3.1522 \(\int (a+b \cos (c+d x))^{5/2} (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^{\frac {3}{2}}(c+d x) \, dx\)

Optimal. Leaf size=707 \[ \frac {\sin (c+d x) \sqrt {\sec (c+d x)} \left (-\left (a^2 (48 A-33 C)\right )+54 a b B+8 b^2 (3 A+2 C)\right ) \sqrt {a+b \cos (c+d x)}}{24 d}-\frac {\sqrt {a+b} \sqrt {\cos (c+d x)} \csc (c+d x) \left (a^2 (48 A-48 B-33 C)-2 a b (72 A+27 B+13 C)-4 b^2 (6 A+3 B+4 C)\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{24 d \sqrt {\sec (c+d x)}}-\frac {(a-b) \sqrt {a+b} \sqrt {\cos (c+d x)} \csc (c+d x) \left (-\left (a^2 (48 A-33 C)\right )+54 a b B+8 b^2 (3 A+2 C)\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{24 a d \sqrt {\sec (c+d x)}}-\frac {\sqrt {a+b} \sqrt {\cos (c+d x)} \csc (c+d x) \left (5 a^3 C+30 a^2 b B+20 a b^2 (2 A+C)+8 b^3 B\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \Pi \left (\frac {a+b}{b};\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{8 b d \sqrt {\sec (c+d x)}}-\frac {b \sin (c+d x) (8 a A-3 a C-2 b B) \sqrt {a+b \cos (c+d x)}}{4 d \sqrt {\sec (c+d x)}}-\frac {b (6 A-C) \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{3 d \sqrt {\sec (c+d x)}}+\frac {2 A \sin (c+d x) \sqrt {\sec (c+d x)} (a+b \cos (c+d x))^{5/2}}{d} \]

[Out]

-1/3*b*(6*A-C)*(a+b*cos(d*x+c))^(3/2)*sin(d*x+c)/d/sec(d*x+c)^(1/2)-1/4*b*(8*A*a-2*B*b-3*C*a)*sin(d*x+c)*(a+b*
cos(d*x+c))^(1/2)/d/sec(d*x+c)^(1/2)+2*A*(a+b*cos(d*x+c))^(5/2)*sin(d*x+c)*sec(d*x+c)^(1/2)/d+1/24*(54*a*b*B-a
^2*(48*A-33*C)+8*b^2*(3*A+2*C))*sin(d*x+c)*(a+b*cos(d*x+c))^(1/2)*sec(d*x+c)^(1/2)/d-1/24*(a-b)*(54*a*b*B-a^2*
(48*A-33*C)+8*b^2*(3*A+2*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/d/s
ec(d*x+c)^(1/2)-1/24*(a^2*(48*A-48*B-33*C)-4*b^2*(6*A+3*B+4*C)-2*a*b*(72*A+27*B+13*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)/d/sec(d*x+c)^(1/2)-1/8*(30*a^2*b*B+8*b^3*B+5*a^3*C+20*a*b
^2*(2*A+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/d/sec(d*x+c
)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 2.59, antiderivative size = 707, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4221, 3047, 3049, 3061, 3053, 2809, 2998, 2816, 2994} \[ \frac {\sin (c+d x) \sqrt {\sec (c+d x)} \left (a^2 (-(48 A-33 C))+54 a b B+8 b^2 (3 A+2 C)\right ) \sqrt {a+b \cos (c+d x)}}{24 d}-\frac {\sqrt {a+b} \sqrt {\cos (c+d x)} \csc (c+d x) \left (a^2 (48 A-48 B-33 C)-2 a b (72 A+27 B+13 C)-4 b^2 (6 A+3 B+4 C)\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{24 d \sqrt {\sec (c+d x)}}-\frac {(a-b) \sqrt {a+b} \sqrt {\cos (c+d x)} \csc (c+d x) \left (a^2 (-(48 A-33 C))+54 a b B+8 b^2 (3 A+2 C)\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{24 a d \sqrt {\sec (c+d x)}}-\frac {\sqrt {a+b} \sqrt {\cos (c+d x)} \csc (c+d x) \left (30 a^2 b B+5 a^3 C+20 a b^2 (2 A+C)+8 b^3 B\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \Pi \left (\frac {a+b}{b};\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{8 b d \sqrt {\sec (c+d x)}}-\frac {b \sin (c+d x) (8 a A-3 a C-2 b B) \sqrt {a+b \cos (c+d x)}}{4 d \sqrt {\sec (c+d x)}}-\frac {b (6 A-C) \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{3 d \sqrt {\sec (c+d x)}}+\frac {2 A \sin (c+d x) \sqrt {\sec (c+d x)} (a+b \cos (c+d x))^{5/2}}{d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^(3/2),x]

