∫ ∘ _________ a+b cos(c+dx) (A+C cos2(c+dx)) _______________________________________________

3.1418 \(\int \frac {\sqrt {a+b \cos (c+d x)} (A+C \cos ^2(c+d x))}{\sec ^{\frac {3}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=698 \[ \frac {\left (5 a^2 C+4 b^2 (4 A+3 C)\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{32 b^2 d \sqrt {\sec (c+d x)}}+\frac {a \left (15 a^2 C+48 A b^2+28 b^2 C\right ) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a+b \cos (c+d x)}}{192 b^3 d}-\frac {(a-b) \sqrt {a+b} \left (15 a^2 C+48 A b^2+28 b^2 C\right ) \sqrt {\cos (c+d x)} \csc (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{192 b^3 d \sqrt {\sec (c+d x)}}+\frac {\sqrt {a+b} \left (5 a^4 C+8 a^2 b^2 (2 A+C)-16 b^4 (4 A+3 C)\right ) \sqrt {\cos (c+d x)} \csc (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \Pi \left (\frac {a+b}{b};\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{64 b^4 d \sqrt {\sec (c+d x)}}+\frac {\sqrt {a+b} \left (15 a^3 C-10 a^2 b C+4 a b^2 (12 A+7 C)+24 b^3 (4 A+3 C)\right ) \sqrt {\cos (c+d x)} \csc (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{192 b^3 d \sqrt {\sec (c+d x)}}-\frac {5 a C \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{24 b^2 d \sqrt {\sec (c+d x)}}+\frac {C \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{4 b d \sec ^{\frac {3}{2}}(c+d x)} \]

[Out]

1/4*C*(a+b*cos(d*x+c))^(3/2)*sin(d*x+c)/b/d/sec(d*x+c)^(3/2)-5/24*a*C*(a+b*cos(d*x+c))^(3/2)*sin(d*x+c)/b^2/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)/b^2/d/sec(d*x+c)^(1/2)+1/192

*a*(48*A*b^2+15*C*a^2+28*C*b^2)*sin(d*x+c)*(a+b*cos(d*x+c))^(1/2)*sec(d*x+c)^(1/2)/b^3/d-1/192*(a-b)*(48*A*b^2

+15*C*a^2+28*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^3/d/sec(d*x+

c)^(1/2)+1/192*(15*a^3*C-10*a^2*b*C+24*b^3*(4*A+3*C)+4*a*b^2*(12*A+7*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^3/d/sec(d*x+c)^(1/2)+1/64*(5*a^4*C+8*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^4/d/sec(d*x+c)^(1/2)

________________________________________________________________________________________

Rubi [A] time = 2.17, antiderivative size = 698, 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} \[ \frac {a \left (15 a^2 C+48 A b^2+28 b^2 C\right ) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a+b \cos (c+d x)}}{192 b^3 d}+\frac {\left (5 a^2 C+4 b^2 (4 A+3 C)\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{32 b^2 d \sqrt {\sec (c+d x)}}+\frac {\sqrt {a+b} \left (-10 a^2 b C+15 a^3 C+4 a b^2 (12 A+7 C)+24 b^3 (4 A+3 C)\right ) \sqrt {\cos (c+d x)} \csc (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{192 b^3 d \sqrt {\sec (c+d x)}}-\frac {(a-b) \sqrt {a+b} \left (15 a^2 C+48 A b^2+28 b^2 C\right ) \sqrt {\cos (c+d x)} \csc (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{192 b^3 d \sqrt {\sec (c+d x)}}+\frac {\sqrt {a+b} \left (8 a^2 b^2 (2 A+C)+5 a^4 C-16 b^4 (4 A+3 C)\right ) \sqrt {\cos (c+d x)} \csc (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \Pi \left (\frac {a+b}{b};\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{64 b^4 d \sqrt {\sec (c+d x)}}-\frac {5 a C \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{24 b^2 d \sqrt {\sec (c+d x)}}+\frac {C \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{4 b d \sec ^{\frac {3}{2}}(c+d x)} \]

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-((a - b)*Sqrt[a + b]*(48*A*b^2 + 15*a^2*C + 28*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^3*d*Sqrt[Sec[c + d*x]]) + (Sqrt[a + b]*(15*a^3*C - 10*a^2*b*C +

24*b^3*(4*A + 3*C) + 4*a*b^2*(12*A + 7*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^3*d*Sqrt[Sec[c + d*x]]) + (Sqrt[a + b]*(5*a^4*C + 8*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^4*d*Sqrt[Sec[c + d*x]]) + (C*(a + b*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(4*b*d*Sec[c + d*x]^(3

/2)) + ((5*a^2*C + 4*b^2*(4*A + 3*C))*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(32*b^2*d*Sqrt[Sec[c + d*x]]) - (

5*a*C*(a + b*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(24*b^2*d*Sqrt[Sec[c + d*x]]) + (a*(48*A*b^2 + 15*a^2*C + 28*b^

