\(\int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^{5/2}} \, dx\) [336]

Optimal result

Integrand size = 31, antiderivative size = 307 \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^{5/2}} \, dx=-\frac {2 \left (a^2 A b+3 A b^3+2 a^3 B-6 a b^2 B\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{3 b^2 \left (a^2-b^2\right )^2 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {2 \left (a A b+2 a^2 B-3 b^2 B\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{3 b^2 \left (a^2-b^2\right ) d \sqrt {a+b \cos (c+d x)}}+\frac {2 a (A b-a B) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}+\frac {2 \left (a^2 A b+3 A b^3+2 a^3 B-6 a b^2 B\right ) \sin (c+d x)}{3 b \left (a^2-b^2\right )^2 d \sqrt {a+b \cos (c+d x)}} \]

[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)+2/3*(A*a^2*b+3*A*b^3+2*B*a^3-6*B*a*b^2)*sin(d*
x+c)/b/(a^2-b^2)^2/d/(a+b*cos(d*x+c))^(1/2)-2/3*(A*a^2*b+3*A*b^3+2*B*a^3-6*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/(a+b))^(1/2))*(a+b*cos(d*x+c))^(1/2)/b^2/(a^2-b^
2)^2/d/((a+b*cos(d*x+c))/(a+b))^(1/2)+2/3*(A*a*b+2*B*a^2-3*B*b^2)*(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/(a+b))^(1/2))*((a+b*cos(d*x+c))/(a+b))^(1/2)/b^2/(a^2-b^2)/d/(a+b*
cos(d*x+c))^(1/2)

Rubi [A] (verified)

Time = 0.60 (sec) , antiderivative size = 307, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.258, Rules used = {3047, 3100, 2833, 2831, 2742, 2740, 2734, 2732} \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^{5/2}} \, dx=\frac {2 a (A b-a B) \sin (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}+\frac {2 \left (2 a^2 B+a A b-3 b^2 B\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{3 b^2 d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}+\frac {2 \left (2 a^3 B+a^2 A b-6 a b^2 B+3 A b^3\right ) \sin (c+d x)}{3 b d \left (a^2-b^2\right )^2 \sqrt {a+b \cos (c+d x)}}-\frac {2 \left (2 a^3 B+a^2 A b-6 a b^2 B+3 A b^3\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{3 b^2 d \left (a^2-b^2\right )^2 \sqrt {\frac {a+b \cos (c+d x)}{a+b}}} \]

[In]

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

[Out]

(-2*(a^2*A*b + 3*A*b^3 + 2*a^3*B - 6*a*b^2*B)*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a + b)])/
(3*b^2*(a^2 - b^2)^2*d*Sqrt[(a + b*Cos[c + d*x])/(a + b)]) + (2*(a*A*b + 2*a^2*B - 3*b^2*B)*Sqrt[(a + b*Cos[c
+ d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)])/(3*b^2*(a^2 - b^2)*d*Sqrt[a + b*Cos[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)) + (2*(a^2*A*b + 3*A*b^3 + 2*a^3*B - 6
*a*b^2*B)*Sin[c + d*x])/(3*b*(a^2 - b^2)^2*d*Sqrt[a + b*Cos[c + d*x]])

Rule 2732

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[2*(Sqrt[a + b]/d)*EllipticE[(1/2)*(c - Pi/2
+ d*x), 2*(b/(a + b))], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2734

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[a + b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c +
d*x])/(a + b)], Int[Sqrt[a/(a + b) + (b/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
b^2, 0] &&  !GtQ[a + b, 0]

Rule 2740

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/(d*Sqrt[a + b]))*EllipticF[(1/2)*(c - P
i/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2742

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[(a + b*Sin[c + d*x])/(a + b)]/Sqrt[a
+ b*Sin[c + d*x]], Int[1/Sqrt[a/(a + b) + (b/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] &&  !GtQ[a + b, 0]

Rule 2831

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

Rule 2833

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

Rule 3047

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

Rule 3100

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(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)/(b*f*(m
+ 1)*(a^2 - b^2))), x] + Dist[1/(b*(m + 1)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(a*A - b*B +
a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b,
e, f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]

