3.1024 \(\int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^2} \, dx\)

Optimal. Leaf size=507 \[ \frac {\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))}-\frac {\sin (c+d x) \left (-\left (a^2 (2 A-5 C)\right )-5 a b B+7 A b^2\right )}{5 a^2 d \left (a^2-b^2\right ) \sec ^{\frac {3}{2}}(c+d x)}+\frac {\sin (c+d x) \left (2 a^3 B-a^2 (4 A b-3 b C)-5 a b^2 B+7 A b^3\right )}{3 a^3 d \left (a^2-b^2\right ) \sqrt {\sec (c+d x)}}-\frac {\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (-2 a^4 (3 A+5 C)+20 a^3 b B-3 a^2 b^2 (8 A-5 C)-25 a b^3 B+35 A b^4\right )}{5 a^4 d \left (a^2-b^2\right )}-\frac {b^2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-5 a^4 C+7 a^3 b B-3 a^2 b^2 (3 A-C)-5 a b^3 B+7 A b^4\right ) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a^5 d (a-b) (a+b)^2}+\frac {\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (2 a^5 B-4 a^4 b (A+3 C)+16 a^3 b^2 B-a^2 b^3 (20 A-9 C)-15 a b^4 B+21 A b^5\right )}{3 a^5 d \left (a^2-b^2\right )} \]

[Out]

-1/5*(7*A*b^2-5*a*b*B-a^2*(2*A-5*C))*sin(d*x+c)/a^2/(a^2-b^2)/d/sec(d*x+c)^(3/2)+(A*b^2-a*(B*b-C*a))*sin(d*x+c
)/a/(a^2-b^2)/d/sec(d*x+c)^(3/2)/(a+b*sec(d*x+c))+1/3*(7*A*b^3+2*a^3*B-5*a*b^2*B-a^2*(4*A*b-3*C*b))*sin(d*x+c)
/a^3/(a^2-b^2)/d/sec(d*x+c)^(1/2)-1/5*(35*A*b^4+20*a^3*b*B-25*a*b^3*B-3*a^2*b^2*(8*A-5*C)-2*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)/a^4/(a^2-b^2)/d+1/3*(21*A*b^5+2*a^5*B+16*a^3*b^2*B-15*a*b^4*B-a^2*b^3*(20*A-9*C)-4*a^4*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)/a^5/(a^2-b^2)/d-b^2*(7*A*b^4+7*a^3*b*B-5*a*b^3*B-3*a^2*b^2*(3*A-C)-5*a^4*C)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/
cos(1/2*d*x+1/2*c)*EllipticPi(sin(1/2*d*x+1/2*c),2*a/(a+b),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/a^5/(a-b
)/(a+b)^2/d

________________________________________________________________________________________

Rubi [A]  time = 1.52, antiderivative size = 507, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 9, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.209, Rules used = {4100, 4104, 4106, 3849, 2805, 3787, 3771, 2639, 2641} \[ \frac {\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))}-\frac {\sin (c+d x) \left (a^2 (-(2 A-5 C))-5 a b B+7 A b^2\right )}{5 a^2 d \left (a^2-b^2\right ) \sec ^{\frac {3}{2}}(c+d x)}+\frac {\sin (c+d x) \left (-a^2 (4 A b-3 b C)+2 a^3 B-5 a b^2 B+7 A b^3\right )}{3 a^3 d \left (a^2-b^2\right ) \sqrt {\sec (c+d x)}}+\frac {\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (-a^2 b^3 (20 A-9 C)-4 a^4 b (A+3 C)+16 a^3 b^2 B+2 a^5 B-15 a b^4 B+21 A b^5\right )}{3 a^5 d \left (a^2-b^2\right )}-\frac {\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (-3 a^2 b^2 (8 A-5 C)-2 a^4 (3 A+5 C)+20 a^3 b B-25 a b^3 B+35 A b^4\right )}{5 a^4 d \left (a^2-b^2\right )}-\frac {b^2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-3 a^2 b^2 (3 A-C)+7 a^3 b B-5 a^4 C-5 a b^3 B+7 A b^4\right ) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a^5 d (a-b) (a+b)^2} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(Sec[c + d*x]^(5/2)*(a + b*Sec[c + d*x])^2),x]

