\(\int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x))^{5/2} \sec ^{\frac {3}{2}}(c+d x)} \, dx\) [629]

Optimal result

Integrand size = 35, antiderivative size = 602 \[ \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x))^{5/2} \sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {2 \left (4 A b^3+3 a^3 B-7 a b^2 B\right ) \sqrt {\cos (c+d x)} \csc (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{3 a (a-b) b^2 (a+b)^{3/2} d \sqrt {\sec (c+d x)}}-\frac {2 \left (3 A b^3+3 a^3 B+a^2 b B-a b^2 (A+6 B)\right ) \sqrt {\cos (c+d x)} \csc (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{3 a (a-b) b^2 (a+b)^{3/2} d \sqrt {\sec (c+d x)}}-\frac {2 \sqrt {a+b} B \sqrt {\cos (c+d x)} \csc (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{b^3 d \sqrt {\sec (c+d x)}}+\frac {2 a (A b-a B) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2} \sqrt {\sec (c+d x)}}-\frac {2 a \left (4 A b^3+3 a^3 B-7 a b^2 B\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 b^2 \left (a^2-b^2\right )^2 d \sqrt {a+b \cos (c+d x)}} \]

[Out]

2/3*a*(A*b-B*a)*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*(4*A*b^3+3*B*a^3-7*B*a*
b^2)*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*(4*A*b^3+3*B*a^3-7*B*a*b^2)*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/(a-b)/b^2/(a+b)^(3/2)/d/sec(d*x+c)^(1/2)-2/3*(3*
A*b^3+3*B*a^3+B*a^2*b-a*b^2*(A+6*B))*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*B*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 = 1.88 (sec) , antiderivative size = 602, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.229, Rules used = {3040, 3068, 3130, 2888, 3072, 3077, 2895, 3073} \[ \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x))^{5/2} \sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {2 \left (3 a^3 B-7 a b^2 B+4 A b^3\right ) \sqrt {\cos (c+d x)} \csc (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{3 a b^2 d (a-b) (a+b)^{3/2} \sqrt {\sec (c+d x)}}+\frac {2 a (A b-a B) \sin (c+d x)}{3 b d \left (a^2-b^2\right ) \sqrt {\sec (c+d x)} (a+b \cos (c+d x))^{3/2}}-\frac {2 \left (3 a^3 B+a^2 b B-a b^2 (A+6 B)+3 A b^3\right ) \sqrt {\cos (c+d x)} \csc (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{3 a b^2 d (a-b) (a+b)^{3/2} \sqrt {\sec (c+d x)}}-\frac {2 a \left (3 a^3 B-7 a b^2 B+4 A b^3\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 b^2 d \left (a^2-b^2\right )^2 \sqrt {a+b \cos (c+d x)}}-\frac {2 B \sqrt {a+b} \sqrt {\cos (c+d x)} \csc (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{b^3 d \sqrt {\sec (c+d x)}} \]

[In]

Int[(A + B*Cos[c + d*x])/((a + b*Cos[c + d*x])^(5/2)*Sec[c + d*x]^(3/2)),x]

[Out]

(2*(4*A*b^3 + 3*a^3*B - 7*a*b^2*B)*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*(a - b)*b^2*(a + b)^(3/2)*d*Sqrt[Sec[c + d*x]]) - (2*(3*A*b^3 + 3*a^3*B + a^2*b*B - a*
b^2*(A + 6*B))*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]*B*Sqrt[Cos[c + d*x]]*Csc[c + d*x]*Ellipti
cPi[(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*(A*b - a*
B)*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*(4*A*b^3 + 3*a^3*B -
7*a*b^2*B)*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 3040

Int[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[(g*Csc[e + f*x])^p*(g*Sin[e + f*x])^p, Int[(a + b*Sin[e + f*x])^m*((
c + d*Sin[e + f*x])^n/(g*Sin[e + f*x])^p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p}, x] && NeQ[b*c - a*d
, 0] &&  !IntegerQ[p] &&  !(IntegerQ[m] && IntegerQ[n])

Rule 3068

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e
_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-(b*c - a*d))*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1
)*((c + d*Sin[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*Si
n[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1)*Simp[b*(b*c - a*d)*(B*c - A*d)*(m - 1) + a*d*(a*A*c + b*B*c -
(A*b + a*B)*d)*(n + 1) + (b*(b*d*(B*c - A*d) + a*(A*c*d + B*(c^2 - 2*d^2)))*(n + 1) - a*(b*c - a*d)*(B*c - A*
d)*(n + 2))*Sin[e + f*x] + b*(d*(A*b*c + a*B*c - a*A*d)*(m + n + 1) - b*B*(c^2*m + d^2*(n + 1)))*Sin[e + f*x]^
2, 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] && GtQ[m, 1] && LtQ[n, -1]

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 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]

