3.1518 \(\int (a+b \cos (c+d x))^{5/2} (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^{\frac {11}{2}}(c+d x) \, dx\)

Optimal. Leaf size=592 \[ \frac {2 \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \left (7 a^2 (7 A+9 C)+90 a b B+15 A b^2\right ) \sqrt {a+b \cos (c+d x)}}{315 d}+\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \left (75 a^3 B+a^2 b (163 A+231 C)+135 a b^2 B+5 A b^3\right ) \sqrt {a+b \cos (c+d x)}}{315 a d}-\frac {2 (a-b) \sqrt {a+b} \sqrt {\cos (c+d x)} \csc (c+d x) \left (3 a^3 (49 A-25 B+63 C)-6 a^2 b (19 A-60 B+28 C)+15 a b^2 (11 A-3 B+21 C)+10 A b^3\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{315 a^2 d \sqrt {\sec (c+d x)}}-\frac {2 (a-b) \sqrt {a+b} \sqrt {\cos (c+d x)} \csc (c+d x) \left (-21 a^4 (7 A+9 C)-435 a^3 b B-3 a^2 b^2 (93 A+161 C)-45 a b^3 B+10 A b^4\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{315 a^3 d \sqrt {\sec (c+d x)}}+\frac {2 (9 a B+5 A b) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}}{63 d}+\frac {2 A \sin (c+d x) \sec ^{\frac {9}{2}}(c+d x) (a+b \cos (c+d x))^{5/2}}{9 d} \]

[Out]

2/63*(5*A*b+9*B*a)*(a+b*cos(d*x+c))^(3/2)*sec(d*x+c)^(7/2)*sin(d*x+c)/d+2/9*A*(a+b*cos(d*x+c))^(5/2)*sec(d*x+c
)^(9/2)*sin(d*x+c)/d+2/315*(5*A*b^3+75*a^3*B+135*a*b^2*B+a^2*b*(163*A+231*C))*sec(d*x+c)^(3/2)*sin(d*x+c)*(a+b
*cos(d*x+c))^(1/2)/a/d+2/315*(15*A*b^2+90*a*b*B+7*a^2*(7*A+9*C))*sec(d*x+c)^(5/2)*sin(d*x+c)*(a+b*cos(d*x+c))^
(1/2)/d-2/315*(a-b)*(10*A*b^4-435*a^3*b*B-45*a*b^3*B-21*a^4*(7*A+9*C)-3*a^2*b^2*(93*A+161*C))*csc(d*x+c)*Ellip
ticE((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^3/d/sec(d*x+c)^(1/2)-2/315*(a-b)*(10*A*b^3+15*a*
b^2*(11*A-3*B+21*C)-6*a^2*b*(19*A-60*B+28*C)+3*a^3*(49*A-25*B+63*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)/a^2/d/sec(d*x+c)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 2.28, antiderivative size = 592, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {4221, 3047, 3055, 2998, 2816, 2994} \[ \frac {2 \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \left (7 a^2 (7 A+9 C)+90 a b B+15 A b^2\right ) \sqrt {a+b \cos (c+d x)}}{315 d}+\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \left (a^2 b (163 A+231 C)+75 a^3 B+135 a b^2 B+5 A b^3\right ) \sqrt {a+b \cos (c+d x)}}{315 a d}-\frac {2 (a-b) \sqrt {a+b} \sqrt {\cos (c+d x)} \csc (c+d x) \left (-6 a^2 b (19 A-60 B+28 C)+3 a^3 (49 A-25 B+63 C)+15 a b^2 (11 A-3 B+21 C)+10 A b^3\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{315 a^2 d \sqrt {\sec (c+d x)}}-\frac {2 (a-b) \sqrt {a+b} \sqrt {\cos (c+d x)} \csc (c+d x) \left (-3 a^2 b^2 (93 A+161 C)-21 a^4 (7 A+9 C)-435 a^3 b B-45 a b^3 B+10 A b^4\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{315 a^3 d \sqrt {\sec (c+d x)}}+\frac {2 (9 a B+5 A b) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}}{63 d}+\frac {2 A \sin (c+d x) \sec ^{\frac {9}{2}}(c+d x) (a+b \cos (c+d x))^{5/2}}{9 d} \]

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]^(11/2),x]

[Out]

