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

Optimal. Leaf size=87 \[ \frac {\tanh ^{-1}\left (\frac {b x}{a}\right )}{4 a^5 b}+\frac {1}{16 a^4 b (a-b x)}-\frac {3}{16 a^4 b (a+b x)}-\frac {1}{8 a^3 b (a+b x)^2}-\frac {1}{12 a^2 b (a+b x)^3} \]

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Rubi [A]  time = 0.05, antiderivative size = 87, normalized size of antiderivative = 1.00, 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} \begin {gather*} \frac {1}{16 a^4 b (a-b x)}-\frac {3}{16 a^4 b (a+b x)}-\frac {1}{8 a^3 b (a+b x)^2}-\frac {1}{12 a^2 b (a+b x)^3}+\frac {\tanh ^{-1}\left (\frac {b x}{a}\right )}{4 a^5 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

1/(16*a^4*b*(a - b*x)) - 1/(12*a^2*b*(a + b*x)^3) - 1/(8*a^3*b*(a + b*x)^2) - 3/(16*a^4*b*(a + b*x)) + ArcTanh
[(b*x)/a]/(4*a^5*b)

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]

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]))

Rubi steps

\begin {align*} \int \frac {1}{(a+b x)^2 \left (a^2-b^2 x^2\right )^2} \, dx &=\int \frac {1}{(a-b x)^2 (a+b x)^4} \, dx\\ &=\int \left (\frac {1}{16 a^4 (a-b x)^2}+\frac {1}{4 a^2 (a+b x)^4}+\frac {1}{4 a^3 (a+b x)^3}+\frac {3}{16 a^4 (a+b x)^2}+\frac {1}{4 a^4 \left (a^2-b^2 x^2\right )}\right ) \, dx\\ &=\frac {1}{16 a^4 b (a-b x)}-\frac {1}{12 a^2 b (a+b x)^3}-\frac {1}{8 a^3 b (a+b x)^2}-\frac {3}{16 a^4 b (a+b x)}+\frac {\int \frac {1}{a^2-b^2 x^2} \, dx}{4 a^4}\\ &=\frac {1}{16 a^4 b (a-b x)}-\frac {1}{12 a^2 b (a+b x)^3}-\frac {1}{8 a^3 b (a+b x)^2}-\frac {3}{16 a^4 b (a+b x)}+\frac {\tanh ^{-1}\left (\frac {b x}{a}\right )}{4 a^5 b}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 75, normalized size = 0.86 \begin {gather*} \frac {\frac {2 a \left (-4 a^3+a^2 b x+6 a b^2 x^2+3 b^3 x^3\right )}{(a-b x) (a+b x)^3}-3 \log (a-b x)+3 \log (a+b x)}{24 a^5 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((2*a*(-4*a^3 + a^2*b*x + 6*a*b^2*x^2 + 3*b^3*x^3))/((a - b*x)*(a + b*x)^3) - 3*Log[a - b*x] + 3*Log[a + b*x])
/(24*a^5*b)

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{(a+b x)^2 \left (a^2-b^2 x^2\right )^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.40, size = 150, normalized size = 1.72 \begin {gather*} -\frac {6 \, a b^{3} x^{3} + 12 \, a^{2} b^{2} x^{2} + 2 \, a^{3} b x - 8 \, a^{4} - 3 \, {\left (b^{4} x^{4} + 2 \, a b^{3} x^{3} - 2 \, a^{3} b x - a^{4}\right )} \log \left (b x + a\right ) + 3 \, {\left (b^{4} x^{4} + 2 \, a b^{3} x^{3} - 2 \, a^{3} b x - a^{4}\right )} \log \left (b x - a\right )}{24 \, {\left (a^{5} b^{5} x^{4} + 2 \, a^{6} b^{4} x^{3} - 2 \, a^{8} b^{2} x - a^{9} b\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(6*a*b^3*x^3 + 12*a^2*b^2*x^2 + 2*a^3*b*x - 8*a^4 - 3*(b^4*x^4 + 2*a*b^3*x^3 - 2*a^3*b*x - a^4)*log(b*x
+ a) + 3*(b^4*x^4 + 2*a*b^3*x^3 - 2*a^3*b*x - a^4)*log(b*x - a))/(a^5*b^5*x^4 + 2*a^6*b^4*x^3 - 2*a^8*b^2*x -
a^9*b)

