3.8.55 \(\int \frac {1}{(a+b x)^3 (a^2-b^2 x^2)} \, dx\) [755]

Optimal. Leaf size=69 \[ -\frac {1}{6 a b (a+b x)^3}-\frac {1}{8 a^2 b (a+b x)^2}-\frac {1}{8 a^3 b (a+b x)}+\frac {\tanh ^{-1}\left (\frac {b x}{a}\right )}{8 a^4 b} \]

[Out]

-1/6/a/b/(b*x+a)^3-1/8/a^2/b/(b*x+a)^2-1/8/a^3/b/(b*x+a)+1/8*arctanh(b*x/a)/a^4/b

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Rubi [A]
time = 0.03, antiderivative size = 69, 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 {\tanh ^{-1}\left (\frac {b x}{a}\right )}{8 a^4 b}-\frac {1}{8 a^3 b (a+b x)}-\frac {1}{8 a^2 b (a+b x)^2}-\frac {1}{6 a b (a+b x)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]
time = 0.02, size = 71, normalized size = 1.03 \begin {gather*} \frac {-2 a \left (10 a^2+9 a b x+3 b^2 x^2\right )-3 (a+b x)^3 \log (a-b x)+3 (a+b x)^3 \log (a+b x)}{48 a^4 b (a+b x)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a*(10*a^2 + 9*a*b*x + 3*b^2*x^2) - 3*(a + b*x)^3*Log[a - b*x] + 3*(a + b*x)^3*Log[a + b*x])/(48*a^4*b*(a +
b*x)^3)

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Maple [A]
time = 0.48, size = 76, normalized size = 1.10

method	result	size
norman	\(\frac {-\frac {5}{12 b a}-\frac {3 x}{8 a^{2}}-\frac {b \,x^{2}}{8 a^{3}}}{\left (b x +a \right )^{3}}-\frac {\ln \left (-b x +a \right )}{16 a^{4} b}+\frac {\ln \left (b x +a \right )}{16 a^{4} b}\)	\(63\)
risch	\(\frac {-\frac {5}{12 b a}-\frac {3 x}{8 a^{2}}-\frac {b \,x^{2}}{8 a^{3}}}{\left (b x +a \right )^{3}}-\frac {\ln \left (-b x +a \right )}{16 a^{4} b}+\frac {\ln \left (b x +a \right )}{16 a^{4} b}\)	\(63\)
default	\(\frac {\ln \left (b x +a \right )}{16 a^{4} b}-\frac {1}{8 a^{3} b \left (b x +a \right )}-\frac {1}{8 a^{2} b \left (b x +a \right )^{2}}-\frac {1}{6 a b \left (b x +a \right )^{3}}-\frac {\ln \left (-b x +a \right )}{16 a^{4} b}\)	\(76\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.33, size = 90, normalized size = 1.30 \begin {gather*} -\frac {3 \, b^{2} x^{2} + 9 \, a b x + 10 \, a^{2}}{24 \, {\left (a^{3} b^{4} x^{3} + 3 \, a^{4} b^{3} x^{2} + 3 \, a^{5} b^{2} x + a^{6} b\right )}} + \frac {\log \left (b x + a\right )}{16 \, a^{4} b} - \frac {\log \left (b x - a\right )}{16 \, a^{4} 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),x, algorithm="maxima")

[Out]

-1/24*(3*b^2*x^2 + 9*a*b*x + 10*a^2)/(a^3*b^4*x^3 + 3*a^4*b^3*x^2 + 3*a^5*b^2*x + a^6*b) + 1/16*log(b*x + a)/(
a^4*b) - 1/16*log(b*x - a)/(a^4*b)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 134 vs. \(2 (61) = 122\).
time = 3.16, size = 134, normalized size = 1.94 \begin {gather*} -\frac {6 \, a b^{2} x^{2} + 18 \, a^{2} b x + 20 \, a^{3} - 3 \, {\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3}\right )} \log \left (b x + a\right ) + 3 \, {\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3}\right )} \log \left (b x - a\right )}{48 \, {\left (a^{4} b^{4} x^{3} + 3 \, a^{5} b^{3} x^{2} + 3 \, a^{6} b^{2} x + a^{7} 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),x, algorithm="fricas")

[Out]

-1/48*(6*a*b^2*x^2 + 18*a^2*b*x + 20*a^3 - 3*(b^3*x^3 + 3*a*b^2*x^2 + 3*a^2*b*x + a^3)*log(b*x + a) + 3*(b^3*x
^3 + 3*a*b^2*x^2 + 3*a^2*b*x + a^3)*log(b*x - a))/(a^4*b^4*x^3 + 3*a^5*b^3*x^2 + 3*a^6*b^2*x + a^7*b)

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Sympy [A]
time = 0.27, size = 83, normalized size = 1.20 \begin {gather*} - \frac {10 a^{2} + 9 a b x + 3 b^{2} x^{2}}{24 a^{6} b + 72 a^{5} b^{2} x + 72 a^{4} b^{3} x^{2} + 24 a^{3} b^{4} x^{3}} - \frac {\frac {\log {\left (- \frac {a}{b} + x \right )}}{16} - \frac {\log {\left (\frac {a}{b} + x \right )}}{16}}{a^{4} 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),x)

[Out]

-(10*a**2 + 9*a*b*x + 3*b**2*x**2)/(24*a**6*b + 72*a**5*b**2*x + 72*a**4*b**3*x**2 + 24*a**3*b**4*x**3) - (log
(-a/b + x)/16 - log(a/b + x)/16)/(a**4*b)

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Giac [A]
time = 1.00, size = 70, normalized size = 1.01 \begin {gather*} \frac {\log \left ({\left | b x + a \right |}\right )}{16 \, a^{4} b} - \frac {\log \left ({\left | b x - a \right |}\right )}{16 \, a^{4} b} - \frac {3 \, a b^{2} x^{2} + 9 \, a^{2} b x + 10 \, a^{3}}{24 \, {\left (b x + a\right )}^{3} a^{4} 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),x, algorithm="giac")

[Out]

1/16*log(abs(b*x + a))/(a^4*b) - 1/16*log(abs(b*x - a))/(a^4*b) - 1/24*(3*a*b^2*x^2 + 9*a^2*b*x + 10*a^3)/((b*
x + a)^3*a^4*b)

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Mupad [B]
time = 0.09, size = 71, normalized size = 1.03 \begin {gather*} \frac {\mathrm {atanh}\left (\frac {b\,x}{a}\right )}{8\,a^4\,b}-\frac {\frac {3\,x}{8\,a^2}+\frac {5}{12\,a\,b}+\frac {b\,x^2}{8\,a^3}}{a^3+3\,a^2\,b\,x+3\,a\,b^2\,x^2+b^3\,x^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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