∫ A+B cos(c+dx)+C cos2(c+dx)______________________ ∘________

3.1531 \(\int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {a+b \cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=653 \[ \frac {\sin (c+d x) \sqrt {\sec (c+d x)} \left (15 a^2 C-18 a b B+24 A b^2+16 b^2 C\right ) \sqrt {a+b \cos (c+d x)}}{24 b^3 d}+\frac {\sqrt {a+b} \sqrt {\cos (c+d x)} \csc (c+d x) \left (15 a^2 C-18 a b B-10 a b C+24 A b^2+12 b^2 B+16 b^2 C\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{24 b^3 d \sqrt {\sec (c+d x)}}-\frac {(a-b) \sqrt {a+b} \sqrt {\cos (c+d x)} \csc (c+d x) \left (15 a^2 C-18 a b B+24 A b^2+16 b^2 C\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{24 a b^3 d \sqrt {\sec (c+d x)}}-\frac {\sqrt {a+b} \sqrt {\cos (c+d x)} \csc (c+d x) \left (-5 a^3 C+6 a^2 b B-4 a b^2 (2 A+C)+8 b^3 B\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \Pi \left (\frac {a+b}{b};\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{8 b^4 d \sqrt {\sec (c+d x)}}+\frac {(6 b B-5 a C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{12 b^2 d \sqrt {\sec (c+d x)}}+\frac {C \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 b d \sec ^{\frac {3}{2}}(c+d x)} \]

[Out]

1/3*C*sin(d*x+c)*(a+b*cos(d*x+c))^(1/2)/b/d/sec(d*x+c)^(3/2)+1/12*(6*B*b-5*C*a)*sin(d*x+c)*(a+b*cos(d*x+c))^(1

/2)/b^2/d/sec(d*x+c)^(1/2)+1/24*(24*A*b^2-18*B*a*b+15*C*a^2+16*C*b^2)*sin(d*x+c)*(a+b*cos(d*x+c))^(1/2)*sec(d*

x+c)^(1/2)/b^3/d-1/24*(a-b)*(24*A*b^2-18*B*a*b+15*C*a^2+16*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)/a/b^3/d/sec(d*x+c)^(1/2)+1/24*(24*A*b^2-18*B*a*b+12*B*b^2+15*C*a^2-10*C*a*b+16

*C*b^2)*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)-1/8

*(6*a^2*b*B+8*b^3*B-5*a^3*C-4*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^4/d/sec(d*x+c)^(1/2)

________________________________________________________________________________________

Rubi [A] time = 2.09, antiderivative size = 653, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.178, Rules used = {4221, 3049, 3061, 3053, 2809, 2998, 2816, 2994} \[ \frac {\sin (c+d x) \sqrt {\sec (c+d x)} \left (15 a^2 C-18 a b B+24 A b^2+16 b^2 C\right ) \sqrt {a+b \cos (c+d x)}}{24 b^3 d}+\frac {\sqrt {a+b} \sqrt {\cos (c+d x)} \csc (c+d x) \left (15 a^2 C-18 a b B-10 a b C+24 A b^2+12 b^2 B+16 b^2 C\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{24 b^3 d \sqrt {\sec (c+d x)}}-\frac {(a-b) \sqrt {a+b} \sqrt {\cos (c+d x)} \csc (c+d x) \left (15 a^2 C-18 a b B+24 A b^2+16 b^2 C\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{24 a b^3 d \sqrt {\sec (c+d x)}}-\frac {\sqrt {a+b} \sqrt {\cos (c+d x)} \csc (c+d x) \left (6 a^2 b B-5 a^3 C-4 a b^2 (2 A+C)+8 b^3 B\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \Pi \left (\frac {a+b}{b};\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{8 b^4 d \sqrt {\sec (c+d x)}}+\frac {(6 b B-5 a C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{12 b^2 d \sqrt {\sec (c+d x)}}+\frac {C \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 b d \sec ^{\frac {3}{2}}(c+d x)} \]

