3.19.72 \(\int \frac {-2-6 x-18 x^2}{1+81 x^4-2 x \log (x^3)+x^2 \log ^2(x^3)+3 x^2 (-6+6 x \log (x^3))} \, dx\)

Optimal. Leaf size=18 \[ \frac {2 x}{-1+9 x^2+x \log \left (x^3\right )} \]

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Rubi [F]  time = 0.32, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {-2-6 x-18 x^2}{1+81 x^4-2 x \log \left (x^3\right )+x^2 \log ^2\left (x^3\right )+3 x^2 \left (-6+6 x \log \left (x^3\right )\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-2 - 6*x - 18*x^2)/(1 + 81*x^4 - 2*x*Log[x^3] + x^2*Log[x^3]^2 + 3*x^2*(-6 + 6*x*Log[x^3])),x]

[Out]

-2*Defer[Int][(-1 + 9*x^2 + x*Log[x^3])^(-2), x] - 6*Defer[Int][x/(-1 + 9*x^2 + x*Log[x^3])^2, x] - 18*Defer[I
nt][x^2/(-1 + 9*x^2 + x*Log[x^3])^2, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2 \left (-1-3 x-9 x^2\right )}{\left (1-9 x^2-x \log \left (x^3\right )\right )^2} \, dx\\ &=2 \int \frac {-1-3 x-9 x^2}{\left (1-9 x^2-x \log \left (x^3\right )\right )^2} \, dx\\ &=2 \int \left (-\frac {1}{\left (-1+9 x^2+x \log \left (x^3\right )\right )^2}-\frac {3 x}{\left (-1+9 x^2+x \log \left (x^3\right )\right )^2}-\frac {9 x^2}{\left (-1+9 x^2+x \log \left (x^3\right )\right )^2}\right ) \, dx\\ &=-\left (2 \int \frac {1}{\left (-1+9 x^2+x \log \left (x^3\right )\right )^2} \, dx\right )-6 \int \frac {x}{\left (-1+9 x^2+x \log \left (x^3\right )\right )^2} \, dx-18 \int \frac {x^2}{\left (-1+9 x^2+x \log \left (x^3\right )\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.10, size = 18, normalized size = 1.00 \begin {gather*} \frac {2 x}{-1+9 x^2+x \log \left (x^3\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2 - 6*x - 18*x^2)/(1 + 81*x^4 - 2*x*Log[x^3] + x^2*Log[x^3]^2 + 3*x^2*(-6 + 6*x*Log[x^3])),x]

[Out]

(2*x)/(-1 + 9*x^2 + x*Log[x^3])

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fricas [A]  time = 1.02, size = 17, normalized size = 0.94 \begin {gather*} \frac {2 \, x}{9 \, x^{2} + 3 \, x \log \relax (x) - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-6*exp(log(3*x)+log(x))-6*x-2)/(9*exp(log(3*x)+log(x))^2+(6*x*log(x^3)-6)*exp(log(3*x)+log(x))+x^2*
log(x^3)^2-2*x*log(x^3)+1),x, algorithm="fricas")

[Out]

2*x/(9*x^2 + 3*x*log(x) - 1)

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giac [A]  time = 0.16, size = 17, normalized size = 0.94 \begin {gather*} \frac {2 \, x}{9 \, x^{2} + 3 \, x \log \relax (x) - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-6*exp(log(3*x)+log(x))-6*x-2)/(9*exp(log(3*x)+log(x))^2+(6*x*log(x^3)-6)*exp(log(3*x)+log(x))+x^2*
log(x^3)^2-2*x*log(x^3)+1),x, algorithm="giac")

[Out]

2*x/(9*x^2 + 3*x*log(x) - 1)

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maple [A]  time = 0.16, size = 19, normalized size = 1.06

method	result	size
norman	\(\frac {2 x}{9 x^{2}+x \ln \left (x^{3}\right )-1}\)	\(19\)
default	\(\frac {2 x}{9 x^{2}+3 x \ln \relax (x )+x \left (\ln \left (x^{3}\right )-3 \ln \relax (x )\right )-1}\)	\(29\)
risch	\(\frac {4 i x}{x \pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 x \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+x \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x^{3}\right )-x \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{3}\right )^{2}+x \pi \mathrm {csgn}\left (i x^{2}\right )^{3}-x \pi \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x^{3}\right )^{2}+x \pi \mathrm {csgn}\left (i x^{3}\right )^{3}+6 i x \ln \relax (x )+18 i x^{2}-2 i}\)	\(141\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-6*exp(ln(3*x)+ln(x))-6*x-2)/(9*exp(ln(3*x)+ln(x))^2+(6*x*ln(x^3)-6)*exp(ln(3*x)+ln(x))+x^2*ln(x^3)^2-2*x
*ln(x^3)+1),x,method=_RETURNVERBOSE)

[Out]

2*x/(9*x^2+x*ln(x^3)-1)

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maxima [A]  time = 0.73, size = 17, normalized size = 0.94 \begin {gather*} \frac {2 \, x}{9 \, x^{2} + 3 \, x \log \relax (x) - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-6*exp(log(3*x)+log(x))-6*x-2)/(9*exp(log(3*x)+log(x))^2+(6*x*log(x^3)-6)*exp(log(3*x)+log(x))+x^2*
log(x^3)^2-2*x*log(x^3)+1),x, algorithm="maxima")

[Out]

2*x/(9*x^2 + 3*x*log(x) - 1)

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mupad [B]  time = 4.03, size = 18, normalized size = 1.00 \begin {gather*} \frac {2\,x}{x\,\ln \left (x^3\right )+9\,x^2-1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(6*x + 6*exp(log(3*x) + log(x)) + 2)/(9*exp(2*log(3*x) + 2*log(x)) - 2*x*log(x^3) + exp(log(3*x) + log(x)
)*(6*x*log(x^3) - 6) + x^2*log(x^3)^2 + 1),x)

[Out]

(2*x)/(x*log(x^3) + 9*x^2 - 1)

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sympy [A]  time = 0.13, size = 15, normalized size = 0.83 \begin {gather*} \frac {2 x}{9 x^{2} + x \log {\left (x^{3} \right )} - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-6*exp(ln(3*x)+ln(x))-6*x-2)/(9*exp(ln(3*x)+ln(x))**2+(6*x*ln(x**3)-6)*exp(ln(3*x)+ln(x))+x**2*ln(x
**3)**2-2*x*ln(x**3)+1),x)

[Out]

2*x/(9*x**2 + x*log(x**3) - 1)

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