3.1432 \(\int (a+b \cos (c+d x))^{5/2} (A+C \cos ^2(c+d x)) \sqrt {\sec (c+d x)} \, dx\)
Optimal. Leaf size=695 \[ \frac {a \left (15 a^2 C+432 A b^2+284 b^2 C\right ) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a+b \cos (c+d x)}}{192 b d}+\frac {\left (5 a^2 C+4 b^2 (4 A+3 C)\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{32 d \sqrt {\sec (c+d x)}}-\frac {(a-b) \sqrt {a+b} \left (15 a^2 C+432 A b^2+284 b^2 C\right ) \sqrt {\cos (c+d x)} \csc (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{192 b d \sqrt {\sec (c+d x)}}+\frac {\sqrt {a+b} \left (5 a^4 C-120 a^2 b^2 (2 A+C)-16 b^4 (4 A+3 C)\right ) \sqrt {\cos (c+d x)} \csc (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \Pi \left (\frac {a+b}{b};\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{64 b^2 d \sqrt {\sec (c+d x)}}+\frac {\sqrt {a+b} \left (15 a^3 C+2 a^2 b (192 A+59 C)+4 a b^2 (108 A+71 C)+24 b^3 (4 A+3 C)\right ) \sqrt {\cos (c+d x)} \csc (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{192 b d \sqrt {\sec (c+d x)}}+\frac {C \sin (c+d x) (a+b \cos (c+d x))^{5/2}}{4 d \sqrt {\sec (c+d x)}}+\frac {5 a C \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{24 d \sqrt {\sec (c+d x)}} \]
[Out]
5/24*a*C*(a+b*cos(d*x+c))^(3/2)*sin(d*x+c)/d/sec(d*x+c)^(1/2)+1/4*C*(a+b*cos(d*x+c))^(5/2)*sin(d*x+c)/d/sec(d*
x+c)^(1/2)+1/32*(5*a^2*C+4*b^2*(4*A+3*C))*sin(d*x+c)*(a+b*cos(d*x+c))^(1/2)/d/sec(d*x+c)^(1/2)+1/192*a*(432*A*
b^2+15*C*a^2+284*C*b^2)*sin(d*x+c)*(a+b*cos(d*x+c))^(1/2)*sec(d*x+c)^(1/2)/b/d-1/192*(a-b)*(432*A*b^2+15*C*a^2
+284*C*b^2)*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)/b/d/sec(d*x+c)^(1/2)+1
/192*(15*a^3*C+24*b^3*(4*A+3*C)+2*a^2*b*(192*A+59*C)+4*a*b^2*(108*A+71*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/d/sec(d*x+c)^(1/2)+1/64*(5*a^4*C-120*a^2*b^2*(2*A+C)-16*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^2/d/sec(d*x+c)^(1/2
)
________________________________________________________________________________________
Rubi [A] time = 2.32, antiderivative size = 695, normalized size of antiderivative = 1.00,
number of steps used = 10, number of rules used = 9, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.243, Rules used
= {4221, 3050, 3049, 3061, 3053, 2809, 2998, 2816, 2994} \[ \frac {a \left (15 a^2 C+432 A b^2+284 b^2 C\right ) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a+b \cos (c+d x)}}{192 b d}+\frac {\left (5 a^2 C+4 b^2 (4 A+3 C)\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{32 d \sqrt {\sec (c+d x)}}+\frac {\sqrt {a+b} \left (2 a^2 b (192 A+59 C)+15 a^3 C+4 a b^2 (108 A+71 C)+24 b^3 (4 A+3 C)\right ) \sqrt {\cos (c+d x)} \csc (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{192 b d \sqrt {\sec (c+d x)}}-\frac {(a-b) \sqrt {a+b} \left (15 a^2 C+432 A b^2+284 b^2 C\right ) \sqrt {\cos (c+d x)} \csc (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{192 b d \sqrt {\sec (c+d x)}}+\frac {\sqrt {a+b} \left (-120 a^2 b^2 (2 A+C)+5 a^4 C-16 b^4 (4 A+3 C)\right ) \sqrt {\cos (c+d x)} \csc (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \Pi \left (\frac {a+b}{b};\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{64 b^2 d \sqrt {\sec (c+d x)}}+\frac {C \sin (c+d x) (a+b \cos (c+d x))^{5/2}}{4 d \sqrt {\sec (c+d x)}}+\frac {5 a C \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{24 d \sqrt {\sec (c+d x)}} \]
Antiderivative was successfully verified.
