\(\int \frac {\cos ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{(a+b \sec (c+d x))^{5/2}} \, dx\) [1368]

Optimal result

Integrand size = 45, antiderivative size = 663 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{5/2}} \, dx=\frac {2 \left (128 A b^5+5 a^5 B+80 a^3 b^2 B-80 a b^4 B-4 a^2 b^3 (29 A-10 C)-a^4 b (17 A+45 C)\right ) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right )}{15 a^5 \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (128 A b^6-40 a^5 b B+140 a^3 b^3 B-80 a b^5 B+5 a^4 b^2 (11 A-15 C)-4 a^2 b^4 (53 A-10 C)+3 a^6 (3 A+5 C)\right ) \sqrt {\cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right ) \sqrt {a+b \sec (c+d x)}}{15 a^5 \left (a^2-b^2\right )^2 d \sqrt {\frac {b+a \cos (c+d x)}{a+b}}}+\frac {2 \left (A b^2-a (b B-a C)\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {2 \left (8 A b^4+9 a^3 b B-5 a b^3 B-2 a^2 b^2 (6 A-C)-6 a^4 C\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}}-\frac {2 \left (64 A b^5-5 a^5 B+65 a^3 b^2 B-40 a b^4 B+2 a^4 b (7 A-20 C)-2 a^2 b^3 (49 A-10 C)\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{15 a^4 \left (a^2-b^2\right )^2 d}+\frac {2 \left (48 A b^4+50 a^3 b B-30 a b^3 B+a^4 (3 A-35 C)-a^2 b^2 (71 A-15 C)\right ) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{15 a^3 \left (a^2-b^2\right )^2 d} \]

[Out]

2/3*(A*b^2-a*(B*b-C*a))*cos(d*x+c)^(3/2)*sin(d*x+c)/a/(a^2-b^2)/d/(a+b*sec(d*x+c))^(3/2)-2/3*(8*A*b^4+9*B*a^3*
b-5*B*a*b^3-2*a^2*b^2*(6*A-C)-6*a^4*C)*cos(d*x+c)^(3/2)*sin(d*x+c)/a^2/(a^2-b^2)^2/d/(a+b*sec(d*x+c))^(1/2)+2/
15*(128*A*b^5+5*a^5*B+80*a^3*b^2*B-80*a*b^4*B-4*a^2*b^3*(29*A-10*C)-a^4*b*(17*A+45*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)*(a/(a+b))^(1/2))*((b+a*cos(d*x+c))/(a+b))^(1/2)/
a^5/(a^2-b^2)/d/cos(d*x+c)^(1/2)/(a+b*sec(d*x+c))^(1/2)+2/15*(48*A*b^4+50*B*a^3*b-30*B*a*b^3+a^4*(3*A-35*C)-a^
2*b^2*(71*A-15*C))*cos(d*x+c)^(3/2)*sin(d*x+c)*(a+b*sec(d*x+c))^(1/2)/a^3/(a^2-b^2)^2/d-2/15*(64*A*b^5-5*a^5*B
+65*a^3*b^2*B-40*a*b^4*B+2*a^4*b*(7*A-20*C)-2*a^2*b^3*(49*A-10*C))*sin(d*x+c)*cos(d*x+c)^(1/2)*(a+b*sec(d*x+c)
)^(1/2)/a^4/(a^2-b^2)^2/d+2/15*(128*A*b^6-40*a^5*b*B+140*a^3*b^3*B-80*a*b^5*B+5*a^4*b^2*(11*A-15*C)-4*a^2*b^4*
(53*A-10*C)+3*a^6*(3*A+5*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)*(a/(a+b))^(1/2))*cos(d*x+c)^(1/2)*(a+b*sec(d*x+c))^(1/2)/a^5/(a^2-b^2)^2/d/((b+a*cos(d*x+c))/(a+b))^(1/2)

Rubi [A] (verified)

Time = 2.85 (sec) , antiderivative size = 663, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {4350, 4185, 4189, 4120, 3941, 2734, 2732, 3943, 2742, 2740} \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{5/2}} \, dx=\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{3 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}+\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (a^4 (3 A-35 C)+50 a^3 b B-a^2 b^2 (71 A-15 C)-30 a b^3 B+48 A b^4\right ) \sqrt {a+b \sec (c+d x)}}{15 a^3 d \left (a^2-b^2\right )^2}-\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (-6 a^4 C+9 a^3 b B-2 a^2 b^2 (6 A-C)-5 a b^3 B+8 A b^4\right )}{3 a^2 d \left (a^2-b^2\right )^2 \sqrt {a+b \sec (c+d x)}}-\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)} \left (-5 a^5 B+2 a^4 b (7 A-20 C)+65 a^3 b^2 B-2 a^2 b^3 (49 A-10 C)-40 a b^4 B+64 A b^5\right ) \sqrt {a+b \sec (c+d x)}}{15 a^4 d \left (a^2-b^2\right )^2}+\frac {2 \left (5 a^5 B-a^4 b (17 A+45 C)+80 a^3 b^2 B-4 a^2 b^3 (29 A-10 C)-80 a b^4 B+128 A b^5\right ) \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right )}{15 a^5 d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {2 \sqrt {\cos (c+d x)} \left (3 a^6 (3 A+5 C)-40 a^5 b B+5 a^4 b^2 (11 A-15 C)+140 a^3 b^3 B-4 a^2 b^4 (53 A-10 C)-80 a b^5 B+128 A b^6\right ) \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{15 a^5 d \left (a^2-b^2\right )^2 \sqrt {\frac {a \cos (c+d x)+b}{a+b}}} \]

[In]

