3.1540 \(\int \frac{(A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^{\frac{3}{2}}(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=535 \[ -\frac{2 \sin (c+d x) \sqrt{\sec (c+d x)} \left (-2 a^2 b^2 (4 A+C)+5 a^3 b B-2 a^4 C-a b^3 B+4 A b^4\right )}{3 a^2 d \left (a^2-b^2\right )^2 \sqrt{a+b \cos (c+d x)}}+\frac{2 \sin (c+d x) \sqrt{\sec (c+d x)} \left (A b^2-a (b B-a C)\right )}{3 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}+\frac{2 \sqrt{\cos (c+d x)} \csc (c+d x) \left (-a^2 b (9 A+3 B+C)-3 a^3 (A-B-C)+2 a b^2 (3 A-B)+8 A b^3\right ) \sqrt{\frac{a (1-\sec (c+d x))}{a+b}} \sqrt{\frac{a (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac{\sqrt{a+b \cos (c+d x)}}{\sqrt{a+b} \sqrt{\cos (c+d x)}}\right )|-\frac{a+b}{a-b}\right )}{3 a^3 d \sqrt{a+b} \left (a^2-b^2\right ) \sqrt{\sec (c+d x)}}+\frac{2 \sqrt{\cos (c+d x)} \csc (c+d x) \left (-a^2 b^2 (15 A+C)+3 a^4 (A-C)+6 a^3 b B-2 a b^3 B+8 A b^4\right ) \sqrt{\frac{a (1-\sec (c+d x))}{a+b}} \sqrt{\frac{a (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \cos (c+d x)}}{\sqrt{a+b} \sqrt{\cos (c+d x)}}\right )|-\frac{a+b}{a-b}\right )}{3 a^4 d \sqrt{a+b} \left (a^2-b^2\right ) \sqrt{\sec (c+d x)}} \]

[Out]

(2*(8*A*b^4 + 6*a^3*b*B - 2*a*b^3*B + 3*a^4*(A - C) - a^2*b^2*(15*A + C))*Sqrt[Cos[c + d*x]]*Csc[c + d*x]*Elli
pticE[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^4*Sqrt[a + b]*(a^2 - b^2)*d*Sqrt[Sec[c + d*x]])
+ (2*(8*A*b^3 + 2*a*b^2*(3*A - B) - 3*a^3*(A - B - C) - a^2*b*(9*A + 3*B + 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)])/(3*a^3*Sqrt[a + b]*(a^2 - b^2)*d*Sqrt[Sec[c +
d*x]]) + (2*(A*b^2 - a*(b*B - a*C))*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(3*a*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^(
3/2)) - (2*(4*A*b^4 + 5*a^3*b*B - a*b^3*B - 2*a^4*C - 2*a^2*b^2*(4*A + C))*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(3
*a^2*(a^2 - b^2)^2*d*Sqrt[a + b*Cos[c + d*x]])

________________________________________________________________________________________

Rubi [A]  time = 1.64606, antiderivative size = 535, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 45, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {4221, 3055, 2998, 2816, 2994} \[ -\frac{2 \sin (c+d x) \sqrt{\sec (c+d x)} \left (-2 a^2 b^2 (4 A+C)+5 a^3 b B-2 a^4 C-a b^3 B+4 A b^4\right )}{3 a^2 d \left (a^2-b^2\right )^2 \sqrt{a+b \cos (c+d x)}}+\frac{2 \sin (c+d x) \sqrt{\sec (c+d x)} \left (A b^2-a (b B-a C)\right )}{3 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}+\frac{2 \sqrt{\cos (c+d x)} \csc (c+d x) \left (-a^2 b (9 A+3 B+C)-3 a^3 (A-B-C)+2 a b^2 (3 A-B)+8 A b^3\right ) \sqrt{\frac{a (1-\sec (c+d x))}{a+b}} \sqrt{\frac{a (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac{\sqrt{a+b \cos (c+d x)}}{\sqrt{a+b} \sqrt{\cos (c+d x)}}\right )|-\frac{a+b}{a-b}\right )}{3 a^3 d \sqrt{a+b} \left (a^2-b^2\right ) \sqrt{\sec (c+d x)}}+\frac{2 \sqrt{\cos (c+d x)} \csc (c+d x) \left (-a^2 b^2 (15 A+C)+3 a^4 (A-C)+6 a^3 b B-2 a b^3 B+8 A b^4\right ) \sqrt{\frac{a (1-\sec (c+d x))}{a+b}} \sqrt{\frac{a (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \cos (c+d x)}}{\sqrt{a+b} \sqrt{\cos (c+d x)}}\right )|-\frac{a+b}{a-b}\right )}{3 a^4 d \sqrt{a+b} \left (a^2-b^2\right ) \sqrt{\sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^(3/2))/(a + b*Cos[c + d*x])^(5/2),x]

[Out]

