3.61.49 \(\int \frac {-2+\log (3)}{(-2 x-x^2+x \log (3)) \log (-\frac {3 x}{-10-5 x+5 \log (3)+(-2-x+\log (3)) \log (5)})} \, dx\)

Optimal. Leaf size=20 \[ \log \left (\log \left (\frac {3 x}{(2+x-\log (3)) (5+\log (5))}\right )\right ) \]

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Rubi [A]  time = 0.16, antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 47, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.106, Rules used = {6, 12, 1593, 2516, 2504} \begin {gather*} \log \left (\log \left (\frac {3 x}{(5+\log (5)) (x+2-\log (3))}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-2 + Log[3])/((-2*x - x^2 + x*Log[3])*Log[(-3*x)/(-10 - 5*x + 5*Log[3] + (-2 - x + Log[3])*Log[5])]),x]

[Out]

Log[Log[(3*x)/((2 + x - Log[3])*(5 + Log[5]))]]

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, 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 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 2504

Int[(u_)/Log[(e_.)*((f_.)*((a_.) + (b_.)*(x_))^(p_.)*((c_.) + (d_.)*(x_))^(q_.))^(r_.)], x_Symbol] :> With[{h
= Simplify[u*(a + b*x)*(c + d*x)]}, Simp[(h*Log[Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]])/(p*r*(b*c - a*d)), x] /
; FreeQ[h, x]] /; FreeQ[{a, b, c, d, e, f, p, q, r}, x] && NeQ[b*c - a*d, 0] && EqQ[p + q, 0]

Rule 2516

Int[Log[(e_.)*((f_.)*(v_)^(p_.)*(w_)^(q_.))^(r_.)]^(s_.)*(u_.), x_Symbol] :> Int[u*Log[e*(f*ExpandToSum[v, x]^
p*ExpandToSum[w, x]^q)^r]^s, x] /; FreeQ[{e, f, p, q, r, s}, x] && LinearQ[{v, w}, x] &&  !LinearMatchQ[{v, w}
, x] && AlgebraicFunctionQ[u, x]

Rubi steps

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

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Mathematica [A]  time = 0.07, size = 34, normalized size = 1.70 \begin {gather*} -\frac {(2-\log (3)) \log \left (\log \left (\frac {3 x}{(2+x-\log (3)) (5+\log (5))}\right )\right )}{-2+\log (3)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2 + Log[3])/((-2*x - x^2 + x*Log[3])*Log[(-3*x)/(-10 - 5*x + 5*Log[3] + (-2 - x + Log[3])*Log[5])]
),x]

[Out]

-(((2 - Log[3])*Log[Log[(3*x)/((2 + x - Log[3])*(5 + Log[5]))]])/(-2 + Log[3]))

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fricas [A]  time = 0.70, size = 26, normalized size = 1.30 \begin {gather*} \log \left (\log \left (\frac {3 \, x}{{\left (x - \log \relax (3) + 2\right )} \log \relax (5) + 5 \, x - 5 \, \log \relax (3) + 10}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log(3)-2)/(x*log(3)-x^2-2*x)/log(-3*x/((log(3)-x-2)*log(5)+5*log(3)-5*x-10)),x, algorithm="fricas")

[Out]

log(log(3*x/((x - log(3) + 2)*log(5) + 5*x - 5*log(3) + 10)))

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giac [A]  time = 0.14, size = 32, normalized size = 1.60 \begin {gather*} \log \left (\log \relax (3) - \log \left (x \log \relax (5) - \log \relax (5) \log \relax (3) + 5 \, x + 2 \, \log \relax (5) - 5 \, \log \relax (3) + 10\right ) + \log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log(3)-2)/(x*log(3)-x^2-2*x)/log(-3*x/((log(3)-x-2)*log(5)+5*log(3)-5*x-10)),x, algorithm="giac")

[Out]

log(log(3) - log(x*log(5) - log(5)*log(3) + 5*x + 2*log(5) - 5*log(3) + 10) + log(x))

