3.147 \(\int \sqrt{a+a \sec (e+f x)} (c+d \sec (e+f x))^4 \, dx\)

Optimal. Leaf size=271 \[ \frac{2 a^{3/2} c^4 \tan (e+f x) \tanh ^{-1}\left (\frac{\sqrt{a-a \sec (e+f x)}}{\sqrt{a}}\right )}{f \sqrt{a-a \sec (e+f x)} \sqrt{a \sec (e+f x)+a}}-\frac{2 d^4 \tan (e+f x) (a-a \sec (e+f x))^3}{7 a^2 f \sqrt{a \sec (e+f x)+a}}-\frac{2 d^2 \left (6 c^2+8 c d+3 d^2\right ) \tan (e+f x) (a-a \sec (e+f x))}{3 f \sqrt{a \sec (e+f x)+a}}+\frac{2 a d (2 c+d) \left (2 c^2+2 c d+d^2\right ) \tan (e+f x)}{f \sqrt{a \sec (e+f x)+a}}+\frac{2 d^3 (4 c+3 d) \tan (e+f x) (a-a \sec (e+f x))^2}{5 a f \sqrt{a \sec (e+f x)+a}} \]

[Out]

(2*a*d*(2*c + d)*(2*c^2 + 2*c*d + d^2)*Tan[e + f*x])/(f*Sqrt[a + a*Sec[e + f*x]]) + (2*a^(3/2)*c^4*ArcTanh[Sqr
t[a - a*Sec[e + f*x]]/Sqrt[a]]*Tan[e + f*x])/(f*Sqrt[a - a*Sec[e + f*x]]*Sqrt[a + a*Sec[e + f*x]]) - (2*d^2*(6
*c^2 + 8*c*d + 3*d^2)*(a - a*Sec[e + f*x])*Tan[e + f*x])/(3*f*Sqrt[a + a*Sec[e + f*x]]) + (2*d^3*(4*c + 3*d)*(
a - a*Sec[e + f*x])^2*Tan[e + f*x])/(5*a*f*Sqrt[a + a*Sec[e + f*x]]) - (2*d^4*(a - a*Sec[e + f*x])^3*Tan[e + f
*x])/(7*a^2*f*Sqrt[a + a*Sec[e + f*x]])

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Rubi [A]  time = 0.172894, antiderivative size = 271, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.148, Rules used = {3940, 88, 63, 206} \[ \frac{2 a^{3/2} c^4 \tan (e+f x) \tanh ^{-1}\left (\frac{\sqrt{a-a \sec (e+f x)}}{\sqrt{a}}\right )}{f \sqrt{a-a \sec (e+f x)} \sqrt{a \sec (e+f x)+a}}-\frac{2 d^4 \tan (e+f x) (a-a \sec (e+f x))^3}{7 a^2 f \sqrt{a \sec (e+f x)+a}}-\frac{2 d^2 \left (6 c^2+8 c d+3 d^2\right ) \tan (e+f x) (a-a \sec (e+f x))}{3 f \sqrt{a \sec (e+f x)+a}}+\frac{2 a d (2 c+d) \left (2 c^2+2 c d+d^2\right ) \tan (e+f x)}{f \sqrt{a \sec (e+f x)+a}}+\frac{2 d^3 (4 c+3 d) \tan (e+f x) (a-a \sec (e+f x))^2}{5 a f \sqrt{a \sec (e+f x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + a*Sec[e + f*x]]*(c + d*Sec[e + f*x])^4,x]

[Out]

(2*a*d*(2*c + d)*(2*c^2 + 2*c*d + d^2)*Tan[e + f*x])/(f*Sqrt[a + a*Sec[e + f*x]]) + (2*a^(3/2)*c^4*ArcTanh[Sqr
t[a - a*Sec[e + f*x]]/Sqrt[a]]*Tan[e + f*x])/(f*Sqrt[a - a*Sec[e + f*x]]*Sqrt[a + a*Sec[e + f*x]]) - (2*d^2*(6
*c^2 + 8*c*d + 3*d^2)*(a - a*Sec[e + f*x])*Tan[e + f*x])/(3*f*Sqrt[a + a*Sec[e + f*x]]) + (2*d^3*(4*c + 3*d)*(
a - a*Sec[e + f*x])^2*Tan[e + f*x])/(5*a*f*Sqrt[a + a*Sec[e + f*x]]) - (2*d^4*(a - a*Sec[e + f*x])^3*Tan[e + f
*x])/(7*a^2*f*Sqrt[a + a*Sec[e + f*x]])