[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]*(432*A*b^2 + 15*a^2*C + 284*b^2*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)])/(192*b*d*Sqrt[Sec[c + d*x]]) + (Sqrt[a + b]*(15*a^3*C + 24*b^3*(4*A +
3*C) + 2*a^2*b*(192*A + 59*C) + 4*a*b^2*(108*A + 71*C))*Sqrt[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)])/(192*b*d*Sqrt[Sec[c + d*x]]) + (Sqrt[a + b]*(5*a^4*C - 120*a^2*b^2*(2
*A + C) - 16*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)])/(64*b^2*d*Sqrt[Sec[c + d*x]]) + ((5*a^2*C + 4*b^2*(4*A + 3*C))*Sqrt[a + b*Cos[c + d*
x]]*Sin[c + d*x])/(32*d*Sqrt[Sec[c + d*x]]) + (5*a*C*(a + b*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(24*d*Sqrt[Sec[c
+ d*x]]) + (C*(a + b*Cos[c + d*x])^(5/2)*Sin[c + d*x])/(4*d*Sqrt[Sec[c + d*x]]) + (a*(432*A*b^2 + 15*a^2*C +
284*b^2*C)*Sqrt[a + b*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(192*b*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 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 3050
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 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+C \cos ^2(c+d x)\right ) \sqrt {\sec (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+C \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {C (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{4 d \sqrt {\sec (c+d x)}}+\frac {1}{4} \left (\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} a (8 A+C)+b (4 A+3 C) \cos (c+d x)+\frac {5}{2} a C \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {5 a C (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{24 d \sqrt {\sec (c+d x)}}+\frac {C (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{4 d \sqrt {\sec (c+d x)}}+\frac {1}{12} \left (\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^2 (48 A+11 C)+\frac {1}{2} a b (48 A+31 C) \cos (c+d x)+\frac {3}{4} \left (5 a^2 C+4 b^2 (4 A+3 C)\right ) \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {\left (5 a^2 C+4 b^2 (4 A+3 C)\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{32 d \sqrt {\sec (c+d x)}}+\frac {5 a C (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{24 d \sqrt {\sec (c+d x)}}+\frac {C (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{4 d \sqrt {\sec (c+d x)}}+\frac {1}{24} \left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {1}{8} a \left (12 b^2 (4 A+3 C)+8 a^2 \left (24 A+\frac {59 C}{8}\right )\right )+\frac {1}{4} b \left (12 b^2 (4 A+3 C)+a^2 (288 A+161 C)\right ) \cos (c+d x)+\frac {1}{8} a \left (432 A b^2+15 a^2 C+284 b^2 C\right ) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}} \, dx\\ &=\frac {\left (5 a^2 C+4 b^2 (4 A+3 C)\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{32 d \sqrt {\sec (c+d x)}}+\frac {5 a C (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{24 d \sqrt {\sec (c+d x)}}+\frac {C (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{4 d \sqrt {\sec (c+d x)}}+\frac {a \left (432 A b^2+15 a^2 C+284 b^2 C\right ) \sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)} \sin (c+d x)}{192 b d}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {-\frac {1}{8} a^2 \left (432 A b^2+15 a^2 C+284 b^2 C\right )+\frac {1}{4} a b \left (12 b^2 (4 A+3 C)+a^2 (192 A+59 C)\right ) \cos (c+d x)-\frac {3}{8} \left (5 a^4 C-120 a^2 b^2 (2 A+C)-16 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}{48 b}\\ &=\frac {\left (5 a^2 C+4 b^2 (4 A+3 C)\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{32 d \sqrt {\sec (c+d x)}}+\frac {5 a C (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{24 d \sqrt {\sec (c+d x)}}+\frac {C (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{4 d \sqrt {\sec (c+d x)}}+\frac {a \left (432 A b^2+15 a^2 C+284 b^2 C\right ) \sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)} \sin (c+d x)}{192 b d}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {-\frac {1}{8} a^2 \left (432 A b^2+15 a^2 C+284 b^2 C\right )+\frac {1}{4} a b \left (12 b^2 (4 A+3 C)+a^2 (192 A+59 C)\right ) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}} \, dx}{48 b}-\frac {\left (\left (5 a^4 C-120 a^2 b^2 (2 A+C)-16 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}{128 b}\\ &=\frac {\sqrt {a+b} \left (5 a^4 C-120 a^2 b^2 (2 A+C)-16 b^4 (4 A+3 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}}}{64 b^2 d \sqrt {\sec (c+d x)}}+\frac {\left (5 a^2 C+4 b^2 (4 A+3 C)\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{32 d \sqrt {\sec (c+d x)}}+\frac {5 a C (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{24 d \sqrt {\sec (c+d x)}}+\frac {C (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{4 d \sqrt {\sec (c+d x)}}+\frac {a \left (432 A b^2+15 a^2 C+284 b^2 C\right ) \sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)} \sin (c+d x)}{192 b d}-\frac {\left (a^2 \left (432 A b^2+15 a^2 C+284 b^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}{384 b}+\frac {\left (a \left (15 a^3 C+24 b^3 (4 A+3 C)+2 a^2 b (192 A+59 C)+4 a b^2 (108 A+71 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}{384 b}\\ &=-\frac {(a-b) \sqrt {a+b} \left (432 A b^2+15 a^2 C+284 b^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}}}{192 b d \sqrt {\sec (c+d x)}}+\frac {\sqrt {a+b} \left (15 a^3 C+24 b^3 (4 A+3 C)+2 a^2 b (192 A+59 C)+4 a b^2 (108 A+71 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}}}{192 b d \sqrt {\sec (c+d x)}}+\frac {\sqrt {a+b} \left (5 a^4 C-120 a^2 b^2 (2 A+C)-16 b^4 (4 A+3 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}}}{64 b^2 d \sqrt {\sec (c+d x)}}+\frac {\left (5 a^2 C+4 b^2 (4 A+3 C)\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{32 d \sqrt {\sec (c+d x)}}+\frac {5 a C (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{24 d \sqrt {\sec (c+d x)}}+\frac {C (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{4 d \sqrt {\sec (c+d x)}}+\frac {a \left (432 A b^2+15 a^2 C+284 b^2 C\right ) \sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)} \sin (c+d x)}{192 b d}\\ \end {align*}
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Mathematica [A] time = 19.93, size = 601, normalized size = 0.86 \[ \frac {\sqrt {\sec (c+d x)} \sqrt {a+b \cos (c+d x)} \left (\frac {1}{192} \left (59 a^2 C+48 A b^2+48 b^2 C\right ) \sin (2 (c+d x))+\frac {17}{96} a b C \sin (c+d x)+\frac {17}{96} a b C \sin (3 (c+d x))+\frac {1}{32} b^2 C \sin (4 (c+d x))\right )}{d}+\frac {\sqrt {\cos (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)} \left (a b \left (15 a^2 C+432 A b^2+284 b^2 C\right ) \tan \left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) \left (\cos (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right )\right )^{3/2} (a+b \cos (c+d x))+a b (a+b) \left (15 a^2 C+432 A b^2+284 b^2 C\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {\frac {\sec ^2\left (\frac {1}{2} (c+d x)\right ) (a+b \cos (c+d x))}{a+b}} E\left (\sin ^{-1}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {b-a}{a+b}\right )-3 \left (5 a^4 C-120 a^2 b^2 (2 A+C)-16 b^4 (4 A+3 C)\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {\frac {\sec ^2\left (\frac {1}{2} (c+d x)\right ) (a+b \cos (c+d x))}{a+b}} \left ((a-b) F\left (\sin ^{-1}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {b-a}{a+b}\right )+2 b \Pi \left (-1;\sin ^{-1}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {b-a}{a+b}\right )\right )+a (a+b) \left (15 a^3 C-30 a^2 b C-4 a b^2 (84 A+53 C)-24 b^3 (4 A+3 C)\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {\frac {\sec ^2\left (\frac {1}{2} (c+d x)\right ) (a+b \cos (c+d x))}{a+b}} F\left (\sin ^{-1}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {b-a}{a+b}\right )\right )}{192 b^2 d \sec ^2\left (\frac {1}{2} (c+d x)\right )^{3/2} \sqrt {a+b \cos (c+d x)}} \]
Warning: Unable to verify antiderivative.
