3.1409 \(\int \frac {A+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^3 \sec ^{\frac {5}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=493 \[ -\frac {\left (a^2 C+A b^2\right ) \sin (c+d x)}{2 b d \left (a^2-b^2\right ) \sec ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^2}+\frac {\left (-7 a^4 C+a^2 b^2 (A+13 C)+5 A b^4\right ) \sin (c+d x)}{4 b^2 d \left (a^2-b^2\right )^2 \sec ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))}-\frac {a \left (35 a^4 C+a^2 b^2 (3 A-65 C)-3 b^4 (3 A-8 C)\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 b^4 d \left (a^2-b^2\right )^2}+\frac {\left (35 a^4 C+a^2 b^2 (3 A-61 C)-b^4 (21 A-8 C)\right ) \sin (c+d x)}{12 b^3 d \left (a^2-b^2\right )^2 \sqrt {\sec (c+d x)}}+\frac {\left (105 a^6 C+a^4 b^2 (9 A-223 C)-a^2 b^4 (15 A-128 C)+8 b^6 (3 A+C)\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{12 b^5 d \left (a^2-b^2\right )^2}-\frac {a \left (35 a^6 C+a^4 b^2 (3 A-86 C)-3 a^2 b^4 (2 A-21 C)+15 A b^6\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{4 b^5 d (a-b)^2 (a+b)^3} \]

[Out]

-1/2*(A*b^2+C*a^2)*sin(d*x+c)/b/(a^2-b^2)/d/(a+b*cos(d*x+c))^2/sec(d*x+c)^(5/2)+1/4*(5*A*b^4-7*a^4*C+a^2*b^2*(
A+13*C))*sin(d*x+c)/b^2/(a^2-b^2)^2/d/(a+b*cos(d*x+c))/sec(d*x+c)^(3/2)+1/12*(a^2*b^2*(3*A-61*C)-b^4*(21*A-8*C
)+35*a^4*C)*sin(d*x+c)/b^3/(a^2-b^2)^2/d/sec(d*x+c)^(1/2)-1/4*a*(a^2*b^2*(3*A-65*C)-3*b^4*(3*A-8*C)+35*a^4*C)*
(cos(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)^2/d+1/12*(a^4*b^2*(9*A-223*C)-a^2*b^4*(15*A-128*C)+105*a^6*C+8*b^6*(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)^2/d-1/4*a*(15*A*b^6+a^4*b^2*(3*A-86*C)-3*a^2*b^4*(2*A-21*C)+35*a^6*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)
^2/b^5/(a+b)^3/d

________________________________________________________________________________________

Rubi [A]  time = 1.87, antiderivative size = 493, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.257, Rules used = {4221, 3048, 3047, 3049, 3059, 2639, 3002, 2641, 2805} \[ \frac {\left (a^2 b^2 (3 A-61 C)+35 a^4 C-b^4 (21 A-8 C)\right ) \sin (c+d x)}{12 b^3 d \left (a^2-b^2\right )^2 \sqrt {\sec (c+d x)}}+\frac {\left (a^2 b^2 (A+13 C)-7 a^4 C+5 A b^4\right ) \sin (c+d x)}{4 b^2 d \left (a^2-b^2\right )^2 \sec ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))}-\frac {\left (a^2 C+A b^2\right ) \sin (c+d x)}{2 b d \left (a^2-b^2\right ) \sec ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^2}+\frac {\left (a^4 b^2 (9 A-223 C)-a^2 b^4 (15 A-128 C)+105 a^6 C+8 b^6 (3 A+C)\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{12 b^5 d \left (a^2-b^2\right )^2}-\frac {a \left (a^2 b^2 (3 A-65 C)+35 a^4 C-3 b^4 (3 A-8 C)\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 b^4 d \left (a^2-b^2\right )^2}-\frac {a \left (a^4 b^2 (3 A-86 C)-3 a^2 b^4 (2 A-21 C)+35 a^6 C+15 A b^6\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{4 b^5 d (a-b)^2 (a+b)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*(a^2*b^2*(3*A - 65*C) - 3*b^4*(3*A - 8*C) + 35*a^4*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Se
c[c + d*x]])/(4*b^4*(a^2 - b^2)^2*d) + ((a^4*b^2*(9*A - 223*C) - a^2*b^4*(15*A - 128*C) + 105*a^6*C + 8*b^6*(3
*A + C))*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(12*b^5*(a^2 - b^2)^2*d) - (a*(15*A*
b^6 + a^4*b^2*(3*A - 86*C) - 3*a^2*b^4*(2*A - 21*C) + 35*a^6*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^5*(a + b)^3*d) - ((A*b^2 + a^2*C)*Sin[c + d*x])/(2*b*(a^2 -
b^2)*d*(a + b*Cos[c + d*x])^2*Sec[c + d*x]^(5/2)) + ((5*A*b^4 - 7*a^4*C + a^2*b^2*(A + 13*C))*Sin[c + d*x])/(4
*b^2*(a^2 - b^2)^2*d*(a + b*Cos[c + d*x])*Sec[c + d*x]^(3/2)) + ((a^2*b^2*(3*A - 61*C) - b^4*(21*A - 8*C) + 35
*a^4*C)*Sin[c + d*x])/(12*b^3*(a^2 - b^2)^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 3048

