\(\int \frac {(a+b \sec (c+d x))^4 (A+B \sec (c+d x)+C \sec ^2(c+d x))}{\sec ^{\frac {7}{2}}(c+d x)} \, dx\) [1008]

Optimal result

Integrand size = 43, antiderivative size = 429 \[ \int \frac {(a+b \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\frac {2 \left (3 a^4 B+30 a^2 b^2 B-5 b^4 B+20 a b^3 (A-C)+4 a^3 b (3 A+5 C)\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {2 \left (28 a^3 b B+84 a b^3 B+7 b^4 (3 A+C)+42 a^2 b^2 (A+3 C)+a^4 (5 A+7 C)\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{21 d}-\frac {2 b \left (217 a^2 b B-105 b^3 B+12 a b^2 (19 A-35 C)+10 a^3 (5 A+7 C)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{105 d}-\frac {2 b^2 \left (98 a b B+b^2 (87 A-35 C)+5 a^2 (5 A+7 C)\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{105 d}+\frac {2 \left (48 A b^2+77 a b B+5 a^2 (5 A+7 C)\right ) (a+b \sec (c+d x))^2 \sin (c+d x)}{105 d \sqrt {\sec (c+d x)}}+\frac {2 (8 A b+7 a B) (a+b \sec (c+d x))^3 \sin (c+d x)}{35 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 A (a+b \sec (c+d x))^4 \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)} \]

[Out]

-2/105*b^2*(98*B*a*b+b^2*(87*A-35*C)+5*a^2*(5*A+7*C))*sec(d*x+c)^(3/2)*sin(d*x+c)/d+2/35*(8*A*b+7*B*a)*(a+b*se
c(d*x+c))^3*sin(d*x+c)/d/sec(d*x+c)^(3/2)+2/7*A*(a+b*sec(d*x+c))^4*sin(d*x+c)/d/sec(d*x+c)^(5/2)+2/105*(48*A*b
^2+77*B*a*b+5*a^2*(5*A+7*C))*(a+b*sec(d*x+c))^2*sin(d*x+c)/d/sec(d*x+c)^(1/2)-2/105*b*(217*B*a^2*b-105*B*b^3+1
2*a*b^2*(19*A-35*C)+10*a^3*(5*A+7*C))*sin(d*x+c)*sec(d*x+c)^(1/2)/d+2/5*(3*B*a^4+30*B*a^2*b^2-5*B*b^4+20*a*b^3
*(A-C)+4*a^3*b*(3*A+5*C))*(cos(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+2/21*(28*B*a^3*b+84*B*a*b^3+7*b^4*(3*A+C)+42*a^2*b^2*(A+3*C)+a^4*(5*A+7*
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 = 1.40 (sec) , antiderivative size = 429, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.163, Rules used = {4179, 4161, 4132, 3856, 2720, 4131, 2719} \[ \int \frac {(a+b \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=-\frac {2 b^2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \left (5 a^2 (5 A+7 C)+98 a b B+b^2 (87 A-35 C)\right )}{105 d}+\frac {2 \sin (c+d x) \left (5 a^2 (5 A+7 C)+77 a b B+48 A b^2\right ) (a+b \sec (c+d x))^2}{105 d \sqrt {\sec (c+d x)}}-\frac {2 b \sin (c+d x) \sqrt {\sec (c+d x)} \left (10 a^3 (5 A+7 C)+217 a^2 b B+12 a b^2 (19 A-35 C)-105 b^3 B\right )}{105 d}+\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (a^4 (5 A+7 C)+28 a^3 b B+42 a^2 b^2 (A+3 C)+84 a b^3 B+7 b^4 (3 A+C)\right )}{21 d}+\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (3 a^4 B+4 a^3 b (3 A+5 C)+30 a^2 b^2 B+20 a b^3 (A-C)-5 b^4 B\right )}{5 d}+\frac {2 (7 a B+8 A b) \sin (c+d x) (a+b \sec (c+d x))^3}{35 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 A \sin (c+d x) (a+b \sec (c+d x))^4}{7 d \sec ^{\frac {5}{2}}(c+d x)} \]

[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 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]

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

Time = 12.88 (sec) , antiderivative size = 394, normalized size of antiderivative = 0.92 \[ \int \frac {(a+b \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\frac {(a+b \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (168 \left (3 a^4 B+30 a^2 b^2 B-5 b^4 B+20 a b^3 (A-C)+4 a^3 b (3 A+5 C)\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+40 \left (28 a^3 b B+84 a b^3 B+7 b^4 (3 A+C)+42 a^2 b^2 (A+3 C)+a^4 (5 A+7 C)\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+168 a^3 A b \sin (c+d x)+42 a^4 B \sin (c+d x)+840 b^4 B \sin (c+d x)+3360 a b^3 C \sin (c+d x)+130 a^4 A \sin (2 (c+d x))+840 a^2 A b^2 \sin (2 (c+d x))+560 a^3 b B \sin (2 (c+d x))+140 a^4 C \sin (2 (c+d x))+168 a^3 A b \sin (3 (c+d x))+42 a^4 B \sin (3 (c+d x))+15 a^4 A \sin (4 (c+d x))+280 b^4 C \tan (c+d x)\right )}{210 d (b+a \cos (c+d x))^4 (A+2 C+2 B \cos (c+d x)+A \cos (2 (c+d x))) \sec ^{\frac {11}{2}}(c+d x)} \]

[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]^(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))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1337\) vs. \(2(449)=898\).

