\(\int \frac {A+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^{5/2} \sqrt {\sec (c+d x)}} \, dx\) [1449]

Optimal result

Integrand size = 37, antiderivative size = 618 \[ \int \frac {A+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^{5/2} \sqrt {\sec (c+d x)}} \, dx=-\frac {2 \left (A b^4-3 a^4 C+a^2 b^2 (3 A+7 C)\right ) \sqrt {\cos (c+d x)} \csc (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{3 a^2 b^2 \sqrt {a+b} \left (a^2-b^2\right ) d \sqrt {\sec (c+d x)}}-\frac {2 \left (A b^3+3 a^3 C+a^2 b C-3 a b^2 (A+2 C)\right ) \sqrt {\cos (c+d x)} \csc (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{3 a (a-b) b^2 (a+b)^{3/2} d \sqrt {\sec (c+d x)}}-\frac {2 \sqrt {a+b} C \sqrt {\cos (c+d x)} \csc (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{b^3 d \sqrt {\sec (c+d x)}}-\frac {2 \left (A b^2+a^2 C\right ) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2} \sqrt {\sec (c+d x)}}+\frac {2 \left (A b^4-3 a^4 C+a^2 b^2 (3 A+7 C)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 b^2 \left (a^2-b^2\right )^2 d \sqrt {a+b \cos (c+d x)}} \]

[Out]

-2/3*(A*b^2+C*a^2)*sin(d*x+c)/b/(a^2-b^2)/d/(a+b*cos(d*x+c))^(3/2)/sec(d*x+c)^(1/2)+2/3*(A*b^4-3*a^4*C+a^2*b^2
*(3*A+7*C))*sin(d*x+c)*sec(d*x+c)^(1/2)/b^2/(a^2-b^2)^2/d/(a+b*cos(d*x+c))^(1/2)-2/3*(A*b^3+3*a^3*C+a^2*b*C-3*
a*b^2*(A+2*C))*csc(d*x+c)*EllipticF((a+b*cos(d*x+c))^(1/2)/(a+b)^(1/2)/cos(d*x+c)^(1/2),((-a-b)/(a-b))^(1/2))*
cos(d*x+c)^(1/2)*(a*(1-sec(d*x+c))/(a+b))^(1/2)*(a*(1+sec(d*x+c))/(a-b))^(1/2)/a/(a-b)/b^2/(a+b)^(3/2)/d/sec(d
*x+c)^(1/2)-2/3*(A*b^4-3*a^4*C+a^2*b^2*(3*A+7*C))*csc(d*x+c)*EllipticE((a+b*cos(d*x+c))^(1/2)/(a+b)^(1/2)/cos(
d*x+c)^(1/2),((-a-b)/(a-b))^(1/2))*cos(d*x+c)^(1/2)*(a*(1-sec(d*x+c))/(a+b))^(1/2)*(a*(1+sec(d*x+c))/(a-b))^(1
/2)/a^2/b^2/(a^2-b^2)/d/(a+b)^(1/2)/sec(d*x+c)^(1/2)-2*C*csc(d*x+c)*EllipticPi((a+b*cos(d*x+c))^(1/2)/(a+b)^(1
/2)/cos(d*x+c)^(1/2),(a+b)/b,((-a-b)/(a-b))^(1/2))*(a+b)^(1/2)*cos(d*x+c)^(1/2)*(a*(1-sec(d*x+c))/(a+b))^(1/2)
*(a*(1+sec(d*x+c))/(a-b))^(1/2)/b^3/d/sec(d*x+c)^(1/2)

Rubi [A] (verified)

Time = 1.92 (sec) , antiderivative size = 618, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.216, Rules used = {4306, 3127, 3130, 2888, 3072, 3077, 2895, 3073} \[ \int \frac {A+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^{5/2} \sqrt {\sec (c+d x)}} \, dx=-\frac {2 \left (a^2 C+A b^2\right ) \sin (c+d x)}{3 b d \left (a^2-b^2\right ) \sqrt {\sec (c+d x)} (a+b \cos (c+d x))^{3/2}}-\frac {2 \left (-3 a^4 C+a^2 b^2 (3 A+7 C)+A b^4\right ) \sqrt {\cos (c+d x)} \csc (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{3 a^2 b^2 d \sqrt {a+b} \left (a^2-b^2\right ) \sqrt {\sec (c+d x)}}+\frac {2 \left (-3 a^4 C+a^2 b^2 (3 A+7 C)+A b^4\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 b^2 d \left (a^2-b^2\right )^2 \sqrt {a+b \cos (c+d x)}}-\frac {2 \left (3 a^3 C+a^2 b C-3 a b^2 (A+2 C)+A b^3\right ) \sqrt {\cos (c+d x)} \csc (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{3 a b^2 d (a-b) (a+b)^{3/2} \sqrt {\sec (c+d x)}}-\frac {2 C \sqrt {a+b} \sqrt {\cos (c+d x)} \csc (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{b^3 d \sqrt {\sec (c+d x)}} \]

