\(\int \frac {-45+30 x-14 x^2+2 x^3}{225-150 x+25 x^2} \, dx\) [7829]

Optimal result

Integrand size = 28, antiderivative size = 27 \[ \int \frac {-45+30 x-14 x^2+2 x^3}{225-150 x+25 x^2} \, dx=-3+\frac {x^3}{25 (-3+x)}+\frac {1}{5} (-1-x-\log (2)) \]

[Out]

-1/5*x-16/5-1/5*ln(2)+x^3/(25*x-75)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {27, 12, 1864} \[ \int \frac {-45+30 x-14 x^2+2 x^3}{225-150 x+25 x^2} \, dx=\frac {x^2}{25}-\frac {2 x}{25}-\frac {27}{25 (3-x)} \]

[In]

Int[(-45 + 30*x - 14*x^2 + 2*x^3)/(225 - 150*x + 25*x^2),x]

[Out]

-27/(25*(3 - x)) - (2*x)/25 + x^2/25

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 1864

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^n)^p, x], x] /; FreeQ[
{a, b, n}, x] && PolyQ[Pq, x] && (IGtQ[p, 0] || EqQ[n, 1])

Rubi steps \begin{align*} \text {integral}& = \int \frac {-45+30 x-14 x^2+2 x^3}{25 (-3+x)^2} \, dx \\ & = \frac {1}{25} \int \frac {-45+30 x-14 x^2+2 x^3}{(-3+x)^2} \, dx \\ & = \frac {1}{25} \int \left (-2-\frac {27}{(-3+x)^2}+2 x\right ) \, dx \\ & = -\frac {27}{25 (3-x)}-\frac {2 x}{25}+\frac {x^2}{25} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.70 \[ \int \frac {-45+30 x-14 x^2+2 x^3}{225-150 x+25 x^2} \, dx=\frac {1}{25} \left (-3+\frac {27}{-3+x}-2 x+x^2\right ) \]

[In]

Integrate[(-45 + 30*x - 14*x^2 + 2*x^3)/(225 - 150*x + 25*x^2),x]

[Out]

(-3 + 27/(-3 + x) - 2*x + x^2)/25

Maple [A] (verified)

Time = 0.19 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.63

method	result	size
default	\(-\frac {2 x}{25}+\frac {x^{2}}{25}+\frac {27}{25 \left (-3+x \right )}\)	\(17\)
risch	\(-\frac {2 x}{25}+\frac {x^{2}}{25}+\frac {27}{25 \left (-3+x \right )}\)	\(17\)
gosper	\(\frac {x^{3}-5 x^{2}+45}{25 x -75}\)	\(18\)
parallelrisch	\(\frac {x^{3}-5 x^{2}+45}{25 x -75}\)	\(18\)
norman	\(\frac {-\frac {1}{5} x^{2}+\frac {1}{25} x^{3}+\frac {9}{5}}{-3+x}\)	\(19\)
meijerg	\(\frac {x}{5-\frac {5 x}{3}}+\frac {3 x \left (-\frac {2}{9} x^{2}-2 x +12\right )}{50 \left (1-\frac {x}{3}\right )}-\frac {14 x \left (-x +6\right )}{75 \left (1-\frac {x}{3}\right )}\)	\(47\)

[In]

int((2*x^3-14*x^2+30*x-45)/(25*x^2-150*x+225),x,method=_RETURNVERBOSE)

[Out]

-2/25*x+1/25*x^2+27/25/(-3+x)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.74 \[ \int \frac {-45+30 x-14 x^2+2 x^3}{225-150 x+25 x^2} \, dx=\frac {x^{3} - 5 \, x^{2} + 6 \, x + 27}{25 \, {\left (x - 3\right )}} \]

[In]

integrate((2*x^3-14*x^2+30*x-45)/(25*x^2-150*x+225),x, algorithm="fricas")

[Out]

1/25*(x^3 - 5*x^2 + 6*x + 27)/(x - 3)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.56 \[ \int \frac {-45+30 x-14 x^2+2 x^3}{225-150 x+25 x^2} \, dx=\frac {x^{2}}{25} - \frac {2 x}{25} + \frac {27}{25 x - 75} \]

[In]

integrate((2*x**3-14*x**2+30*x-45)/(25*x**2-150*x+225),x)

[Out]

x**2/25 - 2*x/25 + 27/(25*x - 75)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.59 \[ \int \frac {-45+30 x-14 x^2+2 x^3}{225-150 x+25 x^2} \, dx=\frac {1}{25} \, x^{2} - \frac {2}{25} \, x + \frac {27}{25 \, {\left (x - 3\right )}} \]

[In]

integrate((2*x^3-14*x^2+30*x-45)/(25*x^2-150*x+225),x, algorithm="maxima")

[Out]

1/25*x^2 - 2/25*x + 27/25/(x - 3)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.59 \[ \int \frac {-45+30 x-14 x^2+2 x^3}{225-150 x+25 x^2} \, dx=\frac {1}{25} \, x^{2} - \frac {2}{25} \, x + \frac {27}{25 \, {\left (x - 3\right )}} \]

[In]

integrate((2*x^3-14*x^2+30*x-45)/(25*x^2-150*x+225),x, algorithm="giac")

[Out]

1/25*x^2 - 2/25*x + 27/25/(x - 3)

Mupad [B] (verification not implemented)

Time = 15.32 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.67 \[ \int \frac {-45+30 x-14 x^2+2 x^3}{225-150 x+25 x^2} \, dx=\frac {27}{25\,\left (x-3\right )}-\frac {2\,x}{25}+\frac {x^2}{25} \]

[In]

int((30*x - 14*x^2 + 2*x^3 - 45)/(25*x^2 - 150*x + 225),x)

[Out]

27/(25*(x - 3)) - (2*x)/25 + x^2/25