Integrand size = 31, antiderivative size = 433 \[ \int \cos (c+d x) (a+b \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\frac {(a-b) \sqrt {a+b} \left (3 a^2 A-6 A b^2-14 a b B\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 b d}+\frac {\sqrt {a+b} \left (2 a b (9 A-7 B)-2 b^2 (3 A-B)+3 a^2 (A+6 B)\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 d}-\frac {a \sqrt {a+b} (5 A b+2 a B) \cot (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{d}+\frac {a A (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{d}-\frac {b (3 a A-2 b B) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{3 d} \]
a*A*(a+b*sec(d*x+c))^(3/2)*sin(d*x+c)/d+1/3*(a-b)*(3*A*a^2-6*A*b^2-14*B*a* b)*cot(d*x+c)*EllipticE((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+1/3*(2*a*b*(9*A-7*B)-2*b^2*(3*A-B)+3*a^2*(A+6*B))*cot(d*x+c)*El lipticF((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)/d-a*(5*A* b+2*B*a)*cot(d*x+c)*EllipticPi((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),(a+b)/a, ((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)/d-1/3*b*(3*A*a-2*B*b)*(a+b*sec(d*x+c))^(1/2)*tan(d*x +c)/d
Leaf count is larger than twice the leaf count of optimal. \(7745\) vs. \(2(433)=866\).
Time = 29.28 (sec) , antiderivative size = 7745, normalized size of antiderivative = 17.89 \[ \int \cos (c+d x) (a+b \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\text {Result too large to show} \]
Time = 1.81 (sec) , antiderivative size = 439, normalized size of antiderivative = 1.01, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.452, Rules used = {3042, 4513, 27, 3042, 4544, 27, 3042, 4546, 3042, 4409, 3042, 4271, 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.
| \(\displaystyle \int \cos (c+d x) (a+b \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx\) |
| \(\Big \downarrow \) 4513 |
| \(\displaystyle \frac {a A \sin (c+d x) (a+b \sec (c+d x))^{3/2}}{d}-\int -\frac {1}{2} \sqrt {a+b \sec (c+d x)} \left (-b (3 a A-2 b B) \sec ^2(c+d x)+2 b (A b+2 a B) \sec (c+d x)+a (5 A b+2 a B)\right )dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{2} \int \sqrt {a+b \sec (c+d x)} \left (-b (3 a A-2 b B) \sec ^2(c+d x)+2 b (A b+2 a B) \sec (c+d x)+a (5 A b+2 a B)\right )dx+\frac {a A \sin (c+d x) (a+b \sec (c+d x))^{3/2}}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \int \sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )} \left (-b (3 a A-2 b B) \csc \left (c+d x+\frac {\pi }{2}\right )^2+2 b (A b+2 a B) \csc \left (c+d x+\frac {\pi }{2}\right )+a (5 A b+2 a B)\right )dx+\frac {a A \sin (c+d x) (a+b \sec (c+d x))^{3/2}}{d}\) |
| \(\Big \downarrow \) 4544 |
| \(\displaystyle \frac {1}{2} \left (\frac {2}{3} \int \frac {3 (5 A b+2 a B) a^2-b \left (3 A a^2-14 b B a-6 A b^2\right ) \sec ^2(c+d x)+2 b \left (9 B a^2+9 A b a+b^2 B\right ) \sec (c+d x)}{2 \sqrt {a+b \sec (c+d x)}}dx-\frac {2 b (3 a A-2 b B) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {a A \sin (c+d x) (a+b \sec (c+d x))^{3/2}}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{3} \int \frac {3 (5 A b+2 a B) a^2-b \left (3 A a^2-14 b B a-6 A b^2\right ) \sec ^2(c+d x)+2 b \left (9 B a^2+9 A b a+b^2 B\right ) \sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx-\frac {2 b (3 a A-2 b B) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {a A \sin (c+d x) (a+b \sec (c+d x))^{3/2}}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{3} \int \frac {3 (5 A b+2 a B) a^2-b \left (3 A a^2-14 b B a-6 A b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2+2 b \left (9 B a^2+9 A b a+b^2 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 b (3 a A-2 b B) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {a A \sin (c+d x) (a+b \sec (c+d x))^{3/2}}{d}\) |
