3.266 \(\int \frac{A+B x^2}{\sqrt{x} (b x^2+c x^4)^{3/2}} \, dx\)

Optimal. Leaf size=368 \[ \frac{3 \sqrt [4]{c} x \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{\frac{b+c x^2}{\left (\sqrt{b}+\sqrt{c} x\right )^2}} (5 b B-7 A c) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right ),\frac{1}{2}\right )}{10 b^{11/4} \sqrt{b x^2+c x^4}}+\frac{3 \sqrt{c} x^{3/2} \left (b+c x^2\right ) (5 b B-7 A c)}{5 b^3 \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{b x^2+c x^4}}+\frac{\sqrt{x} (5 b B-7 A c)}{5 b^2 \sqrt{b x^2+c x^4}}-\frac{3 \sqrt{b x^2+c x^4} (5 b B-7 A c)}{5 b^3 x^{3/2}}-\frac{3 \sqrt [4]{c} x \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{\frac{b+c x^2}{\left (\sqrt{b}+\sqrt{c} x\right )^2}} (5 b B-7 A c) E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )|\frac{1}{2}\right )}{5 b^{11/4} \sqrt{b x^2+c x^4}}-\frac{2 A}{5 b x^{3/2} \sqrt{b x^2+c x^4}} \]

[Out]

(-2*A)/(5*b*x^(3/2)*Sqrt[b*x^2 + c*x^4]) + ((5*b*B - 7*A*c)*Sqrt[x])/(5*b^2*Sqrt[b*x^2 + c*x^4]) + (3*Sqrt[c]*
(5*b*B - 7*A*c)*x^(3/2)*(b + c*x^2))/(5*b^3*(Sqrt[b] + Sqrt[c]*x)*Sqrt[b*x^2 + c*x^4]) - (3*(5*b*B - 7*A*c)*Sq
rt[b*x^2 + c*x^4])/(5*b^3*x^(3/2)) - (3*c^(1/4)*(5*b*B - 7*A*c)*x*(Sqrt[b] + Sqrt[c]*x)*Sqrt[(b + c*x^2)/(Sqrt
[b] + Sqrt[c]*x)^2]*EllipticE[2*ArcTan[(c^(1/4)*Sqrt[x])/b^(1/4)], 1/2])/(5*b^(11/4)*Sqrt[b*x^2 + c*x^4]) + (3
*c^(1/4)*(5*b*B - 7*A*c)*x*(Sqrt[b] + Sqrt[c]*x)*Sqrt[(b + c*x^2)/(Sqrt[b] + Sqrt[c]*x)^2]*EllipticF[2*ArcTan[
(c^(1/4)*Sqrt[x])/b^(1/4)], 1/2])/(10*b^(11/4)*Sqrt[b*x^2 + c*x^4])

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Rubi [A]  time = 0.450657, antiderivative size = 368, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {2038, 2023, 2025, 2032, 329, 305, 220, 1196} \[ \frac{3 \sqrt{c} x^{3/2} \left (b+c x^2\right ) (5 b B-7 A c)}{5 b^3 \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{b x^2+c x^4}}+\frac{\sqrt{x} (5 b B-7 A c)}{5 b^2 \sqrt{b x^2+c x^4}}-\frac{3 \sqrt{b x^2+c x^4} (5 b B-7 A c)}{5 b^3 x^{3/2}}+\frac{3 \sqrt [4]{c} x \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{\frac{b+c x^2}{\left (\sqrt{b}+\sqrt{c} x\right )^2}} (5 b B-7 A c) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )|\frac{1}{2}\right )}{10 b^{11/4} \sqrt{b x^2+c x^4}}-\frac{3 \sqrt [4]{c} x \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{\frac{b+c x^2}{\left (\sqrt{b}+\sqrt{c} x\right )^2}} (5 b B-7 A c) E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )|\frac{1}{2}\right )}{5 b^{11/4} \sqrt{b x^2+c x^4}}-\frac{2 A}{5 b x^{3/2} \sqrt{b x^2+c x^4}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x^2)/(Sqrt[x]*(b*x^2 + c*x^4)^(3/2)),x]

[Out]

