\(\int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))^3} \, dx\) [473]

Optimal result

Integrand size = 43, antiderivative size = 193 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))^3} \, dx=\frac {(9 A+B-C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^3 d}+\frac {(3 A+B+C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{6 a^3 d}-\frac {(A-B+C) \sqrt {\cos (c+d x)} \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(6 A-B-4 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {(9 A+B-C) \sqrt {\cos (c+d x)} \sin (c+d x)}{10 d \left (a^3+a^3 \cos (c+d x)\right )} \]

[Out]

1/10*(9*A+B-C)*(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
*(3*A+B+C)*(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+C)*sin(d*x+c)*cos(d*x+c)^(1/2)/d/(a+a*cos(d*x+c))^3-1/15*(6*A-B-4*C)*sin(d*x+c)*cos(d*x+c)^(1/2)/a/d/(a+a*co
s(d*x+c))^2-1/10*(9*A+B-C)*sin(d*x+c)*cos(d*x+c)^(1/2)/d/(a^3+a^3*cos(d*x+c))

Rubi [A] (verified)

Time = 0.60 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.116, Rules used = {3120, 3057, 2827, 2720, 2719} \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))^3} \, dx=\frac {(3 A+B+C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{6 a^3 d}+\frac {(9 A+B-C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^3 d}-\frac {(9 A+B-C) \sin (c+d x) \sqrt {\cos (c+d x)}}{10 d \left (a^3 \cos (c+d x)+a^3\right )}-\frac {(6 A-B-4 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{15 a d (a \cos (c+d x)+a)^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\cos (c+d x)}}{5 d (a \cos (c+d x)+a)^3} \]

[In]

Int[(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)/(Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])^3),x]

[Out]

((9*A + B - C)*EllipticE[(c + d*x)/2, 2])/(10*a^3*d) + ((3*A + B + C)*EllipticF[(c + d*x)/2, 2])/(6*a^3*d) - (
(A - B + C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(5*d*(a + a*Cos[c + d*x])^3) - ((6*A - B - 4*C)*Sqrt[Cos[c + d*x]
]*Sin[c + d*x])/(15*a*d*(a + a*Cos[c + d*x])^2) - ((9*A + B - C)*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 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])

Rule 3120

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

Rubi steps \begin{align*} \text {integral}& = -\frac {(A-B+C) \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 (9 A+B-C)-\frac {1}{2} a (3 A-3 B-7 C) \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))^2} \, dx}{5 a^2} \\ & = -\frac {(A-B+C) \sqrt {\cos (c+d x)} \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(6 A-B-4 C) \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 (21 A+4 B+C)-\frac {1}{2} a^2 (6 A-B-4 C) \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))} \, dx}{15 a^4} \\ & = -\frac {(A-B+C) \sqrt {\cos (c+d x)} \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(6 A-B-4 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {(9 A+B-C) \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 (3 A+B+C)+\frac {3}{4} a^3 (9 A+B-C) \cos (c+d x)}{\sqrt {\cos (c+d x)}} \, dx}{15 a^6} \\ & = -\frac {(A-B+C) \sqrt {\cos (c+d x)} \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(6 A-B-4 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {(9 A+B-C) \sqrt {\cos (c+d x)} \sin (c+d x)}{10 d \left (a^3+a^3 \cos (c+d x)\right )}+\frac {(9 A+B-C) \int \sqrt {\cos (c+d x)} \, dx}{20 a^3}+\frac {(3 A+B+C) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{12 a^3} \\ & = \frac {(9 A+B-C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^3 d}+\frac {(3 A+B+C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{6 a^3 d}-\frac {(A-B+C) \sqrt {\cos (c+d x)} \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(6 A-B-4 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {(9 A+B-C) \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] (warning: unable to verify)

