3.455 \(\int \frac{(a+b \sec (c+d x))^{5/2} (A+B \sec (c+d x))}{\sec ^{\frac{11}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=519 \[ \frac{2 \left (a^2-b^2\right ) \left (285 a^2 A b^2+675 a^4 A+1254 a^3 b B-110 a b^3 B+40 A b^4\right ) \sqrt{\sec (c+d x)} \sqrt{\frac{a \cos (c+d x)+b}{a+b}} \text{EllipticF}\left (\frac{1}{2} (c+d x),\frac{2 a}{a+b}\right )}{3465 a^3 d \sqrt{a+b \sec (c+d x)}}+\frac{2 \left (1145 a^2 A b+539 a^3 B+825 a b^2 B+15 A b^3\right ) \sin (c+d x) \sqrt{a+b \sec (c+d x)}}{3465 a d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (81 a^2 A+209 a b B+113 A b^2\right ) \sin (c+d x) \sqrt{a+b \sec (c+d x)}}{693 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 \left (1025 a^2 A b^2+675 a^4 A+1793 a^3 b B+55 a b^3 B-20 A b^4\right ) \sin (c+d x) \sqrt{a+b \sec (c+d x)}}{3465 a^2 d \sqrt{\sec (c+d x)}}+\frac{2 \left (255 a^2 A b^3+3705 a^4 A b+3069 a^3 b^2 B+1617 a^5 B-110 a b^4 B+40 A b^5\right ) \sqrt{a+b \sec (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{3465 a^3 d \sqrt{\sec (c+d x)} \sqrt{\frac{a \cos (c+d x)+b}{a+b}}}+\frac{2 a (11 a B+14 A b) \sin (c+d x) \sqrt{a+b \sec (c+d x)}}{99 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{2 a A \sin (c+d x) (a+b \sec (c+d x))^{3/2}}{11 d \sec ^{\frac{9}{2}}(c+d x)} \]

[Out]

(2*(a^2 - b^2)*(675*a^4*A + 285*a^2*A*b^2 + 40*A*b^4 + 1254*a^3*b*B - 110*a*b^3*B)*Sqrt[(b + a*Cos[c + d*x])/(
a + b)]*EllipticF[(c + d*x)/2, (2*a)/(a + b)]*Sqrt[Sec[c + d*x]])/(3465*a^3*d*Sqrt[a + b*Sec[c + d*x]]) + (2*(
3705*a^4*A*b + 255*a^2*A*b^3 + 40*A*b^5 + 1617*a^5*B + 3069*a^3*b^2*B - 110*a*b^4*B)*EllipticE[(c + d*x)/2, (2
*a)/(a + b)]*Sqrt[a + b*Sec[c + d*x]])/(3465*a^3*d*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*Sqrt[Sec[c + d*x]]) + (2
*a*(14*A*b + 11*a*B)*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(99*d*Sec[c + d*x]^(7/2)) + (2*(81*a^2*A + 113*A*b
^2 + 209*a*b*B)*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(693*d*Sec[c + d*x]^(5/2)) + (2*(1145*a^2*A*b + 15*A*b^
3 + 539*a^3*B + 825*a*b^2*B)*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(3465*a*d*Sec[c + d*x]^(3/2)) + (2*(675*a^
4*A + 1025*a^2*A*b^2 - 20*A*b^4 + 1793*a^3*b*B + 55*a*b^3*B)*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(3465*a^2*
d*Sqrt[Sec[c + d*x]]) + (2*a*A*(a + b*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(11*d*Sec[c + d*x]^(9/2))

