3.13.13 \(\int \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [1213]

Optimal. Leaf size=267 \[ \frac {56 a^4 (A-C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {8 a^4 (4 A+5 B+4 C) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}+\frac {4 a^4 (A-25 B-41 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{15 d}+\frac {2 a (5 B+8 C) (a+a \cos (c+d x))^3 \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 C (a+a \cos (c+d x))^4 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 (5 A+15 B+19 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}-\frac {4 (6 A+25 B+34 C) \sqrt {\cos (c+d x)} \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{15 d} \]

[Out]

56/5*a^4*(A-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+8/3*a^4
*(4*A+5*B+4*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*
(5*B+8*C)*(a+a*cos(d*x+c))^3*sin(d*x+c)/d/cos(d*x+c)^(3/2)+2/5*C*(a+a*cos(d*x+c))^4*sin(d*x+c)/d/cos(d*x+c)^(5
/2)+2/5*(5*A+15*B+19*C)*(a^2+a^2*cos(d*x+c))^2*sin(d*x+c)/d/cos(d*x+c)^(1/2)+4/15*a^4*(A-25*B-41*C)*sin(d*x+c)
*cos(d*x+c)^(1/2)/d-4/15*(6*A+25*B+34*C)*(a^4+a^4*cos(d*x+c))*sin(d*x+c)*cos(d*x+c)^(1/2)/d

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Rubi [A]
time = 0.60, antiderivative size = 267, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.209, Rules used = {4197, 3122, 3054, 3055, 3047, 3102, 2827, 2720, 2719} \begin {gather*} \frac {8 a^4 (4 A+5 B+4 C) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}+\frac {4 a^4 (A-25 B-41 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{15 d}-\frac {4 (6 A+25 B+34 C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^4 \cos (c+d x)+a^4\right )}{15 d}+\frac {56 a^4 (A-C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 (5 A+15 B+19 C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{5 d \sqrt {\cos (c+d x)}}+\frac {2 a (5 B+8 C) \sin (c+d x) (a \cos (c+d x)+a)^3}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^4}{5 d \cos ^{\frac {5}{2}}(c+d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^(5/2)*(a + a*Sec[c + d*x])^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(56*a^4*(A - C)*EllipticE[(c + d*x)/2, 2])/(5*d) + (8*a^4*(4*A + 5*B + 4*C)*EllipticF[(c + d*x)/2, 2])/(3*d) +
(4*a^4*(A - 25*B - 41*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(15*d) + (2*a*(5*B + 8*C)*(a + a*Cos[c + d*x])^3*Si
n[c + d*x])/(15*d*Cos[c + d*x]^(3/2)) + (2*C*(a + a*Cos[c + d*x])^4*Sin[c + d*x])/(5*d*Cos[c + d*x]^(5/2)) + (
2*(5*A + 15*B + 19*C)*(a^2 + a^2*Cos[c + d*x])^2*Sin[c + d*x])/(5*d*Sqrt[Cos[c + d*x]]) - (4*(6*A + 25*B + 34*
C)*Sqrt[Cos[c + d*x]]*(a^4 + a^4*Cos[c + d*x])*Sin[c + d*x])/(15*d)

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 3047

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x
]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]

Rule 3054

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b^2)*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d
*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(b*c + a*d))), x] - Dist[b/(d*(n + 1)*(b*c + a*d)), Int[(a + b*Sin[e + f*x
])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[a*A*d*(m - n - 2) - B*(a*c*(m - 1) + b*d*(n + 1)) - (A*b*d*(m + n
+ 1) - B*(b*c*m - a*d*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a
*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1/2] && LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*
n] || EqQ[c, 0])

Rule 3055

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[e + f*x
])^(n + 1)/(d*f*(m + n + 1))), x] + Dist[1/(d*(m + n + 1)), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*
x])^n*Simp[a*A*d*(m + n + 1) + B*(a*c*(m - 1) + b*d*(n + 1)) + (A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))
*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] &
& NeQ[c^2 - d^2, 0] && GtQ[m, 1/2] &&  !LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])

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 3122

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/(b*d*(n +
1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(a*d*m + b*c*(n + 1)) + (c*C
- B*d)*(a*c*m + b*d*(n + 1)) + b*(d*(B*c - A*d)*(m + n + 2) - C*(c^2*(m + 1) + d^2*(n + 1)))*Sin[e + f*x], x]
, x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^
2, 0] &&  !LtQ[m, -2^(-1)] && (LtQ[n, -1] || EqQ[m + n + 2, 0])

