3.1008 \(\int \frac{(a+b \sec (c+d x))^4 (A+B \sec (c+d x)+C \sec ^2(c+d x))}{\sec ^{\frac{7}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=429 \[ \frac{2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right ) \left (42 a^2 b^2 (A+3 C)+a^4 (5 A+7 C)+28 a^3 b B+84 a b^3 B+7 b^4 (3 A+C)\right )}{21 d}-\frac{2 b^2 \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x) \left (5 a^2 (5 A+7 C)+98 a b B+b^2 (87 A-35 C)\right )}{105 d}+\frac{2 \sin (c+d x) \left (5 a^2 (5 A+7 C)+77 a b B+48 A b^2\right ) (a+b \sec (c+d x))^2}{105 d \sqrt{\sec (c+d x)}}-\frac{2 b \sin (c+d x) \sqrt{\sec (c+d x)} \left (10 a^3 (5 A+7 C)+217 a^2 b B+12 a b^2 (19 A-35 C)-105 b^3 B\right )}{105 d}+\frac{2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (4 a^3 b (3 A+5 C)+30 a^2 b^2 B+3 a^4 B+20 a b^3 (A-C)-5 b^4 B\right )}{5 d}+\frac{2 (7 a B+8 A b) \sin (c+d x) (a+b \sec (c+d x))^3}{35 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 A \sin (c+d x) (a+b \sec (c+d x))^4}{7 d \sec ^{\frac{5}{2}}(c+d x)} \]

[Out]

(2*(3*a^4*B + 30*a^2*b^2*B - 5*b^4*B + 20*a*b^3*(A - C) + 4*a^3*b*(3*A + 5*C))*Sqrt[Cos[c + d*x]]*EllipticE[(c
+ d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (2*(28*a^3*b*B + 84*a*b^3*B + 7*b^4*(3*A + C) + 42*a^2*b^2*(A + 3*C)
+ a^4*(5*A + 7*C))*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(21*d) - (2*b*(217*a^2*b*
B - 105*b^3*B + 12*a*b^2*(19*A - 35*C) + 10*a^3*(5*A + 7*C))*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(105*d) - (2*b^2
*(98*a*b*B + b^2*(87*A - 35*C) + 5*a^2*(5*A + 7*C))*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(105*d) + (2*(48*A*b^2 +
77*a*b*B + 5*a^2*(5*A + 7*C))*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(105*d*Sqrt[Sec[c + d*x]]) + (2*(8*A*b + 7*
a*B)*(a + b*Sec[c + d*x])^3*Sin[c + d*x])/(35*d*Sec[c + d*x]^(3/2)) + (2*A*(a + b*Sec[c + d*x])^4*Sin[c + d*x]
)/(7*d*Sec[c + d*x]^(5/2))

________________________________________________________________________________________

Rubi [A]  time = 1.30936, antiderivative size = 429, 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, 4076, 4047, 3771, 2641, 4046, 2639} \[ -\frac{2 b^2 \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x) \left (5 a^2 (5 A+7 C)+98 a b B+b^2 (87 A-35 C)\right )}{105 d}+\frac{2 \sin (c+d x) \left (5 a^2 (5 A+7 C)+77 a b B+48 A b^2\right ) (a+b \sec (c+d x))^2}{105 d \sqrt{\sec (c+d x)}}-\frac{2 b \sin (c+d x) \sqrt{\sec (c+d x)} \left (10 a^3 (5 A+7 C)+217 a^2 b B+12 a b^2 (19 A-35 C)-105 b^3 B\right )}{105 d}+\frac{2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (42 a^2 b^2 (A+3 C)+a^4 (5 A+7 C)+28 a^3 b B+84 a b^3 B+7 b^4 (3 A+C)\right )}{21 d}+\frac{2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (4 a^3 b (3 A+5 C)+30 a^2 b^2 B+3 a^4 B+20 a b^3 (A-C)-5 b^4 B\right )}{5 d}+\frac{2 (7 a B+8 A b) \sin (c+d x) (a+b \sec (c+d x))^3}{35 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 A \sin (c+d x) (a+b \sec (c+d x))^4}{7 d \sec ^{\frac{5}{2}}(c+d x)} \]

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]^(7/2),x]

[Out]

