\(\int \frac {(a+b \cos (c+d x))^{5/2} (A+C \cos ^2(c+d x))}{\sqrt {\sec (c+d x)}} \, dx\) [1433]

Optimal result

Integrand size = 37, antiderivative size = 806 \[ \int \frac {(a+b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=\frac {(a-b) \sqrt {a+b} \left (45 a^4 C-256 b^4 (5 A+4 C)-12 a^2 b^2 (220 A+141 C)\right ) \sqrt {\cos (c+d x)} \csc (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{1920 a b^2 d \sqrt {\sec (c+d x)}}-\frac {\sqrt {a+b} \left (45 a^4 C-30 a^3 b C-256 b^4 (5 A+4 C)-12 a^2 b^2 (220 A+141 C)-8 a b^3 (260 A+193 C)\right ) \sqrt {\cos (c+d x)} \csc (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{1920 b^2 d \sqrt {\sec (c+d x)}}-\frac {a \sqrt {a+b} \left (3 a^4 C+40 a^2 b^2 (2 A+C)+80 b^4 (4 A+3 C)\right ) \sqrt {\cos (c+d x)} \csc (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{128 b^3 d \sqrt {\sec (c+d x)}}+\frac {a \left (240 A b^2-15 a^2 C+172 b^2 C\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{320 b d \sqrt {\sec (c+d x)}}-\frac {\left (15 a^2 C-16 b^2 (5 A+4 C)\right ) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{240 b d \sqrt {\sec (c+d x)}}-\frac {3 a C (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{40 b d \sqrt {\sec (c+d x)}}+\frac {C (a+b \cos (c+d x))^{7/2} \sin (c+d x)}{5 b d \sqrt {\sec (c+d x)}}-\frac {\left (45 a^4 C-256 b^4 (5 A+4 C)-12 a^2 b^2 (220 A+141 C)\right ) \sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)} \sin (c+d x)}{1920 b^2 d} \]

[Out]

-1/240*(15*a^2*C-16*b^2*(5*A+4*C))*(a+b*cos(d*x+c))^(3/2)*sin(d*x+c)/b/d/sec(d*x+c)^(1/2)-3/40*a*C*(a+b*cos(d*
x+c))^(5/2)*sin(d*x+c)/b/d/sec(d*x+c)^(1/2)+1/5*C*(a+b*cos(d*x+c))^(7/2)*sin(d*x+c)/b/d/sec(d*x+c)^(1/2)+1/320
*a*(240*A*b^2-15*C*a^2+172*C*b^2)*sin(d*x+c)*(a+b*cos(d*x+c))^(1/2)/b/d/sec(d*x+c)^(1/2)-1/1920*(45*a^4*C-256*
b^4*(5*A+4*C)-12*a^2*b^2*(220*A+141*C))*sin(d*x+c)*(a+b*cos(d*x+c))^(1/2)*sec(d*x+c)^(1/2)/b^2/d+1/1920*(a-b)*
(45*a^4*C-256*b^4*(5*A+4*C)-12*a^2*b^2*(220*A+141*C))*csc(d*x+c)*EllipticE((a+b*cos(d*x+c))^(1/2)/(a+b)^(1/2)/
cos(d*x+c)^(1/2),((-a-b)/(a-b))^(1/2))*(a+b)^(1/2)*cos(d*x+c)^(1/2)*(a*(1-sec(d*x+c))/(a+b))^(1/2)*(a*(1+sec(d
*x+c))/(a-b))^(1/2)/a/b^2/d/sec(d*x+c)^(1/2)-1/1920*(45*a^4*C-30*a^3*b*C-256*b^4*(5*A+4*C)-12*a^2*b^2*(220*A+1
41*C)-8*a*b^3*(260*A+193*C))*csc(d*x+c)*EllipticF((a+b*cos(d*x+c))^(1/2)/(a+b)^(1/2)/cos(d*x+c)^(1/2),((-a-b)/
(a-b))^(1/2))*(a+b)^(1/2)*cos(d*x+c)^(1/2)*(a*(1-sec(d*x+c))/(a+b))^(1/2)*(a*(1+sec(d*x+c))/(a-b))^(1/2)/b^2/d
/sec(d*x+c)^(1/2)-1/128*a*(3*a^4*C+40*a^2*b^2*(2*A+C)+80*b^4*(4*A+3*C))*csc(d*x+c)*EllipticPi((a+b*cos(d*x+c))
^(1/2)/(a+b)^(1/2)/cos(d*x+c)^(1/2),(a+b)/b,((-a-b)/(a-b))^(1/2))*(a+b)^(1/2)*cos(d*x+c)^(1/2)*(a*(1-sec(d*x+c
))/(a+b))^(1/2)*(a*(1+sec(d*x+c))/(a-b))^(1/2)/b^3/d/sec(d*x+c)^(1/2)

