\(\int \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^4 (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [1315]
Optimal result
Integrand size = 43, antiderivative size = 388 \[
\int \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 \left (20 a^3 b B-20 a b^3 B+30 a^2 b^2 (A-C)-b^4 (5 A+3 C)+a^4 (3 A+5 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 \left (a^4 B+18 a^2 b^2 B+b^4 B+4 a b^3 (3 A+C)+4 a^3 b (A+3 C)\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 a \left (5 a^3 B-105 a b^2 B+4 a^2 b (5 A-33 C)-6 b^3 (5 A+3 C)\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{15 d}-\frac {2 a^2 \left (50 a b B-a^2 (3 A-59 C)+3 b^2 (5 A+3 C)\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 \left (5 A b^2+15 a b B+16 a^2 C+3 b^2 C\right ) (b+a \cos (c+d x))^2 \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {2 (5 b B+8 a C) (b+a \cos (c+d x))^3 \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 C (b+a \cos (c+d x))^4 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}
\]
[Out]
2/5*(20*B*a^3*b-20*B*a*b^3+30*a^2*b^2*(A-C)-b^4*(5*A+3*C)+a^4*(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))/d+2/3*(B*a^4+18*B*a^2*b^2+B*b^4+4*a*b^3*(3*A+C)+4*a^3*b*(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))/d-2/15*a^2*(50*B*a*b
-a^2*(3*A-59*C)+3*b^2*(5*A+3*C))*cos(d*x+c)^(3/2)*sin(d*x+c)/d+2/15*(5*B*b+8*C*a)*(b+a*cos(d*x+c))^3*sin(d*x+c
)/d/cos(d*x+c)^(3/2)+2/5*C*(b+a*cos(d*x+c))^4*sin(d*x+c)/d/cos(d*x+c)^(5/2)+2/5*(5*A*b^2+15*B*a*b+16*C*a^2+3*C
*b^2)*(b+a*cos(d*x+c))^2*sin(d*x+c)/d/cos(d*x+c)^(1/2)+2/15*a*(5*B*a^3-105*B*a*b^2+4*a^2*b*(5*A-33*C)-6*b^3*(5
*A+3*C))*sin(d*x+c)*cos(d*x+c)^(1/2)/d
Rubi [A] (verified)
Time = 1.42 (sec) , antiderivative size = 388, normalized size of antiderivative = 1.00, number of
steps used = 9, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.163, Rules used = {4197, 3126, 3112, 3102, 2827,
2720, 2719} \[
\int \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {2 a^2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (-\left (a^2 (3 A-59 C)\right )+50 a b B+3 b^2 (5 A+3 C)\right )}{15 d}+\frac {2 \sin (c+d x) \left (16 a^2 C+15 a b B+5 A b^2+3 b^2 C\right ) (a \cos (c+d x)+b)^2}{5 d \sqrt {\cos (c+d x)}}+\frac {2 a \sin (c+d x) \sqrt {\cos (c+d x)} \left (5 a^3 B+4 a^2 b (5 A-33 C)-105 a b^2 B-6 b^3 (5 A+3 C)\right )}{15 d}+\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (a^4 B+4 a^3 b (A+3 C)+18 a^2 b^2 B+4 a b^3 (3 A+C)+b^4 B\right )}{3 d}+\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^4 (3 A+5 C)+20 a^3 b B+30 a^2 b^2 (A-C)-20 a b^3 B-b^4 (5 A+3 C)\right )}{5 d}+\frac {2 (8 a C+5 b B) \sin (c+d x) (a \cos (c+d x)+b)^3}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^4}{5 d \cos ^{\frac {5}{2}}(c+d x)}
