Integrand size = 33, antiderivative size = 137 \[ \int \frac {\sqrt {\cos (c+d x)} (A+B \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx=\frac {(4 A-B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d}-\frac {(5 A-2 B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 a^2 d}-\frac {(5 A-2 B) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac {(A-B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2} \]
(4*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^2/d-1/3*(5*A-2*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^2/d-1/3*(A-B)*cos(d* x+c)^(3/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^2-1/3*(5*A-2*B)*sin(d*x+c)*cos(d* x+c)^(1/2)/a^2/d/(1+cos(d*x+c))
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 8.11 (sec) , antiderivative size = 1079, normalized size of antiderivative = 7.88 \[ \int \frac {\sqrt {\cos (c+d x)} (A+B \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx =\text {Too large to display} \]
(10*A*Cos[c/2 + (d*x)/2]^4*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, S in[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*Sec[c + d*x]*(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[Co t[c]]]])/(3*d*(B + A*Cos[c + d*x])*Sqrt[1 + Cot[c]^2]*(a + a*Sec[c + d*x]) ^2) - (4*B*Cos[c/2 + (d*x)/2]^4*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/ 4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*Sec[c + d*x]*(A + B*Sec[c + d*x] )*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqr t[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcT an[Cot[c]]]])/(3*d*(B + A*Cos[c + d*x])*Sqrt[1 + Cot[c]^2]*(a + a*Sec[c + d*x])^2) + (Cos[c/2 + (d*x)/2]^4*(A + B*Sec[c + d*x])*((-4*(2*A - B + 2*A* Cos[c])*Csc[c])/d + (4*Sec[c/2]*Sec[c/2 + (d*x)/2]*(-2*A*Sin[(d*x)/2] + B* Sin[(d*x)/2]))/d - (2*Sec[c/2]*Sec[c/2 + (d*x)/2]^3*(-(A*Sin[(d*x)/2]) + B *Sin[(d*x)/2]))/(3*d) - (2*(-A + B)*Sec[c/2 + (d*x)/2]^2*Tan[c/2])/(3*d))) /(Sqrt[Cos[c + d*x]]*(B + A*Cos[c + d*x])*(a + a*Sec[c + d*x])^2) - (4*A*C os[c/2 + (d*x)/2]^4*Csc[c/2]*Sec[c/2]*Sec[c + d*x]*(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])/Sq...
Time = 0.90 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.05, 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, 3456, 3042, 3227, 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.
| \(\displaystyle \int \frac {\sqrt {\cos (c+d x)} (A+B \sec (c+d x))}{(a \sec (c+d x)+a)^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^2}dx\) |
| \(\Big \downarrow \) 3433 |
| \(\displaystyle \int \frac {\cos ^{\frac {3}{2}}(c+d x) (A \cos (c+d x)+B)}{(a \cos (c+d x)+a)^2}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (A \sin \left (c+d x+\frac {\pi }{2}\right )+B\right )}{\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^2}dx\) |
| \(\Big \downarrow \) 3456 |
| \(\displaystyle \frac {\int -\frac {\sqrt {\cos (c+d x)} (3 a (A-B)-a (7 A-B) \cos (c+d x))}{2 (\cos (c+d x) a+a)}dx}{3 a^2}-\frac {(A-B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\int \frac {\sqrt {\cos (c+d x)} (3 a (A-B)-a (7 A-B) \cos (c+d x))}{\cos (c+d x) a+a}dx}{6 a^2}-\frac {(A-B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (3 a (A-B)-a (7 A-B) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{6 a^2}-\frac {(A-B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3456 |
| \(\displaystyle -\frac {\frac {\int \frac {a^2 (5 A-2 B)-3 a^2 (4 A-B) \cos (c+d x)}{\sqrt {\cos (c+d x)}}dx}{a^2}+\frac {2 (5 A-2 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{d (\cos (c+d x)+1)}}{6 a^2}-\frac {(A-B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\frac {\int \frac {a^2 (5 A-2 B)-3 a^2 (4 A-B) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a^2}+\frac {2 (5 A-2 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{d (\cos (c+d x)+1)}}{6 a^2}-\frac {(A-B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle -\frac {\frac {a^2 (5 A-2 B) \int \frac {1}{\sqrt {\cos (c+d x)}}dx-3 a^2 (4 A-B) \int \sqrt {\cos (c+d x)}dx}{a^2}+\frac {2 (5 A-2 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{d (\cos (c+d x)+1)}}{6 a^2}-\frac {(A-B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\frac {a^2 (5 A-2 B) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-3 a^2 (4 A-B) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}+\frac {2 (5 A-2 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{d (\cos (c+d x)+1)}}{6 a^2}-\frac {(A-B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle -\frac {\frac {a^2 (5 A-2 B) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {6 a^2 (4 A-B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{a^2}+\frac {2 (5 A-2 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{d (\cos (c+d x)+1)}}{6 a^2}-\frac {(A-B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle -\frac {\frac {\frac {2 a^2 (5 A-2 B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}-\frac {6 a^2 (4 A-B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{a^2}+\frac {2 (5 A-2 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{d (\cos (c+d x)+1)}}{6 a^2}-\frac {(A-B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
-1/3*((A - B)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(d*(a + a*Cos[c + d*x])^2) - (((-6*a^2*(4*A - B)*EllipticE[(c + d*x)/2, 2])/d + (2*a^2*(5*A - 2*B)*El lipticF[(c + d*x)/2, 2])/d)/a^2 + (2*(5*A - 2*B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(d*(1 + Cos[c + d*x])))/(6*a^2)
3.6.4.3.1 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[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])
Leaf count of result is larger than twice the leaf count of optimal. \(420\) vs. \(2(179)=358\).
