3.1101 \(\int \cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^3 (A+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=215 \[ \frac {4 a^3 (13 A+35 C) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}+\frac {4 a^3 (7 A+5 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {4 a^3 (41 A-35 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{105 d}+\frac {2 (11 A-35 C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^3 \cos (c+d x)+a^3\right )}{35 d}+\frac {2 (A-7 C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^2 \cos (c+d x)+a^2\right )^2}{7 a d}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^3}{d \sqrt {\cos (c+d x)}} \]

[Out]

4/5*a^3*(7*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+4/21
*a^3*(13*A+35*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*C*(
a+a*cos(d*x+c))^3*sin(d*x+c)/d/cos(d*x+c)^(1/2)+4/105*a^3*(41*A-35*C)*sin(d*x+c)*cos(d*x+c)^(1/2)/d+2/7*(A-7*C
)*(a^2+a^2*cos(d*x+c))^2*sin(d*x+c)*cos(d*x+c)^(1/2)/a/d+2/35*(11*A-35*C)*(a^3+a^3*cos(d*x+c))*sin(d*x+c)*cos(
d*x+c)^(1/2)/d

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Rubi [A]  time = 0.63, antiderivative size = 215, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.229, Rules used = {4114, 3044, 2976, 2968, 3023, 2748, 2641, 2639} \[ \frac {4 a^3 (13 A+35 C) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}+\frac {4 a^3 (7 A+5 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {4 a^3 (41 A-35 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{105 d}+\frac {2 (11 A-35 C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^3 \cos (c+d x)+a^3\right )}{35 d}+\frac {2 (A-7 C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^2 \cos (c+d x)+a^2\right )^2}{7 a d}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^3}{d \sqrt {\cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^(7/2)*(a + a*Sec[c + d*x])^3*(A + C*Sec[c + d*x]^2),x]

[Out]

(4*a^3*(7*A + 5*C)*EllipticE[(c + d*x)/2, 2])/(5*d) + (4*a^3*(13*A + 35*C)*EllipticF[(c + d*x)/2, 2])/(21*d) +
 (4*a^3*(41*A - 35*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(105*d) + (2*C*(a + a*Cos[c + d*x])^3*Sin[c + d*x])/(d*
Sqrt[Cos[c + d*x]]) + (2*(A - 7*C)*Sqrt[Cos[c + d*x]]*(a^2 + a^2*Cos[c + d*x])^2*Sin[c + d*x])/(7*a*d) + (2*(1
1*A - 35*C)*Sqrt[Cos[c + d*x]]*(a^3 + a^3*Cos[c + d*x])*Sin[c + d*x])/(35*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 3044

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((c^2*C + 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*(a*c*m + b*d*(n + 1)) - b*(A*d^2*(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, 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 4114

Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (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 + A
*Cos[e + f*x]^2), x], x] /; FreeQ[{a, b, d, e, f, A, C, n}, x] &&  !IntegerQ[n] && IntegerQ[m]

