3.46.16 \(\int \frac {e (-24-8 x)+42 x^2+(-18 x^2+e (24+8 x)) \log (x)+e (-6-2 x) \log ^2(x)}{4 e-4 e \log (x)+e \log ^2(x)} \, dx\)

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

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Rubi [A]  time = 0.28, antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 58, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {6741, 12, 6742, 2306, 2309, 2178} \begin {gather*} \frac {6 x^3}{e (2-\log (x))}-(x+3)^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E*(-24 - 8*x) + 42*x^2 + (-18*x^2 + E*(24 + 8*x))*Log[x] + E*(-6 - 2*x)*Log[x]^2)/(4*E - 4*E*Log[x] + E*L
og[x]^2),x]

[Out]

-(3 + x)^2 + (6*x^3)/(E*(2 - Log[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 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 2306

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log
[c*x^n])^(p + 1))/(b*d*n*(p + 1)), x] - Dist[(m + 1)/(b*n*(p + 1)), Int[(d*x)^m*(a + b*Log[c*x^n])^(p + 1), x]
, x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1] && LtQ[p, -1]

Rule 2309

Int[((a_.) + Log[(c_.)*(x_)]*(b_.))^(p_)*(x_)^(m_.), x_Symbol] :> Dist[1/c^(m + 1), Subst[Int[E^((m + 1)*x)*(a
+ b*x)^p, x], x, Log[c*x]], x] /; FreeQ[{a, b, c, p}, x] && IntegerQ[m]

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 (-24-8 x)+42 x^2+\left (-18 x^2+e (24+8 x)\right ) \log (x)+e (-6-2 x) \log ^2(x)}{e (2-\log (x))^2} \, dx\\ &=\frac {\int \frac {e (-24-8 x)+42 x^2+\left (-18 x^2+e (24+8 x)\right ) \log (x)+e (-6-2 x) \log ^2(x)}{(2-\log (x))^2} \, dx}{e}\\ &=\frac {\int \left (-2 e (3+x)+\frac {6 x^2}{(-2+\log (x))^2}-\frac {18 x^2}{-2+\log (x)}\right ) \, dx}{e}\\ &=-(3+x)^2+\frac {6 \int \frac {x^2}{(-2+\log (x))^2} \, dx}{e}-\frac {18 \int \frac {x^2}{-2+\log (x)} \, dx}{e}\\ &=-(3+x)^2+\frac {6 x^3}{e (2-\log (x))}+\frac {18 \int \frac {x^2}{-2+\log (x)} \, dx}{e}-\frac {18 \operatorname {Subst}\left (\int \frac {e^{3 x}}{-2+x} \, dx,x,\log (x)\right )}{e}\\ &=-(3+x)^2-18 e^5 \text {Ei}(-3 (2-\log (x)))+\frac {6 x^3}{e (2-\log (x))}+\frac {18 \operatorname {Subst}\left (\int \frac {e^{3 x}}{-2+x} \, dx,x,\log (x)\right )}{e}\\ &=-(3+x)^2+\frac {6 x^3}{e (2-\log (x))}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.10, size = 26, normalized size = 1.08 \begin {gather*} -\frac {2 \left (\frac {1}{2} e x (6+x)+\frac {3 x^3}{-2+\log (x)}\right )}{e} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E*(-24 - 8*x) + 42*x^2 + (-18*x^2 + E*(24 + 8*x))*Log[x] + E*(-6 - 2*x)*Log[x]^2)/(4*E - 4*E*Log[x]
+ E*Log[x]^2),x]

[Out]

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

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fricas [B]  time = 0.57, size = 43, normalized size = 1.79 \begin {gather*} -\frac {6 \, x^{3} + {\left (x^{2} + 6 \, x\right )} e \log \relax (x) - 2 \, {\left (x^{2} + 6 \, x\right )} e}{e \log \relax (x) - 2 \, e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x-6)*exp(1)*log(x)^2+((8*x+24)*exp(1)-18*x^2)*log(x)+(-8*x-24)*exp(1)+42*x^2)/(exp(1)*log(x)^2-
4*exp(1)*log(x)+4*exp(1)),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.17, size = 94, normalized size = 3.92 \begin {gather*} -\frac {x^{2} e \log \relax (x)}{e \log \relax (x) - 2 \, e} - \frac {6 \, x^{3}}{e \log \relax (x) - 2 \, e} + \frac {2 \, x^{2} e}{e \log \relax (x) - 2 \, e} - \frac {6 \, x e \log \relax (x)}{e \log \relax (x) - 2 \, e} + \frac {12 \, x e}{e \log \relax (x) - 2 \, e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x-6)*exp(1)*log(x)^2+((8*x+24)*exp(1)-18*x^2)*log(x)+(-8*x-24)*exp(1)+42*x^2)/(exp(1)*log(x)^2-
4*exp(1)*log(x)+4*exp(1)),x, algorithm="giac")

