3.3.37 \(\int \frac {1}{x^2 (a-b x^2)^2} \, dx\) [237]

Optimal. Leaf size=58 \[ -\frac {3}{2 a^2 x}+\frac {1}{2 a x \left (a-b x^2\right )}+\frac {3 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 a^{5/2}} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {296, 331, 214} \begin {gather*} \frac {3 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 a^{5/2}}-\frac {3}{2 a^2 x}+\frac {1}{2 a x \left (a-b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^2*(a - b*x^2)^2),x]

[Out]

-3/(2*a^2*x) + 1/(2*a*x*(a - b*x^2)) + (3*Sqrt[b]*ArcTanh[(Sqrt[b]*x)/Sqrt[a]])/(2*a^(5/2))

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 296

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-(c*x)^(m + 1))*((a + b*x^n)^(p + 1)/
(a*c*n*(p + 1))), x] + Dist[(m + n*(p + 1) + 1)/(a*n*(p + 1)), Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; Free
Q[{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 331

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c
*(m + 1))), x] - Dist[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rubi steps

\begin {align*} \int \frac {1}{x^2 \left (a-b x^2\right )^2} \, dx &=\frac {1}{2 a x \left (a-b x^2\right )}+\frac {3 \int \frac {1}{x^2 \left (a-b x^2\right )} \, dx}{2 a}\\ &=-\frac {3}{2 a^2 x}+\frac {1}{2 a x \left (a-b x^2\right )}+\frac {(3 b) \int \frac {1}{a-b x^2} \, dx}{2 a^2}\\ &=-\frac {3}{2 a^2 x}+\frac {1}{2 a x \left (a-b x^2\right )}+\frac {3 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 a^{5/2}}\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 56, normalized size = 0.97 \begin {gather*} -\frac {1}{a^2 x}-\frac {b x}{2 a^2 \left (-a+b x^2\right )}+\frac {3 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 a^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^2*(a - b*x^2)^2),x]

[Out]

-(1/(a^2*x)) - (b*x)/(2*a^2*(-a + b*x^2)) + (3*Sqrt[b]*ArcTanh[(Sqrt[b]*x)/Sqrt[a]])/(2*a^(5/2))

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Maple [A]
time = 0.04, size = 45, normalized size = 0.78

method	result	size
default	\(\frac {b \left (\frac {x}{-2 b \,x^{2}+2 a}+\frac {3 \arctanh \left (\frac {b x}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{a^{2}}-\frac {1}{a^{2} x}\)	\(45\)
risch	\(\frac {\frac {3 b \,x^{2}}{2 a^{2}}-\frac {1}{a}}{x \left (-b \,x^{2}+a \right )}+\frac {3 \left (\munderset {\textit {\_R} =\RootOf \left (a^{5} \textit {\_Z}^{2}-b \right )}{\sum }\textit {\_R} \ln \left (\left (3 \textit {\_R}^{2} a^{5}-2 b \right ) x +a^{3} \textit {\_R} \right )\right )}{4}\)	\(71\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.48, size = 65, normalized size = 1.12 \begin {gather*} -\frac {3 \, b x^{2} - 2 \, a}{2 \, {\left (a^{2} b x^{3} - a^{3} x\right )}} - \frac {3 \, b \log \left (\frac {b x - \sqrt {a b}}{b x + \sqrt {a b}}\right )}{4 \, \sqrt {a b} a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.70, size = 140, normalized size = 2.41 \begin {gather*} \left [-\frac {6 \, b x^{2} - 3 \, {\left (b x^{3} - a x\right )} \sqrt {\frac {b}{a}} \log \left (\frac {b x^{2} + 2 \, a x \sqrt {\frac {b}{a}} + a}{b x^{2} - a}\right ) - 4 \, a}{4 \, {\left (a^{2} b x^{3} - a^{3} x\right )}}, -\frac {3 \, b x^{2} + 3 \, {\left (b x^{3} - a x\right )} \sqrt {-\frac {b}{a}} \arctan \left (x \sqrt {-\frac {b}{a}}\right ) - 2 \, a}{2 \, {\left (a^{2} b x^{3} - a^{3} x\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/4*(6*b*x^2 - 3*(b*x^3 - a*x)*sqrt(b/a)*log((b*x^2 + 2*a*x*sqrt(b/a) + a)/(b*x^2 - a)) - 4*a)/(a^2*b*x^3 -
a^3*x), -1/2*(3*b*x^2 + 3*(b*x^3 - a*x)*sqrt(-b/a)*arctan(x*sqrt(-b/a)) - 2*a)/(a^2*b*x^3 - a^3*x)]

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Sympy [A]
time = 0.13, size = 83, normalized size = 1.43 \begin {gather*} - \frac {3 \sqrt {\frac {b}{a^{5}}} \log {\left (- \frac {a^{3} \sqrt {\frac {b}{a^{5}}}}{b} + x \right )}}{4} + \frac {3 \sqrt {\frac {b}{a^{5}}} \log {\left (\frac {a^{3} \sqrt {\frac {b}{a^{5}}}}{b} + x \right )}}{4} + \frac {2 a - 3 b x^{2}}{- 2 a^{3} x + 2 a^{2} b x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 4.51, size = 50, normalized size = 0.86 \begin {gather*} -\frac {3 \, b \arctan \left (\frac {b x}{\sqrt {-a b}}\right )}{2 \, \sqrt {-a b} a^{2}} - \frac {3 \, b x^{2} - 2 \, a}{2 \, {\left (b x^{3} - a x\right )} a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/2*b*arctan(b*x/sqrt(-a*b))/(sqrt(-a*b)*a^2) - 1/2*(3*b*x^2 - 2*a)/((b*x^3 - a*x)*a^2)

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Mupad [B]
time = 4.63, size = 45, normalized size = 0.78 \begin {gather*} \frac {3\,\sqrt {b}\,\mathrm {atanh}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{2\,a^{5/2}}-\frac {\frac {1}{a}-\frac {3\,b\,x^2}{2\,a^2}}{a\,x-b\,x^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2*(a - b*x^2)^2),x)

[Out]

(3*b^(1/2)*atanh((b^(1/2)*x)/a^(1/2)))/(2*a^(5/2)) - (1/a - (3*b*x^2)/(2*a^2))/(a*x - b*x^3)

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