3.103.57 \(\int \frac {816+5 x-18 x^2+x^3+(-96-7 x+x^2) \log (x)}{384 x-72 x^2+3 x^3+(-48 x+3 x^2) \log (x)} \, dx\)

Optimal. Leaf size=25 \[ \log \left (\frac {4 e^{x/3} (16-x) x^2}{-8+x+\log (x)}\right ) \]

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Rubi [A]  time = 0.46, antiderivative size = 28, normalized size of antiderivative = 1.12, number of steps used = 7, number of rules used = 5, integrand size = 53, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.094, Rules used = {6741, 12, 6742, 893, 6684} \begin {gather*} \frac {x}{3}+\log (16-x)+2 \log (x)-\log (-x-\log (x)+8) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(816 + 5*x - 18*x^2 + x^3 + (-96 - 7*x + x^2)*Log[x])/(384*x - 72*x^2 + 3*x^3 + (-48*x + 3*x^2)*Log[x]),x]

[Out]

x/3 + Log[16 - x] + 2*Log[x] - Log[8 - x - 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 893

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] &
& NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && I
ntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rule 6684

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rule 6741

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

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 {816+5 x-18 x^2+x^3+\left (-96-7 x+x^2\right ) \log (x)}{3 (16-x) x (8-x-\log (x))} \, dx\\ &=\frac {1}{3} \int \frac {816+5 x-18 x^2+x^3+\left (-96-7 x+x^2\right ) \log (x)}{(16-x) x (8-x-\log (x))} \, dx\\ &=\frac {1}{3} \int \left (\frac {-96-7 x+x^2}{(-16+x) x}-\frac {3 (1+x)}{x (-8+x+\log (x))}\right ) \, dx\\ &=\frac {1}{3} \int \frac {-96-7 x+x^2}{(-16+x) x} \, dx-\int \frac {1+x}{x (-8+x+\log (x))} \, dx\\ &=-\log (8-x-\log (x))+\frac {1}{3} \int \left (1+\frac {3}{-16+x}+\frac {6}{x}\right ) \, dx\\ &=\frac {x}{3}+\log (16-x)+2 \log (x)-\log (8-x-\log (x))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.13, size = 30, normalized size = 1.20 \begin {gather*} \frac {1}{3} (x+3 \log (16-x)+6 \log (x)-3 \log (8-x-\log (x))) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(816 + 5*x - 18*x^2 + x^3 + (-96 - 7*x + x^2)*Log[x])/(384*x - 72*x^2 + 3*x^3 + (-48*x + 3*x^2)*Log[
x]),x]

[Out]

(x + 3*Log[16 - x] + 6*Log[x] - 3*Log[8 - x - Log[x]])/3

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fricas [A]  time = 0.81, size = 20, normalized size = 0.80 \begin {gather*} \frac {1}{3} \, x - \log \left (x + \log \relax (x) - 8\right ) + \log \left (x - 16\right ) + 2 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2-7*x-96)*log(x)+x^3-18*x^2+5*x+816)/((3*x^2-48*x)*log(x)+3*x^3-72*x^2+384*x),x, algorithm="fric
as")

[Out]

1/3*x - log(x + log(x) - 8) + log(x - 16) + 2*log(x)

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giac [A]  time = 0.15, size = 24, normalized size = 0.96 \begin {gather*} \frac {1}{3} \, x + \log \left (x - 16\right ) + 2 \, \log \relax (x) - \log \left (-x - \log \relax (x) + 8\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2-7*x-96)*log(x)+x^3-18*x^2+5*x+816)/((3*x^2-48*x)*log(x)+3*x^3-72*x^2+384*x),x, algorithm="giac
")

[Out]

1/3*x + log(x - 16) + 2*log(x) - log(-x - log(x) + 8)

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maple [A]  time = 0.04, size = 21, normalized size = 0.84

method	result	size
norman	\(\frac {x}{3}+2 \ln \relax (x )-\ln \left (\ln \relax (x )-8+x \right )+\ln \left (x -16\right )\)	\(21\)
risch	\(\frac {x}{3}+2 \ln \relax (x )-\ln \left (\ln \relax (x )-8+x \right )+\ln \left (x -16\right )\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^2-7*x-96)*ln(x)+x^3-18*x^2+5*x+816)/((3*x^2-48*x)*ln(x)+3*x^3-72*x^2+384*x),x,method=_RETURNVERBOSE)

[Out]

1/3*x+2*ln(x)-ln(ln(x)-8+x)+ln(x-16)

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maxima [A]  time = 0.41, size = 20, normalized size = 0.80 \begin {gather*} \frac {1}{3} \, x - \log \left (x + \log \relax (x) - 8\right ) + \log \left (x - 16\right ) + 2 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2-7*x-96)*log(x)+x^3-18*x^2+5*x+816)/((3*x^2-48*x)*log(x)+3*x^3-72*x^2+384*x),x, algorithm="maxi
ma")

[Out]

1/3*x - log(x + log(x) - 8) + log(x - 16) + 2*log(x)

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mupad [B]  time = 6.69, size = 20, normalized size = 0.80 \begin {gather*} \frac {x}{3}-\ln \left (x+\ln \relax (x)-8\right )+\ln \left (x-16\right )+2\,\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*x - log(x)*(7*x - x^2 + 96) - 18*x^2 + x^3 + 816)/(384*x - log(x)*(48*x - 3*x^2) - 72*x^2 + 3*x^3),x)

[Out]

x/3 - log(x + log(x) - 8) + log(x - 16) + 2*log(x)

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sympy [A]  time = 0.26, size = 20, normalized size = 0.80 \begin {gather*} \frac {x}{3} + 2 \log {\relax (x )} + \log {\left (x - 16 \right )} - \log {\left (x + \log {\relax (x )} - 8 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x**2-7*x-96)*ln(x)+x**3-18*x**2+5*x+816)/((3*x**2-48*x)*ln(x)+3*x**3-72*x**2+384*x),x)

[Out]

x/3 + 2*log(x) + log(x - 16) - log(x + log(x) - 8)

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