[Out]

-x^2*e*log(x)/(e*log(x) - 2*e) - 6*x^3/(e*log(x) - 2*e) + 2*x^2*e/(e*log(x) - 2*e) - 6*x*e*log(x)/(e*log(x) -
2*e) + 12*x*e/(e*log(x) - 2*e)

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maple [A]  time = 0.06, size = 23, normalized size = 0.96

method	result	size
risch	\(-x^{2}-6 x -\frac {6 x^{3} {\mathrm e}^{-1}}{\ln \relax (x )-2}\)	\(23\)
norman	\(\frac {12 x +2 x^{2}-6 x \ln \relax (x )-x^{2} \ln \relax (x )-6 x^{3} {\mathrm e}^{-1}}{\ln \relax (x )-2}\)	\(38\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*x-6)*exp(1)*ln(x)^2+((8*x+24)*exp(1)-18*x^2)*ln(x)+(-8*x-24)*exp(1)+42*x^2)/(exp(1)*ln(x)^2-4*exp(1)*
ln(x)+4*exp(1)),x,method=_RETURNVERBOSE)

[Out]

-x^2-6*x-6*x^3*exp(-1)/(ln(x)-2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {24 \, e^{2} E_{2}\left (-\log \relax (x) + 2\right )}{\log \relax (x) - 2} + \frac {8 \, e^{4} E_{2}\left (-2 \, \log \relax (x) + 4\right )}{\log \relax (x) - 2} - \frac {42 \, e^{5} E_{2}\left (-3 \, \log \relax (x) + 6\right )}{\log \relax (x) - 2} + \frac {36 \, x^{3} - 6 \, x^{2} e - 12 \, x e - {\left (x^{2} e + 6 \, x e\right )} \log \relax (x)}{e \log \relax (x) - 2 \, e} - 2 \, \int \frac {63 \, x^{2} - 8 \, x e - 12 \, e}{e \log \relax (x) - 2 \, e}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x-6)*exp(1)*log(x)^2+((8*x+24)*exp(1)-18*x^2)*log(x)+(-8*x-24)*exp(1)+42*x^2)/(exp(1)*log(x)^2-
4*exp(1)*log(x)+4*exp(1)),x, algorithm="maxima")

[Out]

24*e^2*exp_integral_e(2, -log(x) + 2)/(log(x) - 2) + 8*e^4*exp_integral_e(2, -2*log(x) + 4)/(log(x) - 2) - 42*
e^5*exp_integral_e(2, -3*log(x) + 6)/(log(x) - 2) + (36*x^3 - 6*x^2*e - 12*x*e - (x^2*e + 6*x*e)*log(x))/(e*lo
g(x) - 2*e) - 2*integrate((63*x^2 - 8*x*e - 12*e)/(e*log(x) - 2*e), x)

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mupad [B]  time = 3.15, size = 42, normalized size = 1.75 \begin {gather*} \frac {6\,x^4}{2\,x\,\mathrm {e}-x\,\mathrm {e}\,\ln \relax (x)}-\frac {{\mathrm {e}}^{-1}\,\left (\mathrm {e}\,x^3+6\,\mathrm {e}\,x^2\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(x)*(18*x^2 - exp(1)*(8*x + 24)) - 42*x^2 + exp(1)*(8*x + 24) + exp(1)*log(x)^2*(2*x + 6))/(4*exp(1)
+ exp(1)*log(x)^2 - 4*exp(1)*log(x)),x)

[Out]

(6*x^4)/(2*x*exp(1) - x*exp(1)*log(x)) - (exp(-1)*(6*x^2*exp(1) + x^3*exp(1)))/x

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sympy [A]  time = 0.11, size = 24, normalized size = 1.00 \begin {gather*} - \frac {6 x^{3}}{e \log {\relax (x )} - 2 e} - x^{2} - 6 x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x-6)*exp(1)*ln(x)**2+((8*x+24)*exp(1)-18*x**2)*ln(x)+(-8*x-24)*exp(1)+42*x**2)/(exp(1)*ln(x)**2
-4*exp(1)*ln(x)+4*exp(1)),x)

[Out]

-6*x**3/(E*log(x) - 2*E) - x**2 - 6*x

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