\(\int \frac {1}{b x+c x^2} \, dx\) [263]

Optimal result

Integrand size = 11, antiderivative size = 18 \[ \int \frac {1}{b x+c x^2} \, dx=\frac {\log (x)}{b}-\frac {\log (b+c x)}{b} \]

[Out]

ln(x)/b-ln(c*x+b)/b

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {629} \[ \int \frac {1}{b x+c x^2} \, dx=\frac {\log (x)}{b}-\frac {\log (b+c x)}{b} \]

[In]

Int[(b*x + c*x^2)^(-1),x]

[Out]

Log[x]/b - Log[b + c*x]/b

Rule 629

Int[((b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[Log[x]/b, x] - Simp[Log[RemoveContent[b + c*x, x]]/b,
x] /; FreeQ[{b, c}, x]

Rubi steps \begin{align*} \text {integral}& = \frac {\log (x)}{b}-\frac {\log (b+c x)}{b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {1}{b x+c x^2} \, dx=\frac {\log (x)}{b}-\frac {\log (b+c x)}{b} \]

[In]

Integrate[(b*x + c*x^2)^(-1),x]

[Out]

Log[x]/b - Log[b + c*x]/b

Maple [A] (verified)

Time = 2.04 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.89

method	result	size
parallelrisch	\(\frac {\ln \left (x \right )-\ln \left (c x +b \right )}{b}\)	\(16\)
default	\(\frac {\ln \left (x \right )}{b}-\frac {\ln \left (c x +b \right )}{b}\)	\(19\)
norman	\(\frac {\ln \left (x \right )}{b}-\frac {\ln \left (c x +b \right )}{b}\)	\(19\)
risch	\(\frac {\ln \left (-x \right )}{b}-\frac {\ln \left (c x +b \right )}{b}\)	\(21\)

[In]

int(1/(c*x^2+b*x),x,method=_RETURNVERBOSE)

[Out]

(ln(x)-ln(c*x+b))/b

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.89 \[ \int \frac {1}{b x+c x^2} \, dx=-\frac {\log \left (c x + b\right ) - \log \left (x\right )}{b} \]

[In]

integrate(1/(c*x^2+b*x),x, algorithm="fricas")

[Out]

-(log(c*x + b) - log(x))/b

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.56 \[ \int \frac {1}{b x+c x^2} \, dx=\frac {\log {\left (x \right )} - \log {\left (\frac {b}{c} + x \right )}}{b} \]

[In]

integrate(1/(c*x**2+b*x),x)

[Out]

(log(x) - log(b/c + x))/b

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {1}{b x+c x^2} \, dx=-\frac {\log \left (c x + b\right )}{b} + \frac {\log \left (x\right )}{b} \]

[In]

integrate(1/(c*x^2+b*x),x, algorithm="maxima")

[Out]

-log(c*x + b)/b + log(x)/b

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.11 \[ \int \frac {1}{b x+c x^2} \, dx=-\frac {\log \left ({\left | c x + b \right |}\right )}{b} + \frac {\log \left ({\left | x \right |}\right )}{b} \]

[In]

integrate(1/(c*x^2+b*x),x, algorithm="giac")

[Out]

-log(abs(c*x + b))/b + log(abs(x))/b

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83 \[ \int \frac {1}{b x+c x^2} \, dx=-\frac {2\,\mathrm {atanh}\left (\frac {2\,c\,x}{b}+1\right )}{b} \]

[In]

int(1/(b*x + c*x^2),x)

[Out]

-(2*atanh((2*c*x)/b + 1))/b