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

[Out]

(2*(128*A*b^5 + 5*a^5*B + 80*a^3*b^2*B - 80*a*b^4*B - 4*a^2*b^3*(29*A - 10*C) - a^4*b*(17*A + 45*C))*Sqrt[(b +
a*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*a)/(a + b)])/(15*a^5*(a^2 - b^2)*d*Sqrt[Cos[c + d*x]]*Sqrt
[a + b*Sec[c + d*x]]) + (2*(128*A*b^6 - 40*a^5*b*B + 140*a^3*b^3*B - 80*a*b^5*B + 5*a^4*b^2*(11*A - 15*C) - 4*
a^2*b^4*(53*A - 10*C) + 3*a^6*(3*A + 5*C))*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*a)/(a + b)]*Sqrt[a + b
*Sec[c + d*x]])/(15*a^5*(a^2 - b^2)^2*d*Sqrt[(b + a*Cos[c + d*x])/(a + b)]) + (2*(A*b^2 - a*(b*B - a*C))*Cos[c
+ d*x]^(3/2)*Sin[c + d*x])/(3*a*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^(3/2)) - (2*(8*A*b^4 + 9*a^3*b*B - 5*a*b^3
*B - 2*a^2*b^2*(6*A - C) - 6*a^4*C)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(3*a^2*(a^2 - b^2)^2*d*Sqrt[a + b*Sec[c +
d*x]]) - (2*(64*A*b^5 - 5*a^5*B + 65*a^3*b^2*B - 40*a*b^4*B + 2*a^4*b*(7*A - 20*C) - 2*a^2*b^3*(49*A - 10*C))
*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(15*a^4*(a^2 - b^2)^2*d) + (2*(48*A*b^4 + 50*a^3*b*
B - 30*a*b^3*B + a^4*(3*A - 35*C) - a^2*b^2*(71*A - 15*C))*Cos[c + d*x]^(3/2)*Sqrt[a + b*Sec[c + d*x]]*Sin[c +
d*x])/(15*a^3*(a^2 - b^2)^2*d)

Rule 2732

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

Rule 2734

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[a + b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c +
d*x])/(a + b)], Int[Sqrt[a/(a + b) + (b/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
b^2, 0] &&  !GtQ[a + b, 0]

Rule 2740

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

Rule 2742

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[(a + b*Sin[c + d*x])/(a + b)]/Sqrt[a
+ b*Sin[c + d*x]], Int[1/Sqrt[a/(a + b) + (b/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] &&  !GtQ[a + b, 0]

Rule 3941

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)], x_Symbol] :> Dist[Sqrt[a +
b*Csc[e + f*x]]/(Sqrt[d*Csc[e + f*x]]*Sqrt[b + a*Sin[e + f*x]]), Int[Sqrt[b + a*Sin[e + f*x]], x], x] /; Free
Q[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 3943

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[Sqrt[d*C
sc[e + f*x]]*(Sqrt[b + a*Sin[e + f*x]]/Sqrt[a + b*Csc[e + f*x]]), Int[1/Sqrt[b + a*Sin[e + f*x]], x], x] /; Fr
eeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 4120

Int[(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_))/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(
b_.) + (a_)]), x_Symbol] :> Dist[A/a, Int[Sqrt[a + b*Csc[e + f*x]]/Sqrt[d*Csc[e + f*x]], x], x] - Dist[(A*b -
a*B)/(a*d), Int[Sqrt[d*Csc[e + f*x]]/Sqrt[a + b*Csc[e + f*x]], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && Ne
Q[A*b - a*B, 0] && NeQ[a^2 - b^2, 0]

Rule 4185

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^
(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(A*b^2 - a*b*B + a^2*C)*Cot[e + f*x]*(a +
b*Csc[e + f*x])^(m + 1)*((d*Csc[e + f*x])^n/(a*f*(m + 1)*(a^2 - b^2))), x] + Dist[1/(a*(m + 1)*(a^2 - b^2)), I
nt[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*Simp[a*(a*A - b*B + a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C)*
(m + n + 1) - a*(A*b - a*B + b*C)*(m + 1)*Csc[e + f*x] + (A*b^2 - a*b*B + a^2*C)*(m + n + 2)*Csc[e + f*x]^2, x
], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, n}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] &&  !(ILtQ[m + 1/2, 0] &
& ILtQ[n, 0])

Rule 4189

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^
(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[A*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1
)*((d*Csc[e + f*x])^n/(a*f*n)), x] + Dist[1/(a*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[
a*B*n - A*b*(m + n + 1) + a*(A + A*n + C*n)*Csc[e + f*x] + A*b*(m + n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ
[{a, b, d, e, f, A, B, C, m}, x] && NeQ[a^2 - b^2, 0] && LeQ[n, -1]

Rule 4350

Int[(cos[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Dist[(c*Cos[a + b*x])^m*(c*Sec[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Sec[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSecantIntegrandQ[
u, x]

