\(\int \frac {\cos ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x))}{a+a \sec (c+d x)} \, dx\) [496]

Optimal result

Integrand size = 33, antiderivative size = 157 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x))}{a+a \sec (c+d x)} \, dx=\frac {3 (7 A-5 B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a d}-\frac {5 (A-B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 a d}-\frac {5 (A-B) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 a d}+\frac {(7 A-5 B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 a d}-\frac {(A-B) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{d (a+a \cos (c+d x))} \]

[Out]

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

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3033, 3056, 2827, 2715, 2720, 2719} \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x))}{a+a \sec (c+d x)} \, dx=-\frac {5 (A-B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 a d}+\frac {3 (7 A-5 B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a d}-\frac {(A-B) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{d (a \cos (c+d x)+a)}+\frac {(7 A-5 B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 a d}-\frac {5 (A-B) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 a d} \]

[In]

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

[Out]

(3*(7*A - 5*B)*EllipticE[(c + d*x)/2, 2])/(5*a*d) - (5*(A - B)*EllipticF[(c + d*x)/2, 2])/(3*a*d) - (5*(A - B)
*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*a*d) + ((7*A - 5*B)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(5*a*d) - ((A - B)*C
os[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 3033

Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.)*((g_.)*sin[(e_.
) + (f_.)*(x_)])^(p_.), x_Symbol] :> Dist[g^(m + n), Int[(g*Sin[e + f*x])^(p - m - n)*(b + a*Sin[e + f*x])^m*(
d + c*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[p] && I
ntegerQ[m] && IntegerQ[n]

Rule 3056

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {\cos ^{\frac {5}{2}}(c+d x) (B+A \cos (c+d x))}{a+a \cos (c+d x)} \, dx \\ & = -\frac {(A-B) \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 {5}{2} a (A-B)+\frac {1}{2} a (7 A-5 B) \cos (c+d x)\right ) \, dx}{a^2} \\ & = -\frac {(A-B) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{d (a+a \cos (c+d x))}+\frac {(7 A-5 B) \int \cos ^{\frac {5}{2}}(c+d x) \, dx}{2 a}-\frac {(5 (A-B)) \int \cos ^{\frac {3}{2}}(c+d x) \, dx}{2 a} \\ & = -\frac {5 (A-B) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 a d}+\frac {(7 A-5 B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 a d}-\frac {(A-B) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{d (a+a \cos (c+d x))}+\frac {(3 (7 A-5 B)) \int \sqrt {\cos (c+d x)} \, dx}{10 a}-\frac {(5 (A-B)) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{6 a} \\ & = \frac {3 (7 A-5 B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a d}-\frac {5 (A-B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 a d}-\frac {5 (A-B) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 a d}+\frac {(7 A-5 B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 a d}-\frac {(A-B) \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 = 8.46 (sec) , antiderivative size = 1056, normalized size of antiderivative = 6.73 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x))}{a+a \sec (c+d x)} \, dx=\frac {\cos ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {\cos (c+d x)} (A+B \sec (c+d x)) \left (\frac {2 (-5 A+5 B-16 A \cos (c)+10 B \cos (c)) \csc (c)}{5 d}+\frac {4 (-A+B) \cos (d x) \sin (c)}{3 d}+\frac {2 A \cos (2 d x) \sin (2 c)}{5 d}+\frac {2 \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 )+B \sin \left (\frac {d x}{2}\right )\right )}{d}+\frac {4 (-A+B) \cos (c) \sin (d x)}{3 d}+\frac {2 A \cos (2 c) \sin (2 d x)}{5 d}\right )}{(B+A \cos (c+d x)) (a+a \sec (c+d x))}+\frac {5 A \cos ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \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 ) (A+B \sec (c+d x)) \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 (B+A \cos (c+d x)) \sqrt {1+\cot ^2(c)} (a+a \sec (c+d x))}-\frac {5 B \cos ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \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 ) (A+B \sec (c+d x)) \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 (B+A \cos (c+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 ) \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right ) (A+B \sec (c+d x)) \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 )}{10 d (B+A \cos (c+d x)) (a+a \sec (c+d x))}+\frac {3 B \cos ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right ) (A+B \sec (c+d x)) \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 )}{2 d (B+A \cos (c+d x)) (a+a \sec (c+d x))} \]

[In]

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

[Out]

