Integrand size = 31, antiderivative size = 384 \[ \int \sec (c+d x) (a+b \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=-\frac {2 (a-b) \sqrt {a+b} \left (161 a^2 A b+63 A b^3+15 a^3 B+145 a b^2 B\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{105 b^2 d}+\frac {2 (a-b) \sqrt {a+b} \left (b^2 (63 A-25 B)-8 a b (7 A-15 B)+15 a^2 (7 A-B)\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{105 b d}+\frac {2 \left (56 a A b+15 a^2 B+25 b^2 B\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{105 d}+\frac {2 (7 A b+5 a B) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{35 d}+\frac {2 B (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{7 d} \]
-2/105*(a-b)*(161*A*a^2*b+63*A*b^3+15*B*a^3+145*B*a*b^2)*cot(d*x+c)*Ellipt icE((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^(1/2))*(a+b)^(1/2)*(b *(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/b^2/d+2/105*( a-b)*(b^2*(63*A-25*B)-8*a*b*(7*A-15*B)+15*a^2*(7*A-B))*cot(d*x+c)*Elliptic F((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^(1/2))*(a+b)^(1/2)*(b*( 1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/b/d+2/35*(7*A*b +5*B*a)*(a+b*sec(d*x+c))^(3/2)*tan(d*x+c)/d+2/7*B*(a+b*sec(d*x+c))^(5/2)*t an(d*x+c)/d+2/105*(56*A*a*b+15*B*a^2+25*B*b^2)*(a+b*sec(d*x+c))^(1/2)*tan( d*x+c)/d
Leaf count is larger than twice the leaf count of optimal. \(3431\) vs. \(2(384)=768\).
Time = 33.26 (sec) , antiderivative size = 3431, normalized size of antiderivative = 8.93 \[ \int \sec (c+d x) (a+b \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\text {Result too large to show} \]
(Cos[c + d*x]^3*(a + b*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x])*((2*(161*a ^2*A*b + 63*A*b^3 + 15*a^3*B + 145*a*b^2*B)*Sin[c + d*x])/(105*b) + (2*Sec [c + d*x]^2*(7*A*b^2*Sin[c + d*x] + 15*a*b*B*Sin[c + d*x]))/35 + (2*Sec[c + d*x]*(77*a*A*b*Sin[c + d*x] + 45*a^2*B*Sin[c + d*x] + 25*b^2*B*Sin[c + d *x]))/105 + (2*b^2*B*Sec[c + d*x]^2*Tan[c + d*x])/7))/(d*(b + a*Cos[c + d* x])^2*(B + A*Cos[c + d*x])) - (2*((-23*a^2*A*b)/(15*Sqrt[b + a*Cos[c + d*x ]]*Sqrt[Sec[c + d*x]]) - (3*A*b^3)/(5*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (a^3*B)/(7*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (29*a *b^2*B)/(21*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (8*a^3*A*Sqrt[S ec[c + d*x]])/(15*Sqrt[b + a*Cos[c + d*x]]) + (8*a*A*b^2*Sqrt[Sec[c + d*x] ])/(15*Sqrt[b + a*Cos[c + d*x]]) - (a^4*B*Sqrt[Sec[c + d*x]])/(7*b*Sqrt[b + a*Cos[c + d*x]]) - (2*a^2*b*B*Sqrt[Sec[c + d*x]])/(21*Sqrt[b + a*Cos[c + d*x]]) + (5*b^3*B*Sqrt[Sec[c + d*x]])/(21*Sqrt[b + a*Cos[c + d*x]]) - (23 *a^3*A*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/(15*Sqrt[b + a*Cos[c + d*x]]) - (3*a*A*b^2*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/(5*Sqrt[b + a*Cos[c + d* x]]) - (a^4*B*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/(7*b*Sqrt[b + a*Cos[c + d*x]]) - (29*a^2*b*B*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/(21*Sqrt[b + a* Cos[c + d*x]]))*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*(a + b*Sec[c + d*x]) ^(5/2)*(A + B*Sec[c + d*x])*(2*(a + b)*(161*a^2*A*b + 63*A*b^3 + 15*a^3*B + 145*a*b^2*B)*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c ...
Time = 1.60 (sec) , antiderivative size = 395, normalized size of antiderivative = 1.03, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.452, Rules used = {3042, 4490, 27, 3042, 4490, 27, 3042, 4490, 27, 3042, 4493, 3042, 4319, 4492}
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.
