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

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

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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 {gather*} \log \left (\log \left (\frac {4 x^3}{\left (x^2+x-\log (x)\right )^2}\right )\right ) \end {gather*}

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 {gather*} \begin {aligned} \text {integral} &=\int \frac {2+x-x^2-3 \log (x)}{x \left (x+x^2-\log (x)\right ) \log \left (\frac {4 x^3}{\left (x+x^2-\log (x)\right )^2}\right )} \, dx\\ &=\log \left (\log \left (\frac {4 x^3}{\left (x+x^2-\log (x)\right )^2}\right )\right )\\ \end {aligned} \end {gather*}

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

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 {gather*} \log \left (\log \left (\frac {4 \, x^{3}}{x^{4} + 2 \, x^{3} + x^{2} - 2 \, {\left (x^{2} + x\right )} \log \relax (x) + \log \relax (x)^{2}}\right )\right ) \end {gather*}

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 {gather*} \log \left (2 \, \log \relax (2) - \log \left (x^{4} + 2 \, x^{3} - 2 \, x^{2} \log \relax (x) + x^{2} - 2 \, x \log \relax (x) + \log \relax (x)^{2}\right ) + 3 \, \log \relax (x)\right ) \end {gather*}

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	\(\ln \left (\ln \left (x^{2}-\ln \relax (x )+x \right )+\frac {i \left (\pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+\pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x^{3}\right )-\pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{3}\right )^{2}+\pi \mathrm {csgn}\left (i x^{2}\right )^{3}-\pi \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x^{3}\right )^{2}+\pi \mathrm {csgn}\left (i x^{3}\right )^{3}+\pi \,\mathrm {csgn}\left (i x^{3}\right ) \mathrm {csgn}\left (\frac {i}{\left (-x^{2}+\ln \relax (x )-x \right )^{2}}\right ) \mathrm {csgn}\left (\frac {i x^{3}}{\left (-x^{2}+\ln \relax (x )-x \right )^{2}}\right )-\pi \,\mathrm {csgn}\left (i x^{3}\right ) \mathrm {csgn}\left (\frac {i x^{3}}{\left (-x^{2}+\ln \relax (x )-x \right )^{2}}\right )^{2}-\pi \,\mathrm {csgn}\left (\frac {i}{\left (-x^{2}+\ln \relax (x )-x \right )^{2}}\right ) \mathrm {csgn}\left (\frac {i x^{3}}{\left (-x^{2}+\ln \relax (x )-x \right )^{2}}\right )^{2}-\pi \mathrm {csgn}\left (i \left (-x^{2}+\ln \relax (x )-x \right )\right )^{2} \mathrm {csgn}\left (i \left (-x^{2}+\ln \relax (x )-x \right )^{2}\right )-2 \pi \,\mathrm {csgn}\left (i \left (-x^{2}+\ln \relax (x )-x \right )\right ) \mathrm {csgn}\left (i \left (-x^{2}+\ln \relax (x )-x \right )^{2}\right )^{2}-\pi \mathrm {csgn}\left (i \left (-x^{2}+\ln \relax (x )-x \right )^{2}\right )^{3}+\pi \mathrm {csgn}\left (\frac {i x^{3}}{\left (-x^{2}+\ln \relax (x )-x \right )^{2}}\right )^{3}+4 i \ln \relax (2)+6 i \ln \relax (x )\right )}{4}\right )\)	\(379\)

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 {gather*} \log \left (-\log \relax (2) + \log \left (-x^{2} - x + \log \relax (x)\right ) - \frac {3}{2} \, \log \relax (x)\right ) \end {gather*}

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 {gather*} \ln \left (\ln \left (\frac {4\,x^3}{{\ln \relax (x)}^2-\ln \relax (x)\,\left (2\,x^2+2\,x\right )+x^2+2\,x^3+x^4}\right )\right ) \end {gather*}

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 {gather*} \log {\left (\log {\left (\frac {4 x^{3}}{x^{4} + 2 x^{3} + x^{2} + \left (- 2 x^{2} - 2 x\right ) \log {\relax (x )} + \log {\relax (x )}^{2}} \right )} \right )} \end {gather*}

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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