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

Optimal. Leaf size=312 \[ \frac{d x (a c+2 b) \sqrt{\frac{a c+a d x^2+b}{c+d x^2}}}{c^2}-\frac{(a c+2 b) \left (c+d x^2\right ) \sqrt{\frac{a c+a d x^2+b}{c+d x^2}}}{c^2 x}-\frac{\sqrt{d} (a c+2 b) \sqrt{\frac{a c+a d x^2+b}{c+d x^2}} E\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|\frac{b}{b+a c}\right )}{c^{3/2} \sqrt{\frac{c \left (a c+a d x^2+b\right )}{(a c+b) \left (c+d x^2\right )}}}+\frac{b \sqrt{\frac{a c+a d x^2+b}{c+d x^2}}}{c x}+\frac{a \sqrt{d} \sqrt{\frac{a c+a d x^2+b}{c+d x^2}} F\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|\frac{b}{b+a c}\right )}{\sqrt{c} \sqrt{\frac{c \left (a c+a d x^2+b\right )}{(a c+b) \left (c+d x^2\right )}}} \]

[Out]

(b*Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)])/(c*x) + ((2*b + a*c)*d*x*Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)])/c^2
- ((2*b + a*c)*(c + d*x^2)*Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)])/(c^2*x) - ((2*b + a*c)*Sqrt[d]*Sqrt[(b + a*c
+ a*d*x^2)/(c + d*x^2)]*EllipticE[ArcTan[(Sqrt[d]*x)/Sqrt[c]], b/(b + a*c)])/(c^(3/2)*Sqrt[(c*(b + a*c + a*d*
x^2))/((b + a*c)*(c + d*x^2))]) + (a*Sqrt[d]*Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)]*EllipticF[ArcTan[(Sqrt[d]*x
)/Sqrt[c]], b/(b + a*c)])/(Sqrt[c]*Sqrt[(c*(b + a*c + a*d*x^2))/((b + a*c)*(c + d*x^2))])

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Rubi [A]  time = 0.663576, antiderivative size = 422, normalized size of antiderivative = 1.35, number of steps used = 8, number of rules used = 8, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.381, Rules used = {6722, 1975, 468, 583, 531, 418, 492, 411} \[ \frac{d x (a c+2 b) \sqrt{a c+a d x^2+b} \sqrt{a+\frac{b}{c+d x^2}}}{c^2 \sqrt{a \left (c+d x^2\right )+b}}-\frac{(a c+2 b) \left (c+d x^2\right ) \sqrt{a c+a d x^2+b} \sqrt{a+\frac{b}{c+d x^2}}}{c^2 x \sqrt{a \left (c+d x^2\right )+b}}-\frac{\sqrt{d} (a c+2 b) \sqrt{a c+a d x^2+b} \sqrt{a+\frac{b}{c+d x^2}} E\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|\frac{b}{b+a c}\right )}{c^{3/2} \sqrt{\frac{c \left (a c+a d x^2+b\right )}{(a c+b) \left (c+d x^2\right )}} \sqrt{a \left (c+d x^2\right )+b}}+\frac{b \sqrt{a c+a d x^2+b} \sqrt{a+\frac{b}{c+d x^2}}}{c x \sqrt{a \left (c+d x^2\right )+b}}+\frac{a \sqrt{d} \sqrt{a c+a d x^2+b} \sqrt{a+\frac{b}{c+d x^2}} F\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|\frac{b}{b+a c}\right )}{\sqrt{c} \sqrt{\frac{c \left (a c+a d x^2+b\right )}{(a c+b) \left (c+d x^2\right )}} \sqrt{a \left (c+d x^2\right )+b}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/(c + d*x^2))^(3/2)/x^2,x]

[Out]

