3.75.85 \(\int \frac {-73-24 x+12 x^2}{15-30 x+15 x^2} \, dx\)

Optimal. Leaf size=35 \[ -2+2 x-\frac {\left (\frac {20}{3}-x\right ) x}{1-x}+\frac {1}{5} (-1-x-\log (4)) \]

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Rubi [A]  time = 0.01, antiderivative size = 17, normalized size of antiderivative = 0.49, 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 {4 x}{5}-\frac {17}{3 (1-x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-73 - 24*x + 12*x^2)/(15 - 30*x + 15*x^2),x]

[Out]

-17/(3*(1 - x)) + (4*x)/5

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 {-73-24 x+12 x^2}{15 (-1+x)^2} \, dx\\ &=\frac {1}{15} \int \frac {-73-24 x+12 x^2}{(-1+x)^2} \, dx\\ &=\frac {1}{15} \int \left (12-\frac {85}{(-1+x)^2}\right ) \, dx\\ &=-\frac {17}{3 (1-x)}+\frac {4 x}{5}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 17, normalized size = 0.49 \begin {gather*} \frac {1}{15} \left (\frac {85}{-1+x}+12 (-1+x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-73 - 24*x + 12*x^2)/(15 - 30*x + 15*x^2),x]

[Out]

(85/(-1 + x) + 12*(-1 + x))/15

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fricas [A]  time = 0.71, size = 17, normalized size = 0.49 \begin {gather*} \frac {12 \, x^{2} - 12 \, x + 85}{15 \, {\left (x - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((12*x^2-24*x-73)/(15*x^2-30*x+15),x, algorithm="fricas")

[Out]

1/15*(12*x^2 - 12*x + 85)/(x - 1)

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giac [A]  time = 0.20, size = 11, normalized size = 0.31 \begin {gather*} \frac {4}{5} \, x + \frac {17}{3 \, {\left (x - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((12*x^2-24*x-73)/(15*x^2-30*x+15),x, algorithm="giac")

[Out]

4/5*x + 17/3/(x - 1)

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maple [A]  time = 0.26, size = 12, normalized size = 0.34

method	result	size
default	\(\frac {4 x}{5}+\frac {17}{3 \left (x -1\right )}\)	\(12\)
risch	\(\frac {4 x}{5}+\frac {17}{3 \left (x -1\right )}\)	\(12\)
norman	\(\frac {\frac {4 x^{2}}{5}+\frac {73}{15}}{x -1}\)	\(14\)
gosper	\(\frac {12 x^{2}+73}{15 x -15}\)	\(15\)
meijerg	\(-\frac {97 x}{15 \left (1-x \right )}+\frac {4 x \left (-3 x +6\right )}{15 \left (1-x \right )}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((12*x^2-24*x-73)/(15*x^2-30*x+15),x,method=_RETURNVERBOSE)

[Out]

4/5*x+17/3/(x-1)

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maxima [A]  time = 0.36, size = 11, normalized size = 0.31 \begin {gather*} \frac {4}{5} \, x + \frac {17}{3 \, {\left (x - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((12*x^2-24*x-73)/(15*x^2-30*x+15),x, algorithm="maxima")

[Out]

4/5*x + 17/3/(x - 1)

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mupad [B]  time = 0.04, size = 13, normalized size = 0.37 \begin {gather*} \frac {4\,x}{5}+\frac {17}{3\,\left (x-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(24*x - 12*x^2 + 73)/(15*x^2 - 30*x + 15),x)

[Out]

(4*x)/5 + 17/(3*(x - 1))

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sympy [A]  time = 0.07, size = 10, normalized size = 0.29 \begin {gather*} \frac {4 x}{5} + \frac {17}{3 x - 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((12*x**2-24*x-73)/(15*x**2-30*x+15),x)

[Out]

4*x/5 + 17/(3*x - 3)

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