\(\int \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^{5/2} (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [1351]

Optimal result

Integrand size = 45, antiderivative size = 419 \[ \int \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {\left (10 a^3 B+20 a b^2 B-b^3 (16 A-15 C)+4 a^2 b (4 A+15 C)\right ) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right )}{15 d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {b^2 (2 b B+5 a C) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right )}{d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {\left (70 a b B+b^2 (46 A-15 C)+6 a^2 (3 A+5 C)\right ) \sqrt {\cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right ) \sqrt {a+b \sec (c+d x)}}{15 d \sqrt {\frac {b+a \cos (c+d x)}{a+b}}}-\frac {b (16 A b+10 a B-15 b C) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{15 d \sqrt {\cos (c+d x)}}+\frac {2 (A b+a B) \sqrt {\cos (c+d x)} (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{3 d}+\frac {2 A \cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^{5/2} \sin (c+d x)}{5 d} \]

[Out]

2/5*A*cos(d*x+c)^(3/2)*(a+b*sec(d*x+c))^(5/2)*sin(d*x+c)/d+2/3*(A*b+B*a)*(a+b*sec(d*x+c))^(3/2)*sin(d*x+c)*cos
(d*x+c)^(1/2)/d+1/15*(10*B*a^3+20*B*a*b^2-b^3*(16*A-15*C)+4*a^2*b*(4*A+15*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)/d/cos(d*x
+c)^(1/2)/(a+b*sec(d*x+c))^(1/2)+b^2*(2*B*b+5*C*a)*(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),2,2^(1/2)*(a/(a+b))^(1/2))*((b+a*cos(d*x+c))/(a+b))^(1/2)/d/cos(d*x+c)^(1/2)/(a+b*sec(d*x+c
))^(1/2)-1/15*b*(16*A*b+10*B*a-15*C*b)*sin(d*x+c)*(a+b*sec(d*x+c))^(1/2)/d/cos(d*x+c)^(1/2)+1/15*(70*B*a*b+b^2
*(46*A-15*C)+6*a^2*(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)/d/((b+a*cos(d*x+c))/(a+b))^(1/2)

Rubi [A] (verified)

Time = 2.18 (sec) , antiderivative size = 419, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.311, Rules used = {4350, 4179, 4181, 4193, 3944, 2886, 2884, 4120, 3941, 2734, 2732, 3943, 2742, 2740} \[ \int \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {\sqrt {\cos (c+d x)} \left (6 a^2 (3 A+5 C)+70 a b B+b^2 (46 A-15 C)\right ) \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{15 d \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}+\frac {\left (10 a^3 B+4 a^2 b (4 A+15 C)+20 a b^2 B-b^3 (16 A-15 C)\right ) \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right )}{15 d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}-\frac {b \sin (c+d x) (10 a B+16 A b-15 b C) \sqrt {a+b \sec (c+d x)}}{15 d \sqrt {\cos (c+d x)}}+\frac {2 (a B+A b) \sin (c+d x) \sqrt {\cos (c+d x)} (a+b \sec (c+d x))^{3/2}}{3 d}+\frac {2 A \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^{5/2}}{5 d}+\frac {b^2 (5 a C+2 b B) \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right )}{d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}} \]

[In]

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

[Out]

((10*a^3*B + 20*a*b^2*B - b^3*(16*A - 15*C) + 4*a^2*b*(4*A + 15*C))*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*Ellipti
cF[(c + d*x)/2, (2*a)/(a + b)])/(15*d*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Sec[c + d*x]]) + (b^2*(2*b*B + 5*a*C)*Sqrt
[(b + a*Cos[c + d*x])/(a + b)]*EllipticPi[2, (c + d*x)/2, (2*a)/(a + b)])/(d*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Sec
[c + d*x]]) + ((70*a*b*B + b^2*(46*A - 15*C) + 6*a^2*(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*d*Sqrt[(b + a*Cos[c + d*x])/(a + b)]) - (b*(16*A*b + 10*a*B - 15*b*
C)*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(15*d*Sqrt[Cos[c + d*x]]) + (2*(A*b + a*B)*Sqrt[Cos[c + d*x]]*(a + b
*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(3*d) + (2*A*Cos[c + d*x]^(3/2)*(a + b*Sec[c + d*x])^(5/2)*Sin[c + d*x])/(5
*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 2884

