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

Optimal. Leaf size=282 \[ -\frac {c^{3/2} \sqrt {\frac {a c+a d x^2+b}{c+d x^2}} F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|\frac {b}{b+a c}\right )}{3 d^{3/2} \sqrt {\frac {c \left (a c+a d x^2+b\right )}{(a c+b) \left (c+d x^2\right )}}}-\frac {\sqrt {c} (b-a c) \sqrt {\frac {a c+a d x^2+b}{c+d x^2}} E\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|\frac {b}{b+a c}\right )}{3 a d^{3/2} \sqrt {\frac {c \left (a c+a d x^2+b\right )}{(a c+b) \left (c+d x^2\right )}}}+\frac {x (b-a c) \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{3 a d}+\frac {x \left (c+d x^2\right ) \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.52, antiderivative size = 370, normalized size of antiderivative = 1.31, number of steps used = 7, number of rules used = 7, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6722, 1975, 478, 531, 418, 492, 411} \[ -\frac {c^{3/2} \sqrt {a c+a d x^2+b} \sqrt {a+\frac {b}{c+d x^2}} F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|\frac {b}{b+a c}\right )}{3 d^{3/2} \sqrt {\frac {c \left (a c+a d x^2+b\right )}{(a c+b) \left (c+d x^2\right )}} \sqrt {a \left (c+d x^2\right )+b}}-\frac {\sqrt {c} (b-a c) \sqrt {a c+a d x^2+b} \sqrt {a+\frac {b}{c+d x^2}} E\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|\frac {b}{b+a c}\right )}{3 a d^{3/2} \sqrt {\frac {c \left (a c+a d x^2+b\right )}{(a c+b) \left (c+d x^2\right )}} \sqrt {a \left (c+d x^2\right )+b}}+\frac {x (b-a c) \sqrt {a c+a d x^2+b} \sqrt {a+\frac {b}{c+d x^2}}}{3 a d \sqrt {a \left (c+d x^2\right )+b}}+\frac {x \left (c+d x^2\right ) \sqrt {a c+a d x^2+b} \sqrt {a+\frac {b}{c+d x^2}}}{3 d \sqrt {a \left (c+d x^2\right )+b}} \]

Antiderivative was successfully verified.

[In]

Int[x^2*Sqrt[a + b/(c + d*x^2)],x]

[Out]

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

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 478

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

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

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Mathematica [C]  time = 0.58, size = 250, normalized size = 0.89 \[ \frac {\sqrt {\frac {a c+a d x^2+b}{c+d x^2}} \left (x \left (c+d x^2\right ) \sqrt {\frac {a d}{a c+b}} \left (a c+a d x^2+b\right )+2 i b c \sqrt {\frac {d x^2}{c}+1} \sqrt {\frac {a c+a d x^2+b}{a c+b}} F\left (i \sinh ^{-1}\left (\sqrt {\frac {a d}{b+a c}} x\right )|\frac {b}{a c}+1\right )+i c (a c-b) \sqrt {\frac {d x^2}{c}+1} \sqrt {\frac {a c+a d x^2+b}{a c+b}} E\left (i \sinh ^{-1}\left (\sqrt {\frac {a d}{b+a c}} x\right )|\frac {b}{a c}+1\right )\right )}{3 d \sqrt {\frac {a d}{a c+b}} \left (a \left (c+d x^2\right )+b\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)]*(Sqrt[(a*d)/(b + a*c)]*x*(c + d*x^2)*(b + a*c + a*d*x^2) + I*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)*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)]))/(3*d*Sqrt[(a*d)/(b + a*c)]*(b + a*(c + d*x^2)))

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fricas [F]  time = 0.83, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x^{2} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^2*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 \[ \int \sqrt {a + \frac {b}{d x^{2} + c}} x^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.02, size = 406, normalized size = 1.44 \[ \frac {\left (\sqrt {-\frac {a d}{a c +b}}\, a \,d^{2} x^{5}+2 \sqrt {-\frac {a d}{a c +b}}\, a c d \,x^{3}+\sqrt {-\frac {a d}{a c +b}}\, b d \,x^{3}+\sqrt {-\frac {a d}{a c +b}}\, a \,c^{2} x -\sqrt {\frac {a d \,x^{2}+a c +b}{a c +b}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, a \,c^{2} \EllipticE \left (\sqrt {-\frac {a d}{a c +b}}\, x , \sqrt {\frac {a c +b}{a c}}\right )+\sqrt {-\frac {a d}{a c +b}}\, b c x +\sqrt {\frac {a d \,x^{2}+a c +b}{a c +b}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, b c \EllipticE \left (\sqrt {-\frac {a d}{a c +b}}\, x , \sqrt {\frac {a c +b}{a c}}\right )-2 \sqrt {\frac {a d \,x^{2}+a c +b}{a c +b}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, b c \EllipticF \left (\sqrt {-\frac {a d}{a c +b}}\, x , \sqrt {\frac {a c +b}{a c}}\right )\right ) \left (d \,x^{2}+c \right ) \sqrt {\frac {a d \,x^{2}+a c +b}{d \,x^{2}+c}}}{3 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}\, \sqrt {-\frac {a d}{a c +b}}\, \sqrt {\left (d \,x^{2}+c \right ) \left (a d \,x^{2}+a c +b \right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*((-1/(a*c+b)*a*d)^(1/2)*x^5*a*d^2+2*(-1/(a*c+b)*a*d)^(1/2)*x^3*a*c*d+(-1/(a*c+b)*a*d)^(1/2)*x^3*b*d-((a*d*
x^2+a*c+b)/(a*c+b))^(1/2)*((d*x^2+c)/c)^(1/2)*EllipticE((-1/(a*c+b)*a*d)^(1/2)*x,((a*c+b)/a/c)^(1/2))*a*c^2+(-
1/(a*c+b)*a*d)^(1/2)*x*a*c^2-2*((a*d*x^2+a*c+b)/(a*c+b))^(1/2)*((d*x^2+c)/c)^(1/2)*EllipticF((-1/(a*c+b)*a*d)^
(1/2)*x,((a*c+b)/a/c)^(1/2))*b*c+((a*d*x^2+a*c+b)/(a*c+b))^(1/2)*((d*x^2+c)/c)^(1/2)*EllipticE((-1/(a*c+b)*a*d
)^(1/2)*x,((a*c+b)/a/c)^(1/2))*b*c+(-1/(a*c+b)*a*d)^(1/2)*x*b*c)*(d*x^2+c)*((a*d*x^2+a*c+b)/(d*x^2+c))^(1/2)/d
/(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 \[ \int \sqrt {a + \frac {b}{d x^{2} + c}} x^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a + b/(c + d*x^2))^(1/2),x)

[Out]

int(x^2*(a + b/(c + d*x^2))^(1/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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