\(\int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^{5/2}} \, dx\) [1373]

Optimal result

Integrand size = 45, antiderivative size = 563 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^{5/2}} \, dx=\frac {\left (2 A b^2-2 a b B+5 a^2 C-3 b^2 C\right ) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right )}{3 b^2 \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {(2 b B-5 a C) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right )}{b^3 d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {\left (8 A b^4+6 a^3 b B-14 a b^3 B-15 a^4 C+26 a^2 b^2 C-3 b^4 C\right ) \sqrt {\cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right ) \sqrt {a+b \sec (c+d x)}}{3 b^3 \left (a^2-b^2\right )^2 d \sqrt {\frac {b+a \cos (c+d x)}{a+b}}}-\frac {2 \left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^{3/2}}+\frac {2 \left (3 A b^4+2 a^3 b B-6 a b^3 B-5 a^4 C+a^2 b^2 (A+9 C)\right ) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right )^2 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}}-\frac {\left (8 A b^4+6 a^3 b B-14 a b^3 B-15 a^4 C+26 a^2 b^2 C-3 b^4 C\right ) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{3 b^3 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)}} \]

[Out]

-2/3*(A*b^2-a*(B*b-C*a))*sin(d*x+c)/b/(a^2-b^2)/d/cos(d*x+c)^(5/2)/(a+b*sec(d*x+c))^(3/2)+2/3*(3*A*b^4+2*B*a^3
*b-6*B*a*b^3-5*a^4*C+a^2*b^2*(A+9*C))*sin(d*x+c)/b^2/(a^2-b^2)^2/d/cos(d*x+c)^(3/2)/(a+b*sec(d*x+c))^(1/2)+1/3
*(2*A*b^2-2*B*a*b+5*C*a^2-3*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)*(a/(a+b))^(1/2))*((b+a*cos(d*x+c))/(a+b))^(1/2)/b^2/(a^2-b^2)/d/cos(d*x+c)^(1/2)/(a+b*sec(d*x+c))^(1
/2)+(2*B*b-5*C*a)*(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),2,2^(1/2)*(a/(
a+b))^(1/2))*((b+a*cos(d*x+c))/(a+b))^(1/2)/b^3/d/cos(d*x+c)^(1/2)/(a+b*sec(d*x+c))^(1/2)-1/3*(8*A*b^4+6*B*a^3
*b-14*B*a*b^3-15*C*a^4+26*C*a^2*b^2-3*C*b^4)*sin(d*x+c)*(a+b*sec(d*x+c))^(1/2)/b^3/(a^2-b^2)^2/d/cos(d*x+c)^(1
/2)+1/3*(8*A*b^4+6*B*a^3*b-14*B*a*b^3-15*C*a^4+26*C*a^2*b^2-3*C*b^4)*(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)*(a/(a+b))^(1/2))*cos(d*x+c)^(1/2)*(a+b*sec(d*x+c))^(1/2)/b^3/(a^2-
b^2)^2/d/((b+a*cos(d*x+c))/(a+b))^(1/2)

Rubi [A] (verified)

Time = 2.49 (sec) , antiderivative size = 563, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.311, Rules used = {4350, 4183, 4187, 4193, 3944, 2886, 2884, 4120, 3941, 2734, 2732, 3943, 2742, 2740} \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^{5/2}} \, dx=-\frac {2 \sin (c+d x) \left (A b^2-a (b B-a C)\right )}{3 b d \left (a^2-b^2\right ) \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^{3/2}}+\frac {\left (5 a^2 C-2 a b B+2 A b^2-3 b^2 C\right ) \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right )}{3 b^2 d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {2 \sin (c+d x) \left (-5 a^4 C+2 a^3 b B+a^2 b^2 (A+9 C)-6 a b^3 B+3 A b^4\right )}{3 b^2 d \left (a^2-b^2\right )^2 \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}}-\frac {\sin (c+d x) \left (-15 a^4 C+6 a^3 b B+26 a^2 b^2 C-14 a b^3 B+8 A b^4-3 b^4 C\right ) \sqrt {a+b \sec (c+d x)}}{3 b^3 d \left (a^2-b^2\right )^2 \sqrt {\cos (c+d x)}}+\frac {\sqrt {\cos (c+d x)} \left (-15 a^4 C+6 a^3 b B+26 a^2 b^2 C-14 a b^3 B+8 A b^4-3 b^4 C\right ) \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{3 b^3 d \left (a^2-b^2\right )^2 \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}+\frac {(2 b B-5 a C) \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right )}{b^3 d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}} \]