Rubi [A] (verified)

Time = 3.41 (sec) , antiderivative size = 806, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.243, Rules used = {4306, 3129, 3128, 3140, 3132, 2888, 3077, 2895, 3073} \[ \int \frac {(a+b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=\frac {C \sin (c+d x) (a+b \cos (c+d x))^{7/2}}{5 b d \sqrt {\sec (c+d x)}}-\frac {3 a C \sin (c+d x) (a+b \cos (c+d x))^{5/2}}{40 b d \sqrt {\sec (c+d x)}}-\frac {\left (15 a^2 C-16 b^2 (5 A+4 C)\right ) \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{240 b d \sqrt {\sec (c+d x)}}-\frac {\left (45 C a^4-12 b^2 (220 A+141 C) a^2-256 b^4 (5 A+4 C)\right ) \sqrt {\sec (c+d x)} \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{1920 b^2 d}+\frac {a \left (-15 C a^2+240 A b^2+172 b^2 C\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{320 b d \sqrt {\sec (c+d x)}}+\frac {(a-b) \sqrt {a+b} \left (45 C a^4-12 b^2 (220 A+141 C) a^2-256 b^4 (5 A+4 C)\right ) \sqrt {\cos (c+d x)} \csc (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}}}{1920 a b^2 d \sqrt {\sec (c+d x)}}-\frac {\sqrt {a+b} \left (45 C a^4-30 b C a^3-12 b^2 (220 A+141 C) a^2-8 b^3 (260 A+193 C) a-256 b^4 (5 A+4 C)\right ) \sqrt {\cos (c+d x)} \csc (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}}}{1920 b^2 d \sqrt {\sec (c+d x)}}-\frac {a \sqrt {a+b} \left (3 C a^4+40 b^2 (2 A+C) a^2+80 b^4 (4 A+3 C)\right ) \sqrt {\cos (c+d x)} \csc (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}}}{128 b^3 d \sqrt {\sec (c+d x)}} \]

[In]

Int[((a + b*Cos[c + d*x])^(5/2)*(A + C*Cos[c + d*x]^2))/Sqrt[Sec[c + d*x]],x]

[Out]

((a - b)*Sqrt[a + b]*(45*a^4*C - 256*b^4*(5*A + 4*C) - 12*a^2*b^2*(220*A + 141*C))*Sqrt[Cos[c + d*x]]*Csc[c +
d*x]*EllipticE[ArcSin[Sqrt[a + b*Cos[c + d*x]]/(Sqrt[a + b]*Sqrt[Cos[c + d*x]])], -((a + b)/(a - b))]*Sqrt[(a*
(1 - Sec[c + d*x]))/(a + b)]*Sqrt[(a*(1 + Sec[c + d*x]))/(a - b)])/(1920*a*b^2*d*Sqrt[Sec[c + d*x]]) - (Sqrt[a
+ b]*(45*a^4*C - 30*a^3*b*C - 256*b^4*(5*A + 4*C) - 12*a^2*b^2*(220*A + 141*C) - 8*a*b^3*(260*A + 193*C))*Sqr
t[Cos[c + d*x]]*Csc[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Cos[c + d*x]]/(Sqrt[a + b]*Sqrt[Cos[c + d*x]])], -((a
+ b)/(a - b))]*Sqrt[(a*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[(a*(1 + Sec[c + d*x]))/(a - b)])/(1920*b^2*d*Sqrt[Se
c[c + d*x]]) - (a*Sqrt[a + b]*(3*a^4*C + 40*a^2*b^2*(2*A + C) + 80*b^4*(4*A + 3*C))*Sqrt[Cos[c + d*x]]*Csc[c +
d*x]*EllipticPi[(a + b)/b, ArcSin[Sqrt[a + b*Cos[c + d*x]]/(Sqrt[a + b]*Sqrt[Cos[c + d*x]])], -((a + b)/(a -
b))]*Sqrt[(a*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[(a*(1 + Sec[c + d*x]))/(a - b)])/(128*b^3*d*Sqrt[Sec[c + d*x]])
+ (a*(240*A*b^2 - 15*a^2*C + 172*b^2*C)*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(320*b*d*Sqrt[Sec[c + d*x]]) -
((15*a^2*C - 16*b^2*(5*A + 4*C))*(a + b*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(240*b*d*Sqrt[Sec[c + d*x]]) - (3*a
*C*(a + b*Cos[c + d*x])^(5/2)*Sin[c + d*x])/(40*b*d*Sqrt[Sec[c + d*x]]) + (C*(a + b*Cos[c + d*x])^(7/2)*Sin[c
+ d*x])/(5*b*d*Sqrt[Sec[c + d*x]]) - ((45*a^4*C - 256*b^4*(5*A + 4*C) - 12*a^2*b^2*(220*A + 141*C))*Sqrt[a + b
*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(1920*b^2*d)

