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

Optimal. Leaf size=104 \[ \frac{1}{32 a^5 b (a-b x)}-\frac{1}{8 a^5 b (a+b x)}-\frac{3}{32 a^4 b (a+b x)^2}-\frac{1}{12 a^3 b (a+b x)^3}-\frac{1}{16 a^2 b (a+b x)^4}+\frac{5 \tanh ^{-1}\left (\frac{b x}{a}\right )}{32 a^6 b} \]

[Out]

1/(32*a^5*b*(a - b*x)) - 1/(16*a^2*b*(a + b*x)^4) - 1/(12*a^3*b*(a + b*x)^3) - 3/(32*a^4*b*(a + b*x)^2) - 1/(8
*a^5*b*(a + b*x)) + (5*ArcTanh[(b*x)/a])/(32*a^6*b)

________________________________________________________________________________________

Rubi [A]  time = 0.0658568, antiderivative size = 104, 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} \[ \frac{1}{32 a^5 b (a-b x)}-\frac{1}{8 a^5 b (a+b x)}-\frac{3}{32 a^4 b (a+b x)^2}-\frac{1}{12 a^3 b (a+b x)^3}-\frac{1}{16 a^2 b (a+b x)^4}+\frac{5 \tanh ^{-1}\left (\frac{b x}{a}\right )}{32 a^6 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

1/(32*a^5*b*(a - b*x)) - 1/(16*a^2*b*(a + b*x)^4) - 1/(12*a^3*b*(a + b*x)^3) - 3/(32*a^4*b*(a + b*x)^2) - 1/(8
*a^5*b*(a + b*x)) + (5*ArcTanh[(b*x)/a])/(32*a^6*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 )^2} \, dx &=\int \frac{1}{(a-b x)^2 (a+b x)^5} \, dx\\ &=\int \left (\frac{1}{32 a^5 (a-b x)^2}+\frac{1}{4 a^2 (a+b x)^5}+\frac{1}{4 a^3 (a+b x)^4}+\frac{3}{16 a^4 (a+b x)^3}+\frac{1}{8 a^5 (a+b x)^2}+\frac{5}{32 a^5 \left (a^2-b^2 x^2\right )}\right ) \, dx\\ &=\frac{1}{32 a^5 b (a-b x)}-\frac{1}{16 a^2 b (a+b x)^4}-\frac{1}{12 a^3 b (a+b x)^3}-\frac{3}{32 a^4 b (a+b x)^2}-\frac{1}{8 a^5 b (a+b x)}+\frac{5 \int \frac{1}{a^2-b^2 x^2} \, dx}{32 a^5}\\ &=\frac{1}{32 a^5 b (a-b x)}-\frac{1}{16 a^2 b (a+b x)^4}-\frac{1}{12 a^3 b (a+b x)^3}-\frac{3}{32 a^4 b (a+b x)^2}-\frac{1}{8 a^5 b (a+b x)}+\frac{5 \tanh ^{-1}\left (\frac{b x}{a}\right )}{32 a^6 b}\\ \end{align*}

Mathematica [A]  time = 0.028483, size = 112, normalized size = 1.08 \[ \frac{70 a^3 b^2 x^2+90 a^2 b^3 x^3-30 a^4 b x-64 a^5+30 a b^4 x^4-15 (a-b x) (a+b x)^4 \log (a-b x)+15 (a-b x) (a+b x)^4 \log (a+b x)}{192 a^6 b (a-b x) (a+b x)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-64*a^5 - 30*a^4*b*x + 70*a^3*b^2*x^2 + 90*a^2*b^3*x^3 + 30*a*b^4*x^4 - 15*(a - b*x)*(a + b*x)^4*Log[a - b*x]
+ 15*(a - b*x)*(a + b*x)^4*Log[a + b*x])/(192*a^6*b*(a - b*x)*(a + b*x)^4)

________________________________________________________________________________________

Maple [A]  time = 0.051, size = 109, normalized size = 1.1 \begin{align*}{\frac{5\,\ln \left ( bx+a \right ) }{64\,{a}^{6}b}}-{\frac{1}{8\,{a}^{5}b \left ( bx+a \right ) }}-{\frac{3}{32\,{a}^{4}b \left ( bx+a \right ) ^{2}}}-{\frac{1}{12\,{a}^{3}b \left ( bx+a \right ) ^{3}}}-{\frac{1}{16\,b{a}^{2} \left ( bx+a \right ) ^{4}}}-{\frac{5\,\ln \left ( bx-a \right ) }{64\,{a}^{6}b}}-{\frac{1}{32\,{a}^{5}b \left ( bx-a \right ) }} \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)^2,x)

[Out]

5/64/a^6/b*ln(b*x+a)-1/8/a^5/b/(b*x+a)-3/32/a^4/b/(b*x+a)^2-1/12/a^3/b/(b*x+a)^3-1/16/a^2/b/(b*x+a)^4-5/64/a^6
/b*ln(b*x-a)-1/32/a^5/b/(b*x-a)