________________________________________________________________________________________

Rubi [A]  time = 1.95957, antiderivative size = 519, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 10, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {4025, 4094, 4104, 4035, 3856, 2655, 2653, 3858, 2663, 2661} \[ \frac{2 \left (1145 a^2 A b+539 a^3 B+825 a b^2 B+15 A b^3\right ) \sin (c+d x) \sqrt{a+b \sec (c+d x)}}{3465 a d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (81 a^2 A+209 a b B+113 A b^2\right ) \sin (c+d x) \sqrt{a+b \sec (c+d x)}}{693 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 \left (1025 a^2 A b^2+675 a^4 A+1793 a^3 b B+55 a b^3 B-20 A b^4\right ) \sin (c+d x) \sqrt{a+b \sec (c+d x)}}{3465 a^2 d \sqrt{\sec (c+d x)}}+\frac{2 \left (a^2-b^2\right ) \left (285 a^2 A b^2+675 a^4 A+1254 a^3 b B-110 a b^3 B+40 A b^4\right ) \sqrt{\sec (c+d x)} \sqrt{\frac{a \cos (c+d x)+b}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{3465 a^3 d \sqrt{a+b \sec (c+d x)}}+\frac{2 \left (255 a^2 A b^3+3705 a^4 A b+3069 a^3 b^2 B+1617 a^5 B-110 a b^4 B+40 A b^5\right ) \sqrt{a+b \sec (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{3465 a^3 d \sqrt{\sec (c+d x)} \sqrt{\frac{a \cos (c+d x)+b}{a+b}}}+\frac{2 a (11 a B+14 A b) \sin (c+d x) \sqrt{a+b \sec (c+d x)}}{99 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{2 a A \sin (c+d x) (a+b \sec (c+d x))^{3/2}}{11 d \sec ^{\frac{9}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x]))/Sec[c + d*x]^(11/2),x]

[Out]

(2*(a^2 - b^2)*(675*a^4*A + 285*a^2*A*b^2 + 40*A*b^4 + 1254*a^3*b*B - 110*a*b^3*B)*Sqrt[(b + a*Cos[c + d*x])/(
a + b)]*EllipticF[(c + d*x)/2, (2*a)/(a + b)]*Sqrt[Sec[c + d*x]])/(3465*a^3*d*Sqrt[a + b*Sec[c + d*x]]) + (2*(
3705*a^4*A*b + 255*a^2*A*b^3 + 40*A*b^5 + 1617*a^5*B + 3069*a^3*b^2*B - 110*a*b^4*B)*EllipticE[(c + d*x)/2, (2
*a)/(a + b)]*Sqrt[a + b*Sec[c + d*x]])/(3465*a^3*d*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*Sqrt[Sec[c + d*x]]) + (2
*a*(14*A*b + 11*a*B)*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(99*d*Sec[c + d*x]^(7/2)) + (2*(81*a^2*A + 113*A*b
^2 + 209*a*b*B)*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(693*d*Sec[c + d*x]^(5/2)) + (2*(1145*a^2*A*b + 15*A*b^
3 + 539*a^3*B + 825*a*b^2*B)*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(3465*a*d*Sec[c + d*x]^(3/2)) + (2*(675*a^
4*A + 1025*a^2*A*b^2 - 20*A*b^4 + 1793*a^3*b*B + 55*a*b^3*B)*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(3465*a^2*
d*Sqrt[Sec[c + d*x]]) + (2*a*A*(a + b*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(11*d*Sec[c + d*x]^(9/2))

Rule 4025

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]
+ Dist[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] /; Free
Q[{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]

Rule 4094

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^
(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*
Csc[e + f*x])^n)/(f*n), x] - Dist[1/(d*n), Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Simp[A*b*
m - a*B*n - (b*B*n + a*(C*n + A*(n + 1)))*Csc[e + f*x] - b*(C*n + A*(m + n + 1))*Csc[e + f*x]^2, x], x], x] /;
FreeQ[{a, b, d, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 0] && LeQ[n, -1]

Rule 4104

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^
(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(A*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m +
1)*(d*Csc[e + f*x])^n)/(a*f*n), x] + Dist[1/(a*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[
a*B*n - A*b*(m + n + 1) + a*(A + A*n + C*n)*Csc[e + f*x] + A*b*(m + n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ
[{a, b, d, e, f, A, B, C, m}, x] && NeQ[a^2 - b^2, 0] && LeQ[n, -1]

