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

Optimal. Leaf size=57 \[ -\frac {\log \left (a-b x^2\right )}{2 a^3}+\frac {\log (x)}{a^3}+\frac {1}{2 a^2 \left (a-b x^2\right )}+\frac {1}{4 a \left (a-b x^2\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 57, 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} \[ \frac {1}{2 a^2 \left (a-b x^2\right )}-\frac {\log \left (a-b x^2\right )}{2 a^3}+\frac {\log (x)}{a^3}+\frac {1}{4 a \left (a-b x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(x*(a - b*x^2)^3),x]

[Out]

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

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

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Mathematica [A]  time = 0.04, size = 45, normalized size = 0.79 \[ \frac {\frac {a \left (3 a-2 b x^2\right )}{\left (a-b x^2\right )^2}-2 \log \left (a-b x^2\right )+4 \log (x)}{4 a^3} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(a - b*x^2)^3),x]

[Out]

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

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fricas [A]  time = 0.73, size = 92, normalized size = 1.61 \[ -\frac {2 \, a b x^{2} - 3 \, a^{2} + 2 \, {\left (b^{2} x^{4} - 2 \, a b x^{2} + a^{2}\right )} \log \left (b x^{2} - a\right ) - 4 \, {\left (b^{2} x^{4} - 2 \, a b x^{2} + a^{2}\right )} \log \relax (x)}{4 \, {\left (a^{3} b^{2} x^{4} - 2 \, a^{4} b x^{2} + a^{5}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.63, size = 63, normalized size = 1.11 \[ \frac {\log \left (x^{2}\right )}{2 \, a^{3}} - \frac {\log \left ({\left | b x^{2} - a \right |}\right )}{2 \, a^{3}} + \frac {3 \, b^{2} x^{4} - 8 \, a b x^{2} + 6 \, a^{2}}{4 \, {\left (b x^{2} - a\right )}^{2} a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 55, normalized size = 0.96 \[ \frac {1}{4 \left (b \,x^{2}-a \right )^{2} a}-\frac {1}{2 \left (b \,x^{2}-a \right ) a^{2}}+\frac {\ln \relax (x )}{a^{3}}-\frac {\ln \left (b \,x^{2}-a \right )}{2 a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.37, size = 62, normalized size = 1.09 \[ -\frac {2 \, b x^{2} - 3 \, a}{4 \, {\left (a^{2} b^{2} x^{4} - 2 \, a^{3} b x^{2} + a^{4}\right )}} - \frac {\log \left (b x^{2} - a\right )}{2 \, a^{3}} + \frac {\log \left (x^{2}\right )}{2 \, a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.06, size = 57, normalized size = 1.00 \[ \frac {\ln \relax (x)}{a^3}+\frac {\frac {3}{4\,a}-\frac {b\,x^2}{2\,a^2}}{a^2-2\,a\,b\,x^2+b^2\,x^4}-\frac {\ln \left (a-b\,x^2\right )}{2\,a^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*(a - b*x^2)^3),x)

[Out]

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

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sympy [A]  time = 0.42, size = 56, normalized size = 0.98 \[ - \frac {- 3 a + 2 b x^{2}}{4 a^{4} - 8 a^{3} b x^{2} + 4 a^{2} b^{2} x^{4}} + \frac {\log {\relax (x )}}{a^{3}} - \frac {\log {\left (- \frac {a}{b} + x^{2} \right )}}{2 a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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