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

Optimal. Leaf size=260 \[ -\frac{x (b-a c) \sqrt{\frac{a c+a d x^2+b}{c+d x^2}}}{c}+\frac{b x \sqrt{\frac{a c+a d x^2+b}{c+d x^2}}}{c}+\frac{a \sqrt{c} \sqrt{\frac{a c+a d x^2+b}{c+d x^2}} F\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|\frac{b}{b+a c}\right )}{\sqrt{d} \sqrt{\frac{c \left (a c+a d x^2+b\right )}{(a c+b) \left (c+d x^2\right )}}}+\frac{(b-a c) \sqrt{\frac{a c+a d x^2+b}{c+d x^2}} E\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|\frac{b}{b+a c}\right )}{\sqrt{c} \sqrt{d} \sqrt{\frac{c \left (a c+a d x^2+b\right )}{(a c+b) \left (c+d x^2\right )}}} \]

[Out]

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

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Rubi [A]  time = 0.291417, antiderivative size = 348, normalized size of antiderivative = 1.34, number of steps used = 7, number of rules used = 7, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.412, Rules used = {6722, 1974, 413, 531, 418, 492, 411} \[ -\frac{x (b-a c) \sqrt{a c+a d x^2+b} \sqrt{a+\frac{b}{c+d x^2}}}{c \sqrt{a \left (c+d x^2\right )+b}}+\frac{b x \sqrt{a c+a d x^2+b} \sqrt{a+\frac{b}{c+d x^2}}}{c \sqrt{a \left (c+d x^2\right )+b}}+\frac{a \sqrt{c} \sqrt{a c+a d x^2+b} \sqrt{a+\frac{b}{c+d x^2}} F\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|\frac{b}{b+a c}\right )}{\sqrt{d} \sqrt{\frac{c \left (a c+a d x^2+b\right )}{(a c+b) \left (c+d x^2\right )}} \sqrt{a \left (c+d x^2\right )+b}}+\frac{(b-a c) \sqrt{a c+a d x^2+b} \sqrt{a+\frac{b}{c+d x^2}} E\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|\frac{b}{b+a c}\right )}{\sqrt{c} \sqrt{d} \sqrt{\frac{c \left (a c+a d x^2+b\right )}{(a c+b) \left (c+d x^2\right )}} \sqrt{a \left (c+d x^2\right )+b}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/(c + d*x^2))^(3/2),x]

[Out]

(b*x*Sqrt[b + a*c + a*d*x^2]*Sqrt[a + b/(c + d*x^2)])/(c*Sqrt[b + a*(c + d*x^2)]) - ((b - a*c)*x*Sqrt[b + a*c
+ a*d*x^2]*Sqrt[a + b/(c + d*x^2)])/(c*Sqrt[b + a*(c + d*x^2)]) + ((b - a*c)*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)])/(Sqrt[c]*Sqrt[d]*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[c]*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[d]*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 1974

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

Rule 413

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

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 \left (a+\frac{b}{c+d x^2}\right )^{3/2} \, 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}}{\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}}{\left (c+d x^2\right )^{3/2}} \, dx}{\sqrt{b+a \left (c+d x^2\right )}}\\ &=\frac{b x \sqrt{b+a c+a d x^2} \sqrt{a+\frac{b}{c+d x^2}}}{c \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) d-a (b-a c) d^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 x \sqrt{b+a c+a d x^2} \sqrt{a+\frac{b}{c+d x^2}}}{c \sqrt{b+a \left (c+d x^2\right )}}+\frac{\left (a (b+a c) \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}{\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{x^2}{\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{b x \sqrt{b+a c+a d x^2} \sqrt{a+\frac{b}{c+d x^2}}}{c \sqrt{b+a \left (c+d x^2\right )}}-\frac{(b-a c) x \sqrt{b+a c+a d x^2} \sqrt{a+\frac{b}{c+d x^2}}}{c \sqrt{b+a \left (c+d x^2\right )}}+\frac{a \sqrt{c} \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{d} \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 ((b-a c) \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}{\sqrt{b+a \left (c+d x^2\right )}}\\ &=\frac{b x \sqrt{b+a c+a d x^2} \sqrt{a+\frac{b}{c+d x^2}}}{c \sqrt{b+a \left (c+d x^2\right )}}-\frac{(b-a c) x \sqrt{b+a c+a d x^2} \sqrt{a+\frac{b}{c+d x^2}}}{c \sqrt{b+a \left (c+d x^2\right )}}+\frac{(b-a c) \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 )}{\sqrt{c} \sqrt{d} \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{c} \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{d} \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 = 0.542941, size = 243, normalized size = 0.93 \[ \frac{\sqrt{\frac{a c+a d x^2+b}{c+d x^2}} \left (b x \sqrt{\frac{a d}{a c+b}} \left (a \left (c+d x^2\right )+b\right )-2 i a b c \sqrt{\frac{d x^2}{c}+1} \sqrt{\frac{a c+a d x^2+b}{a c+b}} F\left (i \sinh ^{-1}\left (\sqrt{\frac{a d}{b+a c}} x\right )|\frac{b}{a c}+1\right )-i a c (a c-b) \sqrt{\frac{d x^2}{c}+1} \sqrt{\frac{a c+a d x^2+b}{a c+b}} E\left (i \sinh ^{-1}\left (\sqrt{\frac{a d}{b+a c}} x\right )|\frac{b}{a c}+1\right )\right )}{c \sqrt{\frac{a d}{a c+b}} \left (a \left (c+d x^2\right )+b\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)]*(b*Sqrt[(a*d)/(b + a*c)]*x*(b + a*(c + d*x^2)) - I*a*c*(-b + a*c)*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)]
- (2*I)*a*b*c*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*Sqrt[(a*d)/(b + a*c)]*(b + a*(c + d*x^2)))