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maple [A]  time = 0.29, size = 27, normalized size = 1.35

method	result	size
norman	\(\ln \left (\ln \left (-\frac {3 x}{\left (\ln \relax (3)-x -2\right ) \ln \relax (5)+5 \ln \relax (3)-5 x -10}\right )\right )\)	\(27\)
risch	\(\ln \left (\ln \left (-\frac {3 x}{\left (\ln \relax (3)-x -2\right ) \ln \relax (5)+5 \ln \relax (3)-5 x -10}\right )\right )\)	\(27\)
default	\(-\frac {3 \left (\ln \relax (3)-2\right )^{2} \left (-\frac {5 \ln \left (\ln \left (\frac {3}{5+\ln \relax (5)}+\frac {3 \ln \relax (3)-6}{\left (2+x -\ln \relax (3)\right ) \left (5+\ln \relax (5)\right )}\right )\right )}{3 \left (\ln \relax (3)^{2}-4 \ln \relax (3)+4\right )}-\frac {\ln \relax (5) \ln \left (\ln \left (\frac {3}{5+\ln \relax (5)}+\frac {3 \ln \relax (3)-6}{\left (2+x -\ln \relax (3)\right ) \left (5+\ln \relax (5)\right )}\right )\right )}{3 \left (\ln \relax (3)^{2}-4 \ln \relax (3)+4\right )}\right )}{5+\ln \relax (5)}\)	\(110\)
derivativedivides	\(\frac {3 \left (2-\ln \relax (3)\right ) \left (\ln \relax (3)-2\right ) \left (-\frac {5 \ln \left (\ln \left (\frac {3}{5+\ln \relax (5)}+\frac {3 \ln \relax (3)-6}{\left (2+x -\ln \relax (3)\right ) \left (5+\ln \relax (5)\right )}\right )\right )}{3 \left (\ln \relax (3)^{2}-4 \ln \relax (3)+4\right )}-\frac {\ln \relax (5) \ln \left (\ln \left (\frac {3}{5+\ln \relax (5)}+\frac {3 \ln \relax (3)-6}{\left (2+x -\ln \relax (3)\right ) \left (5+\ln \relax (5)\right )}\right )\right )}{3 \left (\ln \relax (3)^{2}-4 \ln \relax (3)+4\right )}\right )}{5+\ln \relax (5)}\)	\(114\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((ln(3)-2)/(x*ln(3)-x^2-2*x)/ln(-3*x/((ln(3)-x-2)*ln(5)+5*ln(3)-5*x-10)),x,method=_RETURNVERBOSE)

[Out]

ln(ln(-3*x/((ln(3)-x-2)*ln(5)+5*ln(3)-5*x-10)))

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maxima [A]  time = 0.44, size = 23, normalized size = 1.15 \begin {gather*} \log \left (-\log \relax (3) + \log \left (x - \log \relax (3) + 2\right ) - \log \relax (x) + \log \left (\log \relax (5) + 5\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log(3)-2)/(x*log(3)-x^2-2*x)/log(-3*x/((log(3)-x-2)*log(5)+5*log(3)-5*x-10)),x, algorithm="maxima")

[Out]

log(-log(3) + log(x - log(3) + 2) - log(x) + log(log(5) + 5))

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mupad [B]  time = 5.26, size = 26, normalized size = 1.30 \begin {gather*} \ln \left (\ln \left (\frac {3\,x}{5\,x-5\,\ln \relax (3)+\ln \relax (5)\,\left (x-\ln \relax (3)+2\right )+10}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(3) - 2)/(log((3*x)/(5*x - 5*log(3) + log(5)*(x - log(3) + 2) + 10))*(2*x - x*log(3) + x^2)),x)

[Out]

log(log((3*x)/(5*x - 5*log(3) + log(5)*(x - log(3) + 2) + 10)))

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sympy [A]  time = 0.26, size = 27, normalized size = 1.35 \begin {gather*} \log {\left (\log {\left (- \frac {3 x}{- 5 x + \left (- x - 2 + \log {\relax (3 )}\right ) \log {\relax (5 )} - 10 + 5 \log {\relax (3 )}} \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((ln(3)-2)/(x*ln(3)-x**2-2*x)/ln(-3*x/((ln(3)-x-2)*ln(5)+5*ln(3)-5*x-10)),x)

[Out]

log(log(-3*x/(-5*x + (-x - 2 + log(3))*log(5) - 10 + 5*log(3))))

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