3.1005 \(\int \frac{(a+b \sec (c+d x))^4 (A+B \sec (c+d x)+C \sec ^2(c+d x))}{\sqrt{\sec (c+d x)}} \, dx\)

Optimal. Leaf size=441 \[ \frac{2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right ) \left (28 a^3 b (3 A+C)+42 a^2 b^2 B+21 a^4 B+4 a b^3 (7 A+5 C)+5 b^4 B\right )}{21 d}+\frac{2 b \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x) \left (261 a^2 b B+64 a^3 C+2 a b^2 (147 A+101 C)+75 b^3 B\right )}{315 d}+\frac{2 \sin (c+d x) \sqrt{\sec (c+d x)} \left (48 a^2 C+117 a b B+63 A b^2+49 b^2 C\right ) (a+b \sec (c+d x))^2}{315 d}+\frac{2 \sin (c+d x) \sqrt{\sec (c+d x)} \left (7 a^2 b^2 (261 A+155 C)+1098 a^3 b B+192 a^4 C+756 a b^3 B+21 b^4 (9 A+7 C)\right )}{315 d}-\frac{2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (18 a^2 b^2 (5 A+3 C)-15 a^4 (A-C)+60 a^3 b B+36 a b^3 B+b^4 (9 A+7 C)\right )}{15 d}+\frac{2 (8 a C+9 b B) \sin (c+d x) \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^3}{63 d}+\frac{2 C \sin (c+d x) \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^4}{9 d} \]

[Out]

(-2*(60*a^3*b*B + 36*a*b^3*B - 15*a^4*(A - C) + 18*a^2*b^2*(5*A + 3*C) + b^4*(9*A + 7*C))*Sqrt[Cos[c + d*x]]*E
llipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(15*d) + (2*(21*a^4*B + 42*a^2*b^2*B + 5*b^4*B + 28*a^3*b*(3*A +
C) + 4*a*b^3*(7*A + 5*C))*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(21*d) + (2*(1098*a
^3*b*B + 756*a*b^3*B + 192*a^4*C + 21*b^4*(9*A + 7*C) + 7*a^2*b^2*(261*A + 155*C))*Sqrt[Sec[c + d*x]]*Sin[c +
d*x])/(315*d) + (2*b*(261*a^2*b*B + 75*b^3*B + 64*a^3*C + 2*a*b^2*(147*A + 101*C))*Sec[c + d*x]^(3/2)*Sin[c +
d*x])/(315*d) + (2*(63*A*b^2 + 117*a*b*B + 48*a^2*C + 49*b^2*C)*Sqrt[Sec[c + d*x]]*(a + b*Sec[c + d*x])^2*Sin[
c + d*x])/(315*d) + (2*(9*b*B + 8*a*C)*Sqrt[Sec[c + d*x]]*(a + b*Sec[c + d*x])^3*Sin[c + d*x])/(63*d) + (2*C*S
qrt[Sec[c + d*x]]*(a + b*Sec[c + d*x])^4*Sin[c + d*x])/(9*d)

________________________________________________________________________________________

Rubi [A]  time = 1.23921, antiderivative size = 441, 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 = {4096, 4076, 4047, 3771, 2641, 4046, 2639} \[ \frac{2 b \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x) \left (261 a^2 b B+64 a^3 C+2 a b^2 (147 A+101 C)+75 b^3 B\right )}{315 d}+\frac{2 \sin (c+d x) \sqrt{\sec (c+d x)} \left (48 a^2 C+117 a b B+63 A b^2+49 b^2 C\right ) (a+b \sec (c+d x))^2}{315 d}+\frac{2 \sin (c+d x) \sqrt{\sec (c+d x)} \left (7 a^2 b^2 (261 A+155 C)+1098 a^3 b B+192 a^4 C+756 a b^3 B+21 b^4 (9 A+7 C)\right )}{315 d}+\frac{2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (28 a^3 b (3 A+C)+42 a^2 b^2 B+21 a^4 B+4 a b^3 (7 A+5 C)+5 b^4 B\right )}{21 d}-\frac{2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (18 a^2 b^2 (5 A+3 C)-15 a^4 (A-C)+60 a^3 b B+36 a b^3 B+b^4 (9 A+7 C)\right )}{15 d}+\frac{2 (8 a C+9 b B) \sin (c+d x) \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^3}{63 d}+\frac{2 C \sin (c+d x) \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^4}{9 d} \]

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))/Sqrt[Sec[c + d*x]],x]

[Out]