(Cos[c/2 + (d*x)/2]^2*Sqrt[Cos[c + d*x]]*(A + B*Sec[c + d*x])*((2*(-5*A + 5*B - 16*A*Cos[c] + 10*B*Cos[c])*Csc
[c])/(5*d) + (4*(-A + B)*Cos[d*x]*Sin[c])/(3*d) + (2*A*Cos[2*d*x]*Sin[2*c])/(5*d) + (2*Sec[c/2]*Sec[c/2 + (d*x
)/2]*(-(A*Sin[(d*x)/2]) + B*Sin[(d*x)/2]))/d + (4*(-A + B)*Cos[c]*Sin[d*x])/(3*d) + (2*A*Cos[2*c]*Sin[2*d*x])/
(5*d)))/((B + A*Cos[c + d*x])*(a + a*Sec[c + d*x])) + (5*A*Cos[c/2 + (d*x)/2]^2*Csc[c/2]*HypergeometricPFQ[{1/
4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*(A + B*Sec[c + d*x])*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 - Arc
Tan[Cot[c]]]])/(3*d*(B + A*Cos[c + d*x])*Sqrt[1 + Cot[c]^2]*(a + a*Sec[c + d*x])) - (5*B*Cos[c/2 + (d*x)/2]^2*
Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*(A + B*Sec[c + d*x])*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])*Sqrt[1 + Cot[c]^2]*(a + a*Sec[c + d*x])
) - (21*A*Cos[c/2 + (d*x)/2]^2*Csc[c/2]*Sec[c/2]*(A + B*Sec[c + d*x])*((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]) - ((S
in[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*(B + A*Cos[c + d*x]
)*(a + a*Sec[c + d*x])) + (3*B*Cos[c/2 + (d*x)/2]^2*Csc[c/2]*Sec[c/2]*(A + B*Sec[c + d*x])*((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 + ArcTa
n[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
*(B + A*Cos[c + d*x])*(a + a*Sec[c + d*x]))

Maple [A] (verified)

Time = 10.28 (sec) , antiderivative size = 282, normalized size of antiderivative = 1.80

method	result	size
default	\(-\frac {\sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (-\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \left (25 A \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+63 A \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-25 B \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-45 B \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )+48 A \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\left (-56 A -40 B \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (-30 A +90 B \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (23 A -35 B \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{15 a \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\)	\(282\)

[In]

int(cos(d*x+c)^(5/2)*(A+B*sec(d*x+c))/(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)*(2*sin(1/2*d*x+1/2*c)^2-1)^
(1/2)*(sin(1/2*d*x+1/2*c)^2)^(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))-25*B*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-45*B*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-40*B)*sin(1/2*d*x+1/2*c)^6+(-30*A+90*B)*sin(1/2*d*x+1/2*c)^4+(23*A-35*B)*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 = 0.11 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.71 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x))}{a+a \sec (c+d x)} \, dx=\frac {2 \, {\left (6 \, A \cos \left (d x + c\right )^{2} - 2 \, {\left (2 \, A - 5 \, B\right )} \cos \left (d x + c\right ) - 25 \, A + 25 \, B\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 25 \, {\left (\sqrt {2} {\left (-i \, A + i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-i \, A + i \, B\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 25 \, {\left (\sqrt {2} {\left (i \, A - i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (i \, A - i \, B\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 9 \, {\left (\sqrt {2} {\left (-7 i \, A + 5 i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-7 i \, A + 5 i \, B\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 9 \, {\left (\sqrt {2} {\left (7 i \, A - 5 i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (7 i \, A - 5 i \, B\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}{30 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}} \]

[In]

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

[Out]

1/30*(2*(6*A*cos(d*x + c)^2 - 2*(2*A - 5*B)*cos(d*x + c) - 25*A + 25*B)*sqrt(cos(d*x + c))*sin(d*x + c) - 25*(
sqrt(2)*(-I*A + I*B)*cos(d*x + c) + sqrt(2)*(-I*A + I*B))*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x
+ c)) - 25*(sqrt(2)*(I*A - I*B)*cos(d*x + c) + sqrt(2)*(I*A - I*B))*weierstrassPInverse(-4, 0, cos(d*x + c) -
I*sin(d*x + c)) - 9*(sqrt(2)*(-7*I*A + 5*I*B)*cos(d*x + c) + sqrt(2)*(-7*I*A + 5*I*B))*weierstrassZeta(-4, 0,
weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 9*(sqrt(2)*(7*I*A - 5*I*B)*cos(d*x + c) + sqrt(2)
*(7*I*A - 5*I*B))*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. \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x))}{a+a \sec (c+d x)} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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