3.65.18 \(\int \frac {e^{2+x+e x^2-x^3} (-1+e^{-2-x-e x^2+x^3}+x+2 e x^2-3 x^3)}{x^2} \, dx\)

Optimal. Leaf size=25 \[ \frac {-1+e^{2+e-(e-x) \left (1-x^2\right )}}{x} \]

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Rubi [B]  time = 1.22, antiderivative size = 51, normalized size of antiderivative = 2.04, number of steps used = 3, number of rules used = 2, integrand size = 49, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.041, Rules used = {6742, 2288} \begin {gather*} \frac {e^{-x^3+e x^2+x+2} \left (-3 x^3+2 e x^2+x\right )}{x^2 \left (-3 x^2+2 e x+1\right )}-\frac {1}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(2 + x + E*x^2 - x^3)*(-1 + E^(-2 - x - E*x^2 + x^3) + x + 2*E*x^2 - 3*x^3))/x^2,x]

[Out]

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

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]

Rule 6742

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

Rubi steps

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

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Mathematica [A]  time = 0.46, size = 21, normalized size = 0.84 \begin {gather*} \frac {-1+e^{2+x+e x^2-x^3}}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(2 + x + E*x^2 - x^3)*(-1 + E^(-2 - x - E*x^2 + x^3) + x + 2*E*x^2 - 3*x^3))/x^2,x]

[Out]

(-1 + E^(2 + x + E*x^2 - x^3))/x

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fricas [A]  time = 0.78, size = 21, normalized size = 0.84 \begin {gather*} \frac {e^{\left (-x^{3} + x^{2} e + x + 2\right )} - 1}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(e^(-x^3 + x^2*e + x + 2) - 1)/x

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giac [A]  time = 0.21, size = 21, normalized size = 0.84 \begin {gather*} \frac {e^{\left (-x^{3} + x^{2} e + x + 2\right )} - 1}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(e^(-x^3 + x^2*e + x + 2) - 1)/x

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maple [A]  time = 0.15, size = 26, normalized size = 1.04

method	result	size
risch	\(-\frac {1}{x}+\frac {{\mathrm e}^{x^{2} {\mathrm e}-x^{3}+x +2}}{x}\)	\(26\)
norman	\(\frac {\left (1-{\mathrm e}^{-x^{2} {\mathrm e}+x^{3}-x -2}\right ) {\mathrm e}^{x^{2} {\mathrm e}-x^{3}+x +2}}{x}\)	\(43\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/x+exp(x^2*exp(1)-x^3+x+2)/x

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maxima [A]  time = 0.44, size = 25, normalized size = 1.00 \begin {gather*} \frac {e^{\left (-x^{3} + x^{2} e + x + 2\right )}}{x} - \frac {1}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^(-x^3 + x^2*e + x + 2)/x - 1/x

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mupad [B]  time = 4.21, size = 27, normalized size = 1.08 \begin {gather*} \frac {{\mathrm {e}}^{x^2\,\mathrm {e}}\,{\mathrm {e}}^2\,{\mathrm {e}}^{-x^3}\,{\mathrm {e}}^x}{x}-\frac {1}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x + x^2*exp(1) - x^3 + 2)*(x + exp(x^3 - x^2*exp(1) - x - 2) + 2*x^2*exp(1) - 3*x^3 - 1))/x^2,x)

[Out]

(exp(x^2*exp(1))*exp(2)*exp(-x^3)*exp(x))/x - 1/x

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sympy [A]  time = 0.15, size = 19, normalized size = 0.76 \begin {gather*} \frac {e^{- x^{3} + e x^{2} + x + 2}}{x} - \frac {1}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(-x**2*exp(1)+x**3-x-2)+2*x**2*exp(1)-3*x**3+x-1)/x**2/exp(-x**2*exp(1)+x**3-x-2),x)

[Out]

exp(-x**3 + E*x**2 + x + 2)/x - 1/x

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