3.54.59 \(\int \frac {270-270 x^3+90 x^6-10 x^9-x^6 \log ^2(x)}{(-54 x+54 x^4-18 x^7+2 x^{10}) \log ^2(x)} \, dx\)

Optimal. Leaf size=26 \[ 5+\frac {x^2}{36 \left (\frac {3}{x^2}-x\right )^2}+\frac {5}{\log (x)} \]

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Rubi [A]  time = 0.38, antiderivative size = 23, normalized size of antiderivative = 0.88, number of steps used = 7, number of rules used = 6, integrand size = 52, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {6688, 12, 6742, 264, 2302, 30} \begin {gather*} \frac {x^6}{36 \left (3-x^3\right )^2}+\frac {5}{\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(270 - 270*x^3 + 90*x^6 - 10*x^9 - x^6*Log[x]^2)/((-54*x + 54*x^4 - 18*x^7 + 2*x^10)*Log[x]^2),x]

[Out]

x^6/(36*(3 - x^3)^2) + 5/Log[x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 264

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a
*c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]

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 6688

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

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {10 \left (-3+x^3\right )^3+x^6 \log ^2(x)}{2 x \left (3-x^3\right )^3 \log ^2(x)} \, dx\\ &=\frac {1}{2} \int \frac {10 \left (-3+x^3\right )^3+x^6 \log ^2(x)}{x \left (3-x^3\right )^3 \log ^2(x)} \, dx\\ &=\frac {1}{2} \int \left (-\frac {x^5}{\left (-3+x^3\right )^3}-\frac {10}{x \log ^2(x)}\right ) \, dx\\ &=-\left (\frac {1}{2} \int \frac {x^5}{\left (-3+x^3\right )^3} \, dx\right )-5 \int \frac {1}{x \log ^2(x)} \, dx\\ &=\frac {x^6}{36 \left (3-x^3\right )^2}-5 \operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log (x)\right )\\ &=\frac {x^6}{36 \left (3-x^3\right )^2}+\frac {5}{\log (x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 29, normalized size = 1.12 \begin {gather*} \frac {1}{2} \left (-\frac {3-2 x^3}{6 \left (-3+x^3\right )^2}+\frac {10}{\log (x)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(270 - 270*x^3 + 90*x^6 - 10*x^9 - x^6*Log[x]^2)/((-54*x + 54*x^4 - 18*x^7 + 2*x^10)*Log[x]^2),x]

[Out]

(-1/6*(3 - 2*x^3)/(-3 + x^3)^2 + 10/Log[x])/2

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fricas [A]  time = 0.69, size = 40, normalized size = 1.54 \begin {gather*} \frac {60 \, x^{6} - 360 \, x^{3} + {\left (2 \, x^{3} - 3\right )} \log \relax (x) + 540}{12 \, {\left (x^{6} - 6 \, x^{3} + 9\right )} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^6*log(x)^2-10*x^9+90*x^6-270*x^3+270)/(2*x^10-18*x^7+54*x^4-54*x)/log(x)^2,x, algorithm="fricas"
)

[Out]

1/12*(60*x^6 - 360*x^3 + (2*x^3 - 3)*log(x) + 540)/((x^6 - 6*x^3 + 9)*log(x))

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giac [A]  time = 0.15, size = 28, normalized size = 1.08 \begin {gather*} \frac {2 \, x^{3} - 3}{12 \, {\left (x^{6} - 6 \, x^{3} + 9\right )}} + \frac {5}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^6*log(x)^2-10*x^9+90*x^6-270*x^3+270)/(2*x^10-18*x^7+54*x^4-54*x)/log(x)^2,x, algorithm="giac")

[Out]

1/12*(2*x^3 - 3)/(x^6 - 6*x^3 + 9) + 5/log(x)

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maple [A]  time = 0.03, size = 26, normalized size = 1.00

method	result	size
default	\(\frac {5}{\ln \relax (x )}+\frac {1}{4 \left (x^{3}-3\right )^{2}}+\frac {1}{6 x^{3}-18}\)	\(26\)
risch	\(\frac {2 x^{3}-3}{12 x^{6}-72 x^{3}+108}+\frac {5}{\ln \relax (x )}\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-x^6*ln(x)^2-10*x^9+90*x^6-270*x^3+270)/(2*x^10-18*x^7+54*x^4-54*x)/ln(x)^2,x,method=_RETURNVERBOSE)

[Out]

5/ln(x)+1/4/(x^3-3)^2+1/6/(x^3-3)

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maxima [A]  time = 0.48, size = 40, normalized size = 1.54 \begin {gather*} \frac {60 \, x^{6} - 360 \, x^{3} + {\left (2 \, x^{3} - 3\right )} \log \relax (x) + 540}{12 \, {\left (x^{6} - 6 \, x^{3} + 9\right )} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^6*log(x)^2-10*x^9+90*x^6-270*x^3+270)/(2*x^10-18*x^7+54*x^4-54*x)/log(x)^2,x, algorithm="maxima"
)

[Out]

1/12*(60*x^6 - 360*x^3 + (2*x^3 - 3)*log(x) + 540)/((x^6 - 6*x^3 + 9)*log(x))

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mupad [B]  time = 3.65, size = 27, normalized size = 1.04 \begin {gather*} \frac {5}{\ln \relax (x)}+\frac {\frac {x^3}{6}-\frac {1}{4}}{x^6-6\,x^3+9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^6*log(x)^2 + 270*x^3 - 90*x^6 + 10*x^9 - 270)/(log(x)^2*(54*x - 54*x^4 + 18*x^7 - 2*x^10)),x)

[Out]

5/log(x) + (x^3/6 - 1/4)/(x^6 - 6*x^3 + 9)

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sympy [A]  time = 0.15, size = 22, normalized size = 0.85 \begin {gather*} - \frac {3 - 2 x^{3}}{12 x^{6} - 72 x^{3} + 108} + \frac {5}{\log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x**6*ln(x)**2-10*x**9+90*x**6-270*x**3+270)/(2*x**10-18*x**7+54*x**4-54*x)/ln(x)**2,x)

[Out]

-(3 - 2*x**3)/(12*x**6 - 72*x**3 + 108) + 5/log(x)

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