\(\int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x))^3 \sec ^{\frac {5}{2}}(c+d x)} \, dx\) [582]

Optimal result

Integrand size = 33, antiderivative size = 427 \[ \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x))^3 \sec ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {\left (3 a^3 A b-9 a A b^3-15 a^4 B+29 a^2 b^2 B-8 b^4 B\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{4 b^3 \left (a^2-b^2\right )^2 d}+\frac {\left (3 a^4 A b-5 a^2 A b^3+8 A b^5-15 a^5 B+33 a^3 b^2 B-24 a b^4 B\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{4 b^4 \left (a^2-b^2\right )^2 d}-\frac {a \left (3 a^4 A b-6 a^2 A b^3+15 A b^5-15 a^5 B+38 a^3 b^2 B-35 a b^4 B\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{4 (a-b)^2 b^4 (a+b)^3 d}+\frac {a (A b-a B) \sqrt {\sec (c+d x)} \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (b+a \sec (c+d x))^2}+\frac {a \left (a^2 A b-7 A b^3-5 a^3 B+11 a b^2 B\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{4 b^2 \left (a^2-b^2\right )^2 d (b+a \sec (c+d x))} \]

[Out]

1/2*a*(A*b-B*a)*sin(d*x+c)*sec(d*x+c)^(1/2)/b/(a^2-b^2)/d/(b+a*sec(d*x+c))^2+1/4*a*(A*a^2*b-7*A*b^3-5*B*a^3+11
*B*a*b^2)*sin(d*x+c)*sec(d*x+c)^(1/2)/b^2/(a^2-b^2)^2/d/(b+a*sec(d*x+c))-1/4*(3*A*a^3*b-9*A*a*b^3-15*B*a^4+29*
B*a^2*b^2-8*B*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))*cos(d
*x+c)^(1/2)*sec(d*x+c)^(1/2)/b^3/(a^2-b^2)^2/d+1/4*(3*A*a^4*b-5*A*a^2*b^3+8*A*b^5-15*B*a^5+33*B*a^3*b^2-24*B*a
*b^4)*(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))*cos(d*x+c)^(1/2)*s
ec(d*x+c)^(1/2)/b^4/(a^2-b^2)^2/d-1/4*a*(3*A*a^4*b-6*A*a^2*b^3+15*A*b^5-15*B*a^5+38*B*a^3*b^2-35*B*a*b^4)*(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*b/(a+b),2^(1/2))*cos(d*x+c)^(1/2)*
sec(d*x+c)^(1/2)/(a-b)^2/b^4/(a+b)^3/d

Rubi [A] (verified)

Time = 1.13 (sec) , antiderivative size = 427, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.303, Rules used = {3039, 4115, 4185, 4191, 3934, 2884, 3872, 3856, 2719, 2720} \[ \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x))^3 \sec ^{\frac {5}{2}}(c+d x)} \, dx=\frac {a (A b-a B) \sin (c+d x) \sqrt {\sec (c+d x)}}{2 b d \left (a^2-b^2\right ) (a \sec (c+d x)+b)^2}+\frac {a \left (-5 a^3 B+a^2 A b+11 a b^2 B-7 A b^3\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{4 b^2 d \left (a^2-b^2\right )^2 (a \sec (c+d x)+b)}-\frac {\left (-15 a^4 B+3 a^3 A b+29 a^2 b^2 B-9 a A b^3-8 b^4 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 b^3 d \left (a^2-b^2\right )^2}+\frac {\left (-15 a^5 B+3 a^4 A b+33 a^3 b^2 B-5 a^2 A b^3-24 a b^4 B+8 A b^5\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{4 b^4 d \left (a^2-b^2\right )^2}-\frac {a \left (-15 a^5 B+3 a^4 A b+38 a^3 b^2 B-6 a^2 A b^3-35 a b^4 B+15 A b^5\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{4 b^4 d (a-b)^2 (a+b)^3} \]

[In]

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

[Out]