Rule 2888

Int[Sqrt[(b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Simp[2*b*(Tan
[e + f*x]/(d*f))*Rt[(c + d)/b, 2]*Sqrt[c*((1 + Csc[e + f*x])/(c - d))]*Sqrt[c*((1 - Csc[e + f*x])/(c + d))]*El
lipticPi[(c + d)/d, ArcSin[Sqrt[c + d*Sin[e + f*x]]/Sqrt[b*Sin[e + f*x]]/Rt[(c + d)/b, 2]], -(c + d)/(c - d)],
x] /; FreeQ[{b, c, d, e, f}, x] && NeQ[c^2 - d^2, 0] && PosQ[(c + d)/b]

Rule 2895

Int[1/(Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[-2*(
Tan[e + f*x]/(a*f))*Rt[(a + b)/d, 2]*Sqrt[a*((1 - Csc[e + f*x])/(a + b))]*Sqrt[a*((1 + Csc[e + f*x])/(a - b))]
*EllipticF[ArcSin[Sqrt[a + b*Sin[e + f*x]]/Sqrt[d*Sin[e + f*x]]/Rt[(a + b)/d, 2]], -(a + b)/(a - b)], x] /; Fr
eeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && PosQ[(a + b)/d]

Rule 3073

Int[((A_) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((b_.)*sin[(e_.) + (f_.)*(x_)])^(3/2)*Sqrt[(c_) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]]), x_Symbol] :> Simp[-2*A*(c - d)*(Tan[e + f*x]/(f*b*c^2))*Rt[(c + d)/b, 2]*Sqrt[c*((1 + Csc[e +
f*x])/(c - d))]*Sqrt[c*((1 - Csc[e + f*x])/(c + d))]*EllipticE[ArcSin[Sqrt[c + d*Sin[e + f*x]]/Sqrt[b*Sin[e +
f*x]]/Rt[(c + d)/b, 2]], -(c + d)/(c - d)], x] /; FreeQ[{b, c, d, e, f, A, B}, x] && NeQ[c^2 - d^2, 0] && EqQ
[A, B] && PosQ[(c + d)/b]

Rule 3077

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(3/2)*Sqrt[(c_) + (d_.)*s
in[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist[(A - B)/(a - b), Int[1/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e
+ f*x]]), x], x] - Dist[(A*b - a*B)/(a - b), Int[(1 + Sin[e + f*x])/((a + b*Sin[e + f*x])^(3/2)*Sqrt[c + d*Sin
[e + f*x]]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2
- d^2, 0] && NeQ[A, B]

Rule 3128

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)
*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*(a + b*Sin[e
+ f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n + 2))), x] + Dist[1/(d*(m + n + 2)), Int[(a + b*Sin[e + f*
x])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (d*(A*b + a*B)*(m + n +
2) - C*(a*c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + n + 2))*Sin[e + f*x]^2, x
], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d
^2, 0] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))

Rule 3129

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (C_.)
*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^
(n + 1)/(d*f*(m + n + 2))), x] + Dist[1/(d*(m + n + 2)), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])
^n*Simp[a*A*d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (A*b*d*(m + n + 2) - C*(a*c - b*d*(m + n + 1)))*Sin[e +
f*x] + C*(a*d*m - b*c*(m + 1))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, n}, x] && NeQ[b*c
- a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a,
0] && NeQ[c, 0])))

