\(\int \frac {A+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^2 \sec ^{\frac {3}{2}}(c+d x)} \, dx\) [1195]

Optimal result

Integrand size = 35, antiderivative size = 201 \[ \int \frac {A+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^2 \sec ^{\frac {3}{2}}(c+d x)} \, dx=-\frac {(A+7 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{a^2 d}+\frac {2 (A+5 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 a^2 d}-\frac {(A+C) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2 \sec ^{\frac {5}{2}}(c+d x)}-\frac {(A+7 C) \sin (c+d x)}{3 a^2 d (1+\cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 (A+5 C) \sin (c+d x)}{3 a^2 d \sqrt {\sec (c+d x)}} \]

[Out]

-1/3*(A+C)*sin(d*x+c)/d/(a+a*cos(d*x+c))^2/sec(d*x+c)^(5/2)-1/3*(A+7*C)*sin(d*x+c)/a^2/d/(1+cos(d*x+c))/sec(d*
x+c)^(3/2)+2/3*(A+5*C)*sin(d*x+c)/a^2/d/sec(d*x+c)^(1/2)-(A+7*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))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/a^2/d+2/3*(A+5*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))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/a^2/d

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4306, 3121, 3056, 2827, 2719, 2715, 2720} \[ \int \frac {A+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^2 \sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {2 (A+5 C) \sin (c+d x)}{3 a^2 d \sqrt {\sec (c+d x)}}-\frac {(A+7 C) \sin (c+d x)}{3 a^2 d (\cos (c+d x)+1) \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 (A+5 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 a^2 d}-\frac {(A+7 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d}-\frac {(A+C) \sin (c+d x)}{3 d \sec ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)^2} \]

[In]

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

[Out]

-(((A + 7*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(a^2*d)) + (2*(A + 5*C)*Sqrt[Cos
[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*a^2*d) - ((A + C)*Sin[c + d*x])/(3*d*(a + a*Cos[c
+ d*x])^2*Sec[c + d*x]^(5/2)) - ((A + 7*C)*Sin[c + d*x])/(3*a^2*d*(1 + Cos[c + d*x])*Sec[c + d*x]^(3/2)) + (2*
(A + 5*C)*Sin[c + d*x])/(3*a^2*d*Sqrt[Sec[c + d*x]])

Rule 2715

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Sin[c + d*x])^(n - 1)/(d*n))
, x] + Dist[b^2*((n - 1)/n), Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integ
erQ[2*n]

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 3121

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[a*(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)) - C*(a*c*m + b*d*(n + 1)) + (a*A*d*(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, 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)]

Rule 4306

Int[(u_)*((c_.)*sec[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Dist[(c*Sec[a + b*x])^m*(c*Cos[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Cos[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSineIntegrandQ[u,
x]

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

Mathematica [C] (verified)

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

Time = 9.52 (sec) , antiderivative size = 762, normalized size of antiderivative = 3.79 \[ \int \frac {A+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^2 \sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {\sqrt {2} A e^{-i d x} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \csc \left (\frac {c}{2}\right ) \left (-3 \sqrt {1+e^{2 i (c+d x)}}+e^{2 i d x} \left (-1+e^{2 i c}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )\right ) \sec \left (\frac {c}{2}\right )}{3 d (a+a \cos (c+d x))^2}+\frac {7 \sqrt {2} C e^{-i d x} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \csc \left (\frac {c}{2}\right ) \left (-3 \sqrt {1+e^{2 i (c+d x)}}+e^{2 i d x} \left (-1+e^{2 i c}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )\right ) \sec \left (\frac {c}{2}\right )}{3 d (a+a \cos (c+d x))^2}+\frac {4 A \cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {\cos (c+d x)} \csc \left (\frac {c}{2}\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sec \left (\frac {c}{2}\right ) \sqrt {\sec (c+d x)} \sin (c)}{3 d (a+a \cos (c+d x))^2}+\frac {20 C \cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {\cos (c+d x)} \csc \left (\frac {c}{2}\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sec \left (\frac {c}{2}\right ) \sqrt {\sec (c+d x)} \sin (c)}{3 d (a+a \cos (c+d x))^2}+\frac {\cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {\sec (c+d x)} \left (\frac {2 (A+5 C+2 C \cos (2 c)) \cos (d x) \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right )}{d}+\frac {4 C \cos (2 d x) \sin (2 c)}{3 d}+\frac {2 \sec \left (\frac {c}{2}\right ) \sec ^3\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (A \sin \left (\frac {d x}{2}\right )+C \sin \left (\frac {d x}{2}\right )\right )}{3 d}-\frac {8 \sec \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}+\frac {d x}{2}\right ) \left (2 A \sin \left (\frac {d x}{2}\right )+5 C \sin \left (\frac {d x}{2}\right )\right )}{3 d}-\frac {16 C \cos (c) \sin (d x)}{d}+\frac {4 C \cos (2 c) \sin (2 d x)}{3 d}-\frac {8 (2 A+5 C) \tan \left (\frac {c}{2}\right )}{3 d}+\frac {2 (A+C) \sec ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \tan \left (\frac {c}{2}\right )}{3 d}\right )}{(a+a \cos (c+d x))^2} \]

[In]

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

[Out]

