {3}{2}\right /x} x\right )} \left (-81+216 x-144 x^2+e^{\left .\frac {3}{2}\right /x} \left (18 x^2-48 x^3+32 x^4\right )\right )\right )}{18 x^2-48 x^3+32 x^4} \, dx={\mathrm {e}}^{-\frac {3}{{\left ({\mathrm {e}}^{1/x}\right )}^{3/2}}}\,{\mathrm {e}}^x-\frac {45}{8\,\left (x-\frac {3}{4}\right )}-10\,x \]