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

Optimal. Leaf size=20 \[ \frac {2+x-4 e^{-2 x} x^2}{5 x} \]

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Rubi [A]  time = 0.18, antiderivative size = 31, normalized size of antiderivative = 1.55, number of steps used = 5, number of rules used = 4, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {12, 6688, 2176, 2194} \begin {gather*} \frac {2}{5} e^{-2 x} (1-2 x)-\frac {2 e^{-2 x}}{5}+\frac {2}{5 x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

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

Rule 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{5} \int \frac {e^{-2 x} \left (-2 e^{2 x}-4 x^2+8 x^3\right )}{x^2} \, dx\\ &=\frac {1}{5} \int \left (-\frac {2}{x^2}+e^{-2 x} (-4+8 x)\right ) \, dx\\ &=\frac {2}{5 x}+\frac {1}{5} \int e^{-2 x} (-4+8 x) \, dx\\ &=\frac {2}{5} e^{-2 x} (1-2 x)+\frac {2}{5 x}+\frac {4}{5} \int e^{-2 x} \, dx\\ &=-\frac {2}{5} e^{-2 x}+\frac {2}{5} e^{-2 x} (1-2 x)+\frac {2}{5 x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 18, normalized size = 0.90 \begin {gather*} \frac {2}{5 x}-\frac {4}{5} e^{-2 x} x \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

2/(5*x) - (4*x)/(5*E^(2*x))

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fricas [A]  time = 0.91, size = 21, normalized size = 1.05 \begin {gather*} -\frac {2 \, {\left (2 \, x^{2} - e^{\left (2 \, x\right )}\right )} e^{\left (-2 \, x\right )}}{5 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-2*exp(x)^2+8*x^3-4*x^2)/exp(x)^2/x^2,x, algorithm="fricas")

[Out]

-2/5*(2*x^2 - e^(2*x))*e^(-2*x)/x

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giac [A]  time = 0.19, size = 16, normalized size = 0.80 \begin {gather*} -\frac {2 \, {\left (2 \, x^{2} e^{\left (-2 \, x\right )} - 1\right )}}{5 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-2*exp(x)^2+8*x^3-4*x^2)/exp(x)^2/x^2,x, algorithm="giac")

[Out]

-2/5*(2*x^2*e^(-2*x) - 1)/x

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maple [A]  time = 0.02, size = 14, normalized size = 0.70

method	result	size
default	\(\frac {2}{5 x}-\frac {4 x \,{\mathrm e}^{-2 x}}{5}\)	\(14\)
risch	\(\frac {2}{5 x}-\frac {4 x \,{\mathrm e}^{-2 x}}{5}\)	\(14\)
norman	\(\frac {\left (-\frac {4 x^{2}}{5}+\frac {2 \,{\mathrm e}^{2 x}}{5}\right ) {\mathrm e}^{-2 x}}{x}\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*(-2*exp(x)^2+8*x^3-4*x^2)/exp(x)^2/x^2,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.50, size = 23, normalized size = 1.15 \begin {gather*} -\frac {2}{5} \, {\left (2 \, x + 1\right )} e^{\left (-2 \, x\right )} + \frac {2}{5 \, x} + \frac {2}{5} \, e^{\left (-2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-2*exp(x)^2+8*x^3-4*x^2)/exp(x)^2/x^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.07, size = 13, normalized size = 0.65 \begin {gather*} \frac {2}{5\,x}-\frac {4\,x\,{\mathrm {e}}^{-2\,x}}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/(5*x) - (4*x*exp(-2*x))/5

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sympy [A]  time = 0.11, size = 14, normalized size = 0.70 \begin {gather*} - \frac {4 x e^{- 2 x}}{5} + \frac {2}{5 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-2*exp(x)**2+8*x**3-4*x**2)/exp(x)**2/x**2,x)

[Out]

-4*x*exp(-2*x)/5 + 2/(5*x)

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