\(\int \frac {\sqrt {\cos (c+d x)} (A+B \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx\) [162]

Optimal result

Integrand size = 33, antiderivative size = 178 \[ \int \frac {\sqrt {\cos (c+d x)} (A+B \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx=\frac {(A-B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^3 d}+\frac {(A+B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{6 a^3 d}+\frac {(A-B) \sqrt {\cos (c+d x)} \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {(A+4 B) \sqrt {\cos (c+d x)} \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {(A-B) \sqrt {\cos (c+d x)} \sin (c+d x)}{10 d \left (a^3+a^3 \cos (c+d x)\right )} \]

[Out]

1/10*(A-B)*(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))/a^3/d+1/6*(A+
B)*(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))/a^3/d+1/5*(A-B)*sin(d
*x+c)*cos(d*x+c)^(1/2)/d/(a+a*cos(d*x+c))^3+1/15*(A+4*B)*sin(d*x+c)*cos(d*x+c)^(1/2)/a/d/(a+a*cos(d*x+c))^2-1/
10*(A-B)*sin(d*x+c)*cos(d*x+c)^(1/2)/d/(a^3+a^3*cos(d*x+c))

Rubi [A] (verified)

Time = 0.76 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {3056, 3057, 2827, 2720, 2719} \[ \int \frac {\sqrt {\cos (c+d x)} (A+B \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx=\frac {(A+B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{6 a^3 d}+\frac {(A-B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^3 d}-\frac {(A-B) \sin (c+d x) \sqrt {\cos (c+d x)}}{10 d \left (a^3 \cos (c+d x)+a^3\right )}+\frac {(A+4 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{15 a d (a \cos (c+d x)+a)^2}+\frac {(A-B) \sin (c+d x) \sqrt {\cos (c+d x)}}{5 d (a \cos (c+d x)+a)^3} \]

[In]

Int[(Sqrt[Cos[c + d*x]]*(A + B*Cos[c + d*x]))/(a + a*Cos[c + d*x])^3,x]

[Out]

((A - B)*EllipticE[(c + d*x)/2, 2])/(10*a^3*d) + ((A + B)*EllipticF[(c + d*x)/2, 2])/(6*a^3*d) + ((A - B)*Sqrt
[Cos[c + d*x]]*Sin[c + d*x])/(5*d*(a + a*Cos[c + d*x])^3) + ((A + 4*B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(15*a*
d*(a + a*Cos[c + d*x])^2) - ((A - B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(10*d*(a^3 + a^3*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 2827

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 3056

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 - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x]
)^n/(a*f*(2*m + 1))), x] - Dist[1/(a*b*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n -
1)*Simp[A*(a*d*n - b*c*(m + 1)) - B*(a*c*m + b*d*n) - d*(a*B*(m - n) + A*b*(m + n + 1))*Sin[e + f*x], x], x],
x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ
[m, -2^(-1)] && GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])

Rule 3057

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*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*
x])^(n + 1)/(a*f*(2*m + 1)*(b*c - a*d))), x] + Dist[1/(a*(2*m + 1)*(b*c - a*d)), Int[(a + b*Sin[e + f*x])^(m +
1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b*d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*
(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2
- b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)] &&  !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c,
0])