[In]

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

[Out]

((2*A*b^2 - 2*a*b*B + 5*a^2*C - 3*b^2*C)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*a)/(a +
b)])/(3*b^2*(a^2 - b^2)*d*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Sec[c + d*x]]) + ((2*b*B - 5*a*C)*Sqrt[(b + a*Cos[c +
d*x])/(a + b)]*EllipticPi[2, (c + d*x)/2, (2*a)/(a + b)])/(b^3*d*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Sec[c + d*x]])
+ ((8*A*b^4 + 6*a^3*b*B - 14*a*b^3*B - 15*a^4*C + 26*a^2*b^2*C - 3*b^4*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*
x)/2, (2*a)/(a + b)]*Sqrt[a + b*Sec[c + d*x]])/(3*b^3*(a^2 - b^2)^2*d*Sqrt[(b + a*Cos[c + d*x])/(a + b)]) - (2
*(A*b^2 - a*(b*B - a*C))*Sin[c + d*x])/(3*b*(a^2 - b^2)*d*Cos[c + d*x]^(5/2)*(a + b*Sec[c + d*x])^(3/2)) + (2*
(3*A*b^4 + 2*a^3*b*B - 6*a*b^3*B - 5*a^4*C + a^2*b^2*(A + 9*C))*Sin[c + d*x])/(3*b^2*(a^2 - b^2)^2*d*Cos[c + d
*x]^(3/2)*Sqrt[a + b*Sec[c + d*x]]) - ((8*A*b^4 + 6*a^3*b*B - 14*a*b^3*B - 15*a^4*C + 26*a^2*b^2*C - 3*b^4*C)*
Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(3*b^3*(a^2 - b^2)^2*d*Sqrt[Cos[c + d*x]])

Rule 2732

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

Rule 2734

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[a + b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c +
d*x])/(a + b)], Int[Sqrt[a/(a + b) + (b/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
b^2, 0] &&  !GtQ[a + b, 0]

Rule 2740

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

Rule 2742

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[(a + b*Sin[c + d*x])/(a + b)]/Sqrt[a
+ b*Sin[c + d*x]], Int[1/Sqrt[a/(a + b) + (b/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] &&  !GtQ[a + b, 0]

Rule 2884

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2/(f*(a + b)*Sqrt[c + d]))*EllipticPi[2*(b/(a + b)), (1/2)*(e - Pi/2 + f*x), 2*(d/(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 2886

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

Rule 3941

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

Rule 3943

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

Rule 3944

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

Rule 4120

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

Rule 4183

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 4187

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 4193

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

Rule 4350

Int[(cos[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Dist[(c*Cos[a + b*x])^m*(c*Sec[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Sec[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSecantIntegrandQ[
u, x]

