3.61.96 \(\int \frac {e^{\frac {1}{2} (5 e^{\frac {1}{x^2}} x-5 x^2-5 x^3)} (2 x+10 x^3+15 x^4+e^{\frac {1}{x^2}} (10-5 x^2))}{2 e^{5 e^{\frac {1}{x^2}} x-5 x^2-5 x^3} x-4 e^{\frac {1}{2} (5 e^{\frac {1}{x^2}} x-5 x^2-5 x^3)} x^2+2 x^3} \, dx\)

Optimal. Leaf size=32 \[ \frac {x}{e^{\frac {5}{2} \left (e^{\frac {1}{x^2}} x-x^2-x^3\right )}-x} \]

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Rubi [A]  time = 4.09, antiderivative size = 34, normalized size of antiderivative = 1.06, number of steps used = 4, number of rules used = 4, integrand size = 115, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.035, Rules used = {6688, 12, 6711, 32} \begin {gather*} -\frac {1}{1-\frac {e^{\frac {5}{2} e^{\frac {1}{x^2}} x-\frac {5}{2} x^2 (x+1)}}{x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((5*E^x^(-2)*x - 5*x^2 - 5*x^3)/2)*(2*x + 10*x^3 + 15*x^4 + E^x^(-2)*(10 - 5*x^2)))/(2*E^(5*E^x^(-2)*x
- 5*x^2 - 5*x^3)*x - 4*E^((5*E^x^(-2)*x - 5*x^2 - 5*x^3)/2)*x^2 + 2*x^3),x]

[Out]

-(1 - E^((5*E^x^(-2)*x)/2 - (5*x^2*(1 + x))/2)/x)^(-1)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rule 6711

Int[(u_)*((a_.)*(v_)^(p_.) + (b_.)*(w_)^(q_.))^(m_.), x_Symbol] :> With[{c = Simplify[u/(p*w*D[v, x] - q*v*D[w
, x])]}, Dist[c*p, Subst[Int[(b + a*x^p)^m, x], x, v*w^(m*q + 1)], x] /; FreeQ[c, x]] /; FreeQ[{a, b, m, p, q}
, x] && EqQ[p + q*(m*p + 1), 0] && IntegerQ[p] && IntegerQ[m]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{\frac {5}{2} x \left (e^{\frac {1}{x^2}}+x+x^2\right )} \left (-5 e^{\frac {1}{x^2}} \left (-2+x^2\right )+x \left (2+10 x^2+15 x^3\right )\right )}{2 x \left (e^{\frac {5}{2} e^{\frac {1}{x^2}} x}-e^{\frac {5}{2} x^2 (1+x)} x\right )^2} \, dx\\ &=\frac {1}{2} \int \frac {e^{\frac {5}{2} x \left (e^{\frac {1}{x^2}}+x+x^2\right )} \left (-5 e^{\frac {1}{x^2}} \left (-2+x^2\right )+x \left (2+10 x^2+15 x^3\right )\right )}{x \left (e^{\frac {5}{2} e^{\frac {1}{x^2}} x}-e^{\frac {5}{2} x^2 (1+x)} x\right )^2} \, dx\\ &=\operatorname {Subst}\left (\int \frac {1}{(1+x)^2} \, dx,x,-\frac {e^{\frac {5}{2} e^{\frac {1}{x^2}} x-\frac {5}{2} x^2 (1+x)}}{x}\right )\\ &=-\frac {1}{1-\frac {e^{\frac {5}{2} e^{\frac {1}{x^2}} x-\frac {5}{2} x^2 (1+x)}}{x}}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 1.12, size = 46, normalized size = 1.44 \begin {gather*} -\frac {e^{\frac {5}{2} x^2 (1+x)} x}{-e^{\frac {5}{2} e^{\frac {1}{x^2}} x}+e^{\frac {5}{2} x^2 (1+x)} x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((5*E^x^(-2)*x - 5*x^2 - 5*x^3)/2)*(2*x + 10*x^3 + 15*x^4 + E^x^(-2)*(10 - 5*x^2)))/(2*E^(5*E^x^(
-2)*x - 5*x^2 - 5*x^3)*x - 4*E^((5*E^x^(-2)*x - 5*x^2 - 5*x^3)/2)*x^2 + 2*x^3),x]

[Out]

-((E^((5*x^2*(1 + x))/2)*x)/(-E^((5*E^x^(-2)*x)/2) + E^((5*x^2*(1 + x))/2)*x))

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fricas [A]  time = 0.53, size = 28, normalized size = 0.88 \begin {gather*} -\frac {x}{x - e^{\left (-\frac {5}{2} \, x^{3} - \frac {5}{2} \, x^{2} + \frac {5}{2} \, x e^{\left (\frac {1}{x^{2}}\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*x^2+10)*exp(1/x^2)+15*x^4+10*x^3+2*x)*exp(5/2*x*exp(1/x^2)-5/2*x^3-5/2*x^2)/(2*x*exp(5/2*x*exp(
1/x^2)-5/2*x^3-5/2*x^2)^2-4*x^2*exp(5/2*x*exp(1/x^2)-5/2*x^3-5/2*x^2)+2*x^3),x, algorithm="fricas")

[Out]

-x/(x - e^(-5/2*x^3 - 5/2*x^2 + 5/2*x*e^(x^(-2))))

