3.1019 \(\int \frac {\sec ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{(a+b \sec (c+d x))^2} \, dx\)

Optimal. Leaf size=447 \[ -\frac {\sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}+\frac {\sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \left (5 a^2 C-3 a b B+3 A b^2-2 b^2 C\right )}{3 b^2 d \left (a^2-b^2\right )}+\frac {\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (5 a^2 C-3 a b B+3 A b^2-2 b^2 C\right )}{3 b^2 d \left (a^2-b^2\right )}+\frac {\sin (c+d x) \sqrt {\sec (c+d x)} \left (-5 a^3 C+3 a^2 b B-a b^2 (A-4 C)-2 b^3 B\right )}{b^3 d \left (a^2-b^2\right )}-\frac {\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (-5 a^3 C+3 a^2 b B-a b^2 (A-4 C)-2 b^3 B\right )}{b^3 d \left (a^2-b^2\right )}-\frac {\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-5 a^4 C+3 a^3 b B-a^2 b^2 (A-7 C)-5 a b^3 B+3 A b^4\right ) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{b^3 d (a-b) (a+b)^2} \]

[Out]

1/3*(3*A*b^2-3*B*a*b+5*C*a^2-2*C*b^2)*sec(d*x+c)^(3/2)*sin(d*x+c)/b^2/(a^2-b^2)/d-(A*b^2-a*(B*b-C*a))*sec(d*x+
c)^(5/2)*sin(d*x+c)/b/(a^2-b^2)/d/(a+b*sec(d*x+c))+(3*a^2*b*B-2*b^3*B-a*b^2*(A-4*C)-5*a^3*C)*sin(d*x+c)*sec(d*
x+c)^(1/2)/b^3/(a^2-b^2)/d-(3*a^2*b*B-2*b^3*B-a*b^2*(A-4*C)-5*a^3*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)/b^3/(a^2-b^2)/d+1/3*(3*A*b^2-3*
B*a*b+5*C*a^2-2*C*b^2)*(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))*c
os(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/b^2/(a^2-b^2)/d-(3*A*b^4+3*a^3*b*B-5*a*b^3*B-a^2*b^2*(A-7*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-b)/b^3/(a+b)^2/d

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Rubi [A]  time = 1.37, antiderivative size = 447, 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 = {4098, 4102, 4106, 3849, 2805, 3787, 3771, 2639, 2641} \[ -\frac {\sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}+\frac {\sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \left (5 a^2 C-3 a b B+3 A b^2-2 b^2 C\right )}{3 b^2 d \left (a^2-b^2\right )}+\frac {\sin (c+d x) \sqrt {\sec (c+d x)} \left (3 a^2 b B-5 a^3 C-a b^2 (A-4 C)-2 b^3 B\right )}{b^3 d \left (a^2-b^2\right )}+\frac {\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (5 a^2 C-3 a b B+3 A b^2-2 b^2 C\right )}{3 b^2 d \left (a^2-b^2\right )}-\frac {\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (3 a^2 b B-5 a^3 C-a b^2 (A-4 C)-2 b^3 B\right )}{b^3 d \left (a^2-b^2\right )}-\frac {\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-a^2 b^2 (A-7 C)+3 a^3 b B-5 a^4 C-5 a b^3 B+3 A b^4\right ) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{b^3 d (a-b) (a+b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((3*a^2*b*B - 2*b^3*B - a*b^2*(A - 4*C) - 5*a^3*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c +
d*x]])/(b^3*(a^2 - b^2)*d)) + ((3*A*b^2 - 3*a*b*B + 5*a^2*C - 2*b^2*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)
/2, 2]*Sqrt[Sec[c + d*x]])/(3*b^2*(a^2 - b^2)*d) - ((3*A*b^4 + 3*a^3*b*B - 5*a*b^3*B - a^2*b^2*(A - 7*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 - b)*b^3*(a + b)^2*
d) + ((3*a^2*b*B - 2*b^3*B - a*b^2*(A - 4*C) - 5*a^3*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(b^3*(a^2 - b^2)*d) +
((3*A*b^2 - 3*a*b*B + 5*a^2*C - 2*b^2*C)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(3*b^2*(a^2 - b^2)*d) - ((A*b^2 - a
*(b*B - a*C))*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(b*(a^2 - b^2)*d*(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 4098

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

Rule 4102

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*d*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m
+ 1)*(d*Csc[e + f*x])^(n - 1))/(b*f*(m + n + 1)), x] + Dist[d/(b*(m + n + 1)), Int[(a + b*Csc[e + f*x])^m*(d*
Csc[e + f*x])^(n - 1)*Simp[a*C*(n - 1) + (A*b*(m + n + 1) + b*C*(m + n))*Csc[e + f*x] + (b*B*(m + n + 1) - a*C
*n)*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] && GtQ[n, 0]

