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

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

[Out]

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

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Rubi [A]  time = 0.135793, antiderivative size = 73, normalized size of antiderivative = 1.03, number of steps used = 5, number of rules used = 5, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.263, Rules used = {6720, 266, 51, 63, 208} \[ \frac{c \sqrt{\frac{c}{a+b x^2}}}{a}-\frac{c \sqrt{a+b x^2} \sqrt{\frac{c}{a+b x^2}} \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{a^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(c/(a + b*x^2))^(3/2)/x,x]

[Out]

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

Rule 6720

Int[(u_.)*((a_.)*(v_)^(m_.))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a*v^m)^FracPart[p])/v^(m*FracPart[p]), Int
[u*v^(m*p), x], x] /; FreeQ[{a, m, p}, x] &&  !IntegerQ[p] &&  !FreeQ[v, x] &&  !(EqQ[a, 1] && EqQ[m, 1]) &&
!(EqQ[v, x] && EqQ[m, 1])

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

Mathematica [C]  time = 0.0089605, size = 38, normalized size = 0.54 \[ \frac{c \sqrt{\frac{c}{a+b x^2}} \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{b x^2}{a}+1\right )}{a} \]

Antiderivative was successfully verified.

[In]

Integrate[(c/(a + b*x^2))^(3/2)/x,x]

[Out]

(c*Sqrt[c/(a + b*x^2)]*Hypergeometric2F1[-1/2, 1, 1/2, 1 + (b*x^2)/a])/a

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Maple [A]  time = 0.005, size = 64, normalized size = 0.9 \begin{align*} -{(b{x}^{2}+a) \left ({\frac{c}{b{x}^{2}+a}} \right ) ^{{\frac{3}{2}}} \left ( \ln \left ( 2\,{\frac{\sqrt{a}\sqrt{b{x}^{2}+a}+a}{x}} \right ) a\sqrt{b{x}^{2}+a}-{a}^{{\frac{3}{2}}} \right ){a}^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(c/(b*x^2+a))^(3/2)*(b*x^2+a)*(ln(2*(a^(1/2)*(b*x^2+a)^(1/2)+a)/x)*a*(b*x^2+a)^(1/2)-a^(3/2))/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((c/(b*x^2+a))^(3/2)/x,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.53063, size = 289, normalized size = 4.07 \begin{align*} \left [\frac{c \sqrt{\frac{c}{a}} \log \left (-\frac{b c x^{2} + 2 \, a c - 2 \,{\left (a b x^{2} + a^{2}\right )} \sqrt{\frac{c}{b x^{2} + a}} \sqrt{\frac{c}{a}}}{x^{2}}\right ) + 2 \, c \sqrt{\frac{c}{b x^{2} + a}}}{2 \, a}, \frac{c \sqrt{-\frac{c}{a}} \arctan \left (\frac{a \sqrt{\frac{c}{b x^{2} + a}} \sqrt{-\frac{c}{a}}}{c}\right ) + c \sqrt{\frac{c}{b x^{2} + a}}}{a}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (\frac{c}{a + b x^{2}}\right )^{\frac{3}{2}}}{x}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((c/(a + b*x**2))**(3/2)/x, x)

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Giac [A]  time = 1.17665, size = 88, normalized size = 1.24 \begin{align*} c^{3}{\left (\frac{\arctan \left (\frac{\sqrt{b c x^{2} + a c}}{\sqrt{-a c}}\right )}{\sqrt{-a c} a c} + \frac{1}{\sqrt{b c x^{2} + a c} a c}\right )} \mathrm{sgn}\left (b x^{2} + a\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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