Rule 3132

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2)/(((a_.) + (b_.)*sin[(e_.) + (f_.
)*(x_)])^(3/2)*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist[C/b^2, Int[Sqrt[a + b*Sin[e + f
*x]]/Sqrt[c + d*Sin[e + f*x]], x], x] + Dist[1/b^2, Int[(A*b^2 - a^2*C + b*(b*B - 2*a*C)*Sin[e + f*x])/((a + b
*Sin[e + f*x])^(3/2)*Sqrt[c + d*Sin[e + f*x]]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a
*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rule 3140

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2)/(Sqrt[(a_.) + (b_.)*sin[(e_.) +
(f_.)*(x_)]]*Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*(Sqrt[c + d*Sin[
e + f*x]]/(d*f*Sqrt[a + b*Sin[e + f*x]])), x] + Dist[1/(2*d), Int[(1/((a + b*Sin[e + f*x])^(3/2)*Sqrt[c + d*Si
n[e + f*x]]))*Simp[2*a*A*d - C*(b*c - a*d) - 2*(a*c*C - d*(A*b + a*B))*Sin[e + f*x] + (2*b*B*d - C*(b*c + a*d)
)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0
] && NeQ[c^2 - d^2, 0]

Rule 4306

Int[(u_)*((c_.)*sec[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Dist[(c*Sec[a + b*x])^m*(c*Cos[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Cos[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSineIntegrandQ[u,
x]