\]
[In]
Int[Cos[c + d*x]^(5/2)*(a + b*Sec[c + d*x])^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
[Out]
(2*(20*a^3*b*B - 20*a*b^3*B + 30*a^2*b^2*(A - C) - b^4*(5*A + 3*C) + a^4*(3*A + 5*C))*EllipticE[(c + d*x)/2, 2
])/(5*d) + (2*(a^4*B + 18*a^2*b^2*B + b^4*B + 4*a*b^3*(3*A + C) + 4*a^3*b*(A + 3*C))*EllipticF[(c + d*x)/2, 2]
)/(3*d) + (2*a*(5*a^3*B - 105*a*b^2*B + 4*a^2*b*(5*A - 33*C) - 6*b^3*(5*A + 3*C))*Sqrt[Cos[c + d*x]]*Sin[c + d
*x])/(15*d) - (2*a^2*(50*a*b*B - a^2*(3*A - 59*C) + 3*b^2*(5*A + 3*C))*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(15*d)
+ (2*(5*A*b^2 + 15*a*b*B + 16*a^2*C + 3*b^2*C)*(b + a*Cos[c + d*x])^2*Sin[c + d*x])/(5*d*Sqrt[Cos[c + d*x]])
+ (2*(5*b*B + 8*a*C)*(b + a*Cos[c + d*x])^3*Sin[c + d*x])/(15*d*Cos[c + d*x]^(3/2)) + (2*C*(b + a*Cos[c + d*x]
)^4*Sin[c + d*x])/(5*d*Cos[c + d*x]^(5/2))
Rule 2719
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{
c, d}, x]
Rule 2720
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]
Rule 2827
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
Rule 3102
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Dist[1/(
b*(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x],
x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && !LtQ[m, -1]
Rule 3112
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])*((A_.) + (B_.)*sin[(e
_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*d*Cos[e + f*x]*Sin[e + f*x]*((a +
b*Sin[e + f*x])^(m + 1)/(b*f*(m + 3))), x] + Dist[1/(b*(m + 3)), Int[(a + b*Sin[e + f*x])^m*Simp[a*C*d + A*b*
c*(m + 3) + b*(B*c*(m + 3) + d*(C*(m + 2) + A*(m + 3)))*Sin[e + f*x] - (2*a*C*d - b*(c*C + B*d)*(m + 3))*Sin[e
+ f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &&
!LtQ[m, -1]
Rule 3126
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(c^2*C - B*c*d + A*d^2))*Cos[e
+ f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 - d^2))), x] + Dist[1/(d*(n + 1)
*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(b*d*m + a*c*(n + 1)) +
(c*C - B*d)*(b*c*m + a*d*(n + 1)) - (d*(A*(a*d*(n + 2) - b*c*(n + 1)) + B*(b*d*(n + 1) - a*c*(n + 2))) - C*(b*
c*d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x] + b*(d*(B*c - A*d)*(m + n + 2) - C*(c^2*(m + 1) + d^2*(n +
1)))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2
, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 0] && LtQ[n, -1]
Rule 4197
Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sec[(e_.)