[In]

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

[Out]

(-2*(A*b^4 - 3*a^4*C + a^2*b^2*(3*A + 7*C))*Sqrt[Cos[c + d*x]]*Csc[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Cos[c
+ d*x]]/(Sqrt[a + b]*Sqrt[Cos[c + d*x]])], -((a + b)/(a - b))]*Sqrt[(a*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[(a*(1
+ Sec[c + d*x]))/(a - b)])/(3*a^2*b^2*Sqrt[a + b]*(a^2 - b^2)*d*Sqrt[Sec[c + d*x]]) - (2*(A*b^3 + 3*a^3*C + a
^2*b*C - 3*a*b^2*(A + 2*C))*Sqrt[Cos[c + d*x]]*Csc[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Cos[c + d*x]]/(Sqrt[a
+ b]*Sqrt[Cos[c + d*x]])], -((a + b)/(a - b))]*Sqrt[(a*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[(a*(1 + Sec[c + d*x])
)/(a - b)])/(3*a*(a - b)*b^2*(a + b)^(3/2)*d*Sqrt[Sec[c + d*x]]) - (2*Sqrt[a + b]*C*Sqrt[Cos[c + d*x]]*Csc[c +
d*x]*EllipticPi[(a + b)/b, ArcSin[Sqrt[a + b*Cos[c + d*x]]/(Sqrt[a + b]*Sqrt[Cos[c + d*x]])], -((a + b)/(a -
b))]*Sqrt[(a*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[(a*(1 + Sec[c + d*x]))/(a - b)])/(b^3*d*Sqrt[Sec[c + d*x]]) - (
2*(A*b^2 + a^2*C)*Sin[c + d*x])/(3*b*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^(3/2)*Sqrt[Sec[c + d*x]]) + (2*(A*b^4
- 3*a^4*C + a^2*b^2*(3*A + 7*C))*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(3*b^2*(a^2 - b^2)^2*d*Sqrt[a + b*Cos[c + d*
x]])

Rule 2888

Int[Sqrt[(b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Simp[2*b*(Tan
[e + f*x]/(d*f))*Rt[(c + d)/b, 2]*Sqrt[c*((1 + Csc[e + f*x])/(c - d))]*Sqrt[c*((1 - Csc[e + f*x])/(c + d))]*El
lipticPi[(c + d)/d, ArcSin[Sqrt[c + d*Sin[e + f*x]]/Sqrt[b*Sin[e + f*x]]/Rt[(c + d)/b, 2]], -(c + d)/(c - d)],
x] /; FreeQ[{b, c, d, e, f}, x] && NeQ[c^2 - d^2, 0] && PosQ[(c + d)/b]

Rule 2895

Int[1/(Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[-2*(
Tan[e + f*x]/(a*f))*Rt[(a + b)/d, 2]*Sqrt[a*((1 - Csc[e + f*x])/(a + b))]*Sqrt[a*((1 + Csc[e + f*x])/(a - b))]
*EllipticF[ArcSin[Sqrt[a + b*Sin[e + f*x]]/Sqrt[d*Sin[e + f*x]]/Rt[(a + b)/d, 2]], -(a + b)/(a - b)], x] /; Fr
eeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && PosQ[(a + b)/d]

Rule 3072

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]]*((a_) + (b_.)*sin[(e_.) + (f_
.)*(x_)])^(3/2)), x_Symbol] :> Simp[2*(A*b - a*B)*(Cos[e + f*x]/(f*(a^2 - b^2)*Sqrt[a + b*Sin[e + f*x]]*Sqrt[d
*Sin[e + f*x]])), x] + Dist[d/(a^2 - b^2), Int[(A*b - a*B + (a*A - b*B)*Sin[e + f*x])/(Sqrt[a + b*Sin[e + f*x]
]*(d*Sin[e + f*x])^(3/2)), x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[a^2 - b^2, 0]