-1/4*((3*a^3*A*b - 9*a*A*b^3 - 15*a^4*B + 29*a^2*b^2*B - 8*b^4*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]
*Sqrt[Sec[c + d*x]])/(b^3*(a^2 - b^2)^2*d) + ((3*a^4*A*b - 5*a^2*A*b^3 + 8*A*b^5 - 15*a^5*B + 33*a^3*b^2*B - 2
4*a*b^4*B)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(4*b^4*(a^2 - b^2)^2*d) - (a*(3*a^
4*A*b - 6*a^2*A*b^3 + 15*A*b^5 - 15*a^5*B + 38*a^3*b^2*B - 35*a*b^4*B)*Sqrt[Cos[c + d*x]]*EllipticPi[(2*b)/(a
+ b), (c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(4*(a - b)^2*b^4*(a + b)^3*d) + (a*(A*b - a*B)*Sqrt[Sec[c + d*x]]*Si
n[c + d*x])/(2*b*(a^2 - b^2)*d*(b + a*Sec[c + d*x])^2) + (a*(a^2*A*b - 7*A*b^3 - 5*a^3*B + 11*a*b^2*B)*Sqrt[Se
c[c + d*x]]*Sin[c + d*x])/(4*b^2*(a^2 - b^2)^2*d*(b + a*Sec[c + d*x]))

Rule 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{
c, d}, x]

Rule 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]

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 3039

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^(m + n), Int[(g*Csc[e + f*x])^(p - m - n)*(b + a*Csc[e + f*x])^m*(
d + c*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[p] && I
ntegerQ[m] && IntegerQ[n]

Rule 3856

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 3872

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 3934

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 4115

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[b*(A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*((d*Csc[e + f*x])^n/(
a*f*(m + 1)*(a^2 - b^2))), x] + Dist[1/(a*(m + 1)*(a^2 - b^2)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*
x])^n*Simp[A*(a^2*(m + 1) - b^2*(m + n + 1)) + a*b*B*n - a*(A*b - a*B)*(m + 1)*Csc[e + f*x] + b*(A*b - a*B)*(m
+ n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b
^2, 0] && LtQ[m, -1] &&  !(ILtQ[m + 1/2, 0] && ILtQ[n, 0])

Rule 4185

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[(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/(a*f*(m + 1)*(a^2 - b^2))), x] + Dist[1/(a*(m + 1)*(a^2 - b^2)), I
nt[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*Simp[a*(a*A - b*B + a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C)*
(m + n + 1) - a*(A*b - a*B + b*C)*(m + 1)*Csc[e + f*x] + (A*b^2 - a*b*B + a^2*C)*(m + n + 2)*Csc[e + f*x]^2, x
], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, n}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] &&  !(ILtQ[m + 1/2, 0] &
& ILtQ[n, 0])

