\(\int \frac {\cos ^{\frac {5}{2}}(c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^2} \, dx\) [371]

Optimal result

Integrand size = 33, antiderivative size = 303 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^2} \, dx=\frac {\left (3 a^2 A b-2 A b^3-5 a^3 B+4 a b^2 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b^3 \left (a^2-b^2\right ) d}-\frac {\left (9 a^3 A b-12 a A b^3-15 a^4 B+16 a^2 b^2 B+2 b^4 B\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 b^4 \left (a^2-b^2\right ) d}+\frac {a^2 \left (3 a^2 A b-5 A b^3-5 a^3 B+7 a b^2 B\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{(a-b) b^4 (a+b)^2 d}-\frac {\left (3 a A b-5 a^2 B+2 b^2 B\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d}+\frac {a (A b-a B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))} \]

[Out]

(3*A*a^2*b-2*A*b^3-5*B*a^3+4*B*a*b^2)*(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))/b^3/(a^2-b^2)/d-1/3*(9*A*a^3*b-12*A*a*b^3-15*B*a^4+16*B*a^2*b^2+2*B*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))/b^4/(a^2-b^2)/d+a^2*(3*A*a^2*b-5*A*b^3-5*B*a^3+
7*B*a*b^2)*(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))/(a
-b)/b^4/(a+b)^2/d+a*(A*b-B*a)*cos(d*x+c)^(3/2)*sin(d*x+c)/b/(a^2-b^2)/d/(a+b*cos(d*x+c))-1/3*(3*A*a*b-5*B*a^2+
2*B*b^2)*sin(d*x+c)*cos(d*x+c)^(1/2)/b^2/(a^2-b^2)/d

Rubi [A] (verified)

Time = 1.02 (sec) , antiderivative size = 303, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {3068, 3128, 3138, 2719, 3081, 2720, 2884} \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^2} \, dx=\frac {a (A b-a B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\left (-5 a^2 B+3 a A b+2 b^2 B\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b^2 d \left (a^2-b^2\right )}+\frac {\left (-5 a^3 B+3 a^2 A b+4 a b^2 B-2 A b^3\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b^3 d \left (a^2-b^2\right )}+\frac {a^2 \left (-5 a^3 B+3 a^2 A b+7 a b^2 B-5 A b^3\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{b^4 d (a-b) (a+b)^2}-\frac {\left (-15 a^4 B+9 a^3 A b+16 a^2 b^2 B-12 a A b^3+2 b^4 B\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 b^4 d \left (a^2-b^2\right )} \]

[In]

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

[Out]

((3*a^2*A*b - 2*A*b^3 - 5*a^3*B + 4*a*b^2*B)*EllipticE[(c + d*x)/2, 2])/(b^3*(a^2 - b^2)*d) - ((9*a^3*A*b - 12
*a*A*b^3 - 15*a^4*B + 16*a^2*b^2*B + 2*b^4*B)*EllipticF[(c + d*x)/2, 2])/(3*b^4*(a^2 - b^2)*d) + (a^2*(3*a^2*A
*b - 5*A*b^3 - 5*a^3*B + 7*a*b^2*B)*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/((a - b)*b^4*(a + b)^2*d) - ((3
*a*A*b - 5*a^2*B + 2*b^2*B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*b^2*(a^2 - b^2)*d) + (a*(A*b - a*B)*Cos[c + d*
x]^(3/2)*Sin[c + d*x])/(b*(a^2 - b^2)*d*(a + b*Cos[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 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 3081

