\(\int \frac {-17+12 x+12 x^2}{3+12 x+12 x^2} \, dx\) [5165]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 23, antiderivative size = 19 \[ \int \frac {-17+12 x+12 x^2}{3+12 x+12 x^2} \, dx=-e^5+x-\frac {20 x}{3 (1+2 x)} \]

[Out]

x-20/3*x/(1+2*x)-exp(5)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.68, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {27, 12, 697} \[ \int \frac {-17+12 x+12 x^2}{3+12 x+12 x^2} \, dx=x+\frac {10}{3 (2 x+1)} \]

[In]

Int[(-17 + 12*x + 12*x^2)/(3 + 12*x + 12*x^2),x]

[Out]

x + 10/(3*(1 + 2*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 697

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d +
 e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[2*c*d - b*e,
 0] && IGtQ[p, 0] &&  !(EqQ[m, 3] && NeQ[p, 1])

Rubi steps \begin{align*} \text {integral}& = \int \frac {-17+12 x+12 x^2}{3 (1+2 x)^2} \, dx \\ & = \frac {1}{3} \int \frac {-17+12 x+12 x^2}{(1+2 x)^2} \, dx \\ & = \frac {1}{3} \int \left (3-\frac {20}{(1+2 x)^2}\right ) \, dx \\ & = x+\frac {10}{3 (1+2 x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05 \[ \int \frac {-17+12 x+12 x^2}{3+12 x+12 x^2} \, dx=\frac {1}{3} \left (\frac {3}{2}+3 x+\frac {10}{1+2 x}\right ) \]

[In]

Integrate[(-17 + 12*x + 12*x^2)/(3 + 12*x + 12*x^2),x]

[Out]

(3/2 + 3*x + 10/(1 + 2*x))/3

Maple [A] (verified)

Time = 0.47 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.53

method result size
risch \(x +\frac {5}{3 \left (\frac {1}{2}+x \right )}\) \(10\)
default \(x +\frac {10}{3 \left (1+2 x \right )}\) \(12\)
gosper \(\frac {x \left (6 x -17\right )}{6 x +3}\) \(16\)
norman \(\frac {-\frac {17}{3} x +2 x^{2}}{1+2 x}\) \(18\)
parallelrisch \(\frac {6 x^{2}-17 x}{6 x +3}\) \(19\)
meijerg \(-\frac {23 x}{3 \left (1+2 x \right )}+\frac {x \left (6+6 x \right )}{6 x +3}\) \(27\)

[In]

int((12*x^2+12*x-17)/(12*x^2+12*x+3),x,method=_RETURNVERBOSE)

[Out]

x+5/3/(1/2+x)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {-17+12 x+12 x^2}{3+12 x+12 x^2} \, dx=\frac {6 \, x^{2} + 3 \, x + 10}{3 \, {\left (2 \, x + 1\right )}} \]

[In]

integrate((12*x^2+12*x-17)/(12*x^2+12*x+3),x, algorithm="fricas")

[Out]

1/3*(6*x^2 + 3*x + 10)/(2*x + 1)

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.37 \[ \int \frac {-17+12 x+12 x^2}{3+12 x+12 x^2} \, dx=x + \frac {10}{6 x + 3} \]

[In]

integrate((12*x**2+12*x-17)/(12*x**2+12*x+3),x)

[Out]

x + 10/(6*x + 3)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.58 \[ \int \frac {-17+12 x+12 x^2}{3+12 x+12 x^2} \, dx=x + \frac {10}{3 \, {\left (2 \, x + 1\right )}} \]

[In]

integrate((12*x^2+12*x-17)/(12*x^2+12*x+3),x, algorithm="maxima")

[Out]

x + 10/3/(2*x + 1)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.58 \[ \int \frac {-17+12 x+12 x^2}{3+12 x+12 x^2} \, dx=x + \frac {10}{3 \, {\left (2 \, x + 1\right )}} \]

[In]

integrate((12*x^2+12*x-17)/(12*x^2+12*x+3),x, algorithm="giac")

[Out]

x + 10/3/(2*x + 1)

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.58 \[ \int \frac {-17+12 x+12 x^2}{3+12 x+12 x^2} \, dx=x+\frac {5}{3\,\left (x+\frac {1}{2}\right )} \]

[In]

int((12*x + 12*x^2 - 17)/(12*x + 12*x^2 + 3),x)

[Out]

x + 5/(3*(x + 1/2))