Integrand size = 98, antiderivative size = 31 \[ \int \frac {e^{-3/x} \left (180-15 x-50 x^2+28 e^{3/x} x^2+e^{\frac {3}{x}+x} \left (16 x^2+8 x^3+x^4\right )+e^{\frac {3}{x}+x^2} \left (32 x^3+16 x^4+2 x^5\right )\right )}{16 x^2+8 x^3+x^4} \, dx=4+e^x+e^{x^2}-\frac {\left (-4+5 e^{-3/x}\right ) (-3+x)}{4+x} \] Output:
exp(x)+exp(x^2)+4-(-3+x)/(4+x)*(5/exp(3/x)-4)
Time = 5.05 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.06 \[ \int \frac {e^{-3/x} \left (180-15 x-50 x^2+28 e^{3/x} x^2+e^{\frac {3}{x}+x} \left (16 x^2+8 x^3+x^4\right )+e^{\frac {3}{x}+x^2} \left (32 x^3+16 x^4+2 x^5\right )\right )}{16 x^2+8 x^3+x^4} \, dx=e^x+e^{x^2}-\frac {28}{4+x}+e^{-3/x} \left (-5+\frac {35}{4+x}\right ) \] Input:
Integrate[(180 - 15*x - 50*x^2 + 28*E^(3/x)*x^2 + E^(3/x + x)*(16*x^2 + 8* x^3 + x^4) + E^(3/x + x^2)*(32*x^3 + 16*x^4 + 2*x^5))/(E^(3/x)*(16*x^2 + 8 *x^3 + x^4)),x]
Output:
E^x + E^x^2 - 28/(4 + x) + (-5 + 35/(4 + x))/E^(3/x)
Time = 2.58 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.26, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.041, Rules used = {2026, 2007, 7293, 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.
\(\displaystyle \int \frac {e^{-3/x} \left (28 e^{3/x} x^2-50 x^2+e^{x+\frac {3}{x}} \left (x^4+8 x^3+16 x^2\right )+e^{x^2+\frac {3}{x}} \left (2 x^5+16 x^4+32 x^3\right )-15 x+180\right )}{x^4+8 x^3+16 x^2} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {e^{-3/x} \left (28 e^{3/x} x^2-50 x^2+e^{x+\frac {3}{x}} \left (x^4+8 x^3+16 x^2\right )+e^{x^2+\frac {3}{x}} \left (2 x^5+16 x^4+32 x^3\right )-15 x+180\right )}{x^2 \left (x^2+8 x+16\right )}dx\) |
\(\Big \downarrow \) 2007 |
\(\displaystyle \int \frac {e^{-3/x} \left (28 e^{3/x} x^2-50 x^2+e^{x+\frac {3}{x}} \left (x^4+8 x^3+16 x^2\right )+e^{x^2+\frac {3}{x}} \left (2 x^5+16 x^4+32 x^3\right )-15 x+180\right )}{x^2 (x+4)^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (2 e^{x^2} x+\frac {180 e^{-3/x}}{(x+4)^2 x^2}+e^x-\frac {50 e^{-3/x}}{(x+4)^2}+\frac {28}{(x+4)^2}-\frac {15 e^{-3/x}}{(x+4)^2 x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle e^{x^2}-5 e^{-3/x}+e^x+\frac {35 e^{-3/x}}{x+4}-\frac {28}{x+4}\) |
Input:
Int[(180 - 15*x - 50*x^2 + 28*E^(3/x)*x^2 + E^(3/x + x)*(16*x^2 + 8*x^3 + x^4) + E^(3/x + x^2)*(32*x^3 + 16*x^4 + 2*x^5))/(E^(3/x)*(16*x^2 + 8*x^3 + x^4)),x]
Output:
-5/E^(3/x) + E^x + E^x^2 - 28/(4 + x) + 35/(E^(3/x)*(4 + x))
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 = 2.36 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00
method | result | size |