Rule 3940

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.), x_Symbol] :> Di
st[(a^2*Cot[e + f*x])/(f*Sqrt[a + b*Csc[e + f*x]]*Sqrt[a - b*Csc[e + f*x]]), Subst[Int[((a + b*x)^(m - 1/2)*(c
+ d*x)^n)/(x*Sqrt[a - b*x]), x], x, Csc[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && NeQ[b*c - a*d,
0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && IntegerQ[m - 1/2]

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \sqrt{a+a \sec (e+f x)} (c+d \sec (e+f x))^4 \, dx &=-\frac{\left (a^2 \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{(c+d x)^4}{x \sqrt{a-a x}} \, dx,x,\sec (e+f x)\right )}{f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=-\frac{\left (a^2 \tan (e+f x)\right ) \operatorname{Subst}\left (\int \left (\frac{d (2 c+d) \left (2 c^2+2 c d+d^2\right )}{\sqrt{a-a x}}+\frac{c^4}{x \sqrt{a-a x}}-\frac{d^2 \left (6 c^2+8 c d+3 d^2\right ) \sqrt{a-a x}}{a}+\frac{d^3 (4 c+3 d) (a-a x)^{3/2}}{a^2}-\frac{d^4 (a-a x)^{5/2}}{a^3}\right ) \, dx,x,\sec (e+f x)\right )}{f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=\frac{2 a d (2 c+d) \left (2 c^2+2 c d+d^2\right ) \tan (e+f x)}{f \sqrt{a+a \sec (e+f x)}}-\frac{2 d^2 \left (6 c^2+8 c d+3 d^2\right ) (a-a \sec (e+f x)) \tan (e+f x)}{3 f \sqrt{a+a \sec (e+f x)}}+\frac{2 d^3 (4 c+3 d) (a-a \sec (e+f x))^2 \tan (e+f x)}{5 a f \sqrt{a+a \sec (e+f x)}}-\frac{2 d^4 (a-a \sec (e+f x))^3 \tan (e+f x)}{7 a^2 f \sqrt{a+a \sec (e+f x)}}-\frac{\left (a^2 c^4 \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a-a x}} \, dx,x,\sec (e+f x)\right )}{f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=\frac{2 a d (2 c+d) \left (2 c^2+2 c d+d^2\right ) \tan (e+f x)}{f \sqrt{a+a \sec (e+f x)}}-\frac{2 d^2 \left (6 c^2+8 c d+3 d^2\right ) (a-a \sec (e+f x)) \tan (e+f x)}{3 f \sqrt{a+a \sec (e+f x)}}+\frac{2 d^3 (4 c+3 d) (a-a \sec (e+f x))^2 \tan (e+f x)}{5 a f \sqrt{a+a \sec (e+f x)}}-\frac{2 d^4 (a-a \sec (e+f x))^3 \tan (e+f x)}{7 a^2 f \sqrt{a+a \sec (e+f x)}}+\frac{\left (2 a c^4 \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{1}{1-\frac{x^2}{a}} \, dx,x,\sqrt{a-a \sec (e+f x)}\right )}{f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=\frac{2 a d (2 c+d) \left (2 c^2+2 c d+d^2\right ) \tan (e+f x)}{f \sqrt{a+a \sec (e+f x)}}+\frac{2 a^{3/2} c^4 \tanh ^{-1}\left (\frac{\sqrt{a-a \sec (e+f x)}}{\sqrt{a}}\right ) \tan (e+f x)}{f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}-\frac{2 d^2 \left (6 c^2+8 c d+3 d^2\right ) (a-a \sec (e+f x)) \tan (e+f x)}{3 f \sqrt{a+a \sec (e+f x)}}+\frac{2 d^3 (4 c+3 d) (a-a \sec (e+f x))^2 \tan (e+f x)}{5 a f \sqrt{a+a \sec (e+f x)}}-\frac{2 d^4 (a-a \sec (e+f x))^3 \tan (e+f x)}{7 a^2 f \sqrt{a+a \sec (e+f x)}}\\ \end{align*}

