\(\int \frac {\cos ^{\frac {5}{2}}(c+d x) (A+C \sec ^2(c+d x))}{a+a \sec (c+d x)} \, dx\) [1108]

Optimal result

Integrand size = 35, antiderivative size = 159 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=\frac {3 (7 A+5 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a d}-\frac {(5 A+3 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 a d}-\frac {(5 A+3 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 a d}+\frac {(7 A+5 C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 a d}-\frac {(A+C) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{d (a+a \cos (c+d x))} \]

[Out]

3/5*(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))/a/d-1/3*(5
*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))/a/d+1/5*(7*A+5*C
)*cos(d*x+c)^(3/2)*sin(d*x+c)/a/d-(A+C)*cos(d*x+c)^(5/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))-1/3*(5*A+3*C)*sin(d*x+c
)*cos(d*x+c)^(1/2)/a/d

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {4199, 3121, 2827, 2715, 2720, 2719} \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=-\frac {(5 A+3 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 a d}+\frac {3 (7 A+5 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a d}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{d (a \cos (c+d x)+a)}+\frac {(7 A+5 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 a d}-\frac {(5 A+3 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 a d} \]

[In]

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

[Out]

(3*(7*A + 5*C)*EllipticE[(c + d*x)/2, 2])/(5*a*d) - ((5*A + 3*C)*EllipticF[(c + d*x)/2, 2])/(3*a*d) - ((5*A +
3*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*a*d) + ((7*A + 5*C)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(5*a*d) - ((A +
C)*Cos[c + d*x]^(5/2)*Sin[c + d*x])/(d*(a + a*Cos[c + d*x]))

Rule 2715

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Sin[c + d*x])^(n - 1)/(d*n))
, x] + Dist[b^2*((n - 1)/n), Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integ
erQ[2*n]