[Out]

-((35*A*b^4 + 20*a^3*b*B - 25*a*b^3*B - 3*a^2*b^2*(8*A - 5*C) - 2*a^4*(3*A + 5*C))*Sqrt[Cos[c + d*x]]*Elliptic
E[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*a^4*(a^2 - b^2)*d) + ((21*A*b^5 + 2*a^5*B + 16*a^3*b^2*B - 15*a*b^4*B
- a^2*b^3*(20*A - 9*C) - 4*a^4*b*(A + 3*C))*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/
(3*a^5*(a^2 - b^2)*d) - (b^2*(7*A*b^4 + 7*a^3*b*B - 5*a*b^3*B - 3*a^2*b^2*(3*A - C) - 5*a^4*C)*Sqrt[Cos[c + d*
x]]*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(a^5*(a - b)*(a + b)^2*d) - ((7*A*b^2 - 5*a*
b*B - a^2*(2*A - 5*C))*Sin[c + d*x])/(5*a^2*(a^2 - b^2)*d*Sec[c + d*x]^(3/2)) + ((7*A*b^3 + 2*a^3*B - 5*a*b^2*
B - a^2*(4*A*b - 3*b*C))*Sin[c + d*x])/(3*a^3*(a^2 - b^2)*d*Sqrt[Sec[c + d*x]]) + ((A*b^2 - a*(b*B - a*C))*Sin
[c + d*x])/(a*(a^2 - b^2)*d*Sec[c + d*x]^(3/2)*(a + b*Sec[c + d*x]))

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]

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 2805

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2*EllipticPi[(2*b)/(a + b), (1*(e - Pi/2 + f*x))/2, (2*d)/(c + d)])/(f*(a + b)*Sqrt[c + d]), x] /; FreeQ[{a,
b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[c + d, 0]

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 3787

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 3849

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(3/2)/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[d*Sqrt[d*S
in[e + f*x]]*Sqrt[d*Csc[e + f*x]], Int[1/(Sqrt[d*Sin[e + f*x]]*(b + a*Sin[e + f*x])), x], x] /; FreeQ[{a, b, d
, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 4100

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*b^2 - a*b*B + a^2*C)*Cot[e + f*x]*(a +
b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n)/(a*f*(m + 1)*(a^2 - b^2)), x] + Dist[1/(a*(m + 1)*(a^2 - b^2)), I
nt[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*Simp[a*(a*A - b*B + a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C)*
(m + n + 1) - a*(A*b - a*B + b*C)*(m + 1)*Csc[e + f*x] + (A*b^2 - a*b*B + a^2*C)*(m + n + 2)*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] && LtQ[m, -1] &&  !(ILtQ[m + 1/2, 0] &
& ILtQ[n, 0])

Rule 4104

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 +
1)*(d*Csc[e + f*x])^n)/(a*f*n), x] + Dist[1/(a*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[
a*B*n - A*b*(m + n + 1) + a*(A + A*n + C*n)*Csc[e + f*x] + A*b*(m + n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ
[{a, b, d, e, f, A, B, C, m}, x] && NeQ[a^2 - b^2, 0] && LeQ[n, -1]

Rule 4106

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(d
_.)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))), x_Symbol] :> Dist[(A*b^2 - a*b*B + a^2*C)/(a^2*d^2), Int[(d*Csc[
e + f*x])^(3/2)/(a + b*Csc[e + f*x]), x], x] + Dist[1/a^2, Int[(a*A - (A*b - a*B)*Csc[e + f*x])/Sqrt[d*Csc[e +
f*x]], x], x] /; FreeQ[{a, b, d, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0]