Rule 4191

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*} \text {integral}& = \int \frac {B+A \sec (c+d x)}{\sqrt {\sec (c+d x)} (b+a \sec (c+d x))^3} \, dx \\ & = \frac {a (A b-a B) \sqrt {\sec (c+d x)} \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (b+a \sec (c+d x))^2}+\frac {\int \frac {\frac {1}{2} \left (-a A b+5 a^2 B-4 b^2 B\right )-2 b (A b-a B) \sec (c+d x)+\frac {3}{2} a (A b-a B) \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (b+a \sec (c+d x))^2} \, dx}{2 b \left (a^2-b^2\right )} \\ & = \frac {a (A b-a B) \sqrt {\sec (c+d x)} \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (b+a \sec (c+d x))^2}+\frac {a \left (a^2 A b-7 A b^3-5 a^3 B+11 a b^2 B\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{4 b^2 \left (a^2-b^2\right )^2 d (b+a \sec (c+d x))}+\frac {\int \frac {\frac {1}{4} \left (-3 a^3 A b+9 a A b^3+15 a^4 B-29 a^2 b^2 B+8 b^4 B\right )+b \left (a^2 A b+2 A b^3+a^3 B-4 a b^2 B\right ) \sec (c+d x)+\frac {1}{4} a \left (a^2 A b-7 A b^3-5 a^3 B+11 a b^2 B\right ) \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (b+a \sec (c+d x))} \, dx}{2 b^2 \left (a^2-b^2\right )^2} \\ & = \frac {a (A b-a B) \sqrt {\sec (c+d x)} \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (b+a \sec (c+d x))^2}+\frac {a \left (a^2 A b-7 A b^3-5 a^3 B+11 a b^2 B\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{4 b^2 \left (a^2-b^2\right )^2 d (b+a \sec (c+d x))}+\frac {\int \frac {\frac {1}{4} b \left (-3 a^3 A b+9 a A b^3+15 a^4 B-29 a^2 b^2 B+8 b^4 B\right )-\left (-b^2 \left (a^2 A b+2 A b^3+a^3 B-4 a b^2 B\right )+\frac {1}{4} a \left (-3 a^3 A b+9 a A b^3+15 a^4 B-29 a^2 b^2 B+8 b^4 B\right )\right ) \sec (c+d x)}{\sqrt {\sec (c+d x)}} \, dx}{2 b^4 \left (a^2-b^2\right )^2}-\frac {\left (a \left (3 a^4 A b-6 a^2 A b^3+15 A b^5-15 a^5 B+38 a^3 b^2 B-35 a b^4 B\right )\right ) \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{b+a \sec (c+d x)} \, dx}{8 b^4 \left (a^2-b^2\right )^2} \\ & = \frac {a (A b-a B) \sqrt {\sec (c+d x)} \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (b+a \sec (c+d x))^2}+\frac {a \left (a^2 A b-7 A b^3-5 a^3 B+11 a b^2 B\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{4 b^2 \left (a^2-b^2\right )^2 d (b+a \sec (c+d x))}-\frac {\left (3 a^3 A b-9 a A b^3-15 a^4 B+29 a^2 b^2 B-8 b^4 B\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx}{8 b^3 \left (a^2-b^2\right )^2}+\frac {\left (3 a^4 A b-5 a^2 A b^3+8 A b^5-15 a^5 B+33 a^3 b^2 B-24 a b^4 B\right ) \int \sqrt {\sec (c+d x)} \, dx}{8 b^4 \left (a^2-b^2\right )^2}-\frac {\left (a \left (3 a^4 A b-6 a^2 A b^3+15 A b^5-15 a^5 B+38 a^3 b^2 B-35 a b^4 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{8 b^4 \left (a^2-b^2\right )^2} \\ & = -\frac {a \left (3 a^4 A b-6 a^2 A b^3+15 A b^5-15 a^5 B+38 a^3 b^2 B-35 a b^4 B\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{4 (a-b)^2 b^4 (a+b)^3 d}+\frac {a (A b-a B) \sqrt {\sec (c+d x)} \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (b+a \sec (c+d x))^2}+\frac {a \left (a^2 A b-7 A b^3-5 a^3 B+11 a b^2 B\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{4 b^2 \left (a^2-b^2\right )^2 d (b+a \sec (c+d x))}-\frac {\left (\left (3 a^3 A b-9 a A b^3-15 a^4 B+29 a^2 b^2 B-8 b^4 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{8 b^3 \left (a^2-b^2\right )^2}+\frac {\left (\left (3 a^4 A b-5 a^2 A b^3+8 A b^5-15 a^5 B+33 a^3 b^2 B-24 a b^4 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{8 b^4 \left (a^2-b^2\right )^2} \\ & = -\frac {\left (3 a^3 A b-9 a A b^3-15 a^4 B+29 a^2 b^2 B-8 b^4 B\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{4 b^3 \left (a^2-b^2\right )^2 d}+\frac {\left (3 a^4 A b-5 a^2 A b^3+8 A b^5-15 a^5 B+33 a^3 b^2 B-24 a b^4 B\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{4 b^4 \left (a^2-b^2\right )^2 d}-\frac {a \left (3 a^4 A b-6 a^2 A b^3+15 A b^5-15 a^5 B+38 a^3 b^2 B-35 a b^4 B\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{4 (a-b)^2 b^4 (a+b)^3 d}+\frac {a (A b-a B) \sqrt {\sec (c+d x)} \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (b+a \sec (c+d x))^2}+\frac {a \left (a^2 A b-7 A b^3-5 a^3 B+11 a b^2 B\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{4 b^2 \left (a^2-b^2\right )^2 d (b+a \sec (c+d x))} \\ \end{align*}

