3.1404 \(\int \frac {(A+C \cos ^2(c+d x)) \sec ^{\frac {5}{2}}(c+d x)}{(a+b \cos (c+d x))^3} \, dx\)

Optimal. Leaf size=554 \[ \frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}-\frac {b \left (3 a^4 (8 A-3 C)-a^2 b^2 (65 A-3 C)+35 A b^4\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{4 a^4 d \left (a^2-b^2\right )^2}-\frac {\left (-5 a^4 C-a^2 b^2 (13 A+C)+7 A b^4\right ) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{4 a^2 d \left (a^2-b^2\right )^2 (a+b \cos (c+d x))}+\frac {b \left (3 a^4 (8 A-3 C)-a^2 b^2 (65 A-3 C)+35 A b^4\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a^4 d \left (a^2-b^2\right )^2}+\frac {\left (15 a^6 C+3 a^4 b^2 (21 A-2 C)-a^2 b^4 (86 A-3 C)+35 A b^6\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a^4 d (a-b)^2 (a+b)^3}+\frac {\left (a^4 (8 A-21 C)-a^2 b^2 (61 A-3 C)+35 A b^4\right ) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{12 a^3 d \left (a^2-b^2\right )^2}+\frac {\left (a^4 (8 A-21 C)-a^2 b^2 (61 A-3 C)+35 A b^4\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{12 a^3 d \left (a^2-b^2\right )^2} \]

[Out]

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

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Rubi [A]  time = 2.24, antiderivative size = 554, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.229, Rules used = {4221, 3056, 3055, 3059, 2639, 3002, 2641, 2805} \[ \frac {\left (-a^2 b^2 (61 A-3 C)+a^4 (8 A-21 C)+35 A b^4\right ) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{12 a^3 d \left (a^2-b^2\right )^2}-\frac {b \left (-a^2 b^2 (65 A-3 C)+3 a^4 (8 A-3 C)+35 A b^4\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{4 a^4 d \left (a^2-b^2\right )^2}-\frac {\left (-a^2 b^2 (13 A+C)-5 a^4 C+7 A b^4\right ) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{4 a^2 d \left (a^2-b^2\right )^2 (a+b \cos (c+d x))}+\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}+\frac {\left (-a^2 b^2 (61 A-3 C)+a^4 (8 A-21 C)+35 A b^4\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{12 a^3 d \left (a^2-b^2\right )^2}+\frac {b \left (-a^2 b^2 (65 A-3 C)+3 a^4 (8 A-3 C)+35 A b^4\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a^4 d \left (a^2-b^2\right )^2}+\frac {\left (3 a^4 b^2 (21 A-2 C)-a^2 b^4 (86 A-3 C)+15 a^6 C+35 A b^6\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a^4 d (a-b)^2 (a+b)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(35*A*b^4 + 3*a^4*(8*A - 3*C) - a^2*b^2*(65*A - 3*C))*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec
[c + d*x]])/(4*a^4*(a^2 - b^2)^2*d) + ((35*A*b^4 + a^4*(8*A - 21*C) - a^2*b^2*(61*A - 3*C))*Sqrt[Cos[c + d*x]]
*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(12*a^3*(a^2 - b^2)^2*d) + ((35*A*b^6 - a^2*b^4*(86*A - 3*C) +
3*a^4*b^2*(21*A - 2*C) + 15*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^4*(a - b)^2*(a + b)^3*d) - (b*(35*A*b^4 + 3*a^4*(8*A - 3*C) - a^2*b^2*(65*A - 3*C))*Sqrt[Sec[c + d*
x]]*Sin[c + d*x])/(4*a^4*(a^2 - b^2)^2*d) + ((35*A*b^4 + a^4*(8*A - 21*C) - a^2*b^2*(61*A - 3*C))*Sec[c + d*x]
^(3/2)*Sin[c + d*x])/(12*a^3*(a^2 - b^2)^2*d) + ((A*b^2 + a^2*C)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(2*a*(a^2 -
b^2)*d*(a + b*Cos[c + d*x])^2) - ((7*A*b^4 - 5*a^4*C - a^2*b^2*(13*A + C))*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(4
*a^2*(a^2 - b^2)^2*d*(a + b*Cos[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 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 3056