Rubi steps \begin{align*} \text {integral}& = \left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^{5/2}} \, dx \\ & = \frac {2 a (A b-a B) \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} a (A b-a B)+\frac {3}{2} b (A b-a B) \cos (c+d x)-\frac {3}{2} \left (a^2-b^2\right ) B \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 a (A b-a B) \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} a b (A b-a B)+\left (\frac {3}{2} a \left (a^2-b^2\right ) B+\frac {3}{2} b^2 (A b-a B)\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 (B \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} B \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 a (A b-a B) \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 a \left (4 A b^3+3 a^3 B-7 a b^2 B\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} a b^2 (A b-a B)-a \left (\frac {3}{2} a \left (a^2-b^2\right ) B+\frac {3}{2} b^2 (A b-a B)\right )+\left (-\frac {1}{2} a^2 b (A b-a B)-b \left (\frac {3}{2} a \left (a^2-b^2\right ) B+\frac {3}{2} b^2 (A b-a B)\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} B \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 a (A b-a B) \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 a \left (4 A b^3+3 a^3 B-7 a b^2 B\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 \left (4 A b^3+3 a^3 B-7 a b^2 B\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 {\left ((a-b) \left (3 A b^3+3 a^3 B+a^2 b B-a b^2 (A+6 B)\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 {2 \left (4 A b^3+3 a^3 B-7 a b^2 B\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 (a-b) b^2 (a+b)^{3/2} d \sqrt {\sec (c+d x)}}-\frac {2 \left (3 A b^3+3 a^3 B+a^2 b B-a b^2 (A+6 B)\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} B \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 a (A b-a B) \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 a \left (4 A b^3+3 a^3 B-7 a b^2 B\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] (verified)

Leaf count is larger than twice the leaf count of optimal. \(1496\) vs. \(2(602)=1204\).

Time = 14.20 (sec) , antiderivative size = 1496, normalized size of antiderivative = 2.49 \[ \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x))^{5/2} \sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {\sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \left (4 A b^3+3 a^3 B-7 a b^2 B\right ) \sin (c+d x)}{3 b^2 \left (-a^2+b^2\right )^2}-\frac {2 \left (-a^2 A b \sin (c+d x)+a^3 B \sin (c+d x)\right )}{3 b^2 \left (-a^2+b^2\right ) (a+b \cos (c+d x))^2}-\frac {2 \left (-a^3 A b \sin (c+d x)+5 a A b^3 \sin (c+d x)+4 a^4 B \sin (c+d x)-8 a^2 b^2 B \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 {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 )}} \left (-4 a A b^3 \tan \left (\frac {1}{2} (c+d x)\right )-4 A b^4 \tan \left (\frac {1}{2} (c+d x)\right )-3 a^4 B \tan \left (\frac {1}{2} (c+d x)\right )-3 a^3 b B \tan \left (\frac {1}{2} (c+d x)\right )+7 a^2 b^2 B \tan \left (\frac {1}{2} (c+d x)\right )+7 a b^3 B \tan \left (\frac {1}{2} (c+d x)\right )+8 A b^4 \tan ^3\left (\frac {1}{2} (c+d x)\right )+6 a^3 b B \tan ^3\left (\frac {1}{2} (c+d x)\right )-14 a b^3 B \tan ^3\left (\frac {1}{2} (c+d x)\right )+4 a A b^3 \tan ^5\left (\frac {1}{2} (c+d x)\right )-4 A b^4 \tan ^5\left (\frac {1}{2} (c+d x)\right )+3 a^4 B \tan ^5\left (\frac {1}{2} (c+d x)\right )-3 a^3 b B \tan ^5\left (\frac {1}{2} (c+d x)\right )-7 a^2 b^2 B \tan ^5\left (\frac {1}{2} (c+d x)\right )+7 a b^3 B \tan ^5\left (\frac {1}{2} (c+d x)\right )+6 a^4 B \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^2 b^2 B \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 b^4 B \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^4 B \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^2 b^2 B \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 b^4 B \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 (4 A b^3+3 a^3 B-7 a b^2 B\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}}+b (a+b) \left (a b (A-3 B)+3 b^2 (A-B)+2 a^2 B\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 b^2 \left (a^2-b^2\right )^2 d \left (-1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \sqrt {\frac {1+\tan ^2\left (\frac {1}{2} (c+d x)\right )}{1-\tan ^2\left (\frac {1}{2} (c+d x)\right )}} \left (b \left (-1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )-a \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )\right )} \]

[In]

Integrate[(A + B*Cos[c + d*x])/((a + b*Cos[c + d*x])^(5/2)*Sec[c + d*x]^(3/2)),x]

[Out]

