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

Optimal. Leaf size=78 \[ \frac {15 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{7/2}}-\frac {15}{8 a^3 x}+\frac {5}{8 a^2 x \left (a-b x^2\right )}+\frac {1}{4 a x \left (a-b x^2\right )^2} \]

[Out]

-15/8/a^3/x+1/4/a/x/(-b*x^2+a)^2+5/8/a^2/x/(-b*x^2+a)+15/8*arctanh(x*b^(1/2)/a^(1/2))*b^(1/2)/a^(7/2)

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Rubi [A]  time = 0.03, antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {290, 325, 208} \[ \frac {5}{8 a^2 x \left (a-b x^2\right )}+\frac {15 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{7/2}}-\frac {15}{8 a^3 x}+\frac {1}{4 a x \left (a-b x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^2*(a - b*x^2)^3),x]

[Out]

-15/(8*a^3*x) + 1/(4*a*x*(a - b*x^2)^2) + 5/(8*a^2*x*(a - b*x^2)) + (15*Sqrt[b]*ArcTanh[(Sqrt[b]*x)/Sqrt[a]])/
(8*a^(7/2))

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 290

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] /; FreeQ[
{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 325

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 )^3} \, dx &=\frac {1}{4 a x \left (a-b x^2\right )^2}+\frac {5 \int \frac {1}{x^2 \left (a-b x^2\right )^2} \, dx}{4 a}\\ &=\frac {1}{4 a x \left (a-b x^2\right )^2}+\frac {5}{8 a^2 x \left (a-b x^2\right )}+\frac {15 \int \frac {1}{x^2 \left (a-b x^2\right )} \, dx}{8 a^2}\\ &=-\frac {15}{8 a^3 x}+\frac {1}{4 a x \left (a-b x^2\right )^2}+\frac {5}{8 a^2 x \left (a-b x^2\right )}+\frac {(15 b) \int \frac {1}{a-b x^2} \, dx}{8 a^3}\\ &=-\frac {15}{8 a^3 x}+\frac {1}{4 a x \left (a-b x^2\right )^2}+\frac {5}{8 a^2 x \left (a-b x^2\right )}+\frac {15 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 69, normalized size = 0.88 \[ \frac {15 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{7/2}}+\frac {-8 a^2+25 a b x^2-15 b^2 x^4}{8 a^3 x \left (a-b x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^2*(a - b*x^2)^3),x]

[Out]

(-8*a^2 + 25*a*b*x^2 - 15*b^2*x^4)/(8*a^3*x*(a - b*x^2)^2) + (15*Sqrt[b]*ArcTanh[(Sqrt[b]*x)/Sqrt[a]])/(8*a^(7
/2))

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fricas [A]  time = 0.87, size = 202, normalized size = 2.59 \[ \left [-\frac {30 \, b^{2} x^{4} - 50 \, a b x^{2} - 15 \, {\left (b^{2} x^{5} - 2 \, a b x^{3} + a^{2} x\right )} \sqrt {\frac {b}{a}} \log \left (\frac {b x^{2} + 2 \, a x \sqrt {\frac {b}{a}} + a}{b x^{2} - a}\right ) + 16 \, a^{2}}{16 \, {\left (a^{3} b^{2} x^{5} - 2 \, a^{4} b x^{3} + a^{5} x\right )}}, -\frac {15 \, b^{2} x^{4} - 25 \, a b x^{2} + 15 \, {\left (b^{2} x^{5} - 2 \, a b x^{3} + a^{2} x\right )} \sqrt {-\frac {b}{a}} \arctan \left (x \sqrt {-\frac {b}{a}}\right ) + 8 \, a^{2}}{8 \, {\left (a^{3} b^{2} x^{5} - 2 \, a^{4} b x^{3} + a^{5} x\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/16*(30*b^2*x^4 - 50*a*b*x^2 - 15*(b^2*x^5 - 2*a*b*x^3 + a^2*x)*sqrt(b/a)*log((b*x^2 + 2*a*x*sqrt(b/a) + a)
/(b*x^2 - a)) + 16*a^2)/(a^3*b^2*x^5 - 2*a^4*b*x^3 + a^5*x), -1/8*(15*b^2*x^4 - 25*a*b*x^2 + 15*(b^2*x^5 - 2*a
*b*x^3 + a^2*x)*sqrt(-b/a)*arctan(x*sqrt(-b/a)) + 8*a^2)/(a^3*b^2*x^5 - 2*a^4*b*x^3 + a^5*x)]