Mathematica [A] (warning: unable to verify)

Time = 7.14 (sec) , antiderivative size = 820, normalized size of antiderivative = 1.92 \[ \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x))^3 \sec ^{\frac {5}{2}}(c+d x)} \, dx=\frac {\frac {2 \left (-a^3 A b-5 a A b^3+5 a^4 B-7 a^2 b^2 B+8 b^4 B\right ) \cos ^2(c+d x) \left (\operatorname {EllipticF}\left (\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right )-\operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right )\right ) (b+a \sec (c+d x)) \sqrt {1-\sec ^2(c+d x)} \sin (c+d x)}{a (a+b \cos (c+d x)) \left (1-\cos ^2(c+d x)\right )}+\frac {2 \left (8 a^2 A b^2+16 A b^4+8 a^3 b B-32 a b^3 B\right ) \cos ^2(c+d x) \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) (b+a \sec (c+d x)) \sqrt {1-\sec ^2(c+d x)} \sin (c+d x)}{b (a+b \cos (c+d x)) \left (1-\cos ^2(c+d x)\right )}+\frac {\left (-3 a^3 A b+9 a A b^3+15 a^4 B-29 a^2 b^2 B+8 b^4 B\right ) \cos (2 (c+d x)) (b+a \sec (c+d x)) \left (-4 a b+4 a b \sec ^2(c+d x)-4 a b E\left (\left .\arcsin \left (\sqrt {\sec (c+d x)}\right )\right |-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}+2 (2 a-b) b \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}-4 a^2 \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}+2 b^2 \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}\right ) \sin (c+d x)}{a b^2 (a+b \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 )}}{16 (a-b)^2 b^2 (a+b)^2 d}+\frac {\sqrt {\sec (c+d x)} \left (-\frac {a \left (-3 a^2 A b+9 A b^3+7 a^3 B-13 a b^2 B\right ) \sin (c+d x)}{4 b^3 \left (a^2-b^2\right )^2}-\frac {a^3 A b \sin (c+d x)-a^4 B \sin (c+d x)}{2 b^3 \left (-a^2+b^2\right ) (a+b \cos (c+d x))^2}+\frac {-5 a^4 A b \sin (c+d x)+11 a^2 A b^3 \sin (c+d x)+9 a^5 B \sin (c+d x)-15 a^3 b^2 B \sin (c+d x)}{4 b^3 \left (-a^2+b^2\right )^2 (a+b \cos (c+d x))}\right )}{d} \]

[In]

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

[Out]

