\(\int \frac {(A+B \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx\) [626]

Optimal result

Integrand size = 35, antiderivative size = 496 \[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=\frac {2 \left (3 a^4 A-15 a^2 A b^2+8 A b^4+6 a^3 b B-2 a b^3 B\right ) \sqrt {\cos (c+d x)} \csc (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{3 a^4 (a-b) (a+b)^{3/2} d \sqrt {\sec (c+d x)}}+\frac {2 \left (8 A b^3-3 a^3 (A-B)+2 a b^2 (3 A-B)-3 a^2 b (3 A+B)\right ) \sqrt {\cos (c+d x)} \csc (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{3 a^3 \sqrt {a+b} \left (a^2-b^2\right ) d \sqrt {\sec (c+d x)}}+\frac {2 b (A b-a B) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}+\frac {2 b \left (8 a^2 A b-4 A b^3-5 a^3 B+a b^2 B\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 a^2 \left (a^2-b^2\right )^2 d \sqrt {a+b \cos (c+d x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 1.49 (sec) , antiderivative size = 496, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {3040, 3079, 3134, 3077, 2895, 3073} \[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=\frac {2 b (A b-a B) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}+\frac {2 \left (-3 a^3 (A-B)-3 a^2 b (3 A+B)+2 a b^2 (3 A-B)+8 A b^3\right ) \sqrt {\cos (c+d x)} \csc (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{3 a^3 d \sqrt {a+b} \left (a^2-b^2\right ) \sqrt {\sec (c+d x)}}+\frac {2 b \left (-5 a^3 B+8 a^2 A b+a b^2 B-4 A b^3\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 a^2 d \left (a^2-b^2\right )^2 \sqrt {a+b \cos (c+d x)}}+\frac {2 \left (3 a^4 A+6 a^3 b B-15 a^2 A b^2-2 a b^3 B+8 A b^4\right ) \sqrt {\cos (c+d x)} \csc (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{3 a^4 d (a-b) (a+b)^{3/2} \sqrt {\sec (c+d x)}} \]

[In]

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

[Out]

(2*(3*a^4*A - 15*a^2*A*b^2 + 8*A*b^4 + 6*a^3*b*B - 2*a*b^3*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^4*(a - b)*(a + b)^(3/2)*d*Sqrt[Sec[c + d*x]]) + (2*(8*A*b^3
- 3*a^3*(A - B) + 2*a*b^2*(3*A - B) - 3*a^2*b*(3*A + B))*Sqrt[Cos[c + d*x]]*Csc[c + d*x]*EllipticF[ArcSin[Sqr
t[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^3*Sqrt[a + b]*(a^2 - b^2)*d*Sqrt[Sec[c + d*x]]) + (2*b*(A*b - a
*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(3*a*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^(3/2)) + (2*b*(8*a^2*A*b - 4*A*b^
3 - 5*a^3*B + a*b^2*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(3*a^2*(a^2 - b^2)^2*d*Sqrt[a + b*Cos[c + d*x]])

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 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 3079

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e
_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-(A*b^2 - a*b*B))*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*((c +
d*Sin[e + f*x])^(1 + n)/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2))), x] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), I
nt[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(a*A - b*B)*(b*c - a*d)*(m + 1) + b*d*(A*b - a*B)*
(m + n + 2) + (A*b - a*B)*(a*d*(m + 1) - b*c*(m + 2))*Sin[e + f*x] - b*d*(A*b - a*B)*(m + n + 3)*Sin[e + f*x]^
2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 -
d^2, 0] && RationalQ[m] && m < -1 && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ[n
, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 0])))

Rule 3134

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 - a*b*B + a^2*C))*Cos[e
+ f*x]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*Sin[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2))), x] + D
ist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*
(b*c - a*d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(
b*c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A*b^2 - a*b*B + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x]
/; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&
LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n]
&&  !IntegerQ[m]) || EqQ[a, 0])))

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

Mathematica [A] (warning: unable to verify)

Time = 18.73 (sec) , antiderivative size = 610, normalized size of antiderivative = 1.23 \[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=\frac {\sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \left (3 a^4 A-15 a^2 A b^2+8 A b^4+6 a^3 b B-2 a b^3 B\right ) \sin (c+d x)}{3 a^3 \left (a^2-b^2\right )^2}-\frac {2 \left (-A b^2 \sin (c+d x)+a b B \sin (c+d x)\right )}{3 a \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}-\frac {2 \left (-8 a^2 A b^2 \sin (c+d x)+4 A b^4 \sin (c+d x)+5 a^3 b B \sin (c+d x)-a b^3 B \sin (c+d x)\right )}{3 a^2 \left (a^2-b^2\right )^2 (a+b \cos (c+d x))}\right )}{d}+\frac {2 \sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)} \left (-2 (a+b) \left (3 a^4 A-15 a^2 A b^2+8 A b^4+6 a^3 b B-2 a b^3 B\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {a+b \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {-a+b}{a+b}\right )+2 a (a+b) \left (8 A b^3+3 a^2 b (-3 A+B)+3 a^3 (A+B)-2 a b^2 (3 A+B)\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {a+b \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {-a+b}{a+b}\right )-\left (3 a^4 A-15 a^2 A b^2+8 A b^4+6 a^3 b B-2 a b^3 B\right ) \cos (c+d x) (a+b \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{3 a^3 \left (a^2-b^2\right )^2 d \sqrt {a+b \cos (c+d x)} \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )}} \]

[In]

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

[Out]

(Sqrt[a + b*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((2*(3*a^4*A - 15*a^2*A*b^2 + 8*A*b^4 + 6*a^3*b*B - 2*a*b^3*B)*Si
n[c + d*x])/(3*a^3*(a^2 - b^2)^2) - (2*(-(A*b^2*Sin[c + d*x]) + a*b*B*Sin[c + d*x]))/(3*a*(a^2 - b^2)*(a + b*C
os[c + d*x])^2) - (2*(-8*a^2*A*b^2*Sin[c + d*x] + 4*A*b^4*Sin[c + d*x] + 5*a^3*b*B*Sin[c + d*x] - a*b^3*B*Sin[
c + d*x]))/(3*a^2*(a^2 - b^2)^2*(a + b*Cos[c + d*x]))))/d + (2*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*(-2*(a +
b)*(3*a^4*A - 15*a^2*A*b^2 + 8*A*b^4 + 6*a^3*b*B - 2*a*b^3*B)*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(a +
b*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticE[ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)] + 2*a*(a +
b)*(8*A*b^3 + 3*a^2*b*(-3*A + B) + 3*a^3*(A + B) - 2*a*b^2*(3*A + B))*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*S
qrt[(a + b*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticF[ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)] -
(3*a^4*A - 15*a^2*A*b^2 + 8*A*b^4 + 6*a^3*b*B - 2*a*b^3*B)*Cos[c + d*x]*(a + b*Cos[c + d*x])*Sec[(c + d*x)/2]
^2*Tan[(c + d*x)/2]))/(3*a^3*(a^2 - b^2)^2*d*Sqrt[a + b*Cos[c + d*x]]*Sqrt[Sec[(c + d*x)/2]^2])

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(6407\) vs. \(2(456)=912\).

Time = 22.02 (sec) , antiderivative size = 6408, normalized size of antiderivative = 12.92

method	result	size
default	\(\text {Expression too large to display}\)	\(6408\)
parts	\(\text {Expression too large to display}\)	\(6566\)

[In]

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

[Out]

result too large to display

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[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)