3.1010 \(\int \frac{(a+b \sec (c+d x))^4 (A+B \sec (c+d x)+C \sec ^2(c+d x))}{\sec ^{\frac{11}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=444 \[ \frac{2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right ) \left (66 a^2 b^2 (5 A+7 C)+5 a^4 (9 A+11 C)+220 a^3 b B+308 a b^3 B+77 b^4 (A+3 C)\right )}{231 d}+\frac{2 \sin (c+d x) \left (3 a^2 (9 A+11 C)+55 a b B+16 A b^2\right ) (a+b \sec (c+d x))^2}{231 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 a \sin (c+d x) \left (2 a^2 b (673 A+891 C)+539 a^3 B+1353 a b^2 B+192 A b^3\right )}{3465 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 \sin (c+d x) \left (9 a^2 b^2 (101 A+143 C)+15 a^4 (9 A+11 C)+660 a^3 b B+682 a b^3 B+64 A b^4\right )}{693 d \sqrt{\sec (c+d x)}}+\frac{2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (4 a^3 b (7 A+9 C)+54 a^2 b^2 B+7 a^4 B+12 a b^3 (3 A+5 C)+15 b^4 B\right )}{15 d}+\frac{2 (11 a B+8 A b) \sin (c+d x) (a+b \sec (c+d x))^3}{99 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{2 A \sin (c+d x) (a+b \sec (c+d x))^4}{11 d \sec ^{\frac{9}{2}}(c+d x)} \]

[Out]

(2*(7*a^4*B + 54*a^2*b^2*B + 15*b^4*B + 12*a*b^3*(3*A + 5*C) + 4*a^3*b*(7*A + 9*C))*Sqrt[Cos[c + d*x]]*Ellipti
cE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(15*d) + (2*(220*a^3*b*B + 308*a*b^3*B + 77*b^4*(A + 3*C) + 66*a^2*b^2*
(5*A + 7*C) + 5*a^4*(9*A + 11*C))*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(231*d) + (
2*a*(192*A*b^3 + 539*a^3*B + 1353*a*b^2*B + 2*a^2*b*(673*A + 891*C))*Sin[c + d*x])/(3465*d*Sec[c + d*x]^(3/2))
+ (2*(64*A*b^4 + 660*a^3*b*B + 682*a*b^3*B + 15*a^4*(9*A + 11*C) + 9*a^2*b^2*(101*A + 143*C))*Sin[c + d*x])/(
693*d*Sqrt[Sec[c + d*x]]) + (2*(16*A*b^2 + 55*a*b*B + 3*a^2*(9*A + 11*C))*(a + b*Sec[c + d*x])^2*Sin[c + d*x])
/(231*d*Sec[c + d*x]^(5/2)) + (2*(8*A*b + 11*a*B)*(a + b*Sec[c + d*x])^3*Sin[c + d*x])/(99*d*Sec[c + d*x]^(7/2
)) + (2*A*(a + b*Sec[c + d*x])^4*Sin[c + d*x])/(11*d*Sec[c + d*x]^(9/2))

________________________________________________________________________________________

Rubi [A]  time = 1.31581, antiderivative size = 444, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 7, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.163, Rules used = {4094, 4074, 4047, 3771, 2639, 4045, 2641} \[ \frac{2 \sin (c+d x) \left (3 a^2 (9 A+11 C)+55 a b B+16 A b^2\right ) (a+b \sec (c+d x))^2}{231 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 a \sin (c+d x) \left (2 a^2 b (673 A+891 C)+539 a^3 B+1353 a b^2 B+192 A b^3\right )}{3465 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 \sin (c+d x) \left (9 a^2 b^2 (101 A+143 C)+15 a^4 (9 A+11 C)+660 a^3 b B+682 a b^3 B+64 A b^4\right )}{693 d \sqrt{\sec (c+d x)}}+\frac{2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (66 a^2 b^2 (5 A+7 C)+5 a^4 (9 A+11 C)+220 a^3 b B+308 a b^3 B+77 b^4 (A+3 C)\right )}{231 d}+\frac{2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (4 a^3 b (7 A+9 C)+54 a^2 b^2 B+7 a^4 B+12 a b^3 (3 A+5 C)+15 b^4 B\right )}{15 d}+\frac{2 (11 a B+8 A b) \sin (c+d x) (a+b \sec (c+d x))^3}{99 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{2 A \sin (c+d x) (a+b \sec (c+d x))^4}{11 d \sec ^{\frac{9}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*Sec[c + d*x])^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/Sec[c + d*x]^(11/2),x]

