3.1931 \(\int \frac {x}{(a+\frac {b}{x^2})^{3/2}} \, dx\)

Optimal. Leaf size=69 \[ -\frac {3 b \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x^2}}}{\sqrt {a}}\right )}{2 a^{5/2}}+\frac {3 b}{2 a^2 \sqrt {a+\frac {b}{x^2}}}+\frac {x^2}{2 a \sqrt {a+\frac {b}{x^2}}} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {266, 51, 63, 208} \[ \frac {3 x^2 \sqrt {a+\frac {b}{x^2}}}{2 a^2}-\frac {3 b \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x^2}}}{\sqrt {a}}\right )}{2 a^{5/2}}-\frac {x^2}{a \sqrt {a+\frac {b}{x^2}}} \]

Antiderivative was successfully verified.

[In]

Int[x/(a + b/x^2)^(3/2),x]

[Out]

-(x^2/(a*Sqrt[a + b/x^2])) + (3*Sqrt[a + b/x^2]*x^2)/(2*a^2) - (3*b*ArcTanh[Sqrt[a + b/x^2]/Sqrt[a]])/(2*a^(5/
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 {x}{\left (a+\frac {b}{x^2}\right )^{3/2}} \, dx &=-\left (\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{x^2 (a+b x)^{3/2}} \, dx,x,\frac {1}{x^2}\right )\right )\\ &=-\frac {x^2}{a \sqrt {a+\frac {b}{x^2}}}-\frac {3 \operatorname {Subst}\left (\int \frac {1}{x^2 \sqrt {a+b x}} \, dx,x,\frac {1}{x^2}\right )}{2 a}\\ &=-\frac {x^2}{a \sqrt {a+\frac {b}{x^2}}}+\frac {3 \sqrt {a+\frac {b}{x^2}} x^2}{2 a^2}+\frac {(3 b) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,\frac {1}{x^2}\right )}{4 a^2}\\ &=-\frac {x^2}{a \sqrt {a+\frac {b}{x^2}}}+\frac {3 \sqrt {a+\frac {b}{x^2}} x^2}{2 a^2}+\frac {3 \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+\frac {b}{x^2}}\right )}{2 a^2}\\ &=-\frac {x^2}{a \sqrt {a+\frac {b}{x^2}}}+\frac {3 \sqrt {a+\frac {b}{x^2}} x^2}{2 a^2}-\frac {3 b \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x^2}}}{\sqrt {a}}\right )}{2 a^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 74, normalized size = 1.07 \[ \frac {\sqrt {a} x \left (a x^2+3 b\right )-3 b^{3/2} \sqrt {\frac {a x^2}{b}+1} \sinh ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {b}}\right )}{2 a^{5/2} x \sqrt {a+\frac {b}{x^2}}} \]

Antiderivative was successfully verified.

[In]

Integrate[x/(a + b/x^2)^(3/2),x]

[Out]

(Sqrt[a]*x*(3*b + a*x^2) - 3*b^(3/2)*Sqrt[1 + (a*x^2)/b]*ArcSinh[(Sqrt[a]*x)/Sqrt[b]])/(2*a^(5/2)*Sqrt[a + b/x
^2]*x)

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fricas [A]  time = 0.63, size = 192, normalized size = 2.78 \[ \left [\frac {3 \, {\left (a b x^{2} + b^{2}\right )} \sqrt {a} \log \left (-2 \, a x^{2} + 2 \, \sqrt {a} x^{2} \sqrt {\frac {a x^{2} + b}{x^{2}}} - b\right ) + 2 \, {\left (a^{2} x^{4} + 3 \, a b x^{2}\right )} \sqrt {\frac {a x^{2} + b}{x^{2}}}}{4 \, {\left (a^{4} x^{2} + a^{3} b\right )}}, \frac {3 \, {\left (a b x^{2} + b^{2}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-a} x^{2} \sqrt {\frac {a x^{2} + b}{x^{2}}}}{a x^{2} + b}\right ) + {\left (a^{2} x^{4} + 3 \, a b x^{2}\right )} \sqrt {\frac {a x^{2} + b}{x^{2}}}}{2 \, {\left (a^{4} x^{2} + a^{3} b\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(3*(a*b*x^2 + b^2)*sqrt(a)*log(-2*a*x^2 + 2*sqrt(a)*x^2*sqrt((a*x^2 + b)/x^2) - b) + 2*(a^2*x^4 + 3*a*b*x
^2)*sqrt((a*x^2 + b)/x^2))/(a^4*x^2 + a^3*b), 1/2*(3*(a*b*x^2 + b^2)*sqrt(-a)*arctan(sqrt(-a)*x^2*sqrt((a*x^2
+ b)/x^2)/(a*x^2 + b)) + (a^2*x^4 + 3*a*b*x^2)*sqrt((a*x^2 + b)/x^2))/(a^4*x^2 + a^3*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(x/(a+b/x^2)^(3/2),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 = 73, normalized size = 1.06 \[ \frac {\left (a \,x^{2}+b \right ) \left (a^{\frac {5}{2}} x^{3}+3 a^{\frac {3}{2}} b x -3 \sqrt {a \,x^{2}+b}\, a b \ln \left (\sqrt {a}\, x +\sqrt {a \,x^{2}+b}\right )\right )}{2 \left (\frac {a \,x^{2}+b}{x^{2}}\right )^{\frac {3}{2}} a^{\frac {7}{2}} x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(a*x^2+b)*(x^3*a^(5/2)+3*a^(3/2)*x*b-3*ln(a^(1/2)*x+(a*x^2+b)^(1/2))*(a*x^2+b)^(1/2)*a*b)/((a*x^2+b)/x^2)^
(3/2)/x^3/a^(7/2)

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maxima [A]  time = 1.82, size = 86, normalized size = 1.25 \[ \frac {3 \, {\left (a + \frac {b}{x^{2}}\right )} b - 2 \, a b}{2 \, {\left ({\left (a + \frac {b}{x^{2}}\right )}^{\frac {3}{2}} a^{2} - \sqrt {a + \frac {b}{x^{2}}} a^{3}\right )}} + \frac {3 \, b \log \left (\frac {\sqrt {a + \frac {b}{x^{2}}} - \sqrt {a}}{\sqrt {a + \frac {b}{x^{2}}} + \sqrt {a}}\right )}{4 \, a^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.61, size = 53, normalized size = 0.77 \[ \frac {3\,b}{2\,a^2\,\sqrt {a+\frac {b}{x^2}}}+\frac {x^2}{2\,a\,\sqrt {a+\frac {b}{x^2}}}-\frac {3\,b\,\mathrm {atanh}\left (\frac {\sqrt {a+\frac {b}{x^2}}}{\sqrt {a}}\right )}{2\,a^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(a + b/x^2)^(3/2),x)

[Out]

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

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sympy [A]  time = 3.57, size = 71, normalized size = 1.03 \[ \frac {x^{3}}{2 a \sqrt {b} \sqrt {\frac {a x^{2}}{b} + 1}} + \frac {3 \sqrt {b} x}{2 a^{2} \sqrt {\frac {a x^{2}}{b} + 1}} - \frac {3 b \operatorname {asinh}{\left (\frac {\sqrt {a} x}{\sqrt {b}} \right )}}{2 a^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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