Rule 3073

Int[((A_) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((b_.)*sin[(e_.) + (f_.)*(x_)])^(3/2)*Sqrt[(c_) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]]), x_Symbol] :> Simp[-2*A*(c - d)*(Tan[e + f*x]/(f*b*c^2))*Rt[(c + d)/b, 2]*Sqrt[c*((1 + Csc[e +
f*x])/(c - d))]*Sqrt[c*((1 - Csc[e + f*x])/(c + d))]*EllipticE[ArcSin[Sqrt[c + d*Sin[e + f*x]]/Sqrt[b*Sin[e +
f*x]]/Rt[(c + d)/b, 2]], -(c + d)/(c - d)], x] /; FreeQ[{b, c, d, e, f, A, B}, x] && NeQ[c^2 - d^2, 0] && EqQ
[A, B] && PosQ[(c + d)/b]

Rule 3077

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(3/2)*Sqrt[(c_) + (d_.)*s
in[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist[(A - B)/(a - b), Int[1/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e
+ f*x]]), x], x] - Dist[(A*b - a*B)/(a - b), Int[(1 + Sin[e + f*x])/((a + b*Sin[e + f*x])^(3/2)*Sqrt[c + d*Sin
[e + f*x]]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2
- d^2, 0] && NeQ[A, B]

Rule 3127

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(c^2*C + A*d^2))*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Si
n[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 - d^2))), x] + Dist[1/(d*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^
(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(b*d*m + a*c*(n + 1)) + c*C*(b*c*m + a*d*(n + 1)) - (A*d*(a*d*(n
+ 2) - b*c*(n + 1)) - C*(b*c*d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x] - b*(A*d^2*(m + n + 2) + C*(c^2
*(m + 1) + d^2*(n + 1)))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C}, x] && NeQ[b*c - a*d, 0]
&& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 0] && LtQ[n, -1]

Rule 3130

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

Rule 4306

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

Rubi steps \begin{align*} \text {integral}& = \left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {\cos (c+d x)} \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^{5/2}} \, dx \\ & = -\frac {2 \left (A b^2+a^2 C\right ) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2} \sqrt {\sec (c+d x)}}-\frac {\left (2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {1}{2} \left (A b^2+a^2 C\right )-\frac {3}{2} a b (A+C) \cos (c+d x)-\frac {3}{2} \left (a^2-b^2\right ) C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))^{3/2}} \, dx}{3 b \left (a^2-b^2\right )} \\ & = -\frac {2 \left (A b^2+a^2 C\right ) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2} \sqrt {\sec (c+d x)}}-\frac {\left (2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {1}{2} b \left (A b^2+a^2 C\right )+\left (\frac {3}{2} a \left (a^2-b^2\right ) C-\frac {3}{2} a b^2 (A+C)\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))^{3/2}} \, dx}{3 b^2 \left (a^2-b^2\right )}+\frac {\left (C \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {\cos (c+d x)}}{\sqrt {a+b \cos (c+d x)}} \, dx}{b^2} \\ & = -\frac {2 \sqrt {a+b} C \sqrt {\cos (c+d x)} \csc (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{b^3 d \sqrt {\sec (c+d x)}}-\frac {2 \left (A b^2+a^2 C\right ) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2} \sqrt {\sec (c+d x)}}+\frac {2 \left (A b^4-3 a^4 C+a^2 b^2 (3 A+7 C)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 b^2 \left (a^2-b^2\right )^2 d \sqrt {a+b \cos (c+d x)}}-\frac {\left (2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {1}{2} b^2 \left (A b^2+a^2 C\right )-a \left (\frac {3}{2} a \left (a^2-b^2\right ) C-\frac {3}{2} a b^2 (A+C)\right )+\left (\frac {1}{2} a b \left (A b^2+a^2 C\right )-b \left (\frac {3}{2} a \left (a^2-b^2\right ) C-\frac {3}{2} a b^2 (A+C)\right )\right ) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}} \, dx}{3 b^2 \left (a^2-b^2\right )^2} \\ & = -\frac {2 \sqrt {a+b} C \sqrt {\cos (c+d x)} \csc (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{b^3 d \sqrt {\sec (c+d x)}}-\frac {2 \left (A b^2+a^2 C\right ) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2} \sqrt {\sec (c+d x)}}+\frac {2 \left (A b^4-3 a^4 C+a^2 b^2 (3 A+7 C)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 b^2 \left (a^2-b^2\right )^2 d \sqrt {a+b \cos (c+d x)}}-\frac {\left ((a-b) \left (A b^3+3 a^3 C+a^2 b C-3 a b^2 (A+2 C)\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}} \, dx}{3 b^2 \left (a^2-b^2\right )^2}+\frac {\left (\left (-A b^4+3 a^4 C-a^2 b^2 (3 A+7 C)\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1+\cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}} \, dx}{3 b^2 \left (a^2-b^2\right )^2} \\ & = -\frac {2 \left (A b^4-3 a^4 C+a^2 b^2 (3 A+7 C)\right ) \sqrt {\cos (c+d x)} \csc (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{3 a^2 (a-b) b^2 (a+b)^{3/2} d \sqrt {\sec (c+d x)}}-\frac {2 \left (A b^3+3 a^3 C+a^2 b C-3 a b^2 (A+2 C)\right ) \sqrt {\cos (c+d x)} \csc (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{3 a (a-b) b^2 (a+b)^{3/2} d \sqrt {\sec (c+d x)}}-\frac {2 \sqrt {a+b} C \sqrt {\cos (c+d x)} \csc (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{b^3 d \sqrt {\sec (c+d x)}}-\frac {2 \left (A b^2+a^2 C\right ) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2} \sqrt {\sec (c+d x)}}+\frac {2 \left (A b^4-3 a^4 C+a^2 b^2 (3 A+7 C)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 b^2 \left (a^2-b^2\right )^2 d \sqrt {a+b \cos (c+d x)}} \\ \end{align*}