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Maple [A]  time = 0.013, size = 515, normalized size = 2. \begin{align*}{\frac{1}{c \left ( ad{x}^{2}+ac+b \right ) } \left ( \sqrt{a{d}^{2}{x}^{4}+2\,acd{x}^{2}+bd{x}^{2}+{c}^{2}a+bc}\sqrt{-{\frac{ad}{ac+b}}}{x}^{3}abd+\sqrt{{\frac{ad{x}^{2}+ac+b}{ac+b}}}\sqrt{{\frac{d{x}^{2}+c}{c}}}{\it EllipticE} \left ( x\sqrt{-{\frac{ad}{ac+b}}},\sqrt{{\frac{ac+b}{ac}}} \right ) \sqrt{ \left ( d{x}^{2}+c \right ) \left ( ad{x}^{2}+ac+b \right ) }{a}^{2}{c}^{2}+2\,\sqrt{{\frac{ad{x}^{2}+ac+b}{ac+b}}}\sqrt{{\frac{d{x}^{2}+c}{c}}}{\it EllipticF} \left ( x\sqrt{-{\frac{ad}{ac+b}}},\sqrt{{\frac{ac+b}{ac}}} \right ) \sqrt{ \left ( d{x}^{2}+c \right ) \left ( ad{x}^{2}+ac+b \right ) }abc-\sqrt{{\frac{ad{x}^{2}+ac+b}{ac+b}}}\sqrt{{\frac{d{x}^{2}+c}{c}}}{\it EllipticE} \left ( x\sqrt{-{\frac{ad}{ac+b}}},\sqrt{{\frac{ac+b}{ac}}} \right ) \sqrt{ \left ( d{x}^{2}+c \right ) \left ( ad{x}^{2}+ac+b \right ) }abc+\sqrt{a{d}^{2}{x}^{4}+2\,acd{x}^{2}+bd{x}^{2}+{c}^{2}a+bc}\sqrt{-{\frac{ad}{ac+b}}}xabc+\sqrt{a{d}^{2}{x}^{4}+2\,acd{x}^{2}+bd{x}^{2}+{c}^{2}a+bc}\sqrt{-{\frac{ad}{ac+b}}}x{b}^{2} \right ) \sqrt{{\frac{ad{x}^{2}+ac+b}{d{x}^{2}+c}}}{\frac{1}{\sqrt{a{d}^{2}{x}^{4}+2\,acd{x}^{2}+bd{x}^{2}+{c}^{2}a+bc}}}{\frac{1}{\sqrt{-{\frac{ad}{ac+b}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((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)*x^3*a*b*d+((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)*
a^2*c^2+2*((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)*a*b*c-((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)*a*b*c+(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)*x*a*b*c+(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)*x*b^2)*((a*d*x^2+a*c+b)/(d*x^2+c))^(1/2)/(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/(a*d*x^2+a*c+b)

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

[Out]

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

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

[Out]

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

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

[Out]

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