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]

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

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Rubi [A]  time = 0.03, antiderivative size = 64, normalized size of antiderivative = 1.00, 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 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 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])

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 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]

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*}

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Mathematica [C]  time = 0.01, 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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fricas [A]  time = 1.00, size = 142, normalized size = 2.22 \[ \left [\frac {3 \, a \sqrt {b} x \log \left (-\frac {a x^{2} - 2 \, \sqrt {b} x \sqrt {\frac {a x^{2} + b}{x^{2}}} + 2 \, b}{x^{2}}\right ) + 2 \, {\left (2 \, a x^{2} - b\right )} \sqrt {\frac {a x^{2} + b}{x^{2}}}}{4 \, x}, \frac {3 \, a \sqrt {-b} x \arctan \left (\frac {\sqrt {-b} x \sqrt {\frac {a x^{2} + b}{x^{2}}}}{a x^{2} + b}\right ) + {\left (2 \, a x^{2} - b\right )} \sqrt {\frac {a x^{2} + b}{x^{2}}}}{2 \, x}\right ] \]

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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giac [A]  time = 0.21, size = 69, normalized size = 1.08 \[ \frac {\frac {3 \, a^{2} b \arctan \left (\frac {\sqrt {a x^{2} + b}}{\sqrt {-b}}\right ) \mathrm {sgn}\relax (x)}{\sqrt {-b}} + 2 \, \sqrt {a x^{2} + b} a^{2} \mathrm {sgn}\relax (x) - \frac {\sqrt {a x^{2} + b} a b \mathrm {sgn}\relax (x)}{x^{2}}}{2 \, a} \]

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

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maple [A]  time = 0.01, size = 100, normalized size = 1.56 \[ -\frac {\left (\frac {a \,x^{2}+b}{x^{2}}\right )^{\frac {3}{2}} \left (3 a \,b^{\frac {3}{2}} x^{2} \ln \left (\frac {2 b +2 \sqrt {a \,x^{2}+b}\, \sqrt {b}}{x}\right )-3 \sqrt {a \,x^{2}+b}\, a b \,x^{2}-\left (a \,x^{2}+b \right )^{\frac {3}{2}} a \,x^{2}+\left (a \,x^{2}+b \right )^{\frac {5}{2}}\right ) x}{2 \left (a \,x^{2}+b \right )^{\frac {3}{2}} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*((a*x^2+b)/x^2)^(3/2)*x*(-(a*x^2+b)^(3/2)*a*x^2+3*b^(3/2)*ln(2*(b+(a*x^2+b)^(1/2)*b^(1/2))/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 [A]  time = 1.94, size = 86, normalized size = 1.34 \[ \sqrt {a + \frac {b}{x^{2}}} a x - \frac {\sqrt {a + \frac {b}{x^{2}}} a b x}{2 \, {\left ({\left (a + \frac {b}{x^{2}}\right )} x^{2} - b\right )}} + \frac {3}{4} \, a \sqrt {b} \log \left (\frac {\sqrt {a + \frac {b}{x^{2}}} x - \sqrt {b}}{\sqrt {a + \frac {b}{x^{2}}} x + \sqrt {b}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.31, size = 36, normalized size = 0.56 \[ \frac {x\,{\left (a\,x^2+b\right )}^{3/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{2},-\frac {1}{2};\ \frac {1}{2};\ -\frac {b}{a\,x^2}\right )}{{\left (\frac {b}{a}+x^2\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/x^2)^(3/2),x)

[Out]

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

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sympy [A]  time = 2.81, size = 88, normalized size = 1.38 \[ \frac {a^{\frac {3}{2}} x}{\sqrt {1 + \frac {b}{a x^{2}}}} + \frac {\sqrt {a} b}{2 x \sqrt {1 + \frac {b}{a x^{2}}}} - \frac {3 a \sqrt {b} \operatorname {asinh}{\left (\frac {\sqrt {b}}{\sqrt {a} x} \right )}}{2} - \frac {b^{2}}{2 \sqrt {a} x^{3} \sqrt {1 + \frac {b}{a x^{2}}}} \]

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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