3.776 $$\int \frac{1}{(a+b x)^2 (a^2-b^2 x^2)^3} \, dx$$

Optimal. Leaf size=122 $\frac{5}{64 a^6 b (a-b x)}-\frac{5}{32 a^6 b (a+b x)}+\frac{1}{64 a^5 b (a-b x)^2}-\frac{3}{32 a^5 b (a+b x)^2}-\frac{1}{16 a^4 b (a+b x)^3}-\frac{1}{32 a^3 b (a+b x)^4}+\frac{15 \tanh ^{-1}\left (\frac{b x}{a}\right )}{64 a^7 b}$

[Out]

1/(64*a^5*b*(a - b*x)^2) + 5/(64*a^6*b*(a - b*x)) - 1/(32*a^3*b*(a + b*x)^4) - 1/(16*a^4*b*(a + b*x)^3) - 3/(3
2*a^5*b*(a + b*x)^2) - 5/(32*a^6*b*(a + b*x)) + (15*ArcTanh[(b*x)/a])/(64*a^7*b)

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Rubi [A]  time = 0.0788462, antiderivative size = 122, 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} $\frac{5}{64 a^6 b (a-b x)}-\frac{5}{32 a^6 b (a+b x)}+\frac{1}{64 a^5 b (a-b x)^2}-\frac{3}{32 a^5 b (a+b x)^2}-\frac{1}{16 a^4 b (a+b x)^3}-\frac{1}{32 a^3 b (a+b x)^4}+\frac{15 \tanh ^{-1}\left (\frac{b x}{a}\right )}{64 a^7 b}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

1/(64*a^5*b*(a - b*x)^2) + 5/(64*a^6*b*(a - b*x)) - 1/(32*a^3*b*(a + b*x)^4) - 1/(16*a^4*b*(a + b*x)^3) - 3/(3
2*a^5*b*(a + b*x)^2) - 5/(32*a^6*b*(a + b*x)) + (15*ArcTanh[(b*x)/a])/(64*a^7*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)^2 \left (a^2-b^2 x^2\right )^3} \, dx &=\int \frac{1}{(a-b x)^3 (a+b x)^5} \, dx\\ &=\int \left (\frac{1}{32 a^5 (a-b x)^3}+\frac{5}{64 a^6 (a-b x)^2}+\frac{1}{8 a^3 (a+b x)^5}+\frac{3}{16 a^4 (a+b x)^4}+\frac{3}{16 a^5 (a+b x)^3}+\frac{5}{32 a^6 (a+b x)^2}+\frac{15}{64 a^6 \left (a^2-b^2 x^2\right )}\right ) \, dx\\ &=\frac{1}{64 a^5 b (a-b x)^2}+\frac{5}{64 a^6 b (a-b x)}-\frac{1}{32 a^3 b (a+b x)^4}-\frac{1}{16 a^4 b (a+b x)^3}-\frac{3}{32 a^5 b (a+b x)^2}-\frac{5}{32 a^6 b (a+b x)}+\frac{15 \int \frac{1}{a^2-b^2 x^2} \, dx}{64 a^6}\\ &=\frac{1}{64 a^5 b (a-b x)^2}+\frac{5}{64 a^6 b (a-b x)}-\frac{1}{32 a^3 b (a+b x)^4}-\frac{1}{16 a^4 b (a+b x)^3}-\frac{3}{32 a^5 b (a+b x)^2}-\frac{5}{32 a^6 b (a+b x)}+\frac{15 \tanh ^{-1}\left (\frac{b x}{a}\right )}{64 a^7 b}\\ \end{align*}

Mathematica [A]  time = 0.0531594, size = 98, normalized size = 0.8 $\frac{\frac{2 a \left (50 a^3 b^2 x^2+10 a^2 b^3 x^3+17 a^4 b x-16 a^5-30 a b^4 x^4-15 b^5 x^5\right )}{(a-b x)^2 (a+b x)^4}-15 \log (a-b x)+15 \log (a+b x)}{128 a^7 b}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

