3.88.33 \(\int \frac {-24+9 x-x^2+(-36+33 x-10 x^2+x^3) \log (4)+(-12+3 x) \log (-4+x)}{(-36+33 x-10 x^2+x^3) \log (4)} \, dx\)

Optimal. Leaf size=20 \[ x-\frac {x (2+\log (-4+x))}{(-3+x) \log (4)} \]

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Rubi [B]  time = 0.17, antiderivative size = 44, normalized size of antiderivative = 2.20, number of steps used = 13, number of rules used = 8, integrand size = 56, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {12, 6742, 44, 77, 88, 2395, 36, 31} \begin {gather*} x-\frac {\log (4-x)}{\log (4)}+\frac {3 \log (x-4)}{(3-x) \log (4)}+\frac {6}{(3-x) \log (4)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-24 + 9*x - x^2 + (-36 + 33*x - 10*x^2 + x^3)*Log[4] + (-12 + 3*x)*Log[-4 + x])/((-36 + 33*x - 10*x^2 + x
^3)*Log[4]),x]

[Out]

x + 6/((3 - x)*Log[4]) - Log[4 - x]/Log[4] + (3*Log[-4 + x])/((3 - 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 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 2395

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))*((f_.) + (g_.)*(x_))^(q_.), x_Symbol] :> Simp[((f + g
*x)^(q + 1)*(a + b*Log[c*(d + e*x)^n]))/(g*(q + 1)), x] - Dist[(b*e*n)/(g*(q + 1)), Int[(f + g*x)^(q + 1)/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, q}, x] && NeQ[e*f - d*g, 0] && NeQ[q, -1]

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {-24+9 x-x^2+\left (-36+33 x-10 x^2+x^3\right ) \log (4)+(-12+3 x) \log (-4+x)}{-36+33 x-10 x^2+x^3} \, dx}{\log (4)}\\ &=\frac {\int \left (-\frac {24}{(-4+x) (-3+x)^2}+\frac {9 x}{(-4+x) (-3+x)^2}-\frac {x^2}{(-4+x) (-3+x)^2}+\log (4)+\frac {3 \log (-4+x)}{(-3+x)^2}\right ) \, dx}{\log (4)}\\ &=x-\frac {\int \frac {x^2}{(-4+x) (-3+x)^2} \, dx}{\log (4)}+\frac {3 \int \frac {\log (-4+x)}{(-3+x)^2} \, dx}{\log (4)}+\frac {9 \int \frac {x}{(-4+x) (-3+x)^2} \, dx}{\log (4)}-\frac {24 \int \frac {1}{(-4+x) (-3+x)^2} \, dx}{\log (4)}\\ &=x+\frac {3 \log (-4+x)}{(3-x) \log (4)}-\frac {\int \left (\frac {16}{-4+x}-\frac {9}{(-3+x)^2}-\frac {15}{-3+x}\right ) \, dx}{\log (4)}+\frac {3 \int \frac {1}{(-4+x) (-3+x)} \, dx}{\log (4)}+\frac {9 \int \left (\frac {4}{-4+x}-\frac {3}{(-3+x)^2}-\frac {4}{-3+x}\right ) \, dx}{\log (4)}-\frac {24 \int \left (\frac {1}{3-x}+\frac {1}{-4+x}-\frac {1}{(-3+x)^2}\right ) \, dx}{\log (4)}\\ &=x+\frac {6}{(3-x) \log (4)}+\frac {3 \log (3-x)}{\log (4)}-\frac {4 \log (4-x)}{\log (4)}+\frac {3 \log (-4+x)}{(3-x) \log (4)}+\frac {3 \int \frac {1}{-4+x} \, dx}{\log (4)}-\frac {3 \int \frac {1}{-3+x} \, dx}{\log (4)}\\ &=x+\frac {6}{(3-x) \log (4)}-\frac {\log (4-x)}{\log (4)}+\frac {3 \log (-4+x)}{(3-x) \log (4)}\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.09, size = 49, normalized size = 2.45 \begin {gather*} \frac {6 \tanh ^{-1}(7-2 x)+x \log (4)+3 \log (3-x)-4 \log (4-x)+\frac {3 (2+\log (-4+x))}{3-x}}{\log (4)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-24 + 9*x - x^2 + (-36 + 33*x - 10*x^2 + x^3)*Log[4] + (-12 + 3*x)*Log[-4 + x])/((-36 + 33*x - 10*x
^2 + x^3)*Log[4]),x]

[Out]

(6*ArcTanh[7 - 2*x] + x*Log[4] + 3*Log[3 - x] - 4*Log[4 - x] + (3*(2 + Log[-4 + x]))/(3 - x))/Log[4]

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fricas [A]  time = 0.47, size = 31, normalized size = 1.55 \begin {gather*} \frac {2 \, {\left (x^{2} - 3 \, x\right )} \log \relax (2) - x \log \left (x - 4\right ) - 6}{2 \, {\left (x - 3\right )} \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((3*x-12)*log(x-4)+2*(x^3-10*x^2+33*x-36)*log(2)-x^2+9*x-24)/(x^3-10*x^2+33*x-36)/log(2),x, algo
rithm="fricas")

[Out]

1/2*(2*(x^2 - 3*x)*log(2) - x*log(x - 4) - 6)/((x - 3)*log(2))

