### 3.754 $$\int \frac{1}{(a+b x)^2 (a^2-b^2 x^2)} \, dx$$

Optimal. Leaf size=52 $-\frac{1}{4 a^2 b (a+b x)}+\frac{\tanh ^{-1}\left (\frac{b x}{a}\right )}{4 a^3 b}-\frac{1}{4 a b (a+b x)^2}$

[Out]

-1/(4*a*b*(a + b*x)^2) - 1/(4*a^2*b*(a + b*x)) + ArcTanh[(b*x)/a]/(4*a^3*b)

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Rubi [A]  time = 0.0373936, antiderivative size = 52, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 22, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.136, Rules used = {627, 44, 208} $-\frac{1}{4 a^2 b (a+b x)}+\frac{\tanh ^{-1}\left (\frac{b x}{a}\right )}{4 a^3 b}-\frac{1}{4 a b (a+b x)^2}$

Antiderivative was successfully veriﬁed.

[In]

Int[1/((a + b*x)^2*(a^2 - b^2*x^2)),x]

[Out]

-1/(4*a*b*(a + b*x)^2) - 1/(4*a^2*b*(a + b*x)) + ArcTanh[(b*x)/a]/(4*a^3*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 44

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin{align*} \int \frac{1}{(a+b x)^2 \left (a^2-b^2 x^2\right )} \, dx &=\int \frac{1}{(a-b x) (a+b x)^3} \, dx\\ &=\int \left (\frac{1}{2 a (a+b x)^3}+\frac{1}{4 a^2 (a+b x)^2}+\frac{1}{4 a^2 \left (a^2-b^2 x^2\right )}\right ) \, dx\\ &=-\frac{1}{4 a b (a+b x)^2}-\frac{1}{4 a^2 b (a+b x)}+\frac{\int \frac{1}{a^2-b^2 x^2} \, dx}{4 a^2}\\ &=-\frac{1}{4 a b (a+b x)^2}-\frac{1}{4 a^2 b (a+b x)}+\frac{\tanh ^{-1}\left (\frac{b x}{a}\right )}{4 a^3 b}\\ \end{align*}

Mathematica [A]  time = 0.0168035, size = 58, normalized size = 1.12 $\frac{-2 a (2 a+b x)+(a+b x)^2 (-\log (a-b x))+(a+b x)^2 \log (a+b x)}{8 a^3 b (a+b x)^2}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[1/((a + b*x)^2*(a^2 - b^2*x^2)),x]

[Out]

(-2*a*(2*a + b*x) - (a + b*x)^2*Log[a - b*x] + (a + b*x)^2*Log[a + b*x])/(8*a^3*b*(a + b*x)^2)

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Maple [A]  time = 0.047, size = 62, normalized size = 1.2 \begin{align*}{\frac{\ln \left ( bx+a \right ) }{8\,b{a}^{3}}}-{\frac{1}{4\,b{a}^{2} \left ( bx+a \right ) }}-{\frac{1}{4\,ab \left ( bx+a \right ) ^{2}}}-{\frac{\ln \left ( bx-a \right ) }{8\,b{a}^{3}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x+a)^2/(-b^2*x^2+a^2),x)

[Out]

1/8/b/a^3*ln(b*x+a)-1/4/a^2/b/(b*x+a)-1/4/a/b/(b*x+a)^2-1/8/b/a^3*ln(b*x-a)

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Maxima [A]  time = 1.14737, size = 90, normalized size = 1.73 \begin{align*} -\frac{b x + 2 \, a}{4 \,{\left (a^{2} b^{3} x^{2} + 2 \, a^{3} b^{2} x + a^{4} b\right )}} + \frac{\log \left (b x + a\right )}{8 \, a^{3} b} - \frac{\log \left (b x - a\right )}{8 \, a^{3} b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(b*x + 2*a)/(a^2*b^3*x^2 + 2*a^3*b^2*x + a^4*b) + 1/8*log(b*x + a)/(a^3*b) - 1/8*log(b*x - a)/(a^3*b)

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Fricas [A]  time = 1.85799, size = 192, normalized size = 3.69 \begin{align*} -\frac{2 \, a b x + 4 \, a^{2} -{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \log \left (b x + a\right ) +{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \log \left (b x - a\right )}{8 \,{\left (a^{3} b^{3} x^{2} + 2 \, a^{4} b^{2} x + a^{5} b\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(2*a*b*x + 4*a^2 - (b^2*x^2 + 2*a*b*x + a^2)*log(b*x + a) + (b^2*x^2 + 2*a*b*x + a^2)*log(b*x - a))/(a^3*
b^3*x^2 + 2*a^4*b^2*x + a^5*b)

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Sympy [A]  time = 0.492546, size = 58, normalized size = 1.12 \begin{align*} - \frac{2 a + b x}{4 a^{4} b + 8 a^{3} b^{2} x + 4 a^{2} b^{3} x^{2}} - \frac{\frac{\log{\left (- \frac{a}{b} + x \right )}}{8} - \frac{\log{\left (\frac{a}{b} + x \right )}}{8}}{a^{3} b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)**2/(-b**2*x**2+a**2),x)

[Out]

-(2*a + b*x)/(4*a**4*b + 8*a**3*b**2*x + 4*a**2*b**3*x**2) - (log(-a/b + x)/8 - log(a/b + x)/8)/(a**3*b)

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Giac [A]  time = 1.20897, size = 69, normalized size = 1.33 \begin{align*} -\frac{\frac{b}{b x + a} + \frac{a b}{{\left (b x + a\right )}^{2}}}{4 \, a^{2} b^{2}} - \frac{\log \left ({\left | -\frac{2 \, a}{b x + a} + 1 \right |}\right )}{8 \, a^{3} b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(b/(b*x + a) + a*b/(b*x + a)^2)/(a^2*b^2) - 1/8*log(abs(-2*a/(b*x + a) + 1))/(a^3*b)