Rubi steps \begin{align*} \text {integral}& = \frac {(A-B) \sqrt {\cos (c+d x)} \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {\int \frac {\frac {1}{2} a (A-B)+\frac {1}{2} a (3 A+7 B) \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))^2} \, dx}{5 a^2} \\ & = \frac {(A-B) \sqrt {\cos (c+d x)} \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {(A+4 B) \sqrt {\cos (c+d x)} \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac {\int \frac {\frac {1}{2} a^2 (4 A+B)+\frac {1}{2} a^2 (A+4 B) \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))} \, dx}{15 a^4} \\ & = \frac {(A-B) \sqrt {\cos (c+d x)} \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {(A+4 B) \sqrt {\cos (c+d x)} \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {(A-B) \sqrt {\cos (c+d x)} \sin (c+d x)}{10 d \left (a^3+a^3 \cos (c+d x)\right )}+\frac {\int \frac {\frac {5}{4} a^3 (A+B)+\frac {3}{4} a^3 (A-B) \cos (c+d x)}{\sqrt {\cos (c+d x)}} \, dx}{15 a^6} \\ & = \frac {(A-B) \sqrt {\cos (c+d x)} \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {(A+4 B) \sqrt {\cos (c+d x)} \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {(A-B) \sqrt {\cos (c+d x)} \sin (c+d x)}{10 d \left (a^3+a^3 \cos (c+d x)\right )}+\frac {(A-B) \int \sqrt {\cos (c+d x)} \, dx}{20 a^3}+\frac {(A+B) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{12 a^3} \\ & = \frac {(A-B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^3 d}+\frac {(A+B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{6 a^3 d}+\frac {(A-B) \sqrt {\cos (c+d x)} \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {(A+4 B) \sqrt {\cos (c+d x)} \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {(A-B) \sqrt {\cos (c+d x)} \sin (c+d x)}{10 d \left (a^3+a^3 \cos (c+d x)\right )} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 6.35 (sec) , antiderivative size = 664, normalized size of antiderivative = 3.73 \[ \int \frac {\sqrt {\cos (c+d x)} (A+B \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx=-\frac {(A-B) \sqrt {\cos (c+d x)} \csc (c+d x)}{15 a^3 d (1-\cos (c+d x)) (1+\cos (c+d x))}+\frac {2 B \sqrt {\cos (c+d x)} \csc (c+d x)}{3 a^3 d (1-\cos (c+d x)) (1+\cos (c+d x))}-\frac {2 B \cos ^{\frac {3}{2}}(c+d x) \csc (c+d x)}{3 a^3 d (1-\cos (c+d x)) (1+\cos (c+d x))}+\frac {2 (A-B) \sqrt {\cos (c+d x)} \csc ^3(c+d x)}{5 a^3 d (1-\cos (c+d x)) (1+\cos (c+d x))}-\frac {6 (A-B) \cos ^{\frac {3}{2}}(c+d x) \csc ^3(c+d x)}{7 a^3 d (1-\cos (c+d x)) (1+\cos (c+d x))}+\frac {2 (A-B) \cos ^{\frac {5}{2}}(c+d x) \csc ^3(c+d x)}{5 a^3 d (1-\cos (c+d x)) (1+\cos (c+d x))}-\frac {(A-B) \sqrt {\cos (c+d x)} \sin (c+d x)}{6 a^3 d (1-\cos (c+d x)) (1+\cos (c+d x))}-\frac {B \sqrt {\cos (c+d x)} \sin (c+d x)}{3 a^3 d (1-\cos (c+d x)) (1+\cos (c+d x))}-\frac {(A-B) \sqrt {\cos (c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},\cos ^2(c+d x)\right ) \sin (c+d x) \sqrt {\sin ^2(c+d x)}}{6 a^3 d (1-\cos (c+d x)) (1+\cos (c+d x))}-\frac {B \sqrt {\cos (c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},\cos ^2(c+d x)\right ) \sin (c+d x) \sqrt {\sin ^2(c+d x)}}{3 a^3 d (1-\cos (c+d x)) (1+\cos (c+d x))}+\frac {4 (A-B) \cos ^{\frac {3}{2}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {7}{2},\frac {7}{4},\cos ^2(c+d x)\right ) \sin (c+d x) \sqrt {\sin ^2(c+d x)}}{21 a^3 d (1-\cos (c+d x)) (1+\cos (c+d x))} \]

[In]

Integrate[(Sqrt[Cos[c + d*x]]*(A + B*Cos[c + d*x]))/(a + a*Cos[c + d*x])^3,x]

[Out]