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*} \int \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\int \frac {(a+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 (a+a \cos (c+d x))^4 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 \int \frac {(a+a \cos (c+d x))^4 \left (\frac {1}{2} a (5 B+8 C)+\frac {5}{2} a (A-C) \cos (c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx}{5 a}\\ &=\frac {2 a (5 B+8 C) (a+a \cos (c+d x))^3 \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 C (a+a \cos (c+d x))^4 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {4 \int \frac {(a+a \cos (c+d x))^3 \left (\frac {3}{4} a^2 (5 A+15 B+19 C)+\frac {5}{4} a^2 (3 A-5 B-11 C) \cos (c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx}{15 a}\\ &=\frac {2 a (5 B+8 C) (a+a \cos (c+d x))^3 \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 C (a+a \cos (c+d x))^4 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 (5 A+15 B+19 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {8 \int \frac {(a+a \cos (c+d x))^2 \left (\frac {5}{4} a^3 (9 A+20 B+23 C)-\frac {5}{4} a^3 (6 A+25 B+34 C) \cos (c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx}{15 a}\\ &=\frac {2 a (5 B+8 C) (a+a \cos (c+d x))^3 \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 C (a+a \cos (c+d x))^4 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 (5 A+15 B+19 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}-\frac {4 (6 A+25 B+34 C) \sqrt {\cos (c+d x)} \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{15 d}+\frac {16 \int \frac {(a+a \cos (c+d x)) \left (\frac {15}{8} a^4 (13 A+25 B+27 C)+\frac {15}{8} a^4 (A-25 B-41 C) \cos (c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx}{75 a}\\ &=\frac {2 a (5 B+8 C) (a+a \cos (c+d x))^3 \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 C (a+a \cos (c+d x))^4 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 (5 A+15 B+19 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}-\frac {4 (6 A+25 B+34 C) \sqrt {\cos (c+d x)} \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{15 d}+\frac {16 \int \frac {\frac {15}{8} a^5 (13 A+25 B+27 C)+\left (\frac {15}{8} a^5 (A-25 B-41 C)+\frac {15}{8} a^5 (13 A+25 B+27 C)\right ) \cos (c+d x)+\frac {15}{8} a^5 (A-25 B-41 C) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)}} \, dx}{75 a}\\ &=\frac {4 a^4 (A-25 B-41 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{15 d}+\frac {2 a (5 B+8 C) (a+a \cos (c+d x))^3 \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 C (a+a \cos (c+d x))^4 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 (5 A+15 B+19 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}-\frac {4 (6 A+25 B+34 C) \sqrt {\cos (c+d x)} \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{15 d}+\frac {32 \int \frac {\frac {75}{8} a^5 (4 A+5 B+4 C)+\frac {315}{8} a^5 (A-C) \cos (c+d x)}{\sqrt {\cos (c+d x)}} \, dx}{225 a}\\ &=\frac {4 a^4 (A-25 B-41 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{15 d}+\frac {2 a (5 B+8 C) (a+a \cos (c+d x))^3 \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 C (a+a \cos (c+d x))^4 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 (5 A+15 B+19 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}-\frac {4 (6 A+25 B+34 C) \sqrt {\cos (c+d x)} \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{15 d}+\frac {1}{5} \left (28 a^4 (A-C)\right ) \int \sqrt {\cos (c+d x)} \, dx+\frac {1}{3} \left (4 a^4 (4 A+5 B+4 C)\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {56 a^4 (A-C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {8 a^4 (4 A+5 B+4 C) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}+\frac {4 a^4 (A-25 B-41 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{15 d}+\frac {2 a (5 B+8 C) (a+a \cos (c+d x))^3 \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 C (a+a \cos (c+d x))^4 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 (5 A+15 B+19 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}-\frac {4 (6 A+25 B+34 C) \sqrt {\cos (c+d x)} \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{15 d}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 6.77, size = 1449, normalized size = 5.43 \begin {gather*} \frac {\cos ^{\frac {13}{2}}(c+d x) \sec ^8\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (-\frac {(23 A-20 B-61 C+33 A \cos (2 c)+20 B \cos (2 c)+5 C \cos (2 c)) \csc (c) \sec (c)}{40 d}+\frac {(4 A+B) \cos (d x) \sin (c)}{12 d}+\frac {A \cos (2 d x) \sin (2 c)}{40 d}+\frac {(4 A+B) \cos (c) \sin (d x)}{12 d}+\frac {C \sec (c) \sec ^3(c+d x) \sin (d x)}{20 d}+\frac {\sec (c) \sec ^2(c+d x) (3 C \sin (c)+5 B \sin (d x)+20 C \sin (d x))}{60 d}+\frac {\sec (c) \sec (c+d x) (5 B \sin (c)+20 C \sin (c)+15 A \sin (d x)+60 B \sin (d x)+99 C \sin (d x))}{60 d}+\frac {A \cos (2 c) \sin (2 d x)}{40 d}\right )}{A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 d x)}-\frac {4 A \cos ^6(c+d x) \csc (c) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2(d x-\text {ArcTan}(\cot (c)))\right ) \sec ^8\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \sec (d x-\text {ArcTan}(\cot (c))) \sqrt {1-\sin (d x-\text {ArcTan}(\cot (c)))} \sqrt {-\sqrt {1+\cot ^2(c)} \sin (c) \sin (d x-\text {ArcTan}(\cot (c)))} \sqrt {1+\sin (d x-\text {ArcTan}(\cot (c)))}}{3 d (A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 d x)) \sqrt {1+\cot ^2(c)}}-\frac {5 B \cos ^6(c+d x) \csc (c) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2(d x-\text {ArcTan}(\cot (c)))\right ) \sec ^8\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \sec (d x-\text {ArcTan}(\cot (c))) \sqrt {1-\sin (d x-\text {ArcTan}(\cot (c)))} \sqrt {-\sqrt {1+\cot ^2(c)} \sin (c) \sin (d x-\text {ArcTan}(\cot (c)))} \sqrt {1+\sin (d x-\text {ArcTan}(\cot (c)))}}{3 d (A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 d x)) \sqrt {1+\cot ^2(c)}}-\frac {4 C \cos ^6(c+d x) \csc (c) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2(d x-\text {ArcTan}(\cot (c)))\right ) \sec ^8\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \sec (d x-\text {ArcTan}(\cot (c))) \sqrt {1-\sin (d x-\text {ArcTan}(\cot (c)))} \sqrt {-\sqrt {1+\cot ^2(c)} \sin (c) \sin (d x-\text {ArcTan}(\cot (c)))} \sqrt {1+\sin (d x-\text {ArcTan}(\cot (c)))}}{3 d (A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 d x)) \sqrt {1+\cot ^2(c)}}-\frac {7 A \cos ^6(c+d x) \csc (c) \sec ^8\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (\frac {\, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2(d x+\text {ArcTan}(\tan (c)))\right ) \sin (d x+\text {ArcTan}(\tan (c))) \tan (c)}{\sqrt {1-\cos (d x+\text {ArcTan}(\tan (c)))} \sqrt {1+\cos (d x+\text {ArcTan}(\tan (c)))} \sqrt {\cos (c) \cos (d x+\text {ArcTan}(\tan (c))) \sqrt {1+\tan ^2(c)}} \sqrt {1+\tan ^2(c)}}-\frac {\frac {\sin (d x+\text {ArcTan}(\tan (c))) \tan (c)}{\sqrt {1+\tan ^2(c)}}+\frac {2 \cos ^2(c) \cos (d x+\text {ArcTan}(\tan (c))) \sqrt {1+\tan ^2(c)}}{\cos ^2(c)+\sin ^2(c)}}{\sqrt {\cos (c) \cos (d x+\text {ArcTan}(\tan (c))) \sqrt {1+\tan ^2(c)}}}\right )}{10 d (A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 d x))}+\frac {7 C \cos ^6(c+d x) \csc (c) \sec ^8\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (\frac {\, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2(d x+\text {ArcTan}(\tan (c)))\right ) \sin (d x+\text {ArcTan}(\tan (c))) \tan (c)}{\sqrt {1-\cos (d x+\text {ArcTan}(\tan (c)))} \sqrt {1+\cos (d x+\text {ArcTan}(\tan (c)))} \sqrt {\cos (c) \cos (d x+\text {ArcTan}(\tan (c))) \sqrt {1+\tan ^2(c)}} \sqrt {1+\tan ^2(c)}}-\frac {\frac {\sin (d x+\text {ArcTan}(\tan (c))) \tan (c)}{\sqrt {1+\tan ^2(c)}}+\frac {2 \cos ^2(c) \cos (d x+\text {ArcTan}(\tan (c))) \sqrt {1+\tan ^2(c)}}{\cos ^2(c)+\sin ^2(c)}}{\sqrt {\cos (c) \cos (d x+\text {ArcTan}(\tan (c))) \sqrt {1+\tan ^2(c)}}}\right )}{10 d (A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 d x))} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[Cos[c + d*x]^(5/2)*(a + a*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)*Sec[c/2 + (d*x)/2]^8*(a + a*Sec[c + d*x])^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(-1/4
0*((23*A - 20*B - 61*C + 33*A*Cos[2*c] + 20*B*Cos[2*c] + 5*C*Cos[2*c])*Csc[c]*Sec[c])/d + ((4*A + B)*Cos[d*x]*
Sin[c])/(12*d) + (A*Cos[2*d*x]*Sin[2*c])/(40*d) + ((4*A + B)*Cos[c]*Sin[d*x])/(12*d) + (C*Sec[c]*Sec[c + d*x]^
3*Sin[d*x])/(20*d) + (Sec[c]*Sec[c + d*x]^2*(3*C*Sin[c] + 5*B*Sin[d*x] + 20*C*Sin[d*x]))/(60*d) + (Sec[c]*Sec[
c + d*x]*(5*B*Sin[c] + 20*C*Sin[c] + 15*A*Sin[d*x] + 60*B*Sin[d*x] + 99*C*Sin[d*x]))/(60*d) + (A*Cos[2*c]*Sin[
2*d*x])/(40*d)))/(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x]) - (4*A*Cos[c + d*x]^6*Csc[c]*Hypergeometric
PFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^8*(a + a*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*(A + 2*C + 2*B*Cos[c + d*x] +
A*Cos[2*c + 2*d*x])*Sqrt[1 + Cot[c]^2]) - (5*B*Cos[c + d*x]^6*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin
[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^8*(a + a*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 - ArcTa
n[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(3*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*Sqrt[
1 + Cot[c]^2]) - (4*C*Cos[c + d*x]^6*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*
Sec[c/2 + (d*x)/2]^8*(a + a*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*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*Sqrt[1 + Cot[c]^2]) - (7*A*Cos[
c + d*x]^6*Csc[c]*Sec[c/2 + (d*x)/2]^8*(a + a*Sec[c + d*x])^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((Hyperg
eometricPFQ[{-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 + Ar
cTan[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]]))/(10*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])) + (7*C*Cos[c + d*x]^6*Csc[c]*Sec[c/2 + (d*x)/2
]^8*(a + a*Sec[c + d*x])^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Co
s[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]]))/(10*d*(A + 2*C + 2*B*Cos[c +
d*x] + A*Cos[2*c + 2*d*x]))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1213\) vs. \(2(295)=590\).
time = 0.38, size = 1214, normalized size = 4.55