Rubi steps \begin{align*} \text {integral}& = \left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^{5/2}} \, dx \\ & = \frac {2 \left (A b^2-a (b B-a C)\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {\left (2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {1}{2} \left (8 A b^2-5 a b B-a^2 (3 A-5 C)\right )+\frac {3}{2} a (A b-a B+b C) \sec (c+d x)-3 \left (A b^2-a (b B-a C)\right ) \sec ^2(c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^{3/2}} \, dx}{3 a \left (a^2-b^2\right )} \\ & = \frac {2 \left (A b^2-a (b B-a C)\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {2 \left (8 A b^4+9 a^3 b B-5 a b^3 B-2 a^2 b^2 (6 A-C)-6 a^4 C\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}}+\frac {\left (4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {1}{4} \left (48 A b^4+50 a^3 b B-30 a b^3 B+a^4 (3 A-35 C)-a^2 b^2 (71 A-15 C)\right )+\frac {1}{4} a \left (2 A b^3+3 a^3 B+a b^2 B-2 a^2 b (3 A+2 C)\right ) \sec (c+d x)-\left (8 A b^4+9 a^3 b B-5 a b^3 B-2 a^2 b^2 (6 A-C)-6 a^4 C\right ) \sec ^2(c+d x)}{\sec ^{\frac {5}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}} \, dx}{3 a^2 \left (a^2-b^2\right )^2} \\ & = \frac {2 \left (A b^2-a (b B-a C)\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {2 \left (8 A b^4+9 a^3 b B-5 a b^3 B-2 a^2 b^2 (6 A-C)-6 a^4 C\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (48 A b^4+50 a^3 b B-30 a b^3 B+a^4 (3 A-35 C)-a^2 b^2 (71 A-15 C)\right ) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{15 a^3 \left (a^2-b^2\right )^2 d}-\frac {\left (8 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {3}{8} \left (64 A b^5-5 a^5 B+65 a^3 b^2 B-40 a b^4 B+2 a^4 b (7 A-20 C)-2 a^2 b^3 (49 A-10 C)\right )+\frac {1}{8} a \left (16 A b^4+30 a^3 b B-10 a b^3 B-3 a^4 (3 A+5 C)-a^2 b^2 (27 A+5 C)\right ) \sec (c+d x)-\frac {1}{4} b \left (48 A b^4+50 a^3 b B-30 a b^3 B+a^4 (3 A-35 C)-a^2 b^2 (71 A-15 C)\right ) \sec ^2(c+d x)}{\sec ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}} \, dx}{15 a^3 \left (a^2-b^2\right )^2} \\ & = \frac {2 \left (A b^2-a (b B-a C)\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {2 \left (8 A b^4+9 a^3 b B-5 a b^3 B-2 a^2 b^2 (6 A-C)-6 a^4 C\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}}-\frac {2 \left (64 A b^5-5 a^5 B+65 a^3 b^2 B-40 a b^4 B+2 a^4 b (7 A-20 C)-2 a^2 b^3 (49 A-10 C)\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{15 a^4 \left (a^2-b^2\right )^2 d}+\frac {2 \left (48 A b^4+50 a^3 b B-30 a b^3 B+a^4 (3 A-35 C)-a^2 b^2 (71 A-15 C)\right ) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{15 a^3 \left (a^2-b^2\right )^2 d}+\frac {\left (16 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {3}{16} \left (128 A b^6-40 a^5 b B+140 a^3 b^3 B-80 a b^5 B+5 a^4 b^2 (11 A-15 C)-4 a^2 b^4 (53 A-10 C)+3 a^6 (3 A+5 C)\right )+\frac {3}{16} a \left (32 A b^5+5 a^5 B+35 a^3 b^2 B-20 a b^4 B-2 a^2 b^3 (22 A-5 C)-2 a^4 b (4 A+15 C)\right ) \sec (c+d x)}{\sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}} \, dx}{45 a^4 \left (a^2-b^2\right )^2} \\ & = \frac {2 \left (A b^2-a (b B-a C)\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {2 \left (8 A b^4+9 a^3 b B-5 a b^3 B-2 a^2 b^2 (6 A-C)-6 a^4 C\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}}-\frac {2 \left (64 A b^5-5 a^5 B+65 a^3 b^2 B-40 a b^4 B+2 a^4 b (7 A-20 C)-2 a^2 b^3 (49 A-10 C)\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{15 a^4 \left (a^2-b^2\right )^2 d}+\frac {2 \left (48 A b^4+50 a^3 b B-30 a b^3 B+a^4 (3 A-35 C)-a^2 b^2 (71 A-15 C)\right ) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{15 a^3 \left (a^2-b^2\right )^2 d}+\frac {\left (\left (128 A b^6-40 a^5 b B+140 a^3 b^3 B-80 a b^5 B+5 a^4 b^2 (11 A-15 C)-4 a^2 b^4 (53 A-10 C)+3 a^6 (3 A+5 C)\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {\sec (c+d x)}} \, dx}{15 a^5 \left (a^2-b^2\right )^2}+\frac {\left (\left (128 A b^5+5 a^5 B+80 a^3 b^2 B-80 a b^4 B-4 a^2 b^3 (29 A-10 C)-a^4 b (17 A+45 C)\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {a+b \sec (c+d x)}} \, dx}{15 a^5 \left (a^2-b^2\right )} \\ & = \frac {2 \left (A b^2-a (b B-a C)\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {2 \left (8 A b^4+9 a^3 b B-5 a b^3 B-2 a^2 b^2 (6 A-C)-6 a^4 C\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}}-\frac {2 \left (64 A b^5-5 a^5 B+65 a^3 b^2 B-40 a b^4 B+2 a^4 b (7 A-20 C)-2 a^2 b^3 (49 A-10 C)\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{15 a^4 \left (a^2-b^2\right )^2 d}+\frac {2 \left (48 A b^4+50 a^3 b B-30 a b^3 B+a^4 (3 A-35 C)-a^2 b^2 (71 A-15 C)\right ) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{15 a^3 \left (a^2-b^2\right )^2 d}+\frac {\left (\left (128 A b^5+5 a^5 B+80 a^3 b^2 B-80 a b^4 B-4 a^2 b^3 (29 A-10 C)-a^4 b (17 A+45 C)\right ) \sqrt {b+a \cos (c+d x)}\right ) \int \frac {1}{\sqrt {b+a \cos (c+d x)}} \, dx}{15 a^5 \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {\left (\left (128 A b^6-40 a^5 b B+140 a^3 b^3 B-80 a b^5 B+5 a^4 b^2 (11 A-15 C)-4 a^2 b^4 (53 A-10 C)+3 a^6 (3 A+5 C)\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}\right ) \int \sqrt {b+a \cos (c+d x)} \, dx}{15 a^5 \left (a^2-b^2\right )^2 \sqrt {b+a \cos (c+d x)}} \\ & = \frac {2 \left (A b^2-a (b B-a C)\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {2 \left (8 A b^4+9 a^3 b B-5 a b^3 B-2 a^2 b^2 (6 A-C)-6 a^4 C\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}}-\frac {2 \left (64 A b^5-5 a^5 B+65 a^3 b^2 B-40 a b^4 B+2 a^4 b (7 A-20 C)-2 a^2 b^3 (49 A-10 C)\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{15 a^4 \left (a^2-b^2\right )^2 d}+\frac {2 \left (48 A b^4+50 a^3 b B-30 a b^3 B+a^4 (3 A-35 C)-a^2 b^2 (71 A-15 C)\right ) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{15 a^3 \left (a^2-b^2\right )^2 d}+\frac {\left (\left (128 A b^5+5 a^5 B+80 a^3 b^2 B-80 a b^4 B-4 a^2 b^3 (29 A-10 C)-a^4 b (17 A+45 C)\right ) \sqrt {\frac {b+a \cos (c+d x)}{a+b}}\right ) \int \frac {1}{\sqrt {\frac {b}{a+b}+\frac {a \cos (c+d x)}{a+b}}} \, dx}{15 a^5 \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {\left (\left (128 A b^6-40 a^5 b B+140 a^3 b^3 B-80 a b^5 B+5 a^4 b^2 (11 A-15 C)-4 a^2 b^4 (53 A-10 C)+3 a^6 (3 A+5 C)\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}\right ) \int \sqrt {\frac {b}{a+b}+\frac {a \cos (c+d x)}{a+b}} \, dx}{15 a^5 \left (a^2-b^2\right )^2 \sqrt {\frac {b+a \cos (c+d x)}{a+b}}} \\ & = \frac {2 \left (128 A b^5+5 a^5 B+80 a^3 b^2 B-80 a b^4 B-4 a^2 b^3 (29 A-10 C)-a^4 b (17 A+45 C)\right ) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right )}{15 a^5 \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (128 A b^6-40 a^5 b B+140 a^3 b^3 B-80 a b^5 B+5 a^4 b^2 (11 A-15 C)-4 a^2 b^4 (53 A-10 C)+3 a^6 (3 A+5 C)\right ) \sqrt {\cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right ) \sqrt {a+b \sec (c+d x)}}{15 a^5 \left (a^2-b^2\right )^2 d \sqrt {\frac {b+a \cos (c+d x)}{a+b}}}+\frac {2 \left (A b^2-a (b B-a C)\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {2 \left (8 A b^4+9 a^3 b B-5 a b^3 B-2 a^2 b^2 (6 A-C)-6 a^4 C\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}}-\frac {2 \left (64 A b^5-5 a^5 B+65 a^3 b^2 B-40 a b^4 B+2 a^4 b (7 A-20 C)-2 a^2 b^3 (49 A-10 C)\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{15 a^4 \left (a^2-b^2\right )^2 d}+\frac {2 \left (48 A b^4+50 a^3 b B-30 a b^3 B+a^4 (3 A-35 C)-a^2 b^2 (71 A-15 C)\right ) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{15 a^3 \left (a^2-b^2\right )^2 d} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 37.22 (sec) , antiderivative size = 4917, normalized size of antiderivative = 7.42 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{5/2}} \, dx=\text {Result too large to show} \]