(-2*(60*a^3*b*B + 36*a*b^3*B - 15*a^4*(A - C) + 18*a^2*b^2*(5*A + 3*C) + b^4*(9*A + 7*C))*Sqrt[Cos[c + d*x]]*E
llipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(15*d) + (2*(21*a^4*B + 42*a^2*b^2*B + 5*b^4*B + 28*a^3*b*(3*A +
C) + 4*a*b^3*(7*A + 5*C))*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(21*d) + (2*(1098*a
^3*b*B + 756*a*b^3*B + 192*a^4*C + 21*b^4*(9*A + 7*C) + 7*a^2*b^2*(261*A + 155*C))*Sqrt[Sec[c + d*x]]*Sin[c +
d*x])/(315*d) + (2*b*(261*a^2*b*B + 75*b^3*B + 64*a^3*C + 2*a*b^2*(147*A + 101*C))*Sec[c + d*x]^(3/2)*Sin[c +
d*x])/(315*d) + (2*(63*A*b^2 + 117*a*b*B + 48*a^2*C + 49*b^2*C)*Sqrt[Sec[c + d*x]]*(a + b*Sec[c + d*x])^2*Sin[
c + d*x])/(315*d) + (2*(9*b*B + 8*a*C)*Sqrt[Sec[c + d*x]]*(a + b*Sec[c + d*x])^3*Sin[c + d*x])/(63*d) + (2*C*S
qrt[Sec[c + d*x]]*(a + b*Sec[c + d*x])^4*Sin[c + d*x])/(9*d)

