\(\int \frac {(a+b x)^2}{a^2-b^2 x^2} \, dx\) [751]

Optimal result

Integrand size = 22, antiderivative size = 17 \[ \int \frac {(a+b x)^2}{a^2-b^2 x^2} \, dx=-x-\frac {2 a \log (a-b x)}{b} \]

[Out]

-x-2*a*ln(-b*x+a)/b

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {641, 45} \[ \int \frac {(a+b x)^2}{a^2-b^2 x^2} \, dx=-\frac {2 a \log (a-b x)}{b}-x \]

[In]

Int[(a + b*x)^2/(a^2 - b^2*x^2),x]

[Out]

-x - (2*a*Log[a - b*x])/b

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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 {a+b x}{a-b x} \, dx \\ & = \int \left (-1+\frac {2 a}{a-b x}\right ) \, dx \\ & = -x-\frac {2 a \log (a-b x)}{b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {(a+b x)^2}{a^2-b^2 x^2} \, dx=-x-\frac {2 a \log (a-b x)}{b} \]

[In]

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

[Out]

-x - (2*a*Log[a - b*x])/b

Maple [A] (verified)

Time = 2.42 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.06

method	result	size
default	\(-x -\frac {2 a \ln \left (-b x +a \right )}{b}\)	\(18\)
norman	\(-x -\frac {2 a \ln \left (-b x +a \right )}{b}\)	\(18\)
risch	\(-x -\frac {2 a \ln \left (-b x +a \right )}{b}\)	\(18\)
parallelrisch	\(-\frac {2 a \ln \left (b x -a \right )+b x}{b}\)	\(21\)

[In]

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

[Out]

-x-2*a*ln(-b*x+a)/b

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.18 \[ \int \frac {(a+b x)^2}{a^2-b^2 x^2} \, dx=-\frac {b x + 2 \, a \log \left (b x - a\right )}{b} \]

[In]

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

[Out]

-(b*x + 2*a*log(b*x - a))/b

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \frac {(a+b x)^2}{a^2-b^2 x^2} \, dx=- \frac {2 a \log {\left (- a + b x \right )}}{b} - x \]

[In]

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

[Out]

-2*a*log(-a + b*x)/b - x

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.06 \[ \int \frac {(a+b x)^2}{a^2-b^2 x^2} \, dx=-x - \frac {2 \, a \log \left (b x - a\right )}{b} \]

[In]

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

[Out]

-x - 2*a*log(b*x - a)/b

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.12 \[ \int \frac {(a+b x)^2}{a^2-b^2 x^2} \, dx=-x - \frac {2 \, a \log \left ({\left | b x - a \right |}\right )}{b} \]

[In]

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

[Out]

-x - 2*a*log(abs(b*x - a))/b

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.06 \[ \int \frac {(a+b x)^2}{a^2-b^2 x^2} \, dx=-x-\frac {2\,a\,\ln \left (b\,x-a\right )}{b} \]

[In]

int((a + b*x)^2/(a^2 - b^2*x^2),x)

[Out]

- x - (2*a*log(b*x - a))/b