3.755 \(\int \frac{1}{(a+b x)^3 (a^2-b^2 x^2)} \, dx\)

Optimal. Leaf size=69 \[ -\frac{1}{8 a^3 b (a+b x)}-\frac{1}{8 a^2 b (a+b x)^2}+\frac{\tanh ^{-1}\left (\frac{b x}{a}\right )}{8 a^4 b}-\frac{1}{6 a b (a+b x)^3} \]

[Out]

-1/(6*a*b*(a + b*x)^3) - 1/(8*a^2*b*(a + b*x)^2) - 1/(8*a^3*b*(a + b*x)) + ArcTanh[(b*x)/a]/(8*a^4*b)

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Rubi [A]  time = 0.0457732, antiderivative size = 69, 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} \[ -\frac{1}{8 a^3 b (a+b x)}-\frac{1}{8 a^2 b (a+b x)^2}+\frac{\tanh ^{-1}\left (\frac{b x}{a}\right )}{8 a^4 b}-\frac{1}{6 a b (a+b x)^3} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x)^3*(a^2 - b^2*x^2)),x]

[Out]

-1/(6*a*b*(a + b*x)^3) - 1/(8*a^2*b*(a + b*x)^2) - 1/(8*a^3*b*(a + b*x)) + 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)^3 \left (a^2-b^2 x^2\right )} \, dx &=\int \frac{1}{(a-b x) (a+b x)^4} \, dx\\ &=\int \left (\frac{1}{2 a (a+b x)^4}+\frac{1}{4 a^2 (a+b x)^3}+\frac{1}{8 a^3 (a+b x)^2}+\frac{1}{8 a^3 \left (a^2-b^2 x^2\right )}\right ) \, dx\\ &=-\frac{1}{6 a b (a+b x)^3}-\frac{1}{8 a^2 b (a+b x)^2}-\frac{1}{8 a^3 b (a+b x)}+\frac{\int \frac{1}{a^2-b^2 x^2} \, dx}{8 a^3}\\ &=-\frac{1}{6 a b (a+b x)^3}-\frac{1}{8 a^2 b (a+b x)^2}-\frac{1}{8 a^3 b (a+b x)}+\frac{\tanh ^{-1}\left (\frac{b x}{a}\right )}{8 a^4 b}\\ \end{align*}

Mathematica [A]  time = 0.0215884, size = 71, normalized size = 1.03 \[ \frac{-2 a \left (10 a^2+9 a b x+3 b^2 x^2\right )-3 (a+b x)^3 \log (a-b x)+3 (a+b x)^3 \log (a+b x)}{48 a^4 b (a+b x)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b*x)^3*(a^2 - b^2*x^2)),x]

[Out]

(-2*a*(10*a^2 + 9*a*b*x + 3*b^2*x^2) - 3*(a + b*x)^3*Log[a - b*x] + 3*(a + b*x)^3*Log[a + b*x])/(48*a^4*b*(a +
b*x)^3)

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Maple [A]  time = 0.049, size = 77, normalized size = 1.1 \begin{align*}{\frac{\ln \left ( bx+a \right ) }{16\,{a}^{4}b}}-{\frac{1}{8\,{a}^{3}b \left ( bx+a \right ) }}-{\frac{1}{8\,b{a}^{2} \left ( bx+a \right ) ^{2}}}-{\frac{1}{6\,ab \left ( bx+a \right ) ^{3}}}-{\frac{\ln \left ( bx-a \right ) }{16\,{a}^{4}b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x+a)^3/(-b^2*x^2+a^2),x)

[Out]

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

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Maxima [A]  time = 1.02723, size = 122, normalized size = 1.77 \begin{align*} -\frac{3 \, b^{2} x^{2} + 9 \, a b x + 10 \, a^{2}}{24 \,{\left (a^{3} b^{4} x^{3} + 3 \, a^{4} b^{3} x^{2} + 3 \, a^{5} b^{2} x + a^{6} b\right )}} + \frac{\log \left (b x + a\right )}{16 \, a^{4} b} - \frac{\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)^3/(-b^2*x^2+a^2),x, algorithm="maxima")

[Out]

-1/24*(3*b^2*x^2 + 9*a*b*x + 10*a^2)/(a^3*b^4*x^3 + 3*a^4*b^3*x^2 + 3*a^5*b^2*x + a^6*b) + 1/16*log(b*x + a)/(
a^4*b) - 1/16*log(b*x - a)/(a^4*b)

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Fricas [B]  time = 1.78484, size = 288, normalized size = 4.17 \begin{align*} -\frac{6 \, a b^{2} x^{2} + 18 \, a^{2} b x + 20 \, a^{3} - 3 \,{\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3}\right )} \log \left (b x + a\right ) + 3 \,{\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3}\right )} \log \left (b x - a\right )}{48 \,{\left (a^{4} b^{4} x^{3} + 3 \, a^{5} b^{3} x^{2} + 3 \, 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)^3/(-b^2*x^2+a^2),x, algorithm="fricas")

[Out]

-1/48*(6*a*b^2*x^2 + 18*a^2*b*x + 20*a^3 - 3*(b^3*x^3 + 3*a*b^2*x^2 + 3*a^2*b*x + a^3)*log(b*x + a) + 3*(b^3*x
^3 + 3*a*b^2*x^2 + 3*a^2*b*x + a^3)*log(b*x - a))/(a^4*b^4*x^3 + 3*a^5*b^3*x^2 + 3*a^6*b^2*x + a^7*b)

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Sympy [A]  time = 0.609055, size = 83, normalized size = 1.2 \begin{align*} - \frac{10 a^{2} + 9 a b x + 3 b^{2} x^{2}}{24 a^{6} b + 72 a^{5} b^{2} x + 72 a^{4} b^{3} x^{2} + 24 a^{3} b^{4} x^{3}} - \frac{\frac{\log{\left (- \frac{a}{b} + x \right )}}{16} - \frac{\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)**3/(-b**2*x**2+a**2),x)

[Out]

-(10*a**2 + 9*a*b*x + 3*b**2*x**2)/(24*a**6*b + 72*a**5*b**2*x + 72*a**4*b**3*x**2 + 24*a**3*b**4*x**3) - (log
(-a/b + x)/16 - log(a/b + x)/16)/(a**4*b)

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Giac [A]  time = 1.24327, size = 95, normalized size = 1.38 \begin{align*} \frac{\log \left ({\left | b x + a \right |}\right )}{16 \, a^{4} b} - \frac{\log \left ({\left | b x - a \right |}\right )}{16 \, a^{4} b} - \frac{3 \, a b^{2} x^{2} + 9 \, a^{2} b x + 10 \, a^{3}}{24 \,{\left (b x + a\right )}^{3} a^{4} b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^3/(-b^2*x^2+a^2),x, algorithm="giac")

[Out]

1/16*log(abs(b*x + a))/(a^4*b) - 1/16*log(abs(b*x - a))/(a^4*b) - 1/24*(3*a*b^2*x^2 + 9*a^2*b*x + 10*a^3)/((b*
x + a)^3*a^4*b)