\(\int \frac {110 x-25 x^2}{121-110 x+25 x^2} \, dx\) [6479]

Optimal result

Integrand size = 22, antiderivative size = 21 \[ \int \frac {110 x-25 x^2}{121-110 x+25 x^2} \, dx=\frac {5}{3}-e^3-\frac {x^2}{-\frac {11}{5}+x} \]

[Out]

5/3-x^2/(x-11/5)-exp(3)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.71, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {27, 697} \[ \int \frac {110 x-25 x^2}{121-110 x+25 x^2} \, dx=\frac {121}{5 (11-5 x)}-x \]

[In]

Int[(110*x - 25*x^2)/(121 - 110*x + 25*x^2),x]

[Out]

121/(5*(11 - 5*x)) - 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 {110 x-25 x^2}{(-11+5 x)^2} \, dx \\ & = \int \left (-1+\frac {121}{(-11+5 x)^2}\right ) \, dx \\ & = \frac {121}{5 (11-5 x)}-x \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86 \[ \int \frac {110 x-25 x^2}{121-110 x+25 x^2} \, dx=\frac {1}{5} \left (11+\frac {121}{11-5 x}-5 x\right ) \]

[In]

Integrate[(110*x - 25*x^2)/(121 - 110*x + 25*x^2),x]

[Out]

(11 + 121/(11 - 5*x) - 5*x)/5

Maple [A] (verified)

Time = 0.57 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.57

method	result	size
risch	\(-x -\frac {121}{25 \left (x -\frac {11}{5}\right )}\)	\(12\)
gosper	\(-\frac {5 x^{2}}{5 x -11}\)	\(13\)
parallelrisch	\(-\frac {5 x^{2}}{5 x -11}\)	\(13\)
default	\(-x -\frac {121}{5 \left (5 x -11\right )}\)	\(14\)
meijerg	\(-\frac {x \left (-\frac {15 x}{11}+6\right )}{3 \left (1-\frac {5 x}{11}\right )}+\frac {2 x}{1-\frac {5 x}{11}}\)	\(27\)

[In]

int((-25*x^2+110*x)/(25*x^2-110*x+121),x,method=_RETURNVERBOSE)

[Out]

-x-121/25/(x-11/5)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {110 x-25 x^2}{121-110 x+25 x^2} \, dx=-\frac {25 \, x^{2} - 55 \, x + 121}{5 \, {\left (5 \, x - 11\right )}} \]

[In]

integrate((-25*x^2+110*x)/(25*x^2-110*x+121),x, algorithm="fricas")

[Out]

-1/5*(25*x^2 - 55*x + 121)/(5*x - 11)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.38 \[ \int \frac {110 x-25 x^2}{121-110 x+25 x^2} \, dx=- x - \frac {121}{25 x - 55} \]

[In]

integrate((-25*x**2+110*x)/(25*x**2-110*x+121),x)

[Out]

-x - 121/(25*x - 55)

Maxima [A] (verification not implemented)

none

Time = 0.17 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.62 \[ \int \frac {110 x-25 x^2}{121-110 x+25 x^2} \, dx=-x - \frac {121}{5 \, {\left (5 \, x - 11\right )}} \]

[In]

integrate((-25*x^2+110*x)/(25*x^2-110*x+121),x, algorithm="maxima")

[Out]

-x - 121/5/(5*x - 11)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.62 \[ \int \frac {110 x-25 x^2}{121-110 x+25 x^2} \, dx=-x - \frac {121}{5 \, {\left (5 \, x - 11\right )}} \]

[In]

integrate((-25*x^2+110*x)/(25*x^2-110*x+121),x, algorithm="giac")

[Out]

-x - 121/5/(5*x - 11)

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.62 \[ \int \frac {110 x-25 x^2}{121-110 x+25 x^2} \, dx=-x-\frac {121}{25\,\left (x-\frac {11}{5}\right )} \]

[In]

int((110*x - 25*x^2)/(25*x^2 - 110*x + 121),x)

[Out]

- x - 121/(25*(x - 11/5))