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

Rule 3128

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

Rule 3138

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

Rubi steps \begin{align*} \text {integral}& = \frac {a (A b-a B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))}-\frac {\int \frac {\sqrt {\cos (c+d x)} \left (-\frac {3}{2} a (A b-a B)+b (A b-a B) \cos (c+d x)+\frac {1}{2} \left (3 a A b-5 a^2 B+2 b^2 B\right ) \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx}{b \left (a^2-b^2\right )} \\ & = -\frac {\left (3 a A b-5 a^2 B+2 b^2 B\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d}+\frac {a (A b-a B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))}-\frac {2 \int \frac {\frac {1}{4} a \left (3 a A b-5 a^2 B+2 b^2 B\right )-\frac {1}{2} b \left (3 a A b-2 a^2 B-b^2 B\right ) \cos (c+d x)-\frac {3}{4} \left (3 a^2 A b-2 A b^3-5 a^3 B+4 a b^2 B\right ) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{3 b^2 \left (a^2-b^2\right )} \\ & = -\frac {\left (3 a A b-5 a^2 B+2 b^2 B\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d}+\frac {a (A b-a B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac {2 \int \frac {-\frac {1}{4} a b \left (3 a A b-5 a^2 B+2 b^2 B\right )-\frac {1}{4} \left (9 a^3 A b-12 a A b^3-15 a^4 B+16 a^2 b^2 B+2 b^4 B\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{3 b^3 \left (a^2-b^2\right )}+\frac {\left (3 a^2 A b-2 A b^3-5 a^3 B+4 a b^2 B\right ) \int \sqrt {\cos (c+d x)} \, dx}{2 b^3 \left (a^2-b^2\right )} \\ & = \frac {\left (3 a^2 A b-2 A b^3-5 a^3 B+4 a b^2 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b^3 \left (a^2-b^2\right ) d}-\frac {\left (3 a A b-5 a^2 B+2 b^2 B\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d}+\frac {a (A b-a B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac {\left (a^2 \left (3 a^2 A b-5 A b^3-5 a^3 B+7 a b^2 B\right )\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{2 b^4 \left (a^2-b^2\right )}-\frac {\left (9 a^3 A b-12 a A b^3-15 a^4 B+16 a^2 b^2 B+2 b^4 B\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{6 b^4 \left (a^2-b^2\right )} \\ & = \frac {\left (3 a^2 A b-2 A b^3-5 a^3 B+4 a b^2 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b^3 \left (a^2-b^2\right ) d}-\frac {\left (9 a^3 A b-12 a A b^3-15 a^4 B+16 a^2 b^2 B+2 b^4 B\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 b^4 \left (a^2-b^2\right ) d}+\frac {a^2 \left (3 a^2 A b-5 A b^3-5 a^3 B+7 a b^2 B\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{(a-b) b^4 (a+b)^2 d}-\frac {\left (3 a A b-5 a^2 B+2 b^2 B\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d}+\frac {a (A b-a B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.38 (sec) , antiderivative size = 318, normalized size of antiderivative = 1.05 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^2} \, dx=\frac {4 \sqrt {\cos (c+d x)} \left (2 B+\frac {3 a^2 (-A b+a B)}{\left (a^2-b^2\right ) (a+b \cos (c+d x))}\right ) \sin (c+d x)-\frac {\frac {2 \left (-3 a^2 A b+6 A b^3+5 a^3 B-8 a b^2 B\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}+\frac {8 \left (-3 a A b+2 a^2 B+b^2 B\right ) \left ((a+b) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-a \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )\right )}{a+b}+\frac {6 \left (-3 a^2 A b+2 A b^3+5 a^3 B-4 a b^2 B\right ) \left (-2 a b E\left (\left .\arcsin \left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 a (a+b) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )+\left (-2 a^2+b^2\right ) \operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )\right ) \sin (c+d x)}{a b^2 \sqrt {\sin ^2(c+d x)}}}{(a-b) (a+b)}}{12 b^2 d} \]

[In]

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

[Out]

(4*Sqrt[Cos[c + d*x]]*(2*B + (3*a^2*(-(A*b) + a*B))/((a^2 - b^2)*(a + b*Cos[c + d*x])))*Sin[c + d*x] - ((2*(-3
*a^2*A*b + 6*A*b^3 + 5*a^3*B - 8*a*b^2*B)*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(a + b) + (8*(-3*a*A*b +
2*a^2*B + b^2*B)*((a + b)*EllipticF[(c + d*x)/2, 2] - a*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2]))/(a + b) +
(6*(-3*a^2*A*b + 2*A*b^3 + 5*a^3*B - 4*a*b^2*B)*(-2*a*b*EllipticE[ArcSin[Sqrt[Cos[c + d*x]]], -1] + 2*a*(a + b
)*EllipticF[ArcSin[Sqrt[Cos[c + d*x]]], -1] + (-2*a^2 + b^2)*EllipticPi[-(b/a), ArcSin[Sqrt[Cos[c + d*x]]], -1
])*Sin[c + d*x])/(a*b^2*Sqrt[Sin[c + d*x]^2]))/((a - b)*(a + b)))/(12*b^2*d)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1065\) vs. \(2(373)=746\).

Time = 19.24 (sec) , antiderivative size = 1066, normalized size of antiderivative = 3.52

method	result	size
default	\(\text {Expression too large to display}\)	\(1066\)

[In]

int(cos(d*x+c)^(5/2)*(A+B*cos(d*x+c))/(a+cos(d*x+c)*b)^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)*(-4*a^2/b^3*(3*A*b-4*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)*Elli
pticPi(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/a*b^2/(a^2-b^2)*cos(1/2*d*x+1/2*c)*(-2*s
in(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(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)
*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)*EllipticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2)))-2/3/
b^4/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(-4*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4*b^2+6*A
*a*b*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+3*A*(
sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*b^2+2*B*cos
(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2*b^2-9*B*a^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)
*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-B*b^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*Ell
ipticF(cos(1/2*d*x+1/2*c),2^(1/2))-6*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE
(cos(1/2*d*x+1/2*c),2^(1/2))*a*b))/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. \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^2} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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