Rule 4035

Int[(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_))/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(
b_.) + (a_)]), x_Symbol] :> Dist[A/a, Int[Sqrt[a + b*Csc[e + f*x]]/Sqrt[d*Csc[e + f*x]], x], x] - Dist[(A*b -
a*B)/(a*d), Int[Sqrt[d*Csc[e + f*x]]/Sqrt[a + b*Csc[e + f*x]], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && Ne
Q[A*b - a*B, 0] && NeQ[a^2 - b^2, 0]

Rule 3856

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)], x_Symbol] :> Dist[Sqrt[a +
b*Csc[e + f*x]]/(Sqrt[d*Csc[e + f*x]]*Sqrt[b + a*Sin[e + f*x]]), Int[Sqrt[b + a*Sin[e + f*x]], x], x] /; Free
Q[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 2655

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[a + b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c +
d*x])/(a + b)], Int[Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
b^2, 0] &&  !GtQ[a + b, 0]

Rule 2653

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*Sqrt[a + b]*EllipticE[(1*(c - Pi/2 + d*x)
)/2, (2*b)/(a + b)])/d, x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 3858

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[(Sqrt[d*
Csc[e + f*x]]*Sqrt[b + a*Sin[e + f*x]])/Sqrt[a + b*Csc[e + f*x]], Int[1/Sqrt[b + a*Sin[e + f*x]], x], x] /; Fr
eeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 2663

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[(a + b*Sin[c + d*x])/(a + b)]/Sqrt[a
+ b*Sin[c + d*x]], Int[1/Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] &&  !GtQ[a + b, 0]

Rule 2661

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, (2*b)
/(a + b)])/(d*Sqrt[a + b]), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rubi steps

