3.9.61 \(\int \frac {16-8 x+65 x^2-8 x^3-x \log (3)}{16 x-8 x^2+x^3} \, dx\)

Optimal. Leaf size=25 \[ \log \left (\frac {5}{4} e^{-\frac {-8 x^2+\log (3)}{4-x}} x\right ) \]

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Rubi [A]  time = 0.06, antiderivative size = 20, normalized size of antiderivative = 0.80, number of steps used = 5, number of rules used = 4, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {6, 1594, 27, 1620} \begin {gather*} -8 x+\log (x)+\frac {128-\log (3)}{4-x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(16 - 8*x + 65*x^2 - 8*x^3 - x*Log[3])/(16*x - 8*x^2 + x^3),x]

[Out]

-8*x + (128 - Log[3])/(4 - x) + Log[x]

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 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 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 1620

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)
^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2]) &&
GtQ[Expon[Px, x], 2]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {16+65 x^2-8 x^3+x (-8-\log (3))}{16 x-8 x^2+x^3} \, dx\\ &=\int \frac {16+65 x^2-8 x^3+x (-8-\log (3))}{x \left (16-8 x+x^2\right )} \, dx\\ &=\int \frac {16+65 x^2-8 x^3+x (-8-\log (3))}{(-4+x)^2 x} \, dx\\ &=\int \left (-8+\frac {1}{x}+\frac {128-\log (3)}{(-4+x)^2}\right ) \, dx\\ &=-8 x+\frac {128-\log (3)}{4-x}+\log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 16, normalized size = 0.64 \begin {gather*} -8 x+\frac {-128+\log (3)}{-4+x}+\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(16 - 8*x + 65*x^2 - 8*x^3 - x*Log[3])/(16*x - 8*x^2 + x^3),x]

[Out]

-8*x + (-128 + Log[3])/(-4 + x) + Log[x]

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fricas [A]  time = 0.57, size = 28, normalized size = 1.12 \begin {gather*} -\frac {8 \, x^{2} - {\left (x - 4\right )} \log \relax (x) - 32 \, x - \log \relax (3) + 128}{x - 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x*log(3)-8*x^3+65*x^2-8*x+16)/(x^3-8*x^2+16*x),x, algorithm="fricas")

[Out]

-(8*x^2 - (x - 4)*log(x) - 32*x - log(3) + 128)/(x - 4)

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giac [A]  time = 0.35, size = 17, normalized size = 0.68 \begin {gather*} -8 \, x + \frac {\log \relax (3) - 128}{x - 4} + \log \left ({\left | x \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x*log(3)-8*x^3+65*x^2-8*x+16)/(x^3-8*x^2+16*x),x, algorithm="giac")

[Out]

-8*x + (log(3) - 128)/(x - 4) + log(abs(x))

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maple [A]  time = 0.04, size = 18, normalized size = 0.72

method	result	size
norman	\(\frac {\ln \relax (3)-8 x^{2}}{x -4}+\ln \relax (x )\)	\(18\)
default	\(-8 x +\ln \relax (x )-\frac {-\ln \relax (3)+128}{x -4}\)	\(20\)
risch	\(-8 x +\frac {\ln \relax (3)}{x -4}-\frac {128}{x -4}+\ln \relax (x )\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-x*ln(3)-8*x^3+65*x^2-8*x+16)/(x^3-8*x^2+16*x),x,method=_RETURNVERBOSE)

[Out]

(ln(3)-8*x^2)/(x-4)+ln(x)

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maxima [A]  time = 0.39, size = 16, normalized size = 0.64 \begin {gather*} -8 \, x + \frac {\log \relax (3) - 128}{x - 4} + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x*log(3)-8*x^3+65*x^2-8*x+16)/(x^3-8*x^2+16*x),x, algorithm="maxima")

[Out]

-8*x + (log(3) - 128)/(x - 4) + log(x)

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mupad [B]  time = 0.06, size = 16, normalized size = 0.64 \begin {gather*} \ln \relax (x)-8\,x+\frac {\ln \relax (3)-128}{x-4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(8*x + x*log(3) - 65*x^2 + 8*x^3 - 16)/(16*x - 8*x^2 + x^3),x)

[Out]

log(x) - 8*x + (log(3) - 128)/(x - 4)

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sympy [A]  time = 0.18, size = 14, normalized size = 0.56 \begin {gather*} - 8 x + \log {\relax (x )} - \frac {128 - \log {\relax (3 )}}{x - 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x*ln(3)-8*x**3+65*x**2-8*x+16)/(x**3-8*x**2+16*x),x)

[Out]

-8*x + log(x) - (128 - log(3))/(x - 4)

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