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(1576\) vs. \(2(618)=1236\).

Time = 16.77 (sec) , antiderivative size = 1576, normalized size of antiderivative = 2.55 \[ \int \frac {A+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^{5/2} \sqrt {\sec (c+d x)}} \, dx=\frac {\sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \left (-3 a^2 A b^2-A b^4+3 a^4 C-7 a^2 b^2 C\right ) \sin (c+d x)}{3 a b^2 \left (a^2-b^2\right )^2}-\frac {2 \left (a A b^2 \sin (c+d x)+a^3 C \sin (c+d x)\right )}{3 b^2 \left (-a^2+b^2\right ) (a+b \cos (c+d x))^2}+\frac {4 \left (a^2 A b^2 \sin (c+d x)+A b^4 \sin (c+d x)-2 a^4 C \sin (c+d x)+4 a^2 b^2 C \sin (c+d x)\right )}{3 b^2 \left (-a^2+b^2\right )^2 (a+b \cos (c+d x))}\right )}{d}+\frac {2 \sqrt {\frac {1}{1-\tan ^2\left (\frac {1}{2} (c+d x)\right )}} \left (3 a^3 A b^2 \tan \left (\frac {1}{2} (c+d x)\right )+3 a^2 A b^3 \tan \left (\frac {1}{2} (c+d x)\right )+a A b^4 \tan \left (\frac {1}{2} (c+d x)\right )+A b^5 \tan \left (\frac {1}{2} (c+d x)\right )-3 a^5 C \tan \left (\frac {1}{2} (c+d x)\right )-3 a^4 b C \tan \left (\frac {1}{2} (c+d x)\right )+7 a^3 b^2 C \tan \left (\frac {1}{2} (c+d x)\right )+7 a^2 b^3 C \tan \left (\frac {1}{2} (c+d x)\right )-6 a^2 A b^3 \tan ^3\left (\frac {1}{2} (c+d x)\right )-2 A b^5 \tan ^3\left (\frac {1}{2} (c+d x)\right )+6 a^4 b C \tan ^3\left (\frac {1}{2} (c+d x)\right )-14 a^2 b^3 C \tan ^3\left (\frac {1}{2} (c+d x)\right )-3 a^3 A b^2 \tan ^5\left (\frac {1}{2} (c+d x)\right )+3 a^2 A b^3 \tan ^5\left (\frac {1}{2} (c+d x)\right )-a A b^4 \tan ^5\left (\frac {1}{2} (c+d x)\right )+A b^5 \tan ^5\left (\frac {1}{2} (c+d x)\right )+3 a^5 C \tan ^5\left (\frac {1}{2} (c+d x)\right )-3 a^4 b C \tan ^5\left (\frac {1}{2} (c+d x)\right )-7 a^3 b^2 C \tan ^5\left (\frac {1}{2} (c+d x)\right )+7 a^2 b^3 C \tan ^5\left (\frac {1}{2} (c+d x)\right )+6 a^5 C \operatorname {EllipticPi}\left (-1,\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {-a+b}{a+b}\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {\frac {a+b+a \tan ^2\left (\frac {1}{2} (c+d x)\right )-b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}-12 a^3 b^2 C \operatorname {EllipticPi}\left (-1,\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {-a+b}{a+b}\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {\frac {a+b+a \tan ^2\left (\frac {1}{2} (c+d x)\right )-b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}+6 a b^4 C \operatorname {EllipticPi}\left (-1,\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {-a+b}{a+b}\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {\frac {a+b+a \tan ^2\left (\frac {1}{2} (c+d x)\right )-b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}+6 a^5 C \operatorname {EllipticPi}\left (-1,\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {-a+b}{a+b}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {\frac {a+b+a \tan ^2\left (\frac {1}{2} (c+d x)\right )-b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}-12 a^3 b^2 C \operatorname {EllipticPi}\left (-1,\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {-a+b}{a+b}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {\frac {a+b+a \tan ^2\left (\frac {1}{2} (c+d x)\right )-b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}+6 a b^4 C \operatorname {EllipticPi}\left (-1,\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {-a+b}{a+b}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {\frac {a+b+a \tan ^2\left (\frac {1}{2} (c+d x)\right )-b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}-(a+b) \left (-A b^4+3 a^4 C-a^2 b^2 (3 A+7 C)\right ) E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {-a+b}{a+b}\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \sqrt {\frac {a+b+a \tan ^2\left (\frac {1}{2} (c+d x)\right )-b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}+a b (a+b) \left (2 a^2 C-3 a b (A+C)-b^2 (A+3 C)\right ) \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {-a+b}{a+b}\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \sqrt {\frac {a+b+a \tan ^2\left (\frac {1}{2} (c+d x)\right )-b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}\right )}{3 a b^2 \left (a^2-b^2\right )^2 d \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^{3/2} \sqrt {\frac {a+b+a \tan ^2\left (\frac {1}{2} (c+d x)\right )-b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{1+\tan ^2\left (\frac {1}{2} (c+d x)\right )}}} \]

[In]

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

[Out]

