3.1007 \(\int \frac{(a+b \sec (c+d x))^4 (A+B \sec (c+d x)+C \sec ^2(c+d x))}{\sec ^{\frac{5}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=409 \[ \frac{2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right ) \left (4 a^3 b (A+3 C)+18 a^2 b^2 B+a^4 B+4 a b^3 (3 A+C)+b^4 B\right )}{3 d}-\frac{2 b^2 \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x) \left (5 a^2 B+14 a b (A-C)-5 b^2 B\right )}{15 d}-\frac{2 b \sin (c+d x) \sqrt{\sec (c+d x)} \left (a^2 b (31 A-87 C)+10 a^3 B-60 a b^2 B-3 b^3 (5 A+3 C)\right )}{15 d}+\frac{2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (30 a^2 b^2 (A-C)+a^4 (3 A+5 C)+20 a^3 b B-20 a b^3 B-b^4 (5 A+3 C)\right )}{5 d}-\frac{2 b \sin (c+d x) \sqrt{\sec (c+d x)} (5 a B+11 A b-3 b C) (a+b \sec (c+d x))^2}{15 d}+\frac{2 (5 a B+8 A b) \sin (c+d x) (a+b \sec (c+d x))^3}{15 d \sqrt{\sec (c+d x)}}+\frac{2 A \sin (c+d x) (a+b \sec (c+d x))^4}{5 d \sec ^{\frac{3}{2}}(c+d x)} \]

[Out]

(2*(20*a^3*b*B - 20*a*b^3*B + 30*a^2*b^2*(A - C) - b^4*(5*A + 3*C) + a^4*(3*A + 5*C))*Sqrt[Cos[c + d*x]]*Ellip
ticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (2*(a^4*B + 18*a^2*b^2*B + b^4*B + 4*a*b^3*(3*A + C) + 4*a^3*
b*(A + 3*C))*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*d) - (2*b*(10*a^3*B - 60*a*b^
2*B + a^2*b*(31*A - 87*C) - 3*b^3*(5*A + 3*C))*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(15*d) - (2*b^2*(5*a^2*B - 5*b
^2*B + 14*a*b*(A - C))*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(15*d) - (2*b*(11*A*b + 5*a*B - 3*b*C)*Sqrt[Sec[c + d*
x]]*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(15*d) + (2*(8*A*b + 5*a*B)*(a + b*Sec[c + d*x])^3*Sin[c + d*x])/(15*
d*Sqrt[Sec[c + d*x]]) + (2*A*(a + b*Sec[c + d*x])^4*Sin[c + d*x])/(5*d*Sec[c + d*x]^(3/2))

________________________________________________________________________________________

Rubi [A]  time = 1.24684, antiderivative size = 409, 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, 4096, 4076, 4047, 3771, 2641, 4046, 2639} \[ -\frac{2 b^2 \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x) \left (5 a^2 B+14 a b (A-C)-5 b^2 B\right )}{15 d}-\frac{2 b \sin (c+d x) \sqrt{\sec (c+d x)} \left (a^2 b (31 A-87 C)+10 a^3 B-60 a b^2 B-3 b^3 (5 A+3 C)\right )}{15 d}+\frac{2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (4 a^3 b (A+3 C)+18 a^2 b^2 B+a^4 B+4 a b^3 (3 A+C)+b^4 B\right )}{3 d}+\frac{2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (30 a^2 b^2 (A-C)+a^4 (3 A+5 C)+20 a^3 b B-20 a b^3 B-b^4 (5 A+3 C)\right )}{5 d}-\frac{2 b \sin (c+d x) \sqrt{\sec (c+d x)} (5 a B+11 A b-3 b C) (a+b \sec (c+d x))^2}{15 d}+\frac{2 (5 a B+8 A b) \sin (c+d x) (a+b \sec (c+d x))^3}{15 d \sqrt{\sec (c+d x)}}+\frac{2 A \sin (c+d x) (a+b \sec (c+d x))^4}{5 d \sec ^{\frac{3}{2}}(c+d x)} \]

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]^(5/2),x]

[Out]

(2*(20*a^3*b*B - 20*a*b^3*B + 30*a^2*b^2*(A - C) - b^4*(5*A + 3*C) + a^4*(3*A + 5*C))*Sqrt[Cos[c + d*x]]*Ellip
ticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (2*(a^4*B + 18*a^2*b^2*B + b^4*B + 4*a*b^3*(3*A + C) + 4*a^3*
b*(A + 3*C))*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*d) - (2*b*(10*a^3*B - 60*a*b^
2*B + a^2*b*(31*A - 87*C) - 3*b^3*(5*A + 3*C))*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(15*d) - (2*b^2*(5*a^2*B - 5*b
^2*B + 14*a*b*(A - C))*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(15*d) - (2*b*(11*A*b + 5*a*B - 3*b*C)*Sqrt[Sec[c + d*
x]]*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(15*d) + (2*(8*A*b + 5*a*B)*(a + b*Sec[c + d*x])^3*Sin[c + d*x])/(15*
d*Sqrt[Sec[c + d*x]]) + (2*A*(a + b*Sec[c + d*x])^4*Sin[c + d*x])/(5*d*Sec[c + d*x]^(3/2))