Rubi steps \begin{align*} \text {integral}& = \left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \, dx \\ & = \frac {C (a+b \cos (c+d x))^{7/2} \sin (c+d x)}{5 b d \sqrt {\sec (c+d x)}}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+b \cos (c+d x))^{5/2} \left (\frac {a C}{2}+b (5 A+4 C) \cos (c+d x)-\frac {3}{2} a C \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx}{5 b} \\ & = -\frac {3 a C (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{40 b d \sqrt {\sec (c+d x)}}+\frac {C (a+b \cos (c+d x))^{7/2} \sin (c+d x)}{5 b d \sqrt {\sec (c+d x)}}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+b \cos (c+d x))^{3/2} \left (\frac {5 a^2 C}{4}+\frac {1}{2} a b (40 A+27 C) \cos (c+d x)-\frac {1}{4} \left (15 a^2 C-16 b^2 (5 A+4 C)\right ) \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx}{20 b} \\ & = -\frac {\left (15 a^2 C-16 b^2 (5 A+4 C)\right ) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{240 b d \sqrt {\sec (c+d x)}}-\frac {3 a C (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{40 b d \sqrt {\sec (c+d x)}}+\frac {C (a+b \cos (c+d x))^{7/2} \sin (c+d x)}{5 b d \sqrt {\sec (c+d x)}}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {a+b \cos (c+d x)} \left (\frac {1}{8} a \left (15 a^2 C+16 b^2 (5 A+4 C)\right )+\frac {1}{4} b \left (32 b^2 (5 A+4 C)+3 a^2 (80 A+49 C)\right ) \cos (c+d x)+\frac {3}{8} a \left (240 A b^2-15 a^2 C+172 b^2 C\right ) \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx}{60 b} \\ & = \frac {a \left (240 A b^2-15 a^2 C+172 b^2 C\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{320 b d \sqrt {\sec (c+d x)}}-\frac {\left (15 a^2 C-16 b^2 (5 A+4 C)\right ) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{240 b d \sqrt {\sec (c+d x)}}-\frac {3 a C (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{40 b d \sqrt {\sec (c+d x)}}+\frac {C (a+b \cos (c+d x))^{7/2} \sin (c+d x)}{5 b d \sqrt {\sec (c+d x)}}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {1}{16} a^2 \left (1040 A b^2+15 a^2 C+772 b^2 C\right )+\frac {1}{8} a b \left (4 b^2 (380 A+289 C)+a^2 (960 A+573 C)\right ) \cos (c+d x)-\frac {1}{16} \left (45 a^4 C-256 b^4 (5 A+4 C)-12 a^2 b^2 (220 A+141 C)\right ) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}} \, dx}{120 b} \\ & = \frac {a \left (240 A b^2-15 a^2 C+172 b^2 C\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{320 b d \sqrt {\sec (c+d x)}}-\frac {\left (15 a^2 C-16 b^2 (5 A+4 C)\right ) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{240 b d \sqrt {\sec (c+d x)}}-\frac {3 a C (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{40 b d \sqrt {\sec (c+d x)}}+\frac {C (a+b \cos (c+d x))^{7/2} \sin (c+d x)}{5 b d \sqrt {\sec (c+d x)}}-\frac {\left (45 a^4 C-256 b^4 (5 A+4 C)-12 a^2 b^2 (220 A+141 C)\right ) \sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)} \sin (c+d x)}{1920 b^2 d}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {1}{16} a \left (45 a^4 C-256 b^4 (5 A+4 C)-12 a^2 b^2 (220 A+141 C)\right )+\frac {1}{8} a^2 b \left (1040 A b^2+15 a^2 C+772 b^2 C\right ) \cos (c+d x)+\frac {15}{16} a \left (3 a^4 C+40 a^2 b^2 (2 A+C)+80 b^4 (4 A+3 C)\right ) \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}} \, dx}{240 b^2} \\ & = \frac {a \left (240 A b^2-15 a^2 C+172 b^2 C\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{320 b d \sqrt {\sec (c+d x)}}-\frac {\left (15 a^2 C-16 b^2 (5 A+4 C)\right ) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{240 b d \sqrt {\sec (c+d x)}}-\frac {3 a C (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{40 b d \sqrt {\sec (c+d x)}}+\frac {C (a+b \cos (c+d x))^{7/2} \sin (c+d x)}{5 b d \sqrt {\sec (c+d x)}}-\frac {\left (45 a^4 C-256 b^4 (5 A+4 C)-12 a^2 b^2 (220 A+141 C)\right ) \sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)} \sin (c+d x)}{1920 b^2 d}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {1}{16} a \left (45 a^4 C-256 b^4 (5 A+4 C)-12 a^2 b^2 (220 A+141 C)\right )+\frac {1}{8} a^2 b \left (1040 A b^2+15 a^2 C+772 b^2 C\right ) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}} \, dx}{240 b^2}+\frac {\left (a \left (3 a^4 C+40 a^2 b^2 (2 A+C)+80 b^4 (4 A+3 C)\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {\cos (c+d x)}}{\sqrt {a+b \cos (c+d x)}} \, dx}{256 b^2} \\ & = -\frac {a \sqrt {a+b} \left (3 a^4 C+40 a^2 b^2 (2 A+C)+80 b^4 (4 A+3 C)\right ) \sqrt {\cos (c+d x)} \csc (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{128 b^3 d \sqrt {\sec (c+d x)}}+\frac {a \left (240 A b^2-15 a^2 C+172 b^2 C\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{320 b d \sqrt {\sec (c+d x)}}-\frac {\left (15 a^2 C-16 b^2 (5 A+4 C)\right ) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{240 b d \sqrt {\sec (c+d x)}}-\frac {3 a C (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{40 b d \sqrt {\sec (c+d x)}}+\frac {C (a+b \cos (c+d x))^{7/2} \sin (c+d x)}{5 b d \sqrt {\sec (c+d x)}}-\frac {\left (45 a^4 C-256 b^4 (5 A+4 C)-12 a^2 b^2 (220 A+141 C)\right ) \sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)} \sin (c+d x)}{1920 b^2 d}+\frac {\left (a \left (45 a^4 C-256 b^4 (5 A+4 C)-12 a^2 b^2 (220 A+141 C)\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1+\cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}} \, dx}{3840 b^2}-\frac {\left (a \left (45 a^4 C-30 a^3 b C-256 b^4 (5 A+4 C)-12 a^2 b^2 (220 A+141 C)-8 a b^3 (260 A+193 C)\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}} \, dx}{3840 b^2} \\ & = \frac {(a-b) \sqrt {a+b} \left (45 a^4 C-256 b^4 (5 A+4 C)-12 a^2 b^2 (220 A+141 C)\right ) \sqrt {\cos (c+d x)} \csc (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{1920 a b^2 d \sqrt {\sec (c+d x)}}-\frac {\sqrt {a+b} \left (45 a^4 C-30 a^3 b C-256 b^4 (5 A+4 C)-12 a^2 b^2 (220 A+141 C)-8 a b^3 (260 A+193 C)\right ) \sqrt {\cos (c+d x)} \csc (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{1920 b^2 d \sqrt {\sec (c+d x)}}-\frac {a \sqrt {a+b} \left (3 a^4 C+40 a^2 b^2 (2 A+C)+80 b^4 (4 A+3 C)\right ) \sqrt {\cos (c+d x)} \csc (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{128 b^3 d \sqrt {\sec (c+d x)}}+\frac {a \left (240 A b^2-15 a^2 C+172 b^2 C\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{320 b d \sqrt {\sec (c+d x)}}-\frac {\left (15 a^2 C-16 b^2 (5 A+4 C)\right ) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{240 b d \sqrt {\sec (c+d x)}}-\frac {3 a C (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{40 b d \sqrt {\sec (c+d x)}}+\frac {C (a+b \cos (c+d x))^{7/2} \sin (c+d x)}{5 b d \sqrt {\sec (c+d x)}}-\frac {\left (45 a^4 C-256 b^4 (5 A+4 C)-12 a^2 b^2 (220 A+141 C)\right ) \sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)} \sin (c+d x)}{1920 b^2 d} \\ \end{align*}

