3.1003 \(\int \frac{(a+b \sec (c+d x))^3 (A+B \sec (c+d x)+C \sec ^2(c+d x))}{\sec ^{\frac{11}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=401 \[ \frac{2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right ) \left (5 a^3 (9 A+11 C)+165 a^2 b B+33 a b^2 (5 A+7 C)+77 b^3 B\right )}{231 d}+\frac{2 \sin (c+d x) \left (33 a^2 b (7 A+9 C)+77 a^3 B+242 a b^2 B+24 A b^3\right )}{495 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 a \sin (c+d x) \left (9 a^2 (9 A+11 C)+143 a b B+24 A b^2\right )}{693 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 \sin (c+d x) \left (5 a^3 (9 A+11 C)+165 a^2 b B+33 a b^2 (5 A+7 C)+77 b^3 B\right )}{231 d \sqrt{\sec (c+d x)}}+\frac{2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (3 a^2 b (7 A+9 C)+7 a^3 B+27 a b^2 B+3 b^3 (3 A+5 C)\right )}{15 d}+\frac{2 (11 a B+6 A b) \sin (c+d x) (a+b \sec (c+d x))^2}{99 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{2 A \sin (c+d x) (a+b \sec (c+d x))^3}{11 d \sec ^{\frac{9}{2}}(c+d x)} \]

[Out]

(2*(7*a^3*B + 27*a*b^2*B + 3*b^3*(3*A + 5*C) + 3*a^2*b*(7*A + 9*C))*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2,
2]*Sqrt[Sec[c + d*x]])/(15*d) + (2*(165*a^2*b*B + 77*b^3*B + 33*a*b^2*(5*A + 7*C) + 5*a^3*(9*A + 11*C))*Sqrt[C
os[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(231*d) + (2*a*(24*A*b^2 + 143*a*b*B + 9*a^2*(9*A +
11*C))*Sin[c + d*x])/(693*d*Sec[c + d*x]^(5/2)) + (2*(24*A*b^3 + 77*a^3*B + 242*a*b^2*B + 33*a^2*b*(7*A + 9*C
))*Sin[c + d*x])/(495*d*Sec[c + d*x]^(3/2)) + (2*(165*a^2*b*B + 77*b^3*B + 33*a*b^2*(5*A + 7*C) + 5*a^3*(9*A +
11*C))*Sin[c + d*x])/(231*d*Sqrt[Sec[c + d*x]]) + (2*(6*A*b + 11*a*B)*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(9
9*d*Sec[c + d*x]^(7/2)) + (2*A*(a + b*Sec[c + d*x])^3*Sin[c + d*x])/(11*d*Sec[c + d*x]^(9/2))

________________________________________________________________________________________

Rubi [A]  time = 0.914722, antiderivative size = 401, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 8, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.186, Rules used = {4094, 4074, 4047, 3769, 3771, 2641, 4045, 2639} \[ \frac{2 \sin (c+d x) \left (33 a^2 b (7 A+9 C)+77 a^3 B+242 a b^2 B+24 A b^3\right )}{495 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 a \sin (c+d x) \left (9 a^2 (9 A+11 C)+143 a b B+24 A b^2\right )}{693 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 \sin (c+d x) \left (5 a^3 (9 A+11 C)+165 a^2 b B+33 a b^2 (5 A+7 C)+77 b^3 B\right )}{231 d \sqrt{\sec (c+d x)}}+\frac{2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (5 a^3 (9 A+11 C)+165 a^2 b B+33 a b^2 (5 A+7 C)+77 b^3 B\right )}{231 d}+\frac{2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (3 a^2 b (7 A+9 C)+7 a^3 B+27 a b^2 B+3 b^3 (3 A+5 C)\right )}{15 d}+\frac{2 (11 a B+6 A b) \sin (c+d x) (a+b \sec (c+d x))^2}{99 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{2 A \sin (c+d x) (a+b \sec (c+d x))^3}{11 d \sec ^{\frac{9}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/Sec[c + d*x]^(11/2),x]

[Out]