Mathematica [C]  time = 14.6181, size = 589, normalized size = 2.17 \[ \frac{\cos ^4(e+f x) \sec \left (\frac{1}{2} (e+f x)\right ) \sqrt{a (\sec (e+f x)+1)} (c+d \sec (e+f x))^4 \left (\frac{8}{105} d \left (105 c^2 d+105 c^3+56 c d^2+12 d^3\right ) \sin \left (\frac{1}{2} (e+f x)\right )+\frac{4}{105} \sec (e+f x) \left (105 c^2 d^2 \sin \left (\frac{1}{2} (e+f x)\right )+56 c d^3 \sin \left (\frac{1}{2} (e+f x)\right )+12 d^4 \sin \left (\frac{1}{2} (e+f x)\right )\right )+\frac{4}{35} \sec ^2(e+f x) \left (14 c d^3 \sin \left (\frac{1}{2} (e+f x)\right )+3 d^4 \sin \left (\frac{1}{2} (e+f x)\right )\right )+\frac{2}{7} d^4 \sin \left (\frac{1}{2} (e+f x)\right ) \sec ^3(e+f x)\right )}{f (c \cos (e+f x)+d)^4}-\frac{8 \left (-3-2 \sqrt{2}\right ) c^4 \cos ^4\left (\frac{1}{4} (e+f x)\right ) \sqrt{\frac{\left (10-7 \sqrt{2}\right ) \cos \left (\frac{1}{2} (e+f x)\right )-5 \sqrt{2}+7}{\cos \left (\frac{1}{2} (e+f x)\right )+1}} \sqrt{\frac{-\left (\sqrt{2}-2\right ) \cos \left (\frac{1}{2} (e+f x)\right )+\sqrt{2}-1}{\cos \left (\frac{1}{2} (e+f x)\right )+1}} \left (\left (\sqrt{2}-2\right ) \cos \left (\frac{1}{2} (e+f x)\right )-\sqrt{2}+1\right ) \cos ^3(e+f x) \sqrt{-\tan ^2\left (\frac{1}{4} (e+f x)\right )-2 \sqrt{2}+3} \sec \left (\frac{1}{2} (e+f x)\right ) \sqrt{a (\sec (e+f x)+1)} \sqrt{\left (\left (2+\sqrt{2}\right ) \cos \left (\frac{1}{2} (e+f x)\right )-\sqrt{2}-1\right ) \sec ^2\left (\frac{1}{4} (e+f x)\right )} (c+d \sec (e+f x))^4 \left (\text{EllipticF}\left (\sin ^{-1}\left (\frac{\tan \left (\frac{1}{4} (e+f x)\right )}{\sqrt{3-2 \sqrt{2}}}\right ),17-12 \sqrt{2}\right )+2 \Pi \left (-3+2 \sqrt{2};-\sin ^{-1}\left (\frac{\tan \left (\frac{1}{4} (e+f x)\right )}{\sqrt{3-2 \sqrt{2}}}\right )|17-12 \sqrt{2}\right )\right )}{f (c \cos (e+f x)+d)^4} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Sqrt[a + a*Sec[e + f*x]]*(c + d*Sec[e + f*x])^4,x]

[Out]

