3.1902 \(\int (a+\frac{b}{x^2})^{3/2} \, dx\)

Optimal. Leaf size=64 \[ x \left (a+\frac{b}{x^2}\right )^{3/2}-\frac{3 b \sqrt{a+\frac{b}{x^2}}}{2 x}-\frac{3}{2} a \sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b}}{x \sqrt{a+\frac{b}{x^2}}}\right ) \]

[Out]

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

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Rubi [A]  time = 0.0263427, antiderivative size = 64, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.454, Rules used = {242, 277, 195, 217, 206} \[ x \left (a+\frac{b}{x^2}\right )^{3/2}-\frac{3 b \sqrt{a+\frac{b}{x^2}}}{2 x}-\frac{3}{2} a \sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b}}{x \sqrt{a+\frac{b}{x^2}}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x^2)^(3/2),x]

[Out]

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

Rule 242

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

Rule 277

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

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
 Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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

Mathematica [C]  time = 0.0107218, size = 47, normalized size = 0.73 \[ \frac{a x^3 \left (a+\frac{b}{x^2}\right )^{3/2} \left (a x^2+b\right ) \, _2F_1\left (2,\frac{5}{2};\frac{7}{2};\frac{a x^2}{b}+1\right )}{5 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x^2)^(3/2),x]

[Out]

(a*(a + b/x^2)^(3/2)*x^3*(b + a*x^2)*Hypergeometric2F1[2, 5/2, 7/2, 1 + (a*x^2)/b])/(5*b^2)

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Maple [A]  time = 0.006, size = 100, normalized size = 1.6 \begin{align*} -{\frac{x}{2\,b} \left ({\frac{a{x}^{2}+b}{{x}^{2}}} \right ) ^{{\frac{3}{2}}} \left ( - \left ( a{x}^{2}+b \right ) ^{{\frac{3}{2}}}{x}^{2}a+3\,{b}^{3/2}\ln \left ( 2\,{\frac{\sqrt{b}\sqrt{a{x}^{2}+b}+b}{x}} \right ){x}^{2}a+ \left ( a{x}^{2}+b \right ) ^{{\frac{5}{2}}}-3\,\sqrt{a{x}^{2}+b}{x}^{2}ab \right ) \left ( a{x}^{2}+b \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.56836, size = 327, normalized size = 5.11 \begin{align*} \left [\frac{3 \, a \sqrt{b} x \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 (2 \, a x^{2} - b\right )} \sqrt{\frac{a x^{2} + b}{x^{2}}}}{4 \, x}, \frac{3 \, a \sqrt{-b} x \arctan \left (\frac{\sqrt{-b} x \sqrt{\frac{a x^{2} + b}{x^{2}}}}{a x^{2} + b}\right ) +{\left (2 \, a x^{2} - b\right )} \sqrt{\frac{a x^{2} + b}{x^{2}}}}{2 \, x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 2.53202, size = 88, normalized size = 1.38 \begin{align*} \frac{a^{\frac{3}{2}} x}{\sqrt{1 + \frac{b}{a x^{2}}}} + \frac{\sqrt{a} b}{2 x \sqrt{1 + \frac{b}{a x^{2}}}} - \frac{3 a \sqrt{b} \operatorname{asinh}{\left (\frac{\sqrt{b}}{\sqrt{a} x} \right )}}{2} - \frac{b^{2}}{2 \sqrt{a} x^{3} \sqrt{1 + \frac{b}{a x^{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**(3/2)*x/sqrt(1 + b/(a*x**2)) + sqrt(a)*b/(2*x*sqrt(1 + b/(a*x**2))) - 3*a*sqrt(b)*asinh(sqrt(b)/(sqrt(a)*x)
)/2 - b**2/(2*sqrt(a)*x**3*sqrt(1 + b/(a*x**2)))

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Giac [A]  time = 1.20065, size = 80, normalized size = 1.25 \begin{align*} \frac{1}{2} \,{\left (\frac{3 \, b \arctan \left (\frac{\sqrt{a x^{2} + b}}{\sqrt{-b}}\right )}{\sqrt{-b}} + 2 \, \sqrt{a x^{2} + b} - \frac{\sqrt{a x^{2} + b} b}{a x^{2}}\right )} a \mathrm{sgn}\left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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