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((c^2*C + 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*c*m + a*d*(n + 1)) - (A*d*(a*d*(n +
2) - b*c*(n + 1)) - C*(b*c*d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x] - b*(A*d^2*(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, 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+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^3 \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+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^3} \, dx\\ &=-\frac {\left (A b^2+a^2 C\right ) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2 \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^2 C\right )-2 a b (A+C) \cos (c+d x)-\frac {1}{2} \left (3 A b^2+7 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^2 C\right ) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2 \sec ^{\frac {5}{2}}(c+d x)}+\frac {\left (5 A b^4-7 a^4 C+a^2 b^2 (A+13 C)\right ) \sin (c+d x)}{4 b^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x)) \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}{4} \left (5 A b^4-7 a^4 C+a^2 b^2 (A+13 C)\right )-a b \left (3 A b^2-\left (a^2-4 b^2\right ) C\right ) \cos (c+d x)+\frac {1}{4} \left (a^2 b^2 (3 A-61 C)-b^4 (21 A-8 C)+35 a^4 C\right ) \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx}{2 b^2 \left (a^2-b^2\right )^2}\\ &=-\frac {\left (A b^2+a^2 C\right ) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2 \sec ^{\frac {5}{2}}(c+d x)}+\frac {\left (5 A b^4-7 a^4 C+a^2 b^2 (A+13 C)\right ) \sin (c+d x)}{4 b^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x)}+\frac {\left (a^2 b^2 (3 A-61 C)-b^4 (21 A-8 C)+35 a^4 C\right ) \sin (c+d x)}{12 b^3 \left (a^2-b^2\right )^2 d \sqrt {\sec (c+d x)}}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {1}{8} a \left (a^2 b^2 (3 A-61 C)-b^4 (21 A-8 C)+35 a^4 C\right )-\frac {1}{2} b \left (7 a^4 C-2 b^4 (3 A+C)-a^2 b^2 (3 A+14 C)\right ) \cos (c+d x)-\frac {3}{8} a \left (a^2 b^2 (3 A-65 C)-3 b^4 (3 A-8 C)+35 a^4 C\right ) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{3 b^3 \left (a^2-b^2\right )^2}\\ &=-\frac {\left (A b^2+a^2 C\right ) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2 \sec ^{\frac {5}{2}}(c+d x)}+\frac {\left (5 A b^4-7 a^4 C+a^2 b^2 (A+13 C)\right ) \sin (c+d x)}{4 b^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x)}+\frac {\left (a^2 b^2 (3 A-61 C)-b^4 (21 A-8 C)+35 a^4 C\right ) \sin (c+d x)}{12 b^3 \left (a^2-b^2\right )^2 d \sqrt {\sec (c+d x)}}-\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {-\frac {1}{8} a b \left (a^2 b^2 (3 A-61 C)-b^4 (21 A-8 C)+35 a^4 C\right )-\frac {1}{8} \left (a^4 b^2 (9 A-223 C)-a^2 b^4 (15 A-128 C)+105 a^6 C+8 b^6 (3 A+C)\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{3 b^4 \left (a^2-b^2\right )^2}-\frac {\left (a \left (a^2 b^2 (3 A-65 C)-3 b^4 (3 A-8 C)+35 a^4 C\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{8 b^4 \left (a^2-b^2\right )^2}\\ &=-\frac {a \left (a^2 b^2 (3 A-65 C)-3 b^4 (3 A-8 C)+35 a^4 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^4 \left (a^2-b^2\right )^2 d}-\frac {\left (A b^2+a^2 C\right ) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2 \sec ^{\frac {5}{2}}(c+d x)}+\frac {\left (5 A b^4-7 a^4 C+a^2 b^2 (A+13 C)\right ) \sin (c+d x)}{4 b^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x)}+\frac {\left (a^2 b^2 (3 A-61 C)-b^4 (21 A-8 C)+35 a^4 C\right ) \sin (c+d x)}{12 b^3 \left (a^2-b^2\right )^2 d \sqrt {\sec (c+d x)}}-\frac {\left (a \left (15 A b^6+a^4 b^2 (3 A-86 C)-3 a^2 b^4 (2 A-21 C)+35 a^6 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^5 \left (a^2-b^2\right )^2}+\frac {\left (\left (a^4 b^2 (9 A-223 C)-a^2 b^4 (15 A-128 C)+105 a^6 C+8 b^6 (3 A+C)\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{24 b^5 \left (a^2-b^2\right )^2}\\ &=-\frac {a \left (a^2 b^2 (3 A-65 C)-3 b^4 (3 A-8 C)+35 a^4 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^4 \left (a^2-b^2\right )^2 d}+\frac {\left (a^4 b^2 (9 A-223 C)-a^2 b^4 (15 A-128 C)+105 a^6 C+8 b^6 (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)}}{12 b^5 \left (a^2-b^2\right )^2 d}-\frac {a \left (15 A b^6+a^4 b^2 (3 A-86 C)-3 a^2 b^4 (2 A-21 C)+35 a^6 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^5 (a+b)^3 d}-\frac {\left (A b^2+a^2 C\right ) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2 \sec ^{\frac {5}{2}}(c+d x)}+\frac {\left (5 A b^4-7 a^4 C+a^2 b^2 (A+13 C)\right ) \sin (c+d x)}{4 b^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x)}+\frac {\left (a^2 b^2 (3 A-61 C)-b^4 (21 A-8 C)+35 a^4 C\right ) \sin (c+d x)}{12 b^3 \left (a^2-b^2\right )^2 d \sqrt {\sec (c+d x)}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 7.30, size = 857, normalized size = 1.74 \[ \frac {\sqrt {\sec (c+d x)} \left (\frac {a \left (11 C a^4+3 A b^2 a^2-17 b^2 C a^2-9 A b^4\right ) \sin (c+d x)}{4 b^4 \left (a^2-b^2\right )^2}-\frac {C \sin (c+d x) a^5+A b^2 \sin (c+d x) a^3}{2 b^4 \left (b^2-a^2\right ) (a+b \cos (c+d x))^2}+\frac {-13 C \sin (c+d x) a^6-5 A b^2 \sin (c+d x) a^4+19 b^2 C \sin (c+d x) a^4+11 A b^4 \sin (c+d x) a^2}{4 b^4 \left (b^2-a^2\right )^2 (a+b \cos (c+d x))}+\frac {C \sin (2 (c+d x))}{3 b^3}\right )}{d}-\frac {\frac {2 \left (35 C a^5+3 A b^2 a^3-73 b^2 C a^3+15 A b^4 a+56 b^4 C a\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 (-48 A b^5-16 C b^5-24 a^2 A b^3-112 a^2 C b^3+56 a^4 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^5+9 A b^2 a^3-195 b^2 C a^3-27 A b^4 a+72 b^4 C a\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 )}}{48 (a-b)^2 b^3 (a+b)^2 d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/48*((2*(3*a^3*A*b^2 + 15*a*A*b^4 + 35*a^5*C - 73*a^3*b^2*C + 56*a*b^4*C)*Cos[c + d*x]^2*(EllipticF[ArcSin[S
qrt[Sec[c + d*x]]], -1] - EllipticPi[-(a/b), ArcSin[Sqrt[Sec[c + d*x]]], -1])*(b + a*Sec[c + d*x])*Sqrt[1 - Se
c[c + d*x]^2]*Sin[c + d*x])/(a*(a + b*Cos[c + d*x])*(1 - Cos[c + d*x]^2)) + (2*(-24*a^2*A*b^3 - 48*A*b^5 + 56*
a^4*b*C - 112*a^2*b^3*C - 16*b^5*C)*Cos[c + d*x]^2*EllipticPi[-(a/b), ArcSin[Sqrt[Sec[c + d*x]]], -1]*(b + a*S
ec[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)) + ((9*a^3*A*
b^2 - 27*a*A*b^4 + 105*a^5*C - 195*a^3*b^2*C + 72*a*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*Ellip
ticPi[-(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)))/((a - b)^2*b^3*(a + b)^2
*d) + (Sqrt[Sec[c + d*x]]*((a*(3*a^2*A*b^2 - 9*A*b^4 + 11*a^4*C - 17*a^2*b^2*C)*Sin[c + d*x])/(4*b^4*(a^2 - b^
2)^2) - (a^3*A*b^2*Sin[c + d*x] + a^5*C*Sin[c + d*x])/(2*b^4*(-a^2 + b^2)*(a + b*Cos[c + d*x])^2) + (-5*a^4*A*
b^2*Sin[c + d*x] + 11*a^2*A*b^4*Sin[c + d*x] - 13*a^6*C*Sin[c + d*x] + 19*a^4*b^2*C*Sin[c + d*x])/(4*b^4*(-a^2
+ b^2)^2*(a + b*Cos[c + d*x])) + (C*Sin[2*(c + d*x)])/(3*b^3)))/d

