3.85.62 $\int \frac{256+16x^4+32x^5+x^9+x^{10}+(256+80x^4+32x^5+x^{10})\log (x)}{(256x+16x^5+32x^6+x^{10}+x^{11})\log (x)}dx$

Optimal. Leaf size=18 $$\log (x+\frac{x}{\frac{16}{x^4}+x})+\log (\log (x))$$

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Rubi [B]  time = 0.69, antiderivative size = 37, normalized size of antiderivative = 2.06, number of steps used = 9, number of rules used = 6, integrand size = 63, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.095, Rules used = {6741, 6742, 1587, 260, 2302, 29} $$\begin{array}{c}-\log (x^5+16)+\log (x^4-x^3+2x^2-4x+8)+\log (x)+\log (x+2)+\log (\log (x))\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(256 + 16*x^4 + 32*x^5 + x^9 + x^10 + (256 + 80*x^4 + 32*x^5 + x^10)*Log[x])/((256*x + 16*x^5 + 32*x^6 + x
^10 + x^11)*Log[x]),x]

[Out]

Log[x] + Log[2 + x] + Log[8 - 4*x + 2*x^2 - x^3 + x^4] - Log[16 + x^5] + Log[Log[x]]

Rule 29

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

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 1587

Int[(Pp_)/(Qq_), x_Symbol] :> With[{p = Expon[Pp, x], q = Expon[Qq, x]}, Simp[(Coeff[Pp, x, p]*Log[RemoveConte
nt[Qq, x]])/(q*Coeff[Qq, x, q]), x] /; EqQ[p, q - 1] && EqQ[Pp, Simplify[(Coeff[Pp, x, p]*D[Qq, x])/(q*Coeff[Q
q, x, q])]]] /; PolyQ[Pp, x] && PolyQ[Qq, 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 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

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{256+16x^4+32x^5+x^9+x^{10}+(256+80x^4+32x^5+x^{10})\log (x)}{x(256+16x^4+32x^5+x^9+x^{10})\log (x)}dx\\ & =\int (\frac{256+80x^4+32x^5+x^{10}}{x(256+16x^4+32x^5+x^9+x^{10})}+\frac{1}{x\log (x)})dx\\ & =\int \frac{256+80x^4+32x^5+x^{10}}{x(256+16x^4+32x^5+x^9+x^{10})}dx+\int \frac{1}{x\log (x)}dx\\ & =\int (\frac{1}{x}+\frac{1}{2+x}+\frac{-4+4x-3x^2+4x^3}{8-4x+2x^2-x^3+x^4}-\frac{5x^4}{16+x^5})dx+\mathrm{Subst}(\int \frac{1}{x}dx,x,\log (x))\\ & =\log (x)+\log (2+x)+\log (\log (x))-5\int \frac{x^4}{16+x^5}dx+\int \frac{-4+4x-3x^2+4x^3}{8-4x+2x^2-x^3+x^4}dx\\ & =\log (x)+\log (2+x)+\log (8-4x+2x^2-x^3+x^4)-\log (16+x^5)+\log (\log (x))\end{array}\end{array}$$

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Mathematica [A]  time = 0.16, size = 23, normalized size = 1.28 $$\begin{array}{c}\log (x)-\log (16+x^5)+\log (16+x^4+x^5)+\log (\log (x))\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(256 + 16*x^4 + 32*x^5 + x^9 + x^10 + (256 + 80*x^4 + 32*x^5 + x^10)*Log[x])/((256*x + 16*x^5 + 32*x
^6 + x^10 + x^11)*Log[x]),x]

[Out]

Log[x] - Log[16 + x^5] + Log[16 + x^4 + x^5] + Log[Log[x]]

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fricas [A]  time = 0.86, size = 23, normalized size = 1.28 $$\begin{array}{c}\log (x^6+x^5+16x)-\log (x^5+16)+\log (\log (x))\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^10+32*x^5+80*x^4+256)*log(x)+x^10+x^9+32*x^5+16*x^4+256)/(x^11+x^10+32*x^6+16*x^5+256*x)/log(x),
x, algorithm="fricas")

[Out]

log(x^6 + x^5 + 16*x) - log(x^5 + 16) + log(log(x))

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giac [A]  time = 0.20, size = 23, normalized size = 1.28 $$\begin{array}{c}\log (x^5+x^4+16)-\log (x^5+16)+\log (x)+\log (\log (x))\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^10+32*x^5+80*x^4+256)*log(x)+x^10+x^9+32*x^5+16*x^4+256)/(x^11+x^10+32*x^6+16*x^5+256*x)/log(x),
x, algorithm="giac")

[Out]

log(x^5 + x^4 + 16) - log(x^5 + 16) + log(x) + log(log(x))

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maple [A]  time = 0.09, size = 24, normalized size = 1.33

method	result	size
risch	$-\ln (x^5+16)+\ln (x^6+x^5+16x)+\ln (\ln (x))$	$24$
default	$\ln (\ln (x))+\ln (x)+\ln (x^4-x^3+2x^2-4x+8)-\ln (x^5+16)+\ln (2+x)$	$38$
norman	$\ln (\ln (x))+\ln (x)+\ln (x^4-x^3+2x^2-4x+8)-\ln (x^5+16)+\ln (2+x)$	$38$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^10+32*x^5+80*x^4+256)*ln(x)+x^10+x^9+32*x^5+16*x^4+256)/(x^11+x^10+32*x^6+16*x^5+256*x)/ln(x),x,method
=_RETURNVERBOSE)

[Out]

-ln(x^5+16)+ln(x^6+x^5+16*x)+ln(ln(x))

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maxima [B]  time = 0.39, size = 37, normalized size = 2.06 $$\begin{array}{c}-\log (x^5+16)+\log (x^4-x^3+2x^2-4x+8)+\log (x+2)+\log (x)+\log (\log (x))\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^10+32*x^5+80*x^4+256)*log(x)+x^10+x^9+32*x^5+16*x^4+256)/(x^11+x^10+32*x^6+16*x^5+256*x)/log(x),
x, algorithm="maxima")

[Out]

-log(x^5 + 16) + log(x^4 - x^3 + 2*x^2 - 4*x + 8) + log(x + 2) + log(x) + log(log(x))

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mupad [B]  time = 5.47, size = 23, normalized size = 1.28 $$\begin{array}{c}\ln (\ln (x))+\ln \left(x(x^5+x^4+16)\right)-\ln (x^5+16)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((16*x^4 + 32*x^5 + x^9 + x^10 + log(x)*(80*x^4 + 32*x^5 + x^10 + 256) + 256)/(log(x)*(256*x + 16*x^5 + 32*
x^6 + x^10 + x^11)),x)

[Out]

log(log(x)) + log(x*(x^4 + x^5 + 16)) - log(x^5 + 16)

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sympy [A]  time = 0.21, size = 22, normalized size = 1.22 $$\begin{array}{c}-\log (x^5+16)+\log (x^6+x^5+16x)+\log (\log (x))\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x**10+32*x**5+80*x**4+256)*ln(x)+x**10+x**9+32*x**5+16*x**4+256)/(x**11+x**10+32*x**6+16*x**5+256*
x)/ln(x),x)

[Out]

-log(x**5 + 16) + log(x**6 + x**5 + 16*x) + log(log(x))

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