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 3121

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

Rule 4199

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*} \text {integral}& = \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (C+A \cos ^2(c+d x)\right )}{a+a \cos (c+d x)} \, dx \\ & = -\frac {(A+C) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{d (a+a \cos (c+d x))}+\frac {\int \cos ^{\frac {3}{2}}(c+d x) \left (-\frac {1}{2} a (5 A+3 C)+\frac {1}{2} a (7 A+5 C) \cos (c+d x)\right ) \, dx}{a^2} \\ & = -\frac {(A+C) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{d (a+a \cos (c+d x))}-\frac {(5 A+3 C) \int \cos ^{\frac {3}{2}}(c+d x) \, dx}{2 a}+\frac {(7 A+5 C) \int \cos ^{\frac {5}{2}}(c+d x) \, dx}{2 a} \\ & = -\frac {(5 A+3 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 a d}+\frac {(7 A+5 C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 a d}-\frac {(A+C) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{d (a+a \cos (c+d x))}-\frac {(5 A+3 C) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{6 a}+\frac {(3 (7 A+5 C)) \int \sqrt {\cos (c+d x)} \, dx}{10 a} \\ & = \frac {3 (7 A+5 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a d}-\frac {(5 A+3 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 a d}-\frac {(5 A+3 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 a d}+\frac {(7 A+5 C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 a d}-\frac {(A+C) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{d (a+a \cos (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 = 8.31 (sec) , antiderivative size = 1107, normalized size of antiderivative = 6.96 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=\frac {\cos ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \cos ^{\frac {3}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right ) \left (-\frac {4 (5 A+5 C+16 A \cos (c)+10 C \cos (c)) \csc (c)}{5 d}-\frac {8 A \cos (d x) \sin (c)}{3 d}+\frac {4 A \cos (2 d x) \sin (2 c)}{5 d}-\frac {4 \sec \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}+\frac {d x}{2}\right ) \left (A \sin \left (\frac {d x}{2}\right )+C \sin \left (\frac {d x}{2}\right )\right )}{d}-\frac {8 A \cos (c) \sin (d x)}{3 d}+\frac {4 A \cos (2 c) \sin (2 d x)}{5 d}\right )}{(A+2 C+A \cos (2 c+2 d x)) (a+a \sec (c+d x))}+\frac {10 A \cos ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \cos (c+d x) \csc \left (\frac {c}{2}\right ) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2(d x-\arctan (\cot (c)))\right ) \sec \left (\frac {c}{2}\right ) \left (A+C \sec ^2(c+d x)\right ) \sec (d x-\arctan (\cot (c))) \sqrt {1-\sin (d x-\arctan (\cot (c)))} \sqrt {-\sqrt {1+\cot ^2(c)} \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 ^2(c)} (a+a \sec (c+d x))}+\frac {2 C \cos ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \cos (c+d x) \csc \left (\frac {c}{2}\right ) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2(d x-\arctan (\cot (c)))\right ) \sec \left (\frac {c}{2}\right ) \left (A+C \sec ^2(c+d x)\right ) \sec (d x-\arctan (\cot (c))) \sqrt {1-\sin (d x-\arctan (\cot (c)))} \sqrt {-\sqrt {1+\cot ^2(c)} \sin (c) \sin (d x-\arctan (\cot (c)))} \sqrt {1+\sin (d x-\arctan (\cot (c)))}}{d (A+2 C+A \cos (2 c+2 d x)) \sqrt {1+\cot ^2(c)} (a+a \sec (c+d x))}-\frac {21 A \cos ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \cos (c+d x) \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right ) \left (A+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+\arctan (\tan (c)))\right ) \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 ^2(c)}} \sqrt {1+\tan ^2(c)}}-\frac {\frac {\sin (d x+\arctan (\tan (c))) \tan (c)}{\sqrt {1+\tan ^2(c)}}+\frac {2 \cos ^2(c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}}{\cos ^2(c)+\sin ^2(c)}}{\sqrt {\cos (c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}}}\right )}{5 d (A+2 C+A \cos (2 c+2 d x)) (a+a \sec (c+d x))}-\frac {3 C \cos ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \cos (c+d x) \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right ) \left (A+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+\arctan (\tan (c)))\right ) \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 ^2(c)}} \sqrt {1+\tan ^2(c)}}-\frac {\frac {\sin (d x+\arctan (\tan (c))) \tan (c)}{\sqrt {1+\tan ^2(c)}}+\frac {2 \cos ^2(c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}}{\cos ^2(c)+\sin ^2(c)}}{\sqrt {\cos (c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}}}\right )}{d (A+2 C+A \cos (2 c+2 d x)) (a+a \sec (c+d x))} \]

[In]

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

[Out]

(Cos[c/2 + (d*x)/2]^2*Cos[c + d*x]^(3/2)*(A + C*Sec[c + d*x]^2)*((-4*(5*A + 5*C + 16*A*Cos[c] + 10*C*Cos[c])*C
sc[c])/(5*d) - (8*A*Cos[d*x]*Sin[c])/(3*d) + (4*A*Cos[2*d*x]*Sin[2*c])/(5*d) - (4*Sec[c/2]*Sec[c/2 + (d*x)/2]*
(A*Sin[(d*x)/2] + C*Sin[(d*x)/2]))/d - (8*A*Cos[c]*Sin[d*x])/(3*d) + (4*A*Cos[2*c]*Sin[2*d*x])/(5*d)))/((A + 2
*C + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])) + (10*A*Cos[c/2 + (d*x)/2]^2*Cos[c + d*x]*Csc[c/2]*Hypergeometr
icPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*(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 + Si
n[d*x - ArcTan[Cot[c]]]])/(3*d*(A + 2*C + A*Cos[2*c + 2*d*x])*Sqrt[1 + Cot[c]^2]*(a + a*Sec[c + d*x])) + (2*C*
Cos[c/2 + (d*x)/2]^2*Cos[c + d*x]*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*S
ec[c/2]*(A + C*Sec[c + d*x]^2)*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + C
ot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(d*(A + 2*C + A*Cos[2*c + 2*d
*x])*Sqrt[1 + Cot[c]^2]*(a + a*Sec[c + d*x])) - (21*A*Cos[c/2 + (d*x)/2]^2*Cos[c + d*x]*Csc[c/2]*Sec[c/2]*(A +
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 + Ar
cTan[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 + ArcT
an[Tan[c]]]*Sqrt[1 + Tan[c]^2]]))/(5*d*(A + 2*C + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])) - (3*C*Cos[c/2 + (
d*x)/2]^2*Cos[c + d*x]*Csc[c/2]*Sec[c/2]*(A + 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*(A + 2*C + A*Cos[2*c + 2*d*x
])*(a + a*Sec[c + d*x]))

