3.89.44 $\int \frac{2+8x+8x^2+(-2-8x-8x^2)\log (x)+4x{\log }^2(x)}{x{\log }^3(x)}dx$

Optimal. Leaf size=20 $$\log (2)-{(1-\frac{1+2x}{\log (x)})}^2$$

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Rubi [A]  time = 0.41, antiderivative size = 36, normalized size of antiderivative = 1.80, number of steps used = 23, number of rules used = 9, integrand size = 38, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.237, Rules used = {6742, 2353, 2297, 2298, 2302, 30, 2306, 2309, 2178} $$\begin{array}{c}-\frac{4x^2}{{\log }^2(x)}-\frac{4x}{{\log }^2(x)}-\frac{1}{{\log }^2(x)}+\frac{4x}{\log (x)}+\frac{2}{\log (x)}\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(2 + 8*x + 8*x^2 + (-2 - 8*x - 8*x^2)*Log[x] + 4*x*Log[x]^2)/(x*Log[x]^3),x]

[Out]

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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 2297

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Simp[(x*(a + b*Log[c*x^n])^(p + 1))/(b*n*(p + 1))
, x] - Dist[1/(b*n*(p + 1)), Int[(a + b*Log[c*x^n])^(p + 1), x], x] /; FreeQ[{a, b, c, n}, x] && LtQ[p, -1] &&
IntegerQ[2*p]

Rule 2298

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

Rule 2302

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

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 2353

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

Rule 6742

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

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int (\frac{2(1+2x{)}^2}{x{\log }^3(x)}-\frac{2(1+2x{)}^2}{x{\log }^2(x)}+\frac{4}{\log (x)})dx\\ & =2\int \frac{(1+2x{)}^2}{x{\log }^3(x)}dx-2\int \frac{(1+2x{)}^2}{x{\log }^2(x)}dx+4\int \frac{1}{\log (x)}dx\\ & =4\text{li}(x)+2\int (\frac{4}{{\log }^3(x)}+\frac{1}{x{\log }^3(x)}+\frac{4x}{{\log }^3(x)})dx-2\int (\frac{4}{{\log }^2(x)}+\frac{1}{x{\log }^2(x)}+\frac{4x}{{\log }^2(x)})dx\\ & =4\text{li}(x)+2\int \frac{1}{x{\log }^3(x)}dx-2\int \frac{1}{x{\log }^2(x)}dx+8\int \frac{1}{{\log }^3(x)}dx+8\int \frac{x}{{\log }^3(x)}dx-8\int \frac{1}{{\log }^2(x)}dx-8\int \frac{x}{{\log }^2(x)}dx\\ & =-\frac{4x}{{\log }^2(x)}-\frac{4x^2}{{\log }^2(x)}+\frac{8x}{\log (x)}+\frac{8x^2}{\log (x)}+4\text{li}(x)+2\mathrm{Subst}(\int \frac{1}{x^3}dx,x,\log (x))-2\mathrm{Subst}(\int \frac{1}{x^2}dx,x,\log (x))+4\int \frac{1}{{\log }^2(x)}dx+8\int \frac{x}{{\log }^2(x)}dx-8\int \frac{1}{\log (x)}dx-16\int \frac{x}{\log (x)}dx\\ & =-\frac{1}{{\log }^2(x)}-\frac{4x}{{\log }^2(x)}-\frac{4x^2}{{\log }^2(x)}+\frac{2}{\log (x)}+\frac{4x}{\log (x)}-4\text{li}(x)+4\int \frac{1}{\log (x)}dx+16\int \frac{x}{\log (x)}dx-16\mathrm{Subst}(\int \frac{e^{2x}}{x}dx,x,\log (x))\\ & =-16\text{Ei}(2\log (x))-\frac{1}{{\log }^2(x)}-\frac{4x}{{\log }^2(x)}-\frac{4x^2}{{\log }^2(x)}+\frac{2}{\log (x)}+\frac{4x}{\log (x)}+16\mathrm{Subst}(\int \frac{e^{2x}}{x}dx,x,\log (x))\\ & =-\frac{1}{{\log }^2(x)}-\frac{4x}{{\log }^2(x)}-\frac{4x^2}{{\log }^2(x)}+\frac{2}{\log (x)}+\frac{4x}{\log (x)}\end{array}\end{array}$$

