3.1203 \(\int \cos ^{\frac {9}{2}}(c+d x) (a+a \sec (c+d x))^3 (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=231 \[ \frac {4 a^3 (11 A+13 B+21 C) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}+\frac {4 a^3 (17 A+21 B+27 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d}+\frac {4 a^3 (32 A+41 B+42 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{105 d}+\frac {2 (73 A+99 B+63 C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^3 \cos (c+d x)+a^3\right )}{315 d}+\frac {2 (2 A+3 B) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^2 \cos (c+d x)+a^2\right )^2}{21 a d}+\frac {2 A \sin (c+d x) \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^3}{9 d} \]

[Out]

4/15*a^3*(17*A+21*B+27*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+4/21*a^3*(11*A+13*B+21*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+4/105*a^3*(32*A+41*B+42*C)*sin(d*x+c)*cos(d*x+c)^(1/2)/d+2/9*A*(a+a*cos(d*x+c))^3*sin(d*x+c)*cos(d*x+c
)^(1/2)/d+2/21*(2*A+3*B)*(a^2+a^2*cos(d*x+c))^2*sin(d*x+c)*cos(d*x+c)^(1/2)/a/d+2/315*(73*A+99*B+63*C)*(a^3+a^
3*cos(d*x+c))*sin(d*x+c)*cos(d*x+c)^(1/2)/d

________________________________________________________________________________________

Rubi [A]  time = 0.72, antiderivative size = 231, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.186, Rules used = {4112, 3045, 2976, 2968, 3023, 2748, 2641, 2639} \[ \frac {4 a^3 (11 A+13 B+21 C) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}+\frac {4 a^3 (17 A+21 B+27 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d}+\frac {4 a^3 (32 A+41 B+42 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{105 d}+\frac {2 (73 A+99 B+63 C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^3 \cos (c+d x)+a^3\right )}{315 d}+\frac {2 (2 A+3 B) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^2 \cos (c+d x)+a^2\right )^2}{21 a d}+\frac {2 A \sin (c+d x) \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^3}{9 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^(9/2)*(a + a*Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(4*a^3*(17*A + 21*B + 27*C)*EllipticE[(c + d*x)/2, 2])/(15*d) + (4*a^3*(11*A + 13*B + 21*C)*EllipticF[(c + d*x
)/2, 2])/(21*d) + (4*a^3*(32*A + 41*B + 42*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(105*d) + (2*A*Sqrt[Cos[c + d*x
]]*(a + a*Cos[c + d*x])^3*Sin[c + d*x])/(9*d) + (2*(2*A + 3*B)*Sqrt[Cos[c + d*x]]*(a^2 + a^2*Cos[c + d*x])^2*S
in[c + d*x])/(21*a*d) + (2*(73*A + 99*B + 63*C)*Sqrt[Cos[c + d*x]]*(a^3 + a^3*Cos[c + d*x])*Sin[c + d*x])/(315
*d)

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rule 2748

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 2968

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 2976

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)))*S
in[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 3023