| \(\Big \downarrow \) 4546 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (\int \frac {3 (5 A b+2 a B) a^2+\left (b \left (3 A a^2-14 b B a-6 A b^2\right )+2 b \left (9 B a^2+9 A b a+b^2 B\right )\right ) \sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx-b \left (3 a^2 A-14 a b B-6 A b^2\right ) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx\right )-\frac {2 b (3 a A-2 b B) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {a A \sin (c+d x) (a+b \sec (c+d x))^{3/2}}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (\int \frac {3 (5 A b+2 a B) a^2+\left (b \left (3 A a^2-14 b B a-6 A b^2\right )+2 b \left (9 B a^2+9 A b a+b^2 B\right )\right ) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-b \left (3 a^2 A-14 a b B-6 A b^2\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )-\frac {2 b (3 a A-2 b B) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {a A \sin (c+d x) (a+b \sec (c+d x))^{3/2}}{d}\) |
| \(\Big \downarrow \) 4409 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (-b \left (3 a^2 A-14 a b B-6 A b^2\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+b \left (3 a^2 (A+6 B)+2 a b (9 A-7 B)-2 b^2 (3 A-B)\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx+3 a^2 (2 a B+5 A b) \int \frac {1}{\sqrt {a+b \sec (c+d x)}}dx\right )-\frac {2 b (3 a A-2 b B) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {a A \sin (c+d x) (a+b \sec (c+d x))^{3/2}}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (b \left (3 a^2 (A+6 B)+2 a b (9 A-7 B)-2 b^2 (3 A-B)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-b \left (3 a^2 A-14 a b B-6 A b^2\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+3 a^2 (2 a B+5 A b) \int \frac {1}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )-\frac {2 b (3 a A-2 b B) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {a A \sin (c+d x) (a+b \sec (c+d x))^{3/2}}{d}\) |
| \(\Big \downarrow \) 4271 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (b \left (3 a^2 (A+6 B)+2 a b (9 A-7 B)-2 b^2 (3 A-B)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-b \left (3 a^2 A-14 a b B-6 A b^2\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {6 a \sqrt {a+b} (2 a B+5 A b) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}\right )-\frac {2 b (3 a A-2 b B) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {a A \sin (c+d x) (a+b \sec (c+d x))^{3/2}}{d}\) |
| \(\Big \downarrow \) 4319 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (-b \left (3 a^2 A-14 a b B-6 A b^2\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sqrt {a+b} \left (3 a^2 (A+6 B)+2 a b (9 A-7 B)-2 b^2 (3 A-B)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}-\frac {6 a \sqrt {a+b} (2 a B+5 A b) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}\right )-\frac {2 b (3 a A-2 b B) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {a A \sin (c+d x) (a+b \sec (c+d x))^{3/2}}{d}\) |
| \(\Big \downarrow \) 4492 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (\frac {2 \sqrt {a+b} \left (3 a^2 (A+6 B)+2 a b (9 A-7 B)-2 b^2 (3 A-B)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}+\frac {2 (a-b) \sqrt {a+b} \left (3 a^2 A-14 a b B-6 A b^2\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b d}-\frac {6 a \sqrt {a+b} (2 a B+5 A b) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}\right )-\frac {2 b (3 a A-2 b B) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {a A \sin (c+d x) (a+b \sec (c+d x))^{3/2}}{d}\) |
(a*A*(a + b*Sec[c + d*x])^(3/2)*Sin[c + d*x])/d + (((2*(a - b)*Sqrt[a + b] *(3*a^2*A - 6*A*b^2 - 14*a*b*B)*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*S ec[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) + (2*Sqrt[a + b]*(2* a*b*(9*A - 7*B) - 2*b^2*(3*A - B) + 3*a^2*(A + 6*B))*Cot[c + d*x]*Elliptic F[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))])/d - ( 6*a*Sqrt[a + b]*(5*A*b + 2*a*B)*Cot[c + d*x]*EllipticPi[(a + b)/a, 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))])/d)/3 - (2*b*(3 *a*A - 2*b*B)*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(3*d))/2
