### 3.764 $$\int \frac{1}{(a+b x) (a^2-b^2 x^2)^2} \, dx$$

Optimal. Leaf size=70 $\frac{1}{8 a^3 b (a-b x)}-\frac{1}{4 a^3 b (a+b x)}-\frac{1}{8 a^2 b (a+b x)^2}+\frac{3 \tanh ^{-1}\left (\frac{b x}{a}\right )}{8 a^4 b}$

[Out]

1/(8*a^3*b*(a - b*x)) - 1/(8*a^2*b*(a + b*x)^2) - 1/(4*a^3*b*(a + b*x)) + (3*ArcTanh[(b*x)/a])/(8*a^4*b)

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Rubi [A]  time = 0.0460265, antiderivative size = 70, 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} $\frac{1}{8 a^3 b (a-b x)}-\frac{1}{4 a^3 b (a+b x)}-\frac{1}{8 a^2 b (a+b x)^2}+\frac{3 \tanh ^{-1}\left (\frac{b x}{a}\right )}{8 a^4 b}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

1/(8*a^3*b*(a - b*x)) - 1/(8*a^2*b*(a + b*x)^2) - 1/(4*a^3*b*(a + b*x)) + (3*ArcTanh[(b*x)/a])/(8*a^4*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) \left (a^2-b^2 x^2\right )^2} \, dx &=\int \frac{1}{(a-b x)^2 (a+b x)^3} \, dx\\ &=\int \left (\frac{1}{8 a^3 (a-b x)^2}+\frac{1}{4 a^2 (a+b x)^3}+\frac{1}{4 a^3 (a+b x)^2}+\frac{3}{8 a^3 \left (a^2-b^2 x^2\right )}\right ) \, dx\\ &=\frac{1}{8 a^3 b (a-b x)}-\frac{1}{8 a^2 b (a+b x)^2}-\frac{1}{4 a^3 b (a+b x)}+\frac{3 \int \frac{1}{a^2-b^2 x^2} \, dx}{8 a^3}\\ &=\frac{1}{8 a^3 b (a-b x)}-\frac{1}{8 a^2 b (a+b x)^2}-\frac{1}{4 a^3 b (a+b x)}+\frac{3 \tanh ^{-1}\left (\frac{b x}{a}\right )}{8 a^4 b}\\ \end{align*}

Mathematica [A]  time = 0.0203476, size = 87, normalized size = 1.24 $-\frac{1}{8 a^3 b (b x-a)}-\frac{1}{4 a^3 b (a+b x)}-\frac{1}{8 a^2 b (a+b x)^2}-\frac{3 \log (a-b x)}{16 a^4 b}+\frac{3 \log (a+b x)}{16 a^4 b}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-1/(8*a^3*b*(-a + b*x)) - 1/(8*a^2*b*(a + b*x)^2) - 1/(4*a^3*b*(a + b*x)) - (3*Log[a - b*x])/(16*a^4*b) + (3*L
og[a + b*x])/(16*a^4*b)

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Maple [A]  time = 0.051, size = 79, normalized size = 1.1 \begin{align*}{\frac{3\,\ln \left ( bx+a \right ) }{16\,{a}^{4}b}}-{\frac{1}{4\,{a}^{3}b \left ( bx+a \right ) }}-{\frac{1}{8\,b{a}^{2} \left ( bx+a \right ) ^{2}}}-{\frac{3\,\ln \left ( bx-a \right ) }{16\,{a}^{4}b}}-{\frac{1}{8\,{a}^{3}b \left ( bx-a \right ) }} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/16/a^4/b*ln(b*x+a)-1/4/a^3/b/(b*x+a)-1/8/a^2/b/(b*x+a)^2-3/16/a^4/b*ln(b*x-a)-1/8/b/a^3/(b*x-a)

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Maxima [A]  time = 1.04205, size = 122, normalized size = 1.74 \begin{align*} -\frac{3 \, b^{2} x^{2} + 3 \, a b x - 2 \, a^{2}}{8 \,{\left (a^{3} b^{4} x^{3} + a^{4} b^{3} x^{2} - a^{5} b^{2} x - a^{6} b\right )}} + \frac{3 \, \log \left (b x + a\right )}{16 \, a^{4} b} - \frac{3 \, \log \left (b x - a\right )}{16 \, a^{4} b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(3*b^2*x^2 + 3*a*b*x - 2*a^2)/(a^3*b^4*x^3 + a^4*b^3*x^2 - a^5*b^2*x - a^6*b) + 3/16*log(b*x + a)/(a^4*b)
- 3/16*log(b*x - a)/(a^4*b)

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Fricas [B]  time = 1.7458, size = 269, normalized size = 3.84 \begin{align*} -\frac{6 \, a b^{2} x^{2} + 6 \, a^{2} b x - 4 \, a^{3} - 3 \,{\left (b^{3} x^{3} + a b^{2} x^{2} - a^{2} b x - a^{3}\right )} \log \left (b x + a\right ) + 3 \,{\left (b^{3} x^{3} + a b^{2} x^{2} - a^{2} b x - a^{3}\right )} \log \left (b x - a\right )}{16 \,{\left (a^{4} b^{4} x^{3} + a^{5} b^{3} x^{2} - a^{6} b^{2} x - a^{7} b\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(6*a*b^2*x^2 + 6*a^2*b*x - 4*a^3 - 3*(b^3*x^3 + a*b^2*x^2 - a^2*b*x - a^3)*log(b*x + a) + 3*(b^3*x^3 + a
*b^2*x^2 - a^2*b*x - a^3)*log(b*x - a))/(a^4*b^4*x^3 + a^5*b^3*x^2 - a^6*b^2*x - a^7*b)

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Sympy [A]  time = 0.578813, size = 85, normalized size = 1.21 \begin{align*} - \frac{- 2 a^{2} + 3 a b x + 3 b^{2} x^{2}}{- 8 a^{6} b - 8 a^{5} b^{2} x + 8 a^{4} b^{3} x^{2} + 8 a^{3} b^{4} x^{3}} + \frac{- \frac{3 \log{\left (- \frac{a}{b} + x \right )}}{16} + \frac{3 \log{\left (\frac{a}{b} + x \right )}}{16}}{a^{4} b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(-2*a**2 + 3*a*b*x + 3*b**2*x**2)/(-8*a**6*b - 8*a**5*b**2*x + 8*a**4*b**3*x**2 + 8*a**3*b**4*x**3) + (-3*log
(-a/b + x)/16 + 3*log(a/b + x)/16)/(a**4*b)

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Giac [A]  time = 1.24312, size = 107, normalized size = 1.53 \begin{align*} \frac{3 \, \log \left ({\left | b x + a \right |}\right )}{16 \, a^{4} b} - \frac{3 \, \log \left ({\left | b x - a \right |}\right )}{16 \, a^{4} b} - \frac{3 \, a b^{2} x^{2} + 3 \, a^{2} b x - 2 \, a^{3}}{8 \,{\left (b x + a\right )}^{2}{\left (b x - a\right )} a^{4} b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/16*log(abs(b*x + a))/(a^4*b) - 3/16*log(abs(b*x - a))/(a^4*b) - 1/8*(3*a*b^2*x^2 + 3*a^2*b*x - 2*a^3)/((b*x
+ a)^2*(b*x - a)*a^4*b)