3.1494 \(\int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^2 \sec ^{\frac {5}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=505 \[ \frac {\sin (c+d x) \left (7 a^2 C-5 a b B+5 A b^2-2 b^2 C\right )}{5 b^2 d \left (a^2-b^2\right ) \sec ^{\frac {3}{2}}(c+d x)}-\frac {\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) \sec ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))}+\frac {\sin (c+d x) \left (-7 a^3 C+5 a^2 b B-a b^2 (3 A-4 C)-2 b^3 B\right )}{3 b^3 d \left (a^2-b^2\right ) \sqrt {\sec (c+d x)}}-\frac {\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (-35 a^4 C+25 a^3 b B-3 a^2 b^2 (5 A-8 C)-20 a b^3 B+2 b^4 (5 A+3 C)\right )}{5 b^4 d \left (a^2-b^2\right )}-\frac {a^2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-7 a^4 C+5 a^3 b B-3 a^2 b^2 (A-3 C)-7 a b^3 B+5 A b^4\right ) \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{b^5 d (a-b) (a+b)^2}+\frac {\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (-21 a^5 C+15 a^4 b B-a^3 b^2 (9 A-20 C)-16 a^2 b^3 B+4 a b^4 (3 A+C)-2 b^5 B\right )}{3 b^5 d \left (a^2-b^2\right )} \]

[Out]

-(A*b^2-a*(B*b-C*a))*sin(d*x+c)/b/(a^2-b^2)/d/(a+b*cos(d*x+c))/sec(d*x+c)^(5/2)+1/5*(5*A*b^2-5*B*a*b+7*C*a^2-2
*C*b^2)*sin(d*x+c)/b^2/(a^2-b^2)/d/sec(d*x+c)^(3/2)+1/3*(5*a^2*b*B-2*b^3*B-a*b^2*(3*A-4*C)-7*a^3*C)*sin(d*x+c)
/b^3/(a^2-b^2)/d/sec(d*x+c)^(1/2)-1/5*(25*a^3*b*B-20*a*b^3*B-3*a^2*b^2*(5*A-8*C)-35*a^4*C+2*b^4*(5*A+3*C))*(co
s(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)
^(1/2)/b^4/(a^2-b^2)/d+1/3*(15*a^4*b*B-16*a^2*b^3*B-2*b^5*B-a^3*b^2*(9*A-20*C)-21*a^5*C+4*a*b^4*(3*A+C))*(cos(
1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(
1/2)/b^5/(a^2-b^2)/d-a^2*(5*A*b^4+5*a^3*b*B-7*a*b^3*B-3*a^2*b^2*(A-3*C)-7*a^4*C)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/
cos(1/2*d*x+1/2*c)*EllipticPi(sin(1/2*d*x+1/2*c),2*b/(a+b),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/(a-b)/b^
5/(a+b)^2/d

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Rubi [A]  time = 1.81, antiderivative size = 505, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.186, Rules used = {4221, 3047, 3049, 3059, 2639, 3002, 2641, 2805} \[ \frac {\sin (c+d x) \left (7 a^2 C-5 a b B+5 A b^2-2 b^2 C\right )}{5 b^2 d \left (a^2-b^2\right ) \sec ^{\frac {3}{2}}(c+d x)}+\frac {\sin (c+d x) \left (5 a^2 b B-7 a^3 C-a b^2 (3 A-4 C)-2 b^3 B\right )}{3 b^3 d \left (a^2-b^2\right ) \sqrt {\sec (c+d x)}}-\frac {\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) \sec ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))}+\frac {\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (-a^3 b^2 (9 A-20 C)-16 a^2 b^3 B+15 a^4 b B-21 a^5 C+4 a b^4 (3 A+C)-2 b^5 B\right )}{3 b^5 d \left (a^2-b^2\right )}-\frac {\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (-3 a^2 b^2 (5 A-8 C)+25 a^3 b B-35 a^4 C-20 a b^3 B+2 b^4 (5 A+3 C)\right )}{5 b^4 d \left (a^2-b^2\right )}-\frac {a^2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-3 a^2 b^2 (A-3 C)+5 a^3 b B-7 a^4 C-7 a b^3 B+5 A b^4\right ) \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{b^5 d (a-b) (a+b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((25*a^3*b*B - 20*a*b^3*B - 3*a^2*b^2*(5*A - 8*C) - 35*a^4*C + 2*b^4*(5*A + 3*C))*Sqrt[Cos[c + d*x]]*Elliptic
E[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*b^4*(a^2 - b^2)*d) + ((15*a^4*b*B - 16*a^2*b^3*B - 2*b^5*B - a^3*b^2*
(9*A - 20*C) - 21*a^5*C + 4*a*b^4*(3*A + C))*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/
(3*b^5*(a^2 - b^2)*d) - (a^2*(5*A*b^4 + 5*a^3*b*B - 7*a*b^3*B - 3*a^2*b^2*(A - 3*C) - 7*a^4*C)*Sqrt[Cos[c + d*
x]]*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/((a - b)*b^5*(a + b)^2*d) - ((A*b^2 - a*(b*B
- a*C))*Sin[c + d*x])/(b*(a^2 - b^2)*d*(a + b*Cos[c + d*x])*Sec[c + d*x]^(5/2)) + ((5*A*b^2 - 5*a*b*B + 7*a^2
*C - 2*b^2*C)*Sin[c + d*x])/(5*b^2*(a^2 - b^2)*d*Sec[c + d*x]^(3/2)) + ((5*a^2*b*B - 2*b^3*B - a*b^2*(3*A - 4*
C) - 7*a^3*C)*Sin[c + d*x])/(3*b^3*(a^2 - b^2)*d*Sqrt[Sec[c + d*x]])