(2*(3*a^4*B + 30*a^2*b^2*B - 5*b^4*B + 20*a*b^3*(A - C) + 4*a^3*b*(3*A + 5*C))*Sqrt[Cos[c + d*x]]*EllipticE[(c
+ d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (2*(28*a^3*b*B + 84*a*b^3*B + 7*b^4*(3*A + C) + 42*a^2*b^2*(A + 3*C)
+ a^4*(5*A + 7*C))*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(21*d) - (2*b*(217*a^2*b*
B - 105*b^3*B + 12*a*b^2*(19*A - 35*C) + 10*a^3*(5*A + 7*C))*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(105*d) - (2*b^2
*(98*a*b*B + b^2*(87*A - 35*C) + 5*a^2*(5*A + 7*C))*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(105*d) + (2*(48*A*b^2 +
77*a*b*B + 5*a^2*(5*A + 7*C))*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(105*d*Sqrt[Sec[c + d*x]]) + (2*(8*A*b + 7*
a*B)*(a + b*Sec[c + d*x])^3*Sin[c + d*x])/(35*d*Sec[c + d*x]^(3/2)) + (2*A*(a + b*Sec[c + d*x])^4*Sin[c + d*x]
)/(7*d*Sec[c + d*x]^(5/2))

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 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{7}{2}}(c+d x)} \, dx &=\frac{2 A (a+b \sec (c+d x))^4 \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2}{7} \int \frac{(a+b \sec (c+d x))^3 \left (\frac{1}{2} (8 A b+7 a B)+\frac{1}{2} (5 a A+7 b B+7 a C) \sec (c+d x)-\frac{1}{2} b (3 A-7 C) \sec ^2(c+d x)\right )}{\sec ^{\frac{5}{2}}(c+d x)} \, dx\\ &=\frac{2 (8 A b+7 a B) (a+b \sec (c+d x))^3 \sin (c+d x)}{35 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 A (a+b \sec (c+d x))^4 \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{4}{35} \int \frac{(a+b \sec (c+d x))^2 \left (\frac{1}{4} \left (48 A b^2+77 a b B+5 a^2 (5 A+7 C)\right )+\frac{1}{4} \left (34 a A b+21 a^2 B+35 b^2 B+70 a b C\right ) \sec (c+d x)-\frac{1}{4} b (39 A b+21 a B-35 b C) \sec ^2(c+d x)\right )}{\sec ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{2 \left (48 A b^2+77 a b B+5 a^2 (5 A+7 C)\right ) (a+b \sec (c+d x))^2 \sin (c+d x)}{105 d \sqrt{\sec (c+d x)}}+\frac{2 (8 A b+7 a B) (a+b \sec (c+d x))^3 \sin (c+d x)}{35 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 A (a+b \sec (c+d x))^4 \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{8}{105} \int \frac{(a+b \sec (c+d x)) \left (\frac{1}{8} \left (192 A b^3+63 a^3 B+413 a b^2 B+a^2 (202 A b+350 b C)\right )+\frac{1}{8} \left (77 a^2 b B+105 b^3 B+5 a^3 (5 A+7 C)+3 a b^2 (11 A+105 C)\right ) \sec (c+d x)-\frac{3}{8} b \left (98 a b B+b^2 (87 A-35 C)+5 a^2 (5 A+7 C)\right ) \sec ^2(c+d x)\right )}{\sqrt{\sec (c+d x)}} \, dx\\ &=-\frac{2 b^2 \left (98 a b B+b^2 (87 A-35 C)+5 a^2 (5 A+7 C)\right ) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{105 d}+\frac{2 \left (48 A b^2+77 a b B+5 a^2 (5 A+7 C)\right ) (a+b \sec (c+d x))^2 \sin (c+d x)}{105 d \sqrt{\sec (c+d x)}}+\frac{2 (8 A b+7 a B) (a+b \sec (c+d x))^3 \sin (c+d x)}{35 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 A (a+b \sec (c+d x))^4 \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{16}{315} \int \frac{\frac{3}{16} a \left (192 A b^3+63 a^3 B+413 a b^2 B+a^2 (202 A b+350 b C)\right )+\frac{15}{16} \left (28 a^3 b B+84 a b^3 B+7 b^4 (3 A+C)+42 a^2 b^2 (A+3 C)+a^4 (5 A+7 C)\right ) \sec (c+d x)-\frac{3}{16} b \left (217 a^2 b B-105 b^3 B+12 a b^2 (19 A-35 C)+10 a^3 (5 A+7 C)\right ) \sec ^2(c+d x)}{\sqrt{\sec (c+d x)}} \, dx\\ &=-\frac{2 b^2 \left (98 a b B+b^2 (87 A-35 C)+5 a^2 (5 A+7 C)\right ) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{105 d}+\frac{2 \left (48 A b^2+77 a b B+5 a^2 (5 A+7 C)\right ) (a+b \sec (c+d x))^2 \sin (c+d x)}{105 d \sqrt{\sec (c+d x)}}+\frac{2 (8 A b+7 a B) (a+b \sec (c+d x))^3 \sin (c+d x)}{35 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 A (a+b \sec (c+d x))^4 \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{16}{315} \int \frac{\frac{3}{16} a \left (192 A b^3+63 a^3 B+413 a b^2 B+a^2 (202 A b+350 b C)\right )-\frac{3}{16} b \left (217 a^2 b B-105 b^3 B+12 a b^2 (19 A-35 C)+10 a^3 (5 A+7 C)\right ) \sec ^2(c+d x)}{\sqrt{\sec (c+d x)}} \, dx+\frac{1}{21} \left (28 a^3 b B+84 a b^3 B+7 b^4 (3 A+C)+42 a^2 b^2 (A+3 C)+a^4 (5 A+7 C)\right ) \int \sqrt{\sec (c+d x)} \, dx\\ &=-\frac{2 b \left (217 a^2 b B-105 b^3 B+12 a b^2 (19 A-35 C)+10 a^3 (5 A+7 C)\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{105 d}-\frac{2 b^2 \left (98 a b B+b^2 (87 A-35 C)+5 a^2 (5 A+7 C)\right ) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{105 d}+\frac{2 \left (48 A b^2+77 a b B+5 a^2 (5 A+7 C)\right ) (a+b \sec (c+d x))^2 \sin (c+d x)}{105 d \sqrt{\sec (c+d x)}}+\frac{2 (8 A b+7 a B) (a+b \sec (c+d x))^3 \sin (c+d x)}{35 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 A (a+b \sec (c+d x))^4 \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{1}{5} \left (3 a^4 B+30 a^2 b^2 B-5 b^4 B+20 a b^3 (A-C)+4 a^3 b (3 A+5 C)\right ) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx+\frac{1}{21} \left (\left (28 a^3 b B+84 a b^3 B+7 b^4 (3 A+C)+42 a^2 b^2 (A+3 C)+a^4 (5 A+7 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 (28 a^3 b B+84 a b^3 B+7 b^4 (3 A+C)+42 a^2 b^2 (A+3 C)+a^4 (5 A+7 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 (217 a^2 b B-105 b^3 B+12 a b^2 (19 A-35 C)+10 a^3 (5 A+7 C)\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{105 d}-\frac{2 b^2 \left (98 a b B+b^2 (87 A-35 C)+5 a^2 (5 A+7 C)\right ) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{105 d}+\frac{2 \left (48 A b^2+77 a b B+5 a^2 (5 A+7 C)\right ) (a+b \sec (c+d x))^2 \sin (c+d x)}{105 d \sqrt{\sec (c+d x)}}+\frac{2 (8 A b+7 a B) (a+b \sec (c+d x))^3 \sin (c+d x)}{35 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 A (a+b \sec (c+d x))^4 \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{1}{5} \left (\left (3 a^4 B+30 a^2 b^2 B-5 b^4 B+20 a b^3 (A-C)+4 a^3 b (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 (3 a^4 B+30 a^2 b^2 B-5 b^4 B+20 a b^3 (A-C)+4 a^3 b (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 (28 a^3 b B+84 a b^3 B+7 b^4 (3 A+C)+42 a^2 b^2 (A+3 C)+a^4 (5 A+7 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 (217 a^2 b B-105 b^3 B+12 a b^2 (19 A-35 C)+10 a^3 (5 A+7 C)\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{105 d}-\frac{2 b^2 \left (98 a b B+b^2 (87 A-35 C)+5 a^2 (5 A+7 C)\right ) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{105 d}+\frac{2 \left (48 A b^2+77 a b B+5 a^2 (5 A+7 C)\right ) (a+b \sec (c+d x))^2 \sin (c+d x)}{105 d \sqrt{\sec (c+d x)}}+\frac{2 (8 A b+7 a B) (a+b \sec (c+d x))^3 \sin (c+d x)}{35 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 A (a+b \sec (c+d x))^4 \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}\\ \end{align*}