(Sqrt[a + b*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((2*(4*A*b^3 + 3*a^3*B - 7*a*b^2*B)*Sin[c + d*x])/(3*b^2*(-a^2 +
b^2)^2) - (2*(-(a^2*A*b*Sin[c + d*x]) + a^3*B*Sin[c + d*x]))/(3*b^2*(-a^2 + b^2)*(a + b*Cos[c + d*x])^2) - (2*
(-(a^3*A*b*Sin[c + d*x]) + 5*a*A*b^3*Sin[c + d*x] + 4*a^4*B*Sin[c + d*x] - 8*a^2*b^2*B*Sin[c + d*x]))/(3*b^2*(
-a^2 + b^2)^2*(a + b*Cos[c + d*x]))))/d + (2*Sqrt[(a + b + a*Tan[(c + d*x)/2]^2 - b*Tan[(c + d*x)/2]^2)/(1 + T
an[(c + d*x)/2]^2)]*(-4*a*A*b^3*Tan[(c + d*x)/2] - 4*A*b^4*Tan[(c + d*x)/2] - 3*a^4*B*Tan[(c + d*x)/2] - 3*a^3
*b*B*Tan[(c + d*x)/2] + 7*a^2*b^2*B*Tan[(c + d*x)/2] + 7*a*b^3*B*Tan[(c + d*x)/2] + 8*A*b^4*Tan[(c + d*x)/2]^3
+ 6*a^3*b*B*Tan[(c + d*x)/2]^3 - 14*a*b^3*B*Tan[(c + d*x)/2]^3 + 4*a*A*b^3*Tan[(c + d*x)/2]^5 - 4*A*b^4*Tan[(
c + d*x)/2]^5 + 3*a^4*B*Tan[(c + d*x)/2]^5 - 3*a^3*b*B*Tan[(c + d*x)/2]^5 - 7*a^2*b^2*B*Tan[(c + d*x)/2]^5 + 7
*a*b^3*B*Tan[(c + d*x)/2]^5 + 6*a^4*B*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^2*b^2*B*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*b^4*B*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^4*B*EllipticP
i[-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)] - 12*a^2*b^2*B*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*b^4*B*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)
*(4*A*b^3 + 3*a^3*B - 7*a*b^2*B)*EllipticE[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)] + b*(a + b)
*(a*b*(A - 3*B) + 3*b^2*(A - B) + 2*a^2*B)*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*b^2*(a^2 - b^2)^2*d*(-1 + Tan[(c + d*x)/2]^2)*Sqrt[(1 + Tan[(c + d*x)/2]^2)/(1 - Tan[(c + d*x)/2]^2)]*(b*(
-1 + Tan[(c + d*x)/2]^2) - a*(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. \(6417\) vs. \(2(550)=1100\).

Time = 19.33 (sec) , antiderivative size = 6418, normalized size of antiderivative = 10.66

method	result	size
default	\(\text {Expression too large to display}\)	\(6418\)
parts	\(\text {Expression too large to display}\)	\(6730\)

[In]

int((A+B*cos(d*x+c))/(a+cos(d*x+c)*b)^(5/2)/sec(d*x+c)^(3/2),x,method=_RETURNVERBOSE)

[Out]

result too large to display

Fricas [F(-1)]

Timed out. \[ \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x))^{5/2} \sec ^{\frac {3}{2}}(c+d x)} \, dx=\text {Timed out} \]

[In]

integrate((A+B*cos(d*x+c))/(a+b*cos(d*x+c))^(5/2)/sec(d*x+c)^(3/2),x, algorithm="fricas")

[Out]

Timed out

Sympy [F(-1)]

Timed out. \[ \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x))^{5/2} \sec ^{\frac {3}{2}}(c+d x)} \, dx=\text {Timed out} \]

[In]

integrate((A+B*cos(d*x+c))/(a+b*cos(d*x+c))**(5/2)/sec(d*x+c)**(3/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x))^{5/2} \sec ^{\frac {3}{2}}(c+d x)} \, dx=\int { \frac {B \cos \left (d x + c\right ) + A}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sec \left (d x + c\right )^{\frac {3}{2}}} \,d x } \]

[In]

integrate((A+B*cos(d*x+c))/(a+b*cos(d*x+c))^(5/2)/sec(d*x+c)^(3/2),x, algorithm="maxima")

[Out]

integrate((B*cos(d*x + c) + A)/((b*cos(d*x + c) + a)^(5/2)*sec(d*x + c)^(3/2)), x)

Giac [F]

\[ \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x))^{5/2} \sec ^{\frac {3}{2}}(c+d x)} \, dx=\int { \frac {B \cos \left (d x + c\right ) + A}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sec \left (d x + c\right )^{\frac {3}{2}}} \,d x } \]

[In]

integrate((A+B*cos(d*x+c))/(a+b*cos(d*x+c))^(5/2)/sec(d*x+c)^(3/2),x, algorithm="giac")

[Out]

integrate((B*cos(d*x + c) + A)/((b*cos(d*x + c) + a)^(5/2)*sec(d*x + c)^(3/2)), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x))^{5/2} \sec ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {A+B\,\cos \left (c+d\,x\right )}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,d x \]

[In]

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

[Out]

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