[Out]

(2*(7*a^4*B + 54*a^2*b^2*B + 15*b^4*B + 12*a*b^3*(3*A + 5*C) + 4*a^3*b*(7*A + 9*C))*Sqrt[Cos[c + d*x]]*Ellipti
cE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(15*d) + (2*(220*a^3*b*B + 308*a*b^3*B + 77*b^4*(A + 3*C) + 66*a^2*b^2*
(5*A + 7*C) + 5*a^4*(9*A + 11*C))*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(231*d) + (
2*a*(192*A*b^3 + 539*a^3*B + 1353*a*b^2*B + 2*a^2*b*(673*A + 891*C))*Sin[c + d*x])/(3465*d*Sec[c + d*x]^(3/2))
+ (2*(64*A*b^4 + 660*a^3*b*B + 682*a*b^3*B + 15*a^4*(9*A + 11*C) + 9*a^2*b^2*(101*A + 143*C))*Sin[c + d*x])/(
693*d*Sqrt[Sec[c + d*x]]) + (2*(16*A*b^2 + 55*a*b*B + 3*a^2*(9*A + 11*C))*(a + b*Sec[c + d*x])^2*Sin[c + d*x])
/(231*d*Sec[c + d*x]^(5/2)) + (2*(8*A*b + 11*a*B)*(a + b*Sec[c + d*x])^3*Sin[c + d*x])/(99*d*Sec[c + d*x]^(7/2
)) + (2*A*(a + b*Sec[c + d*x])^4*Sin[c + d*x])/(11*d*Sec[c + d*x]^(9/2))

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 4074

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_)), x_Symbol] :> Simp[(A*a*Cot[e + f*x]*(d*Csc[e + f*x])^n)/(f*n), x]
+ Dist[1/(d*n), Int[(d*Csc[e + f*x])^(n + 1)*Simp[n*(B*a + A*b) + (n*(a*C + B*b) + A*a*(n + 1))*Csc[e + f*x]
+ b*C*n*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C}, x] && LtQ[n, -1]

Rule 4047

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(
C_.)), x_Symbol] :> Dist[B/b, Int[(b*Csc[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x
]^2), x] /; FreeQ[{b, e, f, A, B, C, m}, x]