Time = 9.07 (sec) , antiderivative size = 1338, normalized size of antiderivative = 3.12

method	result	size
parts	\(\text {Expression too large to display}\)	\(1338\)
default	\(\text {Expression too large to display}\)	\(2507\)

[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,method=_RETURNVERBOSE)

[Out]

-2/5*(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)*(-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(c
os(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*(B*b^4+4*C*a*b^3)*(-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(c
os(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*(A*b^4+4*B*a*b^3+6*C*a^2*b^2)*((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)*EllipticF(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*(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)*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-2/3*(6*A*a^2*b^2+4*B*a^3*b+C*a^4)*((2*c
os(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-2/21*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)*(48*cos(1/2*d*x+1/2*c)^9-120*
cos(1/2*d*x+1/2*c)^7+128*cos(1/2*d*x+1/2*c)^5-72*cos(1/2*d*x+1/2*c)^3+5*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1
/2*d*x+1/2*c)^2+1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+16*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)/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d-2/3*C*b^4*(-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)*Elliptic
F(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

Fricas [C] (verification not implemented)

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

Time = 0.17 (sec) , antiderivative size = 447, normalized size of antiderivative = 1.04 \[ \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 {5 \, \sqrt {2} {\left (i \, {\left (5 \, A + 7 \, C\right )} a^{4} + 28 i \, B a^{3} b + 42 i \, {\left (A + 3 \, C\right )} a^{2} b^{2} + 84 i \, B a b^{3} + 7 i \, {\left (3 \, A + C\right )} b^{4}\right )} \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, \sqrt {2} {\left (-i \, {\left (5 \, A + 7 \, C\right )} a^{4} - 28 i \, B a^{3} b - 42 i \, {\left (A + 3 \, C\right )} a^{2} b^{2} - 84 i \, B a b^{3} - 7 i \, {\left (3 \, A + C\right )} b^{4}\right )} \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 21 \, \sqrt {2} {\left (-3 i \, B a^{4} - 4 i \, {\left (3 \, A + 5 \, C\right )} a^{3} b - 30 i \, B a^{2} b^{2} - 20 i \, {\left (A - C\right )} a b^{3} + 5 i \, B b^{4}\right )} \cos \left (d x + c\right ) {\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 ) + 21 \, \sqrt {2} {\left (3 i \, B a^{4} + 4 i \, {\left (3 \, A + 5 \, C\right )} a^{3} b + 30 i \, B a^{2} b^{2} + 20 i \, {\left (A - C\right )} a b^{3} - 5 i \, B b^{4}\right )} \cos \left (d x + c\right ) {\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 (15 \, A a^{4} \cos \left (d x + c\right )^{4} + 35 \, C b^{4} + 21 \, {\left (B a^{4} + 4 \, A a^{3} b\right )} \cos \left (d x + c\right )^{3} + 5 \, {\left ({\left (5 \, A + 7 \, C\right )} a^{4} + 28 \, B a^{3} b + 42 \, A a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2} + 105 \, {\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 )}}}{105 \, d \cos \left (d x + c\right )} \]

[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]

-1/105*(5*sqrt(2)*(I*(5*A + 7*C)*a^4 + 28*I*B*a^3*b + 42*I*(A + 3*C)*a^2*b^2 + 84*I*B*a*b^3 + 7*I*(3*A + C)*b^
4)*cos(d*x + c)*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*sqrt(2)*(-I*(5*A + 7*C)*a^4 - 28
*I*B*a^3*b - 42*I*(A + 3*C)*a^2*b^2 - 84*I*B*a*b^3 - 7*I*(3*A + C)*b^4)*cos(d*x + c)*weierstrassPInverse(-4, 0
, cos(d*x + c) - I*sin(d*x + c)) + 21*sqrt(2)*(-3*I*B*a^4 - 4*I*(3*A + 5*C)*a^3*b - 30*I*B*a^2*b^2 - 20*I*(A -
C)*a*b^3 + 5*I*B*b^4)*cos(d*x + c)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x
+ c))) + 21*sqrt(2)*(3*I*B*a^4 + 4*I*(3*A + 5*C)*a^3*b + 30*I*B*a^2*b^2 + 20*I*(A - C)*a*b^3 - 5*I*B*b^4)*cos
(d*x + c)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - 2*(15*A*a^4*cos(
d*x + c)^4 + 35*C*b^4 + 21*(B*a^4 + 4*A*a^3*b)*cos(d*x + c)^3 + 5*((5*A + 7*C)*a^4 + 28*B*a^3*b + 42*A*a^2*b^2
)*cos(d*x + c)^2 + 105*(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))

Sympy [F]

\[ \int \frac {(a+b \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\int \frac {\left (a + b \sec {\left (c + d x \right )}\right )^{4} \left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right )}{\sec ^{\frac {7}{2}}{\left (c + d x \right )}}\, dx \]

[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]

Integral((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)

Maxima [F(-1)]

Timed out. \[ \int \frac {(a+b \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\text {Timed out} \]

[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)^(7/2),x, algorithm="maxima")

[Out]

Timed out

Giac [F]

\[ \int \frac {(a+b \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{4}}{\sec \left (d x + c\right )^{\frac {7}{2}}} \,d x } \]

[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)^(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)

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\int \frac {{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^4\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{7/2}} \,d x \]

[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))^(7/2),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))^(7/2), x)