Integrand size = 33, antiderivative size = 172 \[ \int \frac {(a+b \sec (c+d x))^2 (A+B \sec (c+d x))}{\sqrt {\cos (c+d x)}} \, dx=-\frac {2 \left (10 a A b+5 a^2 B+3 b^2 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 \left (3 a^2 A+A b^2+2 a b B\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 b^2 B \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 b (A b+2 a B) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (10 a A b+5 a^2 B+3 b^2 B\right ) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}} \]
-2/5*(10*A*a*b+5*B*a^2+3*B*b^2)*(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+2/3*(3*A*a^2+A*b^2+2*B*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))/d+2/5*b^2*B*sin(d*x+c)/d/cos(d*x+c)^(5/2)+2/3*b*(A*b+2*B*a)*si n(d*x+c)/d/cos(d*x+c)^(3/2)+2/5*(10*A*a*b+5*B*a^2+3*B*b^2)*sin(d*x+c)/d/co s(d*x+c)^(1/2)
Time = 2.26 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.02 \[ \int \frac {(a+b \sec (c+d x))^2 (A+B \sec (c+d x))}{\sqrt {\cos (c+d x)}} \, dx=\frac {-6 \left (10 a A b+5 a^2 B+3 b^2 B\right ) \cos ^{\frac {3}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+10 \left (3 a^2 A+A b^2+2 a b B\right ) \cos ^{\frac {3}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+10 A b^2 \sin (c+d x)+20 a b B \sin (c+d x)+30 a A b \sin (2 (c+d x))+15 a^2 B \sin (2 (c+d x))+9 b^2 B \sin (2 (c+d x))+6 b^2 B \tan (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)} \]
(-6*(10*a*A*b + 5*a^2*B + 3*b^2*B)*Cos[c + d*x]^(3/2)*EllipticE[(c + d*x)/ 2, 2] + 10*(3*a^2*A + A*b^2 + 2*a*b*B)*Cos[c + d*x]^(3/2)*EllipticF[(c + d *x)/2, 2] + 10*A*b^2*Sin[c + d*x] + 20*a*b*B*Sin[c + d*x] + 30*a*A*b*Sin[2 *(c + d*x)] + 15*a^2*B*Sin[2*(c + d*x)] + 9*b^2*B*Sin[2*(c + d*x)] + 6*b^2 *B*Tan[c + d*x])/(15*d*Cos[c + d*x]^(3/2))
Time = 0.94 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.94, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {3042, 3433, 3042, 3467, 27, 3042, 3500, 27, 3042, 3227, 3042, 3116, 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.
| \(\displaystyle \int \frac {(a+b \sec (c+d x))^2 (A+B \sec (c+d x))}{\sqrt {\cos (c+d x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^2 \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx\) |
| \(\Big \downarrow \) 3433 |
| \(\displaystyle \int \frac {(a \cos (c+d x)+b)^2 (A \cos (c+d x)+B)}{\cos ^{\frac {7}{2}}(c+d x)}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+b\right )^2 \left (A \sin \left (c+d x+\frac {\pi }{2}\right )+B\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{7/2}}dx\) |
| \(\Big \downarrow \) 3467 |
| \(\displaystyle \frac {2 b^2 B \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}-\frac {2}{5} \int -\frac {5 a^2 A \cos ^2(c+d x)+\left (5 B a^2+10 A b a+3 b^2 B\right ) \cos (c+d x)+5 b (A b+2 a B)}{2 \cos ^{\frac {5}{2}}(c+d x)}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \int \frac {5 a^2 A \cos ^2(c+d x)+\left (5 B a^2+10 A b a+3 b^2 B\right ) \cos (c+d x)+5 b (A b+2 a B)}{\cos ^{\frac {5}{2}}(c+d x)}dx+\frac {2 b^2 B \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \int \frac {5 a^2 A \sin \left (c+d x+\frac {\pi }{2}\right )^2+\left (5 B a^2+10 A b a+3 b^2 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+5 b (A b+2 a B)}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx+\frac {2 b^2 B \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3500 |
| \(\displaystyle \frac {1}{5} \left (\frac {2}{3} \int \frac {3 \left (5 B a^2+10 A b a+3 b^2 B\right )+5 \left (3 A a^2+2 b B a+A b^2\right ) \cos (c+d x)}{2 \cos ^{\frac {3}{2}}(c+d x)}dx+\frac {10 b (2 a B+A b) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 b^2 B \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \int \frac {3 \left (5 B a^2+10 A b a+3 b^2 B\right )+5 \left (3 A a^2+2 b B a+A b^2\right ) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x)}dx+\frac {10 b (2 a B+A b) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 b^2 B \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \int \frac {3 \left (5 B a^2+10 A b a+3 b^2 B\right )+5 \left (3 A a^2+2 b B a+A b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+\frac {10 b (2 a B+A b) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 b^2 B \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (3 \left (5 a^2 B+10 a A b+3 b^2 B\right ) \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x)}dx+5 \left (3 a^2 A+2 a b B+A b^2\right ) \int \frac {1}{\sqrt {\cos (c+d x)}}dx\right )+\frac {10 b (2 a B+A b) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 b^2 B \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (3 \left (5 a^2 B+10 a A b+3 b^2 B\right ) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+5 \left (3 a^2 A+2 a b B+A b^2\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\frac {10 b (2 a B+A b) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 b^2 B \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3116 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (5 \left (3 a^2 A+2 a b B+A b^2\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+3 \left (5 a^2 B+10 a A b+3 b^2 B\right ) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\int \sqrt {\cos (c+d x)}dx\right )\right )+\frac {10 b (2 a B+A b) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 b^2 B \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (5 \left (3 a^2 A+2 a b B+A b^2\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+3 \left (5 a^2 B+10 a A b+3 b^2 B\right ) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )\right )+\frac {10 b (2 a B+A b) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 b^2 B \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (5 \left (3 a^2 A+2 a b B+A b^2\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+3 \left (5 a^2 B+10 a A b+3 b^2 B\right ) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )\right )+\frac {10 b (2 a B+A b) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 b^2 B \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\frac {10 \left (3 a^2 A+2 a b B+A b^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}+3 \left (5 a^2 B+10 a A b+3 b^2 B\right ) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )\right )+\frac {10 b (2 a B+A b) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 b^2 B \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
(2*b^2*B*Sin[c + d*x])/(5*d*Cos[c + d*x]^(5/2)) + ((10*b*(A*b + 2*a*B)*Sin [c + d*x])/(3*d*Cos[c + d*x]^(3/2)) + ((10*(3*a^2*A + A*b^2 + 2*a*b*B)*Ell ipticF[(c + d*x)/2, 2])/d + 3*(10*a*A*b + 5*a^2*B + 3*b^2*B)*((-2*Elliptic E[(c + d*x)/2, 2])/d + (2*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]])))/3)/5
3.6.72.3.1 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[Cos[c + d*x]*(( b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1))), x] + Simp[(n + 2)/(b^2*(n + 1)) I nt[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[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_)])^2*((A_.) + (B_.)*sin[(e_.) + (f _.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[ (B*c - A*d)*(b*c - a*d)^2*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(f*d^2 *(n + 1)*(c^2 - d^2))), x] - Simp[1/(d^2*(n + 1)*(c^2 - d^2)) Int[(c + d* Sin[e + f*x])^(n + 1)*Simp[d*(n + 1)*(B*(b*c - a*d)^2 - A*d*(a^2*c + b^2*c - 2*a*b*d)) - ((B*c - A*d)*(a^2*d^2*(n + 2) + b^2*(c^2 + d^2*(n + 1))) + 2* a*b*d*(A*c*d*(n + 2) - B*(c^2 + d^2*(n + 1))))*Sin[e + f*x] - b^2*B*d*(n + 1)*(c^2 - d^2)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B} , x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[ n, -1]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 - a*b*B + a^2*C))*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 1)* (a^2 - b^2))), x] + Simp[1/(b*(m + 1)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x ])^(m + 1)*Simp[b*(a*A - b*B + a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A *b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(722\) vs. \(2(208)=416\).
