3.1943 \(\int \frac{1}{(a+\frac{b}{x^2})^{3/2} x^8} \, dx\)

Optimal. Leaf size=95 \[ -\frac{15 a^2 \tanh ^{-1}\left (\frac{\sqrt{b}}{x \sqrt{a+\frac{b}{x^2}}}\right )}{8 b^{7/2}}+\frac{15 a \sqrt{a+\frac{b}{x^2}}}{8 b^3 x}-\frac{5 \sqrt{a+\frac{b}{x^2}}}{4 b^2 x^3}+\frac{1}{b x^5 \sqrt{a+\frac{b}{x^2}}} \]

[Out]

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

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Rubi [A]  time = 0.0477794, antiderivative size = 95, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {335, 288, 321, 217, 206} \[ -\frac{15 a^2 \tanh ^{-1}\left (\frac{\sqrt{b}}{x \sqrt{a+\frac{b}{x^2}}}\right )}{8 b^{7/2}}+\frac{15 a \sqrt{a+\frac{b}{x^2}}}{8 b^3 x}-\frac{5 \sqrt{a+\frac{b}{x^2}}}{4 b^2 x^3}+\frac{1}{b x^5 \sqrt{a+\frac{b}{x^2}}} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b/x^2)^(3/2)*x^8),x]

[Out]

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

Rule 335

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x] /;
FreeQ[{a, b, p}, x] && ILtQ[n, 0] && IntegerQ[m]

Rule 288

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

Rule 321

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

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{1}{\left (a+\frac{b}{x^2}\right )^{3/2} x^8} \, dx &=-\operatorname{Subst}\left (\int \frac{x^6}{\left (a+b x^2\right )^{3/2}} \, dx,x,\frac{1}{x}\right )\\ &=\frac{1}{b \sqrt{a+\frac{b}{x^2}} x^5}-\frac{5 \operatorname{Subst}\left (\int \frac{x^4}{\sqrt{a+b x^2}} \, dx,x,\frac{1}{x}\right )}{b}\\ &=\frac{1}{b \sqrt{a+\frac{b}{x^2}} x^5}-\frac{5 \sqrt{a+\frac{b}{x^2}}}{4 b^2 x^3}+\frac{(15 a) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{a+b x^2}} \, dx,x,\frac{1}{x}\right )}{4 b^2}\\ &=\frac{1}{b \sqrt{a+\frac{b}{x^2}} x^5}-\frac{5 \sqrt{a+\frac{b}{x^2}}}{4 b^2 x^3}+\frac{15 a \sqrt{a+\frac{b}{x^2}}}{8 b^3 x}-\frac{\left (15 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^2}} \, dx,x,\frac{1}{x}\right )}{8 b^3}\\ &=\frac{1}{b \sqrt{a+\frac{b}{x^2}} x^5}-\frac{5 \sqrt{a+\frac{b}{x^2}}}{4 b^2 x^3}+\frac{15 a \sqrt{a+\frac{b}{x^2}}}{8 b^3 x}-\frac{\left (15 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{1}{\sqrt{a+\frac{b}{x^2}} x}\right )}{8 b^3}\\ &=\frac{1}{b \sqrt{a+\frac{b}{x^2}} x^5}-\frac{5 \sqrt{a+\frac{b}{x^2}}}{4 b^2 x^3}+\frac{15 a \sqrt{a+\frac{b}{x^2}}}{8 b^3 x}-\frac{15 a^2 \tanh ^{-1}\left (\frac{\sqrt{b}}{\sqrt{a+\frac{b}{x^2}} x}\right )}{8 b^{7/2}}\\ \end{align*}

Mathematica [C]  time = 0.0109007, size = 39, normalized size = 0.41 \[ \frac{a^2 \, _2F_1\left (-\frac{1}{2},3;\frac{1}{2};\frac{a x^2}{b}+1\right )}{b^3 x \sqrt{a+\frac{b}{x^2}}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b/x^2)^(3/2)*x^8),x]

[Out]