+ (f_.)*(x_)] + (C_.)*sec[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[d^(m + 2), Int[(b + a*Cos[e + f*x])^m*(d*
Cos[e + f*x])^(n - m - 2)*(C + B*Cos[e + f*x] + A*Cos[e + f*x]^2), x], x] /; FreeQ[{a, b, d, e, f, A, B, C, n}
, x] && !IntegerQ[n] && IntegerQ[m]
Rubi steps \begin{align*}
\text {integral}& = \int \frac {(b+a \cos (c+d x))^4 \left (C+B \cos (c+d x)+A \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx \\ & = \frac {2 C (b+a \cos (c+d x))^4 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2}{5} \int \frac {(b+a \cos (c+d x))^3 \left (\frac {1}{2} (5 b B+8 a C)+\frac {1}{2} (5 A b+5 a B+3 b C) \cos (c+d x)+\frac {5}{2} a (A-C) \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx \\ & = \frac {2 (5 b B+8 a C) (b+a \cos (c+d x))^3 \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 C (b+a \cos (c+d x))^4 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {4}{15} \int \frac {(b+a \cos (c+d x))^2 \left (\frac {3}{4} \left (5 A b^2+15 a b B+16 a^2 C+3 b^2 C\right )+\frac {1}{4} \left (15 a^2 B+5 b^2 B+2 a b (15 A+C)\right ) \cos (c+d x)+\frac {5}{4} a (3 a A-5 b B-11 a C) \cos ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx \\ & = \frac {2 \left (5 A b^2+15 a b B+16 a^2 C+3 b^2 C\right ) (b+a \cos (c+d x))^2 \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {2 (5 b B+8 a C) (b+a \cos (c+d x))^3 \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 C (b+a \cos (c+d x))^4 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {8}{15} \int \frac {(b+a \cos (c+d x)) \left (\frac {1}{8} \left (195 a^2 b B+5 b^3 B+192 a^3 C+2 a b^2 (45 A+19 C)\right )+\frac {1}{8} \left (15 a^3 B-65 a b^2 B+a^2 b (45 A-101 C)-3 b^3 (5 A+3 C)\right ) \cos (c+d x)-\frac {5}{8} a \left (50 a b B-a^2 (3 A-59 C)+3 b^2 (5 A+3 C)\right ) \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx \\ & = -\frac {2 a^2 \left (50 a b B-a^2 (3 A-59 C)+3 b^2 (5 A+3 C)\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 \left (5 A b^2+15 a b B+16 a^2 C+3 b^2 C\right ) (b+a \cos (c+d x))^2 \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {2 (5 b B+8 a C) (b+a \cos (c+d x))^3 \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 C (b+a \cos (c+d x))^4 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {16}{75} \int \frac {\frac {5}{16} b \left (195 a^2 b B+5 b^3 B+192 a^3 C+2 a b^2 (45 A+19 C)\right )+\frac {15}{16} \left (20 a^3 b B-20 a b^3 B+30 a^2 b^2 (A-C)-b^4 (5 A+3 C)+a^4 (3 A+5 C)\right ) \cos (c+d x)+\frac {15}{16} a \left (5 a^3 B-105 a b^2 B+4 a^2 b (5 A-33 C)-6 b^3 (5 A+3 C)\right ) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)}} \, dx \\ & = \frac {2 a \left (5 a^3 B-105 a b^2 B+4 a^2 b (5 A-33 C)-6 b^3 (5 A+3 C)\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{15 d}-\frac {2 a^2 \left (50 a b B-a^2 (3 A-59 C)+3 b^2 (5 A+3 C)\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 \left (5 A b^2+15 a b B+16 a^2 C+3 b^2 C\right ) (b+a \cos (c+d x))^2 \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {2 (5 b B+8 a C) (b+a \cos (c+d x))^3 \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 C (b+a \cos (c+d x))^4 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {32}{225} \int \frac {\frac {75}{32} \left (a^4 B+18 a^2 b^2 B+b^4 B+4 a b^3 (3 A+C)+4 a^3 b (A+3 C)\right )+\frac {45}{32} \left (20 a^3 b B-20 a b^3 B+30 a^2 b^2 (A-C)-b^4 (5 A+3 C)+a^4 (3 A+5 C)\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)}} \, dx \\ & = \frac {2 a \left (5 a^3 B-105 a b^2 B+4 a^2 b (5 A-33 C)-6 b^3 (5 A+3 C)\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{15 