[Out]

-((a - b)*Sqrt[a + b]*(54*a*b*B - a^2*(48*A - 33*C) + 8*b^2*(3*A + 2*C))*Sqrt[Cos[c + d*x]]*Csc[c + d*x]*Ellip
ticE[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)])/(24*a*d*Sqrt[Sec[c + d*x]]) - (Sqrt[a + b]*(a^2*(48*A
- 48*B - 33*C) - 4*b^2*(6*A + 3*B + 4*C) - 2*a*b*(72*A + 27*B + 13*C))*Sqrt[Cos[c + d*x]]*Csc[c + d*x]*Ellipt
icF[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)])/(24*d*Sqrt[Sec[c + d*x]]) - (Sqrt[a + b]*(30*a^2*b*B +
8*b^3*B + 5*a^3*C + 20*a*b^2*(2*A + 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)]*S
qrt[(a*(1 + Sec[c + d*x]))/(a - b)])/(8*b*d*Sqrt[Sec[c + d*x]]) - (b*(8*a*A - 2*b*B - 3*a*C)*Sqrt[a + b*Cos[c
+ d*x]]*Sin[c + d*x])/(4*d*Sqrt[Sec[c + d*x]]) - (b*(6*A - C)*(a + b*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(3*d*Sq
rt[Sec[c + d*x]]) + ((54*a*b*B - a^2*(48*A - 33*C) + 8*b^2*(3*A + 2*C))*Sqrt[a + b*Cos[c + d*x]]*Sqrt[Sec[c +
d*x]]*Sin[c + d*x])/(24*d) + (2*A*(a + b*Cos[c + d*x])^(5/2)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/d

Rule 2809

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

Rule 2816

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

Rule 2994

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]*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))])/(f*b*c^2), x] /; FreeQ[{b, c, d, e, f, A, B}, x] && NeQ[c^2 - d^2, 0] &&
EqQ[A, B] && PosQ[(c + d)/b]

Rule 2998

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 3047

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((c^2*C - B*c*d + A*d^2)*Cos[e +
f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(n + 1)*(c^2 - d^2)), x] + Dist[1/(d*(n + 1)*(
c^2 - d^2)), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(b*d*m + a*c*(n + 1)) + (c
*C - B*d)*(b*c*m + a*d*(n + 1)) - (d*(A*(a*d*(n + 2) - b*c*(n + 1)) + B*(b*d*(n + 1) - a*c*(n + 2))) - C*(b*c*
d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x] + b*(d*(B*c - A*d)*(m + n + 2) - C*(c^2*(m + 1) + d^2*(n + 1)
))*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] && GtQ[m, 0] && LtQ[n, -1]

Rule 3049

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 3053

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 3061

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*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])/((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 4221