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 )}{\sqrt{\sec (c+d x)}} \, dx &=\frac{2 C \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^4 \sin (c+d x)}{9 d}+\frac{2}{9} \int \frac{(a+b \sec (c+d x))^3 \left (\frac{1}{2} a (9 A-C)+\frac{1}{2} (9 A b+9 a B+7 b C) \sec (c+d x)+\frac{1}{2} (9 b B+8 a C) \sec ^2(c+d x)\right )}{\sqrt{\sec (c+d x)}} \, dx\\ &=\frac{2 (9 b B+8 a C) \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^3 \sin (c+d x)}{63 d}+\frac{2 C \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^4 \sin (c+d x)}{9 d}+\frac{4}{63} \int \frac{(a+b \sec (c+d x))^2 \left (\frac{3}{4} a (21 a A-3 b B-5 a C)+\frac{1}{4} \left (126 a A b+63 a^2 B+45 b^2 B+82 a b C\right ) \sec (c+d x)+\frac{1}{4} \left (63 A b^2+117 a b B+48 a^2 C+49 b^2 C\right ) \sec ^2(c+d x)\right )}{\sqrt{\sec (c+d x)}} \, dx\\ &=\frac{2 \left (63 A b^2+117 a b B+48 a^2 C+49 b^2 C\right ) \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^2 \sin (c+d x)}{315 d}+\frac{2 (9 b B+8 a C) \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^3 \sin (c+d x)}{63 d}+\frac{2 C \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^4 \sin (c+d x)}{9 d}+\frac{8}{315} \int \frac{(a+b \sec (c+d x)) \left (-\frac{1}{8} a \left (162 a b B-3 a^2 (105 A-41 C)+7 b^2 (9 A+7 C)\right )+\frac{1}{8} \left (315 a^3 B+531 a b^2 B+21 b^3 (9 A+7 C)+a^2 b (945 A+479 C)\right ) \sec (c+d x)+\frac{3}{8} \left (261 a^2 b B+75 b^3 B+64 a^3 C+2 a b^2 (147 A+101 C)\right ) \sec ^2(c+d x)\right )}{\sqrt{\sec (c+d x)}} \, dx\\ &=\frac{2 b \left (261 a^2 b B+75 b^3 B+64 a^3 C+2 a b^2 (147 A+101 C)\right ) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{315 d}+\frac{2 \left (63 A b^2+117 a b B+48 a^2 C+49 b^2 C\right ) \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^2 \sin (c+d x)}{315 d}+\frac{2 (9 b B+8 a C) \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^3 \sin (c+d x)}{63 d}+\frac{2 C \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^4 \sin (c+d x)}{9 d}+\frac{16}{945} \int \frac{-\frac{3}{16} a^2 \left (162 a b B-a^2 (315 A-123 C)+7 b^2 (9 A+7 C)\right )+\frac{45}{16} \left (21 a^4 B+42 a^2 b^2 B+5 b^4 B+28 a^3 b (3 A+C)+4 a b^3 (7 A+5 C)\right ) \sec (c+d x)+\frac{3}{16} \left (1098 a^3 b B+756 a b^3 B+192 a^4 C+21 b^4 (9 A+7 C)+7 a^2 b^2 (261 A+155 C)\right ) \sec ^2(c+d x)}{\sqrt{\sec (c+d x)}} \, dx\\ &=\frac{2 b \left (261 a^2 b B+75 b^3 B+64 a^3 C+2 a b^2 (147 A+101 C)\right ) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{315 d}+\frac{2 \left (63 A b^2+117 a b B+48 a^2 C+49 b^2 C\right ) \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^2 \sin (c+d x)}{315 d}+\frac{2 (9 b B+8 a C) \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^3 \sin (c+d x)}{63 d}+\frac{2 C \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^4 \sin (c+d x)}{9 d}+\frac{16}{945} \int \frac{-\frac{3}{16} a^2 \left (162 a b B-a^2 (315 A-123 C)+7 b^2 (9 A+7 C)\right )+\frac{3}{16} \left (1098 a^3 b B+756 a b^3 B+192 a^4 C+21 b^4 (9 A+7 C)+7 a^2 b^2 (261 A+155 C)\right ) \sec ^2(c+d x)}{\sqrt{\sec (c+d x)}} \, dx+\frac{1}{21} \left (21 a^4 B+42 a^2 b^2 B+5 b^4 B+28 a^3 b (3 A+C)+4 a b^3 (7 A+5 C)\right ) \int \sqrt{\sec (c+d x)} \, dx\\ &=\frac{2 \left (1098 a^3 b B+756 a b^3 B+192 a^4 C+21 b^4 (9 A+7 C)+7 a^2 b^2 (261 A+155 C)\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{315 d}+\frac{2 b \left (261 a^2 b B+75 b^3 B+64 a^3 C+2 a b^2 (147 A+101 C)\right ) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{315 d}+\frac{2 \left (63 A b^2+117 a b B+48 a^2 C+49 b^2 C\right ) \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^2 \sin (c+d x)}{315 d}+\frac{2 (9 b B+8 a C) \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^3 \sin (c+d x)}{63 d}+\frac{2 C \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^4 \sin (c+d x)}{9 d}+\frac{1}{15} \left (-60 a^3 b B-36 a b^3 B+15 a^4 (A-C)-18 a^2 b^2 (5 A+3 C)-b^4 (9 A+7 C)\right ) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx+\frac{1}{21} \left (\left (21 a^4 B+42 a^2 b^2 B+5 b^4 B+28 a^3 b (3 A+C)+4 a b^3 (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 (21 a^4 B+42 a^2 b^2 B+5 b^4 B+28 a^3 b (3 A+C)+4 a b^3 (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 \left (1098 a^3 b B+756 a b^3 B+192 a^4 C+21 b^4 (9 A+7 C)+7 a^2 b^2 (261 A+155 C)\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{315 d}+\frac{2 b \left (261 a^2 b B+75 b^3 B+64 a^3 C+2 a b^2 (147 A+101 C)\right ) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{315 d}+\frac{2 \left (63 A b^2+117 a b B+48 a^2 C+49 b^2 C\right ) \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^2 \sin (c+d x)}{315 d}+\frac{2 (9 b B+8 a C) \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^3 \sin (c+d x)}{63 d}+\frac{2 C \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^4 \sin (c+d x)}{9 d}+\frac{1}{15} \left (\left (-60 a^3 b B-36 a b^3 B+15 a^4 (A-C)-18 a^2 b^2 (5 A+3 C)-b^4 (9 A+7 C)\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx\\ &=-\frac{2 \left (60 a^3 b B+36 a b^3 B-15 a^4 (A-C)+18 a^2 b^2 (5 A+3 C)+b^4 (9 A+7 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 (21 a^4 B+42 a^2 b^2 B+5 b^4 B+28 a^3 b (3 A+C)+4 a b^3 (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 \left (1098 a^3 b B+756 a b^3 B+192 a^4 C+21 b^4 (9 A+7 C)+7 a^2 b^2 (261 A+155 C)\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{315 d}+\frac{2 b \left (261 a^2 b B+75 b^3 B+64 a^3 C+2 a b^2 (147 A+101 C)\right ) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{315 d}+\frac{2 \left (63 A b^2+117 a b B+48 a^2 C+49 b^2 C\right ) \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^2 \sin (c+d x)}{315 d}+\frac{2 (9 b B+8 a C) \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^3 \sin (c+d x)}{63 d}+\frac{2 C \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^4 \sin (c+d x)}{9 d}\\ \end{align*}

