3.57.78 \(\int \frac {e^{-2 x^2+2 x^3+e^x (x^2-x^3)} (4+16 x^2-24 x^3+e^x (-8 x^2+8 x^3+4 x^4))}{3 x^2} \, dx\)

Optimal. Leaf size=29 \[ \frac {4}{3} \left (1-\frac {e^{(1-x) x \left (-2 x+e^x x\right )}}{x}\right ) \]

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Rubi [B]  time = 0.48, antiderivative size = 104, normalized size of antiderivative = 3.59, number of steps used = 2, number of rules used = 2, integrand size = 65, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.031, Rules used = {12, 2288} \begin {gather*} -\frac {4 e^{2 x^3-2 x^2+e^x \left (x^2-x^3\right )} \left (-6 x^3+4 x^2-e^x \left (-x^4-2 x^3+2 x^2\right )\right )}{3 x^2 \left (-6 x^2-e^x \left (2 x-3 x^2\right )-e^x \left (x^2-x^3\right )+4 x\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(-2*x^2 + 2*x^3 + E^x*(x^2 - x^3))*(4 + 16*x^2 - 24*x^3 + E^x*(-8*x^2 + 8*x^3 + 4*x^4)))/(3*x^2),x]

[Out]

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

Rule 12

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

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 {1}{3} \int \frac {e^{-2 x^2+2 x^3+e^x \left (x^2-x^3\right )} \left (4+16 x^2-24 x^3+e^x \left (-8 x^2+8 x^3+4 x^4\right )\right )}{x^2} \, dx\\ &=-\frac {4 e^{-2 x^2+2 x^3+e^x \left (x^2-x^3\right )} \left (4 x^2-6 x^3-e^x \left (2 x^2-2 x^3-x^4\right )\right )}{3 x^2 \left (4 x-6 x^2-e^x \left (2 x-3 x^2\right )-e^x \left (x^2-x^3\right )\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.27, size = 22, normalized size = 0.76 \begin {gather*} -\frac {4 e^{-\left (\left (-2+e^x\right ) (-1+x) x^2\right )}}{3 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(-2*x^2 + 2*x^3 + E^x*(x^2 - x^3))*(4 + 16*x^2 - 24*x^3 + E^x*(-8*x^2 + 8*x^3 + 4*x^4)))/(3*x^2),
x]

[Out]

-4/(3*E^((-2 + E^x)*(-1 + x)*x^2)*x)

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fricas [A]  time = 1.18, size = 30, normalized size = 1.03 \begin {gather*} -\frac {4 \, e^{\left (2 \, x^{3} - 2 \, x^{2} - {\left (x^{3} - x^{2}\right )} e^{x}\right )}}{3 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((4*x^4+8*x^3-8*x^2)*exp(x)-24*x^3+16*x^2+4)*exp((-x^3+x^2)*exp(x)+2*x^3-2*x^2)/x^2,x, algorithm
="fricas")

[Out]

-4/3*e^(2*x^3 - 2*x^2 - (x^3 - x^2)*e^x)/x

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {4 \, {\left (6 \, x^{3} - 4 \, x^{2} - {\left (x^{4} + 2 \, x^{3} - 2 \, x^{2}\right )} e^{x} - 1\right )} e^{\left (2 \, x^{3} - 2 \, x^{2} - {\left (x^{3} - x^{2}\right )} e^{x}\right )}}{3 \, x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((4*x^4+8*x^3-8*x^2)*exp(x)-24*x^3+16*x^2+4)*exp((-x^3+x^2)*exp(x)+2*x^3-2*x^2)/x^2,x, algorithm
="giac")

[Out]

integrate(-4/3*(6*x^3 - 4*x^2 - (x^4 + 2*x^3 - 2*x^2)*e^x - 1)*e^(2*x^3 - 2*x^2 - (x^3 - x^2)*e^x)/x^2, x)

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maple [A]  time = 0.07, size = 19, normalized size = 0.66

method	result	size
risch	\(-\frac {4 \,{\mathrm e}^{-x^{2} \left (x -1\right ) \left ({\mathrm e}^{x}-2\right )}}{3 x}\)	\(19\)
norman	\(-\frac {4 \,{\mathrm e}^{\left (-x^{3}+x^{2}\right ) {\mathrm e}^{x}+2 x^{3}-2 x^{2}}}{3 x}\)	\(30\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/3*((4*x^4+8*x^3-8*x^2)*exp(x)-24*x^3+16*x^2+4)*exp((-x^3+x^2)*exp(x)+2*x^3-2*x^2)/x^2,x,method=_RETURNVE
RBOSE)

[Out]

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

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maxima [A]  time = 0.46, size = 30, normalized size = 1.03 \begin {gather*} -\frac {4 \, e^{\left (-x^{3} e^{x} + 2 \, x^{3} + x^{2} e^{x} - 2 \, x^{2}\right )}}{3 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((4*x^4+8*x^3-8*x^2)*exp(x)-24*x^3+16*x^2+4)*exp((-x^3+x^2)*exp(x)+2*x^3-2*x^2)/x^2,x, algorithm
="maxima")

[Out]

-4/3*e^(-x^3*e^x + 2*x^3 + x^2*e^x - 2*x^2)/x

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mupad [B]  time = 3.67, size = 32, normalized size = 1.10 \begin {gather*} -\frac {4\,{\mathrm {e}}^{x^2\,{\mathrm {e}}^x}\,{\mathrm {e}}^{-x^3\,{\mathrm {e}}^x}\,{\mathrm {e}}^{-2\,x^2}\,{\mathrm {e}}^{2\,x^3}}{3\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(exp(x)*(x^2 - x^3) - 2*x^2 + 2*x^3)*(exp(x)*(8*x^3 - 8*x^2 + 4*x^4) + 16*x^2 - 24*x^3 + 4))/(3*x^2),x
)

[Out]

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

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sympy [A]  time = 0.20, size = 27, normalized size = 0.93 \begin {gather*} - \frac {4 e^{2 x^{3} - 2 x^{2} + \left (- x^{3} + x^{2}\right ) e^{x}}}{3 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((4*x**4+8*x**3-8*x**2)*exp(x)-24*x**3+16*x**2+4)*exp((-x**3+x**2)*exp(x)+2*x**3-2*x**2)/x**2,x)

[Out]

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

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