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

Mathematica [A] (verified)

Time = 1.69 (sec) , antiderivative size = 224, normalized size of antiderivative = 0.73 \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^{5/2}} \, dx=\frac {2 \left (-\frac {\left (\frac {a+b \cos (c+d x)}{a+b}\right )^{3/2} \left (\left (a^2 A b+3 A b^3+2 a^3 B-6 a b^2 B\right ) E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )-(a-b) \left (a A b+2 a^2 B-3 b^2 B\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )\right )}{(a-b)^2}+\frac {b \left (a \left (2 a^2 A b+2 A b^3+a^3 B-5 a b^2 B\right )+b \left (a^2 A b+3 A b^3+2 a^3 B-6 a b^2 B\right ) \cos (c+d x)\right ) \sin (c+d x)}{\left (a^2-b^2\right )^2}\right )}{3 b^2 d (a+b \cos (c+d x))^{3/2}} \]

[In]

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

[Out]

(2*(-((((a + b*Cos[c + d*x])/(a + b))^(3/2)*((a^2*A*b + 3*A*b^3 + 2*a^3*B - 6*a*b^2*B)*EllipticE[(c + d*x)/2,
(2*b)/(a + b)] - (a - b)*(a*A*b + 2*a^2*B - 3*b^2*B)*EllipticF[(c + d*x)/2, (2*b)/(a + b)]))/(a - b)^2) + (b*(
a*(2*a^2*A*b + 2*A*b^3 + a^3*B - 5*a*b^2*B) + b*(a^2*A*b + 3*A*b^3 + 2*a^3*B - 6*a*b^2*B)*Cos[c + d*x])*Sin[c
+ d*x])/(a^2 - b^2)^2))/(3*b^2*d*(a + b*Cos[c + d*x])^(3/2))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(863\) vs. \(2(345)=690\).

Time = 21.46 (sec) , antiderivative size = 864, normalized size of antiderivative = 2.81

method	result	size
default	\(\text {Expression too large to display}\)	\(864\)
parts	\(\text {Expression too large to display}\)	\(1598\)

[In]

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

[Out]

-(-(-2*b*cos(1/2*d*x+1/2*c)^2-a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*B/b^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*b*cos
(1/2*d*x+1/2*c)^2+a-b)/(a-b))^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos
(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))-2/b^2*(A*b-2*B*a)/sin(1/2*d*x+1/2*c)^2/(2*b*sin(1/2*d*x+1/2*c)^2-a-b)/(a^2
-b^2)*(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2*
b+EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*(sin(1/
2*d*x+1/2*c)^2)^(1/2)*a-EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b
)/(a-b))^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*b)-2*a*(A*b-B*a)/b^2*(1/6/b/(a-b)/(a+b)*cos(1/2*d*x+1/2*c)*(-2*sin
(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2+1/2*(a-b)/b)^2+8/3*sin(1/2*d*x+1/2
*c)^2*b/(a-b)^2/(a+b)^2*cos(1/2*d*x+1/2*c)*a/(-(-2*b*cos(1/2*d*x+1/2*c)^2-a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)+(3*
a-b)/(3*a^3+3*a^2*b-3*a*b^2-3*b^3)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*b*cos(1/2*d*x+1/2*c)^2+a-b)/(a-b))^(1/2)/(
-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))-4
/3*a/(a-b)/(a+b)^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*b*cos(1/2*d*x+1/2*c)^2+a-b)/(a-b))^(1/2)/(-2*sin(1/2*d*x+1
/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))-EllipticE(cos(1/
2*d*x+1/2*c),(-2*b/(a-b))^(1/2)))))/sin(1/2*d*x+1/2*c)/(-2*b*sin(1/2*d*x+1/2*c)^2+a+b)^(1/2)/d

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.16 (sec) , antiderivative size = 1076, normalized size of antiderivative = 3.50 \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^{5/2}} \, dx=\text {Too large to display} \]

[In]

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

[Out]