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Mathematica [A]  time = 0.08, size = 20, normalized size = 1.00 $$\begin{array}{c}-\frac{(1+2x)(1+2x-2\log (x))}{{\log }^2(x)}\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(2 + 8*x + 8*x^2 + (-2 - 8*x - 8*x^2)*Log[x] + 4*x*Log[x]^2)/(x*Log[x]^3),x]

[Out]

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

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fricas [A]  time = 0.45, size = 25, normalized size = 1.25 $$\begin{array}{c}-\frac{4x^2-2(2x+1)\log (x)+4x+1}{\log (x{)}^2}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x*log(x)^2+(-8*x^2-8*x-2)*log(x)+8*x^2+8*x+2)/x/log(x)^3,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.14, size = 25, normalized size = 1.25 $$\begin{array}{c}-\frac{4x^2-4x\log (x)+4x-2\log (x)+1}{\log (x{)}^2}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x*log(x)^2+(-8*x^2-8*x-2)*log(x)+8*x^2+8*x+2)/x/log(x)^3,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.03, size = 25, normalized size = 1.25

method	result	size
norman	$\frac{-1-4x-4x^2+4x\ln (x)+2\ln (x)}{\ln (x{)}^2}$	$25$
risch	$-\frac{4x^2-4x\ln (x)+4x-2\ln (x)+1}{\ln (x{)}^2}$	$26$
default	$\frac{4x}{\ln (x)}-\frac{4x^2}{\ln (x{)}^2}+\frac{2}{\ln (x)}-\frac{4x}{\ln (x{)}^2}-\frac{1}{\ln (x{)}^2}$	$37$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x*ln(x)^2+(-8*x^2-8*x-2)*ln(x)+8*x^2+8*x+2)/x/ln(x)^3,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [C]  time = 0.41, size = 50, normalized size = 2.50 $$\begin{array}{c}\frac{2}{\log (x)}-\frac{1}{\log (x{)}^2}+4\mathrm{E}\mathrm{i}(\log (x))-8\mathrm{\Gamma }(-1,-\log (x))-16\mathrm{\Gamma }(-1,-2\log (x))-8\mathrm{\Gamma }(-2,-\log (x))-32\mathrm{\Gamma }(-2,-2\log (x))\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x*log(x)^2+(-8*x^2-8*x-2)*log(x)+8*x^2+8*x+2)/x/log(x)^3,x, algorithm="maxima")

[Out]

2/log(x) - 1/log(x)^2 + 4*Ei(log(x)) - 8*gamma(-1, -log(x)) - 16*gamma(-1, -2*log(x)) - 8*gamma(-2, -log(x)) -
32*gamma(-2, -2*log(x))

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mupad [B]  time = 5.12, size = 24, normalized size = 1.20 $$\begin{array}{c}\frac{4x+2}{\ln (x)}-\frac{{(2x+1)}^2}{{\ln (x)}^2}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*x + 4*x*log(x)^2 - log(x)*(8*x + 8*x^2 + 2) + 8*x^2 + 2)/(x*log(x)^3),x)

[Out]

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

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sympy [A]  time = 0.10, size = 22, normalized size = 1.10 $$\begin{array}{c}\frac{-4x^2-4x+(4x+2)\log (x)-1}{\log {(x)}^2}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x*ln(x)**2+(-8*x**2-8*x-2)*ln(x)+8*x**2+8*x+2)/x/ln(x)**3,x)

[Out]

(-4*x**2 - 4*x + (4*x + 2)*log(x) - 1)/log(x)**2

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