(b*Sqrt[b + a*c + a*d*x^2]*Sqrt[a + b/(c + d*x^2)])/(c*x*Sqrt[b + a*(c + d*x^2)]) + ((2*b + a*c)*d*x*Sqrt[b +
a*c + a*d*x^2]*Sqrt[a + b/(c + d*x^2)])/(c^2*Sqrt[b + a*(c + d*x^2)]) - ((2*b + a*c)*(c + d*x^2)*Sqrt[b + a*c
+ a*d*x^2]*Sqrt[a + b/(c + d*x^2)])/(c^2*x*Sqrt[b + a*(c + d*x^2)]) - ((2*b + a*c)*Sqrt[d]*Sqrt[b + a*c + a*d*
x^2]*Sqrt[a + b/(c + d*x^2)]*EllipticE[ArcTan[(Sqrt[d]*x)/Sqrt[c]], b/(b + a*c)])/(c^(3/2)*Sqrt[(c*(b + a*c +
a*d*x^2))/((b + a*c)*(c + d*x^2))]*Sqrt[b + a*(c + d*x^2)]) + (a*Sqrt[d]*Sqrt[b + a*c + a*d*x^2]*Sqrt[a + b/(c
+ d*x^2)]*EllipticF[ArcTan[(Sqrt[d]*x)/Sqrt[c]], b/(b + a*c)])/(Sqrt[c]*Sqrt[(c*(b + a*c + a*d*x^2))/((b + a*
c)*(c + d*x^2))]*Sqrt[b + a*(c + d*x^2)])

Rule 6722

Int[(u_.)*((a_.) + (b_.)*(v_)^(n_))^(p_), x_Symbol] :> Dist[(a + b*v^n)^FracPart[p]/(v^(n*FracPart[p])*(b + a/
v^n)^FracPart[p]), Int[u*v^(n*p)*(b + a/v^n)^p, x], x] /; FreeQ[{a, b, p}, x] &&  !IntegerQ[p] && ILtQ[n, 0] &
& BinomialQ[v, x] &&  !LinearQ[v, x]

Rule 1975

Int[(u_)^(p_.)*(v_)^(q_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Int[(e*x)^m*ExpandToSum[u, x]^p*ExpandToSum[v, x]^q
, x] /; FreeQ[{e, m, p, q}, x] && BinomialQ[{u, v}, x] && EqQ[BinomialDegree[u, x] - BinomialDegree[v, x], 0]
&&  !BinomialMatchQ[{u, v}, x]

Rule 468

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[((c*b -
a*d)*(e*x)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 1))/(a*b*e*n*(p + 1)), x] + Dist[1/(a*b*n*(p + 1)), I
nt[(e*x)^m*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 2)*Simp[c*(c*b*n*(p + 1) + (c*b - a*d)*(m + 1)) + d*(c*b*n*(p
+ 1) + (c*b - a*d)*(m + n*(q - 1) + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] &
& IGtQ[n, 0] && LtQ[p, -1] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 583

Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> Simp[(e*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*c*g*(m + 1)), x] + Dist[1/(a*c*
g^n*(m + 1)), Int[(g*x)^(m + n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[a*f*c*(m + 1) - e*(b*c + a*d)*(m + n + 1) - e
*n*(b*c*p + a*d*q) - b*e*d*(m + n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p, q}, x] &&
IGtQ[n, 0] && LtQ[m, -1]

Rule 531

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> Dist[
e, Int[(a + b*x^n)^p*(c + d*x^n)^q, x], x] + Dist[f, Int[x^n*(a + b*x^n)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a,
b, c, d, e, f, n, p, q}, x]

Rule 418

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(Sqrt[a + b*x^2]*EllipticF[ArcT
an[Rt[d/c, 2]*x], 1 - (b*c)/(a*d)])/(a*Rt[d/c, 2]*Sqrt[c + d*x^2]*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]), x] /
; FreeQ[{a, b, c, d}, x] && PosQ[d/c] && PosQ[b/a] &&  !SimplerSqrtQ[b/a, d/c]

Rule 492

Int[(x_)^2/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(x*Sqrt[a + b*x^2])/(b*Sqr
t[c + d*x^2]), x] - Dist[c/b, Int[Sqrt[a + b*x^2]/(c + d*x^2)^(3/2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b
*c - a*d, 0] && PosQ[b/a] && PosQ[d/c] &&  !SimplerSqrtQ[b/a, d/c]