(2*(8*A*b^4 + 6*a^3*b*B - 2*a*b^3*B + 3*a^4*(A - C) - a^2*b^2*(15*A + C))*Sqrt[Cos[c + d*x]]*Csc[c + d*x]*Elli
pticE[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^4*Sqrt[a + b]*(a^2 - b^2)*d*Sqrt[Sec[c + d*x]])
+ (2*(8*A*b^3 + 2*a*b^2*(3*A - B) - 3*a^3*(A - B - C) - a^2*b*(9*A + 3*B + 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)])/(3*a^3*Sqrt[a + b]*(a^2 - b^2)*d*Sqrt[Sec[c +
d*x]]) + (2*(A*b^2 - a*(b*B - a*C))*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(3*a*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^(
3/2)) - (2*(4*A*b^4 + 5*a^3*b*B - a*b^3*B - 2*a^4*C - 2*a^2*b^2*(4*A + C))*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(3
*a^2*(a^2 - b^2)^2*d*Sqrt[a + b*Cos[c + d*x]])

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]

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 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 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]

Rubi steps

\begin{align*} \int \frac{\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac{3}{2}}(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)}{\cos ^{\frac{3}{2}}(c+d x) (a+b \cos (c+d x))^{5/2}} \, dx\\ &=\frac{2 \left (A b^2-a (b B-a C)\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}+\frac{\left (2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\frac{1}{2} \left (-4 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)+\left (A b^2-a (b B-a C)\right ) \cos ^2(c+d x)}{\cos ^{\frac{3}{2}}(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 ) \sqrt{\sec (c+d x)} \sin (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}-\frac{2 \left (4 A b^4+5 a^3 b B-a b^3 B-2 a^4 C-2 a^2 b^2 (4 A+C)\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{3 a^2 \left (a^2-b^2\right )^2 d \sqrt{a+b \cos (c+d x)}}+\frac{\left (4 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\frac{1}{4} \left (8 A b^4+6 a^3 b B-2 a b^3 B+3 a^4 (A-C)-a^2 b^2 (15 A+C)\right )+\frac{1}{4} a \left (2 A b^3+3 a^3 B+a b^2 B-2 a^2 b (3 A+2 C)\right ) \cos (c+d x)}{\cos ^{\frac{3}{2}}(c+d x) \sqrt{a+b \cos (c+d x)}} \, dx}{3 a^2 \left (a^2-b^2\right )^2}\\ &=\frac{2 \left (A b^2-a (b B-a C)\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}-\frac{2 \left (4 A b^4+5 a^3 b B-a b^3 B-2 a^4 C-2 a^2 b^2 (4 A+C)\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{3 a^2 \left (a^2-b^2\right )^2 d \sqrt{a+b \cos (c+d x)}}-\frac{\left (\left (-8 A b^4-6 a^3 b B+2 a b^3 B-3 a^4 (A-C)+a^2 b^2 (15 A+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^2 \left (a^2-b^2\right )^2}+\frac{\left ((a-b) \left (8 A b^3+2 a b^2 (3 A-B)-3 a^3 (A-B-C)-a^2 b (9 A+3 B+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^2 \left (a^2-b^2\right )^2}\\ &=\frac{2 \left (8 A b^4+6 a^3 b B-2 a b^3 B+3 a^4 (A-C)-a^2 b^2 (15 A+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^4 (a-b) (a+b)^{3/2} d \sqrt{\sec (c+d x)}}+\frac{2 \left (8 A b^3+2 a b^2 (3 A-B)-3 a^3 (A-B-C)-a^2 b (9 A+3 B+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^3 (a-b) (a+b)^{3/2} d \sqrt{\sec (c+d x)}}+\frac{2 \left (A b^2-a (b B-a C)\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}-\frac{2 \left (4 A b^4+5 a^3 b B-a b^3 B-2 a^4 C-2 a^2 b^2 (4 A+C)\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{3 a^2 \left (a^2-b^2\right )^2 d \sqrt{a+b \cos (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 21.0117, size = 790, normalized size = 1.48 \[ \frac{\sqrt{\sec (c+d x)} \sqrt{a+b \cos (c+d x)} \left (\frac{2 \sin (c+d x) \left (-15 a^2 A b^2+3 a^4 A-a^2 b^2 C+6 a^3 b B-3 a^4 C-2 a b^3 B+8 A b^4\right )}{3 a^3 \left (a^2-b^2\right )^2}+\frac{2 \left (a^2 C \sin (c+d x)-a b B \sin (c+d x)+A b^2 \sin (c+d x)\right )}{3 a \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}+\frac{2 \left (8 a^2 A b^2 \sin (c+d x)+2 a^2 b^2 C \sin (c+d x)-5 a^3 b B \sin (c+d x)+2 a^4 C \sin (c+d x)+a b^3 B \sin (c+d x)-4 A b^4 \sin (c+d x)\right )}{3 a^2 \left (a^2-b^2\right )^2 (a+b \cos (c+d x))}\right )}{d}+\frac{2 \sqrt{\frac{1}{1-\tan ^2\left (\frac{1}{2} (c+d x)\right )}} \left (\tan \left (\frac{1}{2} (c+d x)\right ) \left (\tan ^2\left (\frac{1}{2} (c+d x)\right )-1\right ) \left (-a^2 b^2 (15 A+C)+3 a^4 (A-C)+6 a^3 b B-2 a b^3 B+8 A b^4\right ) \left (a \tan ^2\left (\frac{1}{2} (c+d x)\right )+a-b \tan ^2\left (\frac{1}{2} (c+d x)\right )+b\right )+a (a+b) \sqrt{1-\tan ^2\left (\frac{1}{2} (c+d x)\right )} \left (\tan ^2\left (\frac{1}{2} (c+d x)\right )+1\right ) \left (-a^2 b (9 A-3 B+C)+3 a^3 (A+B-C)-2 a b^2 (3 A+B)+8 A b^3\right ) \sqrt{\frac{a \tan ^2\left (\frac{1}{2} (c+d x)\right )+a-b \tan ^2\left (\frac{1}{2} (c+d x)\right )+b}{a+b}} F\left (\sin ^{-1}\left (\tan \left (\frac{1}{2} (c+d x)\right )\right )|\frac{b-a}{a+b}\right )-(a+b) \sqrt{1-\tan ^2\left (\frac{1}{2} (c+d x)\right )} \left (\tan ^2\left (\frac{1}{2} (c+d x)\right )+1\right ) \left (-a^2 b^2 (15 A+C)+3 a^4 (A-C)+6 a^3 b B-2 a b^3 B+8 A b^4\right ) \sqrt{\frac{a \tan ^2\left (\frac{1}{2} (c+d x)\right )+a-b \tan ^2\left (\frac{1}{2} (c+d x)\right )+b}{a+b}} E\left (\sin ^{-1}\left (\tan \left (\frac{1}{2} (c+d x)\right )\right )|\frac{b-a}{a+b}\right )\right )}{3 a^3 d \left (a^2-b^2\right )^2 \left (\tan ^2\left (\frac{1}{2} (c+d x)\right )+1\right )^{3/2} \sqrt{\frac{a \tan ^2\left (\frac{1}{2} (c+d x)\right )+a-b \tan ^2\left (\frac{1}{2} (c+d x)\right )+b}{\tan ^2\left (\frac{1}{2} (c+d x)\right )+1}}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[((A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^(3/2))/(a + b*Cos[c + d*x])^(5/2),x]

