3.54.2 \(\int \frac {e^{5-2 x+x^2-4 x^3} (-4-8 x+8 x^2-48 x^3)}{x^2} \, dx\)

Optimal. Leaf size=23 \[ \frac {4 e^{5+x-(3-x) x-4 x^3}}{x} \]

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Rubi [A]  time = 0.07, antiderivative size = 44, normalized size of antiderivative = 1.91, number of steps used = 1, number of rules used = 1, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.029, Rules used = {2288} \begin {gather*} \frac {4 e^{-4 x^3+x^2-2 x+5} \left (6 x^3-x^2+x\right )}{x^2 \left (6 x^2-x+1\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(5 - 2*x + x^2 - 4*x^3)*(-4 - 8*x + 8*x^2 - 48*x^3))/x^2,x]

[Out]

(4*E^(5 - 2*x + x^2 - 4*x^3)*(x - x^2 + 6*x^3))/(x^2*(1 - x + 6*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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {4 e^{5-2 x+x^2-4 x^3} \left (x-x^2+6 x^3\right )}{x^2 \left (1-x+6 x^2\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 20, normalized size = 0.87 \begin {gather*} \frac {4 e^{5-2 x+x^2-4 x^3}}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(5 - 2*x + x^2 - 4*x^3)*(-4 - 8*x + 8*x^2 - 48*x^3))/x^2,x]

[Out]

(4*E^(5 - 2*x + x^2 - 4*x^3))/x

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fricas [A]  time = 0.72, size = 19, normalized size = 0.83 \begin {gather*} \frac {4 \, e^{\left (-4 \, x^{3} + x^{2} - 2 \, x + 5\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-48*x^3+8*x^2-8*x-4)*exp(5)/x^2/exp(-x^2+2*x)/exp(4*x^3),x, algorithm="fricas")

[Out]

4*e^(-4*x^3 + x^2 - 2*x + 5)/x

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giac [A]  time = 0.24, size = 19, normalized size = 0.83 \begin {gather*} \frac {4 \, e^{\left (-4 \, x^{3} + x^{2} - 2 \, x + 5\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-48*x^3+8*x^2-8*x-4)*exp(5)/x^2/exp(-x^2+2*x)/exp(4*x^3),x, algorithm="giac")

[Out]

4*e^(-4*x^3 + x^2 - 2*x + 5)/x

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maple [A]  time = 0.07, size = 22, normalized size = 0.96

method	result	size
risch	\(\frac {4 \,{\mathrm e}^{-\left (x -1\right ) \left (4 x^{2}+3 x +5\right )}}{x}\)	\(22\)
gosper	\(\frac {4 \,{\mathrm e}^{5} {\mathrm e}^{-4 x^{3}} {\mathrm e}^{x^{2}-2 x}}{x}\)	\(28\)
norman	\(\frac {4 \,{\mathrm e}^{5} {\mathrm e}^{-4 x^{3}} {\mathrm e}^{x^{2}-2 x}}{x}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-48*x^3+8*x^2-8*x-4)*exp(5)/x^2/exp(-x^2+2*x)/exp(4*x^3),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.42, size = 19, normalized size = 0.83 \begin {gather*} \frac {4 \, e^{\left (-4 \, x^{3} + x^{2} - 2 \, x + 5\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-48*x^3+8*x^2-8*x-4)*exp(5)/x^2/exp(-x^2+2*x)/exp(4*x^3),x, algorithm="maxima")

[Out]

4*e^(-4*x^3 + x^2 - 2*x + 5)/x

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mupad [B]  time = 3.50, size = 21, normalized size = 0.91 \begin {gather*} \frac {4\,{\mathrm {e}}^{-2\,x}\,{\mathrm {e}}^{x^2}\,{\mathrm {e}}^5\,{\mathrm {e}}^{-4\,x^3}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(5)*exp(x^2 - 2*x)*exp(-4*x^3)*(8*x - 8*x^2 + 48*x^3 + 4))/x^2,x)

[Out]

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

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sympy [A]  time = 0.30, size = 20, normalized size = 0.87 \begin {gather*} \frac {4 e^{5} e^{- 4 x^{3}} e^{x^{2} - 2 x}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-48*x**3+8*x**2-8*x-4)*exp(5)/x**2/exp(-x**2+2*x)/exp(4*x**3),x)

[Out]

4*exp(5)*exp(-4*x**3)*exp(x**2 - 2*x)/x

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