3.6.35 $\int \frac{-16-16x+5x^2+(-20x+5x^2)\log (x)+(-16-12x+10x^2)\log (x)\log (\frac{1}{9}(-4-5x)\log (x))}{(36+45x)\log (x)}dx$

Optimal. Leaf size=25 $$\frac{1}{9}(-4+x)x\log ((-x+\frac{1}{9}(-4+4x))\log (x))$$

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Rubi [A]  time = 0.56, antiderivative size = 37, normalized size of antiderivative = 1.48, number of steps used = 22, number of rules used = 8, integrand size = 59, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.136, Rules used = {6742, 77, 2330, 2298, 2309, 2178, 2555, 12} $$\begin{array}{c}\frac{1}{9}{x}^2\log (-\frac{1}{9}(5x+4)\log (x))-\frac{4}{9}x\log (-\frac{1}{9}(5x+4)\log (x))\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(-16 - 16*x + 5*x^2 + (-20*x + 5*x^2)*Log[x] + (-16 - 12*x + 10*x^2)*Log[x]*Log[((-4 - 5*x)*Log[x])/9])/((
36 + 45*x)*Log[x]),x]

[Out]

(-4*x*Log[-1/9*((4 + 5*x)*Log[x])])/9 + (x^2*Log[-1/9*((4 + 5*x)*Log[x])])/9

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

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 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 2555

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[w*Simplify
[D[u, x]/u], x], x] /; InverseFunctionFreeQ[w, x]] /; ProductQ[u]

Rule 6742

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

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int (\frac{(-4+x)(4+5x+5x\log (x))}{9(4+5x)\log (x)}+\frac{2}{9}(-2+x)\log (-\frac{1}{9}(4+5x)\log (x)))dx\\ & =\frac{1}{9}\int \frac{(-4+x)(4+5x+5x\log (x))}{(4+5x)\log (x)}dx+\frac{2}{9}\int (-2+x)\log (-\frac{1}{9}(4+5x)\log (x))dx\\ & =-\frac{4}{9}x\log (-\frac{1}{9}(4+5x)\log (x))+\frac{1}{9}{x}^2\log (-\frac{1}{9}(4+5x)\log (x))+\frac{1}{9}\int (\frac{5(-4+x)x}{4+5x}+\frac{-4+x}{\log (x)})dx-\frac{2}{9}\int \frac{(4-x)(-4-5x-5x\log (x))}{2(4+5x)\log (x)}dx\\ & =-\frac{4}{9}x\log (-\frac{1}{9}(4+5x)\log (x))+\frac{1}{9}{x}^2\log (-\frac{1}{9}(4+5x)\log (x))+\frac{1}{9}\int \frac{-4+x}{\log (x)}dx-\frac{1}{9}\int \frac{(4-x)(-4-5x-5x\log (x))}{(4+5x)\log (x)}dx+\frac{5}{9}\int \frac{(-4+x)x}{4+5x}dx\\ & =-\frac{4}{9}x\log (-\frac{1}{9}(4+5x)\log (x))+\frac{1}{9}{x}^2\log (-\frac{1}{9}(4+5x)\log (x))-\frac{1}{9}\int (\frac{5(-4+x)x}{4+5x}+\frac{-4+x}{\log (x)})dx+\frac{1}{9}\int (-\frac{4}{\log (x)}+\frac{x}{\log (x)})dx+\frac{5}{9}\int (-\frac{24}{25}+\frac{x}{5}+\frac{96}{25(4+5x)})dx\\ & =-\frac{8x}{15}+\frac{x^2}{18}+\frac{32}{75}\log (4+5x)-\frac{4}{9}x\log (-\frac{1}{9}(4+5x)\log (x))+\frac{1}{9}{x}^2\log (-\frac{1}{9}(4+5x)\log (x))-\frac{1}{9}\int \frac{-4+x}{\log (x)}dx+\frac{1}{9}\int \frac{x}{\log (x)}dx-\frac{4}{9}\int \frac{1}{\log (x)}dx-\frac{5}{9}\int \frac{(-4+x)x}{4+5x}dx\\ & =-\frac{8x}{15}+\frac{x^2}{18}+\frac{32}{75}\log (4+5x)-\frac{4}{9}x\log (-\frac{1}{9}(4+5x)\log (x))+\frac{1}{9}{x}^2\log (-\frac{1}{9}(4+5x)\log (x))-\frac{4\text{li}(x)}{9}-\frac{1}{9}\int (-\frac{4}{\log (x)}+\frac{x}{\log (x)})dx+\frac{1}{9}\mathrm{Subst}(\int \frac{e^{2x}}{x}dx,x,\log (x))-\frac{5}{9}\int (-\frac{24}{25}+\frac{x}{5}+\frac{96}{25(4+5x)})dx\\ & =\frac{1}{9}\text{Ei}(2\log (x))-\frac{4}{9}x\log (-\frac{1}{9}(4+5x)\log (x))+\frac{1}{9}{x}^2\log (-\frac{1}{9}(4+5x)\log (x))-\frac{4\text{li}(x)}{9}-\frac{1}{9}\int \frac{x}{\log (x)}dx+\frac{4}{9}\int \frac{1}{\log (x)}dx\\ & =\frac{1}{9}\text{Ei}(2\log (x))-\frac{4}{9}x\log (-\frac{1}{9}(4+5x)\log (x))+\frac{1}{9}{x}^2\log (-\frac{1}{9}(4+5x)\log (x))-\frac{1}{9}\mathrm{Subst}(\int \frac{e^{2x}}{x}dx,x,\log (x))\\ & =-\frac{4}{9}x\log (-\frac{1}{9}(4+5x)\log (x))+\frac{1}{9}{x}^2\log (-\frac{1}{9}(4+5x)\log (x))\end{array}\end{array}$$