(2*(7*a^3*B + 27*a*b^2*B + 3*b^3*(3*A + 5*C) + 3*a^2*b*(7*A + 9*C))*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2,
2]*Sqrt[Sec[c + d*x]])/(15*d) + (2*(165*a^2*b*B + 77*b^3*B + 33*a*b^2*(5*A + 7*C) + 5*a^3*(9*A + 11*C))*Sqrt[C
os[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(231*d) + (2*a*(24*A*b^2 + 143*a*b*B + 9*a^2*(9*A +
11*C))*Sin[c + d*x])/(693*d*Sec[c + d*x]^(5/2)) + (2*(24*A*b^3 + 77*a^3*B + 242*a*b^2*B + 33*a^2*b*(7*A + 9*C
))*Sin[c + d*x])/(495*d*Sec[c + d*x]^(3/2)) + (2*(165*a^2*b*B + 77*b^3*B + 33*a*b^2*(5*A + 7*C) + 5*a^3*(9*A +
11*C))*Sin[c + d*x])/(231*d*Sqrt[Sec[c + d*x]]) + (2*(6*A*b + 11*a*B)*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(9
9*d*Sec[c + d*x]^(7/2)) + (2*A*(a + b*Sec[c + d*x])^3*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 3769

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Csc[c + d*x])^(n + 1))/(b*d*n), x
] + Dist[(n + 1)/(b^2*n), Int[(b*Csc[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && Integer
Q[2*n]

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 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]

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 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]

Rubi steps

\begin{align*} \int \frac{(a+b \sec (c+d x))^3 \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))^3 \sin (c+d x)}{11 d \sec ^{\frac{9}{2}}(c+d x)}+\frac{2}{11} \int \frac{(a+b \sec (c+d x))^2 \left (\frac{1}{2} (6 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 (3 A+11 C) \sec ^2(c+d x)\right )}{\sec ^{\frac{9}{2}}(c+d x)} \, dx\\ &=\frac{2 (6 A b+11 a B) (a+b \sec (c+d x))^2 \sin (c+d x)}{99 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{2 A (a+b \sec (c+d x))^3 \sin (c+d x)}{11 d \sec ^{\frac{9}{2}}(c+d x)}+\frac{4}{99} \int \frac{(a+b \sec (c+d x)) \left (\frac{1}{4} \left (24 A b^2+143 a b B+9 a^2 (9 A+11 C)\right )+\frac{1}{4} \left (150 a A b+77 a^2 B+99 b^2 B+198 a b C\right ) \sec (c+d x)+\frac{3}{4} b (15 A b+11 a B+33 b C) \sec ^2(c+d x)\right )}{\sec ^{\frac{7}{2}}(c+d x)} \, dx\\ &=\frac{2 a \left (24 A b^2+143 a b B+9 a^2 (9 A+11 C)\right ) \sin (c+d x)}{693 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 (6 A b+11 a B) (a+b \sec (c+d x))^2 \sin (c+d x)}{99 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{2 A (a+b \sec (c+d x))^3 \sin (c+d x)}{11 d \sec ^{\frac{9}{2}}(c+d x)}-\frac{8}{693} \int \frac{-\frac{7}{8} \left (24 A b^3+77 a^3 B+242 a b^2 B+33 a^2 b (7 A+9 C)\right )-\frac{9}{8} \left (165 a^2 b B+77 b^3 B+33 a b^2 (5 A+7 C)+5 a^3 (9 A+11 C)\right ) \sec (c+d x)-\frac{21}{8} b^2 (15 A b+11 a B+33 b C) \sec ^2(c+d x)}{\sec ^{\frac{5}{2}}(c+d x)} \, dx\\ &=\frac{2 a \left (24 A b^2+143 a b B+9 a^2 (9 A+11 C)\right ) \sin (c+d x)}{693 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 (6 A b+11 a B) (a+b \sec (c+d x))^2 \sin (c+d x)}{99 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{2 A (a+b \sec (c+d x))^3 \sin (c+d x)}{11 d \sec ^{\frac{9}{2}}(c+d x)}-\frac{8}{693} \int \frac{-\frac{7}{8} \left (24 A b^3+77 a^3 B+242 a b^2 B+33 a^2 b (7 A+9 C)\right )-\frac{21}{8} b^2 (15 A b+11 a B+33 b C) \sec ^2(c+d x)}{\sec ^{\frac{5}{2}}(c+d x)} \, dx-\frac{1}{77} \left (-165 a^2 b B-77 b^3 B-33 a b^2 (5 A+7 C)-5 a^3 (9 A+11 C)\right ) \int \frac{1}{\sec ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{2 a \left (24 A b^2+143 a b B+9 a^2 (9 A+11 C)\right ) \sin (c+d x)}{693 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 \left (24 A b^3+77 a^3 B+242 a b^2 B+33 a^2 b (7 A+9 C)\right ) \sin (c+d x)}{495 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (165 a^2 b B+77 b^3 B+33 a b^2 (5 A+7 C)+5 a^3 (9 A+11 C)\right ) \sin (c+d x)}{231 d \sqrt{\sec (c+d x)}}+\frac{2 (6 A b+11 a B) (a+b \sec (c+d x))^2 \sin (c+d x)}{99 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{2 A (a+b \sec (c+d x))^3 \sin (c+d x)}{11 d \sec ^{\frac{9}{2}}(c+d x)}-\frac{1}{15} \left (-7 a^3 B-27 a b^2 B-3 b^3 (3 A+5 C)-3 a^2 b (7 A+9 C)\right ) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx-\frac{1}{231} \left (-165 a^2 b B-77 b^3 B-33 a b^2 (5 A+7 C)-5 a^3 (9 A+11 C)\right ) \int \sqrt{\sec (c+d x)} \, dx\\ &=\frac{2 a \left (24 A b^2+143 a b B+9 a^2 (9 A+11 C)\right ) \sin (c+d x)}{693 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 \left (24 A b^3+77 a^3 B+242 a b^2 B+33 a^2 b (7 A+9 C)\right ) \sin (c+d x)}{495 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (165 a^2 b B+77 b^3 B+33 a b^2 (5 A+7 C)+5 a^3 (9 A+11 C)\right ) \sin (c+d x)}{231 d \sqrt{\sec (c+d x)}}+\frac{2 (6 A b+11 a B) (a+b \sec (c+d x))^2 \sin (c+d x)}{99 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{2 A (a+b \sec (c+d x))^3 \sin (c+d x)}{11 d \sec ^{\frac{9}{2}}(c+d x)}-\frac{1}{15} \left (\left (-7 a^3 B-27 a b^2 B-3 b^3 (3 A+5 C)-3 a^2 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{1}{231} \left (\left (-165 a^2 b B-77 b^3 B-33 a b^2 (5 A+7 C)-5 a^3 (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^3 B+27 a b^2 B+3 b^3 (3 A+5 C)+3 a^2 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 (165 a^2 b B+77 b^3 B+33 a b^2 (5 A+7 C)+5 a^3 (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 (24 A b^2+143 a b B+9 a^2 (9 A+11 C)\right ) \sin (c+d x)}{693 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 \left (24 A b^3+77 a^3 B+242 a b^2 B+33 a^2 b (7 A+9 C)\right ) \sin (c+d x)}{495 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (165 a^2 b B+77 b^3 B+33 a b^2 (5 A+7 C)+5 a^3 (9 A+11 C)\right ) \sin (c+d x)}{231 d \sqrt{\sec (c+d x)}}+\frac{2 (6 A b+11 a B) (a+b \sec (c+d x))^2 \sin (c+d x)}{99 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{2 A (a+b \sec (c+d x))^3 \sin (c+d x)}{11 d \sec ^{\frac{9}{2}}(c+d x)}\\ \end{align*}