(-2*(a - b)*Sqrt[a + b]*(10*A*b^4 - 435*a^3*b*B - 45*a*b^3*B - 21*a^4*(7*A + 9*C) - 3*a^2*b^2*(93*A + 161*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)])/(315*a^3*d*Sqrt[
Sec[c + d*x]]) - (2*(a - b)*Sqrt[a + b]*(10*A*b^3 + 15*a*b^2*(11*A - 3*B + 21*C) - 6*a^2*b*(19*A - 60*B + 28*C
) + 3*a^3*(49*A - 25*B + 63*C))*Sqrt[Cos[c + d*x]]*Csc[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Cos[c + d*x]]/(Sqr
t[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)])/(315*a^2*d*Sqrt[Sec[c + d*x]]) + (2*(5*A*b^3 + 75*a^3*B + 135*a*b^2*B + a^2*b*(163*A + 231*C))
*Sqrt[a + b*Cos[c + d*x]]*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(315*a*d) + (2*(15*A*b^2 + 90*a*b*B + 7*a^2*(7*A +
9*C))*Sqrt[a + b*Cos[c + d*x]]*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(315*d) + (2*(5*A*b + 9*a*B)*(a + b*Cos[c + d*
x])^(3/2)*Sec[c + d*x]^(7/2)*Sin[c + d*x])/(63*d) + (2*A*(a + b*Cos[c + d*x])^(5/2)*Sec[c + d*x]^(9/2)*Sin[c +
d*x])/(9*d)

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 3055

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[((A*b^2 - a*b*B + a^2*C)*Cos[e +
f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2)), x] + Dis
t[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*(b
*c - a*d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(b*
c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A*b^2 - a*b*B + a^2*C)*(m + n + 3)*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] && Lt
Q[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&
!IntegerQ[m]) || EqQ[a, 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 {11}{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 {11}{2}}(c+d x)} \, dx\\ &=\frac {2 A (a+b \cos (c+d x))^{5/2} \sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{9 d}+\frac {1}{9} \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+9 a B)+\frac {1}{2} (7 a A+9 b B+9 a C) \cos (c+d x)+\frac {1}{2} b (2 A+9 C) \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx\\ &=\frac {2 (5 A b+9 a B) (a+b \cos (c+d x))^{3/2} \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{63 d}+\frac {2 A (a+b \cos (c+d x))^{5/2} \sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{9 d}+\frac {1}{63} \left (4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {a+b \cos (c+d x)} \left (\frac {1}{4} \left (15 A b^2+90 a b B+7 a^2 (7 A+9 C)\right )+\frac {1}{4} \left (88 a A b+45 a^2 B+63 b^2 B+126 a b C\right ) \cos (c+d x)+\frac {3}{4} b (8 A b+6 a B+21 b C) \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx\\ &=\frac {2 \left (15 A b^2+90 a b B+7 a^2 (7 A+9 C)\right ) \sqrt {a+b \cos (c+d x)} \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{315 d}+\frac {2 (5 A b+9 a B) (a+b \cos (c+d x))^{3/2} \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{63 d}+\frac {2 A (a+b \cos (c+d x))^{5/2} \sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{9 d}+\frac {1}{315} \left (8 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {3}{8} \left (5 A b^3+75 a^3 B+135 a b^2 B+a^2 b (163 A+231 C)\right )+\frac {1}{8} \left (585 a^2 b B+315 b^3 B+21 a^3 (7 A+9 C)+5 a b^2 (121 A+189 C)\right ) \cos (c+d x)+\frac {1}{8} b \left (270 a b B+14 a^2 (7 A+9 C)+15 b^2 (10 A+21 C)\right ) \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}} \, dx\\ &=\frac {2 \left (5 A b^3+75 a^3 B+135 a b^2 B+a^2 b (163 A+231 C)\right ) \sqrt {a+b \cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{315 a d}+\frac {2 \left (15 A b^2+90 a b B+7 a^2 (7 A+9 C)\right ) \sqrt {a+b \cos (c+d x)} \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{315 d}+\frac {2 (5 A b+9 a B) (a+b \cos (c+d x))^{3/2} \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{63 d}+\frac {2 A (a+b \cos (c+d x))^{5/2} \sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{9 d}+\frac {\left (16 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {-\frac {3}{16} \left (10 A b^4-435 a^3 b B-45 a b^3 B-21 a^4 (7 A+9 C)-3 a^2 b^2 (93 A+161 C)\right )+\frac {3}{16} a \left (75 a^3 B+405 a b^2 B+5 b^3 (31 A+63 C)+3 a^2 b (87 A+119 C)\right ) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}} \, dx}{945 a}\\ &=\frac {2 \left (5 A b^3+75 a^3 B+135 a b^2 B+a^2 b (163 A+231 C)\right ) \sqrt {a+b \cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{315 a d}+\frac {2 \left (15 A b^2+90 a b B+7 a^2 (7 A+9 C)\right ) \sqrt {a+b \cos (c+d x)} \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{315 d}+\frac {2 (5 A b+9 a B) (a+b \cos (c+d x))^{3/2} \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{63 d}+\frac {2 A (a+b \cos (c+d x))^{5/2} \sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{9 d}-\frac {\left ((a-b) \left (10 A b^3+15 a b^2 (11 A-3 B+21 C)-6 a^2 b (19 A-60 B+28 C)+3 a^3 (49 A-25 B+63 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}{315 a}-\frac {\left (\left (10 A b^4-435 a^3 b B-45 a b^3 B-21 a^4 (7 A+9 C)-3 a^2 b^2 (93 A+161 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}{315 a}\\ &=-\frac {2 (a-b) \sqrt {a+b} \left (10 A b^4-435 a^3 b B-45 a b^3 B-21 a^4 (7 A+9 C)-3 a^2 b^2 (93 A+161 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}}}{315 a^3 d \sqrt {\sec (c+d x)}}-\frac {2 (a-b) \sqrt {a+b} \left (10 A b^3+15 a b^2 (11 A-3 B+21 C)-6 a^2 b (19 A-60 B+28 C)+3 a^3 (49 A-25 B+63 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}}}{315 a^2 d \sqrt {\sec (c+d x)}}+\frac {2 \left (5 A b^3+75 a^3 B+135 a b^2 B+a^2 b (163 A+231 C)\right ) \sqrt {a+b \cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{315 a d}+\frac {2 \left (15 A b^2+90 a b B+7 a^2 (7 A+9 C)\right ) \sqrt {a+b \cos (c+d x)} \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{315 d}+\frac {2 (5 A b+9 a B) (a+b \cos (c+d x))^{3/2} \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{63 d}+\frac {2 A (a+b \cos (c+d x))^{5/2} \sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{9 d}\\ \end {align*}