[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]]*((17*a*b*C*Sin[c + d*x])/96 + ((48*A*b^2 + 59*a^2*C + 48*b^2*C)*S
in[2*(c + d*x)])/192 + (17*a*b*C*Sin[3*(c + d*x)])/96 + (b^2*C*Sin[4*(c + d*x)])/32))/d + (Sqrt[Cos[c + d*x]*S
ec[(c + d*x)/2]^2]*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*(a*b*(a + b)*(432*A*b^2 + 15*a^2*C + 284*b^2*C)*Ellip
ticE[ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Sec[(c + d*x)/2]^2*Sqrt[((a + b*Cos[c + d*x])*Sec[(c + d*x)/2
]^2)/(a + b)] + a*(a + b)*(15*a^3*C - 30*a^2*b*C - 24*b^3*(4*A + 3*C) - 4*a*b^2*(84*A + 53*C))*EllipticF[ArcSi
n[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Sec[(c + d*x)/2]^2*Sqrt[((a + b*Cos[c + d*x])*Sec[(c + d*x)/2]^2)/(a +
b)] - 3*(5*a^4*C - 120*a^2*b^2*(2*A + C) - 16*b^4*(4*A + 3*C))*((a - b)*EllipticF[ArcSin[Tan[(c + d*x)/2]], (-
a + b)/(a + b)] + 2*b*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)])*Sec[(c + d*x)/2]^2*Sqrt[((a
+ b*Cos[c + d*x])*Sec[(c + d*x)/2]^2)/(a + b)] + a*b*(432*A*b^2 + 15*a^2*C + 284*b^2*C)*(a + b*Cos[c + d*x])*(
Cos[c + d*x]*Sec[(c + d*x)/2]^2)^(3/2)*Sec[c + d*x]*Tan[(c + d*x)/2]))/(192*b^2*d*Sqrt[a + b*Cos[c + d*x]]*(Se
c[(c + d*x)/2]^2)^(3/2))
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fricas [F] time = 4.72, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C b^{2} \cos \left (d x + c\right )^{4} + 2 \, C a b \cos \left (d x + c\right )^{3} + 2 \, A a b \cos \left (d x + c\right ) + A a^{2} + {\left (C a^{2} + A b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {b \cos \left (d x + c\right ) + a} \sqrt {\sec \left (d x + c\right )}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[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]
integral((C*b^2*cos(d*x + c)^4 + 2*C*a*b*cos(d*x + c)^3 + 2*A*a*b*cos(d*x + c) + A*a^2 + (C*a^2 + A*b^2)*cos(d
*x + c)^2)*sqrt(b*cos(d*x + c) + a)*sqrt(sec(d*x + c)), 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+C*cos(d*x+c)^2)*sec(d*x+c)^(1/2),x, algorithm="giac")
[Out]
Timed out
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maple [B] time = 0.83, size = 3993, normalized size = 5.75 \[ \text {output too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+b*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)*sec(d*x+c)^(1/2),x)
[Out]
-1/192/d*(1/cos(d*x+c))^(1/2)/(a+b*cos(d*x+c))^(1/2)*(48*C*cos(d*x+c)^6*b^4-192*A*(cos(d*x+c)/(1+cos(d*x+c)))^
(1/2)*((a+b*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))
*b^4*sin(d*x+c)-72*C*cos(d*x+c)^2*b^4-15*C*cos(d*x+c)*a^4-432*A*cos(d*x+c)^2*a*b^3-432*A*cos(d*x+c)*a^2*b^2-96
*A*cos(d*x+c)*a*b^3+24*C*cos(d*x+c)^4*b^4+172*C*cos(d*x+c)^3*a*b^3-15*C*cos(d*x+c)^2*a^3*b+30*C*cos(d*x+c)^2*a
^2*b^2-284*C*cos(d*x+c)^2*a*b^3-118*C*cos(d*x+c)*a^3*b-284*C*cos(d*x+c)*a^2*b^2-72*C*cos(d*x+c)*a*b^3+184*C*co
s(d*x+c)^5*a*b^3-96*A*cos(d*x+c)^2*b^4+528*A*cos(d*x+c)^3*a*b^3+432*A*cos(d*x+c)^2*a^2*b^2+432*A*sin(d*x+c)*(c
os(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+b*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*
x+c),(-(a-b)/(a+b))^(1/2))*a^2*b^2+432*A*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+b*cos(d*x+c))/(1+cos
(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*a*b^3-1152*A*sin(d*x+c)*(cos(
d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+b*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c
