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

Optimal. Leaf size=213 \[ x \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}+\frac {\sqrt {c} \sqrt {\frac {a c+a d x^2+b}{c+d x^2}} F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|\frac {b}{b+a c}\right )}{\sqrt {d} \sqrt {\frac {c \left (a c+a d x^2+b\right )}{(a c+b) \left (c+d x^2\right )}}}-\frac {\sqrt {c} \sqrt {\frac {a c+a d x^2+b}{c+d x^2}} E\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|\frac {b}{b+a c}\right )}{\sqrt {d} \sqrt {\frac {c \left (a c+a d x^2+b\right )}{(a c+b) \left (c+d x^2\right )}}} \]

[Out]

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

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Rubi [A]  time = 0.20, antiderivative size = 279, normalized size of antiderivative = 1.31, number of steps used = 6, number of rules used = 6, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {6722, 1974, 422, 418, 492, 411} \[ \frac {x \sqrt {a c+a d x^2+b} \sqrt {a+\frac {b}{c+d x^2}}}{\sqrt {a \left (c+d x^2\right )+b}}+\frac {\sqrt {c} \sqrt {a c+a d x^2+b} \sqrt {a+\frac {b}{c+d x^2}} F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|\frac {b}{b+a c}\right )}{\sqrt {d} \sqrt {\frac {c \left (a c+a d x^2+b\right )}{(a c+b) \left (c+d x^2\right )}} \sqrt {a \left (c+d x^2\right )+b}}-\frac {\sqrt {c} \sqrt {a c+a d x^2+b} \sqrt {a+\frac {b}{c+d x^2}} E\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|\frac {b}{b+a c}\right )}{\sqrt {d} \sqrt {\frac {c \left (a c+a d x^2+b\right )}{(a c+b) \left (c+d x^2\right )}} \sqrt {a \left (c+d x^2\right )+b}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b/(c + d*x^2)],x]

[Out]

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

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 422

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

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 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 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]

Rubi steps

\begin {align*} \int \sqrt {a+\frac {b}{c+d x^2}} \, dx &=\frac {\left (\sqrt {c+d x^2} \sqrt {a+\frac {b}{c+d x^2}}\right ) \int \frac {\sqrt {b+a \left (c+d x^2\right )}}{\sqrt {c+d x^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 {\sqrt {b+a c+a d x^2}}{\sqrt {c+d x^2}} \, dx}{\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 {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 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}{\sqrt {b+a \left (c+d x^2\right )}}\\ &=\frac {x \sqrt {b+a c+a d x^2} \sqrt {a+\frac {b}{c+d x^2}}}{\sqrt {b+a \left (c+d x^2\right )}}+\frac {\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 (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 {x \sqrt {b+a c+a d x^2} \sqrt {a+\frac {b}{c+d x^2}}}{\sqrt {b+a \left (c+d x^2\right )}}-\frac {\sqrt {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 {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 {\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*}

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Mathematica [A]  time = 0.06, size = 98, normalized size = 0.46 \[ \frac {\sqrt {\frac {c+d x^2}{c}} \sqrt {\frac {a c+a d x^2+b}{c+d x^2}} E\left (\sin ^{-1}\left (\sqrt {-\frac {d}{c}} x\right )|\frac {a c}{b+a c}\right )}{\sqrt {-\frac {d}{c}} \sqrt {\frac {a c+a d x^2+b}{a c+b}}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + b/(c + d*x^2)],x]

[Out]

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

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fricas [F]  time = 0.76, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {a + \frac {b}{d x^{2} + c}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(a + b/(d*x^2 + c)), x)

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maple [A]  time = 0.02, size = 199, normalized size = 0.93 \[ \frac {\left (a c \EllipticE \left (\sqrt {-\frac {a d}{a c +b}}\, x , \sqrt {\frac {a c +b}{a c}}\right )+b \EllipticF \left (\sqrt {-\frac {a d}{a c +b}}\, x , \sqrt {\frac {a c +b}{a c}}\right )\right ) \sqrt {\frac {d \,x^{2}+c}{c}}\, \sqrt {\frac {a d \,x^{2}+a c +b}{a c +b}}\, \left (d \,x^{2}+c \right ) \sqrt {\frac {a d \,x^{2}+a c +b}{d \,x^{2}+c}}}{\sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}\, \sqrt {-\frac {a d}{a c +b}}\, \sqrt {\left (d \,x^{2}+c \right ) \left (a d \,x^{2}+a c +b \right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {a + \frac {b}{d x^{2} + c}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/(d*x^2+c))^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(a + b/(d*x^2 + c)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \sqrt {a+\frac {b}{d\,x^2+c}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/(c + d*x^2))^(1/2),x)

[Out]

int((a + b/(c + d*x^2))^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {a + \frac {b}{c + d x^{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(a + b/(c + d*x**2)), x)

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