\(\int \frac {3-48 x+32 x^2}{3 x-19 x^2+16 x^3} \, dx\) [4932]

Optimal result

Integrand size = 27, antiderivative size = 16 \[ \int \frac {3-48 x+32 x^2}{3 x-19 x^2+16 x^3} \, dx=\log \left (\frac {25 (3-16 x)^2 x}{-1+x}\right ) \]

[Out]

ln(5*x*(3-16*x)^2/(1/5*x-1/5))

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.19, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {1608, 1642} \[ \int \frac {3-48 x+32 x^2}{3 x-19 x^2+16 x^3} \, dx=2 \log (3-16 x)-\log (1-x)+\log (x) \]

[In]

Int[(3 - 48*x + 32*x^2)/(3*x - 19*x^2 + 16*x^3),x]

[Out]

2*Log[3 - 16*x] - Log[1 - x] + Log[x]

Rule 1608

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 1642

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegra
nd[(d + e*x)^m*Pq*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rubi steps \begin{align*} \text {integral}& = \int \frac {3-48 x+32 x^2}{x \left (3-19 x+16 x^2\right )} \, dx \\ & = \int \left (\frac {1}{1-x}+\frac {1}{x}+\frac {32}{-3+16 x}\right ) \, dx \\ & = 2 \log (3-16 x)-\log (1-x)+\log (x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.19 \[ \int \frac {3-48 x+32 x^2}{3 x-19 x^2+16 x^3} \, dx=2 \log (3-16 x)-\log (1-x)+\log (x) \]

[In]

Integrate[(3 - 48*x + 32*x^2)/(3*x - 19*x^2 + 16*x^3),x]

[Out]

2*Log[3 - 16*x] - Log[1 - x] + Log[x]

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00

method	result	size
parallelrisch	\(\ln \left (x \right )-\ln \left (-1+x \right )+2 \ln \left (x -\frac {3}{16}\right )\)	\(16\)
default	\(2 \ln \left (16 x -3\right )+\ln \left (x \right )-\ln \left (-1+x \right )\)	\(18\)
norman	\(2 \ln \left (16 x -3\right )+\ln \left (x \right )-\ln \left (-1+x \right )\)	\(18\)
risch	\(2 \ln \left (16 x -3\right )+\ln \left (x \right )-\ln \left (-1+x \right )\)	\(18\)

[In]

int((32*x^2-48*x+3)/(16*x^3-19*x^2+3*x),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.06 \[ \int \frac {3-48 x+32 x^2}{3 x-19 x^2+16 x^3} \, dx=2 \, \log \left (16 \, x - 3\right ) - \log \left (x - 1\right ) + \log \left (x\right ) \]

[In]

integrate((32*x^2-48*x+3)/(16*x^3-19*x^2+3*x),x, algorithm="fricas")

[Out]

2*log(16*x - 3) - log(x - 1) + log(x)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94 \[ \int \frac {3-48 x+32 x^2}{3 x-19 x^2+16 x^3} \, dx=\log {\left (x \right )} - \log {\left (x - 1 \right )} + 2 \log {\left (x - \frac {3}{16} \right )} \]

[In]

integrate((32*x**2-48*x+3)/(16*x**3-19*x**2+3*x),x)

[Out]

log(x) - log(x - 1) + 2*log(x - 3/16)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.06 \[ \int \frac {3-48 x+32 x^2}{3 x-19 x^2+16 x^3} \, dx=2 \, \log \left (16 \, x - 3\right ) - \log \left (x - 1\right ) + \log \left (x\right ) \]

[In]

integrate((32*x^2-48*x+3)/(16*x^3-19*x^2+3*x),x, algorithm="maxima")

[Out]

2*log(16*x - 3) - log(x - 1) + log(x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.25 \[ \int \frac {3-48 x+32 x^2}{3 x-19 x^2+16 x^3} \, dx=2 \, \log \left ({\left | 16 \, x - 3 \right |}\right ) - \log \left ({\left | x - 1 \right |}\right ) + \log \left ({\left | x \right |}\right ) \]

[In]

integrate((32*x^2-48*x+3)/(16*x^3-19*x^2+3*x),x, algorithm="giac")

[Out]

2*log(abs(16*x - 3)) - log(abs(x - 1)) + log(abs(x))

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.31 \[ \int \frac {3-48 x+32 x^2}{3 x-19 x^2+16 x^3} \, dx=2\,\ln \left (x-\frac {3}{16}\right )-2\,\mathrm {atanh}\left (\frac {4563}{9088\,\left (\frac {71\,x}{64}+\frac {27}{128}\right )}-\frac {98}{71}\right ) \]

[In]

int((32*x^2 - 48*x + 3)/(3*x - 19*x^2 + 16*x^3),x)

[Out]

2*log(x - 3/16) - 2*atanh(4563/(9088*((71*x)/64 + 27/128)) - 98/71)