\begin{align*} \int \frac{(a+b \sec (c+d x))^{5/2} (A+B \sec (c+d x))}{\sec ^{\frac{11}{2}}(c+d x)} \, dx &=\frac{2 a A (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{11 d \sec ^{\frac{9}{2}}(c+d x)}-\frac{2}{11} \int \frac{\sqrt{a+b \sec (c+d x)} \left (-\frac{1}{2} a (14 A b+11 a B)-\frac{1}{2} \left (9 a^2 A+11 A b^2+22 a b B\right ) \sec (c+d x)-\frac{1}{2} b (6 a A+11 b B) \sec ^2(c+d x)\right )}{\sec ^{\frac{9}{2}}(c+d x)} \, dx\\ &=\frac{2 a (14 A b+11 a B) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{99 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{2 a A (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{11 d \sec ^{\frac{9}{2}}(c+d x)}-\frac{4}{99} \int \frac{-\frac{1}{4} a \left (81 a^2 A+113 A b^2+209 a b B\right )-\frac{1}{4} \left (233 a^2 A b+99 A b^3+77 a^3 B+297 a b^2 B\right ) \sec (c+d x)-\frac{3}{4} b \left (46 a A b+22 a^2 B+33 b^2 B\right ) \sec ^2(c+d x)}{\sec ^{\frac{7}{2}}(c+d x) \sqrt{a+b \sec (c+d x)}} \, dx\\ &=\frac{2 a (14 A b+11 a B) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{99 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{2 \left (81 a^2 A+113 A b^2+209 a b B\right ) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{693 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 a A (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{11 d \sec ^{\frac{9}{2}}(c+d x)}+\frac{8 \int \frac{\frac{1}{8} a \left (1145 a^2 A b+15 A b^3+539 a^3 B+825 a b^2 B\right )+\frac{1}{8} a \left (405 a^3 A+1531 a A b^2+1507 a^2 b B+693 b^3 B\right ) \sec (c+d x)+\frac{1}{2} a b \left (81 a^2 A+113 A b^2+209 a b B\right ) \sec ^2(c+d x)}{\sec ^{\frac{5}{2}}(c+d x) \sqrt{a+b \sec (c+d x)}} \, dx}{693 a}\\ &=\frac{2 a (14 A b+11 a B) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{99 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{2 \left (81 a^2 A+113 A b^2+209 a b B\right ) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{693 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 \left (1145 a^2 A b+15 A b^3+539 a^3 B+825 a b^2 B\right ) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{3465 a d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 a A (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{11 d \sec ^{\frac{9}{2}}(c+d x)}-\frac{16 \int \frac{-\frac{3}{16} a \left (675 a^4 A+1025 a^2 A b^2-20 A b^4+1793 a^3 b B+55 a b^3 B\right )-\frac{1}{16} a^2 \left (5055 a^2 A b+2305 A b^3+1617 a^3 B+6655 a b^2 B\right ) \sec (c+d x)-\frac{1}{8} a b \left (1145 a^2 A b+15 A b^3+539 a^3 B+825 a b^2 B\right ) \sec ^2(c+d x)}{\sec ^{\frac{3}{2}}(c+d x) \sqrt{a+b \sec (c+d x)}} \, dx}{3465 a^2}\\ &=\frac{2 a (14 A b+11 a B) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{99 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{2 \left (81 a^2 A+113 A b^2+209 a b B\right ) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{693 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 \left (1145 a^2 A b+15 A b^3+539 a^3 B+825 a b^2 B\right ) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{3465 a d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (675 a^4 A+1025 a^2 A b^2-20 A b^4+1793 a^3 b B+55 a b^3 B\right ) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{3465 a^2 d \sqrt{\sec (c+d x)}}+\frac{2 a A (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{11 d \sec ^{\frac{9}{2}}(c+d x)}+\frac{32 \int \frac{\frac{3}{32} a \left (3705 a^4 A b+255 a^2 A b^3+40 A b^5+1617 a^5 B+3069 a^3 b^2 B-110 a b^4 B\right )+\frac{3}{32} a^2 \left (675 a^4 A+3315 a^2 A b^2+10 A b^4+2871 a^3 b B+1705 a b^3 B\right ) \sec (c+d x)}{\sqrt{\sec (c+d x)} \sqrt{a+b \sec (c+d x)}} \, dx}{10395 a^3}\\ &=\frac{2 a (14 A b+11 a B) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{99 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{2 \left (81 a^2 A+113 A b^2+209 a b B\right ) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{693 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 \left (1145 a^2 A b+15 A b^3+539 a^3 B+825 a b^2 B\right ) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{3465 a d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (675 a^4 A+1025 a^2 A b^2-20 A b^4+1793 a^3 b B+55 a b^3 B\right ) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{3465 a^2 d \sqrt{\sec (c+d x)}}+\frac{2 a A (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{11 d \sec ^{\frac{9}{2}}(c+d x)}+\frac{\left (\left (a^2-b^2\right ) \left (675 a^4 A+285 a^2 A b^2+40 A b^4+1254 a^3 b B-110 a b^3 B\right )\right ) \int \frac{\sqrt{\sec (c+d x)}}{\sqrt{a+b \sec (c+d x)}} \, dx}{3465 a^3}+\frac{\left (3705 a^4 A b+255 a^2 A b^3+40 A b^5+1617 a^5 B+3069 a^3 b^2 B-110 a b^4 B\right ) \int \frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{\sec (c+d x)}} \, dx}{3465 a^3}\\ &=\frac{2 a (14 A b+11 a B) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{99 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{2 \left (81 a^2 A+113 A b^2+209 a b B\right ) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{693 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 \left (1145 a^2 A b+15 A b^3+539 a^3 B+825 a b^2 B\right ) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{3465 a d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (675 a^4 A+1025 a^2 A b^2-20 A b^4+1793 a^3 b B+55 a b^3 B\right ) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{3465 a^2 d \sqrt{\sec (c+d x)}}+\frac{2 a A (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{11 d \sec ^{\frac{9}{2}}(c+d x)}+\frac{\left (\left (a^2-b^2\right ) \left (675 a^4 A+285 a^2 A b^2+40 A b^4+1254 a^3 b B-110 a b^3 B\right ) \sqrt{b+a \cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{b+a \cos (c+d x)}} \, dx}{3465 a^3 \sqrt{a+b \sec (c+d x)}}+\frac{\left (\left (3705 a^4 A b+255 a^2 A b^3+40 A b^5+1617 a^5 B+3069 a^3 b^2 B-110 a b^4 B\right ) \sqrt{a+b \sec (c+d x)}\right ) \int \sqrt{b+a \cos (c+d x)} \, dx}{3465 a^3 \sqrt{b+a \cos (c+d x)} \sqrt{\sec (c+d x)}}\\ &=\frac{2 a (14 A b+11 a B) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{99 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{2 \left (81 a^2 A+113 A b^2+209 a b B\right ) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{693 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 \left (1145 a^2 A b+15 A b^3+539 a^3 B+825 a b^2 B\right ) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{3465 a d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (675 a^4 A+1025 a^2 A b^2-20 A b^4+1793 a^3 b B+55 a b^3 B\right ) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{3465 a^2 d \sqrt{\sec (c+d x)}}+\frac{2 a A (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{11 d \sec ^{\frac{9}{2}}(c+d x)}+\frac{\left (\left (a^2-b^2\right ) \left (675 a^4 A+285 a^2 A b^2+40 A b^4+1254 a^3 b B-110 a b^3 B\right ) \sqrt{\frac{b+a \cos (c+d x)}{a+b}} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\frac{b}{a+b}+\frac{a \cos (c+d x)}{a+b}}} \, dx}{3465 a^3 \sqrt{a+b \sec (c+d x)}}+\frac{\left (\left (3705 a^4 A b+255 a^2 A b^3+40 A b^5+1617 a^5 B+3069 a^3 b^2 B-110 a b^4 B\right ) \sqrt{a+b \sec (c+d x)}\right ) \int \sqrt{\frac{b}{a+b}+\frac{a \cos (c+d x)}{a+b}} \, dx}{3465 a^3 \sqrt{\frac{b+a \cos (c+d x)}{a+b}} \sqrt{\sec (c+d x)}}\\ &=\frac{2 \left (a^2-b^2\right ) \left (675 a^4 A+285 a^2 A b^2+40 A b^4+1254 a^3 b B-110 a b^3 B\right ) \sqrt{\frac{b+a \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right ) \sqrt{\sec (c+d x)}}{3465 a^3 d \sqrt{a+b \sec (c+d x)}}+\frac{2 \left (3705 a^4 A b+255 a^2 A b^3+40 A b^5+1617 a^5 B+3069 a^3 b^2 B-110 a b^4 B\right ) E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right ) \sqrt{a+b \sec (c+d x)}}{3465 a^3 d \sqrt{\frac{b+a \cos (c+d x)}{a+b}} \sqrt{\sec (c+d x)}}+\frac{2 a (14 A b+11 a B) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{99 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{2 \left (81 a^2 A+113 A b^2+209 a b B\right ) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{693 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 \left (1145 a^2 A b+15 A b^3+539 a^3 B+825 a b^2 B\right ) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{3465 a d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (675 a^4 A+1025 a^2 A b^2-20 A b^4+1793 a^3 b B+55 a b^3 B\right ) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{3465 a^2 d \sqrt{\sec (c+d x)}}+\frac{2 a A (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{11 d \sec ^{\frac{9}{2}}(c+d x)}\\ \end{align*}

