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

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

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {266, 44} \[ \frac {b}{a^3 \left (a-b x^2\right )}+\frac {b}{4 a^2 \left (a-b x^2\right )^2}-\frac {3 b \log \left (a-b x^2\right )}{2 a^4}+\frac {3 b \log (x)}{a^4}-\frac {1}{2 a^3 x^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(a - b*x^2)^3),x]

[Out]

-1/(2*a^3*x^2) + b/(4*a^2*(a - b*x^2)^2) + b/(a^3*(a - b*x^2)) + (3*b*Log[x])/a^4 - (3*b*Log[a - b*x^2])/(2*a^
4)

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 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps

\begin {align*} \int \frac {1}{x^3 \left (a-b x^2\right )^3} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{x^2 (a-b x)^3} \, dx,x,x^2\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \left (\frac {1}{a^3 x^2}+\frac {3 b}{a^4 x}+\frac {b^2}{a^2 (a-b x)^3}+\frac {2 b^2}{a^3 (a-b x)^2}+\frac {3 b^2}{a^4 (a-b x)}\right ) \, dx,x,x^2\right )\\ &=-\frac {1}{2 a^3 x^2}+\frac {b}{4 a^2 \left (a-b x^2\right )^2}+\frac {b}{a^3 \left (a-b x^2\right )}+\frac {3 b \log (x)}{a^4}-\frac {3 b \log \left (a-b x^2\right )}{2 a^4}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 60, normalized size = 0.87 \[ \frac {\frac {a \left (-2 a^2+9 a b x^2-6 b^2 x^4\right )}{\left (a x-b x^3\right )^2}-6 b \log \left (a-b x^2\right )+12 b \log (x)}{4 a^4} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*(a - b*x^2)^3),x]

[Out]

((a*(-2*a^2 + 9*a*b*x^2 - 6*b^2*x^4))/(a*x - b*x^3)^2 + 12*b*Log[x] - 6*b*Log[a - b*x^2])/(4*a^4)

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fricas [A]  time = 1.15, size = 121, normalized size = 1.75 \[ -\frac {6 \, a b^{2} x^{4} - 9 \, a^{2} b x^{2} + 2 \, a^{3} + 6 \, {\left (b^{3} x^{6} - 2 \, a b^{2} x^{4} + a^{2} b x^{2}\right )} \log \left (b x^{2} - a\right ) - 12 \, {\left (b^{3} x^{6} - 2 \, a b^{2} x^{4} + a^{2} b x^{2}\right )} \log \relax (x)}{4 \, {\left (a^{4} b^{2} x^{6} - 2 \, a^{5} b x^{4} + a^{6} x^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(6*a*b^2*x^4 - 9*a^2*b*x^2 + 2*a^3 + 6*(b^3*x^6 - 2*a*b^2*x^4 + a^2*b*x^2)*log(b*x^2 - a) - 12*(b^3*x^6 -
2*a*b^2*x^4 + a^2*b*x^2)*log(x))/(a^4*b^2*x^6 - 2*a^5*b*x^4 + a^6*x^2)

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giac [A]  time = 0.63, size = 84, normalized size = 1.22 \[ \frac {3 \, b \log \left (x^{2}\right )}{2 \, a^{4}} - \frac {3 \, b \log \left ({\left | b x^{2} - a \right |}\right )}{2 \, a^{4}} + \frac {9 \, b^{3} x^{4} - 22 \, a b^{2} x^{2} + 14 \, a^{2} b}{4 \, {\left (b x^{2} - a\right )}^{2} a^{4}} - \frac {3 \, b x^{2} + a}{2 \, a^{4} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/2*b*log(x^2)/a^4 - 3/2*b*log(abs(b*x^2 - a))/a^4 + 1/4*(9*b^3*x^4 - 22*a*b^2*x^2 + 14*a^2*b)/((b*x^2 - a)^2*
a^4) - 1/2*(3*b*x^2 + a)/(a^4*x^2)

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maple [A]  time = 0.01, size = 68, normalized size = 0.99 \[ \frac {b}{4 \left (b \,x^{2}-a \right )^{2} a^{2}}-\frac {b}{\left (b \,x^{2}-a \right ) a^{3}}+\frac {3 b \ln \relax (x )}{a^{4}}-\frac {3 b \ln \left (b \,x^{2}-a \right )}{2 a^{4}}-\frac {1}{2 a^{3} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.31, size = 79, normalized size = 1.14 \[ -\frac {6 \, b^{2} x^{4} - 9 \, a b x^{2} + 2 \, a^{2}}{4 \, {\left (a^{3} b^{2} x^{6} - 2 \, a^{4} b x^{4} + a^{5} x^{2}\right )}} - \frac {3 \, b \log \left (b x^{2} - a\right )}{2 \, a^{4}} + \frac {3 \, b \log \left (x^{2}\right )}{2 \, a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 4.61, size = 76, normalized size = 1.10 \[ \frac {3\,b\,\ln \relax (x)}{a^4}-\frac {3\,b\,\ln \left (a-b\,x^2\right )}{2\,a^4}-\frac {\frac {1}{2\,a}-\frac {9\,b\,x^2}{4\,a^2}+\frac {3\,b^2\,x^4}{2\,a^3}}{a^2\,x^2-2\,a\,b\,x^4+b^2\,x^6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3*(a - b*x^2)^3),x)

[Out]

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

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sympy [A]  time = 0.51, size = 78, normalized size = 1.13 \[ - \frac {2 a^{2} - 9 a b x^{2} + 6 b^{2} x^{4}}{4 a^{5} x^{2} - 8 a^{4} b x^{4} + 4 a^{3} b^{2} x^{6}} + \frac {3 b \log {\relax (x )}}{a^{4}} - \frac {3 b \log {\left (- \frac {a}{b} + x^{2} \right )}}{2 a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*a**2 - 9*a*b*x**2 + 6*b**2*x**4)/(4*a**5*x**2 - 8*a**4*b*x**4 + 4*a**3*b**2*x**6) + 3*b*log(x)/a**4 - 3*b*
log(-a/b + x**2)/(2*a**4)

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