\(\int \frac {A+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^2 \sqrt {\sec (c+d x)}} \, dx\) [1194]

Optimal result

Integrand size = 35, antiderivative size = 166 \[ \int \frac {A+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^2 \sqrt {\sec (c+d x)}} \, dx=\frac {4 C \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{a^2 d}+\frac {(A-5 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 a^2 d}-\frac {(A+C) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2 \sec ^{\frac {3}{2}}(c+d x)}+\frac {(A-5 C) \sin (c+d x)}{3 a^2 d (1+\cos (c+d x)) \sqrt {\sec (c+d x)}} \]

[Out]

-1/3*(A+C)*sin(d*x+c)/d/(a+a*cos(d*x+c))^2/sec(d*x+c)^(3/2)+1/3*(A-5*C)*sin(d*x+c)/a^2/d/(1+cos(d*x+c))/sec(d*
x+c)^(1/2)+4*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+1/3*(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 = 0.44 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {4306, 3121, 3056, 2827, 2720, 2719} \[ \int \frac {A+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^2 \sqrt {\sec (c+d x)}} \, dx=\frac {(A-5 C) \sin (c+d x)}{3 a^2 d (\cos (c+d x)+1) \sqrt {\sec (c+d x)}}+\frac {(A-5 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 a^2 d}+\frac {4 C \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d}-\frac {(A+C) \sin (c+d x)}{3 d \sec ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^2} \]

[In]

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

[Out]

(4*C*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(a^2*d) + ((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*S
ec[c + d*x]^(3/2)) + ((A - 5*C)*Sin[c + d*x])/(3*a^2*d*(1 + Cos[c + d*x])*Sqrt[Sec[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 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 {\sqrt {\cos (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 {3}{2}}(c+d x)}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {\cos (c+d x)} \left (\frac {3}{2} a (A-C)+\frac {1}{2} a (A+7 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 {3}{2}}(c+d x)}+\frac {(A-5 C) \sin (c+d x)}{3 a^2 d (1+\cos (c+d x)) \sqrt {\sec (c+d x)}}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {1}{2} a^2 (A-5 C)+6 a^2 C \cos (c+d x)}{\sqrt {\cos (c+d x)}} \, dx}{3 a^4} \\ & = -\frac {(A+C) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2 \sec ^{\frac {3}{2}}(c+d x)}+\frac {(A-5 C) \sin (c+d x)}{3 a^2 d (1+\cos (c+d x)) \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}{6 a^2}+\frac {\left (2 C \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{a^2} \\ & = \frac {4 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-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 {3}{2}}(c+d x)}+\frac {(A-5 C) \sin (c+d x)}{3 a^2 d (1+\cos (c+d x)) \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 = 3.76 (sec) , antiderivative size = 267, normalized size of antiderivative = 1.61 \[ \int \frac {A+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^2 \sqrt {\sec (c+d x)}} \, dx=\frac {e^{-3 i (c+d x)} \left (1+e^{i (c+d x)}\right ) \left (i \left (1+e^{2 i (c+d x)}\right ) \left (-A e^{i (c+d x)} \left (-1+e^{i (c+d x)}\right )+C \left (3+16 e^{i (c+d x)}+20 e^{2 i (c+d x)}+9 e^{3 i (c+d x)}\right )\right )+(A-5 C) e^{i (c+d x)} \left (1+e^{i (c+d x)}\right )^3 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-4 i C e^{2 i (c+d x)} \left (1+e^{i (c+d x)}\right )^3 \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )\right ) \sqrt {\sec (c+d x)}}{12 a^2 d (1+\cos (c+d x))^2} \]

[In]

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

[Out]

((1 + E^(I*(c + d*x)))*(I*(1 + E^((2*I)*(c + d*x)))*(-(A*E^(I*(c + d*x))*(-1 + E^(I*(c + d*x)))) + C*(3 + 16*E
^(I*(c + d*x)) + 20*E^((2*I)*(c + d*x)) + 9*E^((3*I)*(c + d*x)))) + (A - 5*C)*E^(I*(c + d*x))*(1 + E^(I*(c + d
*x)))^3*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] - (4*I)*C*E^((2*I)*(c + d*x))*(1 + E^(I*(c + d*x)))^3*Sqr
t[1 + E^((2*I)*(c + d*x))]*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))])*Sqrt[Sec[c + d*x]])/(12*a^2
*d*E^((3*I)*(c + d*x))*(1 + Cos[c + d*x])^2)

Maple [A] (verified)

Time = 3.20 (sec) , antiderivative size = 348, normalized size of antiderivative = 2.10

method	result	size
default	\(-\frac {\sqrt {\left (-1+2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (2 A \left (\cos ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-24 C \left (\cos ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-10 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \left (\cos ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-24 \left (\cos ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+2 \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A +38 C \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-3 A \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-15 C \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+A +C \right )}{6 a^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-1+2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}\)	\(348\)

[In]

int((A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^2/sec(d*x+c)^(1/2),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)*(2*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))-24*C*cos(1/2*d*x+1/2*c)^6-10*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)^3-24*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))+2*cos(1/2*d*x+1/2*c)^4*A+38*C*cos(1/2*d*x+1/2*c)^4-3*A*cos(1/2*d*x+1/2*c)^2-15*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 = 0.14 (sec) , antiderivative size = 325, normalized size of antiderivative = 1.96 \[ \int \frac {A+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^2 \sqrt {\sec (c+d x)}} \, dx=\frac {{\left (\sqrt {2} {\left (-i \, A + 5 i \, C\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} {\left (i \, A - 5 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-i \, A + 5 i \, C\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + {\left (\sqrt {2} {\left (i \, A - 5 i \, C\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} {\left (-i \, A + 5 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (i \, A - 5 i \, C\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 12 \, {\left (-i \, \sqrt {2} C \cos \left (d x + c\right )^{2} - 2 i \, \sqrt {2} C \cos \left (d x + c\right ) - i \, \sqrt {2} C\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 12 \, {\left (i \, \sqrt {2} C \cos \left (d x + c\right )^{2} + 2 i \, \sqrt {2} C \cos \left (d x + c\right ) + i \, \sqrt {2} C\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {2 \, {\left (6 \, C \cos \left (d x + c\right )^{2} - {\left (A - 5 \, C\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{6 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \]

[In]

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

[Out]

1/6*((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))*w
eierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + (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))
- 12*(-I*sqrt(2)*C*cos(d*x + c)^2 - 2*I*sqrt(2)*C*cos(d*x + c) - I*sqrt(2)*C)*weierstrassZeta(-4, 0, weierstra
ssPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 12*(I*sqrt(2)*C*cos(d*x + c)^2 + 2*I*sqrt(2)*C*cos(d*x + c
) + I*sqrt(2)*C)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - 2*(6*C*co
s(d*x + c)^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]

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

[In]

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

[Out]

(Integral(A/(cos(c + d*x)**2*sqrt(sec(c + d*x)) + 2*cos(c + d*x)*sqrt(sec(c + d*x)) + sqrt(sec(c + d*x))), x)
+ Integral(C*cos(c + d*x)**2/(cos(c + d*x)**2*sqrt(sec(c + d*x)) + 2*cos(c + d*x)*sqrt(sec(c + d*x)) + sqrt(se
c(c + d*x))), x))/a**2

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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