3.1006 \(\int \frac{(a+b \sec (c+d x))^4 (A+B \sec (c+d x)+C \sec ^2(c+d x))}{\sec ^{\frac{3}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=419 \[ \frac{2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right ) \left (42 a^2 b^2 (3 A+C)+7 a^4 (A+3 C)+84 a^3 b B+28 a b^3 B+b^4 (7 A+5 C)\right )}{21 d}+\frac{2 b^2 \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x) \left (a^2 (-(35 A-87 C))+98 a b B+5 b^2 (7 A+5 C)\right )}{105 d}+\frac{2 b \sin (c+d x) \sqrt{\sec (c+d x)} \left (a^3 (-(70 A-366 C))+609 a^2 b B+84 a b^2 (5 A+3 C)+63 b^3 B\right )}{105 d}+\frac{2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (20 a^3 b (A-C)-30 a^2 b^2 B+5 a^4 B-4 a b^3 (5 A+3 C)-3 b^4 B\right )}{5 d}-\frac{2 b \sin (c+d x) \sqrt{\sec (c+d x)} (35 a A-39 a C-21 b B) (a+b \sec (c+d x))^2}{105 d}-\frac{2 b (7 A-3 C) \sin (c+d x) \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^3}{21 d}+\frac{2 A \sin (c+d x) (a+b \sec (c+d x))^4}{3 d \sqrt{\sec (c+d x)}} \]

[Out]

(2*(5*a^4*B - 30*a^2*b^2*B - 3*b^4*B + 20*a^3*b*(A - C) - 4*a*b^3*(5*A + 3*C))*Sqrt[Cos[c + d*x]]*EllipticE[(c
+ d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (2*(84*a^3*b*B + 28*a*b^3*B + 42*a^2*b^2*(3*A + C) + 7*a^4*(A + 3*C)
+ b^4*(7*A + 5*C))*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(21*d) + (2*b*(609*a^2*b*
B + 63*b^3*B - a^3*(70*A - 366*C) + 84*a*b^2*(5*A + 3*C))*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(105*d) + (2*b^2*(9
8*a*b*B - a^2*(35*A - 87*C) + 5*b^2*(7*A + 5*C))*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(105*d) - (2*b*(35*a*A - 21*
b*B - 39*a*C)*Sqrt[Sec[c + d*x]]*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(105*d) - (2*b*(7*A - 3*C)*Sqrt[Sec[c +
d*x]]*(a + b*Sec[c + d*x])^3*Sin[c + d*x])/(21*d) + (2*A*(a + b*Sec[c + d*x])^4*Sin[c + d*x])/(3*d*Sqrt[Sec[c
+ d*x]])

________________________________________________________________________________________

Rubi [A]  time = 1.22565, antiderivative size = 419, 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} \[ \frac{2 b^2 \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x) \left (a^2 (-(35 A-87 C))+98 a b B+5 b^2 (7 A+5 C)\right )}{105 d}+\frac{2 b \sin (c+d x) \sqrt{\sec (c+d x)} \left (a^3 (-(70 A-366 C))+609 a^2 b B+84 a b^2 (5 A+3 C)+63 b^3 B\right )}{105 d}+\frac{2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (42 a^2 b^2 (3 A+C)+7 a^4 (A+3 C)+84 a^3 b B+28 a b^3 B+b^4 (7 A+5 C)\right )}{21 d}+\frac{2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (20 a^3 b (A-C)-30 a^2 b^2 B+5 a^4 B-4 a b^3 (5 A+3 C)-3 b^4 B\right )}{5 d}-\frac{2 b \sin (c+d x) \sqrt{\sec (c+d x)} (35 a A-39 a C-21 b B) (a+b \sec (c+d x))^2}{105 d}-\frac{2 b (7 A-3 C) \sin (c+d x) \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^3}{21 d}+\frac{2 A \sin (c+d x) (a+b \sec (c+d x))^4}{3 d \sqrt{\sec (c+d x)}} \]

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]^(3/2),x]

[Out]

