∫ (A+B cos(c+dx)+C cos2(c+dx))∘ _______ sec (c+dx) ___________________________________________________________

3.1541 \(\int \frac {(A+B \cos (c+d x)+C \cos ^2(c+d x)) \sqrt {\sec (c+d x)}}{(a+b \cos (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=495 \[ \frac {2 \sin (c+d x) \left (A b^2-a (b B-a C)\right )}{3 a d \left (a^2-b^2\right ) \sqrt {\sec (c+d x)} (a+b \cos (c+d x))^{3/2}}-\frac {2 \sqrt {\cos (c+d x)} \csc (c+d x) \left (-\left (a^2 (3 A+3 B+C)\right )+a b (3 A+B+3 C)+2 A b^2\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{3 a^2 d \sqrt {a+b} \left (a^2-b^2\right ) \sqrt {\sec (c+d x)}}+\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)} \left (3 a^3 B-2 a^2 b (3 A+2 C)+a b^2 B+2 A b^3\right )}{3 a d \left (a^2-b^2\right )^2 \sqrt {a+b \cos (c+d x)}}-\frac {2 \sqrt {\cos (c+d x)} \csc (c+d x) \left (3 a^3 B-2 a^2 b (3 A+2 C)+a b^2 B+2 A b^3\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{3 a^3 d (a-b) (a+b)^{3/2} \sqrt {\sec (c+d x)}} \]

[Out]

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

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

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

b)/(a-b))^(1/2))*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/(a-b)/(a+b

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

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Rubi [A] time = 1.38, antiderivative size = 495, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {4221, 3055, 2993, 2998, 2816, 2994} \[ \frac {2 \sin (c+d x) \sqrt {\sec (c+d x)} \left (-2 a^2 b (3 A+2 C)+3 a^3 B+a b^2 B+2 A b^3\right )}{3 a d \left (a^2-b^2\right )^2 \sqrt {a+b \cos (c+d x)}}+\frac {2 \sin (c+d x) \left (A b^2-a (b B-a C)\right )}{3 a d \left (a^2-b^2\right ) \sqrt {\sec (c+d x)} (a+b \cos (c+d x))^{3/2}}-\frac {2 \sqrt {\cos (c+d x)} \csc (c+d x) \left (a^2 (-(3 A+3 B+C))+a b (3 A+B+3 C)+2 A b^2\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{3 a^2 d \sqrt {a+b} \left (a^2-b^2\right ) \sqrt {\sec (c+d x)}}-\frac {2 \sqrt {\cos (c+d x)} \csc (c+d x) \left (-2 a^2 b (3 A+2 C)+3 a^3 B+a b^2 B+2 A b^3\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{3 a^3 d (a-b) (a+b)^{3/2} \sqrt {\sec (c+d x)}} \]

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(-2*(2*A*b^3 + 3*a^3*B + a*b^2*B - 2*a^2*b*(3*A + 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)])/(3*a^3*(a - b)*(a + b)^(3/2)*d*Sqrt[Sec[c + d*x]]) - (2*(2*A*b^2 - a^2

*(3*A + 3*B + C) + a*b*(3*A + B + 3*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 + S

ec[c + d*x]))/(a - b)])/(3*a^2*Sqrt[a + b]*(a^2 - b^2)*d*Sqrt[Sec[c + d*x]]) + (2*(A*b^2 - a*(b*B - a*C))*Sin[

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

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

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 2993

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]]*((a_) + (b_.)*sin[(e_.) + (f_

.)*(x_)])^(3/2)), x_Symbol] :> Simp[(2*(A*b - a*B)*Cos[e + f*x])/(f*(a^2 - b^2)*Sqrt[a + b*Sin[e + f*x]]*Sqrt[

d*Sin[e + f*x]]), x] + Dist[d/(a^2 - b^2), Int[(A*b - a*B + (a*A - b*B)*Sin[e + f*x])/(Sqrt[a + b*Sin[e + f*x]

]*(d*Sin[e + f*x])^(3/2)), x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[a^2 - b^2, 0]

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 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,

