### 3.763 $$\int \frac{a+b x}{(a^2-b^2 x^2)^2} \, dx$$

Optimal. Leaf size=36 $\frac{\tanh ^{-1}\left (\frac{b x}{a}\right )}{2 a^2 b}+\frac{1}{2 a b (a-b x)}$

[Out]

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

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Rubi [A]  time = 0.0261313, antiderivative size = 36, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 20, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.15, Rules used = {627, 44, 208} $\frac{\tanh ^{-1}\left (\frac{b x}{a}\right )}{2 a^2 b}+\frac{1}{2 a b (a-b x)}$

Antiderivative was successfully veriﬁed.

[In]

Int[(a + b*x)/(a^2 - b^2*x^2)^2,x]

[Out]

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

Mathematica [A]  time = 0.0099478, size = 50, normalized size = 1.39 $\frac{(b x-a) \log (a-b x)+(a-b x) \log (a+b x)+2 a}{4 a^2 b (a-b x)}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(2*a + (-a + b*x)*Log[a - b*x] + (a - b*x)*Log[a + b*x])/(4*a^2*b*(a - b*x))

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Maple [A]  time = 0.047, size = 49, normalized size = 1.4 \begin{align*}{\frac{\ln \left ( bx+a \right ) }{4\,b{a}^{2}}}-{\frac{\ln \left ( bx-a \right ) }{4\,b{a}^{2}}}-{\frac{1}{2\,ab \left ( bx-a \right ) }} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.06567, size = 65, normalized size = 1.81 \begin{align*} -\frac{1}{2 \,{\left (a b^{2} x - a^{2} b\right )}} + \frac{\log \left (b x + a\right )}{4 \, a^{2} b} - \frac{\log \left (b x - a\right )}{4 \, a^{2} b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.8425, size = 109, normalized size = 3.03 \begin{align*} \frac{{\left (b x - a\right )} \log \left (b x + a\right ) -{\left (b x - a\right )} \log \left (b x - a\right ) - 2 \, a}{4 \,{\left (a^{2} b^{2} x - a^{3} b\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.391543, size = 37, normalized size = 1.03 \begin{align*} - \frac{1}{- 2 a^{2} b + 2 a b^{2} x} + \frac{- \frac{\log{\left (- \frac{a}{b} + x \right )}}{4} + \frac{\log{\left (\frac{a}{b} + x \right )}}{4}}{a^{2} b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.23356, size = 68, normalized size = 1.89 \begin{align*} \frac{\log \left ({\left | b x + a \right |}\right )}{4 \, a^{2} b} - \frac{\log \left ({\left | b x - a \right |}\right )}{4 \, a^{2} b} - \frac{1}{2 \,{\left (b x - a\right )} a b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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