Rubi steps

\begin {align*} \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^2} \, dx &=\frac {\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{a \left (a^2-b^2\right ) d \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))}-\frac {\int \frac {\frac {1}{2} \left (7 A b^2-5 a b B-a^2 (2 A-5 C)\right )+a (A b-a B+b C) \sec (c+d x)-\frac {5}{2} \left (A b^2-a (b B-a C)\right ) \sec ^2(c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))} \, dx}{a \left (a^2-b^2\right )}\\ &=-\frac {\left (7 A b^2-5 a b B-a^2 (2 A-5 C)\right ) \sin (c+d x)}{5 a^2 \left (a^2-b^2\right ) d \sec ^{\frac {3}{2}}(c+d x)}+\frac {\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{a \left (a^2-b^2\right ) d \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))}+\frac {2 \int \frac {\frac {5}{4} \left (7 A b^3+2 a^3 B-5 a b^2 B-a^2 (4 A b-3 b C)\right )+\frac {1}{2} a \left (2 A b^2-5 a b B+a^2 (3 A+5 C)\right ) \sec (c+d x)-\frac {3}{4} b \left (7 A b^2-5 a b B-a^2 (2 A-5 C)\right ) \sec ^2(c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))} \, dx}{5 a^2 \left (a^2-b^2\right )}\\ &=-\frac {\left (7 A b^2-5 a b B-a^2 (2 A-5 C)\right ) \sin (c+d x)}{5 a^2 \left (a^2-b^2\right ) d \sec ^{\frac {3}{2}}(c+d x)}+\frac {\left (7 A b^3+2 a^3 B-5 a b^2 B-a^2 (4 A b-3 b C)\right ) \sin (c+d x)}{3 a^3 \left (a^2-b^2\right ) d \sqrt {\sec (c+d x)}}+\frac {\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{a \left (a^2-b^2\right ) d \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))}-\frac {4 \int \frac {\frac {3}{8} \left (35 A b^4+20 a^3 b B-25 a b^3 B-3 a^2 b^2 (8 A-5 C)-2 a^4 (3 A+5 C)\right )+\frac {1}{4} a \left (14 A b^3-5 a^3 B-10 a b^2 B+a^2 b (A+15 C)\right ) \sec (c+d x)-\frac {5}{8} b \left (7 A b^3+2 a^3 B-5 a b^2 B-a^2 (4 A b-3 b C)\right ) \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (a+b \sec (c+d x))} \, dx}{15 a^3 \left (a^2-b^2\right )}\\ &=-\frac {\left (7 A b^2-5 a b B-a^2 (2 A-5 C)\right ) \sin (c+d x)}{5 a^2 \left (a^2-b^2\right ) d \sec ^{\frac {3}{2}}(c+d x)}+\frac {\left (7 A b^3+2 a^3 B-5 a b^2 B-a^2 (4 A b-3 b C)\right ) \sin (c+d x)}{3 a^3 \left (a^2-b^2\right ) d \sqrt {\sec (c+d x)}}+\frac {\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{a \left (a^2-b^2\right ) d \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))}-\frac {4 \int \frac {\frac {3}{8} a \left (35 A b^4+20 a^3 b B-25 a b^3 B-3 a^2 b^2 (8 A-5 C)-2 a^4 (3 A+5 C)\right )-\left (\frac {3}{8} b \left (35 A b^4+20 a^3 b B-25 a b^3 B-3 a^2 b^2 (8 A-5 C)-2 a^4 (3 A+5 C)\right )-\frac {1}{4} a^2 \left (14 A b^3-5 a^3 B-10 a b^2 B+a^2 b (A+15 C)\right )\right ) \sec (c+d x)}{\sqrt {\sec (c+d x)}} \, dx}{15 a^5 \left (a^2-b^2\right )}-\frac {\left (b^2 \left (7 A b^4+7 a^3 b B-5 a b^3 B-3 a^2 b^2 (3 A-C)-5 a^4 C\right )\right ) \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{a+b \sec (c+d x)} \, dx}{2 a^5 \left (a^2-b^2\right )}\\ &=-\frac {\left (7 A b^2-5 a b B-a^2 (2 A-5 C)\right ) \sin (c+d x)}{5 a^2 \left (a^2-b^2\right ) d \sec ^{\frac {3}{2}}(c+d x)}+\frac {\left (7 A b^3+2 a^3 B-5 a b^2 B-a^2 (4 A b-3 b C)\right ) \sin (c+d x)}{3 a^3 \left (a^2-b^2\right ) d \sqrt {\sec (c+d x)}}+\frac {\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{a \left (a^2-b^2\right ) d \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))}+\frac {\left (21 A b^5+2 a^5 B+16 a^3 b^2 B-15 a b^4 B-a^2 b^3 (20 A-9 C)-4 a^4 b (A+3 C)\right ) \int \sqrt {\sec (c+d x)} \, dx}{6 a^5 \left (a^2-b^2\right )}-\frac {\left (35 A b^4+20 a^3 b B-25 a b^3 B-3 a^2 b^2 (8 A-5 C)-2 a^4 (3 A+5 C)\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx}{10 a^4 \left (a^2-b^2\right )}-\frac {\left (b^2 \left (7 A b^4+7 a^3 b B-5 a b^3 B-3 a^2 b^2 (3 A-C)-5 a^4 C\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{2 a^5 \left (a^2-b^2\right )}\\ &=-\frac {b^2 \left (7 A b^4+7 a^3 b B-5 a b^3 B-3 a^2 b^2 (3 A-C)-5 a^4 C\right ) \sqrt {\cos (c+d x)} \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{a^5 (a-b) (a+b)^2 d}-\frac {\left (7 A b^2-5 a b B-a^2 (2 A-5 C)\right ) \sin (c+d x)}{5 a^2 \left (a^2-b^2\right ) d \sec ^{\frac {3}{2}}(c+d x)}+\frac {\left (7 A b^3+2 a^3 B-5 a b^2 B-a^2 (4 A b-3 b C)\right ) \sin (c+d x)}{3 a^3 \left (a^2-b^2\right ) d \sqrt {\sec (c+d x)}}+\frac {\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{a \left (a^2-b^2\right ) d \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))}+\frac {\left (\left (21 A b^5+2 a^5 B+16 a^3 b^2 B-15 a b^4 B-a^2 b^3 (20 A-9 C)-4 a^4 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}{6 a^5 \left (a^2-b^2\right )}-\frac {\left (\left (35 A b^4+20 a^3 b B-25 a b^3 B-3 a^2 b^2 (8 A-5 C)-2 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}{10 a^4 \left (a^2-b^2\right )}\\ &=-\frac {\left (35 A b^4+20 a^3 b B-25 a b^3 B-3 a^2 b^2 (8 A-5 C)-2 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 a^4 \left (a^2-b^2\right ) d}+\frac {\left (21 A b^5+2 a^5 B+16 a^3 b^2 B-15 a b^4 B-a^2 b^3 (20 A-9 C)-4 a^4 b (A+3 C)\right ) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{3 a^5 \left (a^2-b^2\right ) d}-\frac {b^2 \left (7 A b^4+7 a^3 b B-5 a b^3 B-3 a^2 b^2 (3 A-C)-5 a^4 C\right ) \sqrt {\cos (c+d x)} \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{a^5 (a-b) (a+b)^2 d}-\frac {\left (7 A b^2-5 a b B-a^2 (2 A-5 C)\right ) \sin (c+d x)}{5 a^2 \left (a^2-b^2\right ) d \sec ^{\frac {3}{2}}(c+d x)}+\frac {\left (7 A