Integrand size = 33, antiderivative size = 124 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \sec (c+d x))}{a+a \sec (c+d x)} \, dx=-\frac {3 (A-B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d}+\frac {(5 A-3 B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 a d}+\frac {(5 A-3 B) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 a d}-\frac {(A-B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{d (a+a \cos (c+d x))} \]
-3*(A-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+1/3*(5*A-3*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-(A-B)*cos(d*x+c)^(3 /2)*sin(d*x+c)/d/(a+a*cos(d*x+c))+1/3*(5*A-3*B)*sin(d*x+c)*cos(d*x+c)^(1/2 )/a/d
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 8.00 (sec) , antiderivative size = 1003, normalized size of antiderivative = 8.09 \[ \int \frac {\cos ^{\frac {3}{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 (-A+B) (1+2 \cos (c)) \csc (c)}{d}+\frac {4 A \cos (d x) \sin (c)}{3 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 \cos (c) \sin (d x)}{3 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 {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)))}}{d (B+A \cos (c+d x)) \sqrt {1+\cot ^2(c)} (a+a \sec (c+d x))}+\frac {3 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 )}{2 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))} \]
(Cos[c/2 + (d*x)/2]^2*Sqrt[Cos[c + d*x]]*(A + B*Sec[c + d*x])*((-2*(-A + B )*(1 + 2*Cos[c])*Csc[c])/d + (4*A*Cos[d*x]*Sin[c])/(3*d) - (2*Sec[c/2]*Sec [c/2 + (d*x)/2]*(-(A*Sin[(d*x)/2]) + B*Sin[(d*x)/2]))/d + (4*A*Cos[c]*Sin[ d*x])/(3*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 - ArcTan[Cot[c]]]])/(3*d*(B + A*Cos[c + d*x])*Sqrt[1 + Cot[c]^2]*(a + a*Sec[c + d*x])) + (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]]]])/(d*(B + A*Cos[c + d*x])*Sqrt[1 + Cot[c]^2]*(a + a*Sec[c + d*x])) + (3*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]) - ((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...
Time = 0.72 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.98, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {3042, 3433, 3042, 3456, 27, 3042, 3227, 3042, 3115, 3042, 3119, 3120}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \sec (c+d x))}{a \sec (c+d x)+a} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{a \csc \left (c+d x+\frac {\pi }{2}\right )+a}dx\) |
| \(\Big \downarrow \) 3433 |
| \(\displaystyle \int \frac {\cos ^{\frac {3}{2}}(c+d x) (A \cos (c+d x)+B)}{a \cos (c+d x)+a}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (A \sin \left (c+d x+\frac {\pi }{2}\right )+B\right )}{a \sin \left (c+d x+\frac {\pi }{2}\right )+a}dx\) |
| \(\Big \downarrow \) 3456 |
| \(\displaystyle \frac {\int -\frac {1}{2} \sqrt {\cos (c+d x)} (3 a (A-B)-a (5 A-3 B) \cos (c+d x))dx}{a^2}-\frac {(A-B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{d (a \cos (c+d x)+a)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\int \sqrt {\cos (c+d x)} (3 a (A-B)-a (5 A-3 B) \cos (c+d x))dx}{2 a^2}-\frac {(A-B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{d (a \cos (c+d x)+a)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (3 a (A-B)-a (5 A-3 B) \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx}{2 a^2}-\frac {(A-B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{d (a \cos (c+d x)+a)}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle -\frac {3 a (A-B) \int \sqrt {\cos (c+d x)}dx-a (5 A-3 B) \int \cos ^{\frac {3}{2}}(c+d x)dx}{2 a^2}-\frac {(A-B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{d (a \cos (c+d x)+a)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {3 a (A-B) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx-a (5 A-3 B) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}dx}{2 a^2}-\frac {(A-B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{d (a \cos (c+d x)+a)}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle -\frac {3 a (A-B) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx-a (5 A-3 B) \left (\frac {1}{3} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )}{2 a^2}-\frac {(A-B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{d (a \cos (c+d x)+a)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {3 a (A-B) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx-a (5 A-3 B) \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )}{2 a^2}-\frac {(A-B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{d (a \cos (c+d x)+a)}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle -\frac {\frac {6 a (A-B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}-a (5 A-3 B) \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )}{2 a^2}-\frac {(A-B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{d (a \cos (c+d x)+a)}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle -\frac {\frac {6 a (A-B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}-a (5 A-3 B) \left (\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )}{2 a^2}-\frac {(A-B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{d (a \cos (c+d x)+a)}\) |