(-2*A)/(5*b*x^(3/2)*Sqrt[b*x^2 + c*x^4]) + ((5*b*B - 7*A*c)*Sqrt[x])/(5*b^2*Sqrt[b*x^2 + c*x^4]) + (3*Sqrt[c]*
(5*b*B - 7*A*c)*x^(3/2)*(b + c*x^2))/(5*b^3*(Sqrt[b] + Sqrt[c]*x)*Sqrt[b*x^2 + c*x^4]) - (3*(5*b*B - 7*A*c)*Sq
rt[b*x^2 + c*x^4])/(5*b^3*x^(3/2)) - (3*c^(1/4)*(5*b*B - 7*A*c)*x*(Sqrt[b] + Sqrt[c]*x)*Sqrt[(b + c*x^2)/(Sqrt
[b] + Sqrt[c]*x)^2]*EllipticE[2*ArcTan[(c^(1/4)*Sqrt[x])/b^(1/4)], 1/2])/(5*b^(11/4)*Sqrt[b*x^2 + c*x^4]) + (3
*c^(1/4)*(5*b*B - 7*A*c)*x*(Sqrt[b] + Sqrt[c]*x)*Sqrt[(b + c*x^2)/(Sqrt[b] + Sqrt[c]*x)^2]*EllipticF[2*ArcTan[
(c^(1/4)*Sqrt[x])/b^(1/4)], 1/2])/(10*b^(11/4)*Sqrt[b*x^2 + c*x^4])

Rule 2038

Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + (d_.)*(x_)^(n_.)), x_Symbol] :> Sim
p[(c*e^(j - 1)*(e*x)^(m - j + 1)*(a*x^j + b*x^(j + n))^(p + 1))/(a*(m + j*p + 1)), x] + Dist[(a*d*(m + j*p + 1
) - b*c*(m + n + p*(j + n) + 1))/(a*e^n*(m + j*p + 1)), Int[(e*x)^(m + n)*(a*x^j + b*x^(j + n))^p, x], x] /; F
reeQ[{a, b, c, d, e, j, p}, x] && EqQ[jn, j + n] &&  !IntegerQ[p] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && (LtQ[m
+ j*p, -1] || (IntegersQ[m - 1/2, p - 1/2] && LtQ[p, 0] && LtQ[m, -(n*p) - 1])) && (GtQ[e, 0] || IntegersQ[j,
n]) && NeQ[m + j*p + 1, 0] && NeQ[m - n + j*p + 1, 0]

Rule 2023

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> -Simp[(c^(j - 1)*(c*x)^(m - j
+ 1)*(a*x^j + b*x^n)^(p + 1))/(a*(n - j)*(p + 1)), x] + Dist[(c^j*(m + n*p + n - j + 1))/(a*(n - j)*(p + 1)),
Int[(c*x)^(m - j)*(a*x^j + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[p] && LtQ[0, j, n] &
& (IntegersQ[j, n] || GtQ[c, 0]) && LtQ[p, -1]

Rule 2025

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(c^(j - 1)*(c*x)^(m - j +
1)*(a*x^j + b*x^n)^(p + 1))/(a*(m + j*p + 1)), x] - Dist[(b*(m + n*p + n - j + 1))/(a*c^(n - j)*(m + j*p + 1)
), Int[(c*x)^(m + n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IntegerQ[p] && LtQ[0, j,
n] && (IntegersQ[j, n] || GtQ[c, 0]) && LtQ[m + j*p + 1, 0]

Rule 2032

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[(c^IntPart[m]*(c*x)^FracP
art[m]*(a*x^j + b*x^n)^FracPart[p])/(x^(FracPart[m] + j*FracPart[p])*(a + b*x^(n - j))^FracPart[p]), Int[x^(m
+ j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[n, j] && PosQ[n
- j]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 305

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, Dist[1/q, Int[1/Sqrt[a + b*x^4], x],
x] - Dist[1/q, Int[(1 - q*x^2)/Sqrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 220

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 1196

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, -Simp[(d*x*Sqrt[a + c
*x^4])/(a*(1 + q^2*x^2)), x] + Simp[(d*(1 + q^2*x^2)*Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]*EllipticE[2*ArcTan[
q*x], 1/2])/(q*Sqrt[a + c*x^4]), x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rubi steps