d}-\frac {2 a^2 \left (50 a b B-a^2 (3 A-59 C)+3 b^2 (5 A+3 C)\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 \left (5 A b^2+15 a b B+16 a^2 C+3 b^2 C\right ) (b+a \cos (c+d x))^2 \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {2 (5 b B+8 a C) (b+a \cos (c+d x))^3 \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 C (b+a \cos (c+d x))^4 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {1}{3} \left (a^4 B+18 a^2 b^2 B+b^4 B+4 a b^3 (3 A+C)+4 a^3 b (A+3 C)\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx+\frac {1}{5} \left (20 a^3 b B-20 a b^3 B+30 a^2 b^2 (A-C)-b^4 (5 A+3 C)+a^4 (3 A+5 C)\right ) \int \sqrt {\cos (c+d x)} \, dx \\ & = \frac {2 \left (20 a^3 b B-20 a b^3 B+30 a^2 b^2 (A-C)-b^4 (5 A+3 C)+a^4 (3 A+5 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 \left (a^4 B+18 a^2 b^2 B+b^4 B+4 a b^3 (3 A+C)+4 a^3 b (A+3 C)\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 a \left (5 a^3 B-105 a b^2 B+4 a^2 b (5 A-33 C)-6 b^3 (5 A+3 C)\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{15 d}-\frac {2 a^2 \left (50 a b B-a^2 (3 A-59 C)+3 b^2 (5 A+3 C)\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 \left (5 A b^2+15 a b B+16 a^2 C+3 b^2 C\right ) (b+a \cos (c+d x))^2 \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {2 (5 b B+8 a C) (b+a \cos (c+d x))^3 \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 C (b+a \cos (c+d x))^4 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)} \\
\end{align*}
Mathematica [C] (warning: unable to verify)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 17.47 (sec) , antiderivative size = 4016, normalized size of antiderivative = 10.35
\[
\int \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Result too large to show}
\]
[In]
Integrate[Cos[c + d*x]^(5/2)*(a + b*Sec[c + d*x])^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
[Out]
(Cos[c + d*x]^(13/2)*(a + b*Sec[c + d*x])^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((-2*(3*a^4*A + 30*a^2*A*b
^2 - 10*A*b^4 + 20*a^3*b*B - 40*a*b^3*B + 5*a^4*C - 60*a^2*b^2*C - 6*b^4*C + 3*a^4*A*Cos[2*c] + 30*a^2*A*b^2*C
os[2*c] + 20*a^3*b*B*Cos[2*c] + 5*a^4*C*Cos[2*c])*Csc[c]*Sec[c])/(5*d) + (4*a^3*(4*A*b + a*B)*Cos[d*x]*Sin[c])
/(3*d) + (2*a^4*A*Cos[2*d*x]*Sin[2*c])/(5*d) + (4*a^3*(4*A*b + a*B)*Cos[c]*Sin[d*x])/(3*d) + (4*b^4*C*Sec[c]*S
ec[c + d*x]^3*Sin[d*x])/(5*d) + (4*Sec[c]*Sec[c + d*x]^2*(3*b^4*C*Sin[c] + 5*b^4*B*Sin[d*x] + 20*a*b^3*C*Sin[d
*x]))/(15*d) + (4*Sec[c]*Sec[c + d*x]*(5*b^4*B*Sin[c] + 20*a*b^3*C*Sin[c] + 15*A*b^4*Sin[d*x] + 60*a*b^3*B*Sin
[d*x] + 90*a^2*b^2*C*Sin[d*x] + 9*b^4*C*Sin[d*x]))/(15*d) + (2*a^4*A*Cos[2*c]*Sin[2*d*x])/(5*d)))/((b + a*Cos[
c + d*x])^4*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])) - (16*a^3*A*b*Cos[c + d*x]^6*Csc[c]*Hypergeomet
ricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*(a + b*Sec[c + d*x])^4*(A + B*Sec[c + d*x] + C*Sec[c +
d*x]^2)*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x
- ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(3*d*(b + a*Cos[c + d*x])^4*(A + 2*C + 2*B*Cos[c + d
*x] + A*Cos[2*c + 2*d*x])*Sqrt[1 + Cot[c]^2]) - (16*a*A*b^3*Cos[c + d*x]^6*Csc[c]*HypergeometricPFQ[{1/4, 1/2}
, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*(a + b*Sec[c + d*x])^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*Sec[d*x -
ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]
])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(d*(b + a*Cos[c + d*x])^4*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2
