3.82.39 \(\int \frac {e^{-x} (-12-36 x-12 x^2-3 e^5 x \log (3)-3 e^5 x \log (4)+12 x \log (x))}{4 x} \, dx\)

Optimal. Leaf size=26 \[ 3 e^{-x} \left (4+x+\frac {1}{4} e^5 (\log (3)+\log (4))-\log (x)\right ) \]

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Rubi [A]  time = 0.51, antiderivative size = 42, normalized size of antiderivative = 1.62, number of steps used = 15, number of rules used = 9, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.209, Rules used = {6, 12, 6741, 6742, 2199, 2178, 2176, 2194, 2554} \begin {gather*} 3 e^{-x} x+3 e^{-x}-3 e^{-x} \log (x)+\frac {3}{4} e^{-x} \left (12+e^5 \log (12)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-12 - 36*x - 12*x^2 - 3*E^5*x*Log[3] - 3*E^5*x*Log[4] + 12*x*Log[x])/(4*E^x*x),x]

[Out]

3/E^x + (3*x)/E^x + (3*(12 + E^5*Log[12]))/(4*E^x) - (3*Log[x])/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 6741

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

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 (-12-12 x^2+x \left (-36-3 e^5 \log (3)\right )-3 e^5 x \log (4)+12 x \log (x)\right )}{4 x} \, dx\\ &=\int \frac {e^{-x} \left (-12-12 x^2+x \left (-36-3 e^5 \log (3)-3 e^5 \log (4)\right )+12 x \log (x)\right )}{4 x} \, dx\\ &=\frac {1}{4} \int \frac {e^{-x} \left (-12-12 x^2+x \left (-36-3 e^5 \log (3)-3 e^5 \log (4)\right )+12 x \log (x)\right )}{x} \, dx\\ &=\frac {1}{4} \int \frac {3 e^{-x} \left (-4-4 x^2-12 x \left (1+\frac {1}{12} e^5 \log (12)\right )+4 x \log (x)\right )}{x} \, dx\\ &=\frac {3}{4} \int \frac {e^{-x} \left (-4-4 x^2-12 x \left (1+\frac {1}{12} e^5 \log (12)\right )+4 x \log (x)\right )}{x} \, dx\\ &=\frac {3}{4} \int \left (\frac {e^{-x} \left (-4-4 x^2-x \left (12+e^5 \log (12)\right )\right )}{x}+4 e^{-x} \log (x)\right ) \, dx\\ &=\frac {3}{4} \int \frac {e^{-x} \left (-4-4 x^2-x \left (12+e^5 \log (12)\right )\right )}{x} \, dx+3 \int e^{-x} \log (x) \, dx\\ &=-3 e^{-x} \log (x)+\frac {3}{4} \int \left (-\frac {4 e^{-x}}{x}-4 e^{-x} x+e^{-x} \left (-12-e^5 \log (12)\right )\right ) \, dx+3 \int \frac {e^{-x}}{x} \, dx\\ &=3 \text {Ei}(-x)-3 e^{-x} \log (x)-3 \int \frac {e^{-x}}{x} \, dx-3 \int e^{-x} x \, dx-\frac {1}{4} \left (3 \left (12+e^5 \log (12)\right )\right ) \int e^{-x} \, dx\\ &=3 e^{-x} x+\frac {3}{4} e^{-x} \left (12+e^5 \log (12)\right )-3 e^{-x} \log (x)-3 \int e^{-x} \, dx\\ &=3 e^{-x}+3 e^{-x} x+\frac {3}{4} e^{-x} \left (12+e^5 \log (12)\right )-3 e^{-x} \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.09, size = 24, normalized size = 0.92 \begin {gather*} \frac {3}{4} e^{-x} \left (16+4 x+e^5 \log (12)-4 \log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-12 - 36*x - 12*x^2 - 3*E^5*x*Log[3] - 3*E^5*x*Log[4] + 12*x*Log[x])/(4*E^x*x),x]

[Out]

(3*(16 + 4*x + E^5*Log[12] - 4*Log[x]))/(4*E^x)

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fricas [A]  time = 0.77, size = 31, normalized size = 1.19 \begin {gather*} \frac {3}{4} \, {\left (e^{5} \log \relax (3) + 2 \, e^{5} \log \relax (2) + 4 \, x + 16\right )} e^{\left (-x\right )} - 3 \, e^{\left (-x\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(12*x*log(x)-6*x*exp(5)*log(2)-3*x*exp(5)*log(3)-12*x^2-36*x-12)/exp(x)/x,x, algorithm="fricas")