Rule 3771

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

Rule 4045

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

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rubi steps

\begin{align*} \int \frac{(a+b \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac{11}{2}}(c+d x)} \, dx &=\frac{2 A (a+b \sec (c+d x))^4 \sin (c+d x)}{11 d \sec ^{\frac{9}{2}}(c+d x)}+\frac{2}{11} \int \frac{(a+b \sec (c+d x))^3 \left (\frac{1}{2} (8 A b+11 a B)+\frac{1}{2} (9 a A+11 b B+11 a C) \sec (c+d x)+\frac{1}{2} b (A+11 C) \sec ^2(c+d x)\right )}{\sec ^{\frac{9}{2}}(c+d x)} \, dx\\ &=\frac{2 (8 A b+11 a B) (a+b \sec (c+d x))^3 \sin (c+d x)}{99 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{2 A (a+b \sec (c+d x))^4 \sin (c+d x)}{11 d \sec ^{\frac{9}{2}}(c+d x)}+\frac{4}{99} \int \frac{(a+b \sec (c+d x))^2 \left (\frac{3}{4} \left (16 A b^2+55 a b B+3 a^2 (9 A+11 C)\right )+\frac{1}{4} \left (146 a A b+77 a^2 B+99 b^2 B+198 a b C\right ) \sec (c+d x)+\frac{1}{4} b (17 A b+11 a B+99 b C) \sec ^2(c+d x)\right )}{\sec ^{\frac{7}{2}}(c+d x)} \, dx\\ &=\frac{2 \left (16 A b^2+55 a b B+3 a^2 (9 A+11 C)\right ) (a+b \sec (c+d x))^2 \sin (c+d x)}{231 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 (8 A b+11 a B) (a+b \sec (c+d x))^3 \sin (c+d x)}{99 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{2 A (a+b \sec (c+d x))^4 \sin (c+d x)}{11 d \sec ^{\frac{9}{2}}(c+d x)}+\frac{8}{693} \int \frac{(a+b \sec (c+d x)) \left (\frac{1}{8} \left (192 A b^3+539 a^3 B+1353 a b^2 B+2 a^2 b (673 A+891 C)\right )+\frac{1}{8} \left (1441 a^2 b B+693 b^3 B+45 a^3 (9 A+11 C)+a b^2 (1381 A+2079 C)\right ) \sec (c+d x)+\frac{1}{8} b \left (242 a b B+9 a^2 (9 A+11 C)+b^2 (167 A+693 C)\right ) \sec ^2(c+d x)\right )}{\sec ^{\frac{5}{2}}(c+d x)} \, dx\\ &=\frac{2 a \left (192 A b^3+539 a^3 B+1353 a b^2 B+2 a^2 b (673 A+891 C)\right ) \sin (c+d x)}{3465 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (16 A b^2+55 a b B+3 a^2 (9 A+11 C)\right ) (a+b \sec (c+d x))^2 \sin (c+d x)}{231 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 (8 A b+11 a B) (a+b \sec (c+d x))^3 \sin (c+d x)}{99 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{2 A (a+b \sec (c+d x))^4 \sin (c+d x)}{11 d \sec ^{\frac{9}{2}}(c+d x)}-\frac{16 \int \frac{-\frac{15}{16} \left (64 A b^4+660 a^3 b B+682 a b^3 B+15 a^4 (9 A+11 C)+9 a^2 b^2 (101 A+143 C)\right )-\frac{231}{16} \left (7 a^4 B+54 a^2 b^2 B+15 b^4 B+12 a b^3 (3 A+5 C)+4 a^3 b (7 A+9 C)\right ) \sec (c+d x)-\frac{5}{16} b^2 \left (242 a b B+9 a^2 (9 A+11 C)+b^2 (167 A+693 C)\right ) \sec ^2(c+d x)}{\sec ^{\frac{3}{2}}(c+d x)} \, dx}{3465}\\ &=\frac{2 a \left (192 A b^3+539 a^3 B+1353 a b^2 B+2 a^2 b (673 A+891 C)\right ) \sin (c+d x)}{3465 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (16 A b^2+55 a b B+3 a^2 (9 A+11 C)\right ) (a+b \sec (c+d x))^2 \sin (c+d x)}{231 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 (8 A b+11 a B) (a+b \sec (c+d x))^3 \sin (c+d x)}{99 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{2 A (a+b \sec (c+d x))^4 \sin (c+d x)}{11 d \sec ^{\frac{9}{2}}(c+d x)}-\frac{16 \int \frac{-\frac{15}{16} \left (64 A b^4+660 a^3 b B+682 a b^3 B+15 a^4 (9 A+11 C)+9 a^2 b^2 (101 A+143 C)\right )-\frac{5}{16} b^2 \left (242 a b B+9 a^2 (9 A+11 C)+b^2 (167 A+693 C)\right ) \sec ^2(c+d x)}{\sec ^{\frac{3}{2}}(c+d x)} \, dx}{3465}-\frac{1}{15} \left (-7 a^4 B-54 a^2 b^2 B-15 b^4 B-12 a b^3 (3 A+5 C)-4 a^3 b (7 A+9 C)\right ) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx\\ &=\frac{2 a \left (192 A b^3+539 a^3 B+1353 a b^2 B+2 a^2 b (673 A+891 C)\right ) \sin (c+d x)}{3465 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (64 A b^4+660 a^3 b B+682 a b^3 B+15 a^4 (9 A+11 C)+9 a^2 b^2 (101 A+143 C)\right ) \sin (c+d x)}{693 d \sqrt{\sec (c+d x)}}+\frac{2 \left (16 A b^2+55 a b B+3 a^2 (9 A+11 C)\right ) (a+b \sec (c+d x))^2 \sin (c+d x)}{231 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 (8 A b+11 a B) (a+b \sec (c+d x))^3 \sin (c+d x)}{99 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{2 A (a+b \sec (c+d x))^4 \sin (c+d x)}{11 d \sec ^{\frac{9}{2}}(c+d x)}-\frac{1}{231} \left (-220 a^3 b B-308 a b^3 B-77 b^4 (A+3 C)-66 a^2 b^2 (5 A+7 C)-5 a^4 (9 A+11 C)\right ) \int \sqrt{\sec (c+d x)} \, dx-\frac{1}{15} \left (\left (-7 a^4 B-54 a^2 b^2 B-15 b^4 B-12 a b^3 (3 A+5 C)-4 a^3 b (7 A+9 C)\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx\\ &=\frac{2 \left (7 a^4 B+54 a^2 b^2 B+15 b^4 B+12 a b^3 (3 A+5 C)+4 a^3 b (7 A+9 C)\right ) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{15 d}+\frac{2 a \left (192 A b^3+539 a^3 B+1353 a b^2 B+2 a^2 b (673 A+891 C)\right ) \sin (c+d x)}{3465 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (64 A b^4+660 a^3 b B+682 a b^3 B+15 a^4 (9 A+11 C)+9 a^2 b^2 (101 A+143 C)\right ) \sin (c+d x)}{693 d \sqrt{\sec (c+d x)}}+\frac{2 \left (16 A b^2+55 a b B+3 a^2 (9 A+11 C)\right ) (a+b \sec (c+d x))^2 \sin (c+d x)}{231 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 (8 A b+11 a B) (a+b \sec (c+d x))^3 \sin (c+d x)}{99 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{2 A (a+b \sec (c+d x))^4 \sin (c+d x)}{11 d \sec ^{\frac{9}{2}}(c+d x)}-\frac{1}{231} \left (\left (-220 a^3 b B-308 a b^3 B-77 b^4 (A+3 C)-66 a^2 b^2 (5 A+7 C)-5 a^4 (9 A+11 C)\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{2 \left (7 a^4 B+54 a^2 b^2 B+15 b^4 B+12 a b^3 (3 A+5 C)+4 a^3 b (7 A+9 C)\right ) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{15 d}+\frac{2 \left (220 a^3 b B+308 a b^3 B+77 b^4 (A+3 C)+66 a^2 b^2 (5 A+7 C)+5 a^4 (9 A+11 C)\right ) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{231 d}+\frac{2 a \left (192 A b^3+539 a^3 B+1353 a b^2 B+2 a^2 b (673 A+891 C)\right ) \sin (c+d x)}{3465 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (64 A b^4+660 a^3 b B+682 a b^3 B+15 a^4 (9 A+11 C)+9 a^2 b^2 (101 A+143 C)\right ) \sin (c+d x)}{693 d \sqrt{\sec (c+d x)}}+\frac{2 \left (16 A b^2+55 a b B+3 a^2 (9 A+11 C)\right ) (a+b \sec (c+d x))^2 \sin (c+d x)}{231 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 (8 A b+11 a B) (a+b \sec (c+d x))^3 \sin (c+d x)}{99 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{2 A (a+b \sec (c+d x))^4 \sin (c+d x)}{11 d \sec ^{\frac{9}{2}}(c+d x)}\\ \end{align*}