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Mathematica [A]  time = 0.13, size = 20, normalized size = 0.80 $$\begin{array}{c}\frac{1}{9}(-4+x)x\log (-\frac{1}{9}(4+5x)\log (x))\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(-16 - 16*x + 5*x^2 + (-20*x + 5*x^2)*Log[x] + (-16 - 12*x + 10*x^2)*Log[x]*Log[((-4 - 5*x)*Log[x])/
9])/((36 + 45*x)*Log[x]),x]

[Out]

((-4 + x)*x*Log[-1/9*((4 + 5*x)*Log[x])])/9

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fricas [A]  time = 0.81, size = 19, normalized size = 0.76 $$\begin{array}{c}\frac{1}{9}(x^2-4x)\log (-\frac{1}{9}(5x+4)\log (x))\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x^2-12*x-16)*log(x)*log(1/9*(-5*x-4)*log(x))+(5*x^2-20*x)*log(x)+5*x^2-16*x-16)/(45*x+36)/log(x
),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.68, size = 33, normalized size = 1.32 $$\begin{array}{c}-\frac{2}{9}{x}^2\log (3)+\frac{8}{9}x\log (3)+\frac{1}{9}(x^2-4x)\log (-5x\log (x)-4\log (x))\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x^2-12*x-16)*log(x)*log(1/9*(-5*x-4)*log(x))+(5*x^2-20*x)*log(x)+5*x^2-16*x-16)/(45*x+36)/log(x
),x, algorithm="giac")

[Out]

-2/9*x^2*log(3) + 8/9*x*log(3) + 1/9*(x^2 - 4*x)*log(-5*x*log(x) - 4*log(x))

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maple [A]  time = 0.33, size = 30, normalized size = 1.20

