3.1941 \(\int \frac{1}{(a+\frac{b}{x^2})^{3/2} x^4} \, dx\)

Optimal. Leaf size=47 \[ \frac{1}{b x \sqrt{a+\frac{b}{x^2}}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{b}}{x \sqrt{a+\frac{b}{x^2}}}\right )}{b^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0231834, antiderivative size = 47, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {335, 288, 217, 206} \[ \frac{1}{b x \sqrt{a+\frac{b}{x^2}}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{b}}{x \sqrt{a+\frac{b}{x^2}}}\right )}{b^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b/x^2)^(3/2)*x^4),x]

[Out]

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

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

Mathematica [C]  time = 0.0090448, size = 36, normalized size = 0.77 \[ \frac{\, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{a x^2}{b}+1\right )}{b x \sqrt{a+\frac{b}{x^2}}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b/x^2)^(3/2)*x^4),x]

[Out]

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

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Maple [A]  time = 0.006, size = 67, normalized size = 1.4 \begin{align*} -{\frac{a{x}^{2}+b}{{x}^{3}} \left ( \ln \left ( 2\,{\frac{\sqrt{b}\sqrt{a{x}^{2}+b}+b}{x}} \right ) b\sqrt{a{x}^{2}+b}-{b}^{{\frac{3}{2}}} \right ) \left ({\frac{a{x}^{2}+b}{{x}^{2}}} \right ) ^{-{\frac{3}{2}}}{b}^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+1/x^2*b)^(3/2)/x^4,x)

[Out]

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

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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^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.57075, size = 354, normalized size = 7.53 \begin{align*} \left [\frac{2 \, b x \sqrt{\frac{a x^{2} + b}{x^{2}}} +{\left (a x^{2} + b\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 (a b^{2} x^{2} + b^{3}\right )}}, \frac{b x \sqrt{\frac{a x^{2} + b}{x^{2}}} +{\left (a x^{2} + b\right )} \sqrt{-b} \arctan \left (\frac{\sqrt{-b} x \sqrt{\frac{a x^{2} + b}{x^{2}}}}{a x^{2} + b}\right )}{a b^{2} x^{2} + b^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x^2)^(3/2)/x^4,x, algorithm="fricas")

[Out]

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

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Sympy [B]  time = 2.29727, size = 184, normalized size = 3.91 \begin{align*} \frac{a b^{2} x^{2} \log{\left (\frac{a x^{2}}{b} \right )}}{2 a b^{\frac{7}{2}} x^{2} + 2 b^{\frac{9}{2}}} - \frac{2 a b^{2} x^{2} \log{\left (\sqrt{\frac{a x^{2}}{b} + 1} + 1 \right )}}{2 a b^{\frac{7}{2}} x^{2} + 2 b^{\frac{9}{2}}} + \frac{2 b^{3} \sqrt{\frac{a x^{2}}{b} + 1}}{2 a b^{\frac{7}{2}} x^{2} + 2 b^{\frac{9}{2}}} + \frac{b^{3} \log{\left (\frac{a x^{2}}{b} \right )}}{2 a b^{\frac{7}{2}} x^{2} + 2 b^{\frac{9}{2}}} - \frac{2 b^{3} \log{\left (\sqrt{\frac{a x^{2}}{b} + 1} + 1 \right )}}{2 a b^{\frac{7}{2}} x^{2} + 2 b^{\frac{9}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x**2)**(3/2)/x**4,x)

[Out]

a*b**2*x**2*log(a*x**2/b)/(2*a*b**(7/2)*x**2 + 2*b**(9/2)) - 2*a*b**2*x**2*log(sqrt(a*x**2/b + 1) + 1)/(2*a*b*
*(7/2)*x**2 + 2*b**(9/2)) + 2*b**3*sqrt(a*x**2/b + 1)/(2*a*b**(7/2)*x**2 + 2*b**(9/2)) + b**3*log(a*x**2/b)/(2
*a*b**(7/2)*x**2 + 2*b**(9/2)) - 2*b**3*log(sqrt(a*x**2/b + 1) + 1)/(2*a*b**(7/2)*x**2 + 2*b**(9/2))

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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^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x^2)^(3/2)/x^4,x, algorithm="giac")

[Out]

integrate(1/((a + b/x^2)^(3/2)*x^4), x)