Rule 411

Int[Sqrt[(a_) + (b_.)*(x_)^2]/((c_) + (d_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(Sqrt[a + b*x^2]*EllipticE[ArcTan
[Rt[d/c, 2]*x], 1 - (b*c)/(a*d)])/(c*Rt[d/c, 2]*Sqrt[c + d*x^2]*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]), x] /;
FreeQ[{a, b, c, d}, x] && PosQ[b/a] && PosQ[d/c]

Rubi steps

\begin{align*} \int \frac{\left (a+\frac{b}{c+d x^2}\right )^{3/2}}{x^2} \, dx &=\frac{\left (\sqrt{c+d x^2} \sqrt{a+\frac{b}{c+d x^2}}\right ) \int \frac{\left (b+a \left (c+d x^2\right )\right )^{3/2}}{x^2 \left (c+d x^2\right )^{3/2}} \, dx}{\sqrt{b+a \left (c+d x^2\right )}}\\ &=\frac{\left (\sqrt{c+d x^2} \sqrt{a+\frac{b}{c+d x^2}}\right ) \int \frac{\left (b+a c+a d x^2\right )^{3/2}}{x^2 \left (c+d x^2\right )^{3/2}} \, dx}{\sqrt{b+a \left (c+d x^2\right )}}\\ &=\frac{b \sqrt{b+a c+a d x^2} \sqrt{a+\frac{b}{c+d x^2}}}{c x \sqrt{b+a \left (c+d x^2\right )}}-\frac{\left (\sqrt{c+d x^2} \sqrt{a+\frac{b}{c+d x^2}}\right ) \int \frac{-(b+a c) (2 b+a c) d-a (b+a c) d^2 x^2}{x^2 \sqrt{c+d x^2} \sqrt{b+a c+a d x^2}} \, dx}{c d \sqrt{b+a \left (c+d x^2\right )}}\\ &=\frac{b \sqrt{b+a c+a d x^2} \sqrt{a+\frac{b}{c+d x^2}}}{c x \sqrt{b+a \left (c+d x^2\right )}}-\frac{(2 b+a c) \left (c+d x^2\right ) \sqrt{b+a c+a d x^2} \sqrt{a+\frac{b}{c+d x^2}}}{c^2 x \sqrt{b+a \left (c+d x^2\right )}}+\frac{\left (\sqrt{c+d x^2} \sqrt{a+\frac{b}{c+d x^2}}\right ) \int \frac{a c (b+a c)^2 d^2+a (b+a c) (2 b+a c) d^3 x^2}{\sqrt{c+d x^2} \sqrt{b+a c+a d x^2}} \, dx}{c^2 (b+a c) d \sqrt{b+a \left (c+d x^2\right )}}\\ &=\frac{b \sqrt{b+a c+a d x^2} \sqrt{a+\frac{b}{c+d x^2}}}{c x \sqrt{b+a \left (c+d x^2\right )}}-\frac{(2 b+a c) \left (c+d x^2\right ) \sqrt{b+a c+a d x^2} \sqrt{a+\frac{b}{c+d x^2}}}{c^2 x \sqrt{b+a \left (c+d x^2\right )}}+\frac{\left (a (b+a c) d \sqrt{c+d x^2} \sqrt{a+\frac{b}{c+d x^2}}\right ) \int \frac{1}{\sqrt{c+d x^2} \sqrt{b+a c+a d x^2}} \, dx}{c \sqrt{b+a \left (c+d x^2\right )}}+\frac{\left (a (2 b+a c) d^2 \sqrt{c+d x^2} \sqrt{a+\frac{b}{c+d x^2}}\right ) \int \frac{x^2}{\sqrt{c+d x^2} \sqrt{b+a c+a d x^2}} \, dx}{c^2 \sqrt{b+a \left (c+d x^2\right )}}\\ &=\frac{b \sqrt{b+a c+a d x^2} \sqrt{a+\frac{b}{c+d x^2}}}{c x \sqrt{b+a \left (c+d x^2\right )}}+\frac{(2 b+a c) d x \sqrt{b+a c+a d x^2} \sqrt{a+\frac{b}{c+d x^2}}}{c^2 \sqrt{b+a \left (c+d x^2\right )}}-\frac{(2 b+a c) \left (c+d x^2\right ) \sqrt{b+a c+a d x^2} \sqrt{a+\frac{b}{c+d x^2}}}{c^2 x \sqrt{b+a \left (c+d x^2\right )}}+\frac{a \sqrt{d} \sqrt{b+a c+a d x^2} \sqrt{a+\frac{b}{c+d x^2}} F\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|\frac{b}{b+a c}\right )}{\sqrt{c} \sqrt{\frac{c \left (b+a c+a d x^2\right )}{(b+a c) \left (c+d x^2\right )}} \sqrt{b+a \left (c+d x^2\right )}}-\frac{\left ((2 b+a c) d \sqrt{c+d x^2} \sqrt{a+\frac{b}{c+d x^2}}\right ) \int \frac{\sqrt{b+a c+a d x^2}}{\left (c+d x^2\right )^{3/2}} \, dx}{c \sqrt{b+a \left (c+d x^2\right )}}\\ &=\frac{b \sqrt{b+a c+a d x^2} \sqrt{a+\frac{b}{c+d x^2}}}{c x \sqrt{b+a \left (c+d x^2\right )}}+\frac{(2 b+a c) d x \sqrt{b+a c+a d x^2} \sqrt{a+\frac{b}{c+d x^2}}}{c^2 \sqrt{b+a \left (c+d x^2\right )}}-\frac{(2 b+a c) \left (c+d x^2\right ) \sqrt{b+a c+a d x^2} \sqrt{a+\frac{b}{c+d x^2}}}{c^2 x \sqrt{b+a \left (c+d x^2\right )}}-\frac{(2 b+a c) \sqrt{d} \sqrt{b+a c+a d x^2} \sqrt{a+\frac{b}{c+d x^2}} E\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|\frac{b}{b+a c}\right )}{c^{3/2} \sqrt{\frac{c \left (b+a c+a d x^2\right )}{(b+a c) \left (c+d x^2\right )}} \sqrt{b+a \left (c+d x^2\right )}}+\frac{a \sqrt{d} \sqrt{b+a c+a d x^2} \sqrt{a+\frac{b}{c+d x^2}} F\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|\frac{b}{b+a c}\right )}{\sqrt{c} \sqrt{\frac{c \left (b+a c+a d x^2\right )}{(b+a c) \left (c+d x^2\right )}} \sqrt{b+a \left (c+d x^2\right )}}\\ \end{align*}

