\(\int \frac {b+2 c x}{-a+b x+c x^2} \, dx\) [113]

Optimal result

Integrand size = 21, antiderivative size = 13 \[ \int \frac {b+2 c x}{-a+b x+c x^2} \, dx=\log \left (a-b x-c x^2\right ) \]

[Out]

ln(-c*x^2-b*x+a)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {642} \[ \int \frac {b+2 c x}{-a+b x+c x^2} \, dx=\log \left (a-b x-c x^2\right ) \]

[In]

Int[(b + 2*c*x)/(-a + b*x + c*x^2),x]

[Out]

Log[a - b*x - c*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]

Rubi steps \begin{align*} \text {integral}& = \log \left (a-b x-c x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.92 \[ \int \frac {b+2 c x}{-a+b x+c x^2} \, dx=\log (-a+x (b+c x)) \]

[In]

Integrate[(b + 2*c*x)/(-a + b*x + c*x^2),x]

[Out]

Log[-a + x*(b + c*x)]

Maple [A] (verified)

Time = 0.89 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.08

method	result	size
derivativedivides	\(\ln \left (c \,x^{2}+b x -a \right )\)	\(14\)
default	\(\ln \left (-c \,x^{2}-b x +a \right )\)	\(14\)
norman	\(\ln \left (-c \,x^{2}-b x +a \right )\)	\(14\)
risch	\(\ln \left (-c \,x^{2}-b x +a \right )\)	\(14\)
parallelrisch	\(\ln \left (c \,x^{2}+b x -a \right )\)	\(14\)

[In]

int((2*c*x+b)/(c*x^2+b*x-a),x,method=_RETURNVERBOSE)

[Out]

ln(c*x^2+b*x-a)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00 \[ \int \frac {b+2 c x}{-a+b x+c x^2} \, dx=\log \left (c x^{2} + b x - a\right ) \]

[In]

integrate((2*c*x+b)/(c*x^2+b*x-a),x, algorithm="fricas")

[Out]

log(c*x^2 + b*x - a)

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.77 \[ \int \frac {b+2 c x}{-a+b x+c x^2} \, dx=\log {\left (- a + b x + c x^{2} \right )} \]

[In]

integrate((2*c*x+b)/(c*x**2+b*x-a),x)

[Out]

log(-a + b*x + c*x**2)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00 \[ \int \frac {b+2 c x}{-a+b x+c x^2} \, dx=\log \left (c x^{2} + b x - a\right ) \]

[In]

integrate((2*c*x+b)/(c*x^2+b*x-a),x, algorithm="maxima")

[Out]

log(c*x^2 + b*x - a)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.08 \[ \int \frac {b+2 c x}{-a+b x+c x^2} \, dx=\log \left ({\left | c x^{2} + b x - a \right |}\right ) \]

[In]

integrate((2*c*x+b)/(c*x^2+b*x-a),x, algorithm="giac")

[Out]

log(abs(c*x^2 + b*x - a))

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00 \[ \int \frac {b+2 c x}{-a+b x+c x^2} \, dx=\ln \left (c\,x^2+b\,x-a\right ) \]

[In]

int((b + 2*c*x)/(b*x - a + c*x^2),x)

[Out]

log(b*x - a + c*x^2)