\begin{align*} \int \frac{A+B x^2}{\sqrt{x} \left (b x^2+c x^4\right )^{3/2}} \, dx &=-\frac{2 A}{5 b x^{3/2} \sqrt{b x^2+c x^4}}-\frac{\left (2 \left (-\frac{5 b B}{2}+\frac{7 A c}{2}\right )\right ) \int \frac{x^{3/2}}{\left (b x^2+c x^4\right )^{3/2}} \, dx}{5 b}\\ &=-\frac{2 A}{5 b x^{3/2} \sqrt{b x^2+c x^4}}+\frac{(5 b B-7 A c) \sqrt{x}}{5 b^2 \sqrt{b x^2+c x^4}}+\frac{(3 (5 b B-7 A c)) \int \frac{1}{\sqrt{x} \sqrt{b x^2+c x^4}} \, dx}{10 b^2}\\ &=-\frac{2 A}{5 b x^{3/2} \sqrt{b x^2+c x^4}}+\frac{(5 b B-7 A c) \sqrt{x}}{5 b^2 \sqrt{b x^2+c x^4}}-\frac{3 (5 b B-7 A c) \sqrt{b x^2+c x^4}}{5 b^3 x^{3/2}}+\frac{(3 c (5 b B-7 A c)) \int \frac{x^{3/2}}{\sqrt{b x^2+c x^4}} \, dx}{10 b^3}\\ &=-\frac{2 A}{5 b x^{3/2} \sqrt{b x^2+c x^4}}+\frac{(5 b B-7 A c) \sqrt{x}}{5 b^2 \sqrt{b x^2+c x^4}}-\frac{3 (5 b B-7 A c) \sqrt{b x^2+c x^4}}{5 b^3 x^{3/2}}+\frac{\left (3 c (5 b B-7 A c) x \sqrt{b+c x^2}\right ) \int \frac{\sqrt{x}}{\sqrt{b+c x^2}} \, dx}{10 b^3 \sqrt{b x^2+c x^4}}\\ &=-\frac{2 A}{5 b x^{3/2} \sqrt{b x^2+c x^4}}+\frac{(5 b B-7 A c) \sqrt{x}}{5 b^2 \sqrt{b x^2+c x^4}}-\frac{3 (5 b B-7 A c) \sqrt{b x^2+c x^4}}{5 b^3 x^{3/2}}+\frac{\left (3 c (5 b B-7 A c) x \sqrt{b+c x^2}\right ) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{b+c x^4}} \, dx,x,\sqrt{x}\right )}{5 b^3 \sqrt{b x^2+c x^4}}\\ &=-\frac{2 A}{5 b x^{3/2} \sqrt{b x^2+c x^4}}+\frac{(5 b B-7 A c) \sqrt{x}}{5 b^2 \sqrt{b x^2+c x^4}}-\frac{3 (5 b B-7 A c) \sqrt{b x^2+c x^4}}{5 b^3 x^{3/2}}+\frac{\left (3 \sqrt{c} (5 b B-7 A c) x \sqrt{b+c x^2}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{b+c x^4}} \, dx,x,\sqrt{x}\right )}{5 b^{5/2} \sqrt{b x^2+c x^4}}-\frac{\left (3 \sqrt{c} (5 b B-7 A c) x \sqrt{b+c x^2}\right ) \operatorname{Subst}\left (\int \frac{1-\frac{\sqrt{c} x^2}{\sqrt{b}}}{\sqrt{b+c x^4}} \, dx,x,\sqrt{x}\right )}{5 b^{5/2} \sqrt{b x^2+c x^4}}\\ &=-\frac{2 A}{5 b x^{3/2} \sqrt{b x^2+c x^4}}+\frac{(5 b B-7 A c) \sqrt{x}}{5 b^2 \sqrt{b x^2+c x^4}}+\frac{3 \sqrt{c} (5 b B-7 A c) x^{3/2} \left (b+c x^2\right )}{5 b^3 \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{b x^2+c x^4}}-\frac{3 (5 b B-7 A c) \sqrt{b x^2+c x^4}}{5 b^3 x^{3/2}}-\frac{3 \sqrt [4]{c} (5 b B-7 A c) x \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{\frac{b+c x^2}{\left (\sqrt{b}+\sqrt{c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )|\frac{1}{2}\right )}{5 b^{11/4} \sqrt{b x^2+c x^4}}+\frac{3 \sqrt [4]{c} (5 b B-7 A c) x \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{\frac{b+c x^2}{\left (\sqrt{b}+\sqrt{c} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )|\frac{1}{2}\right )}{10 b^{11/4} \sqrt{b x^2+c x^4}}\\ \end{align*}

Mathematica [C]  time = 0.0431953, size = 79, normalized size = 0.21 \[ \frac{2 x^2 \sqrt{\frac{c x^2}{b}+1} (7 A c-5 b B) \, _2F_1\left (-\frac{1}{4},\frac{3}{2};\frac{3}{4};-\frac{c x^2}{b}\right )-2 A b}{5 b^2 x^{3/2} \sqrt{x^2 \left (b+c x^2\right )}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x^2)/(Sqrt[x]*(b*x^2 + c*x^4)^(3/2)),x]

[Out]

(-2*A*b + 2*(-5*b*B + 7*A*c)*x^2*Sqrt[1 + (c*x^2)/b]*Hypergeometric2F1[-1/4, 3/2, 3/4, -((c*x^2)/b)])/(5*b^2*x
^(3/2)*Sqrt[x^2*(b + c*x^2)])