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 4096

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[(C*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d
*Csc[e + f*x])^n)/(f*(m + n + 1)), x] + Dist[1/(m + n + 1), Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^
n*Simp[a*A*(m + n + 1) + a*C*n + ((A*b + a*B)*(m + n + 1) + b*C*(m + n))*Csc[e + f*x] + (b*B*(m + n + 1) + a*C
*m)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, n}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 0] &&
!LeQ[n, -1]

Rule 4076

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[(b*C*Csc[e + f*x]*Cot[e + f*x]*(d*Csc[e + f*x
])^n)/(f*(n + 2)), x] + Dist[1/(n + 2), Int[(d*Csc[e + f*x])^n*Simp[A*a*(n + 2) + (B*a*(n + 2) + b*(C*(n + 1)
+ A*(n + 2)))*Csc[e + f*x] + (a*C + B*b)*(n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C,
n}, 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 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 4046

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> -Simp[(C*Cot[
e + f*x]*(b*Csc[e + f*x])^m)/(f*(m + 1)), x] + Dist[(C*m + A*(m + 1))/(m + 1), Int[(b*Csc[e + f*x])^m, x], x]
/; FreeQ[{b, e, f, A, C, m}, 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))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac{5}{2}}(c+d x)} \, dx &=\frac{2 A (a+b \sec (c+d x))^4 \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2}{5} \int \frac{(a+b \sec (c+d x))^3 \left (\frac{1}{2} (8 A b+5 a B)+\frac{1}{2} (3 a A+5 b B+5 a C) \sec (c+d x)-\frac{5}{2} b (A-C) \sec ^2(c+d x)\right )}{\sec ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{2 (8 A b+5 a B) (a+b \sec (c+d x))^3 \sin (c+d x)}{15 d \sqrt{\sec (c+d x)}}+\frac{2 A (a+b \sec (c+d x))^4 \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{4}{15} \int \frac{(a+b \sec (c+d x))^2 \left (\frac{3}{4} \left (16 A b^2+15 a b B+a^2 (3 A+5 C)\right )+\frac{1}{4} \left (5 a^2 B+15 b^2 B+2 a b (A+15 C)\right ) \sec (c+d x)-\frac{5}{4} b (11 A b+5 a B-3 b C) \sec ^2(c+d x)\right )}{\sqrt{\sec (c+d x)}} \, dx\\ &=-\frac{2 b (11 A b+5 a B-3 b C) \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^2 \sin (c+d x)}{15 d}+\frac{2 (8 A b+5 a B) (a+b \sec (c+d x))^3 \sin (c+d x)}{15 d \sqrt{\sec (c+d x)}}+\frac{2 A (a+b \sec (c+d x))^4 \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{8}{75} \int \frac{(a+b \sec (c+d x)) \left (\frac{5}{8} a \left (50 a b B+b^2 (59 A-3 C)+3 a^2 (3 A+5 C)\right )+\frac{5}{8} \left (5 a^3 B+45 a b^2 B+3 b^3 (5 A+3 C)+a^2 b (11 A+45 C)\right ) \sec (c+d x)-\frac{15}{8} b \left (5 a^2 B-5 b^2 B+14 a b (A-C)\right ) \sec ^2(c+d x)\right )}{\sqrt{\sec (c+d x)}} \, dx\\ &=-\frac{2 b^2 \left (5 a^2 B-5 b^2 B+14 a b (A-C)\right ) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{15 d}-\frac{2 b (11 A b+5 a B-3 b C) \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^2 \sin (c+d x)}{15 d}+\frac{2 (8 A b+5 a B) (a+b \sec (c+d x))^3 \sin (c+d x)}{15 d \sqrt{\sec (c+d x)}}+\frac{2 A (a+b \sec (c+d x))^4 \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{16}{225} \int \frac{\frac{15}{16} a^2 \left (50 a b B+b^2 (59 A-3 C)+3 a^2 (3 A+5 C)\right )+\frac{75}{16} \left (a^4 B+18 a^2 b^2 B+b^4 B+4 a b^3 (3 A+C)+4 a^3 b (A+3 C)\right ) \sec (c+d x)-\frac{15}{16} b \left (10 a^3 