(a^2*Hypergeometric2F1[-1/2, 3, 1/2, 1 + (a*x^2)/b])/(b^3*Sqrt[a + b/x^2]*x)

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Maple [A]  time = 0.007, size = 94, normalized size = 1. \begin{align*} -{\frac{a{x}^{2}+b}{8\,{x}^{7}} \left ( -15\,{b}^{3/2}{x}^{4}{a}^{2}+15\,\ln \left ( 2\,{\frac{\sqrt{b}\sqrt{a{x}^{2}+b}+b}{x}} \right ) \sqrt{a{x}^{2}+b}{x}^{4}{a}^{2}b-5\,{b}^{5/2}{x}^{2}a+2\,{b}^{7/2} \right ) \left ({\frac{a{x}^{2}+b}{{x}^{2}}} \right ) ^{-{\frac{3}{2}}}{b}^{-{\frac{9}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+1/x^2*b)^(3/2)/x^8,x)

[Out]

-1/8*(a*x^2+b)*(-15*b^(3/2)*x^4*a^2+15*ln(2*(b^(1/2)*(a*x^2+b)^(1/2)+b)/x)*(a*x^2+b)^(1/2)*x^4*a^2*b-5*b^(5/2)
*x^2*a+2*b^(7/2))/((a*x^2+b)/x^2)^(3/2)/x^7/b^(9/2)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x^2)^(3/2)/x^8,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.8394, size = 495, normalized size = 5.21 \begin{align*} \left [\frac{15 \,{\left (a^{3} x^{5} + a^{2} b x^{3}\right )} \sqrt{b} \log \left (-\frac{a x^{2} - 2 \, \sqrt{b} x \sqrt{\frac{a x^{2} + b}{x^{2}}} + 2 \, b}{x^{2}}\right ) + 2 \,{\left (15 \, a^{2} b x^{4} + 5 \, a b^{2} x^{2} - 2 \, b^{3}\right )} \sqrt{\frac{a x^{2} + b}{x^{2}}}}{16 \,{\left (a b^{4} x^{5} + b^{5} x^{3}\right )}}, \frac{15 \,{\left (a^{3} x^{5} + a^{2} b x^{3}\right )} \sqrt{-b} \arctan \left (\frac{\sqrt{-b} x \sqrt{\frac{a x^{2} + b}{x^{2}}}}{a x^{2} + b}\right ) +{\left (15 \, a^{2} b x^{4} + 5 \, a b^{2} x^{2} - 2 \, b^{3}\right )} \sqrt{\frac{a x^{2} + b}{x^{2}}}}{8 \,{\left (a b^{4} x^{5} + b^{5} x^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x^2)^(3/2)/x^8,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 6.67125, size = 102, normalized size = 1.07 \begin{align*} \frac{15 a^{\frac{3}{2}}}{8 b^{3} x \sqrt{1 + \frac{b}{a x^{2}}}} + \frac{5 \sqrt{a}}{8 b^{2} x^{3} \sqrt{1 + \frac{b}{a x^{2}}}} - \frac{15 a^{2} \operatorname{asinh}{\left (\frac{\sqrt{b}}{\sqrt{a} x} \right )}}{8 b^{\frac{7}{2}}} - \frac{1}{4 \sqrt{a} b x^{5} \sqrt{1 + \frac{b}{a x^{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x**2)**(3/2)/x**8,x)

[Out]

15*a**(3/2)/(8*b**3*x*sqrt(1 + b/(a*x**2))) + 5*sqrt(a)/(8*b**2*x**3*sqrt(1 + b/(a*x**2))) - 15*a**2*asinh(sqr
t(b)/(sqrt(a)*x))/(8*b**(7/2)) - 1/(4*sqrt(a)*b*x**5*sqrt(1 + b/(a*x**2)))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (a + \frac{b}{x^{2}}\right )}^{\frac{3}{2}} x^{8}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x^2)^(3/2)/x^8,x, algorithm="giac")

[Out]

integrate(1/((a + b/x^2)^(3/2)*x^8), x)