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 {\sec ^{\frac {5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx &=-\frac {\left (A b^2-a (b B-a C)\right ) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac {\int \frac {\sec ^{\frac {3}{2}}(c+d x) \left (\frac {3}{2} \left (A b^2-a (b B-a C)\right )+b (b B-a (A+C)) \sec (c+d x)-\frac {1}{2} \left (3 A b^2-3 a b B+5 a^2 C-2 b^2 C\right ) \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{b \left (a^2-b^2\right )}\\ &=\frac {\left (3 A b^2-3 a b B+5 a^2 C-2 b^2 C\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d}-\frac {\left (A b^2-a (b B-a C)\right ) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac {2 \int \frac {\sqrt {\sec (c+d x)} \left (-\frac {1}{4} a \left (3 A b^2-3 a b B+5 a^2 C-2 b^2 C\right )+\frac {1}{2} b \left (3 A b^2-3 a b B+2 a^2 C+b^2 C\right ) \sec (c+d x)-\frac {3}{4} \left (3 a^2 b B-2 b^3 B-a b^2 (A-4 C)-5 a^3 C\right ) \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{3 b^2 \left (a^2-b^2\right )}\\ &=\frac {\left (3 a^2 b B-2 b^3 B-a b^2 (A-4 C)-5 a^3 C\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{b^3 \left (a^2-b^2\right ) d}+\frac {\left (3 A b^2-3 a b B+5 a^2 C-2 b^2 C\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d}-\frac {\left (A b^2-a (b B-a C)\right ) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac {4 \int \frac {\frac {3}{8} a \left (3 a^2 b B-2 b^3 B-a b^2 (A-4 C)-5 a^3 C\right )+\frac {1}{4} b \left (6 a^2 b B-3 b^3 B-a b^2 (3 A-7 C)-10 a^3 C\right ) \sec (c+d x)+\frac {1}{8} \left (9 a^3 b B-12 a b^3 B-a^2 b^2 (3 A-16 C)-15 a^4 C+2 b^4 (3 A+C)\right ) \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (a+b \sec (c+d x))} \, dx}{3 b^3 \left (a^2-b^2\right )}\\ &=\frac {\left (3 a^2 b B-2 b^3 B-a b^2 (A-4 C)-5 a^3 C\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{b^3 \left (a^2-b^2\right ) d}+\frac {\left (3 A b^2-3 a b B+5 a^2 C-2 b^2 C\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d}-\frac {\left (A b^2-a (b B-a C)\right ) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac {4 \int \frac {\frac {3}{8} a^2 \left (3 a^2 b B-2 b^3 B-a b^2 (A-4 C)-5 a^3 C\right )-\left (-\frac {1}{4} a b \left (6 a^2 b B-3 b^3 B-a b^2 (3 A-7 C)-10 a^3 C\right )+\frac {3}{8} a b \left (3 a^2 b B-2 b^3 B-a b^2 (A-4 C)-5 a^3 C\right )\right ) \sec (c+d x)}{\sqrt {\sec (c+d x)}} \, dx}{3 a^2 b^3 \left (a^2-b^2\right )}-\frac {\left (3 A b^4+3 a^3 b B-5 a b^3 B-a^2 b^2 (A-7 C)-5 a^4 C\right ) \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{a+b \sec (c+d x)} \, dx}{2 b^3 \left (a^2-b^2\right )}\\ &=\frac {\left (3 a^2 b B-2 b^3 B-a b^2 (A-4 C)-5 a^3 C\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{b^3 \left (a^2-b^2\right ) d}+\frac {\left (3 A b^2-3 a b B+5 a^2 C-2 b^2 C\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d}-\frac {\left (A b^2-a (b B-a C)\right ) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac {\left (3 a^2 b B-2 b^3 B-a b^2 (A-4 C)-5 a^3 C\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx}{2 b^3 \left (a^2-b^2\right )}+\frac {\left (3 A b^2-3 a b B+5 a^2 C-2 b^2 C\right ) \int \sqrt {\sec (c+d x)} \, dx}{6 b^2 \left (a^2-b^2\right )}-\frac {\left (\left (3 A b^4+3 a^3 b B-5 a b^3 B-a^2 b^2 (A-7 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 b^3 \left (a^2-b^2\right )}\\ &=-\frac {\left (3 A b^4+3 a^3 b B-5 a b^3 B-a^2 b^2 (A-7 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-b) b^3 (a+b)^2 d}+\frac {\left (3 a^2 b B-2 b^3 B-a b^2 (A-4 C)-5 a^3 C\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{b^3 \left (a^2-b^2\right ) d}+\frac {\left (3 A b^2-3 a b B+5 a^2 C-2 b^2 C\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d}-\frac {\left (A b^2-a (b B-a C)\right ) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac {\left (\left (3 a^2 b B-2 b^3 B-a b^2 (A-4 C)-5 a^3 C\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{2 b^3 \left (a^2-b^2\right )}+\frac {\left (\left (3 A b^2-3 a b B+5 a^2 C-2 b^2 C\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{6 b^2 \left (a^2-b^2\right )}\\ &=-\frac {\left (3 a^2 b B-2 b^3 B-a b^2 (A-4 C)-5 a^3 C\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{b^3 \left (a^2-b^2\right ) d}+\frac {\left (3 A b^2-3 a b B+5 a^2 C-2 b^2 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 b^2 \left (a^2-b^2\right ) d}-\frac {\left (3 A b^4+3 a^3 b B-5 a b^3 B-a^2 b^2 (A-7 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-b) b^3 (a+b)^2 d}+\frac {\left (3 a^2 b B-2 b^3 B-a b^2 (A-4 C)-5 a^3 C\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{b^3 \left (a^2-b^2\right ) d}+\frac {\left (3 A b^2-3 a b B+5 a^2 C-2 b^2 C\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d}-\frac {\left (A b^2-a (b B-a C)\right ) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))}\\ \end {align*}