B-60 a b^2 B-3 b^3 (5 A+3 C)+a^2 (31 A b-87 b C)\right ) \sec ^2(c+d x)}{\sqrt{\sec (c+d x)}} \, dx\\ &=-\frac{2 b^2 \left (5 a^2 B-5 b^2 B+14 a b (A-C)\right ) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{15 d}-\frac{2 b (11 A b+5 a B-3 b C) \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^2 \sin (c+d x)}{15 d}+\frac{2 (8 A b+5 a B) (a+b \sec (c+d x))^3 \sin (c+d x)}{15 d \sqrt{\sec (c+d x)}}+\frac{2 A (a+b \sec (c+d x))^4 \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{16}{225} \int \frac{\frac{15}{16} a^2 \left (50 a b B+b^2 (59 A-3 C)+3 a^2 (3 A+5 C)\right )-\frac{15}{16} b \left (10 a^3 B-60 a b^2 B-3 b^3 (5 A+3 C)+a^2 (31 A b-87 b C)\right ) \sec ^2(c+d x)}{\sqrt{\sec (c+d x)}} \, dx+\frac{1}{3} \left (a^4 B+18 a^2 b^2 B+b^4 B+4 a b^3 (3 A+C)+4 a^3 b (A+3 C)\right ) \int \sqrt{\sec (c+d x)} \, dx\\ &=-\frac{2 b \left (10 a^3 B-60 a b^2 B+a^2 b (31 A-87 C)-3 b^3 (5 A+3 C)\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{15 d}-\frac{2 b^2 \left (5 a^2 B-5 b^2 B+14 a b (A-C)\right ) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{15 d}-\frac{2 b (11 A b+5 a B-3 b C) \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^2 \sin (c+d x)}{15 d}+\frac{2 (8 A b+5 a B) (a+b \sec (c+d x))^3 \sin (c+d x)}{15 d \sqrt{\sec (c+d x)}}+\frac{2 A (a+b \sec (c+d x))^4 \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{1}{5} \left (20 a^3 b B-20 a b^3 B+30 a^2 b^2 (A-C)-b^4 (5 A+3 C)+a^4 (3 A+5 C)\right ) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx+\frac{1}{3} \left (\left (a^4 B+18 a^2 b^2 B+b^4 B+4 a b^3 (3 A+C)+4 a^3 b (A+3 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 (a^4 B+18 a^2 b^2 B+b^4 B+4 a b^3 (3 A+C)+4 a^3 b (A+3 C)\right ) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{3 d}-\frac{2 b \left (10 a^3 B-60 a b^2 B+a^2 b (31 A-87 C)-3 b^3 (5 A+3 C)\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{15 d}-\frac{2 b^2 \left (5 a^2 B-5 b^2 B+14 a b (A-C)\right ) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{15 d}-\frac{2 b (11 A b+5 a B-3 b C) \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^2 \sin (c+d x)}{15 d}+\frac{2 (8 A b+5 a B) (a+b \sec (c+d x))^3 \sin (c+d x)}{15 d \sqrt{\sec (c+d x)}}+\frac{2 A (a+b \sec (c+d x))^4 \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{1}{5} \left (\left (20 a^3 b B-20 a b^3 B+30 a^2 b^2 (A-C)-b^4 (5 A+3 C)+a^4 (3 A+5 C)\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx\\ &=\frac{2 \left (20 a^3 b B-20 a b^3 B+30 a^2 b^2 (A-C)-b^4 (5 A+3 C)+a^4 (3 A+5 C)\right ) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{5 d}+\frac{2 \left (a^4 B+18 a^2 b^2 B+b^4 B+4 a b^3 (3 A+C)+4 a^3 b (A+3 C)\right ) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{3 d}-\frac{2 b \left (10 a^3 B-60 a b^2 B+a^2 b (31 A-87 C)-3 b^3 (5 A+3 C)\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{15 d}-\frac{2 b^2 \left (5 a^2 B-5 b^2 B+14 a b (A-C)\right ) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{15 d}-\frac{2 b (11 A b+5 a B-3 b C) \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^2 \sin (c+d x)}{15 d}+\frac{2 (8 A b+5 a B) (a+b \sec (c+d x))^3 \sin (c+d x)}{15 d \sqrt{\sec (c+d x)}}+\frac{2 A (a+b \sec (c+d x))^4 \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}\\ \end{align*}

