3.1.69 \(\int \frac {-43+40 x-10 x^2}{12-12 x+3 x^2} \, dx\)

Optimal. Leaf size=26 \[ 5-5 e^3-\frac {1}{2-x}+\frac {5 x}{3}-5 (5+x) \]

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Rubi [A]  time = 0.01, antiderivative size = 15, normalized size of antiderivative = 0.58, 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*} -\frac {10 x}{3}-\frac {1}{2-x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-43 + 40*x - 10*x^2)/(12 - 12*x + 3*x^2),x]

[Out]

-(2 - x)^(-1) - (10*x)/3

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 {-43+40 x-10 x^2}{3 (-2+x)^2} \, dx\\ &=\frac {1}{3} \int \frac {-43+40 x-10 x^2}{(-2+x)^2} \, dx\\ &=\frac {1}{3} \int \left (-10-\frac {3}{(-2+x)^2}\right ) \, dx\\ &=-\frac {1}{2-x}-\frac {10 x}{3}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 17, normalized size = 0.65 \begin {gather*} \frac {1}{3} \left (\frac {3}{-2+x}-10 (-2+x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-43 + 40*x - 10*x^2)/(12 - 12*x + 3*x^2),x]

[Out]

(3/(-2 + x) - 10*(-2 + x))/3

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fricas [A]  time = 0.75, size = 17, normalized size = 0.65 \begin {gather*} -\frac {10 \, x^{2} - 20 \, x - 3}{3 \, {\left (x - 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-10*x^2+40*x-43)/(3*x^2-12*x+12),x, algorithm="fricas")

[Out]

-1/3*(10*x^2 - 20*x - 3)/(x - 2)

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giac [A]  time = 0.34, size = 9, normalized size = 0.35 \begin {gather*} -\frac {10}{3} \, x + \frac {1}{x - 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-10*x^2+40*x-43)/(3*x^2-12*x+12),x, algorithm="giac")

[Out]

-10/3*x + 1/(x - 2)

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maple [A]  time = 0.08, size = 10, normalized size = 0.38

method	result	size
default	\(-\frac {10 x}{3}+\frac {1}{x -2}\)	\(10\)
risch	\(-\frac {10 x}{3}+\frac {1}{x -2}\)	\(10\)
norman	\(\frac {-\frac {10 x^{2}}{3}+\frac {43}{3}}{x -2}\)	\(14\)
gosper	\(-\frac {10 x^{2}-43}{3 \left (x -2\right )}\)	\(15\)
meijerg	\(\frac {37 x}{12 \left (1-\frac {x}{2}\right )}-\frac {10 x \left (-\frac {3 x}{2}+6\right )}{9 \left (1-\frac {x}{2}\right )}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-10*x^2+40*x-43)/(3*x^2-12*x+12),x,method=_RETURNVERBOSE)

[Out]

-10/3*x+1/(x-2)

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maxima [A]  time = 0.69, size = 9, normalized size = 0.35 \begin {gather*} -\frac {10}{3} \, x + \frac {1}{x - 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-10*x^2+40*x-43)/(3*x^2-12*x+12),x, algorithm="maxima")

[Out]

-10/3*x + 1/(x - 2)

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mupad [B]  time = 0.21, size = 9, normalized size = 0.35 \begin {gather*} \frac {1}{x-2}-\frac {10\,x}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(10*x^2 - 40*x + 43)/(3*x^2 - 12*x + 12),x)

[Out]

1/(x - 2) - (10*x)/3

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sympy [A]  time = 0.06, size = 8, normalized size = 0.31 \begin {gather*} - \frac {10 x}{3} + \frac {1}{x - 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-10*x**2+40*x-43)/(3*x**2-12*x+12),x)

[Out]

-10*x/3 + 1/(x - 2)

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