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Mathematica [A]  time = 20.86, size = 809, normalized size = 1.37 \[ \frac {2 \sqrt {\frac {1}{1-\tan ^2\left (\frac {1}{2} (c+d x)\right )}} \left (-\left ((a+b) \left (21 (7 A+9 C) a^4+435 b B a^3+3 b^2 (93 A+161 C) a^2+45 b^3 B a-10 A b^4\right ) E\left (\sin ^{-1}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {b-a}{a+b}\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {\frac {a \tan ^2\left (\frac {1}{2} (c+d x)\right )-b \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}{a+b}} \left (\tan ^2\left (\frac {1}{2} (c+d x)\right )+1\right )\right )+a (a+b) \left (3 (49 A+25 B+63 C) a^3+6 b (19 A+60 B+28 C) a^2+15 b^2 (11 A+3 (B+7 C)) a-10 A b^3\right ) F\left (\sin ^{-1}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {b-a}{a+b}\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {\frac {a \tan ^2\left (\frac {1}{2} (c+d x)\right )-b \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}{a+b}} \left (\tan ^2\left (\frac {1}{2} (c+d x)\right )+1\right )+\left (21 (7 A+9 C) a^4+435 b B a^3+3 b^2 (93 A+161 C) a^2+45 b^3 B a-10 A b^4\right ) \tan \left (\frac {1}{2} (c+d x)\right ) \left (\tan ^2\left (\frac {1}{2} (c+d x)\right )-1\right ) \left (a \tan ^2\left (\frac {1}{2} (c+d x)\right )-b \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b\right )\right )}{315 a^2 d \left (\tan ^2\left (\frac {1}{2} (c+d x)\right )+1\right )^{3/2} \sqrt {\frac {a \tan ^2\left (\frac {1}{2} (c+d x)\right )-b \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}{\tan ^2\left (\frac {1}{2} (c+d x)\right )+1}}}+\frac {\sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2}{63} \left (9 B \sin (c+d x) a^2+19 A b \sin (c+d x) a\right ) \sec ^3(c+d x)+\frac {2}{9} a^2 A \tan (c+d x) \sec ^3(c+d x)+\frac {2}{315} \left (49 A \sin (c+d x) a^2+63 C \sin (c+d x) a^2+135 b B \sin (c+d x) a+75 A b^2 \sin (c+d x)\right ) \sec ^2(c+d x)+\frac {2 \left (75 B \sin (c+d x) a^3+163 A b \sin (c+d x) a^2+231 b C \sin (c+d x) a^2+135 b^2 B \sin (c+d x) a+5 A b^3 \sin (c+d x)\right ) \sec (c+d x)}{315 a}-\frac {2 \left (-147 A a^4-189 C a^4-435 b B a^3-279 A b^2 a^2-483 b^2 C a^2-45 b^3 B a+10 A b^4\right ) \sin (c+d x)}{315 a^2}\right )}{d} \]

