\(\int \frac {e^{-\frac {1}{x^2}} (40+10 x+20 x^2-5 x^3)}{64 x^2+48 x^3+12 x^4+x^5} \, dx\) [7731]

Optimal result

Integrand size = 44, antiderivative size = 15 \[ \int \frac {e^{-\frac {1}{x^2}} \left (40+10 x+20 x^2-5 x^3\right )}{64 x^2+48 x^3+12 x^4+x^5} \, dx=\frac {5 e^{-\frac {1}{x^2}} x}{(4+x)^2} \]

[Out]

5*x/exp(1/x^2)/(4+x)^2

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(36\) vs. \(2(15)=30\).

Time = 0.04 (sec) , antiderivative size = 36, normalized size of antiderivative = 2.40, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.023, Rules used = {2326} \[ \int \frac {e^{-\frac {1}{x^2}} \left (40+10 x+20 x^2-5 x^3\right )}{64 x^2+48 x^3+12 x^4+x^5} \, dx=\frac {5 e^{-\frac {1}{x^2}} x^3 (x+4)}{x^5+12 x^4+48 x^3+64 x^2} \]

[In]

Int[(40 + 10*x + 20*x^2 - 5*x^3)/(E^x^(-2)*(64*x^2 + 48*x^3 + 12*x^4 + x^5)),x]

[Out]

(5*x^3*(4 + x))/(E^x^(-2)*(64*x^2 + 48*x^3 + 12*x^4 + x^5))

Rule 2326

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, x]))}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps \begin{align*} \text {integral}& = \frac {5 e^{-\frac {1}{x^2}} x^3 (4+x)}{64 x^2+48 x^3+12 x^4+x^5} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-\frac {1}{x^2}} \left (40+10 x+20 x^2-5 x^3\right )}{64 x^2+48 x^3+12 x^4+x^5} \, dx=\frac {5 e^{-\frac {1}{x^2}} x}{(4+x)^2} \]

[In]

Integrate[(40 + 10*x + 20*x^2 - 5*x^3)/(E^x^(-2)*(64*x^2 + 48*x^3 + 12*x^4 + x^5)),x]

[Out]

(5*x)/(E^x^(-2)*(4 + x)^2)

Maple [A] (verified)

Time = 0.28 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00

method	result	size
norman	\(\frac {5 x \,{\mathrm e}^{-\frac {1}{x^{2}}}}{\left (4+x \right )^{2}}\)	\(15\)
risch	\(\frac {5 x \,{\mathrm e}^{-\frac {1}{x^{2}}}}{\left (4+x \right )^{2}}\)	\(15\)
gosper	\(\frac {5 x \,{\mathrm e}^{-\frac {1}{x^{2}}}}{x^{2}+8 x +16}\)	\(20\)
parallelrisch	\(\frac {5 x \,{\mathrm e}^{-\frac {1}{x^{2}}}}{x^{2}+8 x +16}\)	\(20\)

[In]

int((-5*x^3+20*x^2+10*x+40)/(x^5+12*x^4+48*x^3+64*x^2)/exp(1/x^2),x,method=_RETURNVERBOSE)

[Out]

5*x/exp(1/x^2)/(4+x)^2

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.27 \[ \int \frac {e^{-\frac {1}{x^2}} \left (40+10 x+20 x^2-5 x^3\right )}{64 x^2+48 x^3+12 x^4+x^5} \, dx=\frac {5 \, x e^{\left (-\frac {1}{x^{2}}\right )}}{x^{2} + 8 \, x + 16} \]

[In]

integrate((-5*x^3+20*x^2+10*x+40)/(x^5+12*x^4+48*x^3+64*x^2)/exp(1/x^2),x, algorithm="fricas")

[Out]

5*x*e^(-1/x^2)/(x^2 + 8*x + 16)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.13 \[ \int \frac {e^{-\frac {1}{x^2}} \left (40+10 x+20 x^2-5 x^3\right )}{64 x^2+48 x^3+12 x^4+x^5} \, dx=\frac {5 x e^{- \frac {1}{x^{2}}}}{x^{2} + 8 x + 16} \]

[In]

integrate((-5*x**3+20*x**2+10*x+40)/(x**5+12*x**4+48*x**3+64*x**2)/exp(1/x**2),x)

[Out]

5*x*exp(-1/x**2)/(x**2 + 8*x + 16)

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.27 \[ \int \frac {e^{-\frac {1}{x^2}} \left (40+10 x+20 x^2-5 x^3\right )}{64 x^2+48 x^3+12 x^4+x^5} \, dx=\frac {5 \, x e^{\left (-\frac {1}{x^{2}}\right )}}{x^{2} + 8 \, x + 16} \]

[In]

integrate((-5*x^3+20*x^2+10*x+40)/(x^5+12*x^4+48*x^3+64*x^2)/exp(1/x^2),x, algorithm="maxima")

[Out]

5*x*e^(-1/x^2)/(x^2 + 8*x + 16)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.27 \[ \int \frac {e^{-\frac {1}{x^2}} \left (40+10 x+20 x^2-5 x^3\right )}{64 x^2+48 x^3+12 x^4+x^5} \, dx=\frac {5 \, x e^{\left (-\frac {1}{x^{2}}\right )}}{x^{2} + 8 \, x + 16} \]

[In]

integrate((-5*x^3+20*x^2+10*x+40)/(x^5+12*x^4+48*x^3+64*x^2)/exp(1/x^2),x, algorithm="giac")

[Out]

5*x*e^(-1/x^2)/(x^2 + 8*x + 16)

Mupad [B] (verification not implemented)

Time = 13.52 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.93 \[ \int \frac {e^{-\frac {1}{x^2}} \left (40+10 x+20 x^2-5 x^3\right )}{64 x^2+48 x^3+12 x^4+x^5} \, dx=\frac {5\,x\,{\mathrm {e}}^{-\frac {1}{x^2}}}{{\left (x+4\right )}^2} \]

[In]

int((exp(-1/x^2)*(10*x + 20*x^2 - 5*x^3 + 40))/(64*x^2 + 48*x^3 + 12*x^4 + x^5),x)

[Out]

(5*x*exp(-1/x^2))/(x + 4)^2