3.3.38 \(\int \frac {1}{x^3 (a-b x^2)^2} \, dx\) [238]

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

[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 = 0.03, 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 = {272, 46} \begin {gather*} -\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} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(a - b*x^2)^2),x]

[Out]

-1/2*1/(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 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 272

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} \text {Subst}\left (\int \frac {1}{x^2 (a-b x)^2} \, dx,x,x^2\right )\\ &=\frac {1}{2} \text {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 = 0.02, size = 44, normalized size = 0.85 \begin {gather*} \frac {-\frac {a}{x^2}+\frac {a b}{a-b x^2}+4 b \log (x)-2 b \log \left (a-b x^2\right )}{2 a^3} \end {gather*}

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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Maple [A]
time = 0.04, size = 56, normalized size = 1.08

method	result	size
risch	\(\frac {\frac {b \,x^{2}}{a^{2}}-\frac {1}{2 a}}{x^{2} \left (-b \,x^{2}+a \right )}+\frac {2 b \ln \left (x \right )}{a^{3}}-\frac {b \ln \left (-b \,x^{2}+a \right )}{a^{3}}\)	\(53\)
norman	\(\frac {\frac {b^{2} x^{4}}{a^{3}}-\frac {1}{2 a}}{x^{2} \left (-b \,x^{2}+a \right )}+\frac {2 b \ln \left (x \right )}{a^{3}}-\frac {b \ln \left (-b \,x^{2}+a \right )}{a^{3}}\)	\(55\)
default	\(\frac {b^{2} \left (-\frac {2 \ln \left (-b \,x^{2}+a \right )}{b}+\frac {a}{b \left (-b \,x^{2}+a \right )}\right )}{2 a^{3}}-\frac {1}{2 a^{2} x^{2}}+\frac {2 b \ln \left (x \right )}{a^{3}}\)	\(56\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.28, size = 57, normalized size = 1.10 \begin {gather*} -\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}} \end {gather*}

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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Fricas [A]
time = 0.61, size = 80, normalized size = 1.54 \begin {gather*} -\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 \left (x\right )}{2 \, {\left (a^{3} b x^{4} - a^{4} x^{2}\right )}} \end {gather*}

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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Sympy [A]
time = 0.16, size = 49, normalized size = 0.94 \begin {gather*} \frac {a - 2 b x^{2}}{- 2 a^{3} x^{2} + 2 a^{2} b x^{4}} + \frac {2 b \log {\left (x \right )}}{a^{3}} - \frac {b \log {\left (- \frac {a}{b} + x^{2} \right )}}{a^{3}} \end {gather*}

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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Giac [A]
time = 1.21, size = 56, normalized size = 1.08 \begin {gather*} \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}} \end {gather*}

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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Mupad [B]
time = 4.59, size = 55, normalized size = 1.06 \begin {gather*} \frac {2\,b\,\ln \left (x\right )}{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} \end {gather*}

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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