3.570 \(\int \frac{A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac{5}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx\)
Optimal. Leaf size=254 \[ -\frac{5 (3 A-2 B+C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{3 a^2 d}-\frac{(3 A-2 B+C) \sin (c+d x)}{a^2 d \sec ^{\frac{3}{2}}(c+d x) (\sec (c+d x)+1)}+\frac{(56 A-35 B+20 C) \sin (c+d x)}{15 a^2 d \sec ^{\frac{3}{2}}(c+d x)}-\frac{5 (3 A-2 B+C) \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)}}+\frac{(56 A-35 B+20 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 a^2 d}-\frac{(A-B+C) \sin (c+d x)}{3 d \sec ^{\frac{3}{2}}(c+d x) (a \sec (c+d x)+a)^2} \]
[Out]
((56*A - 35*B + 20*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*a^2*d) - (5*(3*A - 2
*B + C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*a^2*d) + ((56*A - 35*B + 20*C)*Sin
[c + d*x])/(15*a^2*d*Sec[c + d*x]^(3/2)) - (5*(3*A - 2*B + C)*Sin[c + d*x])/(3*a^2*d*Sqrt[Sec[c + d*x]]) - ((3
*A - 2*B + C)*Sin[c + d*x])/(a^2*d*Sec[c + d*x]^(3/2)*(1 + Sec[c + d*x])) - ((A - B + C)*Sin[c + d*x])/(3*d*Se
c[c + d*x]^(3/2)*(a + a*Sec[c + d*x])^2)
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Rubi [A] time = 0.425377, antiderivative size = 254, normalized size of antiderivative = 1.,
number of steps used = 9, number of rules used = 7, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.163, Rules used =
{4084, 4020, 3787, 3769, 3771, 2639, 2641} \[ -\frac{(3 A-2 B+C) \sin (c+d x)}{a^2 d \sec ^{\frac{3}{2}}(c+d x) (\sec (c+d x)+1)}+\frac{(56 A-35 B+20 C) \sin (c+d x)}{15 a^2 d \sec ^{\frac{3}{2}}(c+d x)}-\frac{5 (3 A-2 B+C) \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)}}-\frac{5 (3 A-2 B+C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a^2 d}+\frac{(56 A-35 B+20 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 a^2 d}-\frac{(A-B+C) \sin (c+d x)}{3 d \sec ^{\frac{3}{2}}(c+d x) (a \sec (c+d x)+a)^2} \]
Antiderivative was successfully verified.
[In]
Int[(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(Sec[c + d*x]^(5/2)*(a + a*Sec[c + d*x])^2),x]
[Out]
((56*A - 35*B + 20*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*a^2*d) - (5*(3*A - 2
*B + C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*a^2*d) + ((56*A - 35*B + 20*C)*Sin
[c + d*x])/(15*a^2*d*Sec[c + d*x]^(3/2)) - (5*(3*A - 2*B + C)*Sin[c + d*x])/(3*a^2*d*Sqrt[Sec[c + d*x]]) - ((3
*A - 2*B + C)*Sin[c + d*x])/(a^2*d*Sec[c + d*x]^(3/2)*(1 + Sec[c + d*x])) - ((A - B + C)*Sin[c + d*x])/(3*d*Se
c[c + d*x]^(3/2)*(a + a*Sec[c + d*x])^2)
Rule 4084
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^
(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[((a*A - b*B + a*C)*Cot[e + f*x]*(a + b*Cs
c[e + f*x])^m*(d*Csc[e + f*x])^n)/(a*f*(2*m + 1)), x] - Dist[1/(a*b*(2*m + 1)), Int[(a + b*Csc[e + f*x])^(m +
1)*(d*Csc[e + f*x])^n*Simp[a*B*n - b*C*n - A*b*(2*m + n + 1) - (b*B*(m + n + 1) - a*(A*(m + n + 1) - C*(m - n)
))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, n}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]
Rule 4020
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> -Simp[((A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n)/(b*f*(2
*m + 1)), x] - Dist[1/(a^2*(2*m + 1)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*Simp[b*B*n - a*A*(2
*m + n + 1) + (A*b - a*B)*(m + n + 1)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*
b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)] && !GtQ[n, 0]
Rule 3787
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]
Rule 3769
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Csc[c + d*x])^(n + 1))/(b*d*n), x
] + Dist[(n + 1)/(b^2*n), Int[(b*Csc[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && Integer
Q[2*n]
Rule 3771
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]
Rule 2639
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]
Rule 2641
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]
Rubi steps