Mathematica [C]  time = 0.703092, size = 278, normalized size = 0.89 \[ -\frac{\sqrt{\frac{a c+a d x^2+b}{c+d x^2}} \left (\sqrt{\frac{a d}{a c+b}} \left (a^2 c \left (c+d x^2\right )^2+2 a b \left (c+d x^2\right )^2+b^2 \left (c+2 d x^2\right )\right )-i a b c d x \sqrt{\frac{d x^2}{c}+1} \sqrt{\frac{a c+a d x^2+b}{a c+b}} F\left (i \sinh ^{-1}\left (\sqrt{\frac{a d}{b+a c}} x\right )|\frac{b}{a c}+1\right )+i a c d x (a c+2 b) \sqrt{\frac{d x^2}{c}+1} \sqrt{\frac{a c+a d x^2+b}{a c+b}} E\left (i \sinh ^{-1}\left (\sqrt{\frac{a d}{b+a c}} x\right )|\frac{b}{a c}+1\right )\right )}{c^2 x \sqrt{\frac{a d}{a c+b}} \left (a \left (c+d x^2\right )+b\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)]*(Sqrt[(a*d)/(b + a*c)]*(2*a*b*(c + d*x^2)^2 + a^2*c*(c + d*x^2)^2 + b
^2*(c + 2*d*x^2)) + I*a*c*(2*b + a*c)*d*x*Sqrt[(b + a*c + a*d*x^2)/(b + a*c)]*Sqrt[1 + (d*x^2)/c]*EllipticE[I*
ArcSinh[Sqrt[(a*d)/(b + a*c)]*x], 1 + b/(a*c)] - I*a*b*c*d*x*Sqrt[(b + a*c + a*d*x^2)/(b + a*c)]*Sqrt[1 + (d*x
^2)/c]*EllipticF[I*ArcSinh[Sqrt[(a*d)/(b + a*c)]*x], 1 + b/(a*c)]))/(c^2*Sqrt[(a*d)/(b + a*c)]*x*(b + a*(c + d
*x^2))))

