3.8.75 \(\int \frac {1}{(a+b x) (a^2-b^2 x^2)^3} \, dx\) [775]

Optimal. Leaf size=105 \[ \frac {1}{32 a^4 b (a-b x)^2}+\frac {1}{8 a^5 b (a-b x)}-\frac {1}{24 a^3 b (a+b x)^3}-\frac {3}{32 a^4 b (a+b x)^2}-\frac {3}{16 a^5 b (a+b x)}+\frac {5 \tanh ^{-1}\left (\frac {b x}{a}\right )}{16 a^6 b} \]

[Out]

1/32/a^4/b/(-b*x+a)^2+1/8/a^5/b/(-b*x+a)-1/24/a^3/b/(b*x+a)^3-3/32/a^4/b/(b*x+a)^2-3/16/a^5/b/(b*x+a)+5/16*arc
tanh(b*x/a)/a^6/b

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Rubi [A]
time = 0.05, antiderivative size = 105, 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 )}{16 a^6 b}+\frac {1}{8 a^5 b (a-b x)}-\frac {3}{16 a^5 b (a+b x)}+\frac {1}{32 a^4 b (a-b x)^2}-\frac {3}{32 a^4 b (a+b x)^2}-\frac {1}{24 a^3 b (a+b x)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

1/(32*a^4*b*(a - b*x)^2) + 1/(8*a^5*b*(a - b*x)) - 1/(24*a^3*b*(a + b*x)^3) - 3/(32*a^4*b*(a + b*x)^2) - 3/(16
*a^5*b*(a + b*x)) + (5*ArcTanh[(b*x)/a])/(16*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) \left (a^2-b^2 x^2\right )^3} \, dx &=\int \frac {1}{(a-b x)^3 (a+b x)^4} \, dx\\ &=\int \left (\frac {1}{16 a^4 (a-b x)^3}+\frac {1}{8 a^5 (a-b x)^2}+\frac {1}{8 a^3 (a+b x)^4}+\frac {3}{16 a^4 (a+b x)^3}+\frac {3}{16 a^5 (a+b x)^2}+\frac {5}{16 a^5 \left (a^2-b^2 x^2\right )}\right ) \, dx\\ &=\frac {1}{32 a^4 b (a-b x)^2}+\frac {1}{8 a^5 b (a-b x)}-\frac {1}{24 a^3 b (a+b x)^3}-\frac {3}{32 a^4 b (a+b x)^2}-\frac {3}{16 a^5 b (a+b x)}+\frac {5 \int \frac {1}{a^2-b^2 x^2} \, dx}{16 a^5}\\ &=\frac {1}{32 a^4 b (a-b x)^2}+\frac {1}{8 a^5 b (a-b x)}-\frac {1}{24 a^3 b (a+b x)^3}-\frac {3}{32 a^4 b (a+b x)^2}-\frac {3}{16 a^5 b (a+b x)}+\frac {5 \tanh ^{-1}\left (\frac {b x}{a}\right )}{16 a^6 b}\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 87, normalized size = 0.83 \begin {gather*} \frac {-\frac {2 a \left (8 a^4-25 a^3 b x-25 a^2 b^2 x^2+15 a b^3 x^3+15 b^4 x^4\right )}{(a-b x)^2 (a+b x)^3}-15 \log (a-b x)+15 \log (a+b x)}{96 a^6 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*a*(8*a^4 - 25*a^3*b*x - 25*a^2*b^2*x^2 + 15*a*b^3*x^3 + 15*b^4*x^4))/((a - b*x)^2*(a + b*x)^3) - 15*Log[a
- b*x] + 15*Log[a + b*x])/(96*a^6*b)