| \(\displaystyle \int \sec (c+d x) (a+b \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx\) |
| \(\Big \downarrow \) 4490 |
| \(\displaystyle \frac {2}{7} \int \frac {1}{2} \sec (c+d x) (a+b \sec (c+d x))^{3/2} (7 a A+5 b B+(7 A b+5 a B) \sec (c+d x))dx+\frac {2 B \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{7} \int \sec (c+d x) (a+b \sec (c+d x))^{3/2} (7 a A+5 b B+(7 A b+5 a B) \sec (c+d x))dx+\frac {2 B \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{7} \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2} \left (7 a A+5 b B+(7 A b+5 a B) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {2 B \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}\) |
| \(\Big \downarrow \) 4490 |
| \(\displaystyle \frac {1}{7} \left (\frac {2}{5} \int \frac {1}{2} \sec (c+d x) \sqrt {a+b \sec (c+d x)} \left (35 A a^2+40 b B a+21 A b^2+\left (15 B a^2+56 A b a+25 b^2 B\right ) \sec (c+d x)\right )dx+\frac {2 (5 a B+7 A b) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )+\frac {2 B \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{7} \left (\frac {1}{5} \int \sec (c+d x) \sqrt {a+b \sec (c+d x)} \left (35 A a^2+40 b B a+21 A b^2+\left (15 B a^2+56 A b a+25 b^2 B\right ) \sec (c+d x)\right )dx+\frac {2 (5 a B+7 A b) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )+\frac {2 B \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{7} \left (\frac {1}{5} \int \csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )} \left (35 A a^2+40 b B a+21 A b^2+\left (15 B a^2+56 A b a+25 b^2 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {2 (5 a B+7 A b) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )+\frac {2 B \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}\) |
| \(\Big \downarrow \) 4490 |
| \(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (\frac {2}{3} \int \frac {\sec (c+d x) \left (105 A a^3+135 b B a^2+119 A b^2 a+25 b^3 B+\left (15 B a^3+161 A b a^2+145 b^2 B a+63 A b^3\right ) \sec (c+d x)\right )}{2 \sqrt {a+b \sec (c+d x)}}dx+\frac {2 \left (15 a^2 B+56 a A b+25 b^2 B\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 (5 a B+7 A b) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )+\frac {2 B \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (\frac {1}{3} \int \frac {\sec (c+d x) \left (105 A a^3+135 b B a^2+119 A b^2 a+25 b^3 B+\left (15 B a^3+161 A b a^2+145 b^2 B a+63 A b^3\right ) \sec (c+d x)\right )}{\sqrt {a+b \sec (c+d x)}}dx+\frac {2 \left (15 a^2 B+56 a A b+25 b^2 B\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 (5 a B+7 A b) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )+\frac {2 B \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (\frac {1}{3} \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (105 A a^3+135 b B a^2+119 A b^2 a+25 b^3 B+\left (15 B a^3+161 A b a^2+145 b^2 B a+63 A b^3\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \left (15 a^2 B+56 a A b+25 b^2 B\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 (5 a B+7 A b) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )+\frac {2 B \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}\) |
| \(\Big \downarrow \) 4493 |
| \(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (\frac {1}{3} \left ((a-b) \left (15 a^2 (7 A-B)-8 a b (7 A-15 B)+b^2 (63 A-25 B)\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx+\left (15 a^3 B+161 a^2 A b+145 a b^2 B+63 A b^3\right ) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx\right )+\frac {2 \left (15 a^2 B+56 a A b+25 b^2 B\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 (5 a B+7 A b) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )+\frac {2 B \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (\frac {1}{3} \left ((a-b) \left (15 a^2 (7 A-B)-8 a b (7 A-15 B)+b^2 (63 A-25 B)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\left (15 a^3 B+161 a^2 A b+145 a b^2 B+63 A b^3\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\frac {2 \left (15 a^2 B+56 a A b+25 b^2 B\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 (5 a B+7 A b) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )+\frac {2 B \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}\) |
| \(\Big \downarrow \) 4319 |
| \(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (\frac {1}{3} \left (\left (15 a^3 B+161 a^2 A b+145 a b^2 B+63 A b^3\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 (a-b) \sqrt {a+b} \left (15 a^2 (7 A-B)-8 a b (7 A-15 B)+b^2 (63 A-25 B)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}\right )+\frac {2 \left (15 a^2 B+56 a A b+25 b^2 B\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 (5 a B+7 A b) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )+\frac {2 B \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}\) |
| \(\Big \downarrow \) 4492 |
| \(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (\frac {2 \left (15 a^2 B+56 a A b+25 b^2 B\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}+\frac {1}{3} \left (\frac {2 (a-b) \sqrt {a+b} \left (15 a^2 (7 A-B)-8 a b (7 A-15 B)+b^2 (63 A-25 B)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}-\frac {2 (a-b) \sqrt {a+b} \left (15 a^3 B+161 a^2 A b+145 a b^2 B+63 A b^3\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b^2 d}\right )\right )+\frac {2 (5 a B+7 A b) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )+\frac {2 B \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}\) |
