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

Optimal. Leaf size=19 \[ -e^5+x-\frac {20 x}{3 (1+2 x)} \]

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Rubi [A]  time = 0.01, antiderivative size = 13, normalized size of antiderivative = 0.68, number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {27, 12, 683} \begin {gather*} x+\frac {10}{3 (2 x+1)} \end {gather*}

Antiderivative was successfully verified.

[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 683

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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 20, normalized size = 1.05 \begin {gather*} \frac {1}{3} \left (\frac {3}{2}+3 x+\frac {10}{1+2 x}\right ) \end {gather*}

Antiderivative was successfully verified.

[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

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fricas [A]  time = 0.75, size = 19, normalized size = 1.00 \begin {gather*} \frac {6 \, x^{2} + 3 \, x + 10}{3 \, {\left (2 \, x + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.12, size = 11, normalized size = 0.58 \begin {gather*} x + \frac {10}{3 \, {\left (2 \, x + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x + 10/3/(2*x + 1)

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maple [A]  time = 0.22, size = 10, normalized size = 0.53

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.35, size = 11, normalized size = 0.58 \begin {gather*} x + \frac {10}{3 \, {\left (2 \, x + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x + 10/3/(2*x + 1)

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mupad [B]  time = 0.04, size = 11, normalized size = 0.58 \begin {gather*} x+\frac {5}{3\,\left (x+\frac {1}{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x + 5/(3*(x + 1/2))

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sympy [A]  time = 0.07, size = 7, normalized size = 0.37 \begin {gather*} x + \frac {10}{6 x + 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x + 10/(6*x + 3)

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