[In]

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

[Out]

((b + a*Cos[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((4*(-14*A*b + 5*a*B)*Sin[c + d*x])/(15*a^4) -
(4*(A*b^5*Sin[c + d*x] - a*b^4*B*Sin[c + d*x] + a^2*b^3*C*Sin[c + d*x]))/(3*a^4*(a^2 - b^2)*(b + a*Cos[c + d*
x])^2) - (4*(-15*a^2*A*b^4*Sin[c + d*x] + 11*A*b^6*Sin[c + d*x] + 12*a^3*b^3*B*Sin[c + d*x] - 8*a*b^5*B*Sin[c
+ d*x] - 9*a^4*b^2*C*Sin[c + d*x] + 5*a^2*b^4*C*Sin[c + d*x]))/(3*a^4*(a^2 - b^2)^2*(b + a*Cos[c + d*x])) + (2
*A*Sin[2*(c + d*x)])/(5*a^3)))/(d*Sqrt[Cos[c + d*x]]*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + b*
Sec[c + d*x])^(5/2)) - (4*Cos[c + d*x]^(3/2)*(b + a*Cos[c + d*x])^2*((6*a^2*A*Sqrt[Cos[c + d*x]])/(5*(a^2 - b^
2)^2*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) + (22*A*b^2*Sqrt[Cos[c + d*x]])/(3*(a^2 - b^2)^2*Sqrt[b + a*
Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (424*A*b^4*Sqrt[Cos[c + d*x]])/(15*a^2*(a^2 - b^2)^2*Sqrt[b + a*Cos[c + d*
x]]*Sqrt[Sec[c + d*x]]) + (256*A*b^6*Sqrt[Cos[c + d*x]])/(15*a^4*(a^2 - b^2)^2*Sqrt[b + a*Cos[c + d*x]]*Sqrt[S
ec[c + d*x]]) - (16*a*b*B*Sqrt[Cos[c + d*x]])/(3*(a^2 - b^2)^2*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) +
(56*b^3*B*Sqrt[Cos[c + d*x]])/(3*a*(a^2 - b^2)^2*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (32*b^5*B*Sqrt
[Cos[c + d*x]])/(3*a^3*(a^2 - b^2)^2*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) + (2*a^2*C*Sqrt[Cos[c + d*x]
])/((a^2 - b^2)^2*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (10*b^2*C*Sqrt[Cos[c + d*x]])/((a^2 - b^2)^2*
Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) + (16*b^4*C*Sqrt[Cos[c + d*x]])/(3*a^2*(a^2 - b^2)^2*Sqrt[b + a*C
os[c + d*x]]*Sqrt[Sec[c + d*x]]) - (16*a*A*b*Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]])/(15*(a^2 - b^2)^2*Sqrt[b +
a*Cos[c + d*x]]) - (88*A*b^3*Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]])/(15*a*(a^2 - b^2)^2*Sqrt[b + a*Cos[c + d*
x]]) + (64*A*b^5*Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]])/(15*a^3*(a^2 - b^2)^2*Sqrt[b + a*Cos[c + d*x]]) + (2*a
^2*B*Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]])/(3*(a^2 - b^2)^2*Sqrt[b + a*Cos[c + d*x]]) + (14*b^2*B*Sqrt[Cos[c
+ d*x]]*Sqrt[Sec[c + d*x]])/(3*(a^2 - b^2)^2*Sqrt[b + a*Cos[c + d*x]]) - (8*b^4*B*Sqrt[Cos[c + d*x]]*Sqrt[Sec[
c + d*x]])/(3*a^2*(a^2 - b^2)^2*Sqrt[b + a*Cos[c + d*x]]) - (4*a*b*C*Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]])/((
a^2 - b^2)^2*Sqrt[b + a*Cos[c + d*x]]) + (4*b^3*C*Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]])/(3*a*(a^2 - b^2)^2*Sq
rt[b + a*Cos[c + d*x]]))*Sqrt[Sec[c + d*x]]*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^(3/2)*(A + B*Sec[c + d*x] + C*Se
c[c + d*x]^2)*((-I)*(a + b)*(128*A*b^6 - 40*a^5*b*B + 140*a^3*b^3*B - 80*a*b^5*B + 5*a^4*b^2*(11*A - 15*C) + 3
*a^6*(3*A + 5*C) + 4*a^2*b^4*(-53*A + 10*C))*EllipticE[I*ArcSinh[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Sec[(c +
d*x)/2]^2*Sqrt[((b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2)/(a + b)] + I*a*(a + b)*(128*A*b^5 - 16*a*b^4*(6*A +
5*B) + 2*a^3*b^2*(36*A + 40*B - 15*C) + 4*a^2*b^3*(-29*A + 15*B + 10*C) - a^4*b*(17*A + 45*(B + C)) + a^5*(9*A