Warning: Unable to verify antiderivative.

[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]^(11/2),x]

[Out]

(2*Sqrt[(1 - Tan[(c + d*x)/2]^2)^(-1)]*((-10*A*b^4 + 435*a^3*b*B + 45*a*b^3*B + 21*a^4*(7*A + 9*C) + 3*a^2*b^2
*(93*A + 161*C))*Tan[(c + d*x)/2]*(-1 + Tan[(c + d*x)/2]^2)*(a + b + a*Tan[(c + d*x)/2]^2 - b*Tan[(c + d*x)/2]
^2) - (a + b)*(-10*A*b^4 + 435*a^3*b*B + 45*a*b^3*B + 21*a^4*(7*A + 9*C) + 3*a^2*b^2*(93*A + 161*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)] + a*(a + b)*(-10*A*b^3 + 6*a^2*b*(19*A + 60*B + 28*C)
+ 3*a^3*(49*A + 25*B + 63*C) + 15*a*b^2*(11*A + 3*(B + 7*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)]))/(315*a^2*d*(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)]) + (Sqrt[a + b*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((-2*(-147*a^4*A - 279*a
^2*A*b^2 + 10*A*b^4 - 435*a^3*b*B - 45*a*b^3*B - 189*a^4*C - 483*a^2*b^2*C)*Sin[c + d*x])/(315*a^2) + (2*Sec[c
+ d*x]^3*(19*a*A*b*Sin[c + d*x] + 9*a^2*B*Sin[c + d*x]))/63 + (2*Sec[c + d*x]^2*(49*a^2*A*Sin[c + d*x] + 75*A
*b^2*Sin[c + d*x] + 135*a*b*B*Sin[c + d*x] + 63*a^2*C*Sin[c + d*x]))/315 + (2*Sec[c + d*x]*(163*a^2*A*b*Sin[c
+ d*x] + 5*A*b^3*Sin[c + d*x] + 75*a^3*B*Sin[c + d*x] + 135*a*b^2*B*Sin[c + d*x] + 231*a^2*b*C*Sin[c + d*x]))/
(315*a) + (2*a^2*A*Sec[c + d*x]^3*Tan[c + d*x])/9))/d

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fricas [F]  time = 0.68, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C b^{2} \cos \left (d x + c\right )^{4} + {\left (2 \, C a b + B b^{2}\right )} \cos \left (d x + c\right )^{3} + A a^{2} + {\left (C a^{2} + 2 \, B a b + A b^{2}\right )} \cos \left (d x + c\right )^{2} + {\left (B a^{2} + 2 \, A a b\right )} \cos \left (d x + c\right )\right )} \sqrt {b \cos \left (d x + c\right ) + a} \sec \left (d x + c\right )^{\frac {11}{2}}, x\right ) \]

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)^(11/2),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)^(11/2), 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)^(11/2),x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 1.08, size = 6184, normalized size = 10.45 \[ \text {output too large to display} \]

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)^(11/2),x)

[Out]

result too large to display

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sec \left (d x + c\right )^{\frac {11}{2}}\,{d x} \]

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)^(11/2),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)^(11/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{11/2}\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{5/2}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1/cos(c + d*x))^(11/2)*(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))^(11/2)*(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)**(11/2),x)

[Out]

Timed out

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