Mathematica [A]  time = 3.5526, size = 380, normalized size = 0.73 \[ \frac{(a+b \sec (c+d x))^{5/2} \left (16 \sqrt{\frac{a \cos (c+d x)+b}{a+b}} \left (a^2 \left (3315 a^2 A b^2+675 a^4 A+2871 a^3 b B+1705 a b^3 B+10 A b^4\right ) \text{EllipticF}\left (\frac{1}{2} (c+d x),\frac{2 a}{a+b}\right )+\left (255 a^2 A b^3+3705 a^4 A b+3069 a^3 b^2 B+1617 a^5 B-110 a b^4 B+40 A b^5\right ) \left ((a+b) E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )-b \text{EllipticF}\left (\frac{1}{2} (c+d x),\frac{2 a}{a+b}\right )\right )\right )+a (a \cos (c+d x)+b) \left (2 \left (9330 a^2 A b^2+6525 a^4 A+16434 a^3 b B+440 a b^3 B-160 A b^4\right ) \sin (c+d x)+a \left (4 \left (3095 a^2 A b+1463 a^3 B+1650 a b^2 B+30 A b^3\right ) \sin (2 (c+d x))+5 a \left (\left (513 a^2 A+836 a b B+452 A b^2\right ) \sin (3 (c+d x))+7 a ((22 a B+46 A b) \sin (4 (c+d x))+9 a A \sin (5 (c+d x)))\right )\right )\right )\right )}{27720 a^3 d \sec ^{\frac{5}{2}}(c+d x) (a \cos (c+d x)+b)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x]))/Sec[c + d*x]^(11/2),x]

