\(\int \frac {a+b x}{a c+(b c+a d) x+b d x^2} \, dx\) [1804]

Optimal result

Integrand size = 27, antiderivative size = 10 \[ \int \frac {a+b x}{a c+(b c+a d) x+b d x^2} \, dx=\frac {\log (c+d x)}{d} \]

[Out]

ln(d*x+c)/d

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {640, 31} \[ \int \frac {a+b x}{a c+(b c+a d) x+b d x^2} \, dx=\frac {\log (c+d x)}{d} \]

[In]

Int[(a + b*x)/(a*c + (b*c + a*d)*x + b*d*x^2),x]

[Out]

Log[c + d*x]/d

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 640

Int[((d_) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a
/d + (c/e)*x)^p, x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &&
IntegerQ[p]

Rubi steps \begin{align*} \text {integral}& = \int \frac {1}{c+d x} \, dx \\ & = \frac {\log (c+d x)}{d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.00 \[ \int \frac {a+b x}{a c+(b c+a d) x+b d x^2} \, dx=\frac {\log (c+d x)}{d} \]

[In]

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

[Out]

Log[c + d*x]/d

Maple [A] (verified)

Time = 2.28 (sec) , antiderivative size = 11, normalized size of antiderivative = 1.10

method	result	size
default	\(\frac {\ln \left (d x +c \right )}{d}\)	\(11\)
norman	\(\frac {\ln \left (d x +c \right )}{d}\)	\(11\)
risch	\(\frac {\ln \left (d x +c \right )}{d}\)	\(11\)
parallelrisch	\(\frac {\ln \left (d x +c \right )}{d}\)	\(11\)

[In]

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

[Out]

ln(d*x+c)/d

Fricas [A] (verification not implemented)

none

Time = 0.66 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.00 \[ \int \frac {a+b x}{a c+(b c+a d) x+b d x^2} \, dx=\frac {\log \left (d x + c\right )}{d} \]

[In]

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

[Out]

log(d*x + c)/d

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.70 \[ \int \frac {a+b x}{a c+(b c+a d) x+b d x^2} \, dx=\frac {\log {\left (c + d x \right )}}{d} \]

[In]

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

[Out]

log(c + d*x)/d

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.00 \[ \int \frac {a+b x}{a c+(b c+a d) x+b d x^2} \, dx=\frac {\log \left (d x + c\right )}{d} \]

[In]

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

[Out]

log(d*x + c)/d

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 11, normalized size of antiderivative = 1.10 \[ \int \frac {a+b x}{a c+(b c+a d) x+b d x^2} \, dx=\frac {\log \left ({\left | d x + c \right |}\right )}{d} \]

[In]

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

[Out]

log(abs(d*x + c))/d

Mupad [B] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.00 \[ \int \frac {a+b x}{a c+(b c+a d) x+b d x^2} \, dx=\frac {\ln \left (c+d\,x\right )}{d} \]

[In]

int((a + b*x)/(a*c + x*(a*d + b*c) + b*d*x^2),x)

[Out]

log(c + d*x)/d