Rubi steps \begin{align*} \text {integral}& = \left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \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))^{5/2}} \, dx \\ & = -\frac {2 \left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^{3/2}}-\frac {\left (2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sec ^{\frac {3}{2}}(c+d x) \left (\frac {3}{2} \left (A b^2-a (b B-a C)\right )+\frac {3}{2} b (b B-a (A+C)) \sec (c+d x)-\frac {1}{2} \left (2 A b^2-2 a b B+5 a^2 C-3 b^2 C\right ) \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx}{3 b \left (a^2-b^2\right )} \\ & = -\frac {2 \left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^{3/2}}+\frac {2 \left (3 A b^4+2 a^3 b B-6 a b^3 B-5 a^4 C+a^2 b^2 (A+9 C)\right ) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right )^2 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}}+\frac {\left (4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {\sec (c+d x)} \left (\frac {1}{4} \left (3 A b^4+2 a^3 b B-6 a b^3 B-5 a^4 C+a^2 b^2 (A+9 C)\right )+\frac {1}{4} b \left (a^2 b B+3 b^3 B+2 a^3 C-2 a b^2 (2 A+3 C)\right ) \sec (c+d x)-\frac {1}{4} \left (8 A b^4+6 a^3 b B-14 a b^3 B-15 a^4 C+26 a^2 b^2 C-3 b^4 C\right ) \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx}{3 b^2 \left (a^2-b^2\right )^2} \\ & = -\frac {2 \left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^{3/2}}+\frac {2 \left (3 A b^4+2 a^3 b B-6 a b^3 B-5 a^4 C+a^2 b^2 (A+9 C)\right ) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right )^2 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}}-\frac {\left (8 A b^4+6 a^3 b B-14 a b^3 B-15 a^4 C+26 a^2 b^2 C-3 b^4 C\right ) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{3 b^3 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)}}+\frac {\left (4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {1}{8} a \left (8 A b^4+6 a^3 b B-14 a b^3 B-15 a^4 C+26 a^2 b^2 C-3 b^4 C\right )+\frac {1}{4} b \left (3 A b^4+2 a^3 b B-6 a b^3 B-5 a^4 C+a^2 b^2 (A+9 C)\right ) \sec (c+d x)+\frac {3}{8} \left (a^2-b^2\right )^2 (2 b B-5 a C) \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}} \, dx}{3 b^3 \left (a^2-b^2\right )^2} \\ & = -\frac {2 \left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^{3/2}}+\frac {2 \left (3 A b^4+2 a^3 b B-6 a b^3 B-5 a^4 C+a^2 b^2 (A+9 C)\right ) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right )^2 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}}-\frac {\left (8 A b^4+6 a^3 b B-14 a b^3 B-15 a^4 C+26 a^2 b^2 C-3 b^4 C\right ) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{3 b^3 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)}}+\frac {\left (4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {1}{8} a \left (8 A b^4+6 a^3 b B-14 a b^3 B-15 a^4 C+26 a^2 b^2 C-3 b^4 C\right )+\frac {1}{4} b \left (3 A b^4+2 a^3 b B-6 a b^3 B-5 a^4 C+a^2 b^2 (A+9 C)\right ) \sec (c+d x)}{\sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}} \, dx}{3 b^3 \left (a^2-b^2\right )^2}+\frac {\left ((2 b B-5 a C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{2 b^3} \\ & = -\frac {2 \left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^{3/2}}+\frac {2 \left (3 A b^4+2 a^3 b B-6 a b^3 B-5 a^4 C+a^2 b^2 (A+9 C)\right ) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right )^2 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}}-\frac {\left (8 A b^4+6 a^3 b B-14 a b^3 B-15 a^4 C+26 a^2 b^2 C-3 b^4 C\right ) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{3 b^3 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)}}+\frac {\left (\left (2 A b^2-2 a b B+5 a^2 C-3 b^2 C\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {a+b \sec (c+d x)}} \, dx}{6 b^2 \left (a^2-b^2\right )}+\frac {\left (\left (8 A b^4+6 a^3 b B-14 a b^3 B-15 a^4 C+26 a^2 b^2 C-3 b^4 C\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {\sec (c+d x)}} \, dx}{6 b^3 \left (a^2-b^2\right )^2}+\frac {\left ((2 b B-5 a C) \sqrt {b+a \cos (c+d x)}\right ) \int \frac {\sec (c+d x)}{\sqrt {b+a \cos (c+d x)}} \, dx}{2 b^3 \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}} \\ & = -\frac {2 \left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^{3/2}}+\frac {2 \left (3 A b^4+2 a^3 b B-6 a b^3 B-5 a^4 C+a^2 b^2 (A+9 C)\right ) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right )^2 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}}-\frac {\left (8 A b^4+6 a^3 b B-14 a b^3 B-15 a^4 C+26 a^2 b^2 C-3 b^4 C\right ) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{3 b^3 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)}}+\frac {\left (\left (2 A b^2-2 a b B+5 a^2 C-3 b^2 C\right ) \sqrt {b+a \cos (c+d x)}\right ) \int \frac {1}{\sqrt {b+a \cos (c+d x)}} \, dx}{6 b^2 \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {\left ((2 b B-5 a C) \sqrt {\frac {b+a \cos (c+d x)}{a+b}}\right ) \int \frac {\sec (c+d x)}{\sqrt {\frac {b}{a+b}+\frac {a \cos (c+d x)}{a+b}}} \, dx}{2 b^3 \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {\left (\left (8 A b^4+6 a^3 b B-14 a b^3 B-15 a^4 C+26 a^2 b^2 C-3 b^4 C\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}\right ) \int \sqrt {b+a \cos (c+d x)} \, dx}{6 b^3 \left (a^2-b^2\right )^2 \sqrt {b+a \cos (c+d x)}} \\ & = \frac {(2 b B-5 a C) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right )}{b^3 d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}-\frac {2 \left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^{3/2}}+\frac {2 \left (3 A b^4+2 a^3 b B-6 a b^3 B-5 a^4 C+a^2 b^2 (A+9 C)\right ) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right )^2 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}}-\frac {\left (8 A b^4+6 a^3 b B-14 a b^3 B-15 a^4 C+26 a^2 b^2 C-3 b^4 C\right ) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{3 b^3 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)}}+\frac {\left (\left (2 A b^2-2 a b B+5 a^2 C-3 b^2 C\right ) \sqrt {\frac {b+a \cos (c+d x)}{a+b}}\right ) \int \frac {1}{\sqrt {\frac {b}{a+b}+\frac {a \cos (c+d x)}{a+b}}} \, dx}{6 b^2 \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {\left (\left (8 A b^4+6 a^3 b B-14 a b^3 B-15 a^4 C+26 a^2 b^2 C-3 b^4 C\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}\right ) \int \sqrt {\frac {b}{a+b}+\frac {a \cos (c+d x)}{a+b}} \, dx}{6 b^3 \left (a^2-b^2\right )^2 \sqrt {\frac {b+a \cos (c+d x)}{a+b}}} \\ & = \frac {\left (2 A b^2-2 a b B+5 a^2 C-3 b^2 C\right ) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right )}{3 b^2 \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {(2 b B-5 a C) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right )}{b^3 d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {\left (8 A b^4+6 a^3 b B-14 a b^3 B-15 a^4 C+26 a^2 b^2 C-3 b^4 C\right ) \sqrt {\cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right ) \sqrt {a+b \sec (c+d x)}}{3 b^3 \left (a^2-b^2\right )^2 d \sqrt {\frac {b+a \cos (c+d x)}{a+b}}}-\frac {2 \left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^{3/2}}+\frac {2 \left (3 A b^4+2 a^3 b B-6 a b^3 B-5 a^4 C+a^2 b^2 (A+9 C)\right ) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right )^2 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}}-\frac {\left (8 A b^4+6 a^3 b B-14 a b^3 B-15 a^4 C+26 a^2 b^2 C-3 b^4 C\right ) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{3 b^3 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)}} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 42.28 (sec) , antiderivative size = 290125, normalized size of antiderivative = 515.32 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^{5/2}} \, dx=\text {Result too large to show} \]

[In]

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

[Out]

Result too large to show

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 13.98 (sec) , antiderivative size = 8475, normalized size of antiderivative = 15.05

method	result	size
default	\(\text {Expression too large to display}\)	\(8475\)

[In]

int((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+b*sec(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

result too large to display

Fricas [F(-1)]

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

[In]

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

[Out]

Timed out

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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