(Sqrt[a + b*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((2*(-3*a^2*A*b^2 - A*b^4 + 3*a^4*C - 7*a^2*b^2*C)*Sin[c + d*x])/
(3*a*b^2*(a^2 - b^2)^2) - (2*(a*A*b^2*Sin[c + d*x] + a^3*C*Sin[c + d*x]))/(3*b^2*(-a^2 + b^2)*(a + b*Cos[c + d
*x])^2) + (4*(a^2*A*b^2*Sin[c + d*x] + A*b^4*Sin[c + d*x] - 2*a^4*C*Sin[c + d*x] + 4*a^2*b^2*C*Sin[c + d*x]))/
(3*b^2*(-a^2 + b^2)^2*(a + b*Cos[c + d*x]))))/d + (2*Sqrt[(1 - Tan[(c + d*x)/2]^2)^(-1)]*(3*a^3*A*b^2*Tan[(c +
d*x)/2] + 3*a^2*A*b^3*Tan[(c + d*x)/2] + a*A*b^4*Tan[(c + d*x)/2] + A*b^5*Tan[(c + d*x)/2] - 3*a^5*C*Tan[(c +
d*x)/2] - 3*a^4*b*C*Tan[(c + d*x)/2] + 7*a^3*b^2*C*Tan[(c + d*x)/2] + 7*a^2*b^3*C*Tan[(c + d*x)/2] - 6*a^2*A*
b^3*Tan[(c + d*x)/2]^3 - 2*A*b^5*Tan[(c + d*x)/2]^3 + 6*a^4*b*C*Tan[(c + d*x)/2]^3 - 14*a^2*b^3*C*Tan[(c + d*x
)/2]^3 - 3*a^3*A*b^2*Tan[(c + d*x)/2]^5 + 3*a^2*A*b^3*Tan[(c + d*x)/2]^5 - a*A*b^4*Tan[(c + d*x)/2]^5 + A*b^5*
Tan[(c + d*x)/2]^5 + 3*a^5*C*Tan[(c + d*x)/2]^5 - 3*a^4*b*C*Tan[(c + d*x)/2]^5 - 7*a^3*b^2*C*Tan[(c + d*x)/2]^
5 + 7*a^2*b^3*C*Tan[(c + d*x)/2]^5 + 6*a^5*C*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Sqrt[1
- Tan[(c + d*x)/2]^2]*Sqrt[(a + b + a*Tan[(c + d*x)/2]^2 - b*Tan[(c + d*x)/2]^2)/(a + b)] - 12*a^3*b^2*C*Elli
pticPi[-1, ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b + a*Tan[(c + d
*x)/2]^2 - b*Tan[(c + d*x)/2]^2)/(a + b)] + 6*a*b^4*C*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b
)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b + a*Tan[(c + d*x)/2]^2 - b*Tan[(c + d*x)/2]^2)/(a + b)] + 6*a^5*C*
EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Tan[(c + d*x)/2]^2*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqr
t[(a + b + a*Tan[(c + d*x)/2]^2 - b*Tan[(c + d*x)/2]^2)/(a + b)] - 12*a^3*b^2*C*EllipticPi[-1, ArcSin[Tan[(c +
d*x)/2]], (-a + b)/(a + b)]*Tan[(c + d*x)/2]^2*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b + a*Tan[(c + d*x)/2]^
2 - b*Tan[(c + d*x)/2]^2)/(a + b)] + 6*a*b^4*C*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Tan[
(c + d*x)/2]^2*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b + a*Tan[(c + d*x)/2]^2 - b*Tan[(c + d*x)/2]^2)/(a + b)
