3.1499 \(\int \frac{A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^3 \sec ^{\frac{3}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=483 \[ -\frac{\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{2 b d \left (a^2-b^2\right ) \sec ^{\frac{3}{2}}(c+d x) (a+b \cos (c+d x))^2}+\frac{\sin (c+d x) \left (a^2 b^2 (3 A+11 C)+a^3 b B-5 a^4 C-7 a b^3 B+3 A b^4\right )}{4 b^2 d \left (a^2-b^2\right )^2 \sqrt{\sec (c+d x)} (a+b \cos (c+d x))}+\frac{\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (a^3 b^2 (A+33 C)-5 a^2 b^3 B+3 a^4 b B-15 a^5 C-a b^4 (7 A+24 C)+8 b^5 B\right )}{4 b^4 d \left (a^2-b^2\right )^2}-\frac{\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (a^2 b^2 (A+29 C)+3 a^3 b B-15 a^4 C-9 a b^3 B+b^4 (5 A-8 C)\right )}{4 b^3 d \left (a^2-b^2\right )^2}+\frac{\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \left (-a^4 b^2 (A+38 C)+5 a^2 b^4 (2 A+7 C)+6 a^3 b^3 B-3 a^5 b B+15 a^6 C-15 a b^5 B+3 A b^6\right ) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{4 b^4 d (a-b)^2 (a+b)^3} \]

[Out]

-((3*a^3*b*B - 9*a*b^3*B + b^4*(5*A - 8*C) - 15*a^4*C + a^2*b^2*(A + 29*C))*Sqrt[Cos[c + d*x]]*EllipticE[(c +
d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(4*b^3*(a^2 - b^2)^2*d) + ((3*a^4*b*B - 5*a^2*b^3*B + 8*b^5*B - 15*a^5*C - a*b^
4*(7*A + 24*C) + a^3*b^2*(A + 33*C))*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(4*b^4*(
a^2 - b^2)^2*d) + ((3*A*b^6 - 3*a^5*b*B + 6*a^3*b^3*B - 15*a*b^5*B + 15*a^6*C + 5*a^2*b^4*(2*A + 7*C) - a^4*b^
2*(A + 38*C))*Sqrt[Cos[c + d*x]]*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(4*(a - b)^2*b^
4*(a + b)^3*d) - ((A*b^2 - a*(b*B - a*C))*Sin[c + d*x])/(2*b*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^2*Sec[c + d*x]
^(3/2)) + ((3*A*b^4 + a^3*b*B - 7*a*b^3*B - 5*a^4*C + a^2*b^2*(3*A + 11*C))*Sin[c + d*x])/(4*b^2*(a^2 - b^2)^2
*d*(a + b*Cos[c + d*x])*Sqrt[Sec[c + d*x]])

________________________________________________________________________________________

Rubi [A]  time = 1.68323, antiderivative size = 483, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.163, Rules used = {4221, 3047, 3059, 2639, 3002, 2641, 2805} \[ -\frac{\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{2 b d \left (a^2-b^2\right ) \sec ^{\frac{3}{2}}(c+d x) (a+b \cos (c+d x))^2}+\frac{\sin (c+d x) \left (a^2 b^2 (3 A+11 C)+a^3 b B-5 a^4 C-7 a b^3 B+3 A b^4\right )}{4 b^2 d \left (a^2-b^2\right )^2 \sqrt{\sec (c+d x)} (a+b \cos (c+d x))}+\frac{\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (a^3 b^2 (A+33 C)-5 a^2 b^3 B+3 a^4 b B-15 a^5 C-a b^4 (7 A+24 C)+8 b^5 B\right )}{4 b^4 d \left (a^2-b^2\right )^2}-\frac{\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (a^2 b^2 (A+29 C)+3 a^3 b B-15 a^4 C-9 a b^3 B+b^4 (5 A-8 C)\right )}{4 b^3 d \left (a^2-b^2\right )^2}+\frac{\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \left (-a^4 b^2 (A+38 C)+5 a^2 b^4 (2 A+7 C)+6 a^3 b^3 B-3 a^5 b B+15 a^6 C-15 a b^5 B+3 A b^6\right ) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{4 b^4 d (a-b)^2 (a+b)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((3*a^3*b*B - 9*a*b^3*B + b^4*(5*A - 8*C) - 15*a^4*C + a^2*b^2*(A + 29*C))*Sqrt[Cos[c + d*x]]*EllipticE[(c +
d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(4*b^3*(a^2 - b^2)^2*d) + ((3*a^4*b*B - 5*a^2*b^3*B + 8*b^5*B - 15*a^5*C - a*b^
4*(7*A + 24*C) + a^3*b^2*(A + 33*C))*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(4*b^4*(
a^2 - b^2)^2*d) + ((3*A*b^6 - 3*a^5*b*B + 6*a^3*b^3*B - 15*a*b^5*B + 15*a^6*C + 5*a^2*b^4*(2*A + 7*C) - a^4*b^
2*(A + 38*C))*Sqrt[Cos[c + d*x]]*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(4*(a - b)^2*b^
4*(a + b)^3*d) - ((A*b^2 - a*(b*B - a*C))*Sin[c + d*x])/(2*b*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^2*Sec[c + d*x]
^(3/2)) + ((3*A*b^4 + a^3*b*B - 7*a*b^3*B - 5*a^4*C + a^2*b^2*(3*A + 11*C))*Sin[c + d*x])/(4*b^2*(a^2 - b^2)^2
*d*(a + b*Cos[c + d*x])*Sqrt[Sec[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 3047