*d*x])*Sqrt[1 + Cot[c]^2]) - (4*a^4*B*Cos[c + d*x]^6*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - Arc
Tan[Cot[c]]]^2]*(a + b*Sec[c + d*x])^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[
1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x -
ArcTan[Cot[c]]]])/(3*d*(b + a*Cos[c + d*x])^4*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*Sqrt[1 + Cot[
c]^2]) - (24*a^2*b^2*B*Cos[c + d*x]^6*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]
*(a + b*Sec[c + d*x])^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - A
rcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]
]])/(d*(b + a*Cos[c + d*x])^4*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*Sqrt[1 + Cot[c]^2]) - (4*b^4*B
*Cos[c + d*x]^6*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*(a + b*Sec[c + d*x])^
4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-
(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(3*d*(b + a*Cos[c
+ d*x])^4*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*Sqrt[1 + Cot[c]^2]) - (16*a^3*b*C*Cos[c + d*x]^6*C
sc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*(a + b*Sec[c + d*x])^4*(A + B*Sec[c +
d*x] + C*Sec[c + d*x]^2)*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^
2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(d*(b + a*Cos[c + d*x])^4*(A + 2*C
+ 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*Sqrt[1 + Cot[c]^2]) - (16*a*b^3*C*Cos[c + d*x]^6*Csc[c]*Hypergeometri
cPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*(a + b*Sec[c + d*x])^4*(A + B*Sec[c + d*x] + C*Sec[c + d*
x]^2)*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x -
ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(3*d*(b + a*Cos[c + d*x])^4*(A + 2*C + 2*B*Cos[c + d*x
] + A*Cos[2*c + 2*d*x])*Sqrt[1 + Cot[c]^2]) - (6*a^4*A*Cos[c + d*x]^6*Csc[c]*(a + b*Sec[c + d*x])^4*(A + B*Sec
[c + d*x] + C*Sec[c + d*x]^2)*((HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x +
ArcTan[Tan[c]]]*Tan[c])/(Sqrt[1 - Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[1 + Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[Cos[c]*C
os[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]*Sqrt[1 + Tan[c]^2]) - ((Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/Sqrt[1
+ Tan[c]^2] + (2*Cos[c]^2*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2])/(Cos[c]^2 + Sin[c]^2))/Sqrt[Cos[c]*Cos
[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]))/(5*d*(b + a*Cos[c + d*x])^4*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2
*c + 2*d*x])) - (12*a^2*A*b^2*Cos[c + d*x]^6*Csc[c]*(a + b*Sec[c + d*x])^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x
]^2)*((HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/(
Sqrt[1 - Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[1 + Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*
Sqrt[1 + Tan[c]^2]]*Sqrt[1 + Tan[c]^2]) - ((Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/Sqrt[1 + Tan[c]^2] + (2*Cos[c]^2
*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2])/(Cos[c]^2 + Sin[c]^2))/Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sq
rt[1 + Tan[c]^2]]))/(d*(b + a*Cos[c + d*x])^4*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])) + (2*A*b^4*Co
s[c + d*x]^6*Csc[c]*(a + b*Sec[c + d*x])^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((HypergeometricPFQ[{-1/2,
-1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/(Sqrt[1 - Cos[d*x + ArcTan[Tan[c]
]]]*Sqrt[1 + Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]*Sqrt[1 + Tan
[c]^2]) - ((Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/Sqrt[1 + Tan[c]^2] + (2*Cos[c]^2*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[
