Integrand size = 33, antiderivative size = 182 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx=\frac {2 \left (3 a^2 A+5 A b^2-5 a b B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a^3 d}-\frac {2 \left (a^2+3 b^2\right ) (A b-a B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 a^4 d}+\frac {2 b^3 (A b-a B) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{a^4 (a+b) d}-\frac {2 (A b-a B) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 a^2 d}+\frac {2 A \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 a d} \]
2/5*(3*A*a^2+5*A*b^2-5*B*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^3/d-2/3*(a^2+3*b^2)*(A*b-B*a)* (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^4/d+2*b^3*(A*b-B*a)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x +1/2*c)*EllipticPi(sin(1/2*d*x+1/2*c),2*a/(a+b),2^(1/2))/a^4/(a+b)/d+2/5*A *cos(d*x+c)^(3/2)*sin(d*x+c)/a/d-2/3*(A*b-B*a)*sin(d*x+c)*cos(d*x+c)^(1/2) /a^2/d
Time = 2.88 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.43 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx=\frac {\frac {2 a^2 \left (9 a^2 A+5 A b^2-5 a b B\right ) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}+2 a^2 (4 A b+5 a B) \left (2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-\frac {2 b \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}\right )+4 a^2 \sqrt {\cos (c+d x)} (-5 A b+5 a B+3 a A \cos (c+d x)) \sin (c+d x)+\frac {6 \left (3 a^2 A+5 A b^2-5 a b B\right ) \left (-2 a b E\left (\left .\arcsin \left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 b (a+b) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )+\left (a^2-2 b^2\right ) \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )\right ) \sin (c+d x)}{b \sqrt {\sin ^2(c+d x)}}}{30 a^4 d} \]
((2*a^2*(9*a^2*A + 5*A*b^2 - 5*a*b*B)*EllipticPi[(2*a)/(a + b), (c + d*x)/ 2, 2])/(a + b) + 2*a^2*(4*A*b + 5*a*B)*(2*EllipticF[(c + d*x)/2, 2] - (2*b *EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/(a + b)) + 4*a^2*Sqrt[Cos[c + d*x]]*(-5*A*b + 5*a*B + 3*a*A*Cos[c + d*x])*Sin[c + d*x] + (6*(3*a^2*A + 5 *A*b^2 - 5*a*b*B)*(-2*a*b*EllipticE[ArcSin[Sqrt[Cos[c + d*x]]], -1] + 2*b* (a + b)*EllipticF[ArcSin[Sqrt[Cos[c + d*x]]], -1] + (a^2 - 2*b^2)*Elliptic Pi[-(a/b), ArcSin[Sqrt[Cos[c + d*x]]], -1])*Sin[c + d*x])/(b*Sqrt[Sin[c + d*x]^2]))/(30*a^4*d)
Time = 1.63 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.10, number of steps used = 17, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.515, Rules used = {3042, 3433, 3042, 3469, 27, 3042, 3528, 27, 3042, 3538, 27, 3042, 3119, 3481, 3042, 3120, 3284}
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.