((2*(-(a^3*A*b) - 5*a*A*b^3 + 5*a^4*B - 7*a^2*b^2*B + 8*b^4*B)*Cos[c + d*x]^2*(EllipticF[ArcSin[Sqrt[Sec[c + d
*x]]], -1] - EllipticPi[-(a/b), ArcSin[Sqrt[Sec[c + d*x]]], -1])*(b + a*Sec[c + d*x])*Sqrt[1 - Sec[c + d*x]^2]
*Sin[c + d*x])/(a*(a + b*Cos[c + d*x])*(1 - Cos[c + d*x]^2)) + (2*(8*a^2*A*b^2 + 16*A*b^4 + 8*a^3*b*B - 32*a*b
^3*B)*Cos[c + d*x]^2*EllipticPi[-(a/b), ArcSin[Sqrt[Sec[c + d*x]]], -1]*(b + a*Sec[c + d*x])*Sqrt[1 - Sec[c +
d*x]^2]*Sin[c + d*x])/(b*(a + b*Cos[c + d*x])*(1 - Cos[c + d*x]^2)) + ((-3*a^3*A*b + 9*a*A*b^3 + 15*a^4*B - 29
*a^2*b^2*B + 8*b^4*B)*Cos[2*(c + d*x)]*(b + a*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*(2*a - b)*b*EllipticF[ArcSin[Sq
rt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]^2] - 4*a^2*EllipticPi[-(a/b), ArcSin[Sqrt[Sec[
c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]^2] + 2*b^2*EllipticPi[-(a/b), ArcSin[Sqrt[Sec[c + d*x
]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]^2])*Sin[c + d*x])/(a*b^2*(a + b*Cos[c + d*x])*(1 - Cos[c + d
*x]^2)*Sqrt[Sec[c + d*x]]*(2 - Sec[c + d*x]^2)))/(16*(a - b)^2*b^2*(a + b)^2*d) + (Sqrt[Sec[c + d*x]]*(-1/4*(a
*(-3*a^2*A*b + 9*A*b^3 + 7*a^3*B - 13*a*b^2*B)*Sin[c + d*x])/(b^3*(a^2 - b^2)^2) - (a^3*A*b*Sin[c + d*x] - a^4
*B*Sin[c + d*x])/(2*b^3*(-a^2 + b^2)*(a + b*Cos[c + d*x])^2) + (-5*a^4*A*b*Sin[c + d*x] + 11*a^2*A*b^3*Sin[c +
d*x] + 9*a^5*B*Sin[c + d*x] - 15*a^3*b^2*B*Sin[c + d*x])/(4*b^3*(-a^2 + b^2)^2*(a + b*Cos[c + d*x]))))/d

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1976\) vs. \(2(479)=958\).

Time = 15.44 (sec) , antiderivative size = 1977, normalized size of antiderivative = 4.63

method	result	size
default	\(\text {Expression too large to display}\)	\(1977\)

[In]

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

[Out]

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2/b^4/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^
2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(A*b*EllipticF(cos(1/2*d*x+1/2*c),2^(1/
2))-3*B*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a-B*b*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))+2*a^2/b^4*(3*A*b-4*
B*a)*(-1/a*b^2/(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*b*cos(1/2*
d*x+1/2*c)^2+a-b)-1/2/a/(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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-1/2/(a^2-b^2)*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)*EllipticF(co
s(1/2*d*x+1/2*c),2^(1/2))+1/2/(a^2-b^2)*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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-3*a/(a^2-b^2)/(-2*a*b+
2*b^2)*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(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2))+1/a/(a^2-b^2)/(-2*a*b+2*b^2)*b^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)*Ellipt
icPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2)))-2*a^3*(A*b-B*a)/b^4*(-1/2/a*b^2/(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*b*cos(1/2*d*x+1/2*c)^2+a-b)^2-3/4*b^2*(3*a^2-b^2)/a^2/(a^2
-b^2)^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*b*cos(1/2*d*x+1/2*c)^2+a-b)
-7/8/(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+s
in(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+1/4/(a+b)/(a^2-b^2)/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)*EllipticF(cos(1/2
*d*x+1/2*c),2^(1/2))*b+3/8/(a+b)/(a^2-b^2)/a^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))*b^2-9/8*b/(a^2-b^2)
^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))+3/8*b^3/a^2/(a^2-b^2)^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))+9/8*b/(a^2-b^2)^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))-3/8*b^3/a^2/(a^2-b^2)^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))-15/4*a^2/(a^2-b^2)^2/(-2*a*b+2*b^2)*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(cos(1/2*d*x+1/2*c),-2*b/(a-b),
2^(1/2))+3/2/(a^2-b^2)^2/(-2*a*b+2*b^2)*b^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*b/(a-b),2^(1/2))-3/4/a^2/(a
^2-b^2)^2/(-2*a*b+2*b^2)*b^5*(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*b/(a-b),2^(1/2)))+12*a/b^3*(A*b-2*B*a)/(-2
*a*b+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)*EllipticPi(cos(1/2*d*x+1/2*c),-2*b/(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

Fricas [F(-1)]

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

[In]

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

[Out]

Timed out

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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