3.73.66 $\int \frac{-2\log (x)+(2+(2+x)\log (x))\log (x+9e^{8}x)}{x\log (x)\log (x+9e^{8}x)}dx$

Optimal. Leaf size=25 $$x-\log \left(\frac{36{\log }^2(x+9e^{8}x)}{x^2{\log }^2(x)}\right)$$

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Rubi [A]  time = 0.41, antiderivative size = 24, normalized size of antiderivative = 0.96, number of steps used = 11, number of rules used = 5, integrand size = 42, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.119, Rules used = {2444, 6742, 43, 2302, 29} $$\begin{array}{c}x+2\log (x)+2\log (\log (x))-2\log (\log \left((1+9e^8)x\right))\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(-2*Log[x] + (2 + (2 + x)*Log[x])*Log[x + 9*E^8*x])/(x*Log[x]*Log[x + 9*E^8*x]),x]

[Out]

x + 2*Log[x] + 2*Log[Log[x]] - 2*Log[Log[(1 + 9*E^8)*x]]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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 2444

Int[((a_.) + Log[(c_.)*(v_)^(n_.)]*(b_.))^(p_.)*(u_.), x_Symbol] :> Int[u*(a + b*Log[c*ExpandToSum[v, x]^n])^p
, x] /; FreeQ[{a, b, c, n, p}, x] && LinearQ[v, x] &&  !LinearMatchQ[v, x] &&  !(EqQ[n, 1] && MatchQ[c*v, (e_.
)*((f_) + (g_.)*x) /; FreeQ[{e, f, g}, x]])

Rule 6742

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

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{-2\log (x)+(2+(2+x)\log (x))\log (x+9e^{8}x)}{x\log (x)\log \left((1+9e^8)x\right)}dx\\ & =\int (\frac{2+2\log (x)+x\log (x)}{x\log (x)}-\frac{2}{x\log \left((1+9e^8)x\right)})dx\\ & =-(2\int \frac{1}{x\log \left((1+9e^8)x\right)}dx)+\int \frac{2+2\log (x)+x\log (x)}{x\log (x)}dx\\ & =-\left(2\mathrm{Subst}(\int \frac{1}{x}dx,x,\log \left((1+9e^8)x\right))\right)+\int (\frac{2+x}{x}+\frac{2}{x\log (x)})dx\\ & =-2\log (\log \left((1+9e^8)x\right))+2\int \frac{1}{x\log (x)}dx+\int \frac{2+x}{x}dx\\ & =-2\log (\log \left((1+9e^8)x\right))+2\mathrm{Subst}(\int \frac{1}{x}dx,x,\log (x))+\int (1+\frac{2}{x})dx\\ & =x+2\log (x)+2\log (\log (x))-2\log (\log \left((1+9e^8)x\right))\end{array}\end{array}$$

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Mathematica [A]  time = 0.03, size = 23, normalized size = 0.92 $$\begin{array}{c}x+2\log (x)+2\log (\log (x))-2\log (\log (x+9e^{8}x))\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(-2*Log[x] + (2 + (2 + x)*Log[x])*Log[x + 9*E^8*x])/(x*Log[x]*Log[x + 9*E^8*x]),x]

[Out]

x + 2*Log[x] + 2*Log[Log[x]] - 2*Log[Log[x + 9*E^8*x]]

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fricas [A]  time = 0.77, size = 24, normalized size = 0.96 $$\begin{array}{c}x+2\log (x)-2\log (\log (x)+\log (9e^8+1))+2\log (\log (x))\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2+x)*log(x)+2)*log(9*x*exp(4)^2+x)-2*log(x))/x/log(x)/log(9*x*exp(4)^2+x),x, algorithm="fricas")

[Out]

x + 2*log(x) - 2*log(log(x) + log(9*e^8 + 1)) + 2*log(log(x))

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giac [A]  time = 0.22, size = 28, normalized size = 1.12 $$\begin{array}{c}x+2\log (x)-2\log (-\log (x)-\log (9e^8+1))+2\log (\log (x))\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2+x)*log(x)+2)*log(9*x*exp(4)^2+x)-2*log(x))/x/log(x)/log(9*x*exp(4)^2+x),x, algorithm="giac")

[Out]

x + 2*log(x) - 2*log(-log(x) - log(9*e^8 + 1)) + 2*log(log(x))

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maple [A]  time = 0.06, size = 7, normalized size = 0.28

method	result	size
risch	$2\ln (x)+x$	$7$
default	$2\ln (x)+x+2\ln (\ln (x))-2\ln (\ln \left((9{\mathrm{e}}^8+1)x\right))$	$26$
norman	$x+2\ln (9x{\mathrm{e}}^8+x)+2\ln (\ln (x))-2\ln (\ln (9x{\mathrm{e}}^8+x))$	$33$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((2+x)*ln(x)+2)*ln(9*x*exp(4)^2+x)-2*ln(x))/x/ln(x)/ln(9*x*exp(4)^2+x),x,method=_RETURNVERBOSE)

[Out]

2*ln(x)+x

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maxima [A]  time = 0.35, size = 22, normalized size = 0.88 $$\begin{array}{c}x+2\log (x)-2\log (\log (9x{e}^8+x))+2\log (\log (x))\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2+x)*log(x)+2)*log(9*x*exp(4)^2+x)-2*log(x))/x/log(x)/log(9*x*exp(4)^2+x),x, algorithm="maxima")

[Out]

x + 2*log(x) - 2*log(log(9*x*e^8 + x)) + 2*log(log(x))

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mupad [B]  time = 4.60, size = 23, normalized size = 0.92 $$\begin{array}{c}x+2\ln (\ln (x))-2\ln (\ln \left(x(9{\mathrm{e}}^8+1)\right))+2\ln (x)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*log(x) - log(x + 9*x*exp(8))*(log(x)*(x + 2) + 2))/(x*log(x + 9*x*exp(8))*log(x)),x)

[Out]

x + 2*log(log(x)) - 2*log(log(x*(9*exp(8) + 1))) + 2*log(x)

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sympy [A]  time = 0.35, size = 27, normalized size = 1.08 $$\begin{array}{c}x+2\log (x)-2\log (\log (x)+\log (1+9e^8))+2\log (\log (x))\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2+x)*ln(x)+2)*ln(9*x*exp(4)**2+x)-2*ln(x))/x/ln(x)/ln(9*x*exp(4)**2+x),x)

[Out]

x + 2*log(x) - 2*log(log(x) + log(1 + 9*exp(8))) + 2*log(log(x))

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