(2*(5*a^4*B - 30*a^2*b^2*B - 3*b^4*B + 20*a^3*b*(A - C) - 4*a*b^3*(5*A + 3*C))*Sqrt[Cos[c + d*x]]*EllipticE[(c
+ d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (2*(84*a^3*b*B + 28*a*b^3*B + 42*a^2*b^2*(3*A + C) + 7*a^4*(A + 3*C)
+ b^4*(7*A + 5*C))*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(21*d) + (2*b*(609*a^2*b*
B + 63*b^3*B - a^3*(70*A - 366*C) + 84*a*b^2*(5*A + 3*C))*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(105*d) + (2*b^2*(9
8*a*b*B - a^2*(35*A - 87*C) + 5*b^2*(7*A + 5*C))*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(105*d) - (2*b*(35*a*A - 21*
b*B - 39*a*C)*Sqrt[Sec[c + d*x]]*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(105*d) - (2*b*(7*A - 3*C)*Sqrt[Sec[c +
d*x]]*(a + b*Sec[c + d*x])^3*Sin[c + d*x])/(21*d) + (2*A*(a + b*Sec[c + d*x])^4*Sin[c + d*x])/(3*d*Sqrt[Sec[c
+ d*x]])

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{3}{2}}(c+d x)} \, dx &=\frac{2 A (a+b \sec (c+d x))^4 \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}}+\frac{2}{3} \int \frac{(a+b \sec (c+d x))^3 \left (\frac{1}{2} (8 A b+3 a B)+\frac{1}{2} (3 b B+a (A+3 C)) \sec (c+d x)-\frac{1}{2} b (7 A-3 C) \sec ^2(c+d x)\right )}{\sqrt{\sec (c+d x)}} \, dx\\ &=-\frac{2 b (7 A-3 C) \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^3 \sin (c+d x)}{21 d}+\frac{2 A (a+b \sec (c+d x))^4 \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}}+\frac{4}{21} \int \frac{(a+b \sec (c+d x))^2 \left (\frac{3}{4} a (21 A b+7 a B-b C)+\frac{1}{4} \left (42 a b B+7 a^2 (A+3 C)+3 b^2 (7 A+5 C)\right ) \sec (c+d x)-\frac{1}{4} b (35 a A-21 b B-39 a C) \sec ^2(c+d x)\right )}{\sqrt{\sec (c+d x)}} \, dx\\ &=-\frac{2 b (35 a A-21 b B-39 a C) \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^2 \sin (c+d x)}{105 d}-\frac{2 b (7 A-3 C) \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^3 \sin (c+d x)}{21 d}+\frac{2 A (a+b \sec (c+d x))^4 \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}}+\frac{8}{105} \int \frac{(a+b \sec (c+d x)) \left (\frac{1}{8} a \left (350 a A b+105 a^2 B-21 b^2 B-54 a b C\right )+\frac{1}{8} \left (315 a^2 b B+63 b^3 B+35 a^3 (A+3 C)+3 a b^2 (105 A+59 C)\right ) \sec (c+d x)+\frac{3}{8} b \left (98 a b B-a^2 (35 A-87 C)+5 b^2 (7 A+5 C)\right ) \sec ^2(c+d x)\right )}{\sqrt{\sec (c+d x)}} \, dx\\ &=\frac{2 b^2 \left (98 a b B-a^2 (35 A-87 C)+5 b^2 (7 A+5 C)\right ) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{105 d}-\frac{2 b (35 a A-21 b B-39 a C) \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^2 \sin (c+d x)}{105 d}-\frac{2 b (7 A-3 C) \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^3 \sin (c+d x)}{21 d}+\frac{2 A (a+b \sec (c+d x))^4 \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}}+\frac{16}{315} \int \frac{\frac{3}{16} a^2 \left (350 a A b+105 a^2 B-21 b^2 B-54 a b C\right )+\frac{15}{16} \left (84 a^3 b B+28 a b^3 B+42 a^2 b^2 (3 A+C)+7 a^4 (A+3 C)+b^4 (7 A+5 C)\right ) \sec (c+d x)+\frac{3}{16} b \left (609 a^2 b B+63 b^3 B-a^3 (70 A-366 C)+84 a b^2 (5 A+3 C)\right ) \sec ^2(c+d x)}{\sqrt{\sec (c+d x)}} \, dx\\ &=\frac{2 b^2 \left (98 a b B-a^2 (35 A-87 C)+5 b^2 (7 A+5 C)\right ) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{105 d}-\frac{2 b (35 a A-21 b B-39 a C) \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^2 \sin (c+d x)}{105 d}-\frac{2 b (7 A-3 C) \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^3 \sin (c+d x)}{21 d}+\frac{2 A (a+b \sec (c+d x))^4 \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}}+\frac{16}{315} \int \frac{\frac{3}{16} a^2 \left (350 a A b+105 a^2 B-21 b^2 B-54 a b C\right )+\frac{3}{16} b \left (609 a^2 b B+63 b^3 B-a^3 (70 A-366 C)+84 a b^2 (5 A+3 C)\right ) \sec ^2(c+d x)}{\sqrt{\sec (c+d x)}} \, dx+\frac{1}{21} \left (84 a^3 b B+28 a b^3 B+42 a^2 b^2 (3 A+C)+7 a^4 (A+3 C)+b^4 (7 A+5 C)\right ) \int \sqrt{\sec (c+d x)} \, dx\\ &=\frac{2 b \left (609 a^2 b B+63 b^3 B-a^3 (70 A-366 C)+84 a b^2 (5 A+3 C)\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{105 d}+\frac{2 b^2 \left (98 a b B-a^2 (35 A-87 C)+5 b^2 (7 A+5 C)\right ) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{105 d}-\frac{2 b (35 a A-21 b B-39 a C) \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^2 \sin (c+d x)}{105 d}-\frac{2 b (7 A-3 C) \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^3 \sin (c+d x)}{21 d}+\frac{2 A (a+b \sec (c+d x))^4 \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}}+\frac{1}{5} \left (5 a^4 B-30 a^2 b^2 B-3 b^4 B+20 a^3 b (A-C)-4 a b^3 (5 A+3 C)\right ) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx+\frac{1}{21} \left (\left (84 a^3 b B+28 a b^3 B+42 a^2 b^2 (3 A+C)+7 a^4 (A+3 C)+b^4 (7 A+5 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 (84 a^3 b B+28 a b^3 B+42 a^2 b^2 (3 A+C)+7 a^4 (A+3 C)+b^4 (7 A+5 C)\right ) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{21 d}+\frac{2 b \left (609 a^2 b B+63 b^3 B-a^3 (70 A-366 C)+84 a b^2 (5 A+3 C)\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{105 d}+\frac{2 b^2 \left (98 a b B-a^2 (35 A-87 C)+5 b^2 (7 A+5 C)\right ) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{105 d}-\frac{2 b (35 a A-21 b B-39 a C) \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^2 \sin (c+d x)}{105 d}-\frac{2 b (7 A-3 C) \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^3 \sin (c+d x)}{21 d}+\frac{2 A (a+b \sec (c+d x))^4 \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}}+\frac{1}{5} \left (\left (5 a^4 B-30 a^2 b^2 B-3 b^4 B+20 a^3 b (A-C)-4 a b^3 (5 A+3 C)\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx\\ &=\frac{2 \left (5 a^4 B-30 a^2 b^2 B-3 b^4 B+20 a^3 b (A-C)-4 a b^3 (5 A+3 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 (84 a^3 b B+28 a b^3 B+42 a^2 b^2 (3 A+C)+7 a^4 (A+3 C)+b^4 (7 A+5 C)\right ) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{21 d}+\frac{2 b \left (609 a^2 b B+63 b^3 B-a^3 (70 A-366 C)+84 a b^2 (5 A+3 C)\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{105 d}+\frac{2 b^2 \left (98 a b B-a^2 (35 A-87 C)+5 b^2 (7 A+5 C)\right ) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{105 d}-\frac{2 b (35 a A-21 b B-39 a C) \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^2 \sin (c+d x)}{105 d}-\frac{2 b (7 A-3 C) \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^3 \sin (c+d x)}{21 d}+\frac{2 A (a+b \sec (c+d x))^4 \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 7.36756, size = 530, normalized size = 1.26 \[ \frac{2 \cos ^6(c+d x) (a+b \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right ) \left (630 a^2 A b^2+35 a^4 A+210 a^2 b^2 C+420 a^3 b B+105 a^4 C+140 a b^3 B+35 A b^4+25 b^4 C\right )+\frac{2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (420 a^3 A b-630 a^2 b^2 B-420 a^3 b C+105 a^4 B-420 a A b^3-252 a b^3 C-63 b^4 B\right )}{\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}}\right )}{105 d (a \cos (c+d x)+b)^4 (A \cos (2 c+2 d x)+A+2 B \cos (c+d x)+2 C)}+\frac{(a+b \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (\frac{4}{5} b \sin (c+d x) \left (30 a^2 b B+20 a^3 C+20 a A b^2+12 a b^2 C+3 b^3 B\right )+\frac{4}{21} \sec (c+d x) \left (42 a^2 b^2 C \sin (c+d x)+28 a b^3 B \sin (c+d x)+7 A b^4 \sin (c+d x)+5 b^4 C \sin (c+d x)\right )+\frac{2}{3} a^4 A \sin (2 (c+d x))+\frac{4}{5} \sec ^2(c+d x) \left (4 a b^3 C \sin (c+d x)+b^4 B \sin (c+d x)\right )+\frac{4}{7} b^4 C \tan (c+d x) \sec ^2(c+d x)\right )}{d \sec ^{\frac{11}{2}}(c+d x) (a \cos (c+d x)+b)^4 (A \cos (2 c+2 d x)+A+2 B \cos (c+d x)+2 C)} \]