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*} \int (a+b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x) \, dx &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\\ &=\frac {2 A (a+b \cos (c+d x))^{5/2} \sqrt {\sec (c+d x)} \sin (c+d x)}{d}+\left (2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+b \cos (c+d x))^{3/2} \left (\frac {1}{2} (5 A b+a B)+\frac {1}{2} (b B-a (A-C)) \cos (c+d x)-\frac {1}{2} b (6 A-C) \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx\\ &=-\frac {b (6 A-C) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {2 A (a+b \cos (c+d x))^{5/2} \sqrt {\sec (c+d x)} \sin (c+d x)}{d}+\frac {1}{3} \left (2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {a+b \cos (c+d x)} \left (\frac {1}{4} a (24 A b+6 a B+b C)+\frac {1}{2} \left (6 a b B-3 a^2 (A-C)+b^2 (3 A+2 C)\right ) \cos (c+d x)-\frac {3}{4} b (8 a A-2 b B-3 a C) \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx\\ &=-\frac {b (8 a A-2 b B-3 a C) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{4 d \sqrt {\sec (c+d x)}}-\frac {b (6 A-C) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {2 A (a+b \cos (c+d x))^{5/2} \sqrt {\sec (c+d x)} \sin (c+d x)}{d}+\frac {1}{3} \left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {1}{8} a \left (24 a^2 B+6 b^2 B+a b (72 A+13 C)\right )+\frac {1}{4} \left (36 a^2 b B+6 b^3 B-12 a^3 (A-C)+a b^2 (36 A+19 C)\right ) \cos (c+d x)+\frac {1}{8} b \left (54 a b B-a^2 (48 A-33 C)+8 b^2 (3 A+2 C)\right ) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}} \, dx\\ &=-\frac {b (8 a A-2 b B-3 a C) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{4 d \sqrt {\sec (c+d x)}}-\frac {b (6 A-C) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {\left (54 a b B-a^2 (48 A-33 C)+8 b^2 (3 A+2 C)\right ) \sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)} \sin (c+d x)}{24 d}+\frac {2 A (a+b \cos (c+d x))^{5/2} \sqrt {\sec (c+d x)} \sin (c+d x)}{d}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {-\frac {1}{8} a b \left (54 a b B-a^2 (48 A-33 C)+8 b^2 (3 A+2 C)\right )+\frac {1}{4} a b \left (24 a^2 B+6 b^2 B+a b (72 A+13 C)\right ) \cos (c+d x)+\frac {3}{8} b \left (30 a^2 b B+8 b^3 B+5 a^3 C+20 a b^2 (2 A+C)\right ) \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}} \, dx}{6 b}\\ &=-\frac {b (8 a A-2 b B-3 a C) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{4 d \sqrt {\sec (c+d x)}}-\frac {b (6 A-C) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {\left (54 a b B-a^2 (48 A-33 C)+8 b^2 (3 A+2 C)\right ) \sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)} \sin (c+d x)}{24 d}+\frac {2 A (a+b \cos (c+d x))^{5/2} \sqrt {\sec (c+d x)} \sin (c+d x)}{d}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {-\frac {1}{8} a b \left (54 a b B-a^2 (48 A-33 C)+8 b^2 (3 A+2 C)\right )+\frac {1}{4} a b \left (24 a^2 B+6 b^2 B+a b (72 A+13 C)\right ) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}} \, dx}{6 b}+\frac {1}{16} \left (\left (30 a^2 b B+8 b^3 B+5 a^3 C+20 a b^2 (2 A+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\\ &=-\frac {\sqrt {a+b} \left (30 a^2 b B+8 b^3 B+5 a^3 C+20 a b^2 (2 A+C)\right ) \sqrt {\cos (c+d x)} \csc (c+d x) \Pi \left (\frac {a+b}{b};\sin ^{-1}\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}}}{8 b d \sqrt {\sec (c+d x)}}-\frac {b (8 a A-2 b B-3 a C) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{4 d \sqrt {\sec (c+d x)}}-\frac {b (6 A-C) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {\left (54 a b B-a^2 (48 A-33 C)+8 b^2 (3 A+2 C)\right ) \sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)} \sin (c+d x)}{24 d}+\frac {2 A (a+b \cos (c+d x))^{5/2} \sqrt {\sec (c+d x)} \sin (c+d x)}{d}-\frac {1}{48} \left (a \left (54 a b B-a^2 (48 A-33 C)+8 b^2 (3 A+2 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-\frac {1}{48} \left (a \left (a^2 (48 A-48 B-33 C)-4 b^2 (6 A+3 B+4 C)-2 a b (72 A+27 B+13 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\\ &=-\frac {(a-b) \sqrt {a+b} \left (54 a b B-a^2 (48 A-33 C)+8 b^2 (3 A+2 C)\right ) \sqrt {\cos (c+d x)} \csc (c+d x) E\left (\sin ^{-1}\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}}}{24 a d \sqrt {\sec (c+d x)}}-\frac {\sqrt {a+b} \left (a^2 (48 A-48 B-33 C)-4 b^2 (6 A+3 B+4 C)-2 a b (72 A+27 B+13 C)\right ) \sqrt {\cos (c+d x)} \csc (c+d x) F\left (\sin ^{-1}\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}}}{24 d \sqrt {\sec (c+d x)}}-\frac {\sqrt {a+b} \left (30 a^2 b B+8 b^3 B+5 a^3 C+20 a b^2 (2 A+C)\right ) \sqrt {\cos (c+d x)} \csc (c+d x) \Pi \left (\frac {a+b}{b};\sin ^{-1}\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}}}{8 b d \sqrt {\sec (c+d x)}}-\frac {b (8 a A-2 b B-3 a C) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{4 d \sqrt {\sec (c+d x)}}-\frac {b (6 A-C) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {\left (54 a b B-a^2 (48 A-33 C)+8 b^2 (3 A+2 C)\right ) \sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)} \sin (c+d x)}{24 d}+\frac {2 A (a+b \cos (c+d x))^{5/2} \sqrt {\sec (c+d x)} \sin (c+d x)}{d}\\ \end {align*}

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Mathematica [B]  time = 20.67, size = 1920, normalized size = 2.72 \[ \text {result too large to display} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^(3/2),x]

[Out]