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giac [A]  time = 0.14, size = 36, normalized size = 1.80 \begin {gather*} \frac {2 \, x \log \relax (2) - \frac {3 \, \log \left (x - 4\right )}{x - 3} - \frac {6}{x - 3} - \log \left (x - 4\right )}{2 \, \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((3*x-12)*log(x-4)+2*(x^3-10*x^2+33*x-36)*log(2)-x^2+9*x-24)/(x^3-10*x^2+33*x-36)/log(2),x, algo
rithm="giac")

[Out]

1/2*(2*x*log(2) - 3*log(x - 4)/(x - 3) - 6/(x - 3) - log(x - 4))/log(2)

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maple [A]  time = 0.10, size = 34, normalized size = 1.70

method	result	size
norman	\(\frac {x^{2}-\frac {3 \left (3 \ln \relax (2)+1\right )}{\ln \relax (2)}-\frac {x \ln \left (x -4\right )}{2 \ln \relax (2)}}{x -3}\)	\(34\)
derivativedivides	\(\frac {\frac {3 \ln \left (x -4\right ) \left (x -4\right )}{x -3}-\frac {6}{x -3}-4 \ln \left (x -4\right )+2 \left (x -4\right ) \ln \relax (2)}{2 \ln \relax (2)}\)	\(42\)
default	\(\frac {\frac {3 \ln \left (x -4\right ) \left (x -4\right )}{x -3}-\frac {6}{x -3}-4 \ln \left (x -4\right )+2 \left (x -4\right ) \ln \relax (2)}{2 \ln \relax (2)}\)	\(42\)
risch	\(-\frac {3 \ln \left (x -4\right )}{2 \ln \relax (2) \left (x -3\right )}+\frac {2 x^{2} \ln \relax (2)-6 x \ln \relax (2)-x \ln \left (x -4\right )+3 \ln \left (x -4\right )-6}{2 \ln \relax (2) \left (x -3\right )}\)	\(55\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*((3*x-12)*ln(x-4)+2*(x^3-10*x^2+33*x-36)*ln(2)-x^2+9*x-24)/(x^3-10*x^2+33*x-36)/ln(2),x,method=_RETURN
VERBOSE)

[Out]

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

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maxima [B]  time = 0.48, size = 102, normalized size = 5.10 \begin {gather*} -\frac {72 \, {\left (\frac {1}{x - 3} - \log \left (x - 3\right ) + \log \left (x - 4\right )\right )} \log \relax (2) + 24 \, {\left (3 \, \log \relax (2) + 1\right )} \log \left (x - 3\right ) - \frac {2 \, x^{2} \log \relax (2) - 6 \, x \log \relax (2) + {\left (x {\left (72 \, \log \relax (2) + 23\right )} - 216 \, \log \relax (2) - 72\right )} \log \left (x - 4\right ) + 72 \, \log \relax (2) + 18}{x - 3} + \frac {24}{x - 3} - 24 \, \log \left (x - 3\right ) + 24 \, \log \left (x - 4\right )}{2 \, \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((3*x-12)*log(x-4)+2*(x^3-10*x^2+33*x-36)*log(2)-x^2+9*x-24)/(x^3-10*x^2+33*x-36)/log(2),x, algo
rithm="maxima")

[Out]

-1/2*(72*(1/(x - 3) - log(x - 3) + log(x - 4))*log(2) + 24*(3*log(2) + 1)*log(x - 3) - (2*x^2*log(2) - 6*x*log
(2) + (x*(72*log(2) + 23) - 216*log(2) - 72)*log(x - 4) + 72*log(2) + 18)/(x - 3) + 24/(x - 3) - 24*log(x - 3)
+ 24*log(x - 4))/log(2)

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mupad [B]  time = 0.37, size = 36, normalized size = 1.80 \begin {gather*} -\frac {\ln \left (\frac {x-4}{2^{2\,x}}\right )}{2\,\ln \relax (2)}-\frac {3\,\left (\ln \left (x-4\right )+2\right )}{2\,\ln \relax (2)\,\left (x-3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((9*x)/2 + log(2)*(33*x - 10*x^2 + x^3 - 36) - x^2/2 + (log(x - 4)*(3*x - 12))/2 - 12)/(log(2)*(33*x - 10*
x^2 + x^3 - 36)),x)

[Out]

- log((x - 4)/2^(2*x))/(2*log(2)) - (3*(log(x - 4) + 2))/(2*log(2)*(x - 3))

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sympy [B]  time = 0.34, size = 41, normalized size = 2.05 \begin {gather*} x - \frac {\log {\left (x - 4 \right )}}{2 \log {\relax (2 )}} - \frac {3 \log {\left (x - 4 \right )}}{2 x \log {\relax (2 )} - 6 \log {\relax (2 )}} - \frac {3}{x \log {\relax (2 )} - 3 \log {\relax (2 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((3*x-12)*ln(x-4)+2*(x**3-10*x**2+33*x-36)*ln(2)-x**2+9*x-24)/(x**3-10*x**2+33*x-36)/ln(2),x)

[Out]

x - log(x - 4)/(2*log(2)) - 3*log(x - 4)/(2*x*log(2) - 6*log(2)) - 3/(x*log(2) - 3*log(2))

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