Mathematica [A]  time = 7.38383, size = 609, normalized size = 1.38 \[ \frac{2 \cos ^6(c+d x) (a+b \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right ) \left (420 a^3 A b+210 a^2 b^2 B+140 a^3 b C+105 a^4 B+140 a A b^3+100 a b^3 C+25 b^4 B\right )+\frac{2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (-630 a^2 A b^2+105 a^4 A-378 a^2 b^2 C-420 a^3 b B-105 a^4 C-252 a b^3 B-63 A b^4-49 b^4 C\right )}{\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}}\right )}{105 d (a \cos (c+d x)+b)^4 (A \cos (2 c+2 d x)+A+2 B \cos (c+d x)+2 C)}+\frac{(a+b \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (\frac{4}{15} \sin (c+d x) \left (90 a^2 A b^2+54 a^2 b^2 C+60 a^3 b B+15 a^4 C+36 a b^3 B+9 A b^4+7 b^4 C\right )+\frac{4}{45} \sec ^2(c+d x) \left (54 a^2 b^2 C \sin (c+d x)+36 a b^3 B \sin (c+d x)+9 A b^4 \sin (c+d x)+7 b^4 C \sin (c+d x)\right )+\frac{4}{21} \sec (c+d x) \left (42 a^2 b^2 B \sin (c+d x)+28 a^3 b C \sin (c+d x)+28 a A b^3 \sin (c+d x)+20 a b^3 C \sin (c+d x)+5 b^4 B \sin (c+d x)\right )+\frac{4}{7} \sec ^3(c+d x) \left (4 a b^3 C \sin (c+d x)+b^4 B \sin (c+d x)\right )+\frac{4}{9} b^4 C \tan (c+d x) \sec ^3(c+d x)\right )}{d \sec ^{\frac{11}{2}}(c+d x) (a \cos (c+d x)+b)^4 (A \cos (2 c+2 d x)+A+2 B \cos (c+d x)+2 C)} \]

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))/Sqrt[Sec[c + d*x]],x]

[Out]

(2*Cos[c + d*x]^6*((2*(105*a^4*A - 630*a^2*A*b^2 - 63*A*b^4 - 420*a^3*b*B - 252*a*b^3*B - 105*a^4*C - 378*a^2*
b^2*C - 49*b^4*C)*EllipticE[(c + d*x)/2, 2])/(Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) + 2*(420*a^3*A*b + 140*a*
A*b^3 + 105*a^4*B + 210*a^2*b^2*B + 25*b^4*B + 140*a^3*b*C + 100*a*b^3*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*(90*a^2*A*b^2 + 9*A*b^4 + 60*a^3*b*B + 36*a*b^3*B + 15*a^4*C + 54*a^2*b^2*C + 7*b^4*C)
*Sin[c + d*x])/15 + (4*Sec[c + d*x]^3*(b^4*B*Sin[c + d*x] + 4*a*b^3*C*Sin[c + d*x]))/7 + (4*Sec[c + d*x]*(28*a
*A*b^3*Sin[c + d*x] + 42*a^2*b^2*B*Sin[c + d*x] + 5*b^4*B*Sin[c + d*x] + 28*a^3*b*C*Sin[c + d*x] + 20*a*b^3*C*
Sin[c + d*x]))/21 + (4*Sec[c + d*x]^2*(9*A*b^4*Sin[c + d*x] + 36*a*b^3*B*Sin[c + d*x] + 54*a^2*b^2*C*Sin[c + d
*x] + 7*b^4*C*Sin[c + d*x]))/45 + (4*b^4*C*Sec[c + d*x]^3*Tan[c + d*x])/9))/(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 = 12.626, size = 1550, normalized size = 3.5 \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)^(1/2),x)

[Out]

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(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
))-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))+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))-2/5*b^2*(A*b^2+4*B*a
*b+6*C*a^2)/(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)*(-
1/56*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*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+5/21*(si
n(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*(-1/144*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)^5-7/180*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)^3-14/15*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)/(-(-2*cos(1/2*d*x+
1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)+7/15*(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))-7/15*(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))))+4*a*b*(2*A*b^2+3*B*a*b+2*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)*El
lipticF(cos(1/2*d*x+1/2*c),2^(1/2)))+2*a^2*(6*A*b^2+4*B*a*b+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)^(1/2),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 )}{\sqrt{\sec \left (d x + c\right )}}, 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)^(1/2),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))/sqrt(sec(d*x + c)), 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)**(1/2),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}}{\sqrt{\sec \left (d x + c\right )}}\,{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)^(1/2),x, algorithm="giac")

[Out]

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