| \(\displaystyle \int \frac {\cos ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx\) |
| \(\Big \downarrow \) 3433 |
| \(\displaystyle \int \frac {\cos ^{\frac {5}{2}}(c+d x) (A \cos (c+d x)+B)}{a \cos (c+d x)+b}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (A \sin \left (c+d x+\frac {\pi }{2}\right )+B\right )}{a \sin \left (c+d x+\frac {\pi }{2}\right )+b}dx\) |
| \(\Big \downarrow \) 3469 |
| \(\displaystyle \frac {2 \int \frac {\sqrt {\cos (c+d x)} \left (-5 (A b-a B) \cos ^2(c+d x)+3 a A \cos (c+d x)+3 A b\right )}{2 (b+a \cos (c+d x))}dx}{5 a}+\frac {2 A \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 a d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {\sqrt {\cos (c+d x)} \left (-5 (A b-a B) \cos ^2(c+d x)+3 a A \cos (c+d x)+3 A b\right )}{b+a \cos (c+d x)}dx}{5 a}+\frac {2 A \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 a d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (-5 (A b-a B) \sin \left (c+d x+\frac {\pi }{2}\right )^2+3 a A \sin \left (c+d x+\frac {\pi }{2}\right )+3 A b\right )}{b+a \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{5 a}+\frac {2 A \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 a d}\) |
| \(\Big \downarrow \) 3528 |
| \(\displaystyle \frac {\frac {2 \int -\frac {-3 \left (3 A a^2-5 b B a+5 A b^2\right ) \cos ^2(c+d x)-a (4 A b+5 a B) \cos (c+d x)+5 b (A b-a B)}{2 \sqrt {\cos (c+d x)} (b+a \cos (c+d x))}dx}{3 a}-\frac {10 (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 a d}}{5 a}+\frac {2 A \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 a d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {-\frac {\int \frac {-3 \left (3 A a^2-5 b B a+5 A b^2\right ) \cos ^2(c+d x)-a (4 A b+5 a B) \cos (c+d x)+5 b (A b-a B)}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))}dx}{3 a}-\frac {10 (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 a d}}{5 a}+\frac {2 A \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 a d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {\int \frac {-3 \left (3 A a^2-5 b B a+5 A b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2-a (4 A b+5 a B) \sin \left (c+d x+\frac {\pi }{2}\right )+5 b (A b-a B)}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{3 a}-\frac {10 (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 a d}}{5 a}+\frac {2 A \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 a d}\) |
| \(\Big \downarrow \) 3538 |
| \(\displaystyle \frac {-\frac {-\frac {3 \left (3 a^2 A-5 a b B+5 A b^2\right ) \int \sqrt {\cos (c+d x)}dx}{a}-\frac {\int -\frac {5 \left (a b (A b-a B)+\left (a^2+3 b^2\right ) \cos (c+d x) (A b-a B)\right )}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))}dx}{a}}{3 a}-\frac {10 (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 a d}}{5 a}+\frac {2 A \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 a d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {-\frac {\frac {5 \int \frac {a b (A b-a B)+\left (a^2+3 b^2\right ) \cos (c+d x) (A b-a B)}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))}dx}{a}-\frac {3 \left (3 a^2 A-5 a b B+5 A b^2\right ) \int \sqrt {\cos (c+d x)}dx}{a}}{3 a}-\frac {10 (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 a d}}{5 a}+\frac {2 A \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 a d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {\frac {5 \int \frac {a b (A b-a B)+\left (a^2+3 b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) (A b-a B)}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}-\frac {3 \left (3 a^2 A-5 a b B+5 A b^2\right ) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a}}{3 a}-\frac {10 (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 a d}}{5 a}+\frac {2 A \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 a d}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {-\frac {\frac {5 \int \frac {a b (A b-a B)+\left (a^2+3 b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) (A b-a B)}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}-\frac {6 \left (3 a^2 A-5 a b B+5 A b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d}}{3 a}-\frac {10 (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 a d}}{5 a}+\frac {2 A \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 a d}\) |
| \(\Big \downarrow \) 3481 |
| \(\displaystyle \frac {-\frac {\frac {5 \left (\frac {\left (a^2+3 b^2\right ) (A b-a B) \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{a}-\frac {3 b^3 (A b-a B) \int \frac {1}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))}dx}{a}\right )}{a}-\frac {6 \left (3 a^2 A-5 a b B+5 A b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d}}{3 a}-\frac {10 (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 a d}}{5 a}+\frac {2 A \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 a d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {\frac {5 \left (\frac {\left (a^2+3 b^2\right ) (A b-a B) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}-\frac {3 b^3 (A b-a B) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}\right )}{a}-\frac {6 \left (3 a^2 A-5 a b B+5 A b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d}}{3 a}-\frac {10 (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 a d}}{5 a}+\frac {2 A \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 a d}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {-\frac {\frac {5 \left (\frac {2 \left (a^2+3 b^2\right ) (A b-a B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{a d}-\frac {3 b^3 (A b-a B) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}\right )}{a}-\frac {6 \left (3 a^2 A-5 a b B+5 A b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d}}{3 a}-\frac {10 (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 a d}}{5 a}+\frac {2 A \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 a d}\) |