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[((c^2*C - B*c*d + A*d^2)*Cos[e +
f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(n + 1)*(c^2 - d^2)), x] + Dist[1/(d*(n + 1)*(
c^2 - d^2)), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(b*d*m + a*c*(n + 1)) + (c
*C - B*d)*(b*c*m + a*d*(n + 1)) - (d*(A*(a*d*(n + 2) - b*c*(n + 1)) + B*(b*d*(n + 1) - a*c*(n + 2))) - C*(b*c*
d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x] + b*(d*(B*c - A*d)*(m + n + 2) - C*(c^2*(m + 1) + d^2*(n + 1)
))*Sin[e + f*x]^2, 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] && GtQ[m, 0] && LtQ[n, -1]

Rule 3059

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2)/(Sqrt[(a_.) + (b_.)*sin[(e_.) +
(f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Dist[C/(b*d), Int[Sqrt[a + b*Sin[e + f*x]]
, x], x] - Dist[1/(b*d), Int[Simp[a*c*C - A*b*d + (b*c*C - b*B*d + a*C*d)*Sin[e + f*x], x]/(Sqrt[a + b*Sin[e +
f*x]]*(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 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

Rule 3002

Int[(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]))/((c_.) + (d_.)*sin[
(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[B/d, Int[(a + b*Sin[e + f*x])^m, x], x] - Dist[(B*c - A*d)/d, Int[(a +
b*Sin[e + f*x])^m/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
&& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rule 2805

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2*EllipticPi[(2*b)/(a + b), (1*(e - Pi/2 + f*x))/2, (2*d)/(c + d)])/(f*(a + b)*Sqrt[c + d]), x] /; FreeQ[{a,
b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[c + d, 0]

