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

Optimal. Leaf size=104 \[ -\frac{3 b c \sqrt{\frac{c}{a+b x^2}}}{2 a^2}+\frac{3 b c \sqrt{\frac{b x^2}{a}+1} \sqrt{\frac{c}{a+b x^2}} \tanh ^{-1}\left (\sqrt{\frac{b x^2}{a}+1}\right )}{2 a^2}-\frac{c \sqrt{\frac{c}{a+b x^2}}}{2 a x^2} \]

[Out]

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

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Rubi [A]  time = 0.152599, antiderivative size = 112, normalized size of antiderivative = 1.08, number of steps used = 6, 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} \[ -\frac{3 c \left (a+b x^2\right ) \sqrt{\frac{c}{a+b x^2}}}{2 a^2 x^2}+\frac{3 b c \sqrt{a+b x^2} \sqrt{\frac{c}{a+b x^2}} \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{2 a^{5/2}}+\frac{c \sqrt{\frac{c}{a+b x^2}}}{a x^2} \]

Antiderivative was successfully verified.

[In]

Int[(c/(a + b*x^2))^(3/2)/x^3,x]

[Out]

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

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

Mathematica [C]  time = 0.0091202, size = 40, normalized size = 0.38 \[ -\frac{b c \sqrt{\frac{c}{a+b x^2}} \, _2F_1\left (-\frac{1}{2},2;\frac{1}{2};\frac{b x^2}{a}+1\right )}{a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(c/(a + b*x^2))^(3/2)/x^3,x]

[Out]

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

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Maple [A]  time = 0.007, size = 79, normalized size = 0.8 \begin{align*} -{\frac{b{x}^{2}+a}{2\,{x}^{2}} \left ({\frac{c}{b{x}^{2}+a}} \right ) ^{{\frac{3}{2}}} \left ( 3\,{a}^{3/2}{x}^{2}b-3\,\ln \left ( 2\,{\frac{\sqrt{a}\sqrt{b{x}^{2}+a}+a}{x}} \right ) \sqrt{b{x}^{2}+a}{x}^{2}ab+{a}^{{\frac{5}{2}}} \right ){a}^{-{\frac{7}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(c/(b*x^2+a))^(3/2)*(b*x^2+a)*(3*a^(3/2)*x^2*b-3*ln(2*(a^(1/2)*(b*x^2+a)^(1/2)+a)/x)*(b*x^2+a)^(1/2)*x^2*
a*b+a^(5/2))/a^(7/2)/x^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^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.59462, size = 382, normalized size = 3.67 \begin{align*} \left [\frac{3 \, b c x^{2} \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 \,{\left (3 \, b c x^{2} + a c\right )} \sqrt{\frac{c}{b x^{2} + a}}}{4 \, a^{2} x^{2}}, -\frac{3 \, b c x^{2} \sqrt{-\frac{c}{a}} \arctan \left (\frac{a \sqrt{\frac{c}{b x^{2} + a}} \sqrt{-\frac{c}{a}}}{c}\right ) +{\left (3 \, b c x^{2} + a c\right )} \sqrt{\frac{c}{b x^{2} + a}}}{2 \, a^{2} x^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(3*b*c*x^2*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*(3
*b*c*x^2 + a*c)*sqrt(c/(b*x^2 + a)))/(a^2*x^2), -1/2*(3*b*c*x^2*sqrt(-c/a)*arctan(a*sqrt(c/(b*x^2 + a))*sqrt(-
c/a)/c) + (3*b*c*x^2 + a*c)*sqrt(c/(b*x^2 + a)))/(a^2*x^2)]

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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^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.19783, size = 135, normalized size = 1.3 \begin{align*} -\frac{1}{2} \, b c^{4}{\left (\frac{3 \, \arctan \left (\frac{\sqrt{b c x^{2} + a c}}{\sqrt{-a c}}\right )}{\sqrt{-a c} a^{2} c^{2}} - \frac{3 \, b c x^{2} + a c}{{\left (\sqrt{b c x^{2} + a c} a c -{\left (b c x^{2} + a c\right )}^{\frac{3}{2}}\right )} a^{2} c^{2}}\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^3,x, algorithm="giac")

[Out]

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