Mathematica [A]  time = 7.06534, size = 580, normalized size = 1.31 \[ \frac{2 \cos ^6(c+d x) (a+b \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right ) \left (1650 a^2 A b^2+225 a^4 A+2310 a^2 b^2 C+1100 a^3 b B+275 a^4 C+1540 a b^3 B+385 A b^4+1155 b^4 C\right )+\frac{2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (2156 a^3 A b+4158 a^2 b^2 B+2772 a^3 b C+539 a^4 B+2772 a A b^3+4620 a b^3 C+1155 b^4 B\right )}{\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}}\right )}{1155 d (a \cos (c+d x)+b)^4 (A \cos (2 c+2 d x)+A+2 B \cos (c+d x)+2 C)}+\frac{(a+b \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (\frac{1}{154} a^2 \sin (4 (c+d x)) \left (16 a^2 A+11 a^2 C+44 a b B+66 A b^2\right )+\frac{1}{90} a \sin (c+d x) \left (76 a^2 A b+72 a^2 b C+19 a^3 B+108 a b^2 B+72 A b^3\right )+\frac{1}{180} a \sin (3 (c+d x)) \left (172 a^2 A b+144 a^2 b C+43 a^3 B+216 a b^2 B+144 A b^3\right )+\frac{\sin (2 (c+d x)) \left (6864 a^2 A b^2+1041 a^4 A+7392 a^2 b^2 C+4576 a^3 b B+1144 a^4 C+4928 a b^3 B+1232 A b^4\right )}{1848}+\frac{1}{36} a^3 (a B+4 A b) \sin (5 (c+d x))+\frac{1}{88} a^4 A \sin (6 (c+d x))\right )}{d \sec ^{\frac{11}{2}}(c+d x) (a \cos (c+d x)+b)^4 (A \cos (2 c+2 d x)+A+2 B \cos (c+d x)+2 C)} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*Sec[c + d*x])^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/Sec[c + d*x]^(11/2),x]