Mathematica [A]  time = 7.36504, size = 485, normalized size = 1.19 \[ \frac{2 \cos ^6(c+d x) (a+b \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right ) \left (20 a^3 A b+90 a^2 b^2 B+60 a^3 b C+5 a^4 B+60 a A b^3+20 a b^3 C+5 b^4 B\right )+\frac{2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (90 a^2 A b^2+9 a^4 A-90 a^2 b^2 C+60 a^3 b B+15 a^4 C-60 a b^3 B-15 A b^4-9 b^4 C\right )}{\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}}\right )}{15 d (a \cos (c+d x)+b)^4 (A \cos (2 c+2 d x)+A+2 B \cos (c+d x)+2 C)}+\frac{(a+b \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (\frac{1}{5} \sin (c+d x) \left (a^4 A+120 a^2 b^2 C+80 a b^3 B+20 A b^4+12 b^4 C\right )+\frac{2}{3} a^3 (a B+4 A b) \sin (2 (c+d x))+\frac{1}{5} a^4 A \sin (3 (c+d x))+\frac{4}{3} \sec (c+d x) \left (4 a b^3 C \sin (c+d x)+b^4 B \sin (c+d x)\right )+\frac{4}{5} b^4 C \tan (c+d x) \sec (c+d x)\right )}{d \sec ^{\frac{11}{2}}(c+d x) (a \cos (c+d x)+b)^4 (A \cos (2 c+2 d x)+A+2 B \cos (c+d x)+2 C)} \]

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]^(5/2),x]

[Out]

(2*Cos[c + d*x]^6*((2*(9*a^4*A + 90*a^2*A*b^2 - 15*A*b^4 + 60*a^3*b*B - 60*a*b^3*B + 15*a^4*C - 90*a^2*b^2*C -
9*b^4*C)*EllipticE[(c + d*x)/2, 2])/(Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) + 2*(20*a^3*A*b + 60*a*A*b^3 + 5*
a^4*B + 90*a^2*b^2*B + 5*b^4*B + 60*a^3*b*C + 20*a*b^3*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Se
c[c + d*x]])*(a + b*Sec[c + d*x])^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(15*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^4*A + 20*A*b^4 + 80*a*b^3*B + 120*a^2*b^2*C + 12*b^4*C)*Sin[c + d*x])/5 + (4*Sec[c + d*x]*(b^4*B*Sin[c
+ d*x] + 4*a*b^3*C*Sin[c + d*x]))/3 + (2*a^3*(4*A*b + a*B)*Sin[2*(c + d*x)])/3 + (a^4*A*Sin[3*(c + d*x)])/5 +
(4*b^4*C*Sec[c + d*x]*Tan[c + d*x])/5))/(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 = 10.602, size = 1884, normalized size = 4.6 \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)^(5/2),x)

[Out]

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(4/5*A*a^4*(-4*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2
*c)+14*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+5*(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))-9*(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)
)*(sin(1/2*d*x+1/2*c)^2)^(1/2)-6*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c))/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x
+1/2*c)^2)^(1/2)+1/3*(-12*A*a^4+16*A*a^3*b+4*B*a^4)*(2*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+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))-3*(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))*(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*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)+(6*A*a^4-16*A*a^3*b+12*A*a^2*b^2-4*B*a^4+8*B*a^
3*b+2*C*a^4)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d
*x+1/2*c)^2)^(1/2)*(EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))-2*A*a^4*(sin(
1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)
*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+8*A*a^3*b*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2
)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-12*A*a^2*b^2*(sin
(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2
)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+8*A*a*b^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/
2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+2*B*a^4*(sin(1/2
*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*El
lipticF(cos(1/2*d*x+1/2*c),2^(1/2))-8*B*a^3*b*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(
-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+12*a^2*b^2*B*(sin(1/
2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*E
llipticF(cos(1/2*d*x+1/2*c),2^(1/2))-2*a^4*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-
2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+8*a^3*b*C*(sin(1/2*d*
x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*Ellip
ticF(cos(1/2*d*x+1/2*c),2^(1/2))-2/5*C*b^4/(8*sin(1/2*d*x+1/2*c)^6-12*sin(1/2*d*x+1/2*c)^4+6*sin(1/2*d*x+1/2*c
)^2-1)/sin(1/2*d*x+1/2*c)^2*(12*(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))*(sin(1/
2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^4-24*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)-12*(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))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^2+24*sin(1/2
*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+3*(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))*(sin
(1/2*d*x+1/2*c)^2)^(1/2)-8*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c))*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c
)^2)^(1/2)+2*b^3*(B*b+4*C*a)*(-1/6*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(co
s(1/2*d*x+1/2*c)^2-1/2)^2+1/3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1
/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)))+2*b^2*(A*b^2+4*B*a*b+6*C*a^2)*(-(si
n(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2
)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+2*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*cos(1/2*d*x+1/2
*c)*sin(1/2*d*x+1/2*c)^2)/sin(1/2*d*x+1/2*c)^2/(2*sin(1/2*d*x+1/2*c)^2-1))/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)^(5/2),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{5}{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)^(5/2),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)^(5/2), 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)**(5/2),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{5}{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)^(5/2),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)^(5/2), x)