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Maple [A]  time = 0.021, size = 420, normalized size = 1.1 \begin{align*} -{\frac{c{x}^{2}+b}{10\,{b}^{3}}\sqrt{x} \left ( 42\,A\sqrt{{\frac{cx+\sqrt{-bc}}{\sqrt{-bc}}}}\sqrt{2}\sqrt{{\frac{-cx+\sqrt{-bc}}{\sqrt{-bc}}}}\sqrt{-{\frac{cx}{\sqrt{-bc}}}}{\it EllipticE} \left ( \sqrt{{\frac{cx+\sqrt{-bc}}{\sqrt{-bc}}}},1/2\,\sqrt{2} \right ){x}^{2}bc-21\,A\sqrt{{\frac{cx+\sqrt{-bc}}{\sqrt{-bc}}}}\sqrt{2}\sqrt{{\frac{-cx+\sqrt{-bc}}{\sqrt{-bc}}}}\sqrt{-{\frac{cx}{\sqrt{-bc}}}}{\it EllipticF} \left ( \sqrt{{\frac{cx+\sqrt{-bc}}{\sqrt{-bc}}}},1/2\,\sqrt{2} \right ){x}^{2}bc-30\,B\sqrt{{\frac{cx+\sqrt{-bc}}{\sqrt{-bc}}}}\sqrt{2}\sqrt{{\frac{-cx+\sqrt{-bc}}{\sqrt{-bc}}}}\sqrt{-{\frac{cx}{\sqrt{-bc}}}}{\it EllipticE} \left ( \sqrt{{\frac{cx+\sqrt{-bc}}{\sqrt{-bc}}}},1/2\,\sqrt{2} \right ){x}^{2}{b}^{2}+15\,B\sqrt{{\frac{cx+\sqrt{-bc}}{\sqrt{-bc}}}}\sqrt{2}\sqrt{{\frac{-cx+\sqrt{-bc}}{\sqrt{-bc}}}}\sqrt{-{\frac{cx}{\sqrt{-bc}}}}{\it EllipticF} \left ( \sqrt{{\frac{cx+\sqrt{-bc}}{\sqrt{-bc}}}},1/2\,\sqrt{2} \right ){x}^{2}{b}^{2}-42\,A{c}^{2}{x}^{4}+30\,B{x}^{4}bc-28\,Abc{x}^{2}+20\,B{x}^{2}{b}^{2}+4\,A{b}^{2} \right ) \left ( c{x}^{4}+b{x}^{2} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)/(c*x^4+b*x^2)^(3/2)/x^(1/2),x)

[Out]

-1/10/(c*x^4+b*x^2)^(3/2)*x^(1/2)*(c*x^2+b)*(42*A*((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2)*2^(1/2)*((-c*x+(-b*c
)^(1/2))/(-b*c)^(1/2))^(1/2)*(-x*c/(-b*c)^(1/2))^(1/2)*EllipticE(((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2),1/2*2
^(1/2))*x^2*b*c-21*A*((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2)*2^(1/2)*((-c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2)*
(-x*c/(-b*c)^(1/2))^(1/2)*EllipticF(((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2),1/2*2^(1/2))*x^2*b*c-30*B*((c*x+(-
b*c)^(1/2))/(-b*c)^(1/2))^(1/2)*2^(1/2)*((-c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2)*(-x*c/(-b*c)^(1/2))^(1/2)*Ell
ipticE(((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2),1/2*2^(1/2))*x^2*b^2+15*B*((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/
2)*2^(1/2)*((-c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2)*(-x*c/(-b*c)^(1/2))^(1/2)*EllipticF(((c*x+(-b*c)^(1/2))/(-
b*c)^(1/2))^(1/2),1/2*2^(1/2))*x^2*b^2-42*A*c^2*x^4+30*B*x^4*b*c-28*A*b*c*x^2+20*B*x^2*b^2+4*A*b^2)/b^3

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{B x^{2} + A}{{\left (c x^{4} + b x^{2}\right )}^{\frac{3}{2}} \sqrt{x}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/(c*x^4+b*x^2)^(3/2)/x^(1/2),x, algorithm="maxima")

[Out]

integrate((B*x^2 + A)/((c*x^4 + b*x^2)^(3/2)*sqrt(x)), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{c x^{4} + b x^{2}}{\left (B x^{2} + A\right )} \sqrt{x}}{c^{2} x^{9} + 2 \, b c x^{7} + b^{2} x^{5}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/(c*x^4+b*x^2)^(3/2)/x^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(c*x^4 + b*x^2)*(B*x^2 + A)*sqrt(x)/(c^2*x^9 + 2*b*c*x^7 + b^2*x^5), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**2+A)/(c*x**4+b*x**2)**(3/2)/x**(1/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{B x^{2} + A}{{\left (c x^{4} + b x^{2}\right )}^{\frac{3}{2}} \sqrt{x}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/(c*x^4+b*x^2)^(3/2)/x^(1/2),x, algorithm="giac")

[Out]

integrate((B*x^2 + A)/((c*x^4 + b*x^2)^(3/2)*sqrt(x)), x)