[Out]

(Sqrt[a + b*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((2*(3*a^4*A - 15*a^2*A*b^2 + 8*A*b^4 + 6*a^3*b*B - 2*a*b^3*B - 3
*a^4*C - a^2*b^2*C)*Sin[c + d*x])/(3*a^3*(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*a*(a^2 - b^2)*(a + b*Cos[c + d*x])^2) + (2*(8*a^2*A*b^2*Sin[c + d*x] - 4*A*b^4*Sin[c + d*x]
- 5*a^3*b*B*Sin[c + d*x] + a*b^3*B*Sin[c + d*x] + 2*a^4*C*Sin[c + d*x] + 2*a^2*b^2*C*Sin[c + d*x]))/(3*a^2*(a^
2 - b^2)^2*(a + b*Cos[c + d*x]))))/d + (2*Sqrt[(1 - Tan[(c + d*x)/2]^2)^(-1)]*((8*A*b^4 + 6*a^3*b*B - 2*a*b^3*
B + 3*a^4*(A - C) - a^2*b^2*(15*A + C))*Tan[(c + d*x)/2]*(-1 + Tan[(c + d*x)/2]^2)*(a + b + a*Tan[(c + d*x)/2]
^2 - b*Tan[(c + d*x)/2]^2) - (a + b)*(8*A*b^4 + 6*a^3*b*B - 2*a*b^3*B + 3*a^4*(A - C) - a^2*b^2*(15*A + C))*El
lipticE[ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*(1 + Tan[(c + d*x)/2]^2)*Sqrt
[(a + b + a*Tan[(c + d*x)/2]^2 - b*Tan[(c + d*x)/2]^2)/(a + b)] + a*(a + b)*(8*A*b^3 - 2*a*b^2*(3*A + B) + 3*a
^3*(A + B - C) - a^2*b*(9*A - 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)]))/(3
*a^3*(a^2 - b^2)^2*d*(1 + Tan[(c + d*x)/2]^2)^(3/2)*Sqrt[(a + b + a*Tan[(c + d*x)/2]^2 - b*Tan[(c + d*x)/2]^2)
/(1 + Tan[(c + d*x)/2]^2)])

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Maple [B]  time = 0.498, size = 8934, normalized size = 16.7 \begin{align*} \text{output too large to display} \end{align*}

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))^(5/2),x)

[Out]

result too large to display

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{\frac{3}{2}}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

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))^(5/2),x, algorithm="maxima")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*sec(d*x + c)^(3/2)/(b*cos(d*x + c) + a)^(5/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\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} \sec \left (d x + c\right )^{\frac{3}{2}}}{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 ) \end{align*}

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))^(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)*sec(d*x + c)^(3/2)/(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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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))**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{\frac{3}{2}}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

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))^(5/2),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*sec(d*x + c)^(3/2)/(b*cos(d*x + c) + a)^(5/2), x)