\(\int \frac {(a+b \sec (c+d x))^4 (A+B \sec (c+d x)+C \sec ^2(c+d x))}{\sec ^{\frac {5}{2}}(c+d x)} \, dx\) [1007]

Optimal result

Integrand size = 43, antiderivative size = 409 \[ \int \frac {(a+b \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\frac {2 \left (20 a^3 b B-20 a b^3 B+30 a^2 b^2 (A-C)-b^4 (5 A+3 C)+a^4 (3 A+5 C)\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {2 \left (a^4 B+18 a^2 b^2 B+b^4 B+4 a b^3 (3 A+C)+4 a^3 b (A+3 C)\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 d}-\frac {2 b \left (10 a^3 B-60 a b^2 B+a^2 b (31 A-87 C)-3 b^3 (5 A+3 C)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d}-\frac {2 b^2 \left (5 a^2 B-5 b^2 B+14 a b (A-C)\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}-\frac {2 b (11 A b+5 a B-3 b C) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2 \sin (c+d x)}{15 d}+\frac {2 (8 A b+5 a B) (a+b \sec (c+d x))^3 \sin (c+d x)}{15 d \sqrt {\sec (c+d x)}}+\frac {2 A (a+b \sec (c+d x))^4 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)} \]

[Out]

-2/15*b^2*(5*B*a^2-5*B*b^2+14*a*b*(A-C))*sec(d*x+c)^(3/2)*sin(d*x+c)/d+2/5*A*(a+b*sec(d*x+c))^4*sin(d*x+c)/d/s
ec(d*x+c)^(3/2)+2/15*(8*A*b+5*B*a)*(a+b*sec(d*x+c))^3*sin(d*x+c)/d/sec(d*x+c)^(1/2)-2/15*b*(10*B*a^3-60*B*a*b^
2+a^2*b*(31*A-87*C)-3*b^3*(5*A+3*C))*sin(d*x+c)*sec(d*x+c)^(1/2)/d-2/15*b*(11*A*b+5*B*a-3*C*b)*(a+b*sec(d*x+c)
)^2*sin(d*x+c)*sec(d*x+c)^(1/2)/d+2/5*(20*B*a^3*b-20*B*a*b^3+30*a^2*b^2*(A-C)-b^4*(5*A+3*C)+a^4*(3*A+5*C))*(co
s(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)
^(1/2)/d+2/3*(B*a^4+18*B*a^2*b^2+B*b^4+4*a*b^3*(3*A+C)+4*a^3*b*(A+3*C))*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d
*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/d

Rubi [A] (verified)

Time = 1.32 (sec) , antiderivative size = 409, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.186, Rules used = {4179, 4181, 4161, 4132, 3856, 2720, 4131, 2719} \[ \int \frac {(a+b \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {2 b^2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \left (5 a^2 B+14 a b (A-C)-5 b^2 B\right )}{15 d}-\frac {2 b \sin (c+d x) \sqrt {\sec (c+d x)} \left (10 a^3 B+a^2 b (31 A-87 C)-60 a b^2 B-3 b^3 (5 A+3 C)\right )}{15 d}+\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (a^4 B+4 a^3 b (A+3 C)+18 a^2 b^2 B+4 a b^3 (3 A+C)+b^4 B\right )}{3 d}+\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^4 (3 A+5 C)+20 a^3 b B+30 a^2 b^2 (A-C)-20 a b^3 B-b^4 (5 A+3 C)\right )}{5 d}-\frac {2 b \sin (c+d x) \sqrt {\sec (c+d x)} (5 a B+11 A b-3 b C) (a+b \sec (c+d x))^2}{15 d}+\frac {2 (5 a B+8 A b) \sin (c+d x) (a+b \sec (c+d x))^3}{15 d \sqrt {\sec (c+d x)}}+\frac {2 A \sin (c+d x) (a+b \sec (c+d x))^4}{5 d \sec ^{\frac {3}{2}}(c+d x)} \]