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giac [A]  time = 0.19, size = 99, normalized size = 1.14 \begin {gather*} -\frac {\log \left ({\left | -\frac {2 \, a}{b x + a} + 1 \right |}\right )}{8 \, a^{5} b} + \frac {1}{32 \, a^{5} b {\left (\frac {2 \, a}{b x + a} - 1\right )}} - \frac {\frac {9 \, a^{2} b^{5}}{b x + a} + \frac {6 \, a^{3} b^{5}}{{\left (b x + a\right )}^{2}} + \frac {4 \, a^{4} b^{5}}{{\left (b x + a\right )}^{3}}}{48 \, a^{6} b^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*log(abs(-2*a/(b*x + a) + 1))/(a^5*b) + 1/32/(a^5*b*(2*a/(b*x + a) - 1)) - 1/48*(9*a^2*b^5/(b*x + a) + 6*a
^3*b^5/(b*x + a)^2 + 4*a^4*b^5/(b*x + a)^3)/(a^6*b^6)

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maple [A]  time = 0.06, size = 94, normalized size = 1.08 \begin {gather*} -\frac {1}{12 \left (b x +a \right )^{3} a^{2} b}-\frac {1}{8 \left (b x +a \right )^{2} a^{3} b}-\frac {1}{16 \left (b x -a \right ) a^{4} b}-\frac {3}{16 \left (b x +a \right ) a^{4} b}-\frac {\ln \left (b x -a \right )}{8 a^{5} b}+\frac {\ln \left (b x +a \right )}{8 a^{5} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8/a^5/b*ln(b*x-a)-1/16/a^4/b/(b*x-a)+1/8/a^5/b*ln(b*x+a)-3/16/(b*x+a)/a^4/b-1/8/(b*x+a)^2/a^3/b-1/12/(b*x+a
)^3/a^2/b

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maxima [A]  time = 1.43, size = 101, normalized size = 1.16 \begin {gather*} -\frac {3 \, b^{3} x^{3} + 6 \, a b^{2} x^{2} + a^{2} b x - 4 \, a^{3}}{12 \, {\left (a^{4} b^{5} x^{4} + 2 \, a^{5} b^{4} x^{3} - 2 \, a^{7} b^{2} x - a^{8} b\right )}} + \frac {\log \left (b x + a\right )}{8 \, a^{5} b} - \frac {\log \left (b x - a\right )}{8 \, a^{5} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(3*b^3*x^3 + 6*a*b^2*x^2 + a^2*b*x - 4*a^3)/(a^4*b^5*x^4 + 2*a^5*b^4*x^3 - 2*a^7*b^2*x - a^8*b) + 1/8*lo
g(b*x + a)/(a^5*b) - 1/8*log(b*x - a)/(a^5*b)

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mupad [B]  time = 0.08, size = 82, normalized size = 0.94 \begin {gather*} \frac {\frac {x}{12\,a^2}-\frac {1}{3\,a\,b}+\frac {b\,x^2}{2\,a^3}+\frac {b^2\,x^3}{4\,a^4}}{a^4+2\,a^3\,b\,x-2\,a\,b^3\,x^3-b^4\,x^4}+\frac {\mathrm {atanh}\left (\frac {b\,x}{a}\right )}{4\,a^5\,b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x/(12*a^2) - 1/(3*a*b) + (b*x^2)/(2*a^3) + (b^2*x^3)/(4*a^4))/(a^4 - b^4*x^4 - 2*a*b^3*x^3 + 2*a^3*b*x) + ata
nh((b*x)/a)/(4*a^5*b)

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sympy [A]  time = 0.55, size = 92, normalized size = 1.06 \begin {gather*} \frac {4 a^{3} - a^{2} b x - 6 a b^{2} x^{2} - 3 b^{3} x^{3}}{- 12 a^{8} b - 24 a^{7} b^{2} x + 24 a^{5} b^{4} x^{3} + 12 a^{4} b^{5} x^{4}} + \frac {- \frac {\log {\left (- \frac {a}{b} + x \right )}}{8} + \frac {\log {\left (\frac {a}{b} + x \right )}}{8}}{a^{5} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(4*a**3 - a**2*b*x - 6*a*b**2*x**2 - 3*b**3*x**3)/(-12*a**8*b - 24*a**7*b**2*x + 24*a**5*b**4*x**3 + 12*a**4*b
**5*x**4) + (-log(-a/b + x)/8 + log(a/b + x)/8)/(a**5*b)

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