-1/15*((A - B)*Sqrt[Cos[c + d*x]]*Csc[c + d*x])/(a^3*d*(1 - Cos[c + d*x])*(1 + Cos[c + d*x])) + (2*B*Sqrt[Cos[
c + d*x]]*Csc[c + d*x])/(3*a^3*d*(1 - Cos[c + d*x])*(1 + Cos[c + d*x])) - (2*B*Cos[c + d*x]^(3/2)*Csc[c + d*x]
)/(3*a^3*d*(1 - Cos[c + d*x])*(1 + Cos[c + d*x])) + (2*(A - B)*Sqrt[Cos[c + d*x]]*Csc[c + d*x]^3)/(5*a^3*d*(1
- Cos[c + d*x])*(1 + Cos[c + d*x])) - (6*(A - B)*Cos[c + d*x]^(3/2)*Csc[c + d*x]^3)/(7*a^3*d*(1 - Cos[c + d*x]
)*(1 + Cos[c + d*x])) + (2*(A - B)*Cos[c + d*x]^(5/2)*Csc[c + d*x]^3)/(5*a^3*d*(1 - Cos[c + d*x])*(1 + Cos[c +
d*x])) - ((A - B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(6*a^3*d*(1 - Cos[c + d*x])*(1 + Cos[c + d*x])) - (B*Sqrt[
Cos[c + d*x]]*Sin[c + d*x])/(3*a^3*d*(1 - Cos[c + d*x])*(1 + Cos[c + d*x])) - ((A - B)*Sqrt[Cos[c + d*x]]*Hype
rgeometric2F1[1/4, 1/2, 5/4, Cos[c + d*x]^2]*Sin[c + d*x]*Sqrt[Sin[c + d*x]^2])/(6*a^3*d*(1 - Cos[c + d*x])*(1
+ Cos[c + d*x])) - (B*Sqrt[Cos[c + d*x]]*Hypergeometric2F1[1/4, 1/2, 5/4, Cos[c + d*x]^2]*Sin[c + d*x]*Sqrt[S
in[c + d*x]^2])/(3*a^3*d*(1 - Cos[c + d*x])*(1 + Cos[c + d*x])) + (4*(A - B)*Cos[c + d*x]^(3/2)*Hypergeometric
2F1[3/4, 7/2, 7/4, Cos[c + d*x]^2]*Sin[c + d*x]*Sqrt[Sin[c + d*x]^2])/(21*a^3*d*(1 - Cos[c + d*x])*(1 + Cos[c
+ d*x]))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(450\) vs. \(2(214)=428\).

Time = 5.43 (sec) , antiderivative size = 451, normalized size of antiderivative = 2.53

method	result	size
default	\(\frac {\sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (12 A \left (\cos ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-10 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \left (\cos ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+6 A \left (\cos ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-12 B \left (\cos ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-10 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \left (\cos ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-6 B \left (\cos ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-22 A \left (\cos ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 B \left (\cos ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+6 A \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+24 B \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+7 A \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-17 B \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-3 A +3 B \right )}{60 a^{3} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\)	\(451\)

[In]

int((A+B*cos(d*x+c))*cos(d*x+c)^(1/2)/(a+cos(d*x+c)*a)^3,x,method=_RETURNVERBOSE)

[Out]

1/60*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(12*A*cos(1/2*d*x+1/2*c)^8-10*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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*cos(1/2*d*x+1/2*c)^5+6*A*c
os(1/2*d*x+1/2*c)^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)*EllipticE(cos(1/2*d*x+1/2*c
),2^(1/2))-12*B*cos(1/2*d*x+1/2*c)^8-10*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)*Ellip
ticF(cos(1/2*d*x+1/2*c),2^(1/2))*cos(1/2*d*x+1/2*c)^5-6*B*cos(1/2*d*x+1/2*c)^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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-22*A*cos(1/2*d*x+1/2*c)^6+2*B*cos(1/2*d*
x+1/2*c)^6+6*A*cos(1/2*d*x+1/2*c)^4+24*B*cos(1/2*d*x+1/2*c)^4+7*A*cos(1/2*d*x+1/2*c)^2-17*B*cos(1/2*d*x+1/2*c)
^2-3*A+3*B)/a^3/cos(1/2*d*x+1/2*c)^5/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^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 [C] (verification not implemented)