[Out]

3/4*(e^5*log(3) + 2*e^5*log(2) + 4*x + 16)*e^(-x) - 3*e^(-x)*log(x)

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giac [A]  time = 0.12, size = 42, normalized size = 1.62 \begin {gather*} 3 \, x e^{\left (-x\right )} + \frac {3}{4} \, e^{\left (-x + 5\right )} \log \relax (3) + \frac {3}{2} \, e^{\left (-x + 5\right )} \log \relax (2) - 3 \, e^{\left (-x\right )} \log \relax (x) + 12 \, e^{\left (-x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(12*x*log(x)-6*x*exp(5)*log(2)-3*x*exp(5)*log(3)-12*x^2-36*x-12)/exp(x)/x,x, algorithm="giac")

[Out]

3*x*e^(-x) + 3/4*e^(-x + 5)*log(3) + 3/2*e^(-x + 5)*log(2) - 3*e^(-x)*log(x) + 12*e^(-x)

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maple [A]  time = 0.06, size = 27, normalized size = 1.04

method	result	size
norman	\(\left (3 x +12-3 \ln \relax (x )+\frac {3 \,{\mathrm e}^{5} \ln \relax (2)}{2}+\frac {3 \,{\mathrm e}^{5} \ln \relax (3)}{4}\right ) {\mathrm e}^{-x}\)	\(27\)
risch	\(-3 \ln \relax (x ) {\mathrm e}^{-x}+\frac {3 \left (2 \,{\mathrm e}^{5} \ln \relax (2)+{\mathrm e}^{5} \ln \relax (3)+4 x +16\right ) {\mathrm e}^{-x}}{4}\)	\(32\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/4*(12*x*ln(x)-6*x*exp(5)*ln(2)-3*x*exp(5)*ln(3)-12*x^2-36*x-12)/exp(x)/x,x,method=_RETURNVERBOSE)

[Out]

(3*x+12-3*ln(x)+3/2*exp(5)*ln(2)+3/4*exp(5)*ln(3))/exp(x)

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maxima [A]  time = 0.39, size = 44, normalized size = 1.69 \begin {gather*} 3 \, {\left (x + 1\right )} e^{\left (-x\right )} + \frac {3}{4} \, e^{\left (-x + 5\right )} \log \relax (3) + \frac {3}{2} \, e^{\left (-x + 5\right )} \log \relax (2) - 3 \, e^{\left (-x\right )} \log \relax (x) + 9 \, e^{\left (-x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(12*x*log(x)-6*x*exp(5)*log(2)-3*x*exp(5)*log(3)-12*x^2-36*x-12)/exp(x)/x,x, algorithm="maxima")

[Out]

3*(x + 1)*e^(-x) + 3/4*e^(-x + 5)*log(3) + 3/2*e^(-x + 5)*log(2) - 3*e^(-x)*log(x) + 9*e^(-x)

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mupad [B]  time = 5.87, size = 26, normalized size = 1.00 \begin {gather*} \frac {3\,{\mathrm {e}}^{-x}\,\left (4\,x-4\,\ln \relax (x)+2\,{\mathrm {e}}^5\,\ln \relax (2)+{\mathrm {e}}^5\,\ln \relax (3)+16\right )}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-x)*(9*x - 3*x*log(x) + 3*x^2 + (3*x*exp(5)*log(2))/2 + (3*x*exp(5)*log(3))/4 + 3))/x,x)

[Out]

(3*exp(-x)*(4*x - 4*log(x) + 2*exp(5)*log(2) + exp(5)*log(3) + 16))/4

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sympy [A]  time = 0.37, size = 31, normalized size = 1.19 \begin {gather*} \frac {\left (12 x - 12 \log {\relax (x )} + 48 + 3 e^{5} \log {\relax (3 )} + 6 e^{5} \log {\relax (2 )}\right ) e^{- x}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(12*x*ln(x)-6*x*exp(5)*ln(2)-3*x*exp(5)*ln(3)-12*x**2-36*x-12)/exp(x)/x,x)

[Out]

(12*x - 12*log(x) + 48 + 3*exp(5)*log(3) + 6*exp(5)*log(2))*exp(-x)/4

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