((2*a*(-16*a^5 + 17*a^4*b*x + 50*a^3*b^2*x^2 + 10*a^2*b^3*x^3 - 30*a*b^4*x^4 - 15*b^5*x^5))/((a - b*x)^2*(a +
b*x)^4) - 15*Log[a - b*x] + 15*Log[a + b*x])/(128*a^7*b)

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Maple [A]  time = 0.049, size = 126, normalized size = 1. \begin{align*}{\frac{15\,\ln \left ( bx+a \right ) }{128\,b{a}^{7}}}-{\frac{5}{32\,{a}^{6}b \left ( bx+a \right ) }}-{\frac{3}{32\,{a}^{5}b \left ( bx+a \right ) ^{2}}}-{\frac{1}{16\,{a}^{4}b \left ( bx+a \right ) ^{3}}}-{\frac{1}{32\,{a}^{3}b \left ( bx+a \right ) ^{4}}}-{\frac{15\,\ln \left ( bx-a \right ) }{128\,b{a}^{7}}}-{\frac{5}{64\,{a}^{6}b \left ( bx-a \right ) }}+{\frac{1}{64\,{a}^{5}b \left ( bx-a \right ) ^{2}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

15/128/a^7/b*ln(b*x+a)-5/32/a^6/b/(b*x+a)-3/32/a^5/b/(b*x+a)^2-1/16/a^4/b/(b*x+a)^3-1/32/a^3/b/(b*x+a)^4-15/12
8/a^7/b*ln(b*x-a)-5/64/a^6/b/(b*x-a)+1/64/a^5/b/(b*x-a)^2

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Maxima [A]  time = 1.31346, size = 211, normalized size = 1.73 \begin{align*} -\frac{15 \, b^{5} x^{5} + 30 \, a b^{4} x^{4} - 10 \, a^{2} b^{3} x^{3} - 50 \, a^{3} b^{2} x^{2} - 17 \, a^{4} b x + 16 \, a^{5}}{64 \,{\left (a^{6} b^{7} x^{6} + 2 \, a^{7} b^{6} x^{5} - a^{8} b^{5} x^{4} - 4 \, a^{9} b^{4} x^{3} - a^{10} b^{3} x^{2} + 2 \, a^{11} b^{2} x + a^{12} b\right )}} + \frac{15 \, \log \left (b x + a\right )}{128 \, a^{7} b} - \frac{15 \, \log \left (b x - a\right )}{128 \, a^{7} b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/64*(15*b^5*x^5 + 30*a*b^4*x^4 - 10*a^2*b^3*x^3 - 50*a^3*b^2*x^2 - 17*a^4*b*x + 16*a^5)/(a^6*b^7*x^6 + 2*a^7
*b^6*x^5 - a^8*b^5*x^4 - 4*a^9*b^4*x^3 - a^10*b^3*x^2 + 2*a^11*b^2*x + a^12*b) + 15/128*log(b*x + a)/(a^7*b) -
15/128*log(b*x - a)/(a^7*b)

