\(\int \frac {-120+11 x-8 x^2}{125+5 x+8 x^2} \, dx\) [7736]

Optimal result

Integrand size = 23, antiderivative size = 19 \[ \int \frac {-120+11 x-8 x^2}{125+5 x+8 x^2} \, dx=-x+\log \left (25+x+4 x \log \left (e^{2 x/5}\right )\right ) \]

[Out]

ln(25+x+4*x*ln(exp(1/5*x)^2))-x

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {1671, 642} \[ \int \frac {-120+11 x-8 x^2}{125+5 x+8 x^2} \, dx=\log \left (8 x^2+5 x+125\right )-x \]

[In]

Int[(-120 + 11*x - 8*x^2)/(125 + 5*x + 8*x^2),x]

[Out]

-x + Log[125 + 5*x + 8*x^2]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1671

Int[(Pq_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x + c*x^2)^p, x
], x] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rubi steps \begin{align*} \text {integral}& = \int \left (-1+\frac {5+16 x}{125+5 x+8 x^2}\right ) \, dx \\ & = -x+\int \frac {5+16 x}{125+5 x+8 x^2} \, dx \\ & = -x+\log \left (125+5 x+8 x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {-120+11 x-8 x^2}{125+5 x+8 x^2} \, dx=-x+\log \left (125+5 x+8 x^2\right ) \]

[In]

Integrate[(-120 + 11*x - 8*x^2)/(125 + 5*x + 8*x^2),x]

[Out]

-x + Log[125 + 5*x + 8*x^2]

Maple [A] (verified)

Time = 1.01 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.74

method	result	size
parallelrisch	\(-x +\ln \left (x^{2}+\frac {5}{8} x +\frac {125}{8}\right )\)	\(14\)
default	\(-x +\ln \left (8 x^{2}+5 x +125\right )\)	\(16\)
norman	\(-x +\ln \left (8 x^{2}+5 x +125\right )\)	\(16\)
risch	\(-x +\ln \left (8 x^{2}+5 x +125\right )\)	\(16\)

[In]

int((-8*x^2+11*x-120)/(8*x^2+5*x+125),x,method=_RETURNVERBOSE)

[Out]

-x+ln(x^2+5/8*x+125/8)

Fricas [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {-120+11 x-8 x^2}{125+5 x+8 x^2} \, dx=-x + \log \left (8 \, x^{2} + 5 \, x + 125\right ) \]

[In]

integrate((-8*x^2+11*x-120)/(8*x^2+5*x+125),x, algorithm="fricas")

[Out]

-x + log(8*x^2 + 5*x + 125)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.63 \[ \int \frac {-120+11 x-8 x^2}{125+5 x+8 x^2} \, dx=- x + \log {\left (8 x^{2} + 5 x + 125 \right )} \]

[In]

integrate((-8*x**2+11*x-120)/(8*x**2+5*x+125),x)

[Out]

-x + log(8*x**2 + 5*x + 125)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {-120+11 x-8 x^2}{125+5 x+8 x^2} \, dx=-x + \log \left (8 \, x^{2} + 5 \, x + 125\right ) \]

[In]

integrate((-8*x^2+11*x-120)/(8*x^2+5*x+125),x, algorithm="maxima")

[Out]

-x + log(8*x^2 + 5*x + 125)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {-120+11 x-8 x^2}{125+5 x+8 x^2} \, dx=-x + \log \left (8 \, x^{2} + 5 \, x + 125\right ) \]

[In]

integrate((-8*x^2+11*x-120)/(8*x^2+5*x+125),x, algorithm="giac")

[Out]

-x + log(8*x^2 + 5*x + 125)

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {-120+11 x-8 x^2}{125+5 x+8 x^2} \, dx=\ln \left (8\,x^2+5\,x+125\right )-x \]

[In]

int(-(8*x^2 - 11*x + 120)/(5*x + 8*x^2 + 125),x)

[Out]

log(5*x + 8*x^2 + 125) - x