3.1904 \(\int \frac {(a+\frac {b}{x^2})^{3/2}}{x^2} \, dx\)

Optimal. Leaf size=71 \[ -\frac {3 a^2 \tanh ^{-1}\left (\frac {\sqrt {b}}{x \sqrt {a+\frac {b}{x^2}}}\right )}{8 \sqrt {b}}-\frac {3 a \sqrt {a+\frac {b}{x^2}}}{8 x}-\frac {\left (a+\frac {b}{x^2}\right )^{3/2}}{4 x} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {335, 195, 217, 206} \[ -\frac {3 a^2 \tanh ^{-1}\left (\frac {\sqrt {b}}{x \sqrt {a+\frac {b}{x^2}}}\right )}{8 \sqrt {b}}-\frac {3 a \sqrt {a+\frac {b}{x^2}}}{8 x}-\frac {\left (a+\frac {b}{x^2}\right )^{3/2}}{4 x} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x^2)^(3/2)/x^2,x]

[Out]

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

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 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]

Rubi steps

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

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Mathematica [A]  time = 0.05, size = 85, normalized size = 1.20 \[ -\frac {\sqrt {a+\frac {b}{x^2}} \left (3 a^2 x^4 \sqrt {\frac {a x^2}{b}+1} \tanh ^{-1}\left (\sqrt {\frac {a x^2}{b}+1}\right )+5 a^2 x^4+7 a b x^2+2 b^2\right )}{8 x^3 \left (a x^2+b\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x^2)^(3/2)/x^2,x]

[Out]

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

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fricas [A]  time = 0.90, size = 163, normalized size = 2.30 \[ \left [\frac {3 \, a^{2} \sqrt {b} x^{3} \log \left (-\frac {a x^{2} - 2 \, \sqrt {b} x \sqrt {\frac {a x^{2} + b}{x^{2}}} + 2 \, b}{x^{2}}\right ) - 2 \, {\left (5 \, a b x^{2} + 2 \, b^{2}\right )} \sqrt {\frac {a x^{2} + b}{x^{2}}}}{16 \, b x^{3}}, \frac {3 \, a^{2} \sqrt {-b} x^{3} \arctan \left (\frac {\sqrt {-b} x \sqrt {\frac {a x^{2} + b}{x^{2}}}}{a x^{2} + b}\right ) - {\left (5 \, a b x^{2} + 2 \, b^{2}\right )} \sqrt {\frac {a x^{2} + b}{x^{2}}}}{8 \, b x^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/16*(3*a^2*sqrt(b)*x^3*log(-(a*x^2 - 2*sqrt(b)*x*sqrt((a*x^2 + b)/x^2) + 2*b)/x^2) - 2*(5*a*b*x^2 + 2*b^2)*s
qrt((a*x^2 + b)/x^2))/(b*x^3), 1/8*(3*a^2*sqrt(-b)*x^3*arctan(sqrt(-b)*x*sqrt((a*x^2 + b)/x^2)/(a*x^2 + b)) -
(5*a*b*x^2 + 2*b^2)*sqrt((a*x^2 + b)/x^2))/(b*x^3)]

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giac [A]  time = 0.27, size = 76, normalized size = 1.07 \[ \frac {\frac {3 \, a^{3} \arctan \left (\frac {\sqrt {a x^{2} + b}}{\sqrt {-b}}\right ) \mathrm {sgn}\relax (x)}{\sqrt {-b}} - \frac {5 \, {\left (a x^{2} + b\right )}^{\frac {3}{2}} a^{3} \mathrm {sgn}\relax (x) - 3 \, \sqrt {a x^{2} + b} a^{3} b \mathrm {sgn}\relax (x)}{a^{2} x^{4}}}{8 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(3*a^3*arctan(sqrt(a*x^2 + b)/sqrt(-b))*sgn(x)/sqrt(-b) - (5*(a*x^2 + b)^(3/2)*a^3*sgn(x) - 3*sqrt(a*x^2 +
b)*a^3*b*sgn(x))/(a^2*x^4))/a

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maple [B]  time = 0.01, size = 125, normalized size = 1.76 \[ -\frac {\left (\frac {a \,x^{2}+b}{x^{2}}\right )^{\frac {3}{2}} \left (3 a^{2} b^{\frac {3}{2}} x^{4} \ln \left (\frac {2 b +2 \sqrt {a \,x^{2}+b}\, \sqrt {b}}{x}\right )-3 \sqrt {a \,x^{2}+b}\, a^{2} b \,x^{4}-\left (a \,x^{2}+b \right )^{\frac {3}{2}} a^{2} x^{4}+\left (a \,x^{2}+b \right )^{\frac {5}{2}} a \,x^{2}+2 \left (a \,x^{2}+b \right )^{\frac {5}{2}} b \right )}{8 \left (a \,x^{2}+b \right )^{\frac {3}{2}} b^{2} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 1.94, size = 113, normalized size = 1.59 \[ \frac {3 \, a^{2} \log \left (\frac {\sqrt {a + \frac {b}{x^{2}}} x - \sqrt {b}}{\sqrt {a + \frac {b}{x^{2}}} x + \sqrt {b}}\right )}{16 \, \sqrt {b}} - \frac {5 \, {\left (a + \frac {b}{x^{2}}\right )}^{\frac {3}{2}} a^{2} x^{3} - 3 \, \sqrt {a + \frac {b}{x^{2}}} a^{2} b x}{8 \, {\left ({\left (a + \frac {b}{x^{2}}\right )}^{2} x^{4} - 2 \, {\left (a + \frac {b}{x^{2}}\right )} b x^{2} + b^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.53, size = 39, normalized size = 0.55 \[ -\frac {{\left (a\,x^2+b\right )}^{3/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{2},\frac {1}{2};\ \frac {3}{2};\ -\frac {b}{a\,x^2}\right )}{x\,{\left (\frac {b}{a}+x^2\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/x^2)^(3/2)/x^2,x)

[Out]

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

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sympy [A]  time = 3.65, size = 71, normalized size = 1.00 \[ - \frac {5 a^{\frac {3}{2}} \sqrt {1 + \frac {b}{a x^{2}}}}{8 x} - \frac {\sqrt {a} b \sqrt {1 + \frac {b}{a x^{2}}}}{4 x^{3}} - \frac {3 a^{2} \operatorname {asinh}{\left (\frac {\sqrt {b}}{\sqrt {a} x} \right )}}{8 \sqrt {b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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