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

Time = 0.11 (sec) , antiderivative size = 465, normalized size of antiderivative = 2.61 \[ \int \frac {\sqrt {\cos (c+d x)} (A+B \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx=-\frac {2 \, {\left (3 \, {\left (A - B\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (2 \, A - 7 \, B\right )} \cos \left (d x + c\right ) - 5 \, A - 5 \, B\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + 5 \, {\left (\sqrt {2} {\left (i \, A + i \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, \sqrt {2} {\left (i \, A + i \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, \sqrt {2} {\left (i \, A + i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (i \, A + i \, B\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, {\left (\sqrt {2} {\left (-i \, A - i \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, \sqrt {2} {\left (-i \, A - i \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, \sqrt {2} {\left (-i \, A - i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-i \, A - i \, B\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 \, {\left (\sqrt {2} {\left (-i \, A + i \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, \sqrt {2} {\left (-i \, A + i \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, \sqrt {2} {\left (-i \, A + i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-i \, A + i \, B\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 \, {\left (\sqrt {2} {\left (i \, A - i \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, \sqrt {2} {\left (i \, A - i \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, \sqrt {2} {\left (i \, A - i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (i \, A - i \, B\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}{60 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \]

[In]

integrate((A+B*cos(d*x+c))*cos(d*x+c)^(1/2)/(a+a*cos(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/60*(2*(3*(A - B)*cos(d*x + c)^2 + 2*(2*A - 7*B)*cos(d*x + c) - 5*A - 5*B)*sqrt(cos(d*x + c))*sin(d*x + c) +
5*(sqrt(2)*(I*A + I*B)*cos(d*x + c)^3 + 3*sqrt(2)*(I*A + I*B)*cos(d*x + c)^2 + 3*sqrt(2)*(I*A + I*B)*cos(d*x
+ c) + sqrt(2)*(I*A + I*B))*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*(sqrt(2)*(-I*A - I*B
)*cos(d*x + c)^3 + 3*sqrt(2)*(-I*A - I*B)*cos(d*x + c)^2 + 3*sqrt(2)*(-I*A - I*B)*cos(d*x + c) + sqrt(2)*(-I*A
- I*B))*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 3*(sqrt(2)*(-I*A + I*B)*cos(d*x + c)^3 +
3*sqrt(2)*(-I*A + I*B)*cos(d*x + c)^2 + 3*sqrt(2)*(-I*A + I*B)*cos(d*x + c) + sqrt(2)*(-I*A + I*B))*weierstras
sZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*(sqrt(2)*(I*A - I*B)*cos(d*x + c)^
3 + 3*sqrt(2)*(I*A - I*B)*cos(d*x + c)^2 + 3*sqrt(2)*(I*A - I*B)*cos(d*x + c) + sqrt(2)*(I*A - I*B))*weierstra
ssZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(
d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d)

Sympy [F(-1)]

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

[In]

integrate((A+B*cos(d*x+c))*cos(d*x+c)**(1/2)/(a+a*cos(d*x+c))**3,x)

[Out]

Timed out

Maxima [F]

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

[In]

integrate((A+B*cos(d*x+c))*cos(d*x+c)^(1/2)/(a+a*cos(d*x+c))^3,x, algorithm="maxima")

[Out]

integrate((B*cos(d*x + c) + A)*sqrt(cos(d*x + c))/(a*cos(d*x + c) + a)^3, x)

Giac [F]

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

[In]

integrate((A+B*cos(d*x+c))*cos(d*x+c)^(1/2)/(a+a*cos(d*x+c))^3,x, algorithm="giac")

[Out]

integrate((B*cos(d*x + c) + A)*sqrt(cos(d*x + c))/(a*cos(d*x + c) + a)^3, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {\cos (c+d x)} (A+B \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx=\int \frac {\sqrt {\cos \left (c+d\,x\right )}\,\left (A+B\,\cos \left (c+d\,x\right )\right )}{{\left (a+a\,\cos \left (c+d\,x\right )\right )}^3} \,d x \]

[In]

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

[Out]

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