[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 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{
c, d}, x]

Rule 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]

Rule 3856

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 4131

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 4132

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 4161

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 4179

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 4181

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]

Rubi steps \begin{align*} \text {integral}& = \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)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),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)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),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] (warning: unable to verify)

Time = 15.58 (sec) , antiderivative size = 377, normalized size of antiderivative = 0.92 \[ \int \frac {(a+b \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\frac {(a+b \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (12 \left (20 a^3 b B-20 a b^3 B+30 a^2 b^2 (A-C)-b^4 (5 A+3 C)+a^4 (3 A+5 C)\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+20 \left (a^4 B+18 a^2 b^2 B+b^4 B+4 a b^3 (3 A+C)+4 a^3 b (A+3 C)\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+3 a^4 A \sin (c+d x)+60 A b^4 \sin (c+d x)+240 a b^3 B \sin (c+d x)+360 a^2 b^2 C \sin (c+d x)+36 b^4 C \sin (c+d x)+40 a^3 A b \sin (2 (c+d x))+10 a^4 B \sin (2 (c+d x))+3 a^4 A \sin (3 (c+d x))+20 b^4 B \tan (c+d x)+80 a b^3 C \tan (c+d x)+12 b^4 C \sec (c+d x) \tan (c+d x)\right )}{15 d (b+a \cos (c+d x))^4 (A+2 C+2 B \cos (c+d x)+A \cos (2 (c+d x))) \sec ^{\frac {11}{2}}(c+d x)} \]

[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]

((a + b*Sec[c + d*x])^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(12*(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]]*EllipticE[(c + d*x)/2, 2] + 20*(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] + 3*a^4*A*Sin
[c + d*x] + 60*A*b^4*Sin[c + d*x] + 240*a*b^3*B*Sin[c + d*x] + 360*a^2*b^2*C*Sin[c + d*x] + 36*b^4*C*Sin[c + d
*x] + 40*a^3*A*b*Sin[2*(c + d*x)] + 10*a^4*B*Sin[2*(c + d*x)] + 3*a^4*A*Sin[3*(c + d*x)] + 20*b^4*B*Tan[c + d*
x] + 80*a*b^3*C*Tan[c + d*x] + 12*b^4*C*Sec[c + d*x]*Tan[c + d*x]))/(15*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))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1496\) vs. \(2(429)=858\).

Time = 8.00 (sec) , antiderivative size = 1497, normalized size of antiderivative = 3.66

method	result	size
parts	\(\text {Expression too large to display}\)	\(1497\)
default	\(\text {Expression too large to display}\)	\(1857\)

[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,method=_RETURNVERBOSE)

[Out]