\begin{align*} \int \frac{A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac{5}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx &=-\frac{(A-B+C) \sin (c+d x)}{3 d \sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^2}+\frac{\int \frac{\frac{1}{2} a (11 A-5 B+5 C)-\frac{1}{2} a (7 A-7 B+C) \sec (c+d x)}{\sec ^{\frac{5}{2}}(c+d x) (a+a \sec (c+d x))} \, dx}{3 a^2}\\ &=-\frac{(3 A-2 B+C) \sin (c+d x)}{a^2 d \sec ^{\frac{3}{2}}(c+d x) (1+\sec (c+d x))}-\frac{(A-B+C) \sin (c+d x)}{3 d \sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^2}+\frac{\int \frac{\frac{1}{2} a^2 (56 A-35 B+20 C)-\frac{15}{2} a^2 (3 A-2 B+C) \sec (c+d x)}{\sec ^{\frac{5}{2}}(c+d x)} \, dx}{3 a^4}\\ &=-\frac{(3 A-2 B+C) \sin (c+d x)}{a^2 d \sec ^{\frac{3}{2}}(c+d x) (1+\sec (c+d x))}-\frac{(A-B+C) \sin (c+d x)}{3 d \sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^2}-\frac{(5 (3 A-2 B+C)) \int \frac{1}{\sec ^{\frac{3}{2}}(c+d x)} \, dx}{2 a^2}+\frac{(56 A-35 B+20 C) \int \frac{1}{\sec ^{\frac{5}{2}}(c+d x)} \, dx}{6 a^2}\\ &=\frac{(56 A-35 B+20 C) \sin (c+d x)}{15 a^2 d \sec ^{\frac{3}{2}}(c+d x)}-\frac{5 (3 A-2 B+C) \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)}}-\frac{(3 A-2 B+C) \sin (c+d x)}{a^2 d \sec ^{\frac{3}{2}}(c+d x) (1+\sec (c+d x))}-\frac{(A-B+C) \sin (c+d x)}{3 d \sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^2}-\frac{(5 (3 A-2 B+C)) \int \sqrt{\sec (c+d x)} \, dx}{6 a^2}+\frac{(56 A-35 B+20 C) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx}{10 a^2}\\ &=\frac{(56 A-35 B+20 C) \sin (c+d x)}{15 a^2 d \sec ^{\frac{3}{2}}(c+d x)}-\frac{5 (3 A-2 B+C) \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)}}-\frac{(3 A-2 B+C) \sin (c+d x)}{a^2 d \sec ^{\frac{3}{2}}(c+d x) (1+\sec (c+d x))}-\frac{(A-B+C) \sin (c+d x)}{3 d \sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^2}-\frac{\left (5 (3 A-2 B+C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{6 a^2}+\frac{\left ((56 A-35 B+20 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx}{10 a^2}\\ &=\frac{(56 A-35 B+20 C) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{5 a^2 d}-\frac{5 (3 A-2 B+C) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{3 a^2 d}+\frac{(56 A-35 B+20 C) \sin (c+d x)}{15 a^2 d \sec ^{\frac{3}{2}}(c+d x)}-\frac{5 (3 A-2 B+C) \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)}}-\frac{(3 A-2 B+C) \sin (c+d x)}{a^2 d \sec ^{\frac{3}{2}}(c+d x) (1+\sec (c+d x))}-\frac{(A-B+C) \sin (c+d x)}{3 d \sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^2}\\ \end{align*}
Mathematica [C] time = 7.22942, size = 1442, normalized size = 5.68 \[ \text{result too large to display} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(Sec[c + d*x]^(5/2)*(a + a*Sec[c + d*x])^2),x]
[Out]
(-112*Sqrt[2]*A*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*Cos[c/2 + (d*x)/
2]^4*Csc[c/2]*(-3*Sqrt[1 + E^((2*I)*(c + d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Hypergeometric2F1[1/2, 3/4,
7/4, -E^((2*I)*(c + d*x))])*Sec[c/2]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(15*d*E^(I*d*x)*(A + 2*C + 2*B*
Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])^2) + (14*Sqrt[2]*B*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*
(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*Cos[c/2 + (d*x)/2]^4*Csc[c/2]*(-3*Sqrt[1 + E^((2*I)*(c + d*x))] + E
^((2*I)*d*x)*(-1 + E^((2*I)*c))*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))])*Sec[c/2]*(A + B*Sec[c
+ d*x] + C*Sec[c + d*x]^2))/(3*d*E^(I*d*x)*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*
x])^2) - (8*Sqrt[2]*C*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*Cos[c/2 +
(d*x)/2]^4*Csc[c/2]*(-3*Sqrt[1 + E^((2*I)*(c + d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Hypergeometric2F1[1/2
, 3/4, 7/4, -E^((2*I)*(c + d*x))])*Sec[c/2]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(3*d*E^(I*d*x)*(A + 2*C +
2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])^2) - (20*A*Cos[c/2 + (d*x)/2]^4*Sqrt[Cos[c + d*x]
