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

Optimal. Leaf size=17 \[ -\frac{2 a \log (a-b x)}{b}-x \]

[Out]

-x - (2*a*Log[a - b*x])/b

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Rubi [A]  time = 0.0142805, antiderivative size = 17, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {627, 43} \[ -\frac{2 a \log (a-b x)}{b}-x \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^2/(a^2 - b^2*x^2),x]

[Out]

-x - (2*a*Log[a - b*x])/b

Rule 627

Int[((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a/d + (c*x)/e)^
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]))

Rule 43

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])

Rubi steps

\begin{align*} \int \frac{(a+b x)^2}{a^2-b^2 x^2} \, dx &=\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]  time = 0.0029428, size = 17, normalized size = 1. \[ -\frac{2 a \log (a-b x)}{b}-x \]

Antiderivative was successfully verified.

[In]

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

[Out]

-x - (2*a*Log[a - b*x])/b

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Maple [A]  time = 0.039, size = 19, normalized size = 1.1 \begin{align*} -x-2\,{\frac{a\ln \left ( bx-a \right ) }{b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.03761, size = 24, normalized size = 1.41 \begin{align*} -x - \frac{2 \, a \log \left (b x - a\right )}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x - 2*a*log(b*x - a)/b

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Fricas [A]  time = 1.83442, size = 39, normalized size = 2.29 \begin{align*} -\frac{b x + 2 \, a \log \left (b x - a\right )}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Sympy [A]  time = 0.269772, size = 14, normalized size = 0.82 \begin{align*} - \frac{2 a \log{\left (- a + b x \right )}}{b} - x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*a*log(-a + b*x)/b - x

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Giac [A]  time = 1.18335, size = 26, normalized size = 1.53 \begin{align*} -x - \frac{2 \, a \log \left ({\left | b x - a \right |}\right )}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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