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(2045\) vs. \(2(806)=1612\).

Time = 20.21 (sec) , antiderivative size = 2045, normalized size of antiderivative = 2.54 \[ \int \frac {(a+b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=\text {Result too large to show} \]

[In]

Integrate[((a + b*Cos[c + d*x])^(5/2)*(A + C*Cos[c + d*x]^2))/Sqrt[Sec[c + d*x]],x]

[Out]

(Sqrt[a + b*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(((80*A*b^2 + 93*a^2*C + 88*b^2*C)*Sin[c + d*x])/960 + (a*(1040*A
*b^2 + 15*a^2*C + 1024*b^2*C)*Sin[2*(c + d*x)])/(1920*b) + ((80*A*b^2 + 93*a^2*C + 100*b^2*C)*Sin[3*(c + d*x)]
)/960 + (21*a*b*C*Sin[4*(c + d*x)])/320 + (b^2*C*Sin[5*(c + d*x)])/80))/d - (Sqrt[(1 - Tan[(c + d*x)/2]^2)^(-1
)]*Sqrt[(a + b + a*Tan[(c + d*x)/2]^2 - b*Tan[(c + d*x)/2]^2)/(1 + Tan[(c + d*x)/2]^2)]*(2640*a^3*A*b^2*Tan[(c
+ d*x)/2] + 2640*a^2*A*b^3*Tan[(c + d*x)/2] + 1280*a*A*b^4*Tan[(c + d*x)/2] + 1280*A*b^5*Tan[(c + d*x)/2] - 4
5*a^5*C*Tan[(c + d*x)/2] - 45*a^4*b*C*Tan[(c + d*x)/2] + 1692*a^3*b^2*C*Tan[(c + d*x)/2] + 1692*a^2*b^3*C*Tan[
(c + d*x)/2] + 1024*a*b^4*C*Tan[(c + d*x)/2] + 1024*b^5*C*Tan[(c + d*x)/2] - 5280*a^2*A*b^3*Tan[(c + d*x)/2]^3
- 2560*A*b^5*Tan[(c + d*x)/2]^3 + 90*a^4*b*C*Tan[(c + d*x)/2]^3 - 3384*a^2*b^3*C*Tan[(c + d*x)/2]^3 - 2048*b^
5*C*Tan[(c + d*x)/2]^3 - 2640*a^3*A*b^2*Tan[(c + d*x)/2]^5 + 2640*a^2*A*b^3*Tan[(c + d*x)/2]^5 - 1280*a*A*b^4*
Tan[(c + d*x)/2]^5 + 1280*A*b^5*Tan[(c + d*x)/2]^5 + 45*a^5*C*Tan[(c + d*x)/2]^5 - 45*a^4*b*C*Tan[(c + d*x)/2]
^5 - 1692*a^3*b^2*C*Tan[(c + d*x)/2]^5 + 1692*a^2*b^3*C*Tan[(c + d*x)/2]^5 - 1024*a*b^4*C*Tan[(c + d*x)/2]^5 +
1024*b^5*C*Tan[(c + d*x)/2]^5 + 2400*a^3*A*b^2*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Sqr
t[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b + a*Tan[(c + d*x)/2]^2 - b*Tan[(c + d*x)/2]^2)/(a + b)] + 9600*a*A*b^4*E
llipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b + a*Tan[(c
+ d*x)/2]^2 - b*Tan[(c + d*x)/2]^2)/(a + b)] + 90*a^5*C*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a +
b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b + a*Tan[(c + d*x)/2]^2 - b*Tan[(c + d*x)/2]^2)/(a + b)] + 1200*a
^3*b^2*C*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b +
a*Tan[(c + d*x)/2]^2 - b*Tan[(c + d*x)/2]^2)/(a + b)] + 7200*a*b^4*C*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]],
(-a + b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b + a*Tan[(c + d*x)/2]^2 - b*Tan[(c + d*x)/2]^2)/(a
+ b)] + 2400*a^3*A*b^2*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Tan[(c + d*x)/2]^2*Sqrt[1 -
Tan[(c + d*x)/2]^2]*Sqrt[(a + b + a*Tan[(c + d*x)/2]^2 - b*Tan[(c + d*x)/2]^2)/(a + b)] + 9600*a*A*b^4*Ellipti
cPi[-1, ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Tan[(c + d*x)/2]^2*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a +
b + a*Tan[(c + d*x)/2]^2 - b*Tan[(c + d*x)/2]^2)/(a + b)] + 90*a^5*C*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]],
(-a + b)/(a + b)]*Tan[(c + d*x)/2]^2*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b + a*Tan[(c + d*x)/2]^2 - b*Tan[(
c + d*x)/2]^2)/(a + b)] + 1200*a^3*b^2*C*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Tan[(c + d
*x)/2]^2*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b + a*Tan[(c + d*x)/2]^2 - b*Tan[(c + d*x)/2]^2)/(a + b)] + 72
00*a*b^4*C*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Tan[(c + d*x)/2]^2*Sqrt[1 - Tan[(c + d*x
)/2]^2]*Sqrt[(a + b + a*Tan[(c + d*x)/2]^2 - b*Tan[(c + d*x)/2]^2)/(a + b)] - (a + b)*(45*a^4*C - 256*b^4*(5*A
+ 4*C) - 12*a^2*b^2*(220*A + 141*C))*EllipticE[ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Sqrt[1 - Tan[(c +
d*x)/2]^2]*(1 + Tan[(c + d*x)/2]^2)*Sqrt[(a + b + a*Tan[(c + d*x)/2]^2 - b*Tan[(c + d*x)/2]^2)/(a + b)] + 2*a*
b*(15*a^3*C - 6*a^2*b*(320*A + 191*C) + 4*a*b^2*(260*A + 193*C) - 8*b^3*(380*A + 289*C))*EllipticF[ArcSin[Tan[
(c + d*x)/2]], (-a + b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*(1 + Tan[(c + d*x)/2]^2)*Sqrt[(a + b + a*Tan[(c
+ d*x)/2]^2 - b*Tan[(c + d*x)/2]^2)/(a + b)]))/(1920*b^2*d*Sqrt[1 + Tan[(c + d*x)/2]^2]*(b*(-1 + Tan[(c + d*x)
/2]^2) - a*(1 + Tan[(c + d*x)/2]^2)))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(6348\) vs. \(2(734)=1468\).

