3.1932 \(\int \frac {1}{(a+\frac {b}{x^2})^{3/2} x} \, dx\)

Optimal. Leaf size=41 \[ \frac {\tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x^2}}}{\sqrt {a}}\right )}{a^{3/2}}-\frac {1}{a \sqrt {a+\frac {b}{x^2}}} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 41, normalized size of antiderivative = 1.00, 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 = {266, 51, 63, 208} \[ \frac {\tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x^2}}}{\sqrt {a}}\right )}{a^{3/2}}-\frac {1}{a \sqrt {a+\frac {b}{x^2}}} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b/x^2)^(3/2)*x),x]

[Out]

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

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

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

Rule 266

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

Rubi steps

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

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Mathematica [A]  time = 0.06, size = 62, normalized size = 1.51 \[ \frac {\sqrt {b} \sqrt {\frac {a x^2}{b}+1} \sinh ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {b}}\right )-\sqrt {a} x}{a^{3/2} x \sqrt {a+\frac {b}{x^2}}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b/x^2)^(3/2)*x),x]

[Out]

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

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fricas [B]  time = 0.83, size = 163, normalized size = 3.98 \[ \left [-\frac {2 \, a x^{2} \sqrt {\frac {a x^{2} + b}{x^{2}}} - {\left (a x^{2} + b\right )} \sqrt {a} \log \left (-2 \, a x^{2} - 2 \, \sqrt {a} x^{2} \sqrt {\frac {a x^{2} + b}{x^{2}}} - b\right )}{2 \, {\left (a^{3} x^{2} + a^{2} b\right )}}, -\frac {a x^{2} \sqrt {\frac {a x^{2} + b}{x^{2}}} + {\left (a x^{2} + b\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-a} x^{2} \sqrt {\frac {a x^{2} + b}{x^{2}}}}{a x^{2} + b}\right )}{a^{3} x^{2} + a^{2} b}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(t_nostep)]Unable to divide, perhaps due to rounding error%%%{%%%{-2,[1]%%%},[2,1,2]%%%}+%%%{%%{[4,0]:[1,0,%%
%{-1,[1]%%%}]%%},[1,1,3]%%%}+%%%{-2,[0,1,4]%%%} / %%%{%%%{1,[2]%%%},[2,0,0]%%%}+%%%{%%{[%%%{-2,[1]%%%},0]:[1,0
,%%%{-1,[1]%%%}]%%},[1,0,1]%%%}+%%%{%%%{1,[1]%%%},[0,0,2]%%%} Error: Bad Argument Value

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maple [A]  time = 0.01, size = 63, normalized size = 1.54 \[ -\frac {\left (a \,x^{2}+b \right ) \left (a^{\frac {3}{2}} x -\sqrt {a \,x^{2}+b}\, a \ln \left (\sqrt {a}\, x +\sqrt {a \,x^{2}+b}\right )\right )}{\left (\frac {a \,x^{2}+b}{x^{2}}\right )^{\frac {3}{2}} a^{\frac {5}{2}} x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.98, size = 52, normalized size = 1.27 \[ -\frac {\log \left (\frac {\sqrt {a + \frac {b}{x^{2}}} - \sqrt {a}}{\sqrt {a + \frac {b}{x^{2}}} + \sqrt {a}}\right )}{2 \, a^{\frac {3}{2}}} - \frac {1}{\sqrt {a + \frac {b}{x^{2}}} a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.32, size = 33, normalized size = 0.80 \[ \frac {\mathrm {atanh}\left (\frac {\sqrt {a+\frac {b}{x^2}}}{\sqrt {a}}\right )}{a^{3/2}}-\frac {1}{a\,\sqrt {a+\frac {b}{x^2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*(a + b/x^2)^(3/2)),x)

[Out]

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

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sympy [B]  time = 2.18, size = 187, normalized size = 4.56 \[ - \frac {2 a^{3} x^{2} \sqrt {1 + \frac {b}{a x^{2}}}}{2 a^{\frac {9}{2}} x^{2} + 2 a^{\frac {7}{2}} b} - \frac {a^{3} x^{2} \log {\left (\frac {b}{a x^{2}} \right )}}{2 a^{\frac {9}{2}} x^{2} + 2 a^{\frac {7}{2}} b} + \frac {2 a^{3} x^{2} \log {\left (\sqrt {1 + \frac {b}{a x^{2}}} + 1 \right )}}{2 a^{\frac {9}{2}} x^{2} + 2 a^{\frac {7}{2}} b} - \frac {a^{2} b \log {\left (\frac {b}{a x^{2}} \right )}}{2 a^{\frac {9}{2}} x^{2} + 2 a^{\frac {7}{2}} b} + \frac {2 a^{2} b \log {\left (\sqrt {1 + \frac {b}{a x^{2}}} + 1 \right )}}{2 a^{\frac {9}{2}} x^{2} + 2 a^{\frac {7}{2}} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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