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 3045

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

Rule 4112

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 {9}{2}}(c+d x) (a+a \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\int \frac {(a+a \cos (c+d x))^3 \left (C+B \cos (c+d x)+A \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {2 A \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^3 \sin (c+d x)}{9 d}+\frac {2 \int \frac {(a+a \cos (c+d x))^3 \left (\frac {1}{2} a (A+9 C)+\frac {3}{2} a (2 A+3 B) \cos (c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx}{9 a}\\ &=\frac {2 A \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^3 \sin (c+d x)}{9 d}+\frac {2 (2 A+3 B) \sqrt {\cos (c+d x)} \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{21 a d}+\frac {4 \int \frac {(a+a \cos (c+d x))^2 \left (\frac {1}{4} a^2 (13 A+9 B+63 C)+\frac {1}{4} a^2 (73 A+99 B+63 C) \cos (c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx}{63 a}\\ &=\frac {2 A \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^3 \sin (c+d x)}{9 d}+\frac {2 (2 A+3 B) \sqrt {\cos (c+d x)} \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{21 a d}+\frac {2 (73 A+99 B+63 C) \sqrt {\cos (c+d x)} \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{315 d}+\frac {8 \int \frac {(a+a \cos (c+d x)) \left (\frac {3}{4} a^3 (23 A+24 B+63 C)+\frac {9}{4} a^3 (32 A+41 B+42 C) \cos (c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx}{315 a}\\ &=\frac {2 A \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^3 \sin (c+d x)}{9 d}+\frac {2 (2 A+3 B) \sqrt {\cos (c+d x)} \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{21 a d}+\frac {2 (73 A+99 B+63 C) \sqrt {\cos (c+d x)} \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{315 d}+\frac {8 \int \frac {\frac {3}{4} a^4 (23 A+24 B+63 C)+\left (\frac {9}{4} a^4 (32 A+41 B+42 C)+\frac {3}{4} a^4 (23 A+24 B+63 C)\right ) \cos (c+d x)+\frac {9}{4} a^4 (32 A+41 B+42 C) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)}} \, dx}{315 a}\\ &=\frac {4 a^3 (32 A+41 B+42 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{105 d}+\frac {2 A \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^3 \sin (c+d x)}{9 d}+\frac {2 (2 A+3 B) \sqrt {\cos (c+d x)} \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{21 a d}+\frac {2 (73 A+99 B+63 C) \sqrt {\cos (c+d x)} \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{315 d}+\frac {16 \int \frac {\frac {45}{8} a^4 (11 A+13 B+21 C)+\frac {63}{8} a^4 (17 A+21 B+27 C) \cos (c+d x)}{\sqrt {\cos (c+d x)}} \, dx}{945 a}\\ &=\frac {4 a^3 (32 A+41 B+42 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{105 d}+\frac {2 A \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^3 \sin (c+d x)}{9 d}+\frac {2 (2 A+3 B) \sqrt {\cos (c+d x)} \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{21 a d}+\frac {2 (73 A+99 B+63 C) \sqrt {\cos (c+d x)} \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{315 d}+\frac {1}{21} \left (2 a^3 (11 A+13 B+21 C)\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx+\frac {1}{15} \left (2 a^3 (17 A+21 B+27 C)\right ) \int \sqrt {\cos (c+d x)} \, dx\\ &=\frac {4 a^3 (17 A+21 B+27 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d}+\frac {4 a^3 (11 A+13 B+21 C) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}+\frac {4 a^3 (32 A+41 B+42 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{105 d}+\frac {2 A \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^3 \sin (c+d x)}{9 d}+\frac {2 (2 A+3 B) \sqrt {\cos (c+d x)} \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{21 a d}+\frac {2 (73 A+99 B+63 C) \sqrt {\cos (c+d x)} \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{315 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 6.52, size = 1697, normalized size = 7.35 \[ \text {result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Cos[c + d*x]^(9/2)*(a + a*Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(Cos[c + d*x]^(11/2)*Sec[c/2 + (d*x)/2]^6*(a + a*Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(-1/1
5*((17*A + 21*B + 27*C)*Cot[c])/d + ((97*A + 107*B + 84*C)*Cos[d*x]*Sin[c])/(168*d) + ((73*A + 54*B + 18*C)*Co
s[2*d*x]*Sin[2*c])/(360*d) + ((3*A + B)*Cos[3*d*x]*Sin[3*c])/(56*d) + (A*Cos[4*d*x]*Sin[4*c])/(144*d) + ((97*A
+ 107*B + 84*C)*Cos[c]*Sin[d*x])/(168*d) + ((73*A + 54*B + 18*C)*Cos[2*c]*Sin[2*d*x])/(360*d) + ((3*A + B)*Co
s[3*c]*Sin[3*d*x])/(56*d) + (A*Cos[4*c]*Sin[4*d*x])/(144*d)))/(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x]
) - (11*A*Cos[c + d*x]^5*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2 + (d
*x)/2]^6*(a + a*Sec[c + d*x])^3*(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]]]])/(21*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*Sqrt[1 + Cot[c]^2]) - (13*B*Cos[c + d*x]^5
*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^6*(a + a*Sec[c +
d*x])^3*(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]]]])/(21*d*(A + 2
*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*Sqrt[1 + Cot[c]^2]) - (C*Cos[c + d*x]^5*Csc[c]*HypergeometricPFQ[{
1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^6*(a + a*Sec[c + d*x])^3*(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]*S
in[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[
2*c + 2*d*x])*Sqrt[1 + Cot[c]^2]) - (17*A*Cos[c + d*x]^5*Csc[c]*Sec[c/2 + (d*x)/2]^6*(a + a*Sec[c + d*x])^3*(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]*Si
n[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[Co
s[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]))/(30*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x]))
- (7*B*Cos[c + d*x]^5*Csc[c]*Sec[c/2 + (d*x)/2]^6*(a + a*Sec[c + d*x])^3*(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])/(Sq
rt[1 - Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[1 + Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sq
rt[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*C
os[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])) - (9*C*Cos[c + d*x]^5*Csc[c]*Sec[c/
2 + (d*x)/2]^6*(a + a*Sec[c + d*x])^3*(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]]]]*S
qrt[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 + T
an[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]))