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2/(f*(a + b)*Sqrt[c + d]))*EllipticPi[2*(b/(a + b)), (1/2)*(e - Pi/2 + f*x), 2*(d/(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 2886

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist
[Sqrt[(c + d*Sin[e + f*x])/(c + d)]/Sqrt[c + d*Sin[e + f*x]], Int[1/((a + b*Sin[e + f*x])*Sqrt[c/(c + d) + (d/
(c + d))*Sin[e + f*x]]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && N
eQ[c^2 - d^2, 0] &&  !GtQ[c + d, 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 3944

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(3/2)/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[d*Sqrt
[d*Csc[e + f*x]]*(Sqrt[b + a*Sin[e + f*x]]/Sqrt[a + b*Csc[e + f*x]]), Int[1/(Sin[e + f*x]*Sqrt[b + a*Sin[e + f
*x]]), x], x] /; FreeQ[{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 4179

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*((d*
Csc[e + f*x])^n/(f*n)), x] - Dist[1/(d*n), Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Simp[A*b*
m - a*B*n - (b*B*n + a*(C*n + A*(n + 1)))*Csc[e + f*x] - b*(C*n + A*(m + n + 1))*Csc[e + f*x]^2, x], x], x] /;
FreeQ[{a, b, d, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 0] && LeQ[n, -1]

Rule 4181

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[(-C)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(
(d*Csc[e + f*x])^n/(f*(m + n + 1))), x] + Dist[1/(m + n + 1), Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x]
)^n*Simp[a*A*(m + n + 1) + a*C*n + ((A*b + a*B)*(m + n + 1) + b*C*(m + n))*Csc[e + f*x] + (b*B*(m + n + 1) + a
*C*m)*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] && GtQ[m, 0] &&
!LeQ[n, -1]

Rule 4193

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(d
_.)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]), x_Symbol] :> Dist[C/d^2, Int[(d*Csc[e + f*x])^(3/2)/Sqrt[a +
b*Csc[e + f*x]], x], x] + Int[(A + B*Csc[e + f*x])/(Sqrt[d*Csc[e + f*x]]*Sqrt[a + b*Csc[e + f*x]]), x] /; Fre
eQ[{a, b, d, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0]