1/9*(6*(B*a^4*b^2 + 2*A*a^3*b^3 - 5*B*a^2*b^4 + 2*A*a*b^5 + (2*B*a^3*b^3 + A*a^2*b^4 - 6*B*a*b^5 + 3*A*b^6)*co
s(d*x + c))*sqrt(b*cos(d*x + c) + a)*sin(d*x + c) + (sqrt(2)*(-4*I*B*a^4*b^2 - 2*I*A*a^3*b^3 + 9*I*B*a^2*b^4 +
6*I*A*a*b^5 - 9*I*B*b^6)*cos(d*x + c)^2 - 2*sqrt(2)*(4*I*B*a^5*b + 2*I*A*a^4*b^2 - 9*I*B*a^3*b^3 - 6*I*A*a^2*
b^4 + 9*I*B*a*b^5)*cos(d*x + c) + sqrt(2)*(-4*I*B*a^6 - 2*I*A*a^5*b + 9*I*B*a^4*b^2 + 6*I*A*a^3*b^3 - 9*I*B*a^
2*b^4))*sqrt(b)*weierstrassPInverse(4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*a^3 - 9*a*b^2)/b^3, 1/3*(3*b*cos(d*x + c
) + 3*I*b*sin(d*x + c) + 2*a)/b) + (sqrt(2)*(4*I*B*a^4*b^2 + 2*I*A*a^3*b^3 - 9*I*B*a^2*b^4 - 6*I*A*a*b^5 + 9*I
*B*b^6)*cos(d*x + c)^2 - 2*sqrt(2)*(-4*I*B*a^5*b - 2*I*A*a^4*b^2 + 9*I*B*a^3*b^3 + 6*I*A*a^2*b^4 - 9*I*B*a*b^5
)*cos(d*x + c) + sqrt(2)*(4*I*B*a^6 + 2*I*A*a^5*b - 9*I*B*a^4*b^2 - 6*I*A*a^3*b^3 + 9*I*B*a^2*b^4))*sqrt(b)*we
ierstrassPInverse(4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*a^3 - 9*a*b^2)/b^3, 1/3*(3*b*cos(d*x + c) - 3*I*b*sin(d*x
+ c) + 2*a)/b) - 3*(sqrt(2)*(2*I*B*a^3*b^3 + I*A*a^2*b^4 - 6*I*B*a*b^5 + 3*I*A*b^6)*cos(d*x + c)^2 + 2*sqrt(2)
*(2*I*B*a^4*b^2 + I*A*a^3*b^3 - 6*I*B*a^2*b^4 + 3*I*A*a*b^5)*cos(d*x + c) + sqrt(2)*(2*I*B*a^5*b + I*A*a^4*b^2
- 6*I*B*a^3*b^3 + 3*I*A*a^2*b^4))*sqrt(b)*weierstrassZeta(4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*a^3 - 9*a*b^2)/b^
3, weierstrassPInverse(4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*a^3 - 9*a*b^2)/b^3, 1/3*(3*b*cos(d*x + c) + 3*I*b*sin
(d*x + c) + 2*a)/b)) - 3*(sqrt(2)*(-2*I*B*a^3*b^3 - I*A*a^2*b^4 + 6*I*B*a*b^5 - 3*I*A*b^6)*cos(d*x + c)^2 + 2*
sqrt(2)*(-2*I*B*a^4*b^2 - I*A*a^3*b^3 + 6*I*B*a^2*b^4 - 3*I*A*a*b^5)*cos(d*x + c) + sqrt(2)*(-2*I*B*a^5*b - I*
A*a^4*b^2 + 6*I*B*a^3*b^3 - 3*I*A*a^2*b^4))*sqrt(b)*weierstrassZeta(4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*a^3 - 9*
a*b^2)/b^3, weierstrassPInverse(4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*a^3 - 9*a*b^2)/b^3, 1/3*(3*b*cos(d*x + c) -
3*I*b*sin(d*x + c) + 2*a)/b)))/((a^4*b^5 - 2*a^2*b^7 + b^9)*d*cos(d*x + c)^2 + 2*(a^5*b^4 - 2*a^3*b^6 + a*b^8)
*d*cos(d*x + c) + (a^6*b^3 - 2*a^4*b^5 + a^2*b^7)*d)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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