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Maple [A]
time = 0.46, size = 108, normalized size = 1.03

method	result	size
norman	\(\frac {\frac {7}{48 b a}-\frac {15 b^{2} x^{3}}{16 a^{4}}+\frac {5 b^{4} x^{5}}{16 a^{6}}+\frac {5 x}{6 a^{2}}-\frac {5 b \,x^{2}}{48 a^{3}}}{\left (-b x +a \right )^{2} \left (b x +a \right )^{3}}-\frac {5 \ln \left (-b x +a \right )}{32 a^{6} b}+\frac {5 \ln \left (b x +a \right )}{32 a^{6} b}\)	\(93\)
risch	\(\frac {-\frac {5 b^{3} x^{4}}{16 a^{5}}-\frac {5 b^{2} x^{3}}{16 a^{4}}+\frac {25 b \,x^{2}}{48 a^{3}}+\frac {25 x}{48 a^{2}}-\frac {1}{6 b a}}{\left (b x +a \right ) \left (-b^{2} x^{2}+a^{2}\right )^{2}}-\frac {5 \ln \left (-b x +a \right )}{32 a^{6} b}+\frac {5 \ln \left (b x +a \right )}{32 a^{6} b}\)	\(99\)
default	\(\frac {5 \ln \left (b x +a \right )}{32 a^{6} b}-\frac {3}{16 a^{5} b \left (b x +a \right )}-\frac {3}{32 a^{4} b \left (b x +a \right )^{2}}-\frac {1}{24 a^{3} b \left (b x +a \right )^{3}}-\frac {5 \ln \left (-b x +a \right )}{32 a^{6} b}+\frac {1}{8 a^{5} b \left (-b x +a \right )}+\frac {1}{32 a^{4} b \left (-b x +a \right )^{2}}\)	\(108\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/32/a^6/b*ln(b*x+a)-3/16/a^5/b/(b*x+a)-3/32/a^4/b/(b*x+a)^2-1/24/a^3/b/(b*x+a)^3-5/32/a^6/b*ln(-b*x+a)+1/8/a^
5/b/(-b*x+a)+1/32/a^4/b/(-b*x+a)^2

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Maxima [A]
time = 0.31, size = 132, normalized size = 1.26 \begin {gather*} -\frac {15 \, b^{4} x^{4} + 15 \, a b^{3} x^{3} - 25 \, a^{2} b^{2} x^{2} - 25 \, a^{3} b x + 8 \, a^{4}}{48 \, {\left (a^{5} b^{6} x^{5} + a^{6} b^{5} x^{4} - 2 \, a^{7} b^{4} x^{3} - 2 \, a^{8} b^{3} x^{2} + a^{9} b^{2} x + a^{10} b\right )}} + \frac {5 \, \log \left (b x + a\right )}{32 \, a^{6} b} - \frac {5 \, \log \left (b x - a\right )}{32 \, a^{6} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48*(15*b^4*x^4 + 15*a*b^3*x^3 - 25*a^2*b^2*x^2 - 25*a^3*b*x + 8*a^4)/(a^5*b^6*x^5 + a^6*b^5*x^4 - 2*a^7*b^4
*x^3 - 2*a^8*b^3*x^2 + a^9*b^2*x + a^10*b) + 5/32*log(b*x + a)/(a^6*b) - 5/32*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. 216 vs. \(2 (95) = 190\).
time = 3.29, size = 216, normalized size = 2.06 \begin {gather*} -\frac {30 \, a b^{4} x^{4} + 30 \, a^{2} b^{3} x^{3} - 50 \, a^{3} b^{2} x^{2} - 50 \, a^{4} b x + 16 \, a^{5} - 15 \, {\left (b^{5} x^{5} + a b^{4} x^{4} - 2 \, a^{2} b^{3} x^{3} - 2 \, a^{3} b^{2} x^{2} + a^{4} b x + a^{5}\right )} \log \left (b x + a\right ) + 15 \, {\left (b^{5} x^{5} + a b^{4} x^{4} - 2 \, a^{2} b^{3} x^{3} - 2 \, a^{3} b^{2} x^{2} + a^{4} b x + a^{5}\right )} \log \left (b x - a\right )}{96 \, {\left (a^{6} b^{6} x^{5} + a^{7} b^{5} x^{4} - 2 \, a^{8} b^{4} x^{3} - 2 \, a^{9} b^{3} x^{2} + 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)/(-b^2*x^2+a^2)^3,x, algorithm="fricas")

[Out]