Rubi steps

\begin{align*} \int \frac{A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^3 \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 )}{(a+b \cos (c+d x))^3} \, dx\\ &=-\frac{\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2 \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}{2} \left (A b^2-a (b B-a C)\right )+2 b (b B-a (A+C)) \cos (c+d x)-\frac{1}{2} \left (A b^2-a b B+5 a^2 C-4 b^2 C\right ) \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx}{2 b \left (a^2-b^2\right )}\\ &=-\frac{\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2 \sec ^{\frac{3}{2}}(c+d x)}+\frac{\left (3 A b^4+a^3 b B-7 a b^3 B-5 a^4 C+a^2 b^2 (3 A+11 C)\right ) \sin (c+d x)}{4 b^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x)) \sqrt{\sec (c+d x)}}+\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\frac{1}{4} \left (3 A b^4+a^3 b B-7 a b^3 B-5 a^4 C+a^2 b^2 (3 A+11 C)\right )+b \left (a^2 b B+2 b^3 B+a^3 C-a b^2 (3 A+4 C)\right ) \cos (c+d x)-\frac{1}{4} \left (3 a^3 b B-9 a b^3 B+b^4 (5 A-8 C)-15 a^4 C+a^2 b^2 (A+29 C)\right ) \cos ^2(c+d x)}{\sqrt{\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{2 b^2 \left (a^2-b^2\right )^2}\\ &=-\frac{\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2 \sec ^{\frac{3}{2}}(c+d x)}+\frac{\left (3 A b^4+a^3 b B-7 a b^3 B-5 a^4 C+a^2 b^2 (3 A+11 C)\right ) \sin (c+d x)}{4 b^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x)) \sqrt{\sec (c+d x)}}-\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{-\frac{1}{4} b \left (3 A b^4+a^3 b B-7 a b^3 B-5 a^4 C+a^2 b^2 (3 A+11 C)\right )-\frac{1}{4} \left (3 a^4 b B-5 a^2 b^3 B+8 b^5 B-15 a^5 C-a b^4 (7 A+24 C)+a^3 b^2 (A+33 C)\right ) \cos (c+d x)}{\sqrt{\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{2 b^3 \left (a^2-b^2\right )^2}-\frac{\left (\left (3 a^3 b B-9 a b^3 B+b^4 (5 A-8 C)-15 a^4 C+a^2 b^2 (A+29 C)\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx}{8 b^3 \left (a^2-b^2\right )^2}\\ &=-\frac{\left (3 a^3 b B-9 a b^3 B+b^4 (5 A-8 C)-15 a^4 C+a^2 b^2 (A+29 C)\right ) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{4 b^3 \left (a^2-b^2\right )^2 d}-\frac{\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2 \sec ^{\frac{3}{2}}(c+d x)}+\frac{\left (3 A b^4+a^3 b B-7 a b^3 B-5 a^4 C+a^2 b^2 (3 A+11 C)\right ) \sin (c+d x)}{4 b^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x)) \sqrt{\sec (c+d x)}}+\frac{\left (\left (3 a^4 b B-5 a^2 b^3 B+8 b^5 B-15 a^5 C-a b^4 (7 A+24 C)+a^3 b^2 (A+33 C)\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{8 b^4 \left (a^2-b^2\right )^2}+\frac{\left (\left (3 A b^6-3 a^5 b B+6 a^3 b^3 B-15 a b^5 B+15 a^6 C+5 a^2 b^4 (2 A+7 C)-a^4 b^2 (A+38 C)\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{8 b^4 \left (a^2-b^2\right )^2}\\ &=-\frac{\left (3 a^3 b B-9 a b^3 B+b^4 (5 A-8 C)-15 a^4 C+a^2 b^2 (A+29 C)\right ) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{4 b^3 \left (a^2-b^2\right )^2 d}+\frac{\left (3 a^4 b B-5 a^2 b^3 B+8 b^5 B-15 a^5 C-a b^4 (7 A+24 C)+a^3 b^2 (A+33 C)\right ) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{4 b^4 \left (a^2-b^2\right )^2 d}+\frac{\left (3 A b^6-3 a^5 b B+6 a^3 b^3 B-15 a b^5 B+15 a^6 C+5 a^2 b^4 (2 A+7 C)-a^4 b^2 (A+38 C)\right ) \sqrt{\cos (c+d x)} \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{4 (a-b)^2 b^4 (a+b)^3 d}-\frac{\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2 \sec ^{\frac{3}{2}}(c+d x)}+\frac{\left (3 A b^4+a^3 b B-7 a b^3 B-5 a^4 C+a^2 b^2 (3 A+11 C)\right ) \sin (c+d x)}{4 b^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x)) \sqrt{\sec (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 7.34853, size = 921, normalized size = 1.91 \[ \frac{-\frac{2 \left (16 B b^4-24 a A b^3-32 a C b^3+8 a^2 B b^2+8 a^3 C b\right ) \Pi \left (-\frac{a}{b};\left .