[Out]

(2*Cos[c + d*x]^6*((2*(2156*a^3*A*b + 2772*a*A*b^3 + 539*a^4*B + 4158*a^2*b^2*B + 1155*b^4*B + 2772*a^3*b*C +
4620*a*b^3*C)*EllipticE[(c + d*x)/2, 2])/(Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) + 2*(225*a^4*A + 1650*a^2*A*b
^2 + 385*A*b^4 + 1100*a^3*b*B + 1540*a*b^3*B + 275*a^4*C + 2310*a^2*b^2*C + 1155*b^4*C)*Sqrt[Cos[c + d*x]]*Ell
ipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])*(a + b*Sec[c + d*x])^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(11
55*d*(b + a*Cos[c + d*x])^4*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])) + ((a + b*Sec[c + d*x])^4*(A +
B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((a*(76*a^2*A*b + 72*A*b^3 + 19*a^3*B + 108*a*b^2*B + 72*a^2*b*C)*Sin[c + d
*x])/90 + ((1041*a^4*A + 6864*a^2*A*b^2 + 1232*A*b^4 + 4576*a^3*b*B + 4928*a*b^3*B + 1144*a^4*C + 7392*a^2*b^2
*C)*Sin[2*(c + d*x)])/1848 + (a*(172*a^2*A*b + 144*A*b^3 + 43*a^3*B + 216*a*b^2*B + 144*a^2*b*C)*Sin[3*(c + d*
x)])/180 + (a^2*(16*a^2*A + 66*A*b^2 + 44*a*b*B + 11*a^2*C)*Sin[4*(c + d*x)])/154 + (a^3*(4*A*b + a*B)*Sin[5*(
c + d*x)])/36 + (a^4*A*Sin[6*(c + d*x)])/88))/(d*(b + a*Cos[c + d*x])^4*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*
c + 2*d*x])*Sec[c + d*x]^(11/2))

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Maple [B]  time = 2.626, size = 1273, normalized size = 2.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sec(d*x+c))^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(11/2),x)

[Out]