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

Time = 8.71 (sec) , antiderivative size = 1448, normalized size of antiderivative = 7.50 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))^3} \, dx=-\frac {2 A \cos ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) \csc \left (\frac {c}{2}\right ) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2(d x-\arctan (\cot (c)))\right ) \sec \left (\frac {c}{2}\right ) \sec (d x-\arctan (\cot (c))) \sqrt {1-\sin (d x-\arctan (\cot (c)))} \sqrt {-\sqrt {1+\cot ^2(c)} \sin (c) \sin (d x-\arctan (\cot (c)))} \sqrt {1+\sin (d x-\arctan (\cot (c)))}}{d (a+a \cos (c+d x))^3 \sqrt {1+\cot ^2(c)}}-\frac {2 B \cos ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) \csc \left (\frac {c}{2}\right ) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2(d x-\arctan (\cot (c)))\right ) \sec \left (\frac {c}{2}\right ) \sec (d x-\arctan (\cot (c))) \sqrt {1-\sin (d x-\arctan (\cot (c)))} \sqrt {-\sqrt {1+\cot ^2(c)} \sin (c) \sin (d x-\arctan (\cot (c)))} \sqrt {1+\sin (d x-\arctan (\cot (c)))}}{3 d (a+a \cos (c+d x))^3 \sqrt {1+\cot ^2(c)}}-\frac {2 C \cos ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) \csc \left (\frac {c}{2}\right ) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2(d x-\arctan (\cot (c)))\right ) \sec \left (\frac {c}{2}\right ) \sec (d x-\arctan (\cot (c))) \sqrt {1-\sin (d x-\arctan (\cot (c)))} \sqrt {-\sqrt {1+\cot ^2(c)} \sin (c) \sin (d x-\arctan (\cot (c)))} \sqrt {1+\sin (d x-\arctan (\cot (c)))}}{3 d (a+a \cos (c+d x))^3 \sqrt {1+\cot ^2(c)}}+\frac {\cos ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {\cos (c+d x)} \left (-\frac {4 (9 A+B-C) \csc (c)}{5 d}-\frac {4 \sec \left (\frac {c}{2}\right ) \sec ^3\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (6 A \sin \left (\frac {d x}{2}\right )-B \sin \left (\frac {d x}{2}\right )-4 C \sin \left (\frac {d x}{2}\right )\right )}{15 d}-\frac {4 \sec \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}+\frac {d x}{2}\right ) \left (9 A \sin \left (\frac {d x}{2}\right )+B \sin \left (\frac {d x}{2}\right )-C \sin \left (\frac {d x}{2}\right )\right )}{5 d}-\frac {2 \sec \left (\frac {c}{2}\right ) \sec ^5\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (A \sin \left (\frac {d x}{2}\right )-B \sin \left (\frac {d x}{2}\right )+C \sin \left (\frac {d x}{2}\right )\right )}{5 d}-\frac {4 (6 A-B-4 C) \sec ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \tan \left (\frac {c}{2}\right )}{15 d}-\frac {2 (A-B+C) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \tan \left (\frac {c}{2}\right )}{5 d}\right )}{(a+a \cos (c+d x))^3}-\frac {9 A \cos ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right ) \left (\frac {\, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2(d x+\arctan (\tan (c)))\right ) \sin (d x+\arctan (\tan (c))) \tan (c)}{\sqrt {1-\cos (d x+\arctan (\tan (c)))} \sqrt {1+\cos (d x+\arctan (\tan (c)))} \sqrt {\cos (c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}} \sqrt {1+\tan ^2(c)}}-\frac {\frac {\sin (d x+\arctan (\tan (c))) \tan (c)}{\sqrt {1+\tan ^2(c)}}+\frac {2 \cos ^2(c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}}{\cos ^2(c)+\sin ^2(c)}}{\sqrt {\cos (c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}}}\right )}{5 d (a+a \cos (c+d x))^3}-\frac {B \cos ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right ) \left (\frac {\, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2(d x+\arctan (\tan (c)))\right ) \sin (d x+\arctan (\tan (c))) \tan (c)}{\sqrt {1-\cos (d x+\arctan (\tan (c)))} \sqrt {1+\cos (d x+\arctan (\tan (c)))} \sqrt {\cos (c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}} \sqrt {1+\tan ^2(c)}}-\frac {\frac {\sin (d x+\arctan (\tan (c))) \tan (c)}{\sqrt {1+\tan ^2(c)}}+\frac {2 \cos ^2(c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}}{\cos ^2(c)+\sin ^2(c)}}{\sqrt {\cos (c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}}}\right )}{5 d (a+a \cos (c+d x))^3}+\frac {C \cos ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right ) \left (\frac {\, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2(d x+\arctan (\tan (c)))\right ) \sin (d x+\arctan (\tan (c))) \tan (c)}{\sqrt {1-\cos (d x+\arctan (\tan (c)))} \sqrt {1+\cos (d x+\arctan (\tan (c)))} \sqrt {\cos (c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}} \sqrt {1+\tan ^2(c)}}-\frac {\frac {\sin (d x+\arctan (\tan (c))) \tan (c)}{\sqrt {1+\tan ^2(c)}}+\frac {2 \cos ^2(c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}}{\cos ^2(c)+\sin ^2(c)}}{\sqrt {\cos (c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}}}\right )}{5 d (a+a \cos (c+d x))^3} \]