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-((a - b)*Sqrt[a + b]*(24*A*b^2 - 18*a*b*B + 15*a^2*C + 16*b^2*C)*Sqrt[Cos[c + d*x]]*Csc[c + d*x]*EllipticE[Ar

cSin[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*b^3*d*Sqrt[Sec[c + d*x]]) + (Sqrt[a + b]*(24*A*b^2 - 1

8*a*b*B + 12*b^2*B + 15*a^2*C - 10*a*b*C + 16*b^2*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)]*S

qrt[(a*(1 + Sec[c + d*x]))/(a - b)])/(24*b^3*d*Sqrt[Sec[c + d*x]]) - (Sqrt[a + b]*(6*a^2*b*B + 8*b^3*B - 5*a^3

*C - 4*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)]*Sqrt[(a*(1 + Sec[c

+ d*x]))/(a - b)])/(8*b^4*d*Sqrt[Sec[c + d*x]]) + (C*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(3*b*d*Sec[c + d*

x]^(3/2)) + ((6*b*B - 5*a*C)*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(12*b^2*d*Sqrt[Sec[c + d*x]]) + ((24*A*b^2

- 18*a*b*B + 15*a^2*C + 16*b^2*C)*Sqrt[a + b*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(24*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 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 {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {a+b \cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x)} \, dx &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {a+b \cos (c+d x)}} \, dx\\ &=\frac {C \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{3 b d \sec ^{\frac {3}{2}}(c+d x)}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {\cos (c+d x)} \left (\frac {3 a C}{2}+b (3 A+2 C) \cos (c+d x)+\frac {1}{2} (6 b B-5 a C) \cos ^2(c+d x)\right )}{\sqrt {a+b \cos (c+d x)}} \, dx}{3 b}\\ &=\frac {C \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{3 b d \sec ^{\frac {3}{2}}(c+d x)}+\frac {(6 b B-5 a C) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{12 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}{4} a (6 b B-5 a C)+\frac {1}{2} b (6 b B+a C) \cos (c+d x)+\frac {1}{4} \left (24 A b^2-18 a b B+15 a^2 C+16 b^2 C\right ) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}} \, dx}{6 b^2}\\ &=\frac {C \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{3 b d \sec ^{\frac {3}{2}}(c+d x)}+\frac {(6 b B-5 a C) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{12 b^2 d \sqrt {\sec (c+d x)}}+\frac {\left (24 A b^2-18 a b B+15 a^2 C+16 b^2 C\right ) \sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)} \sin (c+d x)}{24 b^3 d}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {-\frac {1}{4} a \left (24 A b^2-18 a b B+15 a^2 C+16 b^2 C\right )+\frac {1}{2} a b (6 b B-5 a C) \cos (c+d x)+\frac {3}{4} \left (6 a^2 b B+8 b^3 B-5 a^3 C-4 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}{12 b^3}\\ &=\frac {C \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{3 b d \sec ^{\frac {3}{2}}(c+d x)}+\frac {(6 b B-5 a C) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{12 b^2 d \sqrt {\sec (c+d x)}}+\frac {\left (24 A b^2-18 a b B+15 a^2 C+16 b^2 C\right ) \sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)} \sin (c+d x)}{24 b^3 d}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {-\frac {1}{4} a \left (24 A b^2-18 a b B+15 a^2 C+16 b^2 C\right )+\frac {1}{2} a b (6 b B-5 a C) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}} \, dx}{12 b^3}+\frac {\left (\left (6 a^2 b B+8 b^3 B-5 a^3 C-4 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}{16 b^3}\\ &=-\frac {\sqrt {a+b} \left (6 a^2 b B+8 b^3 B-5 a^3 C-4 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^4 d \sqrt {\sec (c+d x)}}+\frac {C \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{3 b d \sec ^{\frac {3}{2}}(c+d x)}+\frac {(6 b B-5 a C) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{12 b^2 d \sqrt {\sec (c+d x)}}+\frac {\left (24 A b^2-18 a b B+15 a^2 C+16 b^2 C\right ) \sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)} \sin (c+d x)}{24 b^3 d}-\frac {\left (a \left (24 A b^2-18 a b B+15 a^2 C+16 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}{48 b^3}+\frac {\left (a \left (24 A b^2-18 a b B+12 b^2 B+15 a^2 C-10 a b C+16 b^2 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}{48 b^3}\\ &=-\frac {(a-b) \sqrt {a+b} \left (24 A b^2-18 a b B+15 a^2 C+16 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}}}{24 a b^3 d \sqrt {\sec (c+d x)}}+\frac {\sqrt {a+b} \left (24 A b^2-18 a b B+12 b^2 B+15 a^2 C-10 a b C+16 b^2 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 b^3 d \sqrt {\sec (c+d x)}}-\frac {\sqrt {a+b} \left (6 a^2 b B+8 b^3 B-5 a^3 C-4 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^4 d \sqrt {\sec (c+d x)}}+\frac {C \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{3 b d \sec ^{\frac {3}{2}}(c+d x)}+\frac {(6 b B-5 a C) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{12 b^2 d \sqrt {\sec (c+d x)}}+\frac {\left (24 A b^2-18 a b B+15 a^2 C+16 b^2 C\right ) \sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)} \sin (c+d x)}{24 b^3 d}\\ \end {align*}