Time = 19.10 (sec) , antiderivative size = 6349, normalized size of antiderivative = 7.88

method	result	size
parts	\(\text {Expression too large to display}\)	\(6349\)
default	\(\text {Expression too large to display}\)	\(6451\)

[In]

int((a+b*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)/sec(d*x+c)^(1/2),x,method=_RETURNVERBOSE)

[Out]

result too large to display

Fricas [F(-1)]

Timed out. \[ \int \frac {(a+b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=\text {Timed out} \]

[In]

integrate((a+b*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)/sec(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

Timed out

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=\text {Timed out} \]

[In]

integrate((a+b*cos(d*x+c))**(5/2)*(A+C*cos(d*x+c)**2)/sec(d*x+c)**(1/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {(a+b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}{\sqrt {\sec \left (d x + c\right )}} \,d x } \]

[In]

integrate((a+b*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)/sec(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

integrate((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c) + a)^(5/2)/sqrt(sec(d*x + c)), x)

Giac [F]

\[ \int \frac {(a+b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}{\sqrt {\sec \left (d x + c\right )}} \,d x } \]

[In]

integrate((a+b*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)/sec(d*x+c)^(1/2),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c) + a)^(5/2)/sqrt(sec(d*x + c)), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=\int \frac {\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{5/2}}{\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}} \,d x \]

[In]

int(((A + C*cos(c + d*x)^2)*(a + b*cos(c + d*x))^(5/2))/(1/cos(c + d*x))^(1/2),x)

[Out]

int(((A + C*cos(c + d*x)^2)*(a + b*cos(c + d*x))^(5/2))/(1/cos(c + d*x))^(1/2), x)