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Fricas [B]  time = 1.70633, size = 547, normalized size = 4.48 \begin{align*} -\frac{30 \, a b^{5} x^{5} + 60 \, a^{2} b^{4} x^{4} - 20 \, a^{3} b^{3} x^{3} - 100 \, a^{4} b^{2} x^{2} - 34 \, a^{5} b x + 32 \, a^{6} - 15 \,{\left (b^{6} x^{6} + 2 \, a b^{5} x^{5} - a^{2} b^{4} x^{4} - 4 \, a^{3} b^{3} x^{3} - a^{4} b^{2} x^{2} + 2 \, a^{5} b x + a^{6}\right )} \log \left (b x + a\right ) + 15 \,{\left (b^{6} x^{6} + 2 \, a b^{5} x^{5} - a^{2} b^{4} x^{4} - 4 \, a^{3} b^{3} x^{3} - a^{4} b^{2} x^{2} + 2 \, a^{5} b x + a^{6}\right )} \log \left (b x - a\right )}{128 \,{\left (a^{7} b^{7} x^{6} + 2 \, a^{8} b^{6} x^{5} - a^{9} b^{5} x^{4} - 4 \, a^{10} b^{4} x^{3} - a^{11} b^{3} x^{2} + 2 \, a^{12} b^{2} x + a^{13} b\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/128*(30*a*b^5*x^5 + 60*a^2*b^4*x^4 - 20*a^3*b^3*x^3 - 100*a^4*b^2*x^2 - 34*a^5*b*x + 32*a^6 - 15*(b^6*x^6 +
2*a*b^5*x^5 - a^2*b^4*x^4 - 4*a^3*b^3*x^3 - a^4*b^2*x^2 + 2*a^5*b*x + a^6)*log(b*x + a) + 15*(b^6*x^6 + 2*a*b
^5*x^5 - a^2*b^4*x^4 - 4*a^3*b^3*x^3 - a^4*b^2*x^2 + 2*a^5*b*x + a^6)*log(b*x - a))/(a^7*b^7*x^6 + 2*a^8*b^6*x
^5 - a^9*b^5*x^4 - 4*a^10*b^4*x^3 - a^11*b^3*x^2 + 2*a^12*b^2*x + a^13*b)

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Sympy [A]  time = 1.04247, size = 158, normalized size = 1.3 \begin{align*} - \frac{16 a^{5} - 17 a^{4} b x - 50 a^{3} b^{2} x^{2} - 10 a^{2} b^{3} x^{3} + 30 a b^{4} x^{4} + 15 b^{5} x^{5}}{64 a^{12} b + 128 a^{11} b^{2} x - 64 a^{10} b^{3} x^{2} - 256 a^{9} b^{4} x^{3} - 64 a^{8} b^{5} x^{4} + 128 a^{7} b^{6} x^{5} + 64 a^{6} b^{7} x^{6}} - \frac{\frac{15 \log{\left (- \frac{a}{b} + x \right )}}{128} - \frac{15 \log{\left (\frac{a}{b} + x \right )}}{128}}{a^{7} b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(16*a**5 - 17*a**4*b*x - 50*a**3*b**2*x**2 - 10*a**2*b**3*x**3 + 30*a*b**4*x**4 + 15*b**5*x**5)/(64*a**12*b +
128*a**11*b**2*x - 64*a**10*b**3*x**2 - 256*a**9*b**4*x**3 - 64*a**8*b**5*x**4 + 128*a**7*b**6*x**5 + 64*a**6
*b**7*x**6) - (15*log(-a/b + x)/128 - 15*log(a/b + x)/128)/(a**7*b)

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Giac [A]  time = 1.20814, size = 169, normalized size = 1.39 \begin{align*} -\frac{15 \, \log \left ({\left | -\frac{2 \, a}{b x + a} + 1 \right |}\right )}{128 \, a^{7} b} + \frac{\frac{24 \, a}{b x + a} - 11}{256 \, a^{7} b{\left (\frac{2 \, a}{b x + a} - 1\right )}^{2}} - \frac{\frac{5 \, a^{6} b^{11}}{b x + a} + \frac{3 \, a^{7} b^{11}}{{\left (b x + a\right )}^{2}} + \frac{2 \, a^{8} b^{11}}{{\left (b x + a\right )}^{3}} + \frac{a^{9} b^{11}}{{\left (b x + a\right )}^{4}}}{32 \, a^{12} b^{12}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-15/128*log(abs(-2*a/(b*x + a) + 1))/(a^7*b) + 1/256*(24*a/(b*x + a) - 11)/(a^7*b*(2*a/(b*x + a) - 1)^2) - 1/3
2*(5*a^6*b^11/(b*x + a) + 3*a^7*b^11/(b*x + a)^2 + 2*a^8*b^11/(b*x + a)^3 + a^9*b^11/(b*x + a)^4)/(a^12*b^12)