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Mathematica [B] time = 21.54, size = 1818, normalized size = 2.78 \[ \text {result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Sqrt[a + b*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((C*Sin[c + d*x])/(12*b) + ((6*b*B - 5*a*C)*Sin[2*(c + d*x)])/(24

*b^2) + (C*Sin[3*(c + d*x)])/(12*b)))/d + (Sqrt[(1 - Tan[(c + d*x)/2]^2)^(-1)]*Sqrt[(a + b + a*Tan[(c + d*x)/2

]^2 - b*Tan[(c + d*x)/2]^2)/(1 + Tan[(c + d*x)/2]^2)]*(-24*a*A*b^2*Tan[(c + d*x)/2] - 24*A*b^3*Tan[(c + d*x)/2

] + 18*a^2*b*B*Tan[(c + d*x)/2] + 18*a*b^2*B*Tan[(c + d*x)/2] - 15*a^3*C*Tan[(c + d*x)/2] - 15*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] + 48*A*b^3*Tan[(c + d*x)/2]^3 - 36*a*b^2*B

*Tan[(c + d*x)/2]^3 + 30*a^2*b*C*Tan[(c + d*x)/2]^3 + 32*b^3*C*Tan[(c + d*x)/2]^3 + 24*a*A*b^2*Tan[(c + d*x)/2

]^5 - 24*A*b^3*Tan[(c + d*x)/2]^5 - 18*a^2*b*B*Tan[(c + d*x)/2]^5 + 18*a*b^2*B*Tan[(c + d*x)/2]^5 + 15*a^3*C*T

an[(c + d*x)/2]^5 - 15*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 + 48*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)] - 36*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]*S

qrt[(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)] + 24*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)] + 48*a*A*b^2*EllipticPi[-1, ArcSin[Ta

n[(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)] - 36*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*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b + a*T

an[(c + d*x)/2]^2 - b*Tan[(c + d*x)/2]^2)/(a + b)] + 24*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]*Sqrt[(a + b + a*Tan[(c + d*x)/2]^2 - b*Tan[(c + d

*x)/2]^2)/(a + b)] - (a + b)*(24*A*b^2 - 18*a*b*B + 15*a^2*C + 16*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*(12*b^2*B + 5*a^2*C + 2*a*b*(-3*B + 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*b^3*d*Sqrt[1 + Tan[(c + d*x)/2]^2]*(b*(-1 + Tan[(c + d*x)/2]^2) - a*(1 +

Tan[(c + d*x)/2]^2)))