]*Csc[c/2]*EllipticF[(c + d*x)/2, 2]*Sec[c/2]*Sqrt[Sec[c + d*x]]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*Sin[c
])/(d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])^2) + (40*B*Cos[c/2 + (d*x)/2]^4*S
qrt[Cos[c + d*x]]*Csc[c/2]*EllipticF[(c + d*x)/2, 2]*Sec[c/2]*Sqrt[Sec[c + d*x]]*(A + B*Sec[c + d*x] + C*Sec[c
+ d*x]^2)*Sin[c])/(3*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])^2) - (20*C*Cos[
c/2 + (d*x)/2]^4*Sqrt[Cos[c + d*x]]*Csc[c/2]*EllipticF[(c + d*x)/2, 2]*Sec[c/2]*Sqrt[Sec[c + d*x]]*(A + B*Sec[
c + d*x] + C*Sec[c + d*x]^2)*Sin[c])/(3*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x
])^2) + (Cos[c/2 + (d*x)/2]^4*Sqrt[Sec[c + d*x]]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(-((151*A - 100*B + 6
0*C + 73*A*Cos[2*c] - 40*B*Cos[2*c] + 20*C*Cos[2*c])*Cos[d*x]*Csc[c/2]*Sec[c/2])/(5*d) - (8*(2*A - B)*Cos[2*d*
x]*Sin[2*c])/(3*d) + (4*A*Cos[3*d*x]*Sin[3*c])/(5*d) - (4*Sec[c/2]*Sec[c/2 + (d*x)/2]^3*(A*Sin[(d*x)/2] - B*Si
n[(d*x)/2] + C*Sin[(d*x)/2]))/(3*d) + (8*Sec[c/2]*Sec[c/2 + (d*x)/2]*(13*A*Sin[(d*x)/2] - 10*B*Sin[(d*x)/2] +
7*C*Sin[(d*x)/2]))/(3*d) + (4*(73*A - 40*B + 20*C)*Cos[c]*Sin[d*x])/(5*d) - (8*(2*A - B)*Cos[2*c]*Sin[2*d*x])/
(3*d) + (4*A*Cos[3*c]*Sin[3*d*x])/(5*d) + (8*(13*A - 10*B + 7*C)*Tan[c/2])/(3*d) - (4*(A - B + C)*Sec[c/2 + (d
*x)/2]^2*Tan[c/2])/(3*d)))/((A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])^2)
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Maple [A] time = 2.749, size = 491, normalized size = 1.9 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(5/2)/(a+a*sec(d*x+c))^2,x)
[Out]
-1/30*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x
+1/2*c)^2)^(1/2)*(75*A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+168*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-50*B*
EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-105*B*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+25*C*EllipticF(cos(1/2*d*x+1
/2*c),2^(1/2))+60*C*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2-2*(2*sin(1/
2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(75*A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+168*A*Ellipti
cE(cos(1/2*d*x+1/2*c),2^(1/2))-50*B*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-105*B*EllipticE(cos(1/2*d*x+1/2*c),2
^(1/2))+25*C*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+60*C*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))*cos(1/2*d*x+1/2
*c)-96*A*sin(1/2*d*x+1/2*c)^10+(128*A+80*B)*sin(1/2*d*x+1/2*c)^8+(328*A-380*B+120*C)*sin(1/2*d*x+1/2*c)^6+(-52
6*A+420*B-170*C)*sin(1/2*d*x+1/2*c)^4+(171*A-125*B+55*C)*sin(1/2*d*x+1/2*c)^2)/a^2/cos(1/2*d*x+1/2*c)^3/(-2*si
n(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d
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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((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(5/2)/(a+a*sec(d*x+c))^2,x, algorithm="maxima")
[Out]
Timed out
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \sqrt{\sec \left (d x + c\right )}}{a^{2} \sec \left (d x + c\right )^{5} + 2 \, a^{2} \sec \left (d x + c\right )^{4} + a^{2} \sec \left (d x + c\right )^{3}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(5/2)/(a+a*sec(d*x+c))^2,x, algorithm="fricas")
[Out]
integral((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*sqrt(sec(d*x + c))/(a^2*sec(d*x + c)^5 + 2*a^2*sec(d*x + c)^4
+ a^2*sec(d*x + c)^3), 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((A+B*sec(d*x+c)+C*sec(d*x+c)**2)/sec(d*x+c)**(5/2)/(a+a*sec(d*x+c))**2,x)
[Out]
Timed out
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A}{{\left (a \sec \left (d x + c\right ) + a\right )}^{2} \sec \left (d x + c\right )^{\frac{5}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(5/2)/(a+a*sec(d*x+c))^2,x, algorithm="giac")
[Out]
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)/((a*sec(d*x + c) + a)^2*sec(d*x + c)^(5/2)), x)