Rubi steps

\begin {align*} \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)}}{(a+b \cos (c+d x))^{5/2}} \, dx &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))^{5/2}} \, dx\\ &=\frac {2 \left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2} \sqrt {\sec (c+d x)}}+\frac {\left (2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {1}{2} \left (-2 A b^2-a b B+a^2 (3 A+C)\right )-\frac {3}{2} a (A b-a B+b C) \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))^{3/2}} \, dx}{3 a \left (a^2-b^2\right )}\\ &=\frac {2 \left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2} \sqrt {\sec (c+d x)}}+\frac {2 \left (2 A b^3+3 a^3 B+a b^2 B-2 a^2 b (3 A+2 C)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 a \left (a^2-b^2\right )^2 d \sqrt {a+b \cos (c+d x)}}+\frac {\left (2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {3}{2} a^2 (A b-a B+b C)+\frac {1}{2} b \left (-2 A b^2-a b B+a^2 (3 A+C)\right )+\left (\frac {3}{2} a b (A b-a B+b C)+\frac {1}{2} a \left (-2 A b^2-a b B+a^2 (3 A+C)\right )\right ) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}} \, dx}{3 a \left (a^2-b^2\right )^2}\\ &=\frac {2 \left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2} \sqrt {\sec (c+d x)}}+\frac {2 \left (2 A b^3+3 a^3 B+a b^2 B-2 a^2 b (3 A+2 C)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 a \left (a^2-b^2\right )^2 d \sqrt {a+b \cos (c+d x)}}-\frac {\left (\left (2 A b^3+3 a^3 B+a b^2 B-2 a^2 b (3 A+2 C)\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1+\cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}} \, dx}{3 a \left (a^2-b^2\right )^2}-\frac {\left ((a-b) \left (2 A b^2-a^2 (3 A+3 B+C)+a b (3 A+B+3 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}{3 a \left (a^2-b^2\right )^2}\\ &=-\frac {2 \left (2 A b^3+3 a^3 B+a b^2 B-2 a^2 b (3 A+2 C)\right ) \sqrt {\cos (c+d x)} \csc (c+d x) E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{3 a^3 (a-b) (a+b)^{3/2} d \sqrt {\sec (c+d x)}}-\frac {2 \left (2 A b^2-a^2 (3 A+3 B+C)+a b (3 A+B+3 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}}}{3 a^2 (a-b) (a+b)^{3/2} d \sqrt {\sec (c+d x)}}+\frac {2 \left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2} \sqrt {\sec (c+d x)}}+\frac {2 \left (2 A b^3+3 a^3 B+a b^2 B-2 a^2 b (3 A+2 C)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 a \left (a^2-b^2\right )^2 d \sqrt {a+b \cos (c+d x)}}\\ \end {align*}

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Mathematica [B] time = 26.59, size = 3853, normalized size = 7.78 \[ \text {Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Sqrt[a + b*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((-2*(-6*a^2*A*b + 2*A*b^3 + 3*a^3*B + a*b^2*B - 4*a^2*b*C)*Sin[c

+ d*x])/(3*a^2*(a^2 - b^2)^2) + (2*(A*b^2*Sin[c + d*x] - a*b*B*Sin[c + d*x] + a^2*C*Sin[c + d*x]))/(3*b*(-a^2

+ b^2)*(a + b*Cos[c + d*x])^2) + (2*(-5*a^2*A*b^2*Sin[c + d*x] + A*b^4*Sin[c + d*x] + 2*a^3*b*B*Sin[c + d*x]

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

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

- b^2)^2*Sqrt[a + b*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) + (a^2*B)/((a^2 - b^2)^2*Sqrt[a + b*Cos[c + d*x]]*Sqrt[

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

^2)^2*Sqrt[a + b*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) + (a^2*A*Sqrt[Sec[c + d*x]])/((a^2 - b^2)^2*Sqrt[a + b*Cos[

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

*x]])/(3*a^2*(a^2 - b^2)^2*Sqrt[a + b*Cos[c + d*x]]) - (a*b*B*Sqrt[Sec[c + d*x]])/(3*(a^2 - b^2)^2*Sqrt[a + b*

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

d*x]])/(3*(a^2 - b^2)^2*Sqrt[a + b*Cos[c + d*x]]) - (b^2*C*Sqrt[Sec[c + d*x]])/(3*(a^2 - b^2)^2*Sqrt[a + b*Co

s[c + d*x]]) - (2*A*b^2*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/((a^2 - b^2)^2*Sqrt[a + b*Cos[c + d*x]]) + (2*A*b

^4*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/(3*a^2*(a^2 - b^2)^2*Sqrt[a + b*Cos[c + d*x]]) + (a*b*B*Cos[2*(c + d*x

)]*Sqrt[Sec[c + d*x]])/((a^2 - b^2)^2*Sqrt[a + b*Cos[c + d*x]]) + (b^3*B*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/

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

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

^2*b*(3*A + 2*C))*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(a + b*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))

]*EllipticE[ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)] + 2*a*(a + b)*(-2*A*b^2 + a^2*(3*A - 3*B + C) + a*b*(3

*A - B + 3*C))*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(a + b*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*E

llipticF[ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)] + (2*A*b^3 + 3*a^3*B + a*b^2*B - 2*a^2*b*(3*A + 2*C))*Cos

[c + d*x]*(a + b*Cos[c + d*x])*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]))/(3*(a^3 - a*b^2)^2*d*Sqrt[a + b*Cos[c + d

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

^3*B + a*b^2*B - 2*a^2*b*(3*A + 2*C))*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(a + b*Cos[c + d*x])/((a + b)

*(1 + Cos[c + d*x]))]*EllipticE[ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)] + 2*a*(a + b)*(-2*A*b^2 + a^2*(3*A

- 3*B + C) + a*b*(3*A - B + 3*C))*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(a + b*Cos[c + d*x])/((a + b)*(1