Time = 9.21 (sec) , antiderivative size = 421, normalized size of antiderivative = 3.07
| method | result | size |
| default | \(\frac {\sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (24 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+10 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+24 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-12 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-4 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-6 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-38 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+20 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+15 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-9 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-A +B \right )}{6 a^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\) | \(421\) |
1/6*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(24*A*cos(1/2* d*x+1/2*c)^6+10*A*cos(1/2*d*x+1/2*c)^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*co s(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+24*A*cos (1/2*d*x+1/2*c)^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1) ^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-12*B*cos(1/2*d*x+1/2*c)^6-4*B *cos(1/2*d*x+1/2*c)^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^ 2+1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-6*B*cos(1/2*d*x+1/2*c)^3* (sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticE(c os(1/2*d*x+1/2*c),2^(1/2))-38*A*cos(1/2*d*x+1/2*c)^4+20*B*cos(1/2*d*x+1/2* c)^4+15*A*cos(1/2*d*x+1/2*c)^2-9*B*cos(1/2*d*x+1/2*c)^2-A+B)/a^2/cos(1/2*d *x+1/2*c)^3/(-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 = 0.10 (sec) , antiderivative size = 356, normalized size of antiderivative = 2.60 \[ \int \frac {\sqrt {\cos (c+d x)} (A+B \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx=-\frac {2 \, {\left (3 \, {\left (2 \, A - B\right )} \cos \left (d x + c\right ) + 5 \, A - 2 \, B\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - {\left (\sqrt {2} {\left (5 i \, A - 2 i \, B\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} {\left (-5 i \, A + 2 i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (5 i \, A - 2 i \, B\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - {\left (\sqrt {2} {\left (-5 i \, A + 2 i \, B\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} {\left (5 i \, A - 2 i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-5 i \, A + 2 i \, B\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 \, {\left (\sqrt {2} {\left (-4 i \, A + i \, B\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} {\left (-4 i \, A + i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-4 i \, A + i \, B\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 \, {\left (\sqrt {2} {\left (4 i \, A - i \, B\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} {\left (4 i \, A - i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (4 i \, A - i \, B\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}{6 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \]
-1/6*(2*(3*(2*A - B)*cos(d*x + c) + 5*A - 2*B)*sqrt(cos(d*x + c))*sin(d*x + c) - (sqrt(2)*(5*I*A - 2*I*B)*cos(d*x + c)^2 - 2*sqrt(2)*(-5*I*A + 2*I*B )*cos(d*x + c) + sqrt(2)*(5*I*A - 2*I*B))*weierstrassPInverse(-4, 0, cos(d *x + c) + I*sin(d*x + c)) - (sqrt(2)*(-5*I*A + 2*I*B)*cos(d*x + c)^2 - 2*s qrt(2)*(5*I*A - 2*I*B)*cos(d*x + c) + sqrt(2)*(-5*I*A + 2*I*B))*weierstras sPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 3*(sqrt(2)*(-4*I*A + I*B )*cos(d*x + c)^2 + 2*sqrt(2)*(-4*I*A + I*B)*cos(d*x + c) + sqrt(2)*(-4*I*A + I*B))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*(sqrt(2)*(4*I*A - I*B)*cos(d*x + c)^2 + 2*sqrt(2)*(4* I*A - I*B)*cos(d*x + c) + sqrt(2)*(4*I*A - I*B))*weierstrassZeta(-4, 0, we ierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)
\[ \int \frac {\sqrt {\cos (c+d x)} (A+B \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx=\frac {\int \frac {A \sqrt {\cos {\left (c + d x \right )}}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {B \sqrt {\cos {\left (c + d x \right )}} \sec {\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx}{a^{2}} \]
(Integral(A*sqrt(cos(c + d*x))/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1), x) + Integral(B*sqrt(cos(c + d*x))*sec(c + d*x)/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1), x))/a**2
\[ \int \frac {\sqrt {\cos (c+d x)} (A+B \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx=\int { \frac {{\left (B \sec \left (d x + c\right ) + A\right )} \sqrt {\cos \left (d x + c\right )}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
\[ \int \frac {\sqrt {\cos (c+d x)} (A+B \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx=\int { \frac {{\left (B \sec \left (d x + c\right ) + A\right )} \sqrt {\cos \left (d x + c\right )}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {\sqrt {\cos (c+d x)} (A+B \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx=\int \frac {\sqrt {\cos \left (c+d\,x\right )}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^2} \,d x \]