),(-(a-b)/(a+b))^(1/2))*a^2*b^2+96*A*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+b*cos(d*x+c))/(1+cos(d*x
+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*a*b^3+254*C*cos(d*x+c)^4*a^2*b^2+
133*C*cos(d*x+c)^3*a^3*b+384*A*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+b*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)
*EllipticPi((-1+cos(d*x+c))/sin(d*x+c),-1,(-(a-b)/(a+b))^(1/2))*b^4*sin(d*x+c)+432*A*sin(d*x+c)*cos(d*x+c)*Ell
ipticE((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+b*cos(d*x+c))/(1
+cos(d*x+c))/(a+b))^(1/2)*a^2*b^2+432*A*sin(d*x+c)*cos(d*x+c)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+
b))^(1/2))*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+b*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*a*b^3-1152*A*sin(d*
x+c)*cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+b*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+
cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*a^2*b^2+96*A*sin(d*x+c)*cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1
/2)*((a+b*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*a
*b^3+384*A*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+b*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*Elliptic
F((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*a^3*b+384*A*sin(d*x+c)*cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)
))^(1/2)*((a+b*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/
2))*a^3*b+15*C*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+b*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+c
os(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*a^4*sin(d*x+c)-144*C*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+b*cos(d
*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*b^4*sin(d*x+c)-3
0*C*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+b*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticPi((-1+cos(d*x+c))
/sin(d*x+c),-1,(-(a-b)/(a+b))^(1/2))*a^4*sin(d*x+c)+288*C*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+b*cos(d*x+c))/
(1+cos(d*x+c))/(a+b))^(1/2)*EllipticPi((-1+cos(d*x+c))/sin(d*x+c),-1,(-(a-b)/(a+b))^(1/2))*b^4*sin(d*x+c)-192*
A*sin(d*x+c)*cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+b*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*Ellipt
icF((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*b^4+384*A*sin(d*x+c)*cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)
))^(1/2)*((a+b*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticPi((-1+cos(d*x+c))/sin(d*x+c),-1,(-(a-b)/(a+b))
^(1/2))*b^4+15*C*sin(d*x+c)*cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+b*cos(d*x+c))/(1+cos(d*x+c))/(a+b
))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*a^4-144*C*sin(d*x+c)*cos(d*x+c)*(cos(d*x+c
)/(1+cos(d*x+c)))^(1/2)*((a+b*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),(-(
a-b)/(a+b))^(1/2))*b^4-30*C*sin(d*x+c)*cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+b*cos(d*x+c))/(1+cos(d
*x+c))/(a+b))^(1/2)*EllipticPi((-1+cos(d*x+c))/sin(d*x+c),-1,(-(a-b)/(a+b))^(1/2))*a^4+288*C*sin(d*x+c)*cos(d*
x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+b*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticPi((-1+cos(d*x+c)
)/sin(d*x+c),-1,(-(a-b)/(a+b))^(1/2))*b^4+1440*A*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+b*cos(d*x+c))/(1+cos(d*