________________________________________________________________________________________

Maxima [A]  time = 1.06657, size = 182, normalized size = 1.75 \begin{align*} -\frac{15 \, b^{4} x^{4} + 45 \, a b^{3} x^{3} + 35 \, a^{2} b^{2} x^{2} - 15 \, a^{3} b x - 32 \, a^{4}}{96 \,{\left (a^{5} b^{6} x^{5} + 3 \, a^{6} b^{5} x^{4} + 2 \, a^{7} b^{4} x^{3} - 2 \, a^{8} b^{3} x^{2} - 3 \, a^{9} b^{2} x - a^{10} b\right )}} + \frac{5 \, \log \left (b x + a\right )}{64 \, a^{6} b} - \frac{5 \, \log \left (b x - a\right )}{64 \, a^{6} 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)^2,x, algorithm="maxima")

[Out]

-1/96*(15*b^4*x^4 + 45*a*b^3*x^3 + 35*a^2*b^2*x^2 - 15*a^3*b*x - 32*a^4)/(a^5*b^6*x^5 + 3*a^6*b^5*x^4 + 2*a^7*
b^4*x^3 - 2*a^8*b^3*x^2 - 3*a^9*b^2*x - a^10*b) + 5/64*log(b*x + a)/(a^6*b) - 5/64*log(b*x - a)/(a^6*b)

________________________________________________________________________________________

Fricas [B]  time = 1.76579, size = 471, normalized size = 4.53 \begin{align*} -\frac{30 \, a b^{4} x^{4} + 90 \, a^{2} b^{3} x^{3} + 70 \, a^{3} b^{2} x^{2} - 30 \, a^{4} b x - 64 \, a^{5} - 15 \,{\left (b^{5} x^{5} + 3 \, a b^{4} x^{4} + 2 \, a^{2} b^{3} x^{3} - 2 \, a^{3} b^{2} x^{2} - 3 \, a^{4} b x - a^{5}\right )} \log \left (b x + a\right ) + 15 \,{\left (b^{5} x^{5} + 3 \, a b^{4} x^{4} + 2 \, a^{2} b^{3} x^{3} - 2 \, a^{3} b^{2} x^{2} - 3 \, a^{4} b x - a^{5}\right )} \log \left (b x - a\right )}{192 \,{\left (a^{6} b^{6} x^{5} + 3 \, a^{7} b^{5} x^{4} + 2 \, a^{8} b^{4} x^{3} - 2 \, a^{9} b^{3} x^{2} - 3 \, a^{10} b^{2} x - a^{11} 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)^2,x, algorithm="fricas")

[Out]

-1/192*(30*a*b^4*x^4 + 90*a^2*b^3*x^3 + 70*a^3*b^2*x^2 - 30*a^4*b*x - 64*a^5 - 15*(b^5*x^5 + 3*a*b^4*x^4 + 2*a
^2*b^3*x^3 - 2*a^3*b^2*x^2 - 3*a^4*b*x - a^5)*log(b*x + a) + 15*(b^5*x^5 + 3*a*b^4*x^4 + 2*a^2*b^3*x^3 - 2*a^3
*b^2*x^2 - 3*a^4*b*x - a^5)*log(b*x - a))/(a^6*b^6*x^5 + 3*a^7*b^5*x^4 + 2*a^8*b^4*x^3 - 2*a^9*b^3*x^2 - 3*a^1
0*b^2*x - a^11*b)

________________________________________________________________________________________

Sympy [A]  time = 0.864606, size = 133, normalized size = 1.28 \begin{align*} - \frac{- 32 a^{4} - 15 a^{3} b x + 35 a^{2} b^{2} x^{2} + 45 a b^{3} x^{3} + 15 b^{4} x^{4}}{- 96 a^{10} b - 288 a^{9} b^{2} x - 192 a^{8} b^{3} x^{2} + 192 a^{7} b^{4} x^{3} + 288 a^{6} b^{5} x^{4} + 96 a^{5} b^{6} x^{5}} + \frac{- \frac{5 \log{\left (- \frac{a}{b} + x \right )}}{64} + \frac{5 \log{\left (\frac{a}{b} + x \right )}}{64}}{a^{6} 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)**2,x)

[Out]

-(-32*a**4 - 15*a**3*b*x + 35*a**2*b**2*x**2 + 45*a*b**3*x**3 + 15*b**4*x**4)/(-96*a**10*b - 288*a**9*b**2*x -
192*a**8*b**3*x**2 + 192*a**7*b**4*x**3 + 288*a**6*b**5*x**4 + 96*a**5*b**6*x**5) + (-5*log(-a/b + x)/64 + 5*
log(a/b + x)/64)/(a**6*b)

________________________________________________________________________________________

Giac [A]  time = 1.18172, size = 136, normalized size = 1.31 \begin{align*} \frac{5 \, \log \left ({\left | b x + a \right |}\right )}{64 \, a^{6} b} - \frac{5 \, \log \left ({\left | b x - a \right |}\right )}{64 \, a^{6} b} - \frac{15 \, a b^{4} x^{4} + 45 \, a^{2} b^{3} x^{3} + 35 \, a^{3} b^{2} x^{2} - 15 \, a^{4} b x - 32 \, a^{5}}{96 \,{\left (b x + a\right )}^{4}{\left (b x - a\right )} a^{6} 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)^2,x, algorithm="giac")

[Out]

5/64*log(abs(b*x + a))/(a^6*b) - 5/64*log(abs(b*x - a))/(a^6*b) - 1/96*(15*a*b^4*x^4 + 45*a^2*b^3*x^3 + 35*a^3
*b^2*x^2 - 15*a^4*b*x - 32*a^5)/((b*x + a)^4*(b*x - a)*a^6*b)