3.1351 \(\int \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^{5/2} (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=419 \[ \frac {\sqrt {\cos (c+d x)} \left (6 a^2 (3 A+5 C)+70 a b B+b^2 (46 A-15 C)\right ) \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{15 d \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}+\frac {\left (10 a^3 B+4 a^2 b (4 A+15 C)+20 a b^2 B-b^3 (16 A-15 C)\right ) \sqrt {\frac {a \cos (c+d x)+b}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{15 d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}-\frac {b \sin (c+d x) (10 a B+16 A b-15 b C) \sqrt {a+b \sec (c+d x)}}{15 d \sqrt {\cos (c+d x)}}+\frac {2 (a B+A b) \sin (c+d x) \sqrt {\cos (c+d x)} (a+b \sec (c+d x))^{3/2}}{3 d}+\frac {2 A \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^{5/2}}{5 d}+\frac {b^2 (5 a C+2 b B) \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \Pi \left (2;\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}} \]

[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*a^3*B+20*a*b^2*B-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*a*b*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]  time = 1.83, antiderivative size = 419, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 14, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.311, Rules used = {4265, 4094, 4096, 4108, 3859, 2807, 2805, 4035, 3856, 2655, 2653, 3858, 2663, 2661} \[ \frac {\left (4 a^2 b (4 A+15 C)+10 a^3 B+20 a b^2 B-b^3 (16 A-15 C)\right ) \sqrt {\frac {a \cos (c+d x)+b}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{15 d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {\sqrt {\cos (c+d x)} \left (6 a^2 (3 A+5 C)+70 a b B+b^2 (46 A-15 C)\right ) \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{15 d \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}-\frac {b \sin (c+d x) (10 a B+16 A b-15 b C) \sqrt {a+b \sec (c+d x)}}{15 d \sqrt {\cos (c+d x)}}+\frac {2 (a B+A b) \sin (c+d x) \sqrt {\cos (c+d x)} (a+b \sec (c+d x))^{3/2}}{3 d}+\frac {2 A \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^{5/2}}{5 d}+\frac {b^2 (5 a C+2 b B) \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \Pi \left (2;\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}} \]

Antiderivative was successfully verified.

[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 2653

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

Rule 2655

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*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
b^2, 0] &&  !GtQ[a + b, 0]

Rule 2661

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

Rule 2663

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*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] &&  !GtQ[a + b, 0]

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 2807

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*
Sin[e + f*x])/(c + d)]), 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 3856

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 3858

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[(Sqrt[d*
Csc[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 3859

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(3/2)/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[(d*Sqr
t[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 4035

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 4094

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 4096

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 4108

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 4265

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*} \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 &=\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}} \Pi \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}} F\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}} \Pi \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*}

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Mathematica [C]  time = 34.41, size = 86542, normalized size = 206.54 \[ \text {Result too large to show} \]

Warning: Unable to verify antiderivative.

[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

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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(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

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [C]  time = 2.45, size = 2893, normalized size = 6.90 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[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]