-2/3*(4*A*a^3*b+B*a^4)*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(4*cos(1/2*d*x+1/2*c)*sin(1/2*d
*x+1/2*c)^4-2*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*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)))/(-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-2/3*(B*b^4+4*C*a*b^3)*(-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))*sin(1/2*d*x+1/2*c)^2-2*sin(1/2*d*x+1/2*c)^2*cos(1/2*
d*x+1/2*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))
)*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(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)
/(2*cos(1/2*d*x+1/2*c)^2-1)^(3/2)/sin(1/2*d*x+1/2*c)/d-2*(A*b^4+4*B*a*b^3+6*C*a^2*b^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)^(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*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-2*(4*A*a*b^3+6*B*a^2*b^2+4*C*a^3*b)*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(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)*Ellipti
cF(cos(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+2*(6*A*a^2*b^2+4*B*a^3*b+
C*a^4)*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/
2*c)^2+1)^(1/2)*EllipticE(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-2/5*A*a^4*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)
^(1/2)*(-8*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+8*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4-2*sin(1/2*d*x+1/2
*c)^2*cos(1/2*d*x+1/2*c)-3*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(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-2/5*C*b^4*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/(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)^3*(24*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c
)-12*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1
/2*d*x+1/2*c)^4-24*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+12*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2
*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^2+8*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/
2*c)-3*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(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)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.16 (sec) , antiderivative size = 455, normalized size of antiderivative = 1.11 \[ \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 {5 \, \sqrt {2} {\left (i \, B a^{4} + 4 i \, {\left (A + 3 \, C\right )} a^{3} b + 18 i \, B a^{2} b^{2} + 4 i \, {\left (3 \, A + C\right )} a b^{3} + i \, B b^{4}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, \sqrt {2} {\left (-i \, B a^{4} - 4 i \, {\left (A + 3 \, C\right )} a^{3} b - 18 i \, B a^{2} b^{2} - 4 i \, {\left (3 \, A + C\right )} a b^{3} - i \, B b^{4}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 \, \sqrt {2} {\left (-i \, {\left (3 \, A + 5 \, C\right )} a^{4} - 20 i \, B a^{3} b - 30 i \, {\left (A - C\right )} a^{2} b^{2} + 20 i \, B a b^{3} + i \, {\left (5 \, A + 3 \, C\right )} b^{4}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 \, \sqrt {2} {\left (i \, {\left (3 \, A + 5 \, C\right )} a^{4} + 20 i \, B a^{3} b + 30 i \, {\left (A - C\right )} a^{2} b^{2} - 20 i \, B a b^{3} - i \, {\left (5 \, A + 3 \, C\right )} b^{4}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {2 \, {\left (3 \, A a^{4} \cos \left (d x + c\right )^{4} + 3 \, C b^{4} + 5 \, {\left (B a^{4} + 4 \, A a^{3} b\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (30 \, C a^{2} b^{2} + 20 \, B a b^{3} + {\left (5 \, A + 3 \, C\right )} b^{4}\right )} \cos \left (d x + c\right )^{2} + 5 \, {\left (4 \, C a b^{3} + B b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{15 \, d \cos \left (d x + c\right )^{2}} \]

[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]

-1/15*(5*sqrt(2)*(I*B*a^4 + 4*I*(A + 3*C)*a^3*b + 18*I*B*a^2*b^2 + 4*I*(3*A + C)*a*b^3 + I*B*b^4)*cos(d*x + c)
^2*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*sqrt(2)*(-I*B*a^4 - 4*I*(A + 3*C)*a^3*b - 18*
I*B*a^2*b^2 - 4*I*(3*A + C)*a*b^3 - I*B*b^4)*cos(d*x + c)^2*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*
x + c)) + 3*sqrt(2)*(-I*(3*A + 5*C)*a^4 - 20*I*B*a^3*b - 30*I*(A - C)*a^2*b^2 + 20*I*B*a*b^3 + I*(5*A + 3*C)*b
^4)*cos(d*x + c)^2*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*sqrt(
2)*(I*(3*A + 5*C)*a^4 + 20*I*B*a^3*b + 30*I*(A - C)*a^2*b^2 - 20*I*B*a*b^3 - I*(5*A + 3*C)*b^4)*cos(d*x + c)^2
*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - 2*(3*A*a^4*cos(d*x + c)^4
+ 3*C*b^4 + 5*(B*a^4 + 4*A*a^3*b)*cos(d*x + c)^3 + 3*(30*C*a^2*b^2 + 20*B*a*b^3 + (5*A + 3*C)*b^4)*cos(d*x +
c)^2 + 5*(4*C*a*b^3 + B*b^4)*cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c)^2)

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+b \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\text {Timed out} \]

[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

Maxima [F(-1)]

Timed out. \[ \int \frac {(a+b \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\text {Timed out} \]

[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

Giac [F]

\[ \int \frac {(a+b \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{4}}{\sec \left (d x + c\right )^{\frac {5}{2}}} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\int \frac {{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^4\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]

[In]

int(((a + b/cos(c + d*x))^4*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/(1/cos(c + d*x))^(5/2),x)

[Out]

int(((a + b/cos(c + d*x))^4*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/(1/cos(c + d*x))^(5/2), x)