________________________________________________________________________________________

fricas [F]  time = 0.49, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C a^{3} \cos \left (d x + c\right )^{4} \sec \left (d x + c\right )^{5} + {\left (B + 3 \, C\right )} a^{3} \cos \left (d x + c\right )^{4} \sec \left (d x + c\right )^{4} + {\left (A + 3 \, B + 3 \, C\right )} a^{3} \cos \left (d x + c\right )^{4} \sec \left (d x + c\right )^{3} + {\left (3 \, A + 3 \, B + C\right )} a^{3} \cos \left (d x + c\right )^{4} \sec \left (d x + c\right )^{2} + {\left (3 \, A + B\right )} a^{3} \cos \left (d x + c\right )^{4} \sec \left (d x + c\right ) + A a^{3} \cos \left (d x + c\right )^{4}\right )} \sqrt {\cos \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(9/2)*(a+a*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

integral((C*a^3*cos(d*x + c)^4*sec(d*x + c)^5 + (B + 3*C)*a^3*cos(d*x + c)^4*sec(d*x + c)^4 + (A + 3*B + 3*C)*
a^3*cos(d*x + c)^4*sec(d*x + c)^3 + (3*A + 3*B + C)*a^3*cos(d*x + c)^4*sec(d*x + c)^2 + (3*A + B)*a^3*cos(d*x
+ c)^4*sec(d*x + c) + A*a^3*cos(d*x + c)^4)*sqrt(cos(d*x + c)), x)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{3} \cos \left (d x + c\right )^{\frac {9}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(9/2)*(a+a*sec(d*x+c))^3*(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)^3*cos(d*x + c)^(9/2), x)

________________________________________________________________________________________

maple [A]  time = 5.54, size = 514, normalized size = 2.23 \[ -\frac {4 \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, a^{3} \left (-560 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (2200 A +360 B \right ) \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-3412 A -1296 B -252 C \right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (2702 A +1806 B +882 C \right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-738 A -624 B -378 C \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+165 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-357 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+195 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-441 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+315 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-567 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{315 \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^(9/2)*(a+a*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)

[Out]

-4/315*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^3*(-560*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*
c)^10+(2200*A+360*B)*sin(1/2*d*x+1/2*c)^8*cos(1/2*d*x+1/2*c)+(-3412*A-1296*B-252*C)*sin(1/2*d*x+1/2*c)^6*cos(1
/2*d*x+1/2*c)+(2702*A+1806*B+882*C)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-738*A-624*B-378*C)*sin(1/2*d*x+1
/2*c)^2*cos(1/2*d*x+1/2*c)+165*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))-357*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))+195*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))-441*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)*EllipticE(cos(1/2*d*x+1/2*c
),2^(1/2))+315*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))-567*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
)))/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d

________________________________________________________________________________________

maxima [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)^(9/2)*(a+a*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

Timed out

________________________________________________________________________________________

mupad [B]  time = 5.49, size = 430, normalized size = 1.86 \[ \frac {2\,\left (B\,a^3\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )+B\,a^3\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )+B\,a^3\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )\right )}{d}+\frac {A\,a^3\,\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 {6\,C\,a^3\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {4\,C\,a^3\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,C\,a^3\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )}{d}-\frac {6\,A\,a^3\,{\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\,a^3\,{\cos \left (c+d\,x\right )}^{9/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {9}{4};\ \frac {13}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{3\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {2\,A\,a^3\,{\cos \left (c+d\,x\right )}^{11/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {11}{4};\ \frac {15}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{11\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {6\,B\,a^3\,{\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\,B\,a^3\,{\cos \left (c+d\,x\right )}^{9/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {9}{4};\ \frac {13}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{9\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {2\,C\,a^3\,{\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}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^(9/2)*(a + a/cos(c + d*x))^3*(A + B/cos(c + d*x) + C/cos(c + d*x)^2),x)

[Out]

(2*(B*a^3*ellipticE(c/2 + (d*x)/2, 2) + B*a^3*ellipticF(c/2 + (d*x)/2, 2) + B*a^3*cos(c + d*x)^(1/2)*sin(c + d
*x)))/d + (A*a^3*((2*cos(c + d*x)^(1/2)*sin(c + d*x))/3 + (2*ellipticF(c/2 + (d*x)/2, 2))/3))/d + (6*C*a^3*ell
ipticE(c/2 + (d*x)/2, 2))/d + (4*C*a^3*ellipticF(c/2 + (d*x)/2, 2))/d + (2*C*a^3*cos(c + d*x)^(1/2)*sin(c + d*
x))/d - (6*A*a^3*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*a^3*cos(c + d*x)^(9/2)*sin(c + d*x)*hypergeom([1/2, 9/4], 13/4, cos(c + d*x)^2))/(3*d*(sin
(c + d*x)^2)^(1/2)) - (2*A*a^3*cos(c + d*x)^(11/2)*sin(c + d*x)*hypergeom([1/2, 11/4], 15/4, cos(c + d*x)^2))/
(11*d*(sin(c + d*x)^2)^(1/2)) - (6*B*a^3*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*B*a^3*cos(c + d*x)^(9/2)*sin(c + d*x)*hypergeom([1/2, 9/4], 13/4, co
s(c + d*x)^2))/(9*d*(sin(c + d*x)^2)^(1/2)) - (2*C*a^3*cos(c + d*x)^(7/2)*sin(c + d*x)*hypergeom([1/2, 7/4], 1
1/4, cos(c + d*x)^2))/(7*d*(sin(c + d*x)^2)^(1/2))

________________________________________________________________________________________

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)**(9/2)*(a+a*sec(d*x+c))**3*(A+B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

Timed out

________________________________________________________________________________________