(2*B*(a + b*Sec[c + d*x])^(5/2)*Tan[c + d*x])/(7*d) + ((2*(7*A*b + 5*a*B)* (a + b*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*d) + (((-2*(a - b)*Sqrt[a + b] *(161*a^2*A*b + 63*A*b^3 + 15*a^3*B + 145*a*b^2*B)*Cot[c + d*x]*EllipticE[ ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(b^2*d) + (2*(a - b)*Sqrt[a + b]*(b^2*(63*A - 25*B) - 8*a*b*(7*A - 15*B) + 15*a^2 *(7*A - B))*Cot[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(b*d))/3 + (2*(56*a*A*b + 15*a^2*B + 25*b^2*B) *Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(3*d))/5)/7
3.4.65.3.1 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[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_S ymbol] :> Simp[-2*(Rt[a + b, 2]/(b*f*Cot[e + f*x]))*Sqrt[(b*(1 - Csc[e + f* x]))/(a + b)]*Sqrt[(-b)*((1 + Csc[e + f*x])/(a - b))]*EllipticF[ArcSin[Sqrt [a + b*Csc[e + f*x]]/Rt[a + b, 2]], (a + b)/(a - b)], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]
Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(cs c[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-B)*Cot[e + f*x]*(( a + b*Csc[e + f*x])^m/(f*(m + 1))), x] + Simp[1/(m + 1) Int[Csc[e + f*x]* (a + b*Csc[e + f*x])^(m - 1)*Simp[b*B*m + a*A*(m + 1) + (a*B*m + A*b*(m + 1 ))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, A, B, e, f}, x] && NeQ[A*b - a* B, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[c sc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*(A*b - a*B)*Rt[a + b*(B/A), 2]*Sqrt[b*((1 - Csc[e + f*x])/(a + b))]*(Sqrt[(-b)*((1 + Csc[e + f*x])/(a - b))]/(b^2*f*Cot[e + f*x]))*EllipticE[ArcSin[Sqrt[a + b*Csc[e + f*x]]/Rt[a + b*(B/A), 2]], (a*A + b*B)/(a*A - b*B)], x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[a^2 - b^2, 0] && EqQ[A^2 - B^2, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[c sc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(A - B) Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] + Simp[B Int[Csc[e + f*x]*((1 + Csc[e + f*x])/Sqrt[a + b*Csc[e + f*x]]), x], x] /; FreeQ[{a, b, e, f, A, B} , x] && NeQ[a^2 - b^2, 0] && NeQ[A^2 - B^2, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(4581\) vs. \(2(350)=700\).
Time = 40.64 (sec) , antiderivative size = 4582, normalized size of antiderivative = 11.93
| method | result | size |
| parts | \(\text {Expression too large to display}\) | \(4582\) |
| default | \(\text {Expression too large to display}\) | \(4633\) |
2/15*A/d*(a+b*sec(d*x+c))^(1/2)/(b+a*cos(d*x+c))/(cos(d*x+c)+1)*(14*a*b^2* sin(d*x+c)+34*a^2*b*sin(d*x+c)-15*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+ b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF(cot(d*x+c)-csc(d*x+c), ((a-b)/(a+b))^(1/2))*a^3*cos(d*x+c)^2+23*EllipticE(cot(d*x+c)-csc(d*x+c),( (a-b)/(a+b))^(1/2))*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d *x+c)/(cos(d*x+c)+1))^(1/2)*a^2*b+9*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b) /(a+b))^(1/2))*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c) /(cos(d*x+c)+1))^(1/2)*a*b^2+23*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^ (1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticE(cot(d*x+c)-csc(d*x+c),(( a-b)/(a+b))^(1/2))*a^3*cos(d*x+c)^2+9*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c )+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticE(cot(d*x+c)-csc(d*x +c),((a-b)/(a+b))^(1/2))*b^3*cos(d*x+c)^2-23*EllipticF(cot(d*x+c)-csc(d*x+ c),((a-b)/(a+b))^(1/2))*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(c os(d*x+c)/(cos(d*x+c)+1))^(1/2)*a^2*b-17*EllipticF(cot(d*x+c)-csc(d*x+c),( (a-b)/(a+b))^(1/2))*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d *x+c)/(cos(d*x+c)+1))^(1/2)*a*b^2-18*EllipticF(cot(d*x+c)-csc(d*x+c),((a-b )/(a+b))^(1/2))*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c )/(cos(d*x+c)+1))^(1/2)*b^3*cos(d*x+c)+46*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d *x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticE(cot(d*x+c)-csc (d*x+c),((a-b)/(a+b))^(1/2))*a^3*cos(d*x+c)-9*EllipticF(cot(d*x+c)-csc(...
\[ \int \sec (c+d x) (a+b \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sec \left (d x + c\right ) \,d x } \]
integral((B*b^2*sec(d*x + c)^4 + A*a^2*sec(d*x + c) + (2*B*a*b + A*b^2)*se c(d*x + c)^3 + (B*a^2 + 2*A*a*b)*sec(d*x + c)^2)*sqrt(b*sec(d*x + c) + a), x)
\[ \int \sec (c+d x) (a+b \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\int \left (A + B \sec {\left (c + d x \right )}\right ) \left (a + b \sec {\left (c + d x \right )}\right )^{\frac {5}{2}} \sec {\left (c + d x \right )}\, dx \]
\[ \int \sec (c+d x) (a+b \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sec \left (d x + c\right ) \,d x } \]
\[ \int \sec (c+d x) (a+b \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sec \left (d x + c\right ) \,d x } \]
Timed out. \[ \int \sec (c+d x) (a+b \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\int \frac {\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/2}}{\cos \left (c+d\,x\right )} \,d x \]