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giac [A]  time = 0.63, size = 61, normalized size = 0.78 \[ -\frac {15 \, b \arctan \left (\frac {b x}{\sqrt {-a b}}\right )}{8 \, \sqrt {-a b} a^{3}} - \frac {7 \, b^{2} x^{3} - 9 \, a b x}{8 \, {\left (b x^{2} - a\right )}^{2} a^{3}} - \frac {1}{a^{3} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-15/8*b*arctan(b*x/sqrt(-a*b))/(sqrt(-a*b)*a^3) - 1/8*(7*b^2*x^3 - 9*a*b*x)/((b*x^2 - a)^2*a^3) - 1/(a^3*x)

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maple [A]  time = 0.01, size = 56, normalized size = 0.72 \[ -\frac {\left (-\frac {15 \arctanh \left (\frac {b x}{\sqrt {a b}}\right )}{8 \sqrt {a b}}+\frac {\frac {7}{8} b \,x^{3}-\frac {9}{8} a x}{\left (b \,x^{2}-a \right )^{2}}\right ) b}{a^{3}}-\frac {1}{a^{3} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/a^3/x-1/a^3*b*((7/8*b*x^3-9/8*a*x)/(b*x^2-a)^2-15/8/(a*b)^(1/2)*arctanh(1/(a*b)^(1/2)*b*x))

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maxima [A]  time = 2.91, size = 86, normalized size = 1.10 \[ -\frac {15 \, b^{2} x^{4} - 25 \, a b x^{2} + 8 \, a^{2}}{8 \, {\left (a^{3} b^{2} x^{5} - 2 \, a^{4} b x^{3} + a^{5} x\right )}} - \frac {15 \, b \log \left (\frac {b x - \sqrt {a b}}{b x + \sqrt {a b}}\right )}{16 \, \sqrt {a b} a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(15*b^2*x^4 - 25*a*b*x^2 + 8*a^2)/(a^3*b^2*x^5 - 2*a^4*b*x^3 + a^5*x) - 15/16*b*log((b*x - sqrt(a*b))/(b*
x + sqrt(a*b)))/(sqrt(a*b)*a^3)

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mupad [B]  time = 4.60, size = 66, normalized size = 0.85 \[ \frac {15\,\sqrt {b}\,\mathrm {atanh}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{8\,a^{7/2}}-\frac {\frac {1}{a}-\frac {25\,b\,x^2}{8\,a^2}+\frac {15\,b^2\,x^4}{8\,a^3}}{a^2\,x-2\,a\,b\,x^3+b^2\,x^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2*(a - b*x^2)^3),x)

[Out]

(15*b^(1/2)*atanh((b^(1/2)*x)/a^(1/2)))/(8*a^(7/2)) - (1/a - (25*b*x^2)/(8*a^2) + (15*b^2*x^4)/(8*a^3))/(a^2*x
+ b^2*x^5 - 2*a*b*x^3)

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sympy [A]  time = 0.43, size = 107, normalized size = 1.37 \[ - \frac {15 \sqrt {\frac {b}{a^{7}}} \log {\left (- \frac {a^{4} \sqrt {\frac {b}{a^{7}}}}{b} + x \right )}}{16} + \frac {15 \sqrt {\frac {b}{a^{7}}} \log {\left (\frac {a^{4} \sqrt {\frac {b}{a^{7}}}}{b} + x \right )}}{16} - \frac {8 a^{2} - 25 a b x^{2} + 15 b^{2} x^{4}}{8 a^{5} x - 16 a^{4} b x^{3} + 8 a^{3} b^{2} x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-15*sqrt(b/a**7)*log(-a**4*sqrt(b/a**7)/b + x)/16 + 15*sqrt(b/a**7)*log(a**4*sqrt(b/a**7)/b + x)/16 - (8*a**2
- 25*a*b*x**2 + 15*b**2*x**4)/(8*a**5*x - 16*a**4*b*x**3 + 8*a**3*b**2*x**5)

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