(Cos[e + f*x]^4*Sec[(e + f*x)/2]*Sqrt[a*(1 + Sec[e + f*x])]*(c + d*Sec[e + f*x])^4*((8*d*(105*c^3 + 105*c^2*d
+ 56*c*d^2 + 12*d^3)*Sin[(e + f*x)/2])/105 + (2*d^4*Sec[e + f*x]^3*Sin[(e + f*x)/2])/7 + (4*Sec[e + f*x]^2*(14
*c*d^3*Sin[(e + f*x)/2] + 3*d^4*Sin[(e + f*x)/2]))/35 + (4*Sec[e + f*x]*(105*c^2*d^2*Sin[(e + f*x)/2] + 56*c*d
^3*Sin[(e + f*x)/2] + 12*d^4*Sin[(e + f*x)/2]))/105))/(f*(d + c*Cos[e + f*x])^4) - (8*(-3 - 2*Sqrt[2])*c^4*Cos
[(e + f*x)/4]^4*Sqrt[(7 - 5*Sqrt[2] + (10 - 7*Sqrt[2])*Cos[(e + f*x)/2])/(1 + Cos[(e + f*x)/2])]*Sqrt[(-1 + Sq
rt[2] - (-2 + Sqrt[2])*Cos[(e + f*x)/2])/(1 + Cos[(e + f*x)/2])]*(1 - Sqrt[2] + (-2 + Sqrt[2])*Cos[(e + f*x)/2
])*Cos[e + f*x]^3*(EllipticF[ArcSin[Tan[(e + f*x)/4]/Sqrt[3 - 2*Sqrt[2]]], 17 - 12*Sqrt[2]] + 2*EllipticPi[-3
+ 2*Sqrt[2], -ArcSin[Tan[(e + f*x)/4]/Sqrt[3 - 2*Sqrt[2]]], 17 - 12*Sqrt[2]])*Sqrt[(-1 - Sqrt[2] + (2 + Sqrt[2
])*Cos[(e + f*x)/2])*Sec[(e + f*x)/4]^2]*Sec[(e + f*x)/2]*Sqrt[a*(1 + Sec[e + f*x])]*(c + d*Sec[e + f*x])^4*Sq
rt[3 - 2*Sqrt[2] - Tan[(e + f*x)/4]^2])/(f*(d + c*Cos[e + f*x])^4)

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Maple [B]  time = 0.413, size = 546, normalized size = 2. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c+d*sec(f*x+e))^4*(a+a*sec(f*x+e))^(1/2),x)

[Out]

-1/840/f*(1/cos(f*x+e)*a*(1+cos(f*x+e)))^(1/2)*(-105*cos(f*x+e)^3*sin(f*x+e)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(7
/2)*arctanh(1/2*2^(1/2)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*sin(f*x+e)/cos(f*x+e))*2^(1/2)*c^4-315*cos(f*x+e)
^2*sin(f*x+e)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(7/2)*arctanh(1/2*2^(1/2)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*si
n(f*x+e)/cos(f*x+e))*2^(1/2)*c^4-315*cos(f*x+e)*sin(f*x+e)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(7/2)*arctanh(1/2*2^
(1/2)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*sin(f*x+e)/cos(f*x+e))*2^(1/2)*c^4-105*(-2*cos(f*x+e)/(1+cos(f*x+e)
))^(7/2)*arctanh(1/2*2^(1/2)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*sin(f*x+e)/cos(f*x+e))*2^(1/2)*c^4*sin(f*x+e
)+6720*cos(f*x+e)^4*c^3*d+6720*cos(f*x+e)^4*c^2*d^2+3584*cos(f*x+e)^4*c*d^3+768*cos(f*x+e)^4*d^4-6720*cos(f*x+
e)^3*c^3*d-3360*cos(f*x+e)^3*c^2*d^2-1792*cos(f*x+e)^3*c*d^3-384*cos(f*x+e)^3*d^4-3360*cos(f*x+e)^2*c^2*d^2-44
8*cos(f*x+e)^2*c*d^3-96*cos(f*x+e)^2*d^4-1344*cos(f*x+e)*c*d^3-48*cos(f*x+e)*d^4-240*d^4)/cos(f*x+e)^3/sin(f*x
+e)