(Sqrt[a + b*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(((24*a^2*A + b^2*C)*Sin[c + d*x])/12 + (b*(6*b*B + 13*a*C)*Sin[2
*(c + d*x)])/24 + (b^2*C*Sin[3*(c + d*x)])/12))/d + (48*a^3*A*Tan[(c + d*x)/2] + 48*a^2*A*b*Tan[(c + d*x)/2] -
24*a*A*b^2*Tan[(c + d*x)/2] - 24*A*b^3*Tan[(c + d*x)/2] - 54*a^2*b*B*Tan[(c + d*x)/2] - 54*a*b^2*B*Tan[(c + d
*x)/2] - 33*a^3*C*Tan[(c + d*x)/2] - 33*a^2*b*C*Tan[(c + d*x)/2] - 16*a*b^2*C*Tan[(c + d*x)/2] - 16*b^3*C*Tan[
(c + d*x)/2] - 96*a^2*A*b*Tan[(c + d*x)/2]^3 + 48*A*b^3*Tan[(c + d*x)/2]^3 + 108*a*b^2*B*Tan[(c + d*x)/2]^3 +
66*a^2*b*C*Tan[(c + d*x)/2]^3 + 32*b^3*C*Tan[(c + d*x)/2]^3 - 48*a^3*A*Tan[(c + d*x)/2]^5 + 48*a^2*A*b*Tan[(c
+ d*x)/2]^5 + 24*a*A*b^2*Tan[(c + d*x)/2]^5 - 24*A*b^3*Tan[(c + d*x)/2]^5 + 54*a^2*b*B*Tan[(c + d*x)/2]^5 - 54
*a*b^2*B*Tan[(c + d*x)/2]^5 + 33*a^3*C*Tan[(c + d*x)/2]^5 - 33*a^2*b*C*Tan[(c + d*x)/2]^5 + 16*a*b^2*C*Tan[(c
+ d*x)/2]^5 - 16*b^3*C*Tan[(c + d*x)/2]^5 - 240*a*A*b^2*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)] - 180*a^
2*b*B*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)] - 48*b^3*B*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)] -
30*a^3*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)] - 120*a*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)] - 240*a*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)] - 180*a^2*b*B*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)] - 48*b^3*B*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)] - 30*a^3*C*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Tan[(c + d*x)/2]^2*S
qrt[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)] - 120*a*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]*Sqr
t[(a + b + a*Tan[(c + d*x)/2]^2 - b*Tan[(c + d*x)/2]^2)/(a + b)] + (a + b)*(-54*a*b*B + a^2*(48*A - 33*C) - 8*
b^2*(3*A + 2*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*(-12*b^3*B + 24*a^3*(A
+ B - C) + a^2*b*(72*A - 72*B + 13*C) - 2*a*b^2*(36*A - 3*B + 19*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)])/(24*d*Sqrt[(1 - Tan[(c + d*x)/2]^2)^(-1)]*(-1 + Tan[(c + d*x)/2]^2)*(1 + Tan[(c + d*
x)/2]^2)^(3/2)*Sqrt[(a + b + a*Tan[(c + d*x)/2]^2 - b*Tan[(c + d*x)/2]^2)/(1 + Tan[(c + d*x)/2]^2)])

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fricas [F]  time = 5.13, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C b^{2} \cos \left (d x + c\right )^{4} + {\left (2 \, C a b + B b^{2}\right )} \cos \left (d x + c\right )^{3} + A a^{2} + {\left (C a^{2} + 2 \, B a b + A b^{2}\right )} \cos \left (d x + c\right )^{2} + {\left (B a^{2} + 2 \, A a b\right )} \cos \left (d x + c\right )\right )} \sqrt {b \cos \left (d x + c\right ) + a} \sec \left (d x + c\right )^{\frac {3}{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(3/2),x, algorithm="fricas")

[Out]

integral((C*b^2*cos(d*x + c)^4 + (2*C*a*b + B*b^2)*cos(d*x + c)^3 + A*a^2 + (C*a^2 + 2*B*a*b + A*b^2)*cos(d*x
+ c)^2 + (B*a^2 + 2*A*a*b)*cos(d*x + c))*sqrt(b*cos(d*x + c) + a)*sec(d*x + c)^(3/2), x)

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(3/2),x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 0.84, size = 5138, normalized size = 7.27 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(3/2),x)

[Out]

result too large to display

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sec \left (d x + c\right )^{\frac {3}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(3/2),x, algorithm="maxima")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(b*cos(d*x + c) + a)^(5/2)*sec(d*x + c)^(3/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{5/2}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1/cos(c + d*x))^(3/2)*(a + b*cos(c + d*x))^(5/2)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2),x)

[Out]

int((1/cos(c + d*x))^(3/2)*(a + b*cos(c + d*x))^(5/2)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))**(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**(3/2),x)

[Out]

Timed out

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