x+c))/(a+b))^(1/2)*EllipticPi((-1+cos(d*x+c))/sin(d*x+c),-1,(-(a-b)/(a+b))^(1/2))*a^2*b^2*sin(d*x+c)+15*C*(cos
(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+b*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+
c),(-(a-b)/(a+b))^(1/2))*a^3*b*sin(d*x+c)+284*C*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+b*cos(d*x+c))/(1+cos(d*x
+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*a^2*b^2*sin(d*x+c)+284*C*(cos(d*x
+c)/(1+cos(d*x+c)))^(1/2)*((a+b*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),(
-(a-b)/(a+b))^(1/2))*a*b^3*sin(d*x+c)+118*C*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+b*cos(d*x+c))/(1+cos(d*x+c))
/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*a^3*b*sin(d*x+c)-644*C*(cos(d*x+c)/(1
+cos(d*x+c)))^(1/2)*((a+b*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)
/(a+b))^(1/2))*a^2*b^2*sin(d*x+c)+72*C*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+b*cos(d*x+c))/(1+cos(d*x+c))/(a+b
))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*a*b^3*sin(d*x+c)+720*C*(cos(d*x+c)/(1+cos(
d*x+c)))^(1/2)*((a+b*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticPi((-1+cos(d*x+c))/sin(d*x+c),-1,(-(a-b)/
(a+b))^(1/2))*a^2*b^2*sin(d*x+c)+1440*A*sin(d*x+c)*cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+b*cos(d*x+
c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticPi((-1+cos(d*x+c))/sin(d*x+c),-1,(-(a-b)/(a+b))^(1/2))*a^2*b^2+15*C*si
n(d*x+c)*cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+b*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE(
(-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*a^3*b+284*C*sin(d*x+c)*cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))
^(1/2)*((a+b*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2)
)*a^2*b^2+284*C*sin(d*x+c)*cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+b*cos(d*x+c))/(1+cos(d*x+c))/(a+b)
)^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*a*b^3+118*C*sin(d*x+c)*cos(d*x+c)*(cos(d*x+
c)/(1+cos(d*x+c)))^(1/2)*((a+b*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),(-
(a-b)/(a+b))^(1/2))*a^3*b-644*C*sin(d*x+c)*cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+b*cos(d*x+c))/(1+c
os(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*a^2*b^2+72*C*sin(d*x+c)*cos
(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+b*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+
c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*a*b^3+720*C*sin(d*x+c)*cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+b
*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticPi((-1+cos(d*x+c))/sin(d*x+c),-1,(-(a-b)/(a+b))^(1/2))*a^2*b^
2+15*C*cos(d*x+c)^2*a^4+96*A*cos(d*x+c)^4*b^4)/sin(d*x+c)/b
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sqrt {\sec \left (d x + c\right )}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[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)
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{5/2} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((A + C*cos(c + d*x)^2)*(1/cos(c + d*x))^(1/2)*(a + b*cos(c + d*x))^(5/2),x)
[Out]
int((A + C*cos(c + d*x)^2)*(1/cos(c + d*x))^(1/2)*(a + b*cos(c + d*x))^(5/2), 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+C*cos(d*x+c)**2)*sec(d*x+c)**(1/2),x)
[Out]
Timed out
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