[In]

Integrate[(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)/(Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])^3),x]

[Out]

(-2*A*Cos[c/2 + (d*x)/2]^6*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]
*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcT
an[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(d*(a + a*Cos[c + d*x])^3*Sqrt[1 + Cot[c]^2]) - (2*B*Cos[c/
2 + (d*x)/2]^6*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*Sec[d*x - A
rcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])
]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(3*d*(a + a*Cos[c + d*x])^3*Sqrt[1 + Cot[c]^2]) - (2*C*Cos[c/2 + (d*x)/
2]^6*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*Sec[d*x - ArcTan[Cot[
c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 +
Sin[d*x - ArcTan[Cot[c]]]])/(3*d*(a + a*Cos[c + d*x])^3*Sqrt[1 + Cot[c]^2]) + (Cos[c/2 + (d*x)/2]^6*Sqrt[Cos[
c + d*x]]*((-4*(9*A + B - C)*Csc[c])/(5*d) - (4*Sec[c/2]*Sec[c/2 + (d*x)/2]^3*(6*A*Sin[(d*x)/2] - B*Sin[(d*x)/
2] - 4*C*Sin[(d*x)/2]))/(15*d) - (4*Sec[c/2]*Sec[c/2 + (d*x)/2]*(9*A*Sin[(d*x)/2] + B*Sin[(d*x)/2] - C*Sin[(d*
x)/2]))/(5*d) - (2*Sec[c/2]*Sec[c/2 + (d*x)/2]^5*(A*Sin[(d*x)/2] - B*Sin[(d*x)/2] + C*Sin[(d*x)/2]))/(5*d) - (
4*(6*A - B - 4*C)*Sec[c/2 + (d*x)/2]^2*Tan[c/2])/(15*d) - (2*(A - B + C)*Sec[c/2 + (d*x)/2]^4*Tan[c/2])/(5*d))
)/(a + a*Cos[c + d*x])^3 - (9*A*Cos[c/2 + (d*x)/2]^6*Csc[c/2]*Sec[c/2]*((HypergeometricPFQ[{-1/2, -1/4}, {3/4}
, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/(Sqrt[1 - Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[1 +
Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]*Sqrt[1 + Tan[c]^2]) - ((
Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/Sqrt[1 + Tan[c]^2] + (2*Cos[c]^2*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2
])/(Cos[c]^2 + Sin[c]^2))/Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]))/(5*d*(a + a*Cos[c + d*x]
)^3) - (B*Cos[c/2 + (d*x)/2]^6*Csc[c/2]*Sec[c/2]*((HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan
[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/(Sqrt[1 - Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[1 + Cos[d*x + ArcTan[Tan[
c]]]]*Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]*Sqrt[1 + Tan[c]^2]) - ((Sin[d*x + ArcTan[Tan[c
]]]*Tan[c])/Sqrt[1 + Tan[c]^2] + (2*Cos[c]^2*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2])/(Cos[c]^2 + Sin[c]^
2))/Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]))/(5*d*(a + a*Cos[c + d*x])^3) + (C*Cos[c/2 + (d
*x)/2]^6*Csc[c/2]*Sec[c/2]*((HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x + Arc
Tan[Tan[c]]]*Tan[c])/(Sqrt[1 - Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[1 + Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[Cos[c]*Cos[
d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]*Sqrt[1 + Tan[c]^2]) - ((Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/Sqrt[1 + T
an[c]^2] + (2*Cos[c]^2*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2])/(Cos[c]^2 + Sin[c]^2))/Sqrt[Cos[c]*Cos[d*
x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]))/(5*d*(a + a*Cos[c + d*x])^3)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(623\) vs. \(2(229)=458\).

