3.1916 \(\int \frac{x}{\sqrt{a+\frac{b}{x^2}}} \, dx\)

Optimal. Leaf size=50 \[ \frac{x^2 \sqrt{a+\frac{b}{x^2}}}{2 a}-\frac{b \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x^2}}}{\sqrt{a}}\right )}{2 a^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0233808, antiderivative size = 50, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.308, Rules used = {266, 51, 63, 208} \[ \frac{x^2 \sqrt{a+\frac{b}{x^2}}}{2 a}-\frac{b \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x^2}}}{\sqrt{a}}\right )}{2 a^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[x/Sqrt[a + b/x^2],x]

[Out]

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

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]]

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]

Rubi steps

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

Mathematica [A]  time = 0.0223497, size = 71, normalized size = 1.42 \[ \frac{\sqrt{a} x \left (a x^2+b\right )-b \sqrt{a x^2+b} \tanh ^{-1}\left (\frac{\sqrt{a} x}{\sqrt{a x^2+b}}\right )}{2 a^{3/2} x \sqrt{a+\frac{b}{x^2}}} \]

Antiderivative was successfully verified.

[In]

Integrate[x/Sqrt[a + b/x^2],x]

[Out]

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

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Maple [A]  time = 0.007, size = 66, normalized size = 1.3 \begin{align*}{\frac{1}{2\,x}\sqrt{a{x}^{2}+b} \left ( x\sqrt{a{x}^{2}+b}{a}^{{\frac{3}{2}}}-b\ln \left ( x\sqrt{a}+\sqrt{a{x}^{2}+b} \right ) a \right ){\frac{1}{\sqrt{{\frac{a{x}^{2}+b}{{x}^{2}}}}}}{a}^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.57155, size = 297, normalized size = 5.94 \begin{align*} \left [\frac{2 \, a x^{2} \sqrt{\frac{a x^{2} + b}{x^{2}}} + \sqrt{a} b \log \left (-2 \, a x^{2} + 2 \, \sqrt{a} x^{2} \sqrt{\frac{a x^{2} + b}{x^{2}}} - b\right )}{4 \, a^{2}}, \frac{a x^{2} \sqrt{\frac{a x^{2} + b}{x^{2}}} + \sqrt{-a} b \arctan \left (\frac{\sqrt{-a} x^{2} \sqrt{\frac{a x^{2} + b}{x^{2}}}}{a x^{2} + b}\right )}{2 \, a^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 3.01197, size = 42, normalized size = 0.84 \begin{align*} \frac{\sqrt{b} x \sqrt{\frac{a x^{2}}{b} + 1}}{2 a} - \frac{b \operatorname{asinh}{\left (\frac{\sqrt{a} x}{\sqrt{b}} \right )}}{2 a^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.2165, size = 90, normalized size = 1.8 \begin{align*} \frac{1}{2} \, b{\left (\frac{\arctan \left (\frac{\sqrt{\frac{a x^{2} + b}{x^{2}}}}{\sqrt{-a}}\right )}{\sqrt{-a} a} - \frac{\sqrt{\frac{a x^{2} + b}{x^{2}}}}{{\left (a - \frac{a x^{2} + b}{x^{2}}\right )} a}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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