1 + Tan[c]^2])/(Cos[c]^2 + Sin[c]^2))/Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]))/(d*(b + a*Co
s[c + d*x])^4*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])) - (8*a^3*b*B*Cos[c + d*x]^6*Csc[c]*(a + b*Sec
[c + d*x])^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan
[Tan[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/(Sqrt[1 - Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[1 + Cos[d*x + ArcTan[
Tan[c]]]]*Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]*Sqrt[1 + Tan[c]^2]) - ((Sin[d*x + ArcTan[T
an[c]]]*Tan[c])/Sqrt[1 + Tan[c]^2] + (2*Cos[c]^2*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2])/(Cos[c]^2 + Sin
[c]^2))/Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]))/(d*(b + a*Cos[c + d*x])^4*(A + 2*C + 2*B*C
os[c + d*x] + A*Cos[2*c + 2*d*x])) + (8*a*b^3*B*Cos[c + d*x]^6*Csc[c]*(a + b*Sec[c + d*x])^4*(A + B*Sec[c + d*
x] + C*Sec[c + d*x]^2)*((HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x + ArcTan[
Tan[c]]]*Tan[c])/(Sqrt[1 - Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[1 + Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[Cos[c]*Cos[d*x
+ ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]*Sqrt[1 + Tan[c]^2]) - ((Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/Sqrt[1 + Tan[c
]^2] + (2*Cos[c]^2*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2])/(Cos[c]^2 + Sin[c]^2))/Sqrt[Cos[c]*Cos[d*x +
ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]))/(d*(b + a*Cos[c + d*x])^4*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*
x])) - (2*a^4*C*Cos[c + d*x]^6*Csc[c]*(a + b*Sec[c + d*x])^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((Hyperge
ometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/(Sqrt[1 - Cos[d
*x + ArcTan[Tan[c]]]]*Sqrt[1 + Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c
]^2]]*Sqrt[1 + Tan[c]^2]) - ((Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/Sqrt[1 + Tan[c]^2] + (2*Cos[c]^2*Cos[d*x + Arc
Tan[Tan[c]]]*Sqrt[1 + Tan[c]^2])/(Cos[c]^2 + Sin[c]^2))/Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^
2]]))/(d*(b + a*Cos[c + d*x])^4*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])) + (12*a^2*b^2*C*Cos[c + d*x
]^6*Csc[c]*(a + b*Sec[c + d*x])^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((HypergeometricPFQ[{-1/2, -1/4}, {3
/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/(Sqrt[1 - Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[
1 + Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]*Sqrt[1 + Tan[c]^2]) -
((Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/Sqrt[1 + Tan[c]^2] + (2*Cos[c]^2*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c
]^2])/(Cos[c]^2 + Sin[c]^2))/Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]))/(d*(b + a*Cos[c + d*x
])^4*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])) + (6*b^4*C*Cos[c + d*x]^6*Csc[c]*(a + b*Sec[c + d*x])^
4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2
]*Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/(Sqrt[1 - Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[1 + Cos[d*x + ArcTan[Tan[c]]]]*S
qrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]*Sqrt[1 + Tan[c]^2]) - ((Sin[d*x + ArcTan[Tan[c]]]*Tan
[c])/Sqrt[1 + Tan[c]^2] + (2*Cos[c]^2*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2])/(Cos[c]^2 + Sin[c]^2))/Sqr
t[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]))/(5*d*(b + a*Cos[c + d*x])^4*(A + 2*C + 2*B*Cos[c + d*
x] + A*Cos[2*c + 2*d*x]))
Maple [F(-1)]
Timed out.