Maple [A] (verified)

Time = 9.35 (sec) , antiderivative size = 277, normalized size of antiderivative = 1.74

method	result	size
default	\(-\frac {\sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (-\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \left (25 A \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+63 A \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+15 C \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+45 C \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )+48 A \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}-56 A \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (-30 A -30 C \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (23 A +15 C \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{15 a \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\)	\(277\)

[In]

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

[Out]

-1/15*((2*cos(1/2*d*x+1/2*c)^2-1)*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)^2)^(1/2
)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(25*A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+63*A*EllipticE(cos(1/2*d*x+1/2*
c),2^(1/2))+15*C*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+45*C*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))+48*A*sin(1/
2*d*x+1/2*c)^8-56*A*sin(1/2*d*x+1/2*c)^6+(-30*A-30*C)*sin(1/2*d*x+1/2*c)^4+(23*A+15*C)*sin(1/2*d*x+1/2*c)^2)/a
/cos(1/2*d*x+1/2*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*cos(1/2*d*x+1/2
*c)^2-1)^(1/2)/d

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.12 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.65 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=\frac {2 \, {\left (6 \, A \cos \left (d x + c\right )^{2} - 4 \, A \cos \left (d x + c\right ) - 25 \, A - 15 \, C\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 5 \, {\left (\sqrt {2} {\left (-5 i \, A - 3 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-5 i \, A - 3 i \, C\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 5 \, {\left (\sqrt {2} {\left (5 i \, A + 3 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (5 i \, A + 3 i \, C\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 9 \, {\left (\sqrt {2} {\left (-7 i \, A - 5 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-7 i \, A - 5 i \, C\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 9 \, {\left (\sqrt {2} {\left (7 i \, A + 5 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (7 i \, A + 5 i \, C\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}{30 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}} \]

[In]

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

[Out]

1/30*(2*(6*A*cos(d*x + c)^2 - 4*A*cos(d*x + c) - 25*A - 15*C)*sqrt(cos(d*x + c))*sin(d*x + c) - 5*(sqrt(2)*(-5
*I*A - 3*I*C)*cos(d*x + c) + sqrt(2)*(-5*I*A - 3*I*C))*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c
)) - 5*(sqrt(2)*(5*I*A + 3*I*C)*cos(d*x + c) + sqrt(2)*(5*I*A + 3*I*C))*weierstrassPInverse(-4, 0, cos(d*x + c
) - I*sin(d*x + c)) - 9*(sqrt(2)*(-7*I*A - 5*I*C)*cos(d*x + c) + sqrt(2)*(-7*I*A - 5*I*C))*weierstrassZeta(-4,
0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 9*(sqrt(2)*(7*I*A + 5*I*C)*cos(d*x + c) + sqr
t(2)*(7*I*A + 5*I*C))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))))/(a*d*
cos(d*x + c) + a*d)

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} \cos \left (d x + c\right )^{\frac {5}{2}}}{a \sec \left (d x + c\right ) + a} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} \cos \left (d x + c\right )^{\frac {5}{2}}}{a \sec \left (d x + c\right ) + a} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^{5/2}\,\left (A+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{a+\frac {a}{\cos \left (c+d\,x\right )}} \,d x \]

[In]

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

[Out]

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