(Sqrt[2]*A*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*Cos[c/2 + (d*x)/2]^4*
Csc[c/2]*(-3*Sqrt[1 + E^((2*I)*(c + d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Hypergeometric2F1[1/2, 3/4, 7/4,
-E^((2*I)*(c + d*x))])*Sec[c/2])/(3*d*E^(I*d*x)*(a + a*Cos[c + d*x])^2) + (7*Sqrt[2]*C*Sqrt[E^(I*(c + d*x))/(
1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*Cos[c/2 + (d*x)/2]^4*Csc[c/2]*(-3*Sqrt[1 + E^((2*I)*(c
+ d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))])*Sec[c/2])
/(3*d*E^(I*d*x)*(a + a*Cos[c + d*x])^2) + (4*A*Cos[c/2 + (d*x)/2]^4*Sqrt[Cos[c + d*x]]*Csc[c/2]*EllipticF[(c +
d*x)/2, 2]*Sec[c/2]*Sqrt[Sec[c + d*x]]*Sin[c])/(3*d*(a + a*Cos[c + d*x])^2) + (20*C*Cos[c/2 + (d*x)/2]^4*Sqrt
[Cos[c + d*x]]*Csc[c/2]*EllipticF[(c + d*x)/2, 2]*Sec[c/2]*Sqrt[Sec[c + d*x]]*Sin[c])/(3*d*(a + a*Cos[c + d*x]
)^2) + (Cos[c/2 + (d*x)/2]^4*Sqrt[Sec[c + d*x]]*((2*(A + 5*C + 2*C*Cos[2*c])*Cos[d*x]*Csc[c/2]*Sec[c/2])/d + (
4*C*Cos[2*d*x]*Sin[2*c])/(3*d) + (2*Sec[c/2]*Sec[c/2 + (d*x)/2]^3*(A*Sin[(d*x)/2] + C*Sin[(d*x)/2]))/(3*d) - (
8*Sec[c/2]*Sec[c/2 + (d*x)/2]*(2*A*Sin[(d*x)/2] + 5*C*Sin[(d*x)/2]))/(3*d) - (16*C*Cos[c]*Sin[d*x])/d + (4*C*C
os[2*c]*Sin[2*d*x])/(3*d) - (8*(2*A + 5*C)*Tan[c/2])/(3*d) + (2*(A + C)*Sec[c/2 + (d*x)/2]^2*Tan[c/2])/(3*d)))
/(a + a*Cos[c + d*x])^2

Maple [A] (verified)

Time = 3.38 (sec) , antiderivative size = 437, normalized size of antiderivative = 2.17

method	result	size
default	\(-\frac {\sqrt {\left (-1+2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (16 C \left (\cos ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+12 \left (\cos ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A +4 A \left (\cos ^{3}\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}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+6 A \left (\cos ^{3}\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 C \left (\cos ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+20 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 ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+42 \left (\cos ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) 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}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-20 \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A -48 C \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+9 A \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+21 C \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-A -C \right )}{6 a^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \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 {-1+2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}\)	\(437\)

[In]

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

[Out]

-1/6*((-1+2*cos(1/2*d*x+1/2*c)^2)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(16*C*cos(1/2*d*x+1/2*c)^8+12*cos(1/2*d*x+1/2*c)
^6*A+4*A*cos(1/2*d*x+1/2*c)^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)*EllipticF(cos(1/2
*d*x+1/2*c),2^(1/2))+6*A*cos(1/2*d*x+1/2*c)^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)*E
llipticE(cos(1/2*d*x+1/2*c),2^(1/2))+12*C*cos(1/2*d*x+1/2*c)^6+20*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)^3+42*cos(1/2*d*x+1/2*c)^3*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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-20*cos(1/2*d*
x+1/2*c)^4*A-48*C*cos(1/2*d*x+1/2*c)^4+9*A*cos(1/2*d*x+1/2*c)^2+21*C*cos(1/2*d*x+1/2*c)^2-A-C)/a^2/cos(1/2*d*x
+1/2*c)^3/(-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)/(-1+2*cos(1/2*d*x+1/2*c)^2)^
(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 = 375, normalized size of antiderivative = 1.87 \[ \int \frac {A+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^2 \sec ^{\frac {3}{2}}(c+d x)} \, dx=-\frac {2 \, {\left (\sqrt {2} {\left (i \, A + 5 i \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} {\left (i \, A + 5 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (i \, A + 5 i \, C\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 2 \, {\left (\sqrt {2} {\left (-i \, A - 5 i \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} {\left (-i \, A - 5 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-i \, A - 5 i \, C\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 \, {\left (\sqrt {2} {\left (i \, A + 7 i \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} {\left (i \, A + 7 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (i \, A + 7 i \, C\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 \, {\left (\sqrt {2} {\left (-i \, A - 7 i \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} {\left (-i \, A - 7 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-i \, A - 7 i \, C\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {2 \, {\left (2 \, C \cos \left (d x + c\right )^{3} + {\left (3 \, A + 13 \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (A + 5 \, C\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{6 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \]

[In]

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

[Out]

-1/6*(2*(sqrt(2)*(I*A + 5*I*C)*cos(d*x + c)^2 + 2*sqrt(2)*(I*A + 5*I*C)*cos(d*x + c) + sqrt(2)*(I*A + 5*I*C))*
weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 2*(sqrt(2)*(-I*A - 5*I*C)*cos(d*x + c)^2 + 2*sqrt(
2)*(-I*A - 5*I*C)*cos(d*x + c) + sqrt(2)*(-I*A - 5*I*C))*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x +
c)) + 3*(sqrt(2)*(I*A + 7*I*C)*cos(d*x + c)^2 + 2*sqrt(2)*(I*A + 7*I*C)*cos(d*x + c) + sqrt(2)*(I*A + 7*I*C))
*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*(sqrt(2)*(-I*A - 7*I*C)
*cos(d*x + c)^2 + 2*sqrt(2)*(-I*A - 7*I*C)*cos(d*x + c) + sqrt(2)*(-I*A - 7*I*C))*weierstrassZeta(-4, 0, weier
strassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - 2*(2*C*cos(d*x + c)^3 + (3*A + 13*C)*cos(d*x + c)^2 +
2*(A + 5*C)*cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*
d)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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