-2/3465*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(20160*A*a^4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/
2*c)^12+(-50400*A*a^4-49280*A*a^3*b-12320*B*a^4)*sin(1/2*d*x+1/2*c)^10*cos(1/2*d*x+1/2*c)+(56880*A*a^4+98560*A
*a^3*b+47520*A*a^2*b^2+24640*B*a^4+31680*B*a^3*b+7920*C*a^4)*sin(1/2*d*x+1/2*c)^8*cos(1/2*d*x+1/2*c)+(-34920*A
*a^4-91168*A*a^3*b-71280*A*a^2*b^2-22176*A*a*b^3-22792*B*a^4-47520*B*a^3*b-33264*B*a^2*b^2-11880*C*a^4-22176*C
*a^3*b)*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+(13860*A*a^4+41888*A*a^3*b+55440*A*a^2*b^2+22176*A*a*b^3+4620*
A*b^4+10472*B*a^4+36960*B*a^3*b+33264*B*a^2*b^2+18480*B*a*b^3+9240*C*a^4+22176*C*a^3*b+27720*C*a^2*b^2)*sin(1/
2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-2790*A*a^4-7392*A*a^3*b-15840*A*a^2*b^2-5544*A*a*b^3-2310*A*b^4-1848*B*a^4
-10560*B*a^3*b-8316*B*a^2*b^2-9240*B*a*b^3-2640*C*a^4-5544*C*a^3*b-13860*C*a^2*b^2)*sin(1/2*d*x+1/2*c)^2*cos(1
/2*d*x+1/2*c)+675*A*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c)
,2^(1/2))*a^4+4950*A*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c
),2^(1/2))*a^2*b^2+1155*A*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+
1/2*c),2^(1/2))*b^4-6468*A*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x
+1/2*c),2^(1/2))*a^3*b-8316*A*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*
d*x+1/2*c),2^(1/2))*a*b^3+3300*B*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1
/2*d*x+1/2*c),2^(1/2))*a^3*b+4620*B*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(co
s(1/2*d*x+1/2*c),2^(1/2))*a*b^3-1617*B*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE
(cos(1/2*d*x+1/2*c),2^(1/2))*a^4-12474*B*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*Ellipti
cE(cos(1/2*d*x+1/2*c),2^(1/2))*a^2*b^2-3465*B*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*El
lipticE(cos(1/2*d*x+1/2*c),2^(1/2))*b^4+825*C*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*El
lipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a^4+6930*C*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*E
llipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a^2*b^2+3465*C*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1
/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*b^4-8316*C*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(
1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a^3*b-13860*C*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^
2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a*b^3)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/sin
(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d

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Maxima [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))^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(11/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{C b^{4} \sec \left (d x + c\right )^{6} +{\left (4 \, C a b^{3} + B b^{4}\right )} \sec \left (d x + c\right )^{5} + A a^{4} +{\left (6 \, C a^{2} b^{2} + 4 \, B a b^{3} + A b^{4}\right )} \sec \left (d x + c\right )^{4} + 2 \,{\left (2 \, C a^{3} b + 3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} \sec \left (d x + c\right )^{3} +{\left (C a^{4} + 4 \, B a^{3} b + 6 \, A a^{2} b^{2}\right )} \sec \left (d x + c\right )^{2} +{\left (B a^{4} + 4 \, A a^{3} b\right )} \sec \left (d x + c\right )}{\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))^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(11/2),x, algorithm="fricas")

[Out]

integral((C*b^4*sec(d*x + c)^6 + (4*C*a*b^3 + B*b^4)*sec(d*x + c)^5 + A*a^4 + (6*C*a^2*b^2 + 4*B*a*b^3 + A*b^4
)*sec(d*x + c)^4 + 2*(2*C*a^3*b + 3*B*a^2*b^2 + 2*A*a*b^3)*sec(d*x + c)^3 + (C*a^4 + 4*B*a^3*b + 6*A*a^2*b^2)*
sec(d*x + c)^2 + (B*a^4 + 4*A*a^3*b)*sec(d*x + c))/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))**4*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/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 (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )}{\left (b \sec \left (d x + c\right ) + a\right )}^{4}}{\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))^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(11/2),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*(b*sec(d*x + c) + a)^4/sec(d*x + c)^(11/2), x)