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))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx \\ & = \frac {2 A \cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^{5/2} \sin (c+d x)}{5 d}+\frac {1}{5} \left (2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+b \sec (c+d x))^{3/2} \left (\frac {5}{2} (A b+a B)+\frac {1}{2} (3 a A+5 b B+5 a C) \sec (c+d x)-\frac {1}{2} b (2 A-5 C) \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx \\ & = \frac {2 (A b+a B) \sqrt {\cos (c+d x)} (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{3 d}+\frac {2 A \cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^{5/2} \sin (c+d x)}{5 d}+\frac {1}{15} \left (4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {a+b \sec (c+d x)} \left (\frac {3}{4} \left (5 A b^2+10 a b B+a^2 (3 A+5 C)\right )+\frac {1}{4} \left (8 a A b+5 a^2 B+15 b^2 B+30 a b C\right ) \sec (c+d x)-\frac {1}{4} b (16 A b+10 a B-15 b C) \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx \\ & = -\frac {b (16 A b+10 a B-15 b C) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{15 d \sqrt {\cos (c+d x)}}+\frac {2 (A b+a B) \sqrt {\cos (c+d x)} (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{3 d}+\frac {2 A \cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^{5/2} \sin (c+d x)}{5 d}+\frac {1}{15} \left (4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {1}{8} a \left (70 a b B+b^2 (46 A-15 C)+6 a^2 (3 A+5 C)\right )+\frac {1}{4} \left (15 A b^3+5 a^3 B+45 a b^2 B+a^2 b (17 A+45 C)\right ) \sec (c+d x)+\frac {15}{8} b^2 (2 b B+5 a C) \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}} \, dx \\ & = -\frac {b (16 A b+10 a B-15 b C) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{15 d \sqrt {\cos (c+d x)}}+\frac {2 (A b+a B) \sqrt {\cos (c+d x)} (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{3 d}+\frac {2 A \cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^{5/2} \sin (c+d x)}{5 d}+\frac {1}{15} \left (4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {1}{8} a \left (70 a b B+b^2 (46 A-15 C)+6 a^2 (3 A+5 C)\right )+\frac {1}{4} \left (15 A b^3+5 a^3 B+45 a b^2 B+a^2 b (17 A+45 C)\right ) \sec (c+d x)}{\sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}} \, dx+\frac {1}{2} \left (b^2 (2 b B+5 a C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx \\ & = -\frac {b (16 A b+10 a B-15 b C) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{15 d \sqrt {\cos (c+d x)}}+\frac {2 (A b+a B) \sqrt {\cos (c+d x)} (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{3 d}+\frac {2 A \cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^{5/2} \sin (c+d x)}{5 d}+\frac {1}{30} \left (\left (70 a b B+b^2 (46 A-15 C)+6 a^2 (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-\frac {1}{30} \left (\left (-10 a^3 B-20 a b^2 B+b^3 (16 A-15 C)-4 a^2 b (4 A+15 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+\frac {\left (b^2 (2 b B+5 a C) \sqrt {b+a \cos (c+d x)}\right ) \int \frac {\sec (c+d x)}{\sqrt {b+a \cos (c+d x)}} \, dx}{2 \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}} \\ & = -\frac {b (16 A b+10 a B-15 b C) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{15 d \sqrt {\cos (c+d x)}}+\frac {2 (A b+a B) \sqrt {\cos (c+d x)} (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{3 d}+\frac {2 A \cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^{5/2} \sin (c+d x)}{5 d}-\frac {\left (\left (-10 a^3 B-20 a b^2 B+b^3 (16 A-15 C)-4 a^2 b (4 A+15 C)\right ) \sqrt {b+a \cos (c+d x)}\right ) \int \frac {1}{\sqrt {b+a \cos (c+d x)}} \, dx}{30 \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {\left (b^2 (2 b B+5 a C) \sqrt {\frac {b+a \cos (c+d x)}{a+b}}\right ) \int \frac {\sec (c+d x)}{\sqrt {\frac {b}{a+b}+\frac {a \cos (c+d x)}{a+b}}} \, dx}{2 \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {\left (\left (70 a b B+b^2 (46 A-15 C)+6 a^2 (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}{30 \sqrt {b+a \cos (c+d x)}} \\ & = \frac {b^2 (2 b B+5 a C) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right )}{d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}-\frac {b (16 A b+10 a B-15 b C) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{15 d \sqrt {\cos (c+d x)}}+\frac {2 (A b+a B) \sqrt {\cos (c+d x)} (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{3 d}+\frac {2 A \cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^{5/2} \sin (c+d x)}{5 d}-\frac {\left (\left (-10 a^3 B-20 a b^2 B+b^3 (16 A-15 C)-4 a^2 b (4 A+15 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}{30 \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {\left (\left (70 a b B+b^2 (46 A-15 C)+6 a^2 (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}{30 \sqrt {\frac {b+a \cos (c+d x)}{a+b}}} \\ & = \frac {\left (10 a^3 B+20 a b^2 B-b^3 (16 A-15 C)+4 a^2 b (4 A+15 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 d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {b^2 (2 b B+5 a C) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right )}{d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {\left (70 a b B+b^2 (46 A-15 C)+6 a^2 (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 d \sqrt {\frac {b+a \cos (c+d x)}{a+b}}}-\frac {b (16 A b+10 a B-15 b C) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{15 d \sqrt {\cos (c+d x)}}+\frac {2 (A b+a B) \sqrt {\cos (c+d x)} (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{3 d}+\frac {2 A \cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^{5/2} \sin (c+d x)}{5 d} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 46.77 (sec) , antiderivative size = 244012, normalized size of antiderivative = 582.37 \[ \int \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Result too large to show} \]

[In]

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

[Out]

Result too large to show

Maple [F(-1)]

Timed out.

hanged

[In]

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

[Out]

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

Fricas [F(-1)]

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

[In]

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

[Out]

Timed out

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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