[Out]

((a + b*Sec[c + d*x])^(5/2)*(16*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*(a^2*(675*a^4*A + 3315*a^2*A*b^2 + 10*A*b^4
+ 2871*a^3*b*B + 1705*a*b^3*B)*EllipticF[(c + d*x)/2, (2*a)/(a + b)] + (3705*a^4*A*b + 255*a^2*A*b^3 + 40*A*b
^5 + 1617*a^5*B + 3069*a^3*b^2*B - 110*a*b^4*B)*((a + b)*EllipticE[(c + d*x)/2, (2*a)/(a + b)] - b*EllipticF[(
c + d*x)/2, (2*a)/(a + b)])) + a*(b + a*Cos[c + d*x])*(2*(6525*a^4*A + 9330*a^2*A*b^2 - 160*A*b^4 + 16434*a^3*
b*B + 440*a*b^3*B)*Sin[c + d*x] + a*(4*(3095*a^2*A*b + 30*A*b^3 + 1463*a^3*B + 1650*a*b^2*B)*Sin[2*(c + d*x)]
+ 5*a*((513*a^2*A + 452*A*b^2 + 836*a*b*B)*Sin[3*(c + d*x)] + 7*a*((46*A*b + 22*a*B)*Sin[4*(c + d*x)] + 9*a*A*
Sin[5*(c + d*x)]))))))/(27720*a^3*d*(b + a*Cos[c + d*x])^3*Sec[c + d*x]^(5/2))

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Maple [B]  time = 1.022, size = 5946, normalized size = 11.5 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c))/sec(d*x+c)^(11/2),x)

[Out]

result too large to display

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\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 )^{\frac{11}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c))/sec(d*x+c)^(11/2),x, algorithm="maxima")

[Out]

integrate((B*sec(d*x + c) + A)*(b*sec(d*x + c) + a)^(5/2)/sec(d*x + c)^(11/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (B b^{2} \sec \left (d x + c\right )^{3} + A a^{2} +{\left (2 \, B a b + A b^{2}\right )} \sec \left (d x + c\right )^{2} +{\left (B a^{2} + 2 \, A a b\right )} \sec \left (d x + c\right )\right )} \sqrt{b \sec \left (d x + c\right ) + a}}{\sec \left (d x + c\right )^{\frac{11}{2}}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c))/sec(d*x+c)^(11/2),x, algorithm="fricas")

[Out]

integral((B*b^2*sec(d*x + c)^3 + A*a^2 + (2*B*a*b + A*b^2)*sec(d*x + c)^2 + (B*a^2 + 2*A*a*b)*sec(d*x + c))*sq
rt(b*sec(d*x + c) + a)/sec(d*x + c)^(11/2), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))**(5/2)*(A+B*sec(d*x+c))/sec(d*x+c)**(11/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\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 )^{\frac{11}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c))/sec(d*x+c)^(11/2),x, algorithm="giac")

[Out]

integrate((B*sec(d*x + c) + A)*(b*sec(d*x + c) + a)^(5/2)/sec(d*x + c)^(11/2), x)