-1/15/d*((b+a*cos(d*x+c))/cos(d*x+c))^(1/2)*(1+cos(d*x+c))^2*(-1+cos(d*x+c))^3*(15*C*((a-b)/(a+b))^(1/2)*b^3*(
1/(1+cos(d*x+c)))^(3/2)*sin(d*x+c)-34*A*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c)
)*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*cos(d*x+c)*a^2*b+70*B*sin(d*x+c)*((a-b)/(a+b))^(1/2)*co
s(d*x+c)*a*b^2*(1/(1+cos(d*x+c)))^(3/2)+30*C*((a-b)/(a+b))^(1/2)*cos(d*x+c)*a^2*b*(1/(1+cos(d*x+c)))^(3/2)*sin
(d*x+c)+28*A*sin(d*x+c)*((a-b)/(a+b))^(1/2)*cos(d*x+c)^3*a^2*b*(1/(1+cos(d*x+c)))^(3/2)+68*A*sin(d*x+c)*((a-b)
/(a+b))^(1/2)*cos(d*x+c)^2*a*b^2*(1/(1+cos(d*x+c)))^(3/2)+80*B*sin(d*x+c)*((a-b)/(a+b))^(1/2)*cos(d*x+c)^2*a^2
*b*(1/(1+cos(d*x+c)))^(3/2)+18*A*cos(d*x+c)^2*sin(d*x+c)*((a-b)/(a+b))^(1/2)*a^3*(1/(1+cos(d*x+c)))^(3/2)+10*B
*cos(d*x+c)^2*sin(d*x+c)*((a-b)/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(3/2)*a^3+6*A*cos(d*x+c)^3*sin(d*x+c)*((a-b)/(
a+b))^(1/2)*a^3*(1/(1+cos(d*x+c)))^(3/2)+60*C*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(
d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*cos(d*x+c)*a*b^2-150*C*((b+a*cos(d*x+c))/(1+cos(d
*x+c))/(a+b))^(1/2)*EllipticPi((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(a+b)/(a-b),I/((a-b)/(a+b))^(1/2
))*cos(d*x+c)*a*b^2+30*C*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))
^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*cos(d*x+c)*a^2*b+15*C*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*El
lipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*cos(d*x+c)*a*b^2+30*C*((a-b)/(a+b
))^(1/2)*cos(d*x+c)^2*a^3*(1/(1+cos(d*x+c)))^(3/2)*sin(d*x+c)+46*A*sin(d*x+c)*((a-b)/(a+b))^(1/2)*cos(d*x+c)*b
^3*(1/(1+cos(d*x+c)))^(3/2)+10*B*((a-b)/(a+b))^(1/2)*cos(d*x+c)^3*sin(d*x+c)*(1/(1+cos(d*x+c)))^(3/2)*a^3+46*A
*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b
)/(a-b))^(1/2))*cos(d*x+c)*a*b^2+18*A*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))*
((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*cos(d*x+c)*a^2*b-46*A*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a
+b))^(1/2)*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*cos(d*x+c)*a*b^2+6*A
*sin(d*x+c)*((a-b)/(a+b))^(1/2)*cos(d*x+c)^4*a^3*(1/(1+cos(d*x+c)))^(3/2)+70*B*EllipticF((-1+cos(d*x+c))*((a-b
)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*cos(d*x+c)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*a^2*b
-90*B*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*cos(d*x+c)*((b+a*cos(d*x+
c))/(1+cos(d*x+c))/(a+b))^(1/2)*a*b^2-70*B*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a
-b))^(1/2))*cos(d*x+c)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*a^2*b+70*B*((b+a*cos(d*x+c))/(1+cos(d*x+c
))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*cos(d*x+c)*a*b^
2-90*C*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),
(-(a+b)/(a-b))^(1/2))*cos(d*x+c)*a^2*b+15*C*cos(d*x+c)*((a-b)/(a+b))^(1/2)*a*b^2*(1/(1+cos(d*x+c)))^(3/2)*sin(
d*x+c)+18*A*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*
x+c),(-(a+b)/(a-b))^(1/2))*cos(d*x+c)*a^3-30*A*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos
(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*cos(d*x+c)*b^3-18*A*((b+a*cos(d*x+c))/(1+cos(d*x
+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*cos(d*x+c)*a^
3+46*A*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),
(-(a+b)/(a-b))^(1/2))*cos(d*x+c)*b^3-10*B*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-
b))^(1/2))*cos(d*x+c)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*a^3+30*B*EllipticF((-1+cos(d*x+c))*((a-b)/
(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*cos(d*x+c)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*b^3-60*
B*EllipticPi((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(a+b)/(a-b),I/((a-b)/(a+b))^(1/2))*cos(d*x+c)*((b+
a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*b^3+30*C*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1
+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*cos(d*x+c)*a^3-30*C*((b+a*cos(d*x+c))/(1+cos
(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*cos(d*x+c
)*a^3-15*C*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x
+c),(-(a+b)/(a-b))^(1/2))*cos(d*x+c)*b^3+28*A*cos(d*x+c)^2*sin(d*x+c)*((a-b)/(a+b))^(1/2)*a^2*b*(1/(1+cos(d*x+
c)))^(3/2)+18*A*cos(d*x+c)*sin(d*x+c)*((a-b)/(a+b))^(1/2)*a^2*b*(1/(1+cos(d*x+c)))^(3/2)+22*A*cos(d*x+c)*sin(d
*x+c)*((a-b)/(a+b))^(1/2)*a*b^2*(1/(1+cos(d*x+c)))^(3/2)+10*B*cos(d*x+c)*sin(d*x+c)*((a-b)/(a+b))^(1/2)*a^2*b*
(1/(1+cos(d*x+c)))^(3/2))/((a-b)/(a+b))^(1/2)/(b+a*cos(d*x+c))/sin(d*x+c)^6/(1/(1+cos(d*x+c)))^(3/2)/cos(d*x+c
)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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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(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

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