3.81.8 \(\int \frac {e^{-x} (6+8 x-e x-2 x^2-3 x \log (32 x^2))}{3 x} \, dx\)

Optimal. Leaf size=24 \[ e^{-x} \left (-2+\frac {e-x}{3}+x+\log \left (32 x^2\right )\right ) \]

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Rubi [A]  time = 0.25, antiderivative size = 46, normalized size of antiderivative = 1.92, number of steps used = 13, number of rules used = 8, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.229, Rules used = {6, 12, 6742, 2199, 2194, 2178, 2176, 2554} \begin {gather*} e^{-x} \log \left (32 x^2\right )+\frac {2 e^{-x} x}{3}-\frac {1}{3} (8-e) e^{-x}+\frac {2 e^{-x}}{3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(6 + 8*x - E*x - 2*x^2 - 3*x*Log[32*x^2])/(3*E^x*x),x]

[Out]

2/(3*E^x) - (8 - E)/(3*E^x) + (2*x)/(3*E^x) + Log[32*x^2]/E^x

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, 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 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$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 2199

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), w*NormalizePo
werOfLinear[u, x]^m, x], x] /; FreeQ[{F, c}, x] && PolynomialQ[w, x] && LinearQ[v, x] && PowerOfLinearQ[u, x]
&& IntegerQ[m] &&  !$UseGamma === True

Rule 2554

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[(w*D[u, x]
)/u, x], x] /; InverseFunctionFreeQ[w, x]] /; InverseFunctionFreeQ[u, 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 \frac {e^{-x} \left (6+(8-e) x-2 x^2-3 x \log \left (32 x^2\right )\right )}{3 x} \, dx\\ &=\frac {1}{3} \int \frac {e^{-x} \left (6+(8-e) x-2 x^2-3 x \log \left (32 x^2\right )\right )}{x} \, dx\\ &=\frac {1}{3} \int \left (\frac {e^{-x} \left (6+(8-e) x-2 x^2\right )}{x}-3 e^{-x} \log \left (32 x^2\right )\right ) \, dx\\ &=\frac {1}{3} \int \frac {e^{-x} \left (6+(8-e) x-2 x^2\right )}{x} \, dx-\int e^{-x} \log \left (32 x^2\right ) \, dx\\ &=e^{-x} \log \left (32 x^2\right )+\frac {1}{3} \int \left ((8-e) e^{-x}+\frac {6 e^{-x}}{x}-2 e^{-x} x\right ) \, dx+\int -\frac {2 e^{-x}}{x} \, dx\\ &=e^{-x} \log \left (32 x^2\right )-\frac {2}{3} \int e^{-x} x \, dx+\frac {1}{3} (8-e) \int e^{-x} \, dx\\ &=-\frac {1}{3} (8-e) e^{-x}+\frac {2 e^{-x} x}{3}+e^{-x} \log \left (32 x^2\right )-\frac {2}{3} \int e^{-x} \, dx\\ &=\frac {2 e^{-x}}{3}-\frac {1}{3} (8-e) e^{-x}+\frac {2 e^{-x} x}{3}+e^{-x} \log \left (32 x^2\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.08, size = 23, normalized size = 0.96 \begin {gather*} \frac {1}{3} e^{-x} \left (-6+e+2 x+3 \log \left (32 x^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(6 + 8*x - E*x - 2*x^2 - 3*x*Log[32*x^2])/(3*E^x*x),x]

[Out]

(-6 + E + 2*x + 3*Log[32*x^2])/(3*E^x)

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fricas [A]  time = 0.65, size = 25, normalized size = 1.04 \begin {gather*} \frac {1}{3} \, {\left (2 \, x + e - 6\right )} e^{\left (-x\right )} + e^{\left (-x\right )} \log \left (32 \, x^{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(-3*x*log(32*x^2)-x*exp(1)-2*x^2+8*x+6)/exp(x)/x,x, algorithm="fricas")

[Out]

1/3*(2*x + e - 6)*e^(-x) + e^(-x)*log(32*x^2)

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giac [A]  time = 0.14, size = 33, normalized size = 1.38 \begin {gather*} \frac {2}{3} \, x e^{\left (-x\right )} + e^{\left (-x\right )} \log \left (32 \, x^{2}\right ) - 2 \, e^{\left (-x\right )} + \frac {1}{3} \, e^{\left (-x + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(-3*x*log(32*x^2)-x*exp(1)-2*x^2+8*x+6)/exp(x)/x,x, algorithm="giac")

[Out]

2/3*x*e^(-x) + e^(-x)*log(32*x^2) - 2*e^(-x) + 1/3*e^(-x + 1)

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maple [A]  time = 0.12, size = 21, normalized size = 0.88

method	result	size
norman	\(\left (\frac {{\mathrm e}}{3}+\frac {2 x}{3}-2+\ln \left (32 x^{2}\right )\right ) {\mathrm e}^{-x}\)	\(21\)
default	\(\frac {\left (2 x -6+{\mathrm e}+3 \ln \left (32 x^{2}\right )\right ) {\mathrm e}^{-x}}{3}\)	\(22\)
risch	\(2 \ln \relax (x ) {\mathrm e}^{-x}+\frac {\left (-12-3 i \pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )+6 i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}-3 i \pi \mathrm {csgn}\left (i x^{2}\right )^{3}+2 \,{\mathrm e}+30 \ln \relax (2)+4 x \right ) {\mathrm e}^{-x}}{6}\)	\(78\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/3*(-3*x*ln(32*x^2)-x*exp(1)-2*x^2+8*x+6)/exp(x)/x,x,method=_RETURNVERBOSE)

[Out]

(1/3*exp(1)+2/3*x-2+ln(32*x^2))/exp(x)

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maxima [A]  time = 0.41, size = 35, normalized size = 1.46 \begin {gather*} \frac {2}{3} \, {\left (x + 1\right )} e^{\left (-x\right )} + e^{\left (-x\right )} \log \left (32 \, x^{2}\right ) - \frac {8}{3} \, e^{\left (-x\right )} + \frac {1}{3} \, e^{\left (-x + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(-3*x*log(32*x^2)-x*exp(1)-2*x^2+8*x+6)/exp(x)/x,x, algorithm="maxima")

[Out]

2/3*(x + 1)*e^(-x) + e^(-x)*log(32*x^2) - 8/3*e^(-x) + 1/3*e^(-x + 1)

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mupad [B]  time = 7.11, size = 21, normalized size = 0.88 \begin {gather*} \frac {{\mathrm {e}}^{-x}\,\left (2\,x+\mathrm {e}+3\,\ln \left (32\,x^2\right )-6\right )}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-x)*((x*exp(1))/3 - (8*x)/3 + x*log(32*x^2) + (2*x^2)/3 - 2))/x,x)

[Out]

(exp(-x)*(2*x + exp(1) + 3*log(32*x^2) - 6))/3

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sympy [A]  time = 0.34, size = 20, normalized size = 0.83 \begin {gather*} \frac {\left (2 x + 3 \log {\left (32 x^{2} \right )} - 6 + e\right ) e^{- x}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(-3*x*ln(32*x**2)-x*exp(1)-2*x**2+8*x+6)/exp(x)/x,x)

[Out]

(2*x + 3*log(32*x**2) - 6 + E)*exp(-x)/3

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