+ 5*(B + 3*C)))*EllipticF[I*ArcSinh[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Sec[(c + d*x)/2]^2*Sqrt[((b + a*Cos[
c + d*x])*Sec[(c + d*x)/2]^2)/(a + b)] - (128*A*b^6 - 40*a^5*b*B + 140*a^3*b^3*B - 80*a*b^5*B + 5*a^4*b^2*(11*
A - 15*C) + 3*a^6*(3*A + 5*C) + 4*a^2*b^4*(-53*A + 10*C))*(b + a*Cos[c + d*x])*(Sec[(c + d*x)/2]^2)^(3/2)*Tan[
(c + d*x)/2]))/(15*a^5*(a^2 - b^2)^2*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + b*Sec[c + d*x])^
(5/2)*((-2*Cos[c + d*x]^(3/2)*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^(3/2)*Sin[c + d*x]*((-I)*(a + b)*(128*A*b^6 -
40*a^5*b*B + 140*a^3*b^3*B - 80*a*b^5*B + 5*a^4*b^2*(11*A - 15*C) + 3*a^6*(3*A + 5*C) + 4*a^2*b^4*(-53*A + 10*
C))*EllipticE[I*ArcSinh[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Sec[(c + d*x)/2]^2*Sqrt[((b + a*Cos[c + d*x])*Sec
[(c + d*x)/2]^2)/(a + b)] + I*a*(a + b)*(128*A*b^5 - 16*a*b^4*(6*A + 5*B) + 2*a^3*b^2*(36*A + 40*B - 15*C) + 4
*a^2*b^3*(-29*A + 15*B + 10*C) - a^4*b*(17*A + 45*(B + C)) + a^5*(9*A + 5*(B + 3*C)))*EllipticF[I*ArcSinh[Tan[
(c + d*x)/2]], (-a + b)/(a + b)]*Sec[(c + d*x)/2]^2*Sqrt[((b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2)/(a + b)] -
(128*A*b^6 - 40*a^5*b*B + 140*a^3*b^3*B - 80*a*b^5*B + 5*a^4*b^2*(11*A - 15*C) + 3*a^6*(3*A + 5*C) + 4*a^2*b^4
*(-53*A + 10*C))*(b + a*Cos[c + d*x])*(Sec[(c + d*x)/2]^2)^(3/2)*Tan[(c + d*x)/2]))/(15*a^4*(a^2 - b^2)^2*(b +
a*Cos[c + d*x])^(3/2)) + (2*Sqrt[Cos[c + d*x]]*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^(3/2)*Sin[c + d*x]*((-I)*(a
+ b)*(128*A*b^6 - 40*a^5*b*B + 140*a^3*b^3*B - 80*a*b^5*B + 5*a^4*b^2*(11*A - 15*C) + 3*a^6*(3*A + 5*C) + 4*a^
2*b^4*(-53*A + 10*C))*EllipticE[I*ArcSinh[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Sec[(c + d*x)/2]^2*Sqrt[((b + a
*Cos[c + d*x])*Sec[(c + d*x)/2]^2)/(a + b)] + I*a*(a + b)*(128*A*b^5 - 16*a*b^4*(6*A + 5*B) + 2*a^3*b^2*(36*A
+ 40*B - 15*C) + 4*a^2*b^3*(-29*A + 15*B + 10*C) - a^4*b*(17*A + 45*(B + C)) + a^5*(9*A + 5*(B + 3*C)))*Ellipt
icF[I*ArcSinh[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Sec[(c + d*x)/2]^2*Sqrt[((b + a*Cos[c + d*x])*Sec[(c + d*x)
/2]^2)/(a + b)] - (128*A*b^6 - 40*a^5*b*B + 140*a^3*b^3*B - 80*a*b^5*B + 5*a^4*b^2*(11*A - 15*C) + 3*a^6*(3*A
+ 5*C) + 4*a^2*b^4*(-53*A + 10*C))*(b + a*Cos[c + d*x])*(Sec[(c + d*x)/2]^2)^(3/2)*Tan[(c + d*x)/2]))/(5*a^5*(
a^2 - b^2)^2*Sqrt[b + a*Cos[c + d*x]]) - (4*Cos[c + d*x]^(3/2)*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^(3/2)*(-1/2*(
(128*A*b^6 - 40*a^5*b*B + 140*a^3*b^3*B - 80*a*b^5*B + 5*a^4*b^2*(11*A - 15*C) + 3*a^6*(3*A + 5*C) + 4*a^2*b^4
*(-53*A + 10*C))*(b + a*Cos[c + d*x])*(Sec[(c + d*x)/2]^2)^(5/2)) - I*(a + b)*(128*A*b^6 - 40*a^5*b*B + 140*a^
3*b^3*B - 80*a*b^5*B + 5*a^4*b^2*(11*A - 15*C) + 3*a^6*(3*A + 5*C) + 4*a^2*b^4*(-53*A + 10*C))*EllipticE[I*Arc
Sinh[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Sec[(c + d*x)/2]^2*Sqrt[((b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2)/(a
+ b)]*Tan[(c + d*x)/2] + I*a*(a + b)*(128*A*b^5 - 16*a*b^4*(6*A + 5*B) + 2*a^3*b^2*(36*A + 40*B - 15*C) + 4*a
^2*b^3*(-29*A + 15*B + 10*C) - a^4*b*(17*A + 45*(B + C)) + a^5*(9*A + 5*(B + 3*C)))*EllipticF[I*ArcSinh[Tan[(c