2*C)*Sqrt[a + b*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(192*b^3*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,

Rubi steps

\begin {align*} \int \frac {\sqrt {a+b \cos (c+d x)} \left (A+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 \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx\\ &=\frac {C (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{4 b d \sec ^{\frac {3}{2}}(c+d x)}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)} \left (\frac {3 a C}{2}+b (4 A+3 C) \cos (c+d x)-\frac {5}{2} a C \cos ^2(c+d x)\right ) \, dx}{4 b}\\ &=\frac {C (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{4 b d \sec ^{\frac {3}{2}}(c+d x)}-\frac {5 a C (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{24 b^2 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 {5 a^2 C}{4}-\frac {1}{2} a b 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}{12 b^2}\\ &=\frac {C (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{4 b d \sec ^{\frac {3}{2}}(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 b^2 d \sqrt {\sec (c+d x)}}-\frac {5 a C (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{24 b^2 d \sqrt {\sec (c+d x)}}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {-\frac {1}{8} a \left (5 a^2 C-12 b^2 (4 A+3 C)\right )+\frac {1}{4} b \left (48 A b^2+\left (a^2+36 b^2\right ) C\right ) \cos (c+d x)+\frac {1}{8} a \left (48 A b^2+15 a^2 C+28 b^2 C\right ) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}} \, dx}{24 b^2}\\ &=\frac {C (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{4 b d \sec ^{\frac {3}{2}}(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 b^2 d \sqrt {\sec (c+d x)}}-\frac {5 a C (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{24 b^2 d \sqrt {\sec (c+d x)}}+\frac {a \left (48 A b^2+15 a^2 C+28 b^2 C\right ) \sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)} \sin (c+d x)}{192 b^3 d}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {-\frac {1}{8} a^2 \left (48 A b^2+15 a^2 C+28 b^2 C\right )-\frac {1}{4} a b \left (5 a^2 C-12 b^2 (4 A+3 C)\right ) \cos (c+d x)-\frac {3}{8} \left (5 a^4 C+8 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^3}\\ &=\frac {C (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{4 b d \sec ^{\frac {3}{2}}(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 b^2 d \sqrt {\sec (c+d x)}}-\frac {5 a C (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{24 b^2 d \sqrt {\sec (c+d x)}}+\frac {a \left (48 A b^2+15 a^2 C+28 b^2 C\right ) \sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)} \sin (c+d x)}{192 b^3 d}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {-\frac {1}{8} a^2 \left (48 A b^2+15 a^2 C+28 b^2 C\right )-\frac {1}{4} a b \left (5 a^2 C-12 b^2 (4 A+3 C)\right ) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}} \, dx}{48 b^3}-\frac {\left (\left (5 a^4 C+8 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^3}\\ &=\frac {\sqrt {a+b} \left (5 a^4 C+8 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^4 d \sqrt {\sec (c+d x)}}+\frac {C (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{4 b d \sec ^{\frac {3}{2}}(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 b^2 d \sqrt {\sec (c+d x)}}-\frac {5 a C (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{24 b^2 d \sqrt {\sec (c+d x)}}+\frac {a \left (48 A b^2+15 a^2 C+28 b^2 C\right ) \sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)} \sin (c+d x)}{192 b^3 d}-\frac {\left (a^2 \left (48 A b^2+15 a^2 C+28 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^3}+\frac {\left (a \left (15 a^3 C-10 a^2 b C+24 b^3 (4 A+3 C)+4 a b^2 (12 A+7 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^3}\\ &=-\frac {(a-b) \sqrt {a+b} \left (48 A b^2+15 a^2 C+28 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^3 d \sqrt {\sec (c+d x)}}+\frac {\sqrt {a+b} \left (15 a^3 C-10 a^2 b C+24 b^3 (4 A+3 C)+4 a b^2 (12 A+7 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^3 d \sqrt {\sec (c+d x)}}+\frac {\sqrt {a+b} \left (5 a^4 C+8 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^4 d \sqrt {\sec (c+d x)}}+\frac {C (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{4 b d \sec ^{\frac {3}{2}}(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 b^2 d \sqrt {\sec (c+d x)}}-\frac {5 a C (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{24 b^2 d \sqrt {\sec (c+d x)}}+\frac {a \left (48 A b^2+15 a^2 C+28 b^2 C\right ) \sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)} \sin (c+d x)}{192 b^3 d}\\ \end {align*}

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Mathematica [B] time = 15.69, size = 1798, normalized size = 2.58 \[ \text {result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(Sqrt[a + b*Cos[c + d*x]]*(A + 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]]*((a*C*Sin[c + d*x])/(96*b) + ((48*A*b^2 - 5*a^2*C + 48*b^2*C)*Sin

[2*(c + d*x)])/(192*b^2) + (a*C*Sin[3*(c + d*x)])/(96*b) + (C*Sin[4*(c + d*x)])/32))/d - (Sqrt[(a + b + a*Tan[