| \(\Big \downarrow \) 3284 |
| \(\displaystyle \frac {-\frac {\frac {5 \left (\frac {2 \left (a^2+3 b^2\right ) (A b-a B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{a d}-\frac {6 b^3 (A b-a B) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{a d (a+b)}\right )}{a}-\frac {6 \left (3 a^2 A-5 a b B+5 A b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d}}{3 a}-\frac {10 (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 a d}}{5 a}+\frac {2 A \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 a d}\) |
(2*A*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(5*a*d) + (-1/3*((-6*(3*a^2*A + 5*A* b^2 - 5*a*b*B)*EllipticE[(c + d*x)/2, 2])/(a*d) + (5*((2*(a^2 + 3*b^2)*(A* b - a*B)*EllipticF[(c + d*x)/2, 2])/(a*d) - (6*b^3*(A*b - a*B)*EllipticPi[ (2*a)/(a + b), (c + d*x)/2, 2])/(a*(a + b)*d)))/a)/a - (10*(A*b - a*B)*Sqr t[Cos[c + d*x]]*Sin[c + d*x])/(3*a*d))/(5*a)
3.6.74.3.1 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[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[(2/(f*(a + b)*Sqrt[c + d]))*EllipticPi[ 2*(b/(a + b)), (1/2)*(e - Pi/2 + f*x), 2*(d/(c + d))], x] /; FreeQ[{a, b, c , d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[c + d, 0]
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] :> Si mp[(-b)*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[e + f*x])^( n + 1)/(d*f*(m + n + 1))), x] + Simp[1/(d*(m + n + 1)) Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^n*Simp[a^2*A*d*(m + n + 1) + b*B*(b*c*( m - 1) + a*d*(n + 1)) + (a*d*(2*A*b + a*B)*(m + n + 1) - b*B*(a*c - b*d*(m + n)))*Sin[e + f*x] + b*(A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*Sin [e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1] && !(IGt Q[n, 1] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))
Int[(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]))/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[ B/d Int[(a + b*Sin[e + f*x])^m, x], x] - Simp[(B*c - A*d)/d Int[(a + b* Sin[e + f*x])^m/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_ .) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*(a + b*Sin[e + f*x ])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n + 2))), x] + Simp[1/(d*(m + n + 2)) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A* d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (d*(A*b + a*B)*(m + n + 2) - C*(a *c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + n + 2))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n} , x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[ m, 0] && !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^ 2)/(Sqrt[(a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp[C/(b*d) Int[Sqrt[a + b*Sin[e + f*x]], x] , x] - Simp[1/(b*d) Int[Simp[a*c*C - A*b*d + (b*c*C - b*B*d + a*C*d)*Sin[ e + f*x], x]/(Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])), x], x] /; Fre eQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0 ] && NeQ[c^2 - d^2, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(1073\) vs. \(2(248)=496\).
Time = 12.50 (sec) , antiderivative size = 1074, normalized size of antiderivative = 5.90
-2/15*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*((-24*A*a^4+ 24*A*a^3*b)*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6+(24*A*a^4-44*A*a^3*b+2 0*A*a^2*b^2+20*B*a^4-20*B*a^3*b)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+( -6*A*a^4+16*A*a^3*b-10*A*a^2*b^2-10*B*a^4+10*B*a^3*b)*sin(1/2*d*x+1/2*c)^2 *cos(1/2*d*x+1/2*c)-5*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c) ^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a^3*b+5*A*(sin(1/2*d*x+1 /2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2* c),2^(1/2))*a^2*b^2-15*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c )^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a*b^3+15*A*(sin(1/2*d*x +1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/ 2*c),2^(1/2))*b^4-9*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2 -1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a^4+9*A*(sin(1/2*d*x+1/2*c )^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2 ^(1/2))*a^3*b-15*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1) ^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a^2*b^2+15*A*(sin(1/2*d*x+1/2 *c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c) ,2^(1/2))*a*b^3-15*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2- 1)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2))*b^4+5*B*(sin(1/2 *d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d* x+1/2*c),2^(1/2))*a^4-5*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1...
Timed out. \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx=\text {Timed out} \]
\[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx=\int { \frac {{\left (B \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{\frac {5}{2}}}{b \sec \left (d x + c\right ) + a} \,d x } \]
\[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx=\int { \frac {{\left (B \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{\frac {5}{2}}}{b \sec \left (d x + c\right ) + a} \,d x } \]
Timed out. \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^{5/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )}{a+\frac {b}{\cos \left (c+d\,x\right )}} \,d x \]