b^3+2 a^3 B-5 a b^2 B-a^2 (4 A b-3 b C)\right ) \sin (c+d x)}{3 a^3 \left (a^2-b^2\right ) d \sqrt {\sec (c+d x)}}+\frac {\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{a \left (a^2-b^2\right ) d \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 7.47, size = 971, normalized size = 1.92 \[ \frac {\left (C \sec ^2(c+d x)+B \sec (c+d x)+A\right ) \left (\frac {2 \left (-18 A a^4-30 C a^4+40 b B a^3-32 A b^2 a^2+15 b^2 C a^2-25 b^3 B a+35 A b^4\right ) \left (F\left (\left .\sin ^{-1}\left (\sqrt {\sec (c+d x)}\right )\right |-1\right )-\Pi \left (-\frac {b}{a};\left .\sin ^{-1}\left (\sqrt {\sec (c+d x)}\right )\right |-1\right )\right ) (a+b \sec (c+d x)) \sqrt {1-\sec ^2(c+d x)} \sin (c+d x) \cos ^2(c+d x)}{b (b+a \cos (c+d x)) \left (1-\cos ^2(c+d x)\right )}+\frac {2 \left (-20 B a^4+4 A b a^3+60 b C a^3-40 b^2 B a^2+56 A b^3 a\right ) \Pi \left (-\frac {b}{a};\left .\sin ^{-1}\left (\sqrt {\sec (c+d x)}\right )\right |-1\right ) (a+b \sec (c+d x)) \sqrt {1-\sec ^2(c+d x)} \sin (c+d x) \cos ^2(c+d x)}{a (b+a \cos (c+d x)) \left (1-\cos ^2(c+d x)\right )}+\frac {\left (-18 A a^4-30 C a^4+60 b B a^3-72 A b^2 a^2+45 b^2 C a^2-75 b^3 B a+105 A b^4\right ) \cos (2 (c+d x)) (a+b \sec (c+d x)) \left (2 \Pi \left (-\frac {b}{a};\left .\sin ^{-1}\left (\sqrt {\sec (c+d x)}\right )\right |-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)} a^2+4 b \sec ^2(c+d x) a-4 b a-4 b E\left (\left .\sin ^{-1}\left (\sqrt {\sec (c+d x)}\right )\right |-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)} a-2 (a-2 b) F\left (\left .\sin ^{-1}\left (\sqrt {\sec (c+d x)}\right )\right |-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)} a-4 b^2 \Pi \left (-\frac {b}{a};\left .\sin ^{-1}\left (\sqrt {\sec (c+d x)}\right )\right |-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}\right ) \sin (c+d x)}{a^2 b (b+a \cos (c+d x)) \left (1-\cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)} \left (2-\sec ^2(c+d x)\right )}\right ) (b+a \cos (c+d x))^2}{30 a^3 (b-a) (a+b) d (\cos (2 c+2 d x) A+A+2 C+2 B \cos (c+d x)) (a+b \sec (c+d x))^2}+\frac {\sqrt {\sec (c+d x)} \left (C \sec ^2(c+d x)+B \sec (c+d x)+A\right ) \left (\frac {\left (A a^4-A b^2 a^2+10 b^2 C a^2-10 b^3 B a+10 A b^4\right ) \sin (c+d x)}{5 a^4 \left (a^2-b^2\right )}-\frac {2 \left (A \sin (c+d x) b^5-a B \sin (c+d x) b^4+a^2 C \sin (c+d x) b^3\right )}{a^4 \left (a^2-b^2\right ) (b+a \cos (c+d x))}+\frac {2 (a B-2 A b) \sin (2 (c+d x))}{3 a^3}+\frac {A \sin (3 (c+d x))}{5 a^2}\right ) (b+a \cos (c+d x))^2}{d (\cos (2 c+2 d x) A+A+2 C+2 B \cos (c+d x)) (a+b \sec (c+d x))^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(Sec[c + d*x]^(5/2)*(a + b*Sec[c + d*x])^2),x]