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[((A*b^2 + 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] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), I
nt[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[a*(m + 1)*(b*c - a*d)*(A + C) + d*(A*b^2 + a^2*C)*
(m + n + 2) - (c*(A*b^2 + a^2*C) + b*(m + 1)*(b*c - a*d)*(A + C))*Sin[e + f*x] - d*(A*b^2 + a^2*C)*(m + n + 3)
*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]
 && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ
[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 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 {\left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x)}{(a+b \cos (c+d x))^3} \, dx &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^3} \, dx\\ &=\frac {\left (A b^2+a^2 C\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {1}{2} \left (-7 A b^2+a^2 (4 A-3 C)\right )-2 a b (A+C) \cos (c+d x)+\frac {5}{2} \left (A b^2+a^2 C\right ) \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^2} \, dx}{2 a \left (a^2-b^2\right )}\\ &=\frac {\left (A b^2+a^2 C\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac {\left (7 A b^4-5 a^4 C-a^2 b^2 (13 A+C)\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {1}{4} \left (35 A b^4+a^4 (8 A-21 C)-a^2 b^2 (61 A-3 C)\right )+a b \left (A b^2-a^2 (4 A+3 C)\right ) \cos (c+d x)-\frac {3}{4} \left (7 A b^4-5 a^4 C-a^2 b^2 (13 A+C)\right ) \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))} \, dx}{2 a^2 \left (a^2-b^2\right )^2}\\ &=\frac {\left (35 A b^4+a^4 (8 A-21 C)-a^2 b^2 (61 A-3 C)\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{12 a^3 \left (a^2-b^2\right )^2 d}+\frac {\left (A b^2+a^2 C\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac {\left (7 A b^4-5 a^4 C-a^2 b^2 (13 A+C)\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {-\frac {3}{8} b \left (35 A b^4+3 a^4 (8 A-3 C)-a^2 b^2 (65 A-3 C)\right )-\frac {1}{2} a \left (7 A b^4-2 a^4 (A+3 C)-a^2 b^2 (14 A+3 C)\right ) \cos (c+d x)+\frac {1}{8} b \left (35 A b^4+a^4 (8 A-21 C)-a^2 b^2 (61 A-3 C)\right ) \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))} \, dx}{3 a^3 \left (a^2-b^2\right )^2}\\ &=-\frac {b \left (35 A b^4+3 a^4 (8 A-3 C)-a^2 b^2 (65 A-3 C)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{4 a^4 \left (a^2-b^2\right )^2 d}+\frac {\left (35 A b^4+a^4 (8 A-21 C)-a^2 b^2 (61 A-3 C)\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{12 a^3 \left (a^2-b^2\right )^2 d}+\frac {\left (A b^2+a^2 C\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac {\left (7 A b^4-5 a^4 C-a^2 b^2 (13 A+C)\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}+\frac {\left (2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {1}{16} \left (105 A b^6+a^4 b^2 (128 A-15 C)-a^2 b^4 (223 A-9 C)+8 a^6 (A+3 C)\right )+\frac {1}{4} a b \left (35 A b^4+4 a^4 (5 A-3 C)-a^2 b^2 (64 A-3 C)\right ) \cos (c+d x)+\frac {3}{16} b^2 \left (35 A b^4+3 a^4 (8 A-3 C)-a^2 b^2 (65 A-3 C)\right ) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{3 a^4 \left (a^2-b^2\right )^2}\\ &=-\frac {b \left (35 A b^4+3 a^4 (8 A-3 C)-a^2 b^2 (65 A-3 C)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{4 a^4 \left (a^2-b^2\right )^2 d}+\frac {\left (35 A b^4+a^4 (8 A-21 C)-a^2 b^2 (61 A-3 C)\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{12 a^3 \left (a^2-b^2\right )^2 d}+\frac {\left (A b^2+a^2 C\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac {\left (7 A b^4-5 a^4 C-a^2 b^2 (13 A+C)\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}-\frac {\left (2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {-\frac {1}{16} b \left (105 A b^6+a^4 b^2 (128 A-15 C)-a^2 b^4 (223 A-9 C)+8 a^6 (A+3 C)\right )-\frac {1}{16} a b^2 \left (35 A b^4+a^4 (8 A-21 C)-a^2 b^2 (61 A-3 C)\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{3 a^4 b \left (a^2-b^2\right )^2}+\frac {\left (b \left (35 A b^4+3 a^4 (8 A-3 C)-a^2 b^2 (65 A-3 C)\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{8 a^4 \left (a^2-b^2\right )^2}\\ &=\frac {b \left (35 A b^4+3 a^4 (8 A-3 C)-a^2 b^2 (65 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)}}{4 a^4 \left (a^2-b^2\right )^2 d}-\frac {b \left (35 A b^4+3 a^4 (8 A-3 C)-a^2 b^2 (65 A-3 C)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{4 a^4 \left (a^2-b^2\right )^2 d}+\frac {\left (35 A b^4+a^4 (8 A-21 C)-a^2 b^2 (61 A-3 C)\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{12 a^3 \left (a^2-b^2\right )^2 d}+\frac {\left (A b^2+a^2 C\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac {\left (7 A b^4-5 a^4 C-a^2 b^2 (13 A+C)\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}+\frac {\left (\left (35 A b^4+a^4 (8 A-21 C)-a^2 b^2 (61 A-3 C)\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{24 a^3 \left (a^2-b^2\right )^2}+\frac {\left (\left (35 A b^6-a^2 b^4 (86 A-3 C)+3 a^4 b^2 (21 A-2 C)+15 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 a^4 \left (a^2-b^2\right )^2}\\ &=\frac {b \left (35 A b^4+3 a^4 (8 A-3 C)-a^2 b^2 (65 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)}}{4 a^4 \left (a^2-b^2\right )^2 d}+\frac {\left (35 A b^4+a^4 (8 A-21 C)-a^2 b^2 (61 A-3 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 a^3 \left (a^2-b^2\right )^2 d}+\frac {\left (35 A b^6-a^2 b^4 (86 A-3 C)+3 a^4 b^2 (21 A-2 C)+15 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^4 (a-b)^2 (a+b)^3 d}-\frac {b \left (35 A b^4+3 a^4 (8 A-3 C)-a^2 b^2 (65 A-3 C)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{4 a^4 \left (a^2-b^2\right )^2 d}+\frac {\left (35 A b^4+a^4 (8 A-21 C)-a^2 b^2 (61 A-3 C)\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{12 a^3 \left (a^2-b^2\right )^2 d}+\frac {\left (A b^2+a^2 C\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac {\left (7 A b^4-5 a^4 C-a^2 b^2 (13 A+C)\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 7.38, size = 880, normalized size = 1.59 \[ \frac {\frac {2 \left (16 A a^6+48 C a^6+328 A b^2 a^4-57 b^2 C a^4-641 A b^4 a^2+27 b^4 C a^2+315 A b^6\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 (160 A b a^5-96 b C a^5-512 A b^3 a^3+24 b^3 C a^3+280 A b^5 a\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 A b^6-195 a^2 A b^4+9 a^2 C b^4+72 a^4 A b^2-27 a^4 C b^2\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^4 (a-b)^2 (a+b)^2 d}+\frac {\sqrt {\sec (c+d x)} \left (-\frac {b \left (24 A a^4-9 C a^4-65 A b^2 a^2+3 b^2 C a^2+35 A b^4\right ) \sin (c+d x)}{4 a^4 \left (a^2-b^2\right )^2}+\frac {-A \sin (c+d x) b^3-a^2 C \sin (c+d x) b}{2 a^2 \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}+\frac {9 A \sin (c+d x) b^5-15 a^2 A \sin (c+d x) b^3+a^2 C \sin (c+d x) b^3-7 a^4 C \sin (c+d x) b}{4 a^3 \left (a^2-b^2\right )^2 (a+b \cos (c+d x))}+\frac {2 A \tan (c+d x)}{3 a^3}\right )}{d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((2*(16*a^6*A + 328*a^4*A*b^2 - 641*a^2*A*b^4 + 315*A*b^6 + 48*a^6*C - 57*a^4*b^2*C + 27*a^2*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*S
ec[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*(160*a^
5*A*b - 512*a^3*A*b^3 + 280*a*A*b^5 - 96*a^5*b*C + 24*a^3*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)) + ((72*a^4*A*b^2 - 195*a^2*A*b^4 + 105*A*b^6 - 27*a^4*b^2*C + 9*a^2*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)))/(48*a^4*(a - b)^2*(a + b)^2*d) + (Sqrt[Sec[c + d*x]]*(-1/4*(b*(24*a^4*A - 65*a^2*A*b^2 + 35*A*b^4 - 9
*a^4*C + 3*a^2*b^2*C)*Sin[c + d*x])/(a^4*(a^2 - b^2)^2) + (-(A*b^3*Sin[c + d*x]) - a^2*b*C*Sin[c + d*x])/(2*a^
2*(a^2 - b^2)*(a + b*Cos[c + d*x])^2) + (-15*a^2*A*b^3*Sin[c + d*x] + 9*A*b^5*Sin[c + d*x] - 7*a^4*b*C*Sin[c +
 d*x] + a^2*b^3*C*Sin[c + d*x])/(4*a^3*(a^2 - b^2)^2*(a + b*Cos[c + d*x])) + (2*A*Tan[c + d*x])/(3*a^3)))/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+C*cos(d*x+c)^2)*sec(d*x+c)^(5/2)/(a+b*cos(d*x+c))^3,x, algorithm="fricas")

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 21.87, size = 2140, normalized size = 3.86 \[ \text {Expression too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+C*cos(d*x+c)^2)*sec(d*x+c)^(5/2)/(a+b*cos(d*x+c))^3,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*A/a^3*(-1/6*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)/(-1/2+cos(1/2*d*x+1/2*c)^2)^2+1/3*(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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/
2)))-12*A*b^3/a^4/(-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^2+C*a^2)/a^2*(
-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*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+s
in(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)*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)*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*(si
n(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)))-6/a^4*b*A*(-(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1
/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+2*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*
d*x+1/2*c)^2)^(1/2)*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2)/sin(1/2*d*x+1/2*c)^2/(2*sin(1/2*d*x+1/2*c)^2-1)+4
*A*b^2/a^3*(-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)*Elliptic
F(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)*El
lipticPi(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+C*cos(d*x+c)^2)*sec(d*x+c)^(5/2)/(a+b*cos(d*x+c))^3,x, algorithm="maxima")

[Out]

Timed out

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

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)

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

[Out]

Timed out

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