3.4.67.3.1 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[1/Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[2*(Rt[a + b, 2]/(a*d*Cot[c + d*x]))*Sqrt[b*((1 - Csc[c + d*x])/(a + b))]*Sqrt[(-b) *((1 + Csc[c + d*x])/(a - b))]*EllipticPi[(a + b)/a, ArcSin[Sqrt[a + b*Csc[ c + d*x]]/Rt[a + b, 2]], (a + b)/(a - b)], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0]
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_)]*(d_.) + (c_))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_ .) + (a_)], x_Symbol] :> Simp[c Int[1/Sqrt[a + b*Csc[e + f*x]], x], x] + Simp[d Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[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[-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_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[a*A*Cot [e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*((d*Csc[e + f*x])^n/(f*n)), x] + Sim p[1/(d*n) Int[(a + b*Csc[e + f*x])^(m - 2)*(d*Csc[e + f*x])^(n + 1)*Simp[ a*(a*B*n - A*b*(m - n - 1)) + (2*a*b*B*n + A*(b^2*n + a^2*(1 + n)))*Csc[e + f*x] + b*(b*B*n + a*A*(m + n))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 1] & & LeQ[n, -1]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Simp[(-C)*Cot [e + f*x]*((a + b*Csc[e + f*x])^m/(f*(m + 1))), x] + Simp[1/(m + 1) Int[( a + b*Csc[e + f*x])^(m - 1)*Simp[a*A*(m + 1) + ((A*b + a*B)*(m + 1) + b*C*m )*Csc[e + f*x] + (b*B*(m + 1) + a*C*m)*Csc[e + f*x]^2, x], x], x] /; FreeQ[ {a, b, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && IGtQ[2*m, 0]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Int[(A + (B - C )*Csc[e + f*x])/Sqrt[a + b*Csc[e + f*x]], x] + Simp[C 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, C}, x] && NeQ[a^2 - b^2, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(4118\) vs. \(2(396)=792\).
Time = 28.08 (sec) , antiderivative size = 4119, normalized size of antiderivative = 9.51
1/3/d*(a+b*sec(d*x+c))^(1/2)/(b+a*cos(d*x+c))/(cos(d*x+c)+1)*(2*B*sin(d*x+ c)*b^3-30*A*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(c os(d*x+c)+1))^(1/2)*EllipticPi(cot(d*x+c)-csc(d*x+c),-1,((a-b)/(a+b))^(1/2 ))*a^2*b*cos(d*x+c)^2+36*A*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^2*b*cos(d*x+c)+18*A*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^2*b*cos(d*x+c)^2-3*A*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^3+6*A*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)*b^3-12*B*EllipticPi(cot(d*x+c)-csc(d*x+c),-1,( (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^3+6*B*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^3+3*A*a^2*b*cos(d*x+c)*sin(d*x+c)+6*A*a*b^2*cos(d *x+c)*sin(d*x+c)+2*B*a*b^2*cos(d*x+c)*sin(d*x+c)+14*B*a^2*b*cos(d*x+c)*sin (d*x+c)-30*A*EllipticPi(cot(d*x+c)-csc(d*x+c),-1,((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-4*B*EllipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(1/(...
\[ \int \cos (c+d x) (a+b \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cos \left (d x + c\right ) \,d x } \]
integral((B*b^2*cos(d*x + c)*sec(d*x + c)^3 + A*a^2*cos(d*x + c) + (2*B*a* b + A*b^2)*cos(d*x + c)*sec(d*x + c)^2 + (B*a^2 + 2*A*a*b)*cos(d*x + c)*se c(d*x + c))*sqrt(b*sec(d*x + c) + a), x)
Timed out. \[ \int \cos (c+d x) (a+b \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\text {Timed out} \]
\[ \int \cos (c+d x) (a+b \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cos \left (d x + c\right ) \,d x } \]
\[ \int \cos (c+d x) (a+b \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cos \left (d x + c\right ) \,d x } \]
Timed out. \[ \int \cos (c+d x) (a+b \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\int \cos \left (c+d\,x\right )\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/2} \,d x \]