[In]
int(cos(d*x+c)^(5/2)*(a+b*sec(d*x+c))^4*(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))^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)
Fricas [C] (verification not implemented)
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.16 (sec) , antiderivative size = 455, normalized size of antiderivative = 1.17
\[
\int \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {5 \, \sqrt {2} {\left (i \, B a^{4} + 4 i \, {\left (A + 3 \, C\right )} a^{3} b + 18 i \, B a^{2} b^{2} + 4 i \, {\left (3 \, A + C\right )} a b^{3} + i \, B b^{4}\right )} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, \sqrt {2} {\left (-i \, B a^{4} - 4 i \, {\left (A + 3 \, C\right )} a^{3} b - 18 i \, B a^{2} b^{2} - 4 i \, {\left (3 \, A + C\right )} a b^{3} - i \, B b^{4}\right )} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 \, \sqrt {2} {\left (-i \, {\left (3 \, A + 5 \, C\right )} a^{4} - 20 i \, B a^{3} b - 30 i \, {\left (A - C\right )} a^{2} b^{2} + 20 i \, B a b^{3} + i \, {\left (5 \, A + 3 \, C\right )} b^{4}\right )} \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 \, \sqrt {2} {\left (i \, {\left (3 \, A + 5 \, C\right )} a^{4} + 20 i \, B a^{3} b + 30 i \, {\left (A - C\right )} a^{2} b^{2} - 20 i \, B a b^{3} - i \, {\left (5 \, A + 3 \, C\right )} b^{4}\right )} \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - 2 \, {\left (3 \, A a^{4} \cos \left (d x + c\right )^{4} + 3 \, C b^{4} + 5 \, {\left (B a^{4} + 4 \, A a^{3} b\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (30 \, C a^{2} b^{2} + 20 \, B a b^{3} + {\left (5 \, A + 3 \, C\right )} b^{4}\right )} \cos \left (d x + c\right )^{2} + 5 \, {\left (4 \, C a b^{3} + B b^{4}\right )} \cos \left (d x + c\right )\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{15 \, d \cos \left (d x + c\right )^{3}}
\]
[In]
integrate(cos(d*x+c)^(5/2)*(a+b*sec(d*x+c))^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")
[Out]
-1/15*(5*sqrt(2)*(I*B*a^4 + 4*I*(A + 3*C)*a^3*b + 18*I*B*a^2*b^2 + 4*I*(3*A + C)*a*b^3 + I*B*b^4)*cos(d*x + c)
^3*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*sqrt(2)*(-I*B*a^4 - 4*I*(A + 3*C)*a^3*b - 18*
I*B*a^2*b^2 - 4*I*(3*A + C)*a*b^3 - I*B*b^4)*cos(d*x + c)^3*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*
x + c)) + 3*sqrt(2)*(-I*(3*A + 5*C)*a^4 - 20*I*B*a^3*b - 30*I*(A - C)*a^2*b^2 + 20*I*B*a*b^3 + I*(5*A + 3*C)*b
^4)*cos(d*x + c)^3*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*sqrt(
2)*(I*(3*A + 5*C)*a^4 + 20*I*B*a^3*b + 30*I*(A - C)*a^2*b^2 - 20*I*B*a*b^3 - I*(5*A + 3*C)*b^4)*cos(d*x + c)^3
*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - 2*(3*A*a^4*cos(d*x + c)^4
+ 3*C*b^4 + 5*(B*a^4 + 4*A*a^3*b)*cos(d*x + c)^3 + 3*(30*C*a^2*b^2 + 20*B*a*b^3 + (5*A + 3*C)*b^4)*cos(d*x +
c)^2 + 5*(4*C*a*b^3 + B*b^4)*cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^3)
Sympy [F(-1)]
Timed out. \[
\int \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out}
\]
[In]
integrate(cos(d*x+c)**(5/2)*(a+b*sec(d*x+c))**4*(A+B*sec(d*x+c)+C*sec(d*x+c)**2),x)
[Out]
Timed out
Maxima [F(-1)]
Timed out. \[
\int \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out}
\]
[In]
integrate(cos(d*x+c)^(5/2)*(a+b*sec(d*x+c))^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")
[Out]
Timed out
Giac [F]
\[
\int \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{4} \cos \left (d x + c\right )^{\frac {5}{2}} \,d x }