Mathematica [A]  time = 6.66818, size = 394, normalized size = 0.92 \[ \frac{(a+b \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (40 \sqrt{\cos (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right ) \left (42 a^2 b^2 (A+3 C)+a^4 (5 A+7 C)+28 a^3 b B+84 a b^3 B+7 b^4 (3 A+C)\right )+168 \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (4 a^3 b (3 A+5 C)+30 a^2 b^2 B+3 a^4 B+20 a b^3 (A-C)-5 b^4 B\right )+840 a^2 A b^2 \sin (2 (c+d x))+168 a^3 A b \sin (c+d x)+168 a^3 A b \sin (3 (c+d x))+130 a^4 A \sin (2 (c+d x))+15 a^4 A \sin (4 (c+d x))+560 a^3 b B \sin (2 (c+d x))+42 a^4 B \sin (c+d x)+42 a^4 B \sin (3 (c+d x))+140 a^4 C \sin (2 (c+d x))+3360 a b^3 C \sin (c+d x)+840 b^4 B \sin (c+d x)+280 b^4 C \tan (c+d x)\right )}{210 d \sec ^{\frac{11}{2}}(c+d x) (a \cos (c+d x)+b)^4 (A \cos (2 (c+d x))+A+2 B \cos (c+d x)+2 C)} \]