Mathematica [A]  time = 6.83624, size = 538, normalized size = 1.34 \[ \frac{2 \cos ^5(c+d x) (a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right ) \left (225 a^3 A+825 a^2 b B+275 a^3 C+825 a A b^2+1155 a b^2 C+385 b^3 B\right )+\frac{2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (1617 a^2 A b+2079 a^2 b C+539 a^3 B+2079 a b^2 B+693 A b^3+1155 b^3 C\right )}{\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}}\right )}{1155 d (a \cos (c+d x)+b)^3 (A \cos (2 c+2 d x)+A+2 B \cos (c+d x)+2 C)}+\frac{(a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (\frac{1}{154} a \sin (4 (c+d x)) \left (16 a^2 A+11 a^2 C+33 a b B+33 A b^2\right )+\frac{1}{90} \sin (c+d x) \left (57 a^2 A b+54 a^2 b C+19 a^3 B+54 a b^2 B+18 A b^3\right )+\frac{\sin (2 (c+d x)) \left (1041 a^3 A+3432 a^2 b B+1144 a^3 C+3432 a A b^2+3696 a b^2 C+1232 b^3 B\right )}{1848}+\frac{1}{180} \sin (3 (c+d x)) \left (129 a^2 A b+108 a^2 b C+43 a^3 B+108 a b^2 B+36 A b^3\right )+\frac{1}{36} a^2 (a B+3 A b) \sin (5 (c+d x))+\frac{1}{88} a^3 A \sin (6 (c+d x))\right )}{d \sec ^{\frac{9}{2}}(c+d x) (a \cos (c+d x)+b)^3 (A \cos (2 c+2 d x)+A+2 B \cos (c+d x)+2 C)} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*Sec[c + d*x])^3*(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]^5*((2*(1617*a^2*A*b + 693*A*b^3 + 539*a^3*B + 2079*a*b^2*B + 2079*a^2*b*C + 1155*b^3*C)*Ellipt
icE[(c + d*x)/2, 2])/(Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) + 2*(225*a^3*A + 825*a*A*b^2 + 825*a^2*b*B + 385*
b^3*B + 275*a^3*C + 1155*a*b^2*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])*(a + b*Sec[
c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(1155*d*(b + a*Cos[c + d*x])^3*(A + 2*C + 2*B*Cos[c + d*x
] + A*Cos[2*c + 2*d*x])) + ((a + b*Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(((57*a^2*A*b + 18*
A*b^3 + 19*a^3*B + 54*a*b^2*B + 54*a^2*b*C)*Sin[c + d*x])/90 + ((1041*a^3*A + 3432*a*A*b^2 + 3432*a^2*b*B + 12
32*b^3*B + 1144*a^3*C + 3696*a*b^2*C)*Sin[2*(c + d*x)])/1848 + ((129*a^2*A*b + 36*A*b^3 + 43*a^3*B + 108*a*b^2
*B + 108*a^2*b*C)*Sin[3*(c + d*x)])/180 + (a*(16*a^2*A + 33*A*b^2 + 33*a*b*B + 11*a^2*C)*Sin[4*(c + d*x)])/154
+ (a^2*(3*A*b + a*B)*Sin[5*(c + d*x)])/36 + (a^3*A*Sin[6*(c + d*x)])/88))/(d*(b + a*Cos[c + d*x])^3*(A + 2*C
+ 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*Sec[c + d*x]^(9/2))