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]^(3/2),x]

[Out]

(2*Cos[c + d*x]^6*((2*(420*a^3*A*b - 420*a*A*b^3 + 105*a^4*B - 630*a^2*b^2*B - 63*b^4*B - 420*a^3*b*C - 252*a*
b^3*C)*EllipticE[(c + d*x)/2, 2])/(Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) + 2*(35*a^4*A + 630*a^2*A*b^2 + 35*A
*b^4 + 420*a^3*b*B + 140*a*b^3*B + 105*a^4*C + 210*a^2*b^2*C + 25*b^4*C)*Sqrt[Cos[c + d*x]]*EllipticF[(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))/(105*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)*((4*b*(20*a*A*b^2 + 30*a^2*b*B + 3*b^3*B + 20*a^3*C + 12*a*b^2*C)*Sin[c + d*x])/5 + (4*Sec[
c + d*x]^2*(b^4*B*Sin[c + d*x] + 4*a*b^3*C*Sin[c + d*x]))/5 + (4*Sec[c + d*x]*(7*A*b^4*Sin[c + d*x] + 28*a*b^3
*B*Sin[c + d*x] + 42*a^2*b^2*C*Sin[c + d*x] + 5*b^4*C*Sin[c + d*x]))/21 + (2*a^4*A*Sin[2*(c + d*x)])/3 + (4*b^
4*C*Sec[c + d*x]^2*Tan[c + d*x])/7))/(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 = 10.528, size = 1624, normalized size = 3.9 \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)^(3/2),x)

[Out]

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(4/3*A*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)+(-4*A*a^4+8*A*a^3*b+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)*(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(co
s(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(c
os(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)*EllipticF(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))+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))+2*C*b^4*(-1/56*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)/(cos(1/2*d*x+1/2*c)^2-1/2)^4-5/42*cos(1/2*d*x+1/2*c)*(-2*s
in(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)^2-1/2)^2+5/21*(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)))-2/5*b^3*(B*b+4*C*a)/(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^2*(A*b^2+4*B*a*b+6*C*a^2)*(-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)/(cos(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)))+4*a*b*(2*A*b^2+3*B*a*b+2*C*a^
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)*(-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)^(3/2),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{3}{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)^(3/2),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)^(3/2), 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)**(3/2),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{3}{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)^(3/2),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)^(3/2), x)