+ d*x)/2]], (-a + b)/(a + b)]*Sec[(c + d*x)/2]^2*Sqrt[((b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2)/(a + b)]*Tan[
(c + d*x)/2] + a*(128*A*b^6 - 40*a^5*b*B + 140*a^3*b^3*B - 80*a*b^5*B + 5*a^4*b^2*(11*A - 15*C) + 3*a^6*(3*A +
5*C) + 4*a^2*b^4*(-53*A + 10*C))*(Sec[(c + d*x)/2]^2)^(3/2)*Sin[c + d*x]*Tan[(c + d*x)/2] - (3*(128*A*b^6 - 4
0*a^5*b*B + 140*a^3*b^3*B - 80*a*b^5*B + 5*a^4*b^2*(11*A - 15*C) + 3*a^6*(3*A + 5*C) + 4*a^2*b^4*(-53*A + 10*C
))*(b + a*Cos[c + d*x])*(Sec[(c + d*x)/2]^2)^(3/2)*Tan[(c + d*x)/2]^2)/2 - ((I/2)*(a + b)*(128*A*b^6 - 40*a^5*
b*B + 140*a^3*b^3*B - 80*a*b^5*B + 5*a^4*b^2*(11*A - 15*C) + 3*a^6*(3*A + 5*C) + 4*a^2*b^4*(-53*A + 10*C))*Ell
ipticE[I*ArcSinh[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Sec[(c + d*x)/2]^2*(-((a*Sec[(c + d*x)/2]^2*Sin[c + d*x]
)/(a + b)) + ((b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2])/(a + b)))/Sqrt[((b + a*Cos[c + d*x])*S
ec[(c + d*x)/2]^2)/(a + b)] + ((I/2)*a*(a + b)*(128*A*b^5 - 16*a*b^4*(6*A + 5*B) + 2*a^3*b^2*(36*A + 40*B - 15
*C) + 4*a^2*b^3*(-29*A + 15*B + 10*C) - a^4*b*(17*A + 45*(B + C)) + a^5*(9*A + 5*(B + 3*C)))*EllipticF[I*ArcSi
nh[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Sec[(c + d*x)/2]^2*(-((a*Sec[(c + d*x)/2]^2*Sin[c + d*x])/(a + b)) + (
(b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2])/(a + b)))/Sqrt[((b + a*Cos[c + d*x])*Sec[(c + d*x)/2
]^2)/(a + b)] - (a*(a + b)*(128*A*b^5 - 16*a*b^4*(6*A + 5*B) + 2*a^3*b^2*(36*A + 40*B - 15*C) + 4*a^2*b^3*(-29
*A + 15*B + 10*C) - a^4*b*(17*A + 45*(B + C)) + a^5*(9*A + 5*(B + 3*C)))*Sec[(c + d*x)/2]^4*Sqrt[((b + a*Cos[c
+ d*x])*Sec[(c + d*x)/2]^2)/(a + b)])/(2*Sqrt[1 + Tan[(c + d*x)/2]^2]*Sqrt[1 + ((-a + b)*Tan[(c + d*x)/2]^2)/
(a + b)]) + ((a + b)*(128*A*b^6 - 40*a^5*b*B + 140*a^3*b^3*B - 80*a*b^5*B + 5*a^4*b^2*(11*A - 15*C) + 3*a^6*(3
*A + 5*C) + 4*a^2*b^4*(-53*A + 10*C))*Sec[(c + d*x)/2]^4*Sqrt[((b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2)/(a + b
)]*Sqrt[1 + ((-a + b)*Tan[(c + d*x)/2]^2)/(a + b)])/(2*Sqrt[1 + Tan[(c + d*x)/2]^2])))/(15*a^5*(a^2 - b^2)^2*S
qrt[b + a*Cos[c + d*x]]) - (2*Cos[c + d*x]^(3/2)*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*((-I)*(a + b)*(128*A*b^
6 - 40*a^5*b*B + 140*a^3*b^3*B - 80*a*b^5*B + 5*a^4*b^2*(11*A - 15*C) + 3*a^6*(3*A + 5*C) + 4*a^2*b^4*(-53*A +
10*C))*EllipticE[I*ArcSinh[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Sec[(c + d*x)/2]^2*Sqrt[((b + a*Cos[c + d*x])
*Sec[(c + d*x)/2]^2)/(a + b)] + I*a*(a + b)*(128*A*b^5 - 16*a*b^4*(6*A + 5*B) + 2*a^3*b^2*(36*A + 40*B - 15*C)
+ 4*a^2*b^3*(-29*A + 15*B + 10*C) - a^4*b*(17*A + 45*(B + C)) + a^5*(9*A + 5*(B + 3*C)))*EllipticF[I*ArcSinh[
Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Sec[(c + d*x)/2]^2*Sqrt[((b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2)/(a + b)
] - (128*A*b^6 - 40*a^5*b*B + 140*a^3*b^3*B - 80*a*b^5*B + 5*a^4*b^2*(11*A - 15*C) + 3*a^6*(3*A + 5*C) + 4*a^2
*b^4*(-53*A + 10*C))*(b + a*Cos[c + d*x])*(Sec[(c + d*x)/2]^2)^(3/2)*Tan[(c + d*x)/2])*(-(Cos[(c + d*x)/2]*Sec
[c + d*x]*Sin[(c + d*x)/2]) + Cos[(c + d*x)/2]^2*Sec[c + d*x]*Tan[c + d*x]))/(5*a^5*(a^2 - b^2)^2*Sqrt[b + a*C
os[c + d*x]])))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(10154\) vs. \(2(681)=1362\).