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Maple [B]  time = 0.016, size = 873, normalized size = 2.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((-a*d/(a*c+b))^(1/2)*((d*x^2+c)*(a*d*x^2+a*c+b))^(1/2)*x^4*a^2*c*d^2+(-a*d/(a*c+b))^(1/2)*(a*d^2*x^4+2*a*c*d
*x^2+b*d*x^2+a*c^2+b*c)^(1/2)*x^4*a*b*d^2+(-a*d/(a*c+b))^(1/2)*((d*x^2+c)*(a*d*x^2+a*c+b))^(1/2)*x^4*a*b*d^2-(
(a*d*x^2+a*c+b)/(a*c+b))^(1/2)*((d*x^2+c)/c)^(1/2)*EllipticE(x*(-a*d/(a*c+b))^(1/2),((a*c+b)/a/c)^(1/2))*((d*x
^2+c)*(a*d*x^2+a*c+b))^(1/2)*x*a^2*c^2*d+2*(-a*d/(a*c+b))^(1/2)*((d*x^2+c)*(a*d*x^2+a*c+b))^(1/2)*x^2*a^2*c^2*
d+((a*d*x^2+a*c+b)/(a*c+b))^(1/2)*((d*x^2+c)/c)^(1/2)*EllipticF(x*(-a*d/(a*c+b))^(1/2),((a*c+b)/a/c)^(1/2))*((
d*x^2+c)*(a*d*x^2+a*c+b))^(1/2)*x*a*b*c*d-2*((a*d*x^2+a*c+b)/(a*c+b))^(1/2)*((d*x^2+c)/c)^(1/2)*EllipticE(x*(-
a*d/(a*c+b))^(1/2),((a*c+b)/a/c)^(1/2))*((d*x^2+c)*(a*d*x^2+a*c+b))^(1/2)*x*a*b*c*d+(-a*d/(a*c+b))^(1/2)*(a*d^
2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^(1/2)*x^2*a*b*c*d+3*(-a*d/(a*c+b))^(1/2)*((d*x^2+c)*(a*d*x^2+a*c+b))^(1/2
)*x^2*a*b*c*d+(-a*d/(a*c+b))^(1/2)*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^(1/2)*x^2*b^2*d+(-a*d/(a*c+b))^(1
/2)*((d*x^2+c)*(a*d*x^2+a*c+b))^(1/2)*x^2*b^2*d+(-a*d/(a*c+b))^(1/2)*((d*x^2+c)*(a*d*x^2+a*c+b))^(1/2)*a^2*c^3
+2*(-a*d/(a*c+b))^(1/2)*((d*x^2+c)*(a*d*x^2+a*c+b))^(1/2)*a*b*c^2+(-a*d/(a*c+b))^(1/2)*((d*x^2+c)*(a*d*x^2+a*c
+b))^(1/2)*b^2*c)*((a*d*x^2+a*c+b)/(d*x^2+c))^(1/2)/x/(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^(1/2)/(-a*d/(a
*c+b))^(1/2)/c^2/(a*d*x^2+a*c+b)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (a d x^{2} + a c + b\right )} \sqrt{\frac{a d x^{2} + a c + b}{d x^{2} + c}}}{d x^{4} + c x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((a*d*x^2 + a*c + b)*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c))/(d*x^4 + c*x^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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