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 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]

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 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 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 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 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]

Rubi steps

\begin {align*} \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^2 \sec ^{\frac {5}{2}}(c+d x)} \, dx &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx\\ &=-\frac {\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x)}-\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (\frac {5}{2} \left (A b^2-a (b B-a C)\right )+b (b B-a (A+C)) \cos (c+d x)-\frac {1}{2} \left (5 A b^2-5 a b B+7 a^2 C-2 b^2 C\right ) \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx}{b \left (a^2-b^2\right )}\\ &=-\frac {\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x)}+\frac {\left (5 A b^2-5 a b B+7 a^2 C-2 b^2 C\right ) \sin (c+d x)}{5 b^2 \left (a^2-b^2\right ) d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\left (2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {\cos (c+d x)} \left (-\frac {3}{4} a \left (5 A b^2-5 a b B+7 a^2 C-2 b^2 C\right )+\frac {1}{2} b \left (5 A b^2-5 a b B+2 a^2 C+3 b^2 C\right ) \cos (c+d x)-\frac {5}{4} \left (5 a^2 b B-2 b^3 B-a b^2 (3 A-4 C)-7 a^3 C\right ) \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx}{5 b^2 \left (a^2-b^2\right )}\\ &=-\frac {\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x)}+\frac {\left (5 A b^2-5 a b B+7 a^2 C-2 b^2 C\right ) \sin (c+d x)}{5 b^2 \left (a^2-b^2\right ) d \sec ^{\frac {3}{2}}(c+d x)}+\frac {\left (5 a^2 b B-2 b^3 B-a b^2 (3 A-4 C)-7 a^3 C\right ) \sin (c+d x)}{3 b^3 \left (a^2-b^2\right ) d \sqrt {\sec (c+d x)}}-\frac {\left (4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {-\frac {5}{8} a \left (5 a^2 b B-2 b^3 B-a b^2 (3 A-4 C)-7 a^3 C\right )+\frac {1}{4} b \left (10 a^2 b B+5 b^3 B-14 a^3 C-a b^2 (15 A+C)\right ) \cos (c+d x)+\frac {3}{8} \left (25 a^3 b B-20 a b^3 B-3 a^2 b^2 (5 A-8 C)-35 a^4 C+2 b^4 (5 A+3 C)\right ) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{15 b^3 \left (a^2-b^2\right )}\\ &=-\frac {\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x)}+\frac {\left (5 A b^2-5 a b B+7 a^2 C-2 b^2 C\right ) \sin (c+d x)}{5 b^2 \left (a^2-b^2\right ) d \sec ^{\frac {3}{2}}(c+d x)}+\frac {\left (5 a^2 b B-2 b^3 B-a b^2 (3 A-4 C)-7 a^3 C\right ) \sin (c+d x)}{3 b^3 \left (a^2-b^2\right ) d \sqrt {\sec (c+d x)}}+\frac {\left (4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {5}{8} a b \left (5 a^2 b B-2 b^3 B-a b^2 (3 A-4 C)-7 a^3 C\right )+\frac {5}{8} \left (15 a^4 b B-16 a^2 b^3 B-2 b^5 B-a^3 b^2 (9 A-20 C)-21 a^5 C+4 a b^4 (3 A+C)\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{15 b^4 \left (a^2-b^2\right )}-\frac {\left (\left (25 a^3 b B-20 a b^3 B-3 a^2 b^2 (5 A-8 C)-35 a^4 C+2 b^4 (5 A+3 C)\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{10 b^4 \left (a^2-b^2\right )}\\ &=-\frac {\left (25 a^3 b B-20 a b^3 B-3 a^2 b^2 (5 A-8 C)-35 a^4 C+2 b^4 (5 A+3 C)\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 b^4 \left (a^2-b^2\right ) d}-\frac {\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x)}+\frac {\left (5 A b^2-5 a b B+7 a^2 C-2 b^2 C\right ) \sin (c+d x)}{5 b^2 \left (a^2-b^2\right ) d \sec ^{\frac {3}{2}}(c+d x)}+\frac {\left (5 a^2 b B-2 b^3 B-a b^2 (3 A-4 C)-7 a^3 C\right ) \sin (c+d x)}{3 b^3 \left (a^2-b^2\right ) d \sqrt {\sec (c+d x)}}-\frac {\left (a^2 \left (5 A b^4+5 a^3 b B-7 a b^3 B-3 a^2 b^2 (A-3 C)-7 a^4 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}{2 b^5 \left (a^2-b^2\right )}+\frac {\left (\left (15 a^4 b B-16 a^2 b^3 B-2 b^5 B-a^3 b^2 (9 A-20 C)-21 a^5 C+4 a b^4 (3 A+C)\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{6 b^5 \left (a^2-b^2\right )}\\ &=-\frac {\left (25 a^3 b B-20 a b^3 B-3 a^2 b^2 (5 A-8 C)-35 a^4 C+2 b^4 (5 A+3 C)\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 b^4 \left (a^2-b^2\right ) d}+\frac {\left (15 a^4 b B-16 a^2 b^3 B-2 b^5 B-a^3 b^2 (9 A-20 C)-21 a^5 C+4 a b^4 (3 A+C)\right ) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{3 b^5 \left (a^2-b^2\right ) d}-\frac {a^2 \left (5 A b^4+5 a^3 b B-7 a b^3 B-3 a^2 b^2 (A-3 C)-7 a^4 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)}}{(a-b) b^5 (a+b)^2 d}-\frac {\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x)}+\frac {\left (5 A b^2-5 a b B+7 a^2 C-2 b^2 C\right ) \sin (c+d x)}{5 b^2 \left (a^2-b^2\right ) d \sec ^{\frac {3}{2}}(c+d x)}+\frac {\left (5 a^2 b B-2 b^3 B-a b^2 (3 A-4 C)-7 a^3 C\right ) \sin (c+d x)}{3 b^3 \left (a^2-b^2\right ) d \sqrt {\sec (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 7.37, size = 831, normalized size = 1.65 \[ \frac {\frac {2 \left (35 C a^4-25 b B a^3+15 A b^2 a^2-32 b^2 C a^2+40 b^3 B a-30 A b^4-18 b^4 C\right ) \left (F\left (\left .\sin ^{-1}\left (\sqrt {\sec (c+d x)}\right )\right |-1\right )-\Pi \left (-\frac {a}{b};\left .\sin ^{-1}\left (\sqrt {\sec (c+d x)}\right )\right |-1\right )\right ) (b+a \sec (c+d x)) \sqrt {1-\sec ^2(c+d x)} \sin (c+d x) \cos ^2(c+d x)}{a (a+b \cos (c+d x)) \left (1-\cos ^2(c+d x)\right )}+\frac {2 \left (-20 B b^4+60 a A b^3+4 a C b^3-40 a^2 B b^2+56 a^3 C b\right ) \Pi \left (-\frac {a}{b};\left .\sin ^{-1}\left (\sqrt {\sec (c+d x)}\right )\right |-1\right ) (b+a \sec (c+d x)) \sqrt {1-\sec ^2(c+d x)} \sin (c+d x) \cos ^2(c+d x)}{b (a+b \cos (c+d x)) \left (1-\cos ^2(c+d x)\right )}+\frac {\left (105 C a^4-75 b B a^3+45 A b^2 a^2-72 b^2 C a^2+60 b^3 B a-30 A b^4-18 b^4 C\right ) \cos (2 (c+d x)) (b+a \sec (c+d x)) \left (-4 \Pi \left (-\frac {a}{b};\left .\sin ^{-1}\left (\sqrt {\sec (c+d x)}\right )\right |-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)} a^2+4 b \sec ^2(c+d x) a-4 b a-4 b E\left (\left .\sin ^{-1}\left (\sqrt {\sec (c+d x)}\right )\right |-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)} a+2 (2 a-b) b F\left (\left .\sin ^{-1}\left (\sqrt {\sec (c+d x)}\right )\right |-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}+2 b^2 \Pi \left (-\frac {a}{b};\left .\sin ^{-1}\left (\sqrt {\sec (c+d x)}\right )\right |-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}\right ) \sin (c+d x)}{a b^2 (a+b \cos (c+d x)) \left (1-\cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)} \left (2-\sec ^2(c+d x)\right )}}{60 (a-b) b^3 (a+b) d}+\frac {\sqrt {\sec (c+d x)} \left (-\frac {\left (10 C a^4-10 b B a^3+10 A b^2 a^2-b^2 C a^2+b^4 C\right ) \sin (c+d x)}{10 b^4 \left (a^2-b^2\right )}-\frac {C \sin (c+d x) a^5-b B \sin (c+d x) a^4+A b^2 \sin (c+d x) a^3}{b^4 \left (b^2-a^2\right ) (a+b \cos (c+d x))}+\frac {(b B-2 a C) \sin (2 (c+d x))}{3 b^3}+\frac {C \sin (3 (c+d x))}{10 b^2}\right )}{d} \]