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Mathematica [B]  time = 7.46, size = 926, normalized size = 2.07 \[ \frac {\left (C \sec ^2(c+d x)+B \sec (c+d x)+A\right ) \left (\frac {2 \left (45 C a^4-27 b B a^3+9 A b^2 a^2-44 b^2 C a^2+30 b^3 B a-12 A b^4-4 b^4 C\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 (12 B b^4+12 a A b^3-28 a C b^3-24 a^2 B b^2+40 a^3 C b\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 (15 C a^4-9 b B a^3+3 A b^2 a^2-12 b^2 C a^2+6 b^3 B a\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}{6 (a-b) b^3 (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 {2 \left (5 C a^3-3 b B a^2+A b^2 a-4 b^2 C a+2 b^3 B\right ) \sin (c+d x)}{b^3 \left (b^2-a^2\right )}-\frac {2 \left (C \sin (c+d x) a^3-b B \sin (c+d x) a^2+A b^2 \sin (c+d x) a\right )}{b^2 \left (b^2-a^2\right ) (b+a \cos (c+d x))}+\frac {4 C \tan (c+d x)}{3 b^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[(Sec[c + d*x]^(5/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^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*(9*a^2*A*b^2 - 12*A*b^4 - 27*a^3*b*B + 30*
a*b^3*B + 45*a^4*C - 44*a^2*b^2*C - 4*b^4*C)*Cos[c + d*x]^2*(EllipticF[ArcSin[Sqrt[Sec[c + d*x]]], -1] - Ellip
ticPi[-(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*(12*a*A*b^3 - 24*a^2*b^2*B + 12*b^4*B + 40*a^3*b*C - 28*a*b^3*
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)) + ((3*a^2*A*b^2 - 9*a^3*b*B + 6*a*b^3*B + 15*
a^4*C - 12*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[A
rcSin[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[Sq
rt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]^2] + 2*a^2*EllipticPi[-(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))))/(6*(a - b)*b^3*(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*Se
c[c + d*x]^2)*((2*(a*A*b^2 - 3*a^2*b*B + 2*b^3*B + 5*a^3*C - 4*a*b^2*C)*Sin[c + d*x])/(b^3*(-a^2 + b^2)) - (2*
(a*A*b^2*Sin[c + d*x] - a^2*b*B*Sin[c + d*x] + a^3*C*Sin[c + d*x]))/(b^2*(-a^2 + b^2)*(b + a*Cos[c + d*x])) +
(4*C*Tan[c + d*x])/(3*b^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)

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maple [B]  time = 22.40, size = 1031, normalized size = 2.31 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^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)*(2*a^2*(B*b-2*C*a)/b^3/(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(c
os(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2))+2*(B*b-2*C*a)/b^3*(-(-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)^(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))+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)+2*C/b^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)/
(-1/2+cos(1/2*d*x+1/2*c)^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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)))+2*(A*b^2-B*a*b+C*a^2)/b^2*(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

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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