] - (a + b)*(-(A*b^4) + 3*a^4*C - a^2*b^2*(3*A + 7*C))*EllipticE[ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*S
qrt[1 - Tan[(c + d*x)/2]^2]*(1 + Tan[(c + d*x)/2]^2)*Sqrt[(a + b + a*Tan[(c + d*x)/2]^2 - b*Tan[(c + d*x)/2]^2
)/(a + b)] + a*b*(a + b)*(2*a^2*C - 3*a*b*(A + C) - b^2*(A + 3*C))*EllipticF[ArcSin[Tan[(c + d*x)/2]], (-a + b
)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*(1 + Tan[(c + d*x)/2]^2)*Sqrt[(a + b + a*Tan[(c + d*x)/2]^2 - b*Tan[(c
+ d*x)/2]^2)/(a + b)]))/(3*a*b^2*(a^2 - b^2)^2*d*(1 + Tan[(c + d*x)/2]^2)^(3/2)*Sqrt[(a + b + a*Tan[(c + d*x)
/2]^2 - b*Tan[(c + d*x)/2]^2)/(1 + Tan[(c + d*x)/2]^2)])

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(6936\) vs. \(2(566)=1132\).

Time = 11.47 (sec) , antiderivative size = 6937, normalized size of antiderivative = 11.22

method	result	size
default	\(\text {Expression too large to display}\)	\(6937\)
parts	\(\text {Expression too large to display}\)	\(7163\)

[In]

int((A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^(5/2)/sec(d*x+c)^(1/2),x,method=_RETURNVERBOSE)

[Out]

result too large to display

Fricas [F]

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

[In]

integrate((A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^(5/2)/sec(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

integral((C*cos(d*x + c)^2 + A)*sqrt(b*cos(d*x + c) + a)/((b^3*cos(d*x + c)^3 + 3*a*b^2*cos(d*x + c)^2 + 3*a^2
*b*cos(d*x + c) + a^3)*sqrt(sec(d*x + c))), x)

Sympy [F(-1)]

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

[In]

integrate((A+C*cos(d*x+c)**2)/(a+b*cos(d*x+c))**(5/2)/sec(d*x+c)**(1/2),x)

[Out]

Timed out

Maxima [F]

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

[In]

integrate((A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^(5/2)/sec(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

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

Giac [F]

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

[In]

integrate((A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^(5/2)/sec(d*x+c)^(1/2),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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