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

Optimal. Leaf size=52 \[ -\frac {b \log \left (a-b x^2\right )}{a^3}+\frac {2 b \log (x)}{a^3}+\frac {b}{2 a^2 \left (a-b x^2\right )}-\frac {1}{2 a^2 x^2} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 52, 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} \[ \frac {b}{2 a^2 \left (a-b x^2\right )}-\frac {b \log \left (a-b x^2\right )}{a^3}+\frac {2 b \log (x)}{a^3}-\frac {1}{2 a^2 x^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(a - b*x^2)^2),x]

[Out]

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

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Mathematica [A]  time = 0.03, size = 44, normalized size = 0.85 \[ \frac {\frac {a b}{a-b x^2}-2 b \log \left (a-b x^2\right )-\frac {a}{x^2}+4 b \log (x)}{2 a^3} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*(a - b*x^2)^2),x]

[Out]

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

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fricas [A]  time = 0.84, size = 80, normalized size = 1.54 \[ -\frac {2 \, a b x^{2} - a^{2} + 2 \, {\left (b^{2} x^{4} - a b x^{2}\right )} \log \left (b x^{2} - a\right ) - 4 \, {\left (b^{2} x^{4} - a b x^{2}\right )} \log \relax (x)}{2 \, {\left (a^{3} b x^{4} - a^{4} x^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.60, size = 56, normalized size = 1.08 \[ \frac {b \log \left (x^{2}\right )}{a^{3}} - \frac {b \log \left ({\left | b x^{2} - a \right |}\right )}{a^{3}} - \frac {2 \, b x^{2} - a}{2 \, {\left (b x^{4} - a x^{2}\right )} a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 51, normalized size = 0.98 \[ -\frac {b}{2 \left (b \,x^{2}-a \right ) a^{2}}+\frac {2 b \ln \relax (x )}{a^{3}}-\frac {b \ln \left (b \,x^{2}-a \right )}{a^{3}}-\frac {1}{2 a^{2} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.31, size = 57, normalized size = 1.10 \[ -\frac {2 \, b x^{2} - a}{2 \, {\left (a^{2} b x^{4} - a^{3} x^{2}\right )}} - \frac {b \log \left (b x^{2} - a\right )}{a^{3}} + \frac {b \log \left (x^{2}\right )}{a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 4.59, size = 55, normalized size = 1.06 \[ \frac {2\,b\,\ln \relax (x)}{a^3}-\frac {b\,\ln \left (a-b\,x^2\right )}{a^3}-\frac {\frac {1}{2\,a}-\frac {b\,x^2}{a^2}}{a\,x^2-b\,x^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3*(a - b*x^2)^2),x)

[Out]

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

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sympy [A]  time = 0.37, size = 49, normalized size = 0.94 \[ \frac {a - 2 b x^{2}}{- 2 a^{3} x^{2} + 2 a^{2} b x^{4}} + \frac {2 b \log {\relax (x )}}{a^{3}} - \frac {b \log {\left (- \frac {a}{b} + x^{2} \right )}}{a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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