(c + d*x)/2]^2 - b*Tan[(c + d*x)/2]^2)/(1 + Tan[(c + d*x)/2]^2)]*(-48*a^2*A*b^2*Tan[(c + d*x)/2] - 48*a*A*b^3*

Tan[(c + d*x)/2] - 15*a^4*C*Tan[(c + d*x)/2] - 15*a^3*b*C*Tan[(c + d*x)/2] - 28*a^2*b^2*C*Tan[(c + d*x)/2] - 2

8*a*b^3*C*Tan[(c + d*x)/2] + 96*a*A*b^3*Tan[(c + d*x)/2]^3 + 30*a^3*b*C*Tan[(c + d*x)/2]^3 + 56*a*b^3*C*Tan[(c

+ d*x)/2]^3 + 48*a^2*A*b^2*Tan[(c + d*x)/2]^5 - 48*a*A*b^3*Tan[(c + d*x)/2]^5 + 15*a^4*C*Tan[(c + d*x)/2]^5 -

15*a^3*b*C*Tan[(c + d*x)/2]^5 + 28*a^2*b^2*C*Tan[(c + d*x)/2]^5 - 28*a*b^3*C*Tan[(c + d*x)/2]^5 + 96*a^2*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)] - 384*A*b^4*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^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)] + 48*a^2*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)] - 288*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)] + 96*a^2*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)] - 384*A*b^4*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^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)] + 48*a^2*b^2*C*EllipticPi[-1, ArcSi

n[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)] - 288*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*(a + b)*(48*A*b^2 + 15*a^2*C + 28*b^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*b*(5*a^3*C + 2*a^2*b*C - 12*a*b^2*(4*A + 3*C) + 24*b^3*(4*A + 3*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)]))/(192*b^3*d*(-1 + Tan[(c + d*x)/2]^2)*Sqrt[(1 + Tan[(c + d*x)/2]^2)/

(1 - Tan[(c + d*x)/2]^2)]*(b*(-1 + Tan[(c + d*x)/2]^2) - a*(1 + Tan[(c + d*x)/2]^2)))

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fricas [F] time = 2.49, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \sqrt {b \cos \left (d x + c\right ) + a}}{\sec \left (d x + c\right )^{\frac {3}{2}}}, x\right ) \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*cos(d*x+c)^2)*(a+b*cos(d*x+c))^(1/2)/sec(d*x+c)^(3/2),x, algorithm="fricas")

[Out]

integral((C*cos(d*x + c)^2 + A)*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+C*cos(d*x+c)^2)*(a+b*cos(d*x+c))^(1/2)/sec(d*x+c)^(3/2),x, algorithm="giac")

[Out]

Timed out

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maple [B] time = 0.78, size = 3615, normalized size = 5.18 \[ \text {output too large to display} \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/192/d*(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*c

os(d*x+c)*a^4-48*A*cos(d*x+c)^2*a*b^3-48*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+44*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^2-28*C*cos(d*x+c)^2*a*b^3+10*C*cos(d*x+c)*

a^3*b-28*C*cos(d*x+c)*a^2*b^2-72*C*cos(d*x+c)*a*b^3+56*C*cos(d*x+c)^5*a*b^3-96*A*cos(d*x+c)^2*b^4+144*A*cos(d*

x+c)^3*a*b^3+48*A*cos(d*x+c)^2*a^2*b^2+48*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^2*b^2+48*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*b^3+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^3-2*C*cos(d*x+c)^4*a^2*b^2+5

*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)*El

lipticPi((-1+cos(d*x+c))/sin(d*x+c),-1,(-(a-b)/(a+b))^(1/2))*b^4*sin(d*x+c)+48*A*sin(d*x+c)*cos(d*x+c)*Ellipti

cE((-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+48*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+96*A*sin(d*x+c)*co

s(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+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^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)-30*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)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*b^4+384*A*sin(d*x+c)*co

s(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)*Ellipt

icF((-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^4-96*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)*El

lipticE((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*a^3*b*sin(d*x+c)+28*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)+28*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

E((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*a*b^3*sin(d*x+c)-10*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*si

n(d*x+c)-4*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

)-48*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)-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)*EllipticPi((-1+cos(d*x+c))/sin(d*x+c),-1,(-(a-b)/(a+b)

)^(1/2))*a^2*b^2+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^3*b+28*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+28*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^3-10*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-4*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^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)*Ell

ipticF((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*a*b^3-48*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)*cos(d*x+c)*(1/cos(d*x+c))^(3/2)/sin(d*x+c)/(a

+b*cos(d*x+c))^(1/2)/b^3

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*cos(d*x+c)^2)*(a+b*cos(d*x+c))^(1/2)/sec(d*x+c)^(3/2),x, algorithm="maxima")

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*cos(d*x+c)**2)*(a+b*cos(d*x+c))**(1/2)/sec(d*x+c)**(3/2),x)

[Out]

Integral((A + C*cos(c + d*x)**2)*sqrt(a + b*cos(c + d*x))/sec(c + d*x)**(3/2), x)

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