Time = 21.69 (sec) , antiderivative size = 723, normalized size of antiderivative = 4.20
-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*A*a^2*(sin(1 /2*d*x+1/2*c)^2)^(1/2)*(-2*cos(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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+ 2/5*b^2*B/(8*sin(1/2*d*x+1/2*c)^6-12*sin(1/2*d*x+1/2*c)^4+6*sin(1/2*d*x+1/ 2*c)^2-1)/sin(1/2*d*x+1/2*c)^2*(24*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c) -12*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2* sin(1/2*d*x+1/2*c)^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^4-24*sin(1/2*d*x+1/2*c)^4 *cos(1/2*d*x+1/2*c)+12*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+ 1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^2+8*co s(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2-3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*Ellip ticE(cos(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)+2*a*(2*A*b+B*a)/sin(1/2*d*x+ 1/2*c)^2/(2*sin(1/2*d*x+1/2*c)^2-1)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1 /2*c)^2)^(1/2)*(2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2-(sin(1/2*d*x+1/2 *c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2 -1)^(1/2))+2*b*(A*b+2*B*a)*(-1/6*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)/(cos(1/2*d*x+1/2*c)^2-1/2)^2+1/3*(sin(1/2*d *x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c )^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))))/si n(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 = 0.12 (sec) , antiderivative size = 286, normalized size of antiderivative = 1.66 \[ \int \frac {(a+b \sec (c+d x))^2 (A+B \sec (c+d x))}{\sqrt {\cos (c+d x)}} \, dx=-\frac {5 \, \sqrt {2} {\left (3 i \, A a^{2} + 2 i \, B a b + i \, A b^{2}\right )} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, \sqrt {2} {\left (-3 i \, A a^{2} - 2 i \, B a b - i \, A b^{2}\right )} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 \, \sqrt {2} {\left (5 i \, B a^{2} + 10 i \, A a b + 3 i \, B b^{2}\right )} \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 \, \sqrt {2} {\left (-5 i \, B a^{2} - 10 i \, A a b - 3 i \, B b^{2}\right )} \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - 2 \, {\left (3 \, B b^{2} + 3 \, {\left (5 \, B a^{2} + 10 \, A a b + 3 \, B b^{2}\right )} \cos \left (d x + c\right )^{2} + 5 \, {\left (2 \, B a b + A b^{2}\right )} \cos \left (d x + c\right )\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{15 \, d \cos \left (d x + c\right )^{3}} \]
-1/15*(5*sqrt(2)*(3*I*A*a^2 + 2*I*B*a*b + I*A*b^2)*cos(d*x + c)^3*weierstr assPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*sqrt(2)*(-3*I*A*a^2 - 2*I*B*a*b - I*A*b^2)*cos(d*x + c)^3*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 3*sqrt(2)*(5*I*B*a^2 + 10*I*A*a*b + 3*I*B*b^2)*cos (d*x + c)^3*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*sqrt(2)*(-5*I*B*a^2 - 10*I*A*a*b - 3*I*B*b^2)*cos( d*x + c)^3*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - 2*(3*B*b^2 + 3*(5*B*a^2 + 10*A*a*b + 3*B*b^2)*cos(d*x + c)^2 + 5*(2*B*a*b + A*b^2)*cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c ))/(d*cos(d*x + c)^3)
\[ \int \frac {(a+b \sec (c+d x))^2 (A+B \sec (c+d x))}{\sqrt {\cos (c+d x)}} \, dx=\int \frac {\left (A + B \sec {\left (c + d x \right )}\right ) \left (a + b \sec {\left (c + d x \right )}\right )^{2}}{\sqrt {\cos {\left (c + d x \right )}}}\, dx \]
\[ \int \frac {(a+b \sec (c+d x))^2 (A+B \sec (c+d x))}{\sqrt {\cos (c+d x)}} \, dx=\int { \frac {{\left (B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{2}}{\sqrt {\cos \left (d x + c\right )}} \,d x } \]
\[ \int \frac {(a+b \sec (c+d x))^2 (A+B \sec (c+d x))}{\sqrt {\cos (c+d x)}} \, dx=\int { \frac {{\left (B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{2}}{\sqrt {\cos \left (d x + c\right )}} \,d x } \]
Time = 17.88 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.32 \[ \int \frac {(a+b \sec (c+d x))^2 (A+B \sec (c+d x))}{\sqrt {\cos (c+d x)}} \, dx=\frac {6\,B\,b^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{4},\frac {1}{2};\ -\frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )+30\,B\,a^2\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )+20\,B\,a\,b\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{15\,d\,{\cos \left (c+d\,x\right )}^{5/2}\,\sqrt {1-{\cos \left (c+d\,x\right )}^2}}+\frac {2\,A\,a^2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,A\,b^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{3\,d\,{\cos \left (c+d\,x\right )}^{3/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {4\,A\,a\,b\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \]
(6*B*b^2*sin(c + d*x)*hypergeom([-5/4, 1/2], -1/4, cos(c + d*x)^2) + 30*B* a^2*cos(c + d*x)^2*sin(c + d*x)*hypergeom([-1/4, 1/2], 3/4, cos(c + d*x)^2 ) + 20*B*a*b*cos(c + d*x)*sin(c + d*x)*hypergeom([-3/4, 1/2], 1/4, cos(c + d*x)^2))/(15*d*cos(c + d*x)^(5/2)*(1 - cos(c + d*x)^2)^(1/2)) + (2*A*a^2* ellipticF(c/2 + (d*x)/2, 2))/d + (2*A*b^2*sin(c + d*x)*hypergeom([-3/4, 1/ 2], 1/4, cos(c + d*x)^2))/(3*d*cos(c + d*x)^(3/2)*(sin(c + d*x)^2)^(1/2)) + (4*A*a*b*sin(c + d*x)*hypergeom([-1/4, 1/2], 3/4, cos(c + d*x)^2))/(d*co s(c + d*x)^(1/2)*(sin(c + d*x)^2)^(1/2))