-(((A - B)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(d*(a + a*Cos[c + d*x]))) - (( 6*a*(A - B)*EllipticE[(c + d*x)/2, 2])/d - a*(5*A - 3*B)*((2*EllipticF[(c + d*x)/2, 2])/(3*d) + (2*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*d)))/(2*a^2)
3.5.97.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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] + Simp[b^2*((n - 1)/n) Int[(b*Sin [c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 2*n]
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]
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]
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[c Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b Int [(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]* (d_.) + (c_))^(n_.)*((g_.)*sin[(e_.) + (f_.)*(x_)])^(p_.), x_Symbol] :> Sim p[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] && IntegerQ[m] && IntegerQ[n]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[(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] - Simp[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] /; Fre eQ[{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] && (In tegerQ[2*n] || EqQ[c, 0])
Time = 8.30 (sec) , antiderivative size = 262, normalized size of antiderivative = 2.11
| method | result | size |
| default | \(-\frac {\sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \left (5 A \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+9 A \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-3 B \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-9 B \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )-8 A \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (18 A -6 B \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (-7 A +3 B \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{3 a \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\) | \(262\) |
-1/3*((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)*(5*A* EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+9*A*EllipticE(cos(1/2*d*x+1/2*c),2^( 1/2))-3*B*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-9*B*EllipticE(cos(1/2*d*x+ 1/2*c),2^(1/2)))-8*A*sin(1/2*d*x+1/2*c)^6+(18*A-6*B)*sin(1/2*d*x+1/2*c)^4+ (-7*A+3*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
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.11 (sec) , antiderivative size = 250, normalized size of antiderivative = 2.02 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \sec (c+d x))}{a+a \sec (c+d x)} \, dx=\frac {2 \, {\left (2 \, A \cos \left (d x + c\right ) + 5 \, A - 3 \, B\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + {\left (\sqrt {2} {\left (-5 i \, A + 3 i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-5 i \, A + 3 i \, B\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + {\left (\sqrt {2} {\left (5 i \, A - 3 i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (5 i \, A - 3 i \, B\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 9 \, {\left (\sqrt {2} {\left (i \, A - i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (i \, A - i \, B\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 9 \, {\left (\sqrt {2} {\left (-i \, A + i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-i \, A + i \, B\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}{6 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}} \]
1/6*(2*(2*A*cos(d*x + c) + 5*A - 3*B)*sqrt(cos(d*x + c))*sin(d*x + c) + (s qrt(2)*(-5*I*A + 3*I*B)*cos(d*x + c) + sqrt(2)*(-5*I*A + 3*I*B))*weierstra ssPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + (sqrt(2)*(5*I*A - 3*I*B )*cos(d*x + c) + sqrt(2)*(5*I*A - 3*I*B))*weierstrassPInverse(-4, 0, cos(d *x + c) - I*sin(d*x + c)) - 9*(sqrt(2)*(I*A - I*B)*cos(d*x + c) + sqrt(2)* (I*A - I*B))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c ) + I*sin(d*x + c))) - 9*(sqrt(2)*(-I*A + I*B)*cos(d*x + c) + sqrt(2)*(-I* A + 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)
Timed out. \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \sec (c+d x))}{a+a \sec (c+d x)} \, dx=\text {Timed out} \]
\[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \sec (c+d x))}{a+a \sec (c+d x)} \, dx=\int { \frac {{\left (B \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{\frac {3}{2}}}{a \sec \left (d x + c\right ) + a} \,d x } \]
\[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \sec (c+d x))}{a+a \sec (c+d x)} \, dx=\int { \frac {{\left (B \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{\frac {3}{2}}}{a \sec \left (d x + c\right ) + a} \,d x } \]
Timed out. \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \sec (c+d x))}{a+a \sec (c+d x)} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^{3/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )}{a+\frac {a}{\cos \left (c+d\,x\right )}} \,d x \]