-1/96*(30*a*b^4*x^4 + 30*a^2*b^3*x^3 - 50*a^3*b^2*x^2 - 50*a^4*b*x + 16*a^5 - 15*(b^5*x^5 + a*b^4*x^4 - 2*a^2*
b^3*x^3 - 2*a^3*b^2*x^2 + a^4*b*x + a^5)*log(b*x + a) + 15*(b^5*x^5 + a*b^4*x^4 - 2*a^2*b^3*x^3 - 2*a^3*b^2*x^
2 + a^4*b*x + a^5)*log(b*x - a))/(a^6*b^6*x^5 + a^7*b^5*x^4 - 2*a^8*b^4*x^3 - 2*a^9*b^3*x^2 + a^10*b^2*x + a^1
1*b)

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Sympy [A]
time = 0.29, size = 134, normalized size = 1.28 \begin {gather*} - \frac {8 a^{4} - 25 a^{3} b x - 25 a^{2} b^{2} x^{2} + 15 a b^{3} x^{3} + 15 b^{4} x^{4}}{48 a^{10} b + 48 a^{9} b^{2} x - 96 a^{8} b^{3} x^{2} - 96 a^{7} b^{4} x^{3} + 48 a^{6} b^{5} x^{4} + 48 a^{5} b^{6} x^{5}} - \frac {\frac {5 \log {\left (- \frac {a}{b} + x \right )}}{32} - \frac {5 \log {\left (\frac {a}{b} + x \right )}}{32}}{a^{6} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(8*a**4 - 25*a**3*b*x - 25*a**2*b**2*x**2 + 15*a*b**3*x**3 + 15*b**4*x**4)/(48*a**10*b + 48*a**9*b**2*x - 96*
a**8*b**3*x**2 - 96*a**7*b**4*x**3 + 48*a**6*b**5*x**4 + 48*a**5*b**6*x**5) - (5*log(-a/b + x)/32 - 5*log(a/b
+ x)/32)/(a**6*b)

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Giac [A]
time = 1.21, size = 101, normalized size = 0.96 \begin {gather*} \frac {5 \, \log \left ({\left | b x + a \right |}\right )}{32 \, a^{6} b} - \frac {5 \, \log \left ({\left | b x - a \right |}\right )}{32 \, a^{6} b} - \frac {15 \, a b^{4} x^{4} + 15 \, a^{2} b^{3} x^{3} - 25 \, a^{3} b^{2} x^{2} - 25 \, a^{4} b x + 8 \, a^{5}}{48 \, {\left (b x + a\right )}^{3} {\left (b x - a\right )}^{2} a^{6} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/32*log(abs(b*x + a))/(a^6*b) - 5/32*log(abs(b*x - a))/(a^6*b) - 1/48*(15*a*b^4*x^4 + 15*a^2*b^3*x^3 - 25*a^3
*b^2*x^2 - 25*a^4*b*x + 8*a^5)/((b*x + a)^3*(b*x - a)^2*a^6*b)

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Mupad [B]
time = 0.11, size = 113, normalized size = 1.08 \begin {gather*} \frac {5\,\mathrm {atanh}\left (\frac {b\,x}{a}\right )}{16\,a^6\,b}-\frac {\frac {1}{6\,a\,b}-\frac {25\,x}{48\,a^2}-\frac {25\,b\,x^2}{48\,a^3}+\frac {5\,b^2\,x^3}{16\,a^4}+\frac {5\,b^3\,x^4}{16\,a^5}}{a^5+a^4\,b\,x-2\,a^3\,b^2\,x^2-2\,a^2\,b^3\,x^3+a\,b^4\,x^4+b^5\,x^5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(5*atanh((b*x)/a))/(16*a^6*b) - (1/(6*a*b) - (25*x)/(48*a^2) - (25*b*x^2)/(48*a^3) + (5*b^2*x^3)/(16*a^4) + (5
*b^3*x^4)/(16*a^5))/(a^5 + b^5*x^5 + a*b^4*x^4 - 2*a^3*b^2*x^2 - 2*a^2*b^3*x^3 + a^4*b*x)

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