[Out]

((b + a*Cos[c + d*x])^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((2*(-18*a^4*A - 32*a^2*A*b^2 + 35*A*b^4 + 40*
a^3*b*B - 25*a*b^3*B - 30*a^4*C + 15*a^2*b^2*C)*Cos[c + d*x]^2*(EllipticF[ArcSin[Sqrt[Sec[c + d*x]]], -1] - El
lipticPi[-(b/a), ArcSin[Sqrt[Sec[c + d*x]]], -1])*(a + b*Sec[c + d*x])*Sqrt[1 - Sec[c + d*x]^2]*Sin[c + d*x])/
(b*(b + a*Cos[c + d*x])*(1 - Cos[c + d*x]^2)) + (2*(4*a^3*A*b + 56*a*A*b^3 - 20*a^4*B - 40*a^2*b^2*B + 60*a^3*
b*C)*Cos[c + d*x]^2*EllipticPi[-(b/a), ArcSin[Sqrt[Sec[c + d*x]]], -1]*(a + b*Sec[c + d*x])*Sqrt[1 - Sec[c + d
*x]^2]*Sin[c + d*x])/(a*(b + a*Cos[c + d*x])*(1 - Cos[c + d*x]^2)) + ((-18*a^4*A - 72*a^2*A*b^2 + 105*A*b^4 +
60*a^3*b*B - 75*a*b^3*B - 30*a^4*C + 45*a^2*b^2*C)*Cos[2*(c + d*x)]*(a + b*Sec[c + d*x])*(-4*a*b + 4*a*b*Sec[c
+ d*x]^2 - 4*a*b*EllipticE[ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]^2] - 2*a*
(a - 2*b)*EllipticF[ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]^2] + 2*a^2*Ellipt
icPi[-(b/a), ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]^2] - 4*b^2*EllipticPi[-(
b/a), ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]^2])*Sin[c + d*x])/(a^2*b*(b + a
*Cos[c + d*x])*(1 - Cos[c + d*x]^2)*Sqrt[Sec[c + d*x]]*(2 - Sec[c + d*x]^2))))/(30*a^3*(-a + b)*(a + b)*d*(A +
2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + b*Sec[c + d*x])^2) + ((b + a*Cos[c + d*x])^2*Sqrt[Sec[c + d
*x]]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(((a^4*A - a^2*A*b^2 + 10*A*b^4 - 10*a*b^3*B + 10*a^2*b^2*C)*Sin[
c + d*x])/(5*a^4*(a^2 - b^2)) - (2*(A*b^5*Sin[c + d*x] - a*b^4*B*Sin[c + d*x] + a^2*b^3*C*Sin[c + d*x]))/(a^4*
(a^2 - b^2)*(b + a*Cos[c + d*x])) + (2*(-2*A*b + a*B)*Sin[2*(c + d*x)])/(3*a^3) + (A*Sin[3*(c + d*x)])/(5*a^2)
))/(d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + b*Sec[c + d*x])^2)

________________________________________________________________________________________

fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(5/2)/(a+b*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

Timed out

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A}{{\left (b \sec \left (d x + c\right ) + a\right )}^{2} \sec \left (d x + c\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(5/2)/(a+b*sec(d*x+c))^2,x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)/((b*sec(d*x + c) + a)^2*sec(d*x + c)^(5/2)), x)

________________________________________________________________________________________

maple [B]  time = 17.31, size = 1377, normalized size = 2.72 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(5/2)/(a+b*sec(d*x+c))^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)*(4/5*A/a^2*(-4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c
)^6+14*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+5*(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))-9*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*Ellipti
cE(cos(1/2*d*x+1/2*c),2^(1/2))-6*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)-4/3/a^3*(3*A*a+2*A*b-B*a)*(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*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(
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*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/a^4*(3*A*a^2+4*A*a*b+3*A*b^2-2*B*a^2-2*B*a*b+C*a^2)*(s
in(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^3+2*A*a^2*b+3*A*a*b^2
+4*A*b^3-B*a^3-2*B*a^2*b-3*B*a*b^2+C*a^3+2*C*a^2*b)/a^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)/(-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^2/a^4*
(5*A*b^2-4*B*a*b+3*C*a^2)/(a^2-a*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)*EllipticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2))-2*b^3*(A*b^2-B*a*b+
C*a^2)/a^5*(a^2/b/(a^2-b^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*a*cos(1
/2*d*x+1/2*c)^2-a+b)-1/2/(a+b)/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))+1/2*a/b/(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
(cos(1/2*d*x+1/2*c),2^(1/2))-1/2*a/b/(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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-1/2/b/(a^2-b^2)/(a^
2-a*b)*a^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)*EllipticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2))+3/2*b/(a^2-b^2)/(a^2-a*b)*a*(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)*EllipticPi
(cos(1/2*d*x+1/2*c),2*a/(a-b),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

________________________________________________________________________________________

maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(5/2)/(a+b*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

Timed out

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^2\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)**2)/sec(d*x+c)**(5/2)/(a+b*sec(d*x+c))**2,x)

[Out]

Timed out

________________________________________________________________________________________