3.48.92 \(\int \frac {3+71 x-48 x^2+8 x^3+x \log (x)}{72 x-48 x^2+8 x^3} \, dx\)

Optimal. Leaf size=16 \[ -21+x+\frac {\log (x)}{8 (3-x)} \]

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Rubi [A]  time = 0.21, antiderivative size = 22, normalized size of antiderivative = 1.38, number of steps used = 13, number of rules used = 8, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {1594, 27, 12, 6742, 44, 43, 2314, 31} \begin {gather*} x+\frac {x \log (x)}{24 (3-x)}+\frac {\log (x)}{24} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(3 + 71*x - 48*x^2 + 8*x^3 + x*Log[x])/(72*x - 48*x^2 + 8*x^3),x]

[Out]

x + Log[x]/24 + (x*Log[x])/(24*(3 - 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 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 31

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

Rule 43

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 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 1594

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

Rule 2314

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

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 {3+71 x-48 x^2+8 x^3+x \log (x)}{x \left (72-48 x+8 x^2\right )} \, dx\\ &=\int \frac {3+71 x-48 x^2+8 x^3+x \log (x)}{8 (-3+x)^2 x} \, dx\\ &=\frac {1}{8} \int \frac {3+71 x-48 x^2+8 x^3+x \log (x)}{(-3+x)^2 x} \, dx\\ &=\frac {1}{8} \int \left (\frac {71}{(-3+x)^2}+\frac {3}{(-3+x)^2 x}-\frac {48 x}{(-3+x)^2}+\frac {8 x^2}{(-3+x)^2}+\frac {\log (x)}{(-3+x)^2}\right ) \, dx\\ &=\frac {71}{8 (3-x)}+\frac {1}{8} \int \frac {\log (x)}{(-3+x)^2} \, dx+\frac {3}{8} \int \frac {1}{(-3+x)^2 x} \, dx-6 \int \frac {x}{(-3+x)^2} \, dx+\int \frac {x^2}{(-3+x)^2} \, dx\\ &=\frac {71}{8 (3-x)}+\frac {x \log (x)}{24 (3-x)}+\frac {1}{24} \int \frac {1}{-3+x} \, dx+\frac {3}{8} \int \left (\frac {1}{3 (-3+x)^2}-\frac {1}{9 (-3+x)}+\frac {1}{9 x}\right ) \, dx-6 \int \left (\frac {3}{(-3+x)^2}+\frac {1}{-3+x}\right ) \, dx+\int \left (1+\frac {9}{(-3+x)^2}+\frac {6}{-3+x}\right ) \, dx\\ &=x+\frac {\log (x)}{24}+\frac {x \log (x)}{24 (3-x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 17, normalized size = 1.06 \begin {gather*} \frac {1}{8} \left (8 x-\frac {\log (x)}{-3+x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(3 + 71*x - 48*x^2 + 8*x^3 + x*Log[x])/(72*x - 48*x^2 + 8*x^3),x]

[Out]

(8*x - Log[x]/(-3 + x))/8

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fricas [A]  time = 0.65, size = 20, normalized size = 1.25 \begin {gather*} \frac {8 \, x^{2} - 24 \, x - \log \relax (x)}{8 \, {\left (x - 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x*log(x)+8*x^3-48*x^2+71*x+3)/(8*x^3-48*x^2+72*x),x, algorithm="fricas")

[Out]

1/8*(8*x^2 - 24*x - log(x))/(x - 3)

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giac [A]  time = 0.15, size = 11, normalized size = 0.69 \begin {gather*} x - \frac {\log \relax (x)}{8 \, {\left (x - 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x*log(x)+8*x^3-48*x^2+71*x+3)/(8*x^3-48*x^2+72*x),x, algorithm="giac")

[Out]

x - 1/8*log(x)/(x - 3)

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maple [A]  time = 0.05, size = 12, normalized size = 0.75

method	result	size
risch	\(-\frac {\ln \relax (x )}{8 \left (x -3\right )}+x\)	\(12\)
norman	\(\frac {x^{2}-\frac {\ln \relax (x )}{8}-9}{x -3}\)	\(16\)
default	\(-\frac {\ln \relax (x ) x}{24 \left (x -3\right )}+x +\frac {\ln \relax (x )}{24}\)	\(17\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*ln(x)+8*x^3-48*x^2+71*x+3)/(8*x^3-48*x^2+72*x),x,method=_RETURNVERBOSE)

[Out]

-1/8/(x-3)*ln(x)+x

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maxima [A]  time = 0.35, size = 11, normalized size = 0.69 \begin {gather*} x - \frac {\log \relax (x)}{8 \, {\left (x - 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x*log(x)+8*x^3-48*x^2+71*x+3)/(8*x^3-48*x^2+72*x),x, algorithm="maxima")

[Out]

x - 1/8*log(x)/(x - 3)

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mupad [B]  time = 3.31, size = 13, normalized size = 0.81 \begin {gather*} x-\frac {\ln \relax (x)}{8\,\left (x-3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((71*x + x*log(x) - 48*x^2 + 8*x^3 + 3)/(72*x - 48*x^2 + 8*x^3),x)

[Out]

x - log(x)/(8*(x - 3))

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sympy [A]  time = 0.12, size = 8, normalized size = 0.50 \begin {gather*} x - \frac {\log {\relax (x )}}{8 x - 24} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x*ln(x)+8*x**3-48*x**2+71*x+3)/(8*x**3-48*x**2+72*x),x)

[Out]

x - log(x)/(8*x - 24)

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