-\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right ) (b+a \sec (c+d x)) \sqrt{1-\sec ^2(c+d x)} \sin (c+d x) \cos ^2(c+d x)}{b (a+b \cos (c+d x)) \left (1-\cos ^2(c+d x)\right )}+\frac{2 \left (5 C a^4-b B a^3+5 A b^2 a^2-7 b^2 C a^2-5 b^3 B a+A b^4+8 b^4 C\right ) \left (F\left (\left .\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )+\Pi \left (-\frac{a}{b};\left .-\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )\right ) (b+a \sec (c+d x)) \sqrt{1-\sec ^2(c+d x)} \sin (c+d x) \cos ^2(c+d x)}{a (a+b \cos (c+d x)) \left (1-\cos ^2(c+d x)\right )}+\frac{\left (15 C a^4-3 b B a^3-A b^2 a^2-29 b^2 C a^2+9 b^3 B a-5 A b^4+8 b^4 C\right ) \cos (2 (c+d x)) (b+a \sec (c+d x)) \left (4 \Pi \left (-\frac{a}{b};\left .-\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right ) \sqrt{\sec (c+d x)} \sqrt{1-\sec ^2(c+d x)} a^2+4 b \sec ^2(c+d x) a-4 b a-4 b E\left (\left .\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right ) \sqrt{\sec (c+d x)} \sqrt{1-\sec ^2(c+d x)} a+2 (2 a-b) b F\left (\left .\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right ) \sqrt{\sec (c+d x)} \sqrt{1-\sec ^2(c+d x)}-2 b^2 \Pi \left (-\frac{a}{b};\left .-\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right ) \sqrt{\sec (c+d x)} \sqrt{1-\sec ^2(c+d x)}\right ) \sin (c+d x)}{a b^2 (a+b \cos (c+d x)) \left (1-\cos ^2(c+d x)\right ) \sqrt{\sec (c+d x)} \left (2-\sec ^2(c+d x)\right )}}{16 (a-b)^2 b^2 (a+b)^2 d}+\frac{\sqrt{\sec (c+d x)} \left (-\frac{\left (7 C a^4-3 b B a^3-A b^2 a^2-13 b^2 C a^2+9 b^3 B a-5 A b^4\right ) \sin (c+d x)}{4 b^3 \left (a^2-b^2\right )^2}-\frac{-C \sin (c+d x) a^4+b B \sin (c+d x) a^3-A b^2 \sin (c+d x) a^2}{2 b^3 \left (b^2-a^2\right ) (a+b \cos (c+d x))^2}+\frac{9 C \sin (c+d x) a^5-5 b B \sin (c+d x) a^4+A b^2 \sin (c+d x) a^3-15 b^2 C \sin (c+d x) a^3+11 b^3 B \sin (c+d x) a^2-7 A b^4 \sin (c+d x) a}{4 b^3 \left (b^2-a^2\right )^2 (a+b \cos (c+d x))}\right )}{d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-2*(-24*a*A*b^3 + 8*a^2*b^2*B + 16*b^4*B + 8*a^3*b*C - 32*a*b^3*C)*Cos[c + d*x]^2*EllipticPi[-(a/b), -ArcSin
[Sqrt[Sec[c + d*x]]], -1]*(b + a*Sec[c + d*x])*Sqrt[1 - Sec[c + d*x]^2]*Sin[c + d*x])/(b*(a + b*Cos[c + d*x])*
(1 - Cos[c + d*x]^2)) + (2*(5*a^2*A*b^2 + A*b^4 - a^3*b*B - 5*a*b^3*B + 5*a^4*C - 7*a^2*b^2*C + 8*b^4*C)*Cos[c
+ d*x]^2*(EllipticF[ArcSin[Sqrt[Sec[c + d*x]]], -1] + EllipticPi[-(a/b), -ArcSin[Sqrt[Sec[c + d*x]]], -1])*(b
+ a*Sec[c + d*x])*Sqrt[1 - Sec[c + d*x]^2]*Sin[c + d*x])/(a*(a + b*Cos[c + d*x])*(1 - Cos[c + d*x]^2)) + ((-(
a^2*A*b^2) - 5*A*b^4 - 3*a^3*b*B + 9*a*b^3*B + 15*a^4*C - 29*a^2*b^2*C + 8*b^4*C)*Cos[2*(c + d*x)]*(b + a*Sec[
c + d*x])*(-4*a*b + 4*a*b*Sec[c + d*x]^2 - 4*a*b*EllipticE[ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*
Sqrt[1 - Sec[c + d*x]^2] + 2*(2*a - b)*b*EllipticF[ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[1 -
Sec[c + d*x]^2] + 4*a^2*EllipticPi[-(a/b), -ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[1 - Sec[c
+ d*x]^2] - 2*b^2*EllipticPi[-(a/b), -ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x
]^2])*Sin[c + d*x])/(a*b^2*(a + b*Cos[c + d*x])*(1 - Cos[c + d*x]^2)*Sqrt[Sec[c + d*x]]*(2 - Sec[c + d*x]^2)))
/(16*(a - b)^2*b^2*(a + b)^2*d) + (Sqrt[Sec[c + d*x]]*(-((-(a^2*A*b^2) - 5*A*b^4 - 3*a^3*b*B + 9*a*b^3*B + 7*a
^4*C - 13*a^2*b^2*C)*Sin[c + d*x])/(4*b^3*(a^2 - b^2)^2) - (-(a^2*A*b^2*Sin[c + d*x]) + a^3*b*B*Sin[c + d*x] -
a^4*C*Sin[c + d*x])/(2*b^3*(-a^2 + b^2)*(a + b*Cos[c + d*x])^2) + (a^3*A*b^2*Sin[c + d*x] - 7*a*A*b^4*Sin[c +
d*x] - 5*a^4*b*B*Sin[c + d*x] + 11*a^2*b^3*B*Sin[c + d*x] + 9*a^5*C*Sin[c + d*x] - 15*a^3*b^2*C*Sin[c + d*x])
/(4*b^3*(-a^2 + b^2)^2*(a + b*Cos[c + d*x]))))/d

