3.8.66 \(\int \frac {1}{(a+b x)^3 (a^2-b^2 x^2)^2} \, dx\) [766]

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

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 104, 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 = {641, 46, 214} \begin {gather*} \frac {5 \tanh ^{-1}\left (\frac {b x}{a}\right )}{32 a^6 b}+\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} \end {gather*}

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 46

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] && Lt
Q[m + n + 2, 0])

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 641

Int[((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a/d + (c/e)*x)^
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)^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*}

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Mathematica [A]
time = 0.02, size = 112, normalized size = 1.08 \begin {gather*} \frac {-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} \end {gather*}

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)

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Maple [A]
time = 0.47, size = 107, normalized size = 1.03

method	result	size
norman	\(\frac {\frac {27 x}{32 a^{2}}+\frac {33 b \,x^{2}}{32 a^{3}}-\frac {19 b^{2} x^{3}}{96 a^{4}}-\frac {27 b^{3} x^{4}}{32 a^{5}}-\frac {b^{4} x^{5}}{3 a^{6}}}{\left (-b x +a \right ) \left (b x +a \right )^{4}}-\frac {5 \ln \left (-b x +a \right )}{64 a^{6} b}+\frac {5 \ln \left (b x +a \right )}{64 a^{6} b}\)	\(96\)
risch	\(\frac {\frac {5 b^{3} x^{4}}{32 a^{5}}+\frac {15 b^{2} x^{3}}{32 a^{4}}+\frac {35 b \,x^{2}}{96 a^{3}}-\frac {5 x}{32 a^{2}}-\frac {1}{3 b a}}{\left (b x +a \right )^{3} \left (-b^{2} x^{2}+a^{2}\right )}-\frac {5 \ln \left (-b x +a \right )}{64 a^{6} b}+\frac {5 \ln \left (b x +a \right )}{64 a^{6} b}\)	\(99\)
default	\(\frac {5 \ln \left (b x +a \right )}{64 a^{6} b}-\frac {1}{8 a^{5} b \left (b x +a \right )}-\frac {3}{32 a^{4} b \left (b x +a \right )^{2}}-\frac {1}{12 a^{3} b \left (b x +a \right )^{3}}-\frac {1}{16 a^{2} b \left (b x +a \right )^{4}}-\frac {5 \ln \left (-b x +a \right )}{64 a^{6} b}+\frac {1}{32 a^{5} b \left (-b x +a \right )}\)	\(107\)

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,method=_RETURNVERBOSE)

[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)

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Maxima [A]
time = 0.35, size = 135, normalized size = 1.30 \begin {gather*} -\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 {gather*}

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)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 227 vs. \(2 (93) = 186\).
time = 2.60, size = 227, normalized size = 2.18 \begin {gather*} -\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 {gather*}

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)

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Sympy [A]
time = 0.30, size = 133, normalized size = 1.28 \begin {gather*} \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 {gather*}

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 - 1
92*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*lo
g(a/b + x)/64)/(a**6*b)

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Giac [A]
time = 1.08, size = 101, normalized size = 0.97 \begin {gather*} \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 {gather*}

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)

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Mupad [B]
time = 0.47, size = 115, normalized size = 1.11 \begin {gather*} \frac {\frac {35\,b\,x^2}{96\,a^3}-\frac {1}{3\,a\,b}-\frac {5\,x}{32\,a^2}+\frac {15\,b^2\,x^3}{32\,a^4}+\frac {5\,b^3\,x^4}{32\,a^5}}{a^5+3\,a^4\,b\,x+2\,a^3\,b^2\,x^2-2\,a^2\,b^3\,x^3-3\,a\,b^4\,x^4-b^5\,x^5}+\frac {5\,\mathrm {atanh}\left (\frac {b\,x}{a}\right )}{32\,a^6\,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)^3),x)

[Out]

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

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