3.25.63 \(\int \frac {e^{-2-2 x} (-1-2 x+48 e^{2+2 x} x^2)}{16 x^2} \, dx\)

Optimal. Leaf size=19 \[ -1+\frac {e^{-2-2 x}}{16 x}+3 x \]

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Rubi [A]  time = 0.23, antiderivative size = 18, normalized size of antiderivative = 0.95, number of steps used = 4, number of rules used = 3, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {12, 6742, 2197} \begin {gather*} 3 x+\frac {e^{-2 x-2}}{16 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(-2 - 2*x)*(-1 - 2*x + 48*E^(2 + 2*x)*x^2))/(16*x^2),x]

[Out]

E^(-2 - 2*x)/(16*x) + 3*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 2197

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> With[{b = Coefficient[v, x, 1], d = Coefficient[u, x, 0],
e = Coefficient[u, x, 1], f = Coefficient[w, x, 0], g = Coefficient[w, x, 1]}, Simp[(g*u^(m + 1)*F^(c*v))/(b*c
*e*Log[F]), x] /; EqQ[e*g*(m + 1) - b*c*(e*f - d*g)*Log[F], 0]] /; FreeQ[{F, c, m}, x] && LinearQ[{u, v, w}, 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} &=\frac {1}{16} \int \frac {e^{-2-2 x} \left (-1-2 x+48 e^{2+2 x} x^2\right )}{x^2} \, dx\\ &=\frac {1}{16} \int \left (48+\frac {e^{-2-2 x} (-1-2 x)}{x^2}\right ) \, dx\\ &=3 x+\frac {1}{16} \int \frac {e^{-2-2 x} (-1-2 x)}{x^2} \, dx\\ &=\frac {e^{-2-2 x}}{16 x}+3 x\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.10, size = 19, normalized size = 1.00 \begin {gather*} \frac {1}{16} \left (\frac {e^{-2-2 x}}{x}+48 x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(-2 - 2*x)*(-1 - 2*x + 48*E^(2 + 2*x)*x^2))/(16*x^2),x]

[Out]

(E^(-2 - 2*x)/x + 48*x)/16

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fricas [A]  time = 0.67, size = 31, normalized size = 1.63 \begin {gather*} \frac {{\left (3 \, x^{2} e^{\left (2 \, x + 4 \, \log \relax (2) + 2\right )} + 1\right )} e^{\left (-2 \, x - 4 \, \log \relax (2) - 2\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(3*x^2*e^(2*x + 4*log(2) + 2) + 1)*e^(-2*x - 4*log(2) - 2)/x

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giac [A]  time = 0.28, size = 19, normalized size = 1.00 \begin {gather*} \frac {{\left (48 \, x^{2} e^{2} + e^{\left (-2 \, x\right )}\right )} e^{\left (-2\right )}}{16 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(48*x^2*e^2 + e^(-2*x))*e^(-2)/x

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maple [A]  time = 0.08, size = 16, normalized size = 0.84

method	result	size
risch	\(3 x +\frac {{\mathrm e}^{-2 x -2}}{16 x}\)	\(16\)
norman	\(\frac {\left (1+48 x^{2} {\mathrm e}^{2 x +2}\right ) {\mathrm e}^{-2 x -2}}{16 x}\)	\(32\)
derivativedivides	\(-\frac {{\mathrm e}^{-4 \ln \relax (2)-2 x -2}}{x}+6 \ln \relax (2)+3 x +3+\frac {2 \,{\mathrm e}^{-4 \ln \relax (2)-2 x -2} \left (1+2 \ln \relax (2)\right )}{x}-\frac {4 \ln \relax (2) {\mathrm e}^{-4 \ln \relax (2)-2 x -2}}{x}\)	\(63\)
default	\(-\frac {{\mathrm e}^{-4 \ln \relax (2)-2 x -2}}{x}+6 \ln \relax (2)+3 x +3+\frac {2 \,{\mathrm e}^{-4 \ln \relax (2)-2 x -2} \left (1+2 \ln \relax (2)\right )}{x}-\frac {4 \ln \relax (2) {\mathrm e}^{-4 \ln \relax (2)-2 x -2}}{x}\)	\(63\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*x+1/16/x*exp(-2*x-2)

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maxima [C]  time = 0.45, size = 21, normalized size = 1.11 \begin {gather*} -\frac {1}{8} \, {\rm Ei}\left (-2 \, x\right ) e^{\left (-2\right )} + \frac {1}{8} \, e^{\left (-2\right )} \Gamma \left (-1, 2 \, x\right ) + 3 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*Ei(-2*x)*e^(-2) + 1/8*e^(-2)*gamma(-1, 2*x) + 3*x

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mupad [B]  time = 0.09, size = 15, normalized size = 0.79 \begin {gather*} 3\,x+\frac {{\mathrm {e}}^{-2\,x}\,{\mathrm {e}}^{-2}}{16\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(- 2*x - 4*log(2) - 2)*(2*x - 3*x^2*exp(2*x + 4*log(2) + 2) + 1))/x^2,x)

[Out]

3*x + (exp(-2*x)*exp(-2))/(16*x)

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sympy [A]  time = 0.11, size = 14, normalized size = 0.74 \begin {gather*} 3 x + \frac {e^{- 2 x - 2}}{16 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*x + exp(-2*x - 2)/(16*x)

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