________________________________________________________________________________________

Maple [B]  time = 7.813, size = 2000, normalized size = 4.1 \begin{align*} \text{result 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)/(a+b*cos(d*x+c))^3/sec(d*x+c)^(3/2),x)

[Out]

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2/b^4/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^
2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(B*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)
)*b-3*C*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a-C*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*b)-4/b^3*(A*b^2-3*B*a*
b+6*C*a^2)/(-2*a*b+2*b^2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c
)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2))-2*a/b^4*(2*A*b^2-3*B*a*b+4*C
*a^2)*(-1/a*b^2/(a^2-b^2)*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(2*cos(1/2*d
*x+1/2*c)^2*b+a-b)-1/2/(a+b)/a*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+
1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-1/2/a*b/(a^2-b^2)*(sin(1/2*d*x+1/2*
c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(c
os(1/2*d*x+1/2*c),2^(1/2))+1/2/a*b/(a^2-b^2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-
2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-3*a/(a^2-b^2)/(-2*a*b
+2*b^2)*b*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+
1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2))+1/a/(a^2-b^2)/(-2*a*b+2*b^2)*b^3*(sin(1/2*d*
x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*Ellip
ticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2)))+2*a^2*(A*b^2-B*a*b+C*a^2)/b^4*(-1/2/a*b^2/(a^2-b^2)*cos(1/2*d*x+
1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(2*cos(1/2*d*x+1/2*c)^2*b+a-b)^2-3/4*b^2*(3*a^2-b^
2)/a^2/(a^2-b^2)^2*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(2*cos(1/2*d*x+1/2*
c)^2*b+a-b)-7/8/(a+b)/(a^2-b^2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x
+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+1/4/(a+b)/(a^2-b^2)/a*(sin(1/2*d*x
+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*Ellipt
icF(cos(1/2*d*x+1/2*c),2^(1/2))*b+3/8/(a+b)/(a^2-b^2)/a^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^
2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*b^2-9/8*
b/(a^2-b^2)^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*
d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+3/8*b^3/a^2/(a^2-b^2)^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)
*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+
1/2*c),2^(1/2))+9/8*b/(a^2-b^2)^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d
*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-3/8*b^3/a^2/(a^2-b^2)^2*(sin(1/2
*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*El
lipticE(cos(1/2*d*x+1/2*c),2^(1/2))-15/4*a^2/(a^2-b^2)^2/(-2*a*b+2*b^2)*b*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos
(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),
-2*b/(a-b),2^(1/2))+3/2/(a^2-b^2)^2/(-2*a*b+2*b^2)*b^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1
)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2))
-3/4/a^2/(a^2-b^2)^2/(-2*a*b+2*b^2)*b^5*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin
(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2))))/sin(1/2*d*x+
1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d

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Maxima [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)/(a+b*cos(d*x+c))^3/sec(d*x+c)^(3/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [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)/(a+b*cos(d*x+c))^3/sec(d*x+c)^(3/2),x, algorithm="fricas")

[Out]

Timed out

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

[Out]

Timed out

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

[Out]

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