Time = 20.93 (sec) , antiderivative size = 10155, normalized size of antiderivative = 15.32

method	result	size
default	\(\text {Expression too large to display}\)	\(10155\)

[In]

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

[Out]

result too large to display

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.34 (sec) , antiderivative size = 1793, normalized size of antiderivative = 2.70 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{5/2}} \, dx=\text {Too large to display} \]

[In]

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

[Out]

1/45*(6*(5*B*a^7*b^2 - 2*(7*A - 20*C)*a^6*b^3 - 65*B*a^5*b^4 + 2*(49*A - 10*C)*a^4*b^5 + 40*B*a^3*b^6 - 64*A*a
^2*b^7 + 3*(A*a^9 - 2*A*a^7*b^2 + A*a^5*b^4)*cos(d*x + c)^3 + (5*B*a^9 - 8*A*a^8*b - 10*B*a^7*b^2 + 16*A*a^6*b
^3 + 5*B*a^5*b^4 - 8*A*a^4*b^5)*cos(d*x + c)^2 + 5*(2*B*a^8*b - (5*A - 9*C)*a^7*b^2 - 16*B*a^6*b^3 + 5*(5*A -
C)*a^5*b^4 + 10*B*a^4*b^5 - 16*A*a^3*b^6)*cos(d*x + c))*sqrt((a*cos(d*x + c) + b)/cos(d*x + c))*sqrt(cos(d*x +
c))*sin(d*x + c) + (sqrt(2)*(-15*I*B*a^9 + 6*I*(7*A + 20*C)*a^8*b - 185*I*B*a^7*b^2 + 2*I*(121*A - 90*C)*a^6*
b^3 + 340*I*B*a^5*b^4 - 40*I*(13*A - 2*C)*a^4*b^5 - 160*I*B*a^3*b^6 + 256*I*A*a^2*b^7)*cos(d*x + c)^2 - 2*sqrt
(2)*(15*I*B*a^8*b - 6*I*(7*A + 20*C)*a^7*b^2 + 185*I*B*a^6*b^3 - 2*I*(121*A - 90*C)*a^5*b^4 - 340*I*B*a^4*b^5
+ 40*I*(13*A - 2*C)*a^3*b^6 + 160*I*B*a^2*b^7 - 256*I*A*a*b^8)*cos(d*x + c) + sqrt(2)*(-15*I*B*a^7*b^2 + 6*I*(
7*A + 20*C)*a^6*b^3 - 185*I*B*a^5*b^4 + 2*I*(121*A - 90*C)*a^4*b^5 + 340*I*B*a^3*b^6 - 40*I*(13*A - 2*C)*a^2*b
^7 - 160*I*B*a*b^8 + 256*I*A*b^9))*sqrt(a)*weierstrassPInverse(-4/3*(3*a^2 - 4*b^2)/a^2, 8/27*(9*a^2*b - 8*b^3
)/a^3, 1/3*(3*a*cos(d*x + c) + 3*I*a*sin(d*x + c) + 2*b)/a) + (sqrt(2)*(15*I*B*a^9 - 6*I*(7*A + 20*C)*a^8*b +
185*I*B*a^7*b^2 - 2*I*(121*A - 90*C)*a^6*b^3 - 340*I*B*a^5*b^4 + 40*I*(13*A - 2*C)*a^4*b^5 + 160*I*B*a^3*b^6 -
256*I*A*a^2*b^7)*cos(d*x + c)^2 - 2*sqrt(2)*(-15*I*B*a^8*b + 6*I*(7*A + 20*C)*a^7*b^2 - 185*I*B*a^6*b^3 + 2*I
*(121*A - 90*C)*a^5*b^4 + 340*I*B*a^4*b^5 - 40*I*(13*A - 2*C)*a^3*b^6 - 160*I*B*a^2*b^7 + 256*I*A*a*b^8)*cos(d
*x + c) + sqrt(2)*(15*I*B*a^7*b^2 - 6*I*(7*A + 20*C)*a^6*b^3 + 185*I*B*a^5*b^4 - 2*I*(121*A - 90*C)*a^4*b^5 -
340*I*B*a^3*b^6 + 40*I*(13*A - 2*C)*a^2*b^7 + 160*I*B*a*b^8 - 256*I*A*b^9))*sqrt(a)*weierstrassPInverse(-4/3*(
3*a^2 - 4*b^2)/a^2, 8/27*(9*a^2*b - 8*b^3)/a^3, 1/3*(3*a*cos(d*x + c) - 3*I*a*sin(d*x + c) + 2*b)/a) - 3*(sqrt
(2)*(-3*I*(3*A + 5*C)*a^9 + 40*I*B*a^8*b - 5*I*(11*A - 15*C)*a^7*b^2 - 140*I*B*a^6*b^3 + 4*I*(53*A - 10*C)*a^5
*b^4 + 80*I*B*a^4*b^5 - 128*I*A*a^3*b^6)*cos(d*x + c)^2 + 2*sqrt(2)*(-3*I*(3*A + 5*C)*a^8*b + 40*I*B*a^7*b^2 -
5*I*(11*A - 15*C)*a^6*b^3 - 140*I*B*a^5*b^4 + 4*I*(53*A - 10*C)*a^4*b^5 + 80*I*B*a^3*b^6 - 128*I*A*a^2*b^7)*c
os(d*x + c) + sqrt(2)*(-3*I*(3*A + 5*C)*a^7*b^2 + 40*I*B*a^6*b^3 - 5*I*(11*A - 15*C)*a^5*b^4 - 140*I*B*a^4*b^5
+ 4*I*(53*A - 10*C)*a^3*b^6 + 80*I*B*a^2*b^7 - 128*I*A*a*b^8))*sqrt(a)*weierstrassZeta(-4/3*(3*a^2 - 4*b^2)/a
^2, 8/27*(9*a^2*b - 8*b^3)/a^3, weierstrassPInverse(-4/3*(3*a^2 - 4*b^2)/a^2, 8/27*(9*a^2*b - 8*b^3)/a^3, 1/3*
(3*a*cos(d*x + c) + 3*I*a*sin(d*x + c) + 2*b)/a)) - 3*(sqrt(2)*(3*I*(3*A + 5*C)*a^9 - 40*I*B*a^8*b + 5*I*(11*A
- 15*C)*a^7*b^2 + 140*I*B*a^6*b^3 - 4*I*(53*A - 10*C)*a^5*b^4 - 80*I*B*a^4*b^5 + 128*I*A*a^3*b^6)*cos(d*x + c
)^2 + 2*sqrt(2)*(3*I*(3*A + 5*C)*a^8*b - 40*I*B*a^7*b^2 + 5*I*(11*A - 15*C)*a^6*b^3 + 140*I*B*a^5*b^4 - 4*I*(5
3*A - 10*C)*a^4*b^5 - 80*I*B*a^3*b^6 + 128*I*A*a^2*b^7)*cos(d*x + c) + sqrt(2)*(3*I*(3*A + 5*C)*a^7*b^2 - 40*I
*B*a^6*b^3 + 5*I*(11*A - 15*C)*a^5*b^4 + 140*I*B*a^4*b^5 - 4*I*(53*A - 10*C)*a^3*b^6 - 80*I*B*a^2*b^7 + 128*I*
A*a*b^8))*sqrt(a)*weierstrassZeta(-4/3*(3*a^2 - 4*b^2)/a^2, 8/27*(9*a^2*b - 8*b^3)/a^3, weierstrassPInverse(-4
/3*(3*a^2 - 4*b^2)/a^2, 8/27*(9*a^2*b - 8*b^3)/a^3, 1/3*(3*a*cos(d*x + c) - 3*I*a*sin(d*x + c) + 2*b)/a)))/((a
^12 - 2*a^10*b^2 + a^8*b^4)*d*cos(d*x + c)^2 + 2*(a^11*b - 2*a^9*b^3 + a^7*b^5)*d*cos(d*x + c) + (a^10*b^2 - 2
*a^8*b^4 + a^6*b^6)*d)

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{5/2}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F(-1)]

Timed out. \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{5/2}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Giac [F]

\[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{5/2}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{\frac {5}{2}}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{5/2}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^{5/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]

[In]

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

[Out]

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