3.42.64 \(\int \frac {x^2+e^{\frac {-2-3 x+5 x^2+x^3}{x^2}} (-4-3 x-x^2-x^3)}{x^2} \, dx\)

Optimal. Leaf size=22 \[ 14+x-e^{5+\frac {(-2+x) (1+x)^2}{x^2}} x \]

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Rubi [A]  time = 0.10, antiderivative size = 44, normalized size of antiderivative = 2.00, number of steps used = 3, number of rules used = 2, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.047, Rules used = {14, 2288} \begin {gather*} x-\frac {e^{-\frac {2}{x^2}+x-\frac {3}{x}+5} \left (x^3+3 x+4\right )}{\left (\frac {4}{x^3}+\frac {3}{x^2}+1\right ) x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

x - (E^(5 - 2/x^2 - 3/x + x)*(4 + 3*x + x^3))/((1 + 4/x^3 + 3/x^2)*x^2)

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
 &&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2288

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 {gather*} \begin {aligned} \text {integral} &=\int \left (1-\frac {e^{5-\frac {2}{x^2}-\frac {3}{x}+x} \left (4+3 x+x^2+x^3\right )}{x^2}\right ) \, dx\\ &=x-\int \frac {e^{5-\frac {2}{x^2}-\frac {3}{x}+x} \left (4+3 x+x^2+x^3\right )}{x^2} \, dx\\ &=x-\frac {e^{5-\frac {2}{x^2}-\frac {3}{x}+x} \left (4+3 x+x^3\right )}{\left (1+\frac {4}{x^3}+\frac {3}{x^2}\right ) x^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 20, normalized size = 0.91 \begin {gather*} x-e^{5-\frac {2}{x^2}-\frac {3}{x}+x} x \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

x - E^(5 - 2/x^2 - 3/x + x)*x

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fricas [A]  time = 0.53, size = 23, normalized size = 1.05 \begin {gather*} -x e^{\left (\frac {x^{3} + 5 \, x^{2} - 3 \, x - 2}{x^{2}}\right )} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.22, size = 23, normalized size = 1.05 \begin {gather*} -x e^{\left (\frac {x^{3} + 5 \, x^{2} - 3 \, x - 2}{x^{2}}\right )} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.11, size = 24, normalized size = 1.09




method result size



risch \(-{\mathrm e}^{\frac {x^{3}+5 x^{2}-3 x -2}{x^{2}}} x +x\) \(24\)
norman \(\frac {x^{2}-{\mathrm e}^{\frac {x^{3}+5 x^{2}-3 x -2}{x^{2}}} x^{2}}{x}\) \(32\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.55, size = 19, normalized size = 0.86 \begin {gather*} -x e^{\left (x - \frac {3}{x} - \frac {2}{x^{2}} + 5\right )} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x*e^(x - 3/x - 2/x^2 + 5) + x

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mupad [B]  time = 3.35, size = 21, normalized size = 0.95 \begin {gather*} x-x\,{\mathrm {e}}^5\,{\mathrm {e}}^{-\frac {3}{x}}\,{\mathrm {e}}^{-\frac {2}{x^2}}\,{\mathrm {e}}^x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-(3*x - 5*x^2 - x^3 + 2)/x^2)*(3*x + x^2 + x^3 + 4) - x^2)/x^2,x)

[Out]

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

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sympy [A]  time = 0.15, size = 20, normalized size = 0.91 \begin {gather*} - x e^{\frac {x^{3} + 5 x^{2} - 3 x - 2}{x^{2}}} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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