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fricas [F] time = 3.75, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A}{\sqrt {b \cos \left (d x + c\right ) + a} \sec \left (d x + c\right )^{\frac {3}{2}}}, x\right ) \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B] time = 0.75, size = 3583, normalized size = 5.49 \[ \text {output too large to display} \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24/d*(15*C*cos(d*x+c)^2*a^3-10*C*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))*cos(d*x+c)*a^2*b+24*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)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),(-(a-

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

/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*a*b^2+12*B*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^2-18*B*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-18*B*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^2+48*B*cos(d*x+c)*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)*EllipticPi((-1+cos(d*x+c))/sin(d*x+c),-1,(-(a-b)/(a+b))^(1/2))*b^3-24*B*cos(d*x+c)*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))*b^3+36*B*(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)*Ellip

ticPi((-1+cos(d*x+c))/sin(d*x+c),-1,(-(a-b)/(a+b))^(1/2))*a^2*b*sin(d*x+c)-24*A*cos(d*x+c)^2*b^3-12*B*cos(d*x+

c)^2*b^3+24*A*cos(d*x+c)^2*a*b^2-24*A*cos(d*x+c)*a*b^2-18*B*cos(d*x+c)^2*a^2*b+18*B*cos(d*x+c)^2*a*b^2+18*B*co

s(d*x+c)*a^2*b-12*B*cos(d*x+c)*a*b^2+24*A*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+b*cos(d*x+c))/(1+co

s(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^2+12*B*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^2-18*B*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-18*B*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^2+18*C*cos(d*x+c)^2*a*b^2-15*C*cos(d*x+c)^2*a^2*b-2*C*cos(d*x+c)^4*a*b^2+10*C*cos(d*x+c)*a^

2*b-16*C*cos(d*x+c)*a*b^2+16*C*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))*b^3-6*B*cos(d*x+c)^3*a*b^2-48*A*cos(d*x

+c)*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)*EllipticPi((-1+

cos(d*x+c))/sin(d*x+c),-1,(-(a-b)/(a+b))^(1/2))*a*b^2-24*C*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)*EllipticPi((-1+cos(d*x+c))/sin(d*x+c),-1,(-(a-b)/(a+b))^(1/2))*cos(d*x

+c)*a*b^2-4*C*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)*Ellip

ticF((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*cos(d*x+c)*a*b^2+24*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)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*sin(d

*x+c)*b^3+16*C*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)*Elli

pticE((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*cos(d*x+c)*b^3-24*C*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)*EllipticPi((-1+cos(d*x+c))/sin(d*x+c),-1,(-(a-b)/(a+b)

)^(1/2))*a*b^2-16*C*cos(d*x+c)^2*b^3+8*C*cos(d*x+c)^5*b^3+8*C*cos(d*x+c)^3*b^3-15*C*cos(d*x+c)*a^3+12*B*cos(d*

x+c)^4*b^3-48*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*b^2*sin(d*x+c)-10*C*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-4*C*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*b^2+15*C*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+16

*C*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+co

s(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*a*b^2-30*C*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)*EllipticPi((-1+cos(d*x+c))/sin(d*x+c),-1,(-(a-b)/(a+b))^(1/2))*cos(d*x+c)*a

^3+15*C*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))*cos(d*x+c)*a^3+15*C*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))*co

s(d*x+c)*a^2*b+16*C*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))*cos(d*x+c)*a*b^2-30*C*sin(d*x+c)*(cos(d*x+c)/(1+co

s(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^3+15*C*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^3-24*B*(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^3*sin

(d*x+c)+48*B*(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+co

s(d*x+c))/sin(d*x+c),-1,(-(a-b)/(a+b))^(1/2))*b^3*sin(d*x+c)+36*B*cos(d*x+c)*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)*EllipticPi((-1+cos(d*x+c))/sin(d*x+c),-1,(-(a-b)/(a+

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + 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 \[ \int \frac {C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}\,\sqrt {a+b\,\cos \left (c+d\,x\right )}} \,d x \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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