method	result	size
default	\(\text {Expression too large to display}\)	\(1214\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^(5/2)*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x,method=_RETURNVERBOSE)

[Out]

-8/15*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^4/(8*sin(1/2*d*x+1/2*c)^6-12*sin(1/2*d*x+1/2
*c)^4+6*sin(1/2*d*x+1/2*c)^2-1)/sin(1/2*d*x+1/2*c)^3*(80*A*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c
)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^2-84*A*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(s
in(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^2+100*B*(2*sin(1/2*d*x+1/2
*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^2+80*C*(2
*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+
1/2*c)^2+84*C*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/
2))*sin(1/2*d*x+1/2*c)^2+24*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^10-128*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/
2*c)^8-102*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+19*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)*A+61*sin(1/2*d
*x+1/2*c)^2*cos(1/2*d*x+1/2*c)*C-80*A*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(
cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^4+84*A*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^
(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^4-100*B*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1
/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^4-80*C*(2*sin(1/2*d*x+1/2*c)^2
-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^4-84*C*(2*sin(
1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c
)^4+186*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-20*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8-20*A*(sin(1/2*d
*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+21*A*(sin(1/2*d*x+1/
2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-25*B*(sin(1/2*d*x+1/2*c)^
2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-20*C*(sin(1/2*d*x+1/2*c)^2)^(1
/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-21*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(
2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+150*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2
*c)^6+198*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-140*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4-218*C*cos(1/
2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+35*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1
/2*d*x+1/2*c)^2)^(1/2)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d

