\(\int \frac {a+b x}{a^2-b^2 x^2} \, dx\) [752]

Optimal result

Integrand size = 20, antiderivative size = 12 \[ \int \frac {a+b x}{a^2-b^2 x^2} \, dx=-\frac {\log (a-b x)}{b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {641, 31} \[ \int \frac {a+b x}{a^2-b^2 x^2} \, dx=-\frac {\log (a-b x)}{b} \]

[In]

Int[(a + b*x)/(a^2 - b^2*x^2),x]

[Out]

-(Log[a - b*x]/b)

Rule 31

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

Rule 641

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {1}{a-b x} \, dx \\ & = -\frac {\log (a-b x)}{b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00 \[ \int \frac {a+b x}{a^2-b^2 x^2} \, dx=-\frac {\log (a-b x)}{b} \]

[In]

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

[Out]

-(Log[a - b*x]/b)

Maple [A] (verified)

Time = 2.63 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.08

method	result	size
default	\(-\frac {\ln \left (-b x +a \right )}{b}\)	\(13\)
norman	\(-\frac {\ln \left (-b x +a \right )}{b}\)	\(13\)
risch	\(-\frac {\ln \left (-b x +a \right )}{b}\)	\(13\)
parallelrisch	\(-\frac {\ln \left (b x -a \right )}{b}\)	\(14\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.08 \[ \int \frac {a+b x}{a^2-b^2 x^2} \, dx=-\frac {\log \left (b x - a\right )}{b} \]

[In]

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

[Out]

-log(b*x - a)/b

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.67 \[ \int \frac {a+b x}{a^2-b^2 x^2} \, dx=- \frac {\log {\left (- a + b x \right )}}{b} \]

[In]

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

[Out]

-log(-a + b*x)/b

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.08 \[ \int \frac {a+b x}{a^2-b^2 x^2} \, dx=-\frac {\log \left (b x - a\right )}{b} \]

[In]

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

[Out]

-log(b*x - a)/b

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.17 \[ \int \frac {a+b x}{a^2-b^2 x^2} \, dx=-\frac {\log \left ({\left | b x - a \right |}\right )}{b} \]

[In]

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

[Out]

-log(abs(b*x - a))/b

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.08 \[ \int \frac {a+b x}{a^2-b^2 x^2} \, dx=-\frac {\ln \left (b\,x-a\right )}{b} \]

[In]

int((a + b*x)/(a^2 - b^2*x^2),x)

[Out]

-log(b*x - a)/b