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giac [A]  time = 0.48, size = 28, normalized size = 0.88 \begin {gather*} -\frac {x}{x - e^{\left (-\frac {5}{2} \, x^{3} - \frac {5}{2} \, x^{2} + \frac {5}{2} \, x e^{\left (\frac {1}{x^{2}}\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*x^2+10)*exp(1/x^2)+15*x^4+10*x^3+2*x)*exp(5/2*x*exp(1/x^2)-5/2*x^3-5/2*x^2)/(2*x*exp(5/2*x*exp(
1/x^2)-5/2*x^3-5/2*x^2)^2-4*x^2*exp(5/2*x*exp(1/x^2)-5/2*x^3-5/2*x^2)+2*x^3),x, algorithm="giac")

[Out]

-x/(x - e^(-5/2*x^3 - 5/2*x^2 + 5/2*x*e^(x^(-2))))

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maple [A]  time = 0.12, size = 25, normalized size = 0.78

method	result	size
risch	\(-\frac {x}{x -{\mathrm e}^{-\frac {5 x \left (x^{2}-{\mathrm e}^{\frac {1}{x^{2}}}+x \right )}{2}}}\)	\(25\)
norman	\(-\frac {{\mathrm e}^{\frac {5 x \,{\mathrm e}^{\frac {1}{x^{2}}}}{2}-\frac {5 x^{3}}{2}-\frac {5 x^{2}}{2}}}{x -{\mathrm e}^{\frac {5 x \,{\mathrm e}^{\frac {1}{x^{2}}}}{2}-\frac {5 x^{3}}{2}-\frac {5 x^{2}}{2}}}\)	\(47\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-5*x^2+10)*exp(1/x^2)+15*x^4+10*x^3+2*x)*exp(5/2*x*exp(1/x^2)-5/2*x^3-5/2*x^2)/(2*x*exp(5/2*x*exp(1/x^2)
-5/2*x^3-5/2*x^2)^2-4*x^2*exp(5/2*x*exp(1/x^2)-5/2*x^3-5/2*x^2)+2*x^3),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.44, size = 42, normalized size = 1.31 \begin {gather*} -\frac {x e^{\left (\frac {5}{2} \, x^{3} + \frac {5}{2} \, x^{2}\right )}}{x e^{\left (\frac {5}{2} \, x^{3} + \frac {5}{2} \, x^{2}\right )} - e^{\left (\frac {5}{2} \, x e^{\left (\frac {1}{x^{2}}\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*x^2+10)*exp(1/x^2)+15*x^4+10*x^3+2*x)*exp(5/2*x*exp(1/x^2)-5/2*x^3-5/2*x^2)/(2*x*exp(5/2*x*exp(
1/x^2)-5/2*x^3-5/2*x^2)^2-4*x^2*exp(5/2*x*exp(1/x^2)-5/2*x^3-5/2*x^2)+2*x^3),x, algorithm="maxima")

[Out]

-x*e^(5/2*x^3 + 5/2*x^2)/(x*e^(5/2*x^3 + 5/2*x^2) - e^(5/2*x*e^(x^(-2))))

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mupad [B]  time = 4.56, size = 95, normalized size = 2.97 \begin {gather*} -\frac {x^4\,\left (2\,x+10\,{\mathrm {e}}^{\frac {1}{x^2}}-5\,x^2\,{\mathrm {e}}^{\frac {1}{x^2}}+10\,x^3+15\,x^4\right )}{\left (x-{\mathrm {e}}^{\frac {5\,x\,{\mathrm {e}}^{\frac {1}{x^2}}}{2}-\frac {5\,x^2}{2}-\frac {5\,x^3}{2}}\right )\,\left (10\,x^3\,{\mathrm {e}}^{\frac {1}{x^2}}-5\,x^5\,{\mathrm {e}}^{\frac {1}{x^2}}+2\,x^4+10\,x^6+15\,x^7\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((5*x*exp(1/x^2))/2 - (5*x^2)/2 - (5*x^3)/2)*(2*x - exp(1/x^2)*(5*x^2 - 10) + 10*x^3 + 15*x^4))/(2*x*e
xp(5*x*exp(1/x^2) - 5*x^2 - 5*x^3) - 4*x^2*exp((5*x*exp(1/x^2))/2 - (5*x^2)/2 - (5*x^3)/2) + 2*x^3),x)

[Out]

-(x^4*(2*x + 10*exp(1/x^2) - 5*x^2*exp(1/x^2) + 10*x^3 + 15*x^4))/((x - exp((5*x*exp(1/x^2))/2 - (5*x^2)/2 - (
5*x^3)/2))*(10*x^3*exp(1/x^2) - 5*x^5*exp(1/x^2) + 2*x^4 + 10*x^6 + 15*x^7))

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sympy [A]  time = 0.35, size = 29, normalized size = 0.91 \begin {gather*} \frac {x}{- x + e^{- \frac {5 x^{3}}{2} - \frac {5 x^{2}}{2} + \frac {5 x e^{\frac {1}{x^{2}}}}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*x**2+10)*exp(1/x**2)+15*x**4+10*x**3+2*x)*exp(5/2*x*exp(1/x**2)-5/2*x**3-5/2*x**2)/(2*x*exp(5/2
*x*exp(1/x**2)-5/2*x**3-5/2*x**2)**2-4*x**2*exp(5/2*x*exp(1/x**2)-5/2*x**3-5/2*x**2)+2*x**3),x)

[Out]

x/(-x + exp(-5*x**3/2 - 5*x**2/2 + 5*x*exp(x**(-2))/2))

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