\]
[In]
integrate(cos(d*x+c)^(5/2)*(a+b*sec(d*x+c))^4*(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)^4*cos(d*x + c)^(5/2), x)
Mupad [B] (verification not implemented)
Time = 24.83 (sec) , antiderivative size = 524, normalized size of antiderivative = 1.35
\[
\int \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2\,\left (B\,a^4\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )+12\,B\,a^3\,b\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )+B\,a^4\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )+18\,B\,a^2\,b^2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )\right )}{3\,d}+\frac {2\,C\,a^4\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {8\,A\,a\,b^3\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {8\,C\,a^3\,b\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {4\,A\,a^3\,b\,\left (\frac {2\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )}{3}+\frac {2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{3}\right )}{d}+\frac {12\,A\,a^2\,b^2\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}-\frac {2\,A\,a^4\,{\cos \left (c+d\,x\right )}^{7/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {11}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {2\,A\,b^4\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {2\,B\,b^4\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{3\,d\,{\cos \left (c+d\,x\right )}^{3/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {2\,C\,b^4\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{4},\frac {1}{2};\ -\frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{5\,d\,{\cos \left (c+d\,x\right )}^{5/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {8\,B\,a\,b^3\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {8\,C\,a\,b^3\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{3\,d\,{\cos \left (c+d\,x\right )}^{3/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {12\,C\,a^2\,b^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}
\]
[In]
int(cos(c + d*x)^(5/2)*(a + b/cos(c + d*x))^4*(A + B/cos(c + d*x) + C/cos(c + d*x)^2),x)
[Out]
(2*(B*a^4*ellipticF(c/2 + (d*x)/2, 2) + 12*B*a^3*b*ellipticE(c/2 + (d*x)/2, 2) + B*a^4*cos(c + d*x)^(1/2)*sin(
c + d*x) + 18*B*a^2*b^2*ellipticF(c/2 + (d*x)/2, 2)))/(3*d) + (2*C*a^4*ellipticE(c/2 + (d*x)/2, 2))/d + (8*A*a
*b^3*ellipticF(c/2 + (d*x)/2, 2))/d + (8*C*a^3*b*ellipticF(c/2 + (d*x)/2, 2))/d + (4*A*a^3*b*((2*cos(c + d*x)^
(1/2)*sin(c + d*x))/3 + (2*ellipticF(c/2 + (d*x)/2, 2))/3))/d + (12*A*a^2*b^2*ellipticE(c/2 + (d*x)/2, 2))/d -
(2*A*a^4*cos(c + d*x)^(7/2)*sin(c + d*x)*hypergeom([1/2, 7/4], 11/4, cos(c + d*x)^2))/(7*d*(sin(c + d*x)^2)^(
1/2)) + (2*A*b^4*sin(c + d*x)*hypergeom([-1/4, 1/2], 3/4, cos(c + d*x)^2))/(d*cos(c + d*x)^(1/2)*(sin(c + d*x)
^2)^(1/2)) + (2*B*b^4*sin(c + d*x)*hypergeom([-3/4, 1/2], 1/4, cos(c + d*x)^2))/(3*d*cos(c + d*x)^(3/2)*(sin(c
+ d*x)^2)^(1/2)) + (2*C*b^4*sin(c + d*x)*hypergeom([-5/4, 1/2], -1/4, cos(c + d*x)^2))/(5*d*cos(c + d*x)^(5/2
)*(sin(c + d*x)^2)^(1/2)) + (8*B*a*b^3*sin(c + d*x)*hypergeom([-1/4, 1/2], 3/4, cos(c + d*x)^2))/(d*cos(c + d*
x)^(1/2)*(sin(c + d*x)^2)^(1/2)) + (8*C*a*b^3*sin(c + d*x)*hypergeom([-3/4, 1/2], 1/4, cos(c + d*x)^2))/(3*d*c
os(c + d*x)^(3/2)*(sin(c + d*x)^2)^(1/2)) + (12*C*a^2*b^2*sin(c + d*x)*hypergeom([-1/4, 1/2], 3/4, cos(c + d*x
)^2))/(d*cos(c + d*x)^(1/2)*(sin(c + d*x)^2)^(1/2))