3.18.60 \(\int \frac {-36+24 x-4 x^2+(36-48 x+12 x^2) \log (x)}{5 \log ^2(x)} \, dx\)

Optimal. Leaf size=18 \[ -3+\frac {(6-2 x)^2 x}{5 \log (x)} \]

________________________________________________________________________________________

Rubi [A]  time = 0.28, antiderivative size = 16, normalized size of antiderivative = 0.89, number of steps used = 25, number of rules used = 9, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.290, Rules used = {12, 6688, 6742, 2320, 2330, 2298, 2309, 2178, 2356} \begin {gather*} \frac {4 (3-x)^2 x}{5 \log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-36 + 24*x - 4*x^2 + (36 - 48*x + 12*x^2)*Log[x])/(5*Log[x]^2),x]

[Out]

(4*(3 - x)^2*x)/(5*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 2298

Int[Log[(c_.)*(x_)]^(-1), x_Symbol] :> Simp[LogIntegral[c*x]/c, x] /; FreeQ[c, x]

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 2320

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_)*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Simp[(x*(d + e*x)^q*(a
+ b*Log[c*x^n])^(p + 1))/(b*n*(p + 1)), x] + (-Dist[(q + 1)/(b*n*(p + 1)), Int[(d + e*x)^q*(a + b*Log[c*x^n])^
(p + 1), x], x] + Dist[(d*q)/(b*n*(p + 1)), Int[(d + e*x)^(q - 1)*(a + b*Log[c*x^n])^(p + 1), x], x]) /; FreeQ
[{a, b, c, d, e, n}, x] && LtQ[p, -1] && GtQ[q, 0]

Rule 2330

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = Expand
Integrand[(a + b*Log[c*x^n])^p, (d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, d, e, n, p, q, r}
, x] && IntegerQ[q] && (GtQ[q, 0] || (IGtQ[p, 0] && IntegerQ[r]))

Rule 2356

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(Polyx_), x_Symbol] :> Int[ExpandIntegrand[Polyx*(a + b*Log[c*
x^n])^p, x], x] /; FreeQ[{a, b, c, n, p}, x] && PolynomialQ[Polyx, x]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, 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} &=\frac {1}{5} \int \frac {-36+24 x-4 x^2+\left (36-48 x+12 x^2\right ) \log (x)}{\log ^2(x)} \, dx\\ &=\frac {1}{5} \int \frac {4 (3-x) (-3+x-3 (-1+x) \log (x))}{\log ^2(x)} \, dx\\ &=\frac {4}{5} \int \frac {(3-x) (-3+x-3 (-1+x) \log (x))}{\log ^2(x)} \, dx\\ &=\frac {4}{5} \int \left (-\frac {(-3+x)^2}{\log ^2(x)}+\frac {3 (-3+x) (-1+x)}{\log (x)}\right ) \, dx\\ &=-\left (\frac {4}{5} \int \frac {(-3+x)^2}{\log ^2(x)} \, dx\right )+\frac {12}{5} \int \frac {(-3+x) (-1+x)}{\log (x)} \, dx\\ &=\frac {4 (3-x)^2 x}{5 \log (x)}+\frac {12}{5} \int \left (\frac {3}{\log (x)}-\frac {4 x}{\log (x)}+\frac {x^2}{\log (x)}\right ) \, dx-\frac {12}{5} \int \frac {(-3+x)^2}{\log (x)} \, dx-\frac {24}{5} \int \frac {-3+x}{\log (x)} \, dx\\ &=\frac {4 (3-x)^2 x}{5 \log (x)}-\frac {12}{5} \int \left (\frac {9}{\log (x)}-\frac {6 x}{\log (x)}+\frac {x^2}{\log (x)}\right ) \, dx+\frac {12}{5} \int \frac {x^2}{\log (x)} \, dx-\frac {24}{5} \int \left (-\frac {3}{\log (x)}+\frac {x}{\log (x)}\right ) \, dx+\frac {36}{5} \int \frac {1}{\log (x)} \, dx-\frac {48}{5} \int \frac {x}{\log (x)} \, dx\\ &=\frac {4 (3-x)^2 x}{5 \log (x)}+\frac {36 \text {li}(x)}{5}-\frac {12}{5} \int \frac {x^2}{\log (x)} \, dx+\frac {12}{5} \operatorname {Subst}\left (\int \frac {e^{3 x}}{x} \, dx,x,\log (x)\right )-\frac {24}{5} \int \frac {x}{\log (x)} \, dx-\frac {48}{5} \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right )+\frac {72}{5} \int \frac {1}{\log (x)} \, dx+\frac {72}{5} \int \frac {x}{\log (x)} \, dx-\frac {108}{5} \int \frac {1}{\log (x)} \, dx\\ &=-\frac {48}{5} \text {Ei}(2 \log (x))+\frac {12}{5} \text {Ei}(3 \log (x))+\frac {4 (3-x)^2 x}{5 \log (x)}-\frac {12}{5} \operatorname {Subst}\left (\int \frac {e^{3 x}}{x} \, dx,x,\log (x)\right )-\frac {24}{5} \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right )+\frac {72}{5} \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right )\\ &=\frac {4 (3-x)^2 x}{5 \log (x)}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.07, size = 14, normalized size = 0.78 \begin {gather*} \frac {4 (-3+x)^2 x}{5 \log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-36 + 24*x - 4*x^2 + (36 - 48*x + 12*x^2)*Log[x])/(5*Log[x]^2),x]