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]^(7/2),x]

[Out]

((a + b*Sec[c + d*x])^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(168*(3*a^4*B + 30*a^2*b^2*B - 5*b^4*B + 20*a*
b^3*(A - C) + 4*a^3*b*(3*A + 5*C))*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2] + 40*(28*a^3*b*B + 84*a*b^3*B
+ 7*b^4*(3*A + C) + 42*a^2*b^2*(A + 3*C) + a^4*(5*A + 7*C))*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] + 168
*a^3*A*b*Sin[c + d*x] + 42*a^4*B*Sin[c + d*x] + 840*b^4*B*Sin[c + d*x] + 3360*a*b^3*C*Sin[c + d*x] + 130*a^4*A
*Sin[2*(c + d*x)] + 840*a^2*A*b^2*Sin[2*(c + d*x)] + 560*a^3*b*B*Sin[2*(c + d*x)] + 140*a^4*C*Sin[2*(c + d*x)]
+ 168*a^3*A*b*Sin[3*(c + d*x)] + 42*a^4*B*Sin[3*(c + d*x)] + 15*a^4*A*Sin[4*(c + d*x)] + 280*b^4*C*Tan[c + d*
x]))/(210*d*(b + a*Cos[c + d*x])^4*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*(c + d*x)])*Sec[c + d*x]^(11/2))

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Maple [B]  time = 9.3, size = 2507, normalized size = 5.8 \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)^(7/2),x)

[Out]

-2/105*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/(4*sin(1/2*d*x+1/2*c)^4-4*sin(1/2*d*x+1/2*c)^
2+1)/sin(1/2*d*x+1/2*c)^3*(-630*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^2*b^2-50*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(c
os(1/2*d*x+1/2*c),2^(1/2))*a^4*sin(1/2*d*x+1/2*c)^2-210*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*sin(1/2*d*x+1/2*c)^2+126*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*sin(1/2*d*x+1/2*c)^2-210*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))*b^4*sin(1/2*d*x+1/2*
c)^2-70*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))*
a^4*sin(1/2*d*x+1/2*c)^2-70*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*sin(1/2*d*x+1/2*c)^2-252*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-420*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+210*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^2-504*B*a^4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-280*C*a
^4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6+440*A*a^4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+252*B*a^4*cos(1/2
*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+420*B*b^4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+280*C*a^4*cos(1/2*d*x+1/2*c
)*sin(1/2*d*x+1/2*c)^4-80*A*a^4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2-42*B*a^4*cos(1/2*d*x+1/2*c)*sin(1/2*d*
x+1/2*c)^2-210*B*b^4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2-70*C*a^4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2-
70*C*b^4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2-480*A*a^4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^10+960*A*a^4*
cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8+336*B*a^4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8-920*A*a^4*cos(1/2*d*
x+1/2*c)*sin(1/2*d*x+1/2*c)^6+504*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(co
s(1/2*d*x+1/2*c),2^(1/2))*a^3*b*sin(1/2*d*x+1/2*c)^2+840*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*sin(1/2*d*x+1/2*c)^2-420*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*sin(1/2*d*x+1/2*c)^2+1260*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^2*b^2*sin(
1/2*d*x+1/2*c)^2-280*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*sin(1/2*d*x+1/2*c)^2-840*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
lipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a*b^3*sin(1/2*d*x+1/2*c)^2+840*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*sin(1/2*d*x+1/2*c)^2-840*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*sin(1/2*d*x+1/2*c)^2-1
260*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))*a^2*
b^2*sin(1/2*d*x+1/2*c)^2-2016*A*a^3*b*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-1680*A*a^2*b^2*cos(1/2*d*x+1/2*c
)*sin(1/2*d*x+1/2*c)^6-1120*B*a^3*b*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6+1008*A*a^3*b*cos(1/2*d*x+1/2*c)*si
n(1/2*d*x+1/2*c)^4+1680*A*a^2*b^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+1120*B*a^3*b*cos(1/2*d*x+1/2*c)*sin(
1/2*d*x+1/2*c)^4+1680*C*a*b^3*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4-168*A*a^3*b*cos(1/2*d*x+1/2*c)*sin(1/2*d
*x+1/2*c)^2-420*A*a^2*b^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2-280*B*a^3*b*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1
/2*c)^2-840*C*a*b^3*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2+25*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+105*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^4-63*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+105*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))*b^4+35*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))*a^4+35*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+1344*A*a^3*b*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8+1
40*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
+420*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*b
^3-420*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+420*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+630*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
))*a^2*b^2)*(-2*sin(1/2*d*x+1/2*c)^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)/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)^(7/2),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{7}{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)^(7/2),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)^(7/2), 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)**(7/2),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{7}{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)^(7/2),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)^(7/2), x)