Rubi steps

\begin {align*} \int \cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx &=\int \frac {(a+a \cos (c+d x))^3 \left (C+A \cos ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\\ &=\frac {2 C (a+a \cos (c+d x))^3 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}+\frac {2 \int \frac {(a+a \cos (c+d x))^3 \left (3 a C+\frac {1}{2} a (A-7 C) \cos (c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx}{a}\\ &=\frac {2 C (a+a \cos (c+d x))^3 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}+\frac {2 (A-7 C) \sqrt {\cos (c+d x)} \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{7 a d}+\frac {4 \int \frac {(a+a \cos (c+d x))^2 \left (\frac {1}{4} a^2 (A+35 C)+\frac {1}{4} a^2 (11 A-35 C) \cos (c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx}{7 a}\\ &=\frac {2 C (a+a \cos (c+d x))^3 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}+\frac {2 (A-7 C) \sqrt {\cos (c+d x)} \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{7 a d}+\frac {2 (11 A-35 C) \sqrt {\cos (c+d x)} \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{35 d}+\frac {8 \int \frac {(a+a \cos (c+d x)) \left (\frac {1}{2} a^3 (4 A+35 C)+\frac {1}{4} a^3 (41 A-35 C) \cos (c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx}{35 a}\\ &=\frac {2 C (a+a \cos (c+d x))^3 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}+\frac {2 (A-7 C) \sqrt {\cos (c+d x)} \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{7 a d}+\frac {2 (11 A-35 C) \sqrt {\cos (c+d x)} \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{35 d}+\frac {8 \int \frac {\frac {1}{2} a^4 (4 A+35 C)+\left (\frac {1}{4} a^4 (41 A-35 C)+\frac {1}{2} a^4 (4 A+35 C)\right ) \cos (c+d x)+\frac {1}{4} a^4 (41 A-35 C) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)}} \, dx}{35 a}\\ &=\frac {4 a^3 (41 A-35 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{105 d}+\frac {2 C (a+a \cos (c+d x))^3 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}+\frac {2 (A-7 C) \sqrt {\cos (c+d x)} \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{7 a d}+\frac {2 (11 A-35 C) \sqrt {\cos (c+d x)} \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{35 d}+\frac {16 \int \frac {\frac {5}{8} a^4 (13 A+35 C)+\frac {21}{8} a^4 (7 A+5 C) \cos (c+d x)}{\sqrt {\cos (c+d x)}} \, dx}{105 a}\\ &=\frac {4 a^3 (41 A-35 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{105 d}+\frac {2 C (a+a \cos (c+d x))^3 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}+\frac {2 (A-7 C) \sqrt {\cos (c+d x)} \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{7 a d}+\frac {2 (11 A-35 C) \sqrt {\cos (c+d x)} \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{35 d}+\frac {1}{5} \left (2 a^3 (7 A+5 C)\right ) \int \sqrt {\cos (c+d x)} \, dx+\frac {1}{21} \left (2 a^3 (13 A+35 C)\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {4 a^3 (7 A+5 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {4 a^3 (13 A+35 C) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}+\frac {4 a^3 (41 A-35 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{105 d}+\frac {2 C (a+a \cos (c+d x))^3 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}+\frac {2 (A-7 C) \sqrt {\cos (c+d x)} \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{7 a d}+\frac {2 (11 A-35 C) \sqrt {\cos (c+d x)} \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{35 d}\\ \end {align*}

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Mathematica [C]  time = 6.61, size = 1108, normalized size = 5.15 \[ \text {result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Cos[c + d*x]^(7/2)*(a + a*Sec[c + d*x])^3*(A + 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 + C*Sec[c + d*x]^2)*(-1/20*((14*A + 5*C +
14*A*Cos[2*c] + 15*C*Cos[2*c])*Csc[c]*Sec[c])/d + ((107*A + 28*C)*Cos[d*x]*Sin[c])/(168*d) + (3*A*Cos[2*d*x]*S
in[2*c])/(20*d) + (A*Cos[3*d*x]*Sin[3*c])/(56*d) + ((107*A + 28*C)*Cos[c]*Sin[d*x])/(168*d) + (C*Sec[c]*Sec[c
+ d*x]*Sin[d*x])/(2*d) + (3*A*Cos[2*c]*Sin[2*d*x])/(20*d) + (A*Cos[3*c]*Sin[3*d*x])/(56*d)))/(A + 2*C + A*Cos[
2*c + 2*d*x]) - (13*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 + 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 + A*Cos[2*c + 2*d*x])*Sqrt[1 + Cot[c]^2]) - (5*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 + 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 + A*Cos[2*c + 2*d*x])*Sqrt[1 + Cot[c]^2]
) - (7*A*Cos[c + d*x]^5*Csc[c]*Sec[c/2 + (d*x)/2]^6*(a + a*Sec[c + d*x])^3*(A + C*Sec[c + d*x]^2)*((Hypergeome
tricPFQ[{-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]]
))/(10*d*(A + 2*C + A*Cos[2*c + 2*d*x])) - (C*Cos[c + d*x]^5*Csc[c]*Sec[c/2 + (d*x)/2]^6*(a + a*Sec[c + d*x])^
3*(A + C*Sec[c + d*x]^2)*((HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x + ArcTa
n[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]]))/(2*d*(A + 2*C + A*Cos[2*c + 2*d*x]))