method	result	size
norman	$-\frac{4x\ln \left(\frac{(-5x-4)\ln (x)}{9}\right)}{9}+\frac{x^2\ln \left(\frac{(-5x-4)\ln (x)}{9}\right)}{9}$	$30$
risch	$\frac{x^2\ln (5)}{9}-\frac{2x^2\ln (3)}{9}-\frac{4x\ln (\ln (x))}{9}+\frac{x^2\ln (\ln (x))}{9}-\frac{4x\ln (5)}{9}+\frac{8x\ln (3)}{9}+\frac{i\pi x^2}{9}-\frac{2i\pi x\mathrm{csgn}{(i\ln (x)(x+\frac{4}{5}))}^3}{9}+\frac{i\pi x^2\mathrm{csgn}{(i\ln (x)(x+\frac{4}{5}))}^3}{18}-\frac{2i\pi x\mathrm{csgn}(i\ln (x))\mathrm{csgn}{(i\ln (x)(x+\frac{4}{5}))}^2}{9}-\frac{i\pi x^2\mathrm{csgn}(i\ln (x))\mathrm{csgn}\left(i(x+\frac{4}{5})\right)\mathrm{csgn}(i\ln (x)(x+\frac{4}{5}))}{18}-\frac{2i\pi x\mathrm{csgn}\left(i(x+\frac{4}{5})\right)\mathrm{csgn}{(i\ln (x)(x+\frac{4}{5}))}^2}{9}+\frac{i\pi x^2\mathrm{csgn}\left(i(x+\frac{4}{5})\right)\mathrm{csgn}{(i\ln (x)(x+\frac{4}{5}))}^2}{18}+\frac{i\pi x^2\mathrm{csgn}(i\ln (x))\mathrm{csgn}{(i\ln (x)(x+\frac{4}{5}))}^2}{18}+\frac{2i\pi x\mathrm{csgn}(i\ln (x))\mathrm{csgn}\left(i(x+\frac{4}{5})\right)\mathrm{csgn}(i\ln (x)(x+\frac{4}{5}))}{9}-\frac{i\pi x^2\mathrm{csgn}{(i\ln (x)(x+\frac{4}{5}))}^2}{9}+(\frac{1}{9}{x}^2-\frac{4}{9}x)\ln (x+\frac{4}{5})+\frac{4i\pi x\mathrm{csgn}{(i\ln (x)(x+\frac{4}{5}))}^2}{9}-\frac{4i\pi x}{9}$	$284$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((10*x^2-12*x-16)*ln(x)*ln(1/9*(-5*x-4)*ln(x))+(5*x^2-20*x)*ln(x)+5*x^2-16*x-16)/(45*x+36)/ln(x),x,method=
_RETURNVERBOSE)

[Out]

-4/9*x*ln(1/9*(-5*x-4)*ln(x))+1/9*x^2*ln(1/9*(-5*x-4)*ln(x))

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maxima [B]  time = 0.67, size = 40, normalized size = 1.60 $$\begin{array}{c}-\frac{2}{9}{x}^2\log (3)+\frac{8}{9}x\log (3)+\frac{1}{9}(x^2-4x)\log (-5x-4)+\frac{1}{9}(x^2-4x)\log (\log (x))\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x^2-12*x-16)*log(x)*log(1/9*(-5*x-4)*log(x))+(5*x^2-20*x)*log(x)+5*x^2-16*x-16)/(45*x+36)/log(x
),x, algorithm="maxima")

[Out]

-2/9*x^2*log(3) + 8/9*x*log(3) + 1/9*(x^2 - 4*x)*log(-5*x - 4) + 1/9*(x^2 - 4*x)*log(log(x))

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mupad [B]  time = 0.94, size = 21, normalized size = 0.84 $$\begin{array}{c}-\ln (-\frac{\ln (x)(5x+4)}{9})(\frac{4x}{9}-\frac{x^2}{9})\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(16*x + log(x)*(20*x - 5*x^2) - 5*x^2 + log(x)*log(-(log(x)*(5*x + 4))/9)*(12*x - 10*x^2 + 16) + 16)/(log
(x)*(45*x + 36)),x)

[Out]

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

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sympy [B]  time = 0.66, size = 46, normalized size = 1.84 $$\begin{array}{c}(\frac{x^2}{9}-\frac{4x}{9}-\frac{16}{225})\log ((-\frac{5x}{9}-\frac{4}{9})\log (x))+\frac{16\log (225x+180)}{225}+\frac{16\log (\log (x))}{225}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x**2-12*x-16)*ln(x)*ln(1/9*(-5*x-4)*ln(x))+(5*x**2-20*x)*ln(x)+5*x**2-16*x-16)/(45*x+36)/ln(x),
x)

[Out]

(x**2/9 - 4*x/9 - 16/225)*log((-5*x/9 - 4/9)*log(x)) + 16*log(225*x + 180)/225 + 16*log(log(x))/225

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