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Maxima [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(5/2)*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.84, size = 280, normalized size = 1.05 \begin {gather*} -\frac {2 \, {\left (10 i \, \sqrt {2} {\left (4 \, A + 5 \, B + 4 \, C\right )} a^{4} \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 ) - 10 i \, \sqrt {2} {\left (4 \, A + 5 \, B + 4 \, C\right )} a^{4} \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 ) - 42 i \, \sqrt {2} {\left (A - C\right )} a^{4} \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 ) + 42 i \, \sqrt {2} {\left (A - C\right )} a^{4} \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 ) - {\left (3 \, A a^{4} \cos \left (d x + c\right )^{4} + 5 \, {\left (4 \, A + B\right )} a^{4} \cos \left (d x + c\right )^{3} + 3 \, {\left (5 \, A + 20 \, B + 33 \, C\right )} a^{4} \cos \left (d x + c\right )^{2} + 5 \, {\left (B + 4 \, C\right )} a^{4} \cos \left (d x + c\right ) + 3 \, C a^{4}\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )\right )}}{15 \, d \cos \left (d x + c\right )^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(5/2)*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

-2/15*(10*I*sqrt(2)*(4*A + 5*B + 4*C)*a^4*cos(d*x + c)^3*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x +
c)) - 10*I*sqrt(2)*(4*A + 5*B + 4*C)*a^4*cos(d*x + c)^3*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x +
c)) - 42*I*sqrt(2)*(A - C)*a^4*cos(d*x + c)^3*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c)
+ I*sin(d*x + c))) + 42*I*sqrt(2)*(A - C)*a^4*cos(d*x + c)^3*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0,
cos(d*x + c) - I*sin(d*x + c))) - (3*A*a^4*cos(d*x + c)^4 + 5*(4*A + B)*a^4*cos(d*x + c)^3 + 3*(5*A + 20*B +
33*C)*a^4*cos(d*x + c)^2 + 5*(B + 4*C)*a^4*cos(d*x + c) + 3*C*a^4)*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x
+ c)^3)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**(5/2)*(a+a*sec(d*x+c))**4*(A+B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(5/2)*(a+a*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)*(a*sec(d*x + c) + a)^4*cos(d*x + c)^(5/2), x)

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Mupad [B]
time = 8.03, size = 525, normalized size = 1.97 \begin {gather*} \frac {2\,\left (12\,B\,a^4\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )+19\,B\,a^4\,\mathrm {F}\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 )\right )}{3\,d}+\frac {2\,\left (C\,a^4\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )+4\,C\,a^4\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )\right )}{d}+\frac {2\,\left (\frac {34\,C\,a^4\,\sin \left (c+d\,x\right )}{\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {C\,a^4\,\sin \left (c+d\,x\right )}{{\cos \left (c+d\,x\right )}^{5/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{5\,d}+\frac {4\,A\,a^4\,\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^4\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {8\,A\,a^4\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,A\,a^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\,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 {8\,B\,a^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\,a^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 {8\,C\,a^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 {8\,C\,a^4\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {7}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{15\,d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^(5/2)*(a + a/cos(c + d*x))^4*(A + B/cos(c + d*x) + C/cos(c + d*x)^2),x)

[Out]

(2*(12*B*a^4*ellipticE(c/2 + (d*x)/2, 2) + 19*B*a^4*ellipticF(c/2 + (d*x)/2, 2) + B*a^4*cos(c + d*x)^(1/2)*sin
(c + d*x)))/(3*d) + (2*(C*a^4*ellipticE(c/2 + (d*x)/2, 2) + 4*C*a^4*ellipticF(c/2 + (d*x)/2, 2)))/d + (2*((34*
C*a^4*sin(c + d*x))/(cos(c + d*x)^(1/2)*(sin(c + d*x)^2)^(1/2)) + (C*a^4*sin(c + d*x))/(cos(c + d*x)^(5/2)*(si
n(c + d*x)^2)^(1/2)))*hypergeom([-1/4, 1/2], 3/4, cos(c + d*x)^2))/(5*d) + (4*A*a^4*((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^4*ellipticE(c/2 + (d*x)/2, 2))/d + (8*A*a^4*ell
ipticF(c/2 + (d*x)/2, 2))/d + (2*A*a^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*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)) + (8*B*a^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*a^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)) + (8*C*a^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)) - (8*C*a^4*sin(c + d*x)*hypergeom([-1/4, 1/2], 7/4,
cos(c + d*x)^2))/(15*d*cos(c + d*x)^(1/2)*(sin(c + d*x)^2)^(1/2))

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