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fricas [F]  time = 0.70, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C a^{3} \cos \left (d x + c\right )^{3} \sec \left (d x + c\right )^{5} + 3 \, C a^{3} \cos \left (d x + c\right )^{3} \sec \left (d x + c\right )^{4} + {\left (A + 3 \, C\right )} a^{3} \cos \left (d x + c\right )^{3} \sec \left (d x + c\right )^{3} + {\left (3 \, A + C\right )} a^{3} \cos \left (d x + c\right )^{3} \sec \left (d x + c\right )^{2} + 3 \, A a^{3} \cos \left (d x + c\right )^{3} \sec \left (d x + c\right ) + A a^{3} \cos \left (d x + c\right )^{3}\right )} \sqrt {\cos \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(7/2)*(a+a*sec(d*x+c))^3*(A+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \sec \left (d x + c\right )^{2} + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{3} \cos \left (d x + c\right )^{\frac {7}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(7/2)*(a+a*sec(d*x+c))^3*(A+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + A)*(a*sec(d*x + c) + a)^3*cos(d*x + c)^(7/2), x)

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maple [B]  time = 5.22, size = 569, normalized size = 2.65 \[ -\frac {4 a^{3} \left (120 A \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 )}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-432 A \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 )}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+14 \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 )}\, \left (43 A +5 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 )-4 \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 )}\, \left (52 A +35 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 )+65 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}\, \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 )}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-147 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}\, \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 )}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+175 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 ) \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 )}-105 C \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 )}\, \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 )}{105 \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)^(7/2)*(a+a*sec(d*x+c))^3*(A+C*sec(d*x+c)^2),x)

[Out]

-4/105*a^3*(120*A*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8
-432*A*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6+14*(-2*sin
(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(43*A+5*C)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)-4*(-2*sin(1/2
*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(52*A+35*C)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+65*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)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*Ellipt
icF(cos(1/2*d*x+1/2*c),2^(1/2))-147*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)*(-2*sin(1/
2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+175*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))*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*
d*x+1/2*c)^2)^(1/2)-105*C*(-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)^(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

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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)^(7/2)*(a+a*sec(d*x+c))^3*(A+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

Timed out

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mupad [B]  time = 5.01, size = 269, normalized size = 1.25 \[ \frac {2\,\left (A\,a^3\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )+A\,a^3\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )+A\,a^3\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )\right )}{d}+\frac {C\,a^3\,\left (\frac {2\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )}{3}+\frac {2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{3}\right )}{d}+\frac {6\,C\,a^3\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {6\,C\,a^3\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}-\frac {6\,A\,a^3\,{\cos \left (c+d\,x\right )}^{7/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {11}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {2\,A\,a^3\,{\cos \left (c+d\,x\right )}^{9/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {9}{4};\ \frac {13}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{9\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {2\,C\,a^3\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^(7/2)*(A + C/cos(c + d*x)^2)*(a + a/cos(c + d*x))^3,x)

[Out]

(2*(A*a^3*ellipticE(c/2 + (d*x)/2, 2) + A*a^3*ellipticF(c/2 + (d*x)/2, 2) + A*a^3*cos(c + d*x)^(1/2)*sin(c + d
*x)))/d + (C*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 + (6*C*a^3*ellipticF(c/2 + (d*x)/2, 2))/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))/(9*d*(sin(c + d*x)^2)^(1/2)) + (2*C*a^3*sin(c + d*x)*hy
pergeom([-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))

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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)**(7/2)*(a+a*sec(d*x+c))**3*(A+C*sec(d*x+c)**2),x)

[Out]

Timed out

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