+ Cos[c + d*x]))]*EllipticF[ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)] + (2*A*b^3 + 3*a^3*B + a*b^2*B - 2*a^

2*b*(3*A + 2*C))*Cos[c + d*x]*(a + b*Cos[c + d*x])*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]))/(3*(a^3 - a*b^2)^2*(a

+ b*Cos[c + d*x])^(3/2)*Sqrt[Sec[(c + d*x)/2]^2]) - (Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*Tan[(c + d*x)/2]*(

2*(a + b)*(2*A*b^3 + 3*a^3*B + a*b^2*B - 2*a^2*b*(3*A + 2*C))*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(a +

b*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticE[ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)] + 2*a*(a +

b)*(-2*A*b^2 + a^2*(3*A - 3*B + C) + a*b*(3*A - B + 3*C))*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(a + b*C

os[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticF[ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)] + (2*A*b^3 +

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

))/(3*(a^3 - a*b^2)^2*Sqrt[a + b*Cos[c + d*x]]*Sqrt[Sec[(c + d*x)/2]^2]) + (2*Sqrt[Cos[(c + d*x)/2]^2*Sec[c +

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

^4)/2 + ((a + b)*(2*A*b^3 + 3*a^3*B + a*b^2*B - 2*a^2*b*(3*A + 2*C))*Sqrt[(a + b*Cos[c + d*x])/((a + b)*(1 + C

os[c + d*x]))]*EllipticE[ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*((Cos[c + d*x]*Sin[c + d*x])/(1 + Cos[c +

d*x])^2 - Sin[c + d*x]/(1 + Cos[c + d*x])))/Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])] + (a*(a + b)*(-2*A*b^2 + a^

2*(3*A - 3*B + C) + a*b*(3*A - B + 3*C))*Sqrt[(a + b*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticF[Arc

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

Cos[c + d*x])))/Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])] + ((a + b)*(2*A*b^3 + 3*a^3*B + a*b^2*B - 2*a^2*b*(3*A

+ 2*C))*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*EllipticE[ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*(-((b*Sin[

c + d*x])/((a + b)*(1 + Cos[c + d*x]))) + ((a + b*Cos[c + d*x])*Sin[c + d*x])/((a + b)*(1 + Cos[c + d*x])^2)))

/Sqrt[(a + b*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))] + (a*(a + b)*(-2*A*b^2 + a^2*(3*A - 3*B + C) + a*b*(3

*A - B + 3*C))*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*EllipticF[ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*(-(

(b*Sin[c + d*x])/((a + b)*(1 + Cos[c + d*x]))) + ((a + b*Cos[c + d*x])*Sin[c + d*x])/((a + b)*(1 + Cos[c + d*x

])^2)))/Sqrt[(a + b*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))] - b*(2*A*b^3 + 3*a^3*B + a*b^2*B - 2*a^2*b*(3*

A + 2*C))*Cos[c + d*x]*Sec[(c + d*x)/2]^2*Sin[c + d*x]*Tan[(c + d*x)/2] - (2*A*b^3 + 3*a^3*B + a*b^2*B - 2*a^2

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

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

+ b)*(-2*A*b^2 + a^2*(3*A - 3*B + C) + a*b*(3*A - B + 3*C))*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(a + b*

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

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

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

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

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

x]/(1 + Cos[c + d*x])]*Sqrt[(a + b*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticE[ArcSin[Tan[(c + d*x)/

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

(1 + Cos[c + d*x])]*Sqrt[(a + b*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticF[ArcSin[Tan[(c + d*x)/2]]

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

c[(c + d*x)/2]^2*Tan[(c + d*x)/2])*(-(Cos[(c + d*x)/2]*Sec[c + d*x]*Sin[(c + d*x)/2]) + Cos[(c + d*x)/2]^2*Sec

[c + d*x]*Tan[c + d*x]))/(3*(a^3 - a*b^2)^2*Sqrt[a + b*Cos[c + d*x]]*Sqrt[Sec[(c + d*x)/2]^2]*Sqrt[Cos[(c + d*

x)/2]^2*Sec[c + d*x]])))

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fricas [F] time = 0.84, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \sqrt {b \cos \left (d x + c\right ) + a} \sqrt {\sec \left (d x + c\right )}}{b^{3} \cos \left (d x + c\right )^{3} + 3 \, a b^{2} \cos \left (d x + c\right )^{2} + 3 \, a^{2} b \cos \left (d x + c\right ) + a^{3}}, x\right ) \]

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)^(1/2)/(a+b*cos(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

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

^3 + 3*a*b^2*cos(d*x + c)^2 + 3*a^2*b*cos(d*x + c) + a^3), x)

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

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)^(1/2)/(a+b*cos(d*x+c))^(5/2),x, algorithm="giac")

[Out]

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

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maple [B] time = 0.68, size = 7005, normalized size = 14.15 \[ \text {output too large to display} \]

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)^(1/2)/(a+b*cos(d*x+c))^(5/2),x)

[Out]

result too large to display

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

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)^(1/2)/(a+b*cos(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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