3.69.91 $\int \frac{-2-x+x^2+3\log (x)}{(-x^2-x^3+x\log (x))\log (\frac{4x^3}{x^2+2x^3+x^4+(-2x-2x^2)\log (x)+{\log }^2(x)})}dx$

Optimal. Leaf size=19 $$\log (\log \left(\frac{4x}{{(x+\frac{x-\log (x)}{x})}^2}\right))$$

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Rubi [A]  time = 0.39, antiderivative size = 18, normalized size of antiderivative = 0.95, number of steps used = 2, number of rules used = 2, integrand size = 68, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.029, Rules used = {6688, 6684} $$\begin{array}{c}\log (\log \left(\frac{4x^3}{{(x^2+x-\log (x))}^2}\right))\end{array}$$

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6684

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rule 6688

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

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{2+x-x^2-3\log (x)}{x(x+x^2-\log (x))\log \left(\frac{4x^3}{{(x+x^2-\log (x))}^2}\right)}dx\\ & =\log (\log \left(\frac{4x^3}{{(x+x^2-\log (x))}^2}\right))\end{array}\end{array}$$

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Mathematica [A]  time = 0.25, size = 18, normalized size = 0.95 $$\begin{array}{c}\log (\log \left(\frac{4x^3}{{(x+x^2-\log (x))}^2}\right))\end{array}$$

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.59, size = 34, normalized size = 1.79 $$\begin{array}{c}\log (\log \left(\frac{4x^3}{x^4+2x^3+x^2-2(x^2+x)\log (x)+\log (x{)}^2}\right))\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.36, size = 41, normalized size = 2.16 $$\begin{array}{c}\log (2\log (2)-\log (x^4+2x^3-2x^2\log (x)+x^2-2x\log (x)+\log (x{)}^2)+3\log (x))\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(2*log(2) - log(x^4 + 2*x^3 - 2*x^2*log(x) + x^2 - 2*x*log(x) + log(x)^2) + 3*log(x))

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maple [C]  time = 0.13, size = 379, normalized size = 19.95

method	result	size
risch	$\ln (\ln (x^2-\ln (x)+x)+\frac{i(\pi \mathrm{csgn}{\left(ix\right)}^2\mathrm{csgn}\left(i{x}^2\right)-2\pi \mathrm{csgn}\left(ix\right)\mathrm{csgn}{\left(i{x}^2\right)}^2+\pi \mathrm{csgn}\left(ix\right)\mathrm{csgn}\left(i{x}^2\right)\mathrm{csgn}\left(i{x}^3\right)-\pi \mathrm{csgn}\left(ix\right)\mathrm{csgn}{\left(i{x}^3\right)}^2+\pi \mathrm{csgn}{\left(i{x}^2\right)}^3-\pi \mathrm{csgn}\left(i{x}^2\right)\mathrm{csgn}{\left(i{x}^3\right)}^2+\pi \mathrm{csgn}{\left(i{x}^3\right)}^3+\pi \mathrm{csgn}\left(i{x}^3\right)\mathrm{csgn}\left(\frac{i}{{(-x^2+\ln (x)-x)}^2}\right)\mathrm{csgn}\left(\frac{i{x}^3}{{(-x^2+\ln (x)-x)}^2}\right)-\pi \mathrm{csgn}\left(i{x}^3\right)\mathrm{csgn}{\left(\frac{i{x}^3}{{(-x^2+\ln (x)-x)}^2}\right)}^2-\pi \mathrm{csgn}\left(\frac{i}{{(-x^2+\ln (x)-x)}^2}\right)\mathrm{csgn}{\left(\frac{i{x}^3}{{(-x^2+\ln (x)-x)}^2}\right)}^2-\pi \mathrm{csgn}{\left(i(-x^2+\ln (x)-x)\right)}^2\mathrm{csgn}\left(i{(-x^2+\ln (x)-x)}^2\right)-2\pi \mathrm{csgn}\left(i(-x^2+\ln (x)-x)\right)\mathrm{csgn}{\left(i{(-x^2+\ln (x)-x)}^2\right)}^2-\pi \mathrm{csgn}{\left(i{(-x^2+\ln (x)-x)}^2\right)}^3+\pi \mathrm{csgn}{\left(\frac{i{x}^3}{{(-x^2+\ln (x)-x)}^2}\right)}^3+4i\ln (2)+6i\ln (x))}{4})$	$379$

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

ln(ln(x^2-ln(x)+x)+1/4*I*(Pi*csgn(I*x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2+Pi*csgn(I*x)*csgn(I*x^2)*csg
n(I*x^3)-Pi*csgn(I*x)*csgn(I*x^3)^2+Pi*csgn(I*x^2)^3-Pi*csgn(I*x^2)*csgn(I*x^3)^2+Pi*csgn(I*x^3)^3+Pi*csgn(I*x
^3)*csgn(I/(-x^2+ln(x)-x)^2)*csgn(I*x^3/(-x^2+ln(x)-x)^2)-Pi*csgn(I*x^3)*csgn(I*x^3/(-x^2+ln(x)-x)^2)^2-Pi*csg
n(I/(-x^2+ln(x)-x)^2)*csgn(I*x^3/(-x^2+ln(x)-x)^2)^2-Pi*csgn(I*(-x^2+ln(x)-x))^2*csgn(I*(-x^2+ln(x)-x)^2)-2*Pi
*csgn(I*(-x^2+ln(x)-x))*csgn(I*(-x^2+ln(x)-x)^2)^2-Pi*csgn(I*(-x^2+ln(x)-x)^2)^3+Pi*csgn(I*x^3/(-x^2+ln(x)-x)^
2)^3+4*I*ln(2)+6*I*ln(x)))

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maxima [A]  time = 0.48, size = 22, normalized size = 1.16 $$\begin{array}{c}\log (-\log (2)+\log (-x^2-x+\log (x))-\frac{3}{2}\log (x))\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(-log(2) + log(-x^2 - x + log(x)) - 3/2*log(x))

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mupad [B]  time = 5.29, size = 38, normalized size = 2.00 $$\begin{array}{c}\ln (\ln \left(\frac{4x^3}{{\ln (x)}^2-\ln (x)(2x^2+2x)+x^2+2x^3+x^4}\right))\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [B]  time = 0.64, size = 37, normalized size = 1.95 $$\begin{array}{c}\log (\log \left(\frac{4x^3}{x^4+2x^3+x^2+(-2x^2-2x)\log (x)+\log {(x)}^2}\right))\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(log(4*x**3/(x**4 + 2*x**3 + x**2 + (-2*x**2 - 2*x)*log(x) + log(x)**2)))

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