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])^2*Sec[c + d*x]^(5/2)),x]

[Out]

((2*(15*a^2*A*b^2 - 30*A*b^4 - 25*a^3*b*B + 40*a*b^3*B + 35*a^4*C - 32*a^2*b^2*C - 18*b^4*C)*Cos[c + d*x]^2*(E
llipticF[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)) + (2*(60*a*A*b^3 -
40*a^2*b^2*B - 20*b^4*B + 56*a^3*b*C + 4*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)
) + ((45*a^2*A*b^2 - 30*A*b^4 - 75*a^3*b*B + 60*a*b^3*B + 105*a^4*C - 72*a^2*b^2*C - 18*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[S
ec[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]]*Sq
rt[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)))/(60*(a - b)*b^3*(a + b)*d) + (Sqrt[Sec[c + d*x]]*(-1/10*((10*a^2*A*b^2 - 10*a^3*b*B + 10*a^4*C - a
^2*b^2*C + b^4*C)*Sin[c + d*x])/(b^4*(a^2 - b^2)) - (a^3*A*b^2*Sin[c + d*x] - a^4*b*B*Sin[c + d*x] + a^5*C*Sin
[c + d*x])/(b^4*(-a^2 + b^2)*(a + b*Cos[c + d*x])) + ((b*B - 2*a*C)*Sin[2*(c + d*x)])/(3*b^3) + (C*Sin[3*(c +
d*x)])/(10*b^2)))/d

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 11.90, size = 1382, normalized size = 2.74 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(4/5/b^2*C*(-4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c
)^6+14*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+5*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)
*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-9*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*Ellipti
cE(cos(1/2*d*x+1/2*c),2^(1/2))-6*sin(1/2*d*x+1/2*c)^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)+4/3/b^3*(B*b-2*C*a-3*C*b)*(2*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+2*(sin(1/2*d*x+1/2*c)^2)
^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(
2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-sin(1/2*d*x+1/2*c)^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/b^4*(A*b^2-2*B*a*b-2*B*b^2+3*C*a^2+4*C*a*b+3*C*b^2)*(s
in(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))-EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))-2*(2*A*a*b^2+A*b^3-3*B*a^2*b
-2*B*a*b^2-B*b^3+4*C*a^3+3*C*a^2*b+2*C*a*b^2+C*b^3)/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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-4*a^2/b^4*
(3*A*b^2-4*B*a*b+5*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*si
n(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^3*(A*b^2-
B*a*b+C*a^2)/b^5*(-b^2/a/(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*b/a/(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)*El
lipticF(cos(1/2*d*x+1/2*c),2^(1/2))+1/2*b/a/(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*(s
in(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 = 0.00, 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)/(a+b*cos(d*x+c))^2/sec(d*x+c)^(5/2),x, algorithm="maxima")

[Out]

Timed out

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

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))^(5/2)*(a + b*cos(c + d*x))^2),x)

[Out]

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

[Out]

Timed out

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