Time = 8.10 (sec) , antiderivative size = 624, normalized size of antiderivative = 3.23

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 (108 A \left (\cos ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-30 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 )+54 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 )-12 \left (\cos ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C -10 C \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 C \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 )-138 A \left (\cos ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-22 B \left (\cos ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 C \left (\cos ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+24 A \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+6 B \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+24 C \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+3 A \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+7 B \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-17 C \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+3 A -3 B +3 C \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}\)	\(624\)

[In]

int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+cos(d*x+c)*a)^3/cos(d*x+c)^(1/2),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)*(108*A*cos(1/2*d*x+1/2*c)^8-30*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+54*A
*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))+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)*Ell
ipticF(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))-12*cos(1/2*d*x+1/2*c)^8*C-10*C*(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*C*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))-138*A*cos(1/2*d*x+1/2*c)^6-22*B*cos(1/2*d*x+1/2*c)^6+2*C*cos(1/2*d*x+1/2*c)^6+24*A*cos(1/
2*d*x+1/2*c)^4+6*B*cos(1/2*d*x+1/2*c)^4+24*C*cos(1/2*d*x+1/2*c)^4+3*A*cos(1/2*d*x+1/2*c)^2+7*B*cos(1/2*d*x+1/2
*c)^2-17*C*cos(1/2*d*x+1/2*c)^2+3*A-3*B+3*C)/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 = 0.12 (sec) , antiderivative size = 522, normalized size of antiderivative = 2.70 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))^3} \, dx=-\frac {2 \, {\left (3 \, {\left (9 \, A + B - C\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (33 \, A + 2 \, B - 7 \, C\right )} \cos \left (d x + c\right ) + 45 \, A - 5 \, B - 5 \, C\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + 5 \, {\left (\sqrt {2} {\left (3 i \, A + i \, B + i \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, \sqrt {2} {\left (3 i \, A + i \, B + i \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, \sqrt {2} {\left (3 i \, A + i \, B + i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (3 i \, A + i \, B + i \, C\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 (-3 i \, A - i \, B - i \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, \sqrt {2} {\left (-3 i \, A - i \, B - i \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, \sqrt {2} {\left (-3 i \, A - i \, B - i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-3 i \, A - i \, B - i \, C\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 (-9 i \, A - i \, B + i \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, \sqrt {2} {\left (-9 i \, A - i \, B + i \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, \sqrt {2} {\left (-9 i \, A - i \, B + i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-9 i \, A - i \, B + i \, C\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 (9 i \, A + i \, B - i \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, \sqrt {2} {\left (9 i \, A + i \, B - i \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, \sqrt {2} {\left (9 i \, A + i \, B - i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (9 i \, A + i \, B - i \, C\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)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^3/cos(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

-1/60*(2*(3*(9*A + B - C)*cos(d*x + c)^2 + 2*(33*A + 2*B - 7*C)*cos(d*x + c) + 45*A - 5*B - 5*C)*sqrt(cos(d*x
+ c))*sin(d*x + c) + 5*(sqrt(2)*(3*I*A + I*B + I*C)*cos(d*x + c)^3 + 3*sqrt(2)*(3*I*A + I*B + I*C)*cos(d*x + c
)^2 + 3*sqrt(2)*(3*I*A + I*B + I*C)*cos(d*x + c) + sqrt(2)*(3*I*A + I*B + I*C))*weierstrassPInverse(-4, 0, cos
(d*x + c) + I*sin(d*x + c)) + 5*(sqrt(2)*(-3*I*A - I*B - I*C)*cos(d*x + c)^3 + 3*sqrt(2)*(-3*I*A - I*B - I*C)*
cos(d*x + c)^2 + 3*sqrt(2)*(-3*I*A - I*B - I*C)*cos(d*x + c) + sqrt(2)*(-3*I*A - I*B - I*C))*weierstrassPInver
se(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 3*(sqrt(2)*(-9*I*A - I*B + I*C)*cos(d*x + c)^3 + 3*sqrt(2)*(-9*I*A
- I*B + I*C)*cos(d*x + c)^2 + 3*sqrt(2)*(-9*I*A - I*B + I*C)*cos(d*x + c) + sqrt(2)*(-9*I*A - I*B + I*C))*weie
rstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*(sqrt(2)*(9*I*A + I*B - I*C)
*cos(d*x + c)^3 + 3*sqrt(2)*(9*I*A + I*B - I*C)*cos(d*x + c)^2 + 3*sqrt(2)*(9*I*A + I*B - I*C)*cos(d*x + c) +
sqrt(2)*(9*I*A + I*B - I*C))*weierstrassZeta(-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. \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))^3} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F(-1)]

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

[In]

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

[Out]

Timed out

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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