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Maple [B]  time = 2.287, size = 1082, normalized size = 2.7 \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))^3*(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^3*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/
2*c)^12+(-50400*A*a^3-36960*A*a^2*b-12320*B*a^3)*sin(1/2*d*x+1/2*c)^10*cos(1/2*d*x+1/2*c)+(56880*A*a^3+73920*A
*a^2*b+23760*A*a*b^2+24640*B*a^3+23760*B*a^2*b+7920*C*a^3)*sin(1/2*d*x+1/2*c)^8*cos(1/2*d*x+1/2*c)+(-34920*A*a
^3-68376*A*a^2*b-35640*A*a*b^2-5544*A*b^3-22792*B*a^3-35640*B*a^2*b-16632*B*a*b^2-11880*C*a^3-16632*C*a^2*b)*s
in(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+(13860*A*a^3+31416*A*a^2*b+27720*A*a*b^2+5544*A*b^3+10472*B*a^3+27720*B
*a^2*b+16632*B*a*b^2+4620*B*b^3+9240*C*a^3+16632*C*a^2*b+13860*C*a*b^2)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c
)+(-2790*A*a^3-5544*A*a^2*b-7920*A*a*b^2-1386*A*b^3-1848*B*a^3-7920*B*a^2*b-4158*B*a*b^2-2310*B*b^3-2640*C*a^3
-4158*C*a^2*b-6930*C*a*b^2)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-4851*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-2079*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))*b^3+675*A*a^3*(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))+2475*A*a*b^2*(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))-1617*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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a^3-6237*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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a*b^2+2475*B*a^2*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))+1155*B*b^3*(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))-6237*C*(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-3465*C*(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))*b^3+825*a^3*C*(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))+3465*C*a*b^2*(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)))/(-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))^3*(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^{3} \sec \left (d x + c\right )^{5} +{\left (3 \, C a b^{2} + B b^{3}\right )} \sec \left (d x + c\right )^{4} + A a^{3} +{\left (3 \, C a^{2} b + 3 \, B a b^{2} + A b^{3}\right )} \sec \left (d x + c\right )^{3} +{\left (C a^{3} + 3 \, B a^{2} b + 3 \, A a b^{2}\right )} \sec \left (d x + c\right )^{2} +{\left (B a^{3} + 3 \, A a^{2} 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))^3*(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^3*sec(d*x + c)^5 + (3*C*a*b^2 + B*b^3)*sec(d*x + c)^4 + A*a^3 + (3*C*a^2*b + 3*B*a*b^2 + A*b^3)*
sec(d*x + c)^3 + (C*a^3 + 3*B*a^2*b + 3*A*a*b^2)*sec(d*x + c)^2 + (B*a^3 + 3*A*a^2*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))**3*(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 )}^{3}}{\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))^3*(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)^3/sec(d*x + c)^(11/2), x)