risch | \(-\frac {28}{4+x}+{\mathrm e}^{x}+{\mathrm e}^{x^{2}}-\frac {5 \left (-3+x \right ) {\mathrm e}^{-\frac {3}{x}}}{4+x}\) | \(31\) |
parts | \(-\frac {28}{4+x}+{\mathrm e}^{x^{2}}-\frac {105 \,{\mathrm e}^{-\frac {3}{x}}}{4 \left (3+\frac {12}{x}\right )}+\frac {15 \,{\mathrm e}^{-\frac {3}{x}}}{4}+{\mathrm e}^{x}\) | \(40\) |
parallelrisch | \(\frac {\left (x^{2} {\mathrm e}^{\frac {3}{x}} {\mathrm e}^{x}+{\mathrm e}^{\frac {3}{x}} x^{2} {\mathrm e}^{x^{2}}+4 \,{\mathrm e}^{\frac {3}{x}} {\mathrm e}^{x} x +4 \,{\mathrm e}^{\frac {3}{x}} x \,{\mathrm e}^{x^{2}}-5 x^{2}-28 x \,{\mathrm e}^{\frac {3}{x}}+15 x \right ) {\mathrm e}^{-\frac {3}{x}}}{x \left (4+x \right )}\) | \(86\) |
orering | \(\text {Expression too large to display}\) | \(6413\) |
Input:
int(((2*x^5+16*x^4+32*x^3)*exp(3/x)*exp(x^2)+(x^4+8*x^3+16*x^2)*exp(3/x)*e xp(x)+28*x^2*exp(3/x)-50*x^2-15*x+180)/(x^4+8*x^3+16*x^2)/exp(3/x),x,metho d=_RETURNVERBOSE)
Output:
-28/(4+x)+exp(x)+exp(x^2)-5*(-3+x)/(4+x)*exp(-3/x)
Time = 0.09 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.71 \[ \int \frac {e^{-3/x} \left (180-15 x-50 x^2+28 e^{3/x} x^2+e^{\frac {3}{x}+x} \left (16 x^2+8 x^3+x^4\right )+e^{\frac {3}{x}+x^2} \left (32 x^3+16 x^4+2 x^5\right )\right )}{16 x^2+8 x^3+x^4} \, dx=\frac {{\left ({\left (x + 4\right )} e^{\left (\frac {x^{3} + 3}{x}\right )} + {\left (x + 4\right )} e^{\left (\frac {x^{2} + 3}{x}\right )} - 5 \, x - 28 \, e^{\frac {3}{x}} + 15\right )} e^{\left (-\frac {3}{x}\right )}}{x + 4} \] Input:
integrate(((2*x^5+16*x^4+32*x^3)*exp(3/x)*exp(x^2)+(x^4+8*x^3+16*x^2)*exp( 3/x)*exp(x)+28*x^2*exp(3/x)-50*x^2-15*x+180)/(x^4+8*x^3+16*x^2)/exp(3/x),x , algorithm="fricas")
Output:
((x + 4)*e^((x^3 + 3)/x) + (x + 4)*e^((x^2 + 3)/x) - 5*x - 28*e^(3/x) + 15 )*e^(-3/x)/(x + 4)
Time = 0.25 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84 \[ \int \frac {e^{-3/x} \left (180-15 x-50 x^2+28 e^{3/x} x^2+e^{\frac {3}{x}+x} \left (16 x^2+8 x^3+x^4\right )+e^{\frac {3}{x}+x^2} \left (32 x^3+16 x^4+2 x^5\right )\right )}{16 x^2+8 x^3+x^4} \, dx=\frac {\left (15 - 5 x\right ) e^{- \frac {3}{x}}}{x + 4} + e^{x} + e^{x^{2}} - \frac {28}{x + 4} \] Input:
integrate(((2*x**5+16*x**4+32*x**3)*exp(3/x)*exp(x**2)+(x**4+8*x**3+16*x** 2)*exp(3/x)*exp(x)+28*x**2*exp(3/x)-50*x**2-15*x+180)/(x**4+8*x**3+16*x**2 )/exp(3/x),x)
Output:
(15 - 5*x)*exp(-3/x)/(x + 4) + exp(x) + exp(x**2) - 28/(x + 4)
Time = 0.08 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.06 \[ \int \frac {e^{-3/x} \left (180-15 x-50 x^2+28 e^{3/x} x^2+e^{\frac {3}{x}+x} \left (16 x^2+8 x^3+x^4\right )+e^{\frac {3}{x}+x^2} \left (32 x^3+16 x^4+2 x^5\right )\right )}{16 x^2+8 x^3+x^4} \, dx=\frac {{\left (x + 4\right )} e^{\left (x^{2}\right )} + {\left (x + 4\right )} e^{x} - 5 \, {\left (x - 3\right )} e^{\left (-\frac {3}{x}\right )} - 28}{x + 4} \] Input:
integrate(((2*x^5+16*x^4+32*x^3)*exp(3/x)*exp(x^2)+(x^4+8*x^3+16*x^2)*exp( 3/x)*exp(x)+28*x^2*exp(3/x)-50*x^2-15*x+180)/(x^4+8*x^3+16*x^2)/exp(3/x),x , algorithm="maxima")
Output:
((x + 4)*e^(x^2) + (x + 4)*e^x - 5*(x - 3)*e^(-3/x) - 28)/(x + 4)
Time = 0.12 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.45 \[ \int \frac {e^{-3/x} \left (180-15 x-50 x^2+28 e^{3/x} x^2+e^{\frac {3}{x}+x} \left (16 x^2+8 x^3+x^4\right )+e^{\frac {3}{x}+x^2} \left (32 x^3+16 x^4+2 x^5\right )\right )}{16 x^2+8 x^3+x^4} \, dx=\frac {x e^{\left (x^{2}\right )} + x e^{x} - 5 \, x e^{\left (-\frac {3}{x}\right )} + 4 \, e^{\left (x^{2}\right )} + 4 \, e^{x} + 15 \, e^{\left (-\frac {3}{x}\right )} - 28}{x + 4} \] Input:
integrate(((2*x^5+16*x^4+32*x^3)*exp(3/x)*exp(x^2)+(x^4+8*x^3+16*x^2)*exp( 3/x)*exp(x)+28*x^2*exp(3/x)-50*x^2-15*x+180)/(x^4+8*x^3+16*x^2)/exp(3/x),x , algorithm="giac")
Output:
(x*e^(x^2) + x*e^x - 5*x*e^(-3/x) + 4*e^(x^2) + 4*e^x + 15*e^(-3/x) - 28)/ (x + 4)
Time = 3.23 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.03 \[ \int \frac {e^{-3/x} \left (180-15 x-50 x^2+28 e^{3/x} x^2+e^{\frac {3}{x}+x} \left (16 x^2+8 x^3+x^4\right )+e^{\frac {3}{x}+x^2} \left (32 x^3+16 x^4+2 x^5\right )\right )}{16 x^2+8 x^3+x^4} \, dx={\mathrm {e}}^{x^2}+{\mathrm {e}}^x-\frac {28}{x+4}-\frac {{\mathrm {e}}^{-\frac {3}{x}}\,\left (5\,x-15\right )}{x+4} \] Input:
int((exp(-3/x)*(28*x^2*exp(3/x) - 15*x - 50*x^2 + exp(x^2)*exp(3/x)*(32*x^ 3 + 16*x^4 + 2*x^5) + exp(3/x)*exp(x)*(16*x^2 + 8*x^3 + x^4) + 180))/(16*x ^2 + 8*x^3 + x^4),x)
Output:
exp(x^2) + exp(x) - 28/(x + 4) - (exp(-3/x)*(5*x - 15))/(x + 4)
Time = 0.18 (sec) , antiderivative size = 82, normalized size of antiderivative = 2.65 \[ \int \frac {e^{-3/x} \left (180-15 x-50 x^2+28 e^{3/x} x^2+e^{\frac {3}{x}+x} \left (16 x^2+8 x^3+x^4\right )+e^{\frac {3}{x}+x^2} \left (32 x^3+16 x^4+2 x^5\right )\right )}{16 x^2+8 x^3+x^4} \, dx=\frac {e^{\frac {x^{3}+3}{x}} x +4 e^{\frac {x^{3}+3}{x}}+e^{\frac {x^{2}+3}{x}} x +4 e^{\frac {x^{2}+3}{x}}+7 e^{\frac {3}{x}} x -5 x +15}{e^{\frac {3}{x}} \left (x +4\right )} \] Input:
int(((2*x^5+16*x^4+32*x^3)*exp(3/x)*exp(x^2)+(x^4+8*x^3+16*x^2)*exp(3/x)*e xp(x)+28*x^2*exp(3/x)-50*x^2-15*x+180)/(x^4+8*x^3+16*x^2)/exp(3/x),x)
Output:
(e**((x**3 + 3)/x)*x + 4*e**((x**3 + 3)/x) + e**((x**2 + 3)/x)*x + 4*e**(( x**2 + 3)/x) + 7*e**(3/x)*x - 5*x + 15)/(e**(3/x)*(x + 4))