[Out]

(4*(-3 + x)^2*x)/(5*Log[x])

________________________________________________________________________________________

fricas [A]  time = 0.94, size = 18, normalized size = 1.00 \begin {gather*} \frac {4 \, {\left (x^{3} - 6 \, x^{2} + 9 \, x\right )}}{5 \, \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((12*x^2-48*x+36)*log(x)-4*x^2+24*x-36)/log(x)^2,x, algorithm="fricas")

[Out]

4/5*(x^3 - 6*x^2 + 9*x)/log(x)

________________________________________________________________________________________

giac [A]  time = 0.23, size = 26, normalized size = 1.44 \begin {gather*} \frac {4 \, x^{3}}{5 \, \log \relax (x)} - \frac {24 \, x^{2}}{5 \, \log \relax (x)} + \frac {36 \, x}{5 \, \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((12*x^2-48*x+36)*log(x)-4*x^2+24*x-36)/log(x)^2,x, algorithm="giac")

[Out]

4/5*x^3/log(x) - 24/5*x^2/log(x) + 36/5*x/log(x)

________________________________________________________________________________________

maple [A]  time = 0.02, size = 16, normalized size = 0.89

method	result	size
risch	\(\frac {4 x \left (x^{2}-6 x +9\right )}{5 \ln \relax (x )}\)	\(16\)
norman	\(\frac {\frac {36}{5} x -\frac {24}{5} x^{2}+\frac {4}{5} x^{3}}{\ln \relax (x )}\)	\(20\)
default	\(\frac {4 x^{3}}{5 \ln \relax (x )}-\frac {24 x^{2}}{5 \ln \relax (x )}+\frac {36 x}{5 \ln \relax (x )}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*((12*x^2-48*x+36)*ln(x)-4*x^2+24*x-36)/ln(x)^2,x,method=_RETURNVERBOSE)

[Out]

4/5*x*(x^2-6*x+9)/ln(x)

________________________________________________________________________________________

maxima [C]  time = 0.48, size = 44, normalized size = 2.44 \begin {gather*} \frac {12}{5} \, {\rm Ei}\left (3 \, \log \relax (x)\right ) - \frac {48}{5} \, {\rm Ei}\left (2 \, \log \relax (x)\right ) + \frac {36}{5} \, {\rm Ei}\left (\log \relax (x)\right ) - \frac {36}{5} \, \Gamma \left (-1, -\log \relax (x)\right ) + \frac {48}{5} \, \Gamma \left (-1, -2 \, \log \relax (x)\right ) - \frac {12}{5} \, \Gamma \left (-1, -3 \, \log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((12*x^2-48*x+36)*log(x)-4*x^2+24*x-36)/log(x)^2,x, algorithm="maxima")

[Out]

12/5*Ei(3*log(x)) - 48/5*Ei(2*log(x)) + 36/5*Ei(log(x)) - 36/5*gamma(-1, -log(x)) + 48/5*gamma(-1, -2*log(x))
- 12/5*gamma(-1, -3*log(x))

________________________________________________________________________________________

mupad [B]  time = 1.15, size = 12, normalized size = 0.67 \begin {gather*} \frac {4\,x\,{\left (x-3\right )}^2}{5\,\ln \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((24*x)/5 + (log(x)*(12*x^2 - 48*x + 36))/5 - (4*x^2)/5 - 36/5)/log(x)^2,x)

[Out]

(4*x*(x - 3)^2)/(5*log(x))

________________________________________________________________________________________

sympy [A]  time = 0.10, size = 17, normalized size = 0.94 \begin {gather*} \frac {4 x^{3} - 24 x^{2} + 36 x}{5 \log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((12*x**2-48*x+36)*ln(x)-4*x**2+24*x-36)/ln(x)**2,x)

[Out]

(4*x**3 - 24*x**2 + 36*x)/(5*log(x))

________________________________________________________________________________________