3.38.7 \(\int \frac {40-180 x^2+360 x^2 \log (x)}{(-2 x+9 x^3) \log (4) \log (x)} \, dx\)

Optimal. Leaf size=23 \[ 4 \left (-8+\frac {5 \log \left (\frac {-2+9 x^2}{\log (x)}\right )}{\log (4)}\right ) \]

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Rubi [A]  time = 0.25, antiderivative size = 24, normalized size of antiderivative = 1.04, number of steps used = 7, number of rules used = 6, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {12, 1593, 6725, 260, 2302, 29} \begin {gather*} \frac {20 \log \left (2-9 x^2\right )}{\log (4)}-\frac {20 \log (\log (x))}{\log (4)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(40 - 180*x^2 + 360*x^2*Log[x])/((-2*x + 9*x^3)*Log[4]*Log[x]),x]

[Out]

(20*Log[2 - 9*x^2])/Log[4] - (20*Log[Log[x]])/Log[4]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

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 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {40-180 x^2+360 x^2 \log (x)}{\left (-2 x+9 x^3\right ) \log (x)} \, dx}{\log (4)}\\ &=\frac {\int \frac {40-180 x^2+360 x^2 \log (x)}{x \left (-2+9 x^2\right ) \log (x)} \, dx}{\log (4)}\\ &=\frac {\int \left (\frac {360 x}{-2+9 x^2}-\frac {20}{x \log (x)}\right ) \, dx}{\log (4)}\\ &=-\frac {20 \int \frac {1}{x \log (x)} \, dx}{\log (4)}+\frac {360 \int \frac {x}{-2+9 x^2} \, dx}{\log (4)}\\ &=\frac {20 \log \left (2-9 x^2\right )}{\log (4)}-\frac {20 \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\log (x)\right )}{\log (4)}\\ &=\frac {20 \log \left (2-9 x^2\right )}{\log (4)}-\frac {20 \log (\log (x))}{\log (4)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 20, normalized size = 0.87 \begin {gather*} \frac {20 \left (\log \left (2-9 x^2\right )-\log (\log (x))\right )}{\log (4)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(40 - 180*x^2 + 360*x^2*Log[x])/((-2*x + 9*x^3)*Log[4]*Log[x]),x]

[Out]

(20*(Log[2 - 9*x^2] - Log[Log[x]]))/Log[4]

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fricas [A]  time = 0.73, size = 20, normalized size = 0.87 \begin {gather*} \frac {10 \, {\left (\log \left (9 \, x^{2} - 2\right ) - \log \left (\log \relax (x)\right )\right )}}{\log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(360*x^2*log(x)-180*x^2+40)/(9*x^3-2*x)/log(2)/log(x),x, algorithm="fricas")

[Out]

10*(log(9*x^2 - 2) - log(log(x)))/log(2)

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giac [A]  time = 0.23, size = 20, normalized size = 0.87 \begin {gather*} \frac {10 \, {\left (\log \left (9 \, x^{2} - 2\right ) - \log \left (\log \relax (x)\right )\right )}}{\log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(360*x^2*log(x)-180*x^2+40)/(9*x^3-2*x)/log(2)/log(x),x, algorithm="giac")

[Out]

10*(log(9*x^2 - 2) - log(log(x)))/log(2)

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maple [A]  time = 0.09, size = 21, normalized size = 0.91

method	result	size
default	\(\frac {-10 \ln \left (\ln \relax (x )\right )+10 \ln \left (9 x^{2}-2\right )}{\ln \relax (2)}\)	\(21\)
norman	\(-\frac {10 \ln \left (\ln \relax (x )\right )}{\ln \relax (2)}+\frac {10 \ln \left (9 x^{2}-2\right )}{\ln \relax (2)}\)	\(25\)
risch	\(-\frac {10 \ln \left (\ln \relax (x )\right )}{\ln \relax (2)}+\frac {10 \ln \left (9 x^{2}-2\right )}{\ln \relax (2)}\)	\(25\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*(360*x^2*ln(x)-180*x^2+40)/(9*x^3-2*x)/ln(2)/ln(x),x,method=_RETURNVERBOSE)

[Out]

10/ln(2)*(-ln(ln(x))+ln(9*x^2-2))

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maxima [A]  time = 0.40, size = 20, normalized size = 0.87 \begin {gather*} \frac {10 \, {\left (\log \left (9 \, x^{2} - 2\right ) - \log \left (\log \relax (x)\right )\right )}}{\log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(360*x^2*log(x)-180*x^2+40)/(9*x^3-2*x)/log(2)/log(x),x, algorithm="maxima")

[Out]

10*(log(9*x^2 - 2) - log(log(x)))/log(2)

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mupad [B]  time = 2.26, size = 18, normalized size = 0.78 \begin {gather*} -\frac {10\,\left (\ln \left (\ln \relax (x)\right )-\ln \left (x^2-\frac {2}{9}\right )\right )}{\ln \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(180*x^2*log(x) - 90*x^2 + 20)/(log(2)*log(x)*(2*x - 9*x^3)),x)

[Out]

-(10*(log(log(x)) - log(x^2 - 2/9)))/log(2)

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sympy [A]  time = 0.20, size = 22, normalized size = 0.96 \begin {gather*} \frac {10 \log {\left (9 x^{2} - 2 \right )}}{\log {\relax (2 )}} - \frac {10 \log {\left (\log {\relax (x )} \right )}}{\log {\relax (2 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(360*x**2*ln(x)-180*x**2+40)/(9*x**3-2*x)/ln(2)/ln(x),x)

[Out]

10*log(9*x**2 - 2)/log(2) - 10*log(log(x))/log(2)

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