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Maxima [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((c+d*sec(f*x+e))^4*(a+a*sec(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 1.69881, size = 1181, normalized size = 4.36 \begin{align*} \left [\frac{105 \,{\left (c^{4} \cos \left (f x + e\right )^{4} + c^{4} \cos \left (f x + e\right )^{3}\right )} \sqrt{-a} \log \left (\frac{2 \, a \cos \left (f x + e\right )^{2} - 2 \, \sqrt{-a} \sqrt{\frac{a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + a \cos \left (f x + e\right ) - a}{\cos \left (f x + e\right ) + 1}\right ) + 2 \,{\left (15 \, d^{4} + 4 \,{\left (105 \, c^{3} d + 105 \, c^{2} d^{2} + 56 \, c d^{3} + 12 \, d^{4}\right )} \cos \left (f x + e\right )^{3} + 2 \,{\left (105 \, c^{2} d^{2} + 56 \, c d^{3} + 12 \, d^{4}\right )} \cos \left (f x + e\right )^{2} + 6 \,{\left (14 \, c d^{3} + 3 \, d^{4}\right )} \cos \left (f x + e\right )\right )} \sqrt{\frac{a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sin \left (f x + e\right )}{105 \,{\left (f \cos \left (f x + e\right )^{4} + f \cos \left (f x + e\right )^{3}\right )}}, -\frac{2 \,{\left (105 \,{\left (c^{4} \cos \left (f x + e\right )^{4} + c^{4} \cos \left (f x + e\right )^{3}\right )} \sqrt{a} \arctan \left (\frac{\sqrt{\frac{a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{\sqrt{a} \sin \left (f x + e\right )}\right ) -{\left (15 \, d^{4} + 4 \,{\left (105 \, c^{3} d + 105 \, c^{2} d^{2} + 56 \, c d^{3} + 12 \, d^{4}\right )} \cos \left (f x + e\right )^{3} + 2 \,{\left (105 \, c^{2} d^{2} + 56 \, c d^{3} + 12 \, d^{4}\right )} \cos \left (f x + e\right )^{2} + 6 \,{\left (14 \, c d^{3} + 3 \, d^{4}\right )} \cos \left (f x + e\right )\right )} \sqrt{\frac{a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sin \left (f x + e\right )\right )}}{105 \,{\left (f \cos \left (f x + e\right )^{4} + f \cos \left (f x + e\right )^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*sec(f*x+e))^4*(a+a*sec(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

[1/105*(105*(c^4*cos(f*x + e)^4 + c^4*cos(f*x + e)^3)*sqrt(-a)*log((2*a*cos(f*x + e)^2 - 2*sqrt(-a)*sqrt((a*co
s(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)*sin(f*x + e) + a*cos(f*x + e) - a)/(cos(f*x + e) + 1)) + 2*(15*d^4
+ 4*(105*c^3*d + 105*c^2*d^2 + 56*c*d^3 + 12*d^4)*cos(f*x + e)^3 + 2*(105*c^2*d^2 + 56*c*d^3 + 12*d^4)*cos(f*x
+ e)^2 + 6*(14*c*d^3 + 3*d^4)*cos(f*x + e))*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sin(f*x + e))/(f*cos(f*x
+ e)^4 + f*cos(f*x + e)^3), -2/105*(105*(c^4*cos(f*x + e)^4 + c^4*cos(f*x + e)^3)*sqrt(a)*arctan(sqrt((a*cos(f
*x + e) + a)/cos(f*x + e))*cos(f*x + e)/(sqrt(a)*sin(f*x + e))) - (15*d^4 + 4*(105*c^3*d + 105*c^2*d^2 + 56*c*
d^3 + 12*d^4)*cos(f*x + e)^3 + 2*(105*c^2*d^2 + 56*c*d^3 + 12*d^4)*cos(f*x + e)^2 + 6*(14*c*d^3 + 3*d^4)*cos(f
*x + e))*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sin(f*x + e))/(f*cos(f*x + e)^4 + f*cos(f*x + e)^3)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{a \left (\sec{\left (e + f x \right )} + 1\right )} \left (c + d \sec{\left (e + f x \right )}\right )^{4}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*sec(f*x+e))**4*(a+a*sec(f*x+e))**(1/2),x)

[Out]

Integral(sqrt(a*(sec(e + f*x) + 1))*(c + d*sec(e + f*x))**4, x)

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Giac [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((c+d*sec(f*x+e))^4*(a+a*sec(f*x+e))^(1/2),x, algorithm="giac")

[Out]

Timed out