________________________________________________________________________________________

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+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^3/sec(d*x+c)^(5/2),x, algorithm="fricas")

[Out]

Timed out

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {C \cos \left (d x + c\right )^{2} + A}{{\left (b \cos \left (d x + c\right ) + a\right )}^{3} \sec \left (d x + c\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [B]  time = 15.09, size = 2240, normalized size = 4.54 \[ \text {Expression too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^3/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)*(4/3/b^3*C*(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*(si
n(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*C/b^4*(3*a+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)*(Elli
pticF(cos(1/2*d*x+1/2*c),2^(1/2))-EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))+2*(A*b^2+6*C*a^2+3*C*a*b+C*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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+4/b^4*a*(3*A*b^2+10*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^3*(A*b^2+C*a^2)/b^5*(-1/2*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)^2-3/4*b^2*(3*a^2-b^2)/a^2/(a^2-b^2)^2*co
s(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)*EllipticF(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)*E
llipticF(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)*EllipticE(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)))+2*a^2/b^5*(3*A*b^2+5*C*a^2)*(-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)*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)*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)*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

________________________________________________________________________________________

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+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^3/sec(d*x+c)^(5/2),x, algorithm="maxima")

[Out]

Timed out

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {C\,{\cos \left (c+d\,x\right )}^2+A}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + C*cos(c + d*x)^2)/((1/cos(c + d*x))^(5/2)*(a + b*cos(c + d*x))^3),x)

[Out]

int((A + C*cos(c + d*x)^2)/((1/cos(c + d*x))^(5/2)*(a + b*cos(c + d*x))^3), x)

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________