\(\int \frac {(a+a \cos (c+d x))^2 (A+C \cos ^2(c+d x))}{\sqrt {\sec (c+d x)}} \, dx\) [1173]

Optimal result

Integrand size = 35, antiderivative size = 237 \[ \int \frac {(a+a \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=\frac {16 a^2 (3 A+2 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{15 d}+\frac {4 a^2 (7 A+5 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{21 d}+\frac {2 a^2 (21 A+19 C) \sin (c+d x)}{105 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 C (a+a \cos (c+d x))^2 \sin (c+d x)}{9 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {8 C \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{63 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {4 a^2 (7 A+5 C) \sin (c+d x)}{21 d \sqrt {\sec (c+d x)}} \]

[Out]

2/105*a^2*(21*A+19*C)*sin(d*x+c)/d/sec(d*x+c)^(3/2)+2/9*C*(a+a*cos(d*x+c))^2*sin(d*x+c)/d/sec(d*x+c)^(3/2)+8/6
3*C*(a^2+a^2*cos(d*x+c))*sin(d*x+c)/d/sec(d*x+c)^(3/2)+4/21*a^2*(7*A+5*C)*sin(d*x+c)/d/sec(d*x+c)^(1/2)+16/15*
a^2*(3*A+2*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)/d+4/21*a^2*(7*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)/d

Rubi [A] (verified)

Time = 0.58 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.257, Rules used = {4306, 3125, 3055, 3047, 3102, 2827, 2719, 2715, 2720} \[ \int \frac {(a+a \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=\frac {2 a^2 (21 A+19 C) \sin (c+d x)}{105 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {4 a^2 (7 A+5 C) \sin (c+d x)}{21 d \sqrt {\sec (c+d x)}}+\frac {4 a^2 (7 A+5 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d}+\frac {16 a^2 (3 A+2 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d}+\frac {8 C \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{63 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^2}{9 d \sec ^{\frac {3}{2}}(c+d x)} \]

[In]

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

[Out]

(16*a^2*(3*A + 2*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(15*d) + (4*a^2*(7*A + 5*
C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(21*d) + (2*a^2*(21*A + 19*C)*Sin[c + d*x]
)/(105*d*Sec[c + d*x]^(3/2)) + (2*C*(a + a*Cos[c + d*x])^2*Sin[c + d*x])/(9*d*Sec[c + d*x]^(3/2)) + (8*C*(a^2
+ a^2*Cos[c + d*x])*Sin[c + d*x])/(63*d*Sec[c + d*x]^(3/2)) + (4*a^2*(7*A + 5*C)*Sin[c + d*x])/(21*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 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 3055

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

Rule 3102

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

Rule 3125

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

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 \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \, dx \\ & = \frac {2 C (a+a \cos (c+d x))^2 \sin (c+d x)}{9 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {\left (2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^2 \left (\frac {3}{2} a (3 A+C)+2 a C \cos (c+d x)\right ) \, dx}{9 a} \\ & = \frac {2 C (a+a \cos (c+d x))^2 \sin (c+d x)}{9 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {8 C \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{63 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {\left (4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} (a+a \cos (c+d x)) \left (\frac {3}{4} a^2 (21 A+11 C)+\frac {3}{4} a^2 (21 A+19 C) \cos (c+d x)\right ) \, dx}{63 a} \\ & = \frac {2 C (a+a \cos (c+d x))^2 \sin (c+d x)}{9 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {8 C \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{63 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {\left (4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \left (\frac {3}{4} a^3 (21 A+11 C)+\left (\frac {3}{4} a^3 (21 A+11 C)+\frac {3}{4} a^3 (21 A+19 C)\right ) \cos (c+d x)+\frac {3}{4} a^3 (21 A+19 C) \cos ^2(c+d x)\right ) \, dx}{63 a} \\ & = \frac {2 a^2 (21 A+19 C) \sin (c+d x)}{105 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 C (a+a \cos (c+d x))^2 \sin (c+d x)}{9 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {8 C \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{63 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {\left (8 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \left (21 a^3 (3 A+2 C)+\frac {45}{4} a^3 (7 A+5 C) \cos (c+d x)\right ) \, dx}{315 a} \\ & = \frac {2 a^2 (21 A+19 C) \sin (c+d x)}{105 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 C (a+a \cos (c+d x))^2 \sin (c+d x)}{9 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {8 C \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{63 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {1}{15} \left (8 a^2 (3 A+2 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx+\frac {1}{7} \left (2 a^2 (7 A+5 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \cos ^{\frac {3}{2}}(c+d x) \, dx \\ & = \frac {16 a^2 (3 A+2 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{15 d}+\frac {2 a^2 (21 A+19 C) \sin (c+d x)}{105 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 C (a+a \cos (c+d x))^2 \sin (c+d x)}{9 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {8 C \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{63 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {4 a^2 (7 A+5 C) \sin (c+d x)}{21 d \sqrt {\sec (c+d x)}}+\frac {1}{21} \left (2 a^2 (7 A+5 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx \\ & = \frac {16 a^2 (3 A+2 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{15 d}+\frac {4 a^2 (7 A+5 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{21 d}+\frac {2 a^2 (21 A+19 C) \sin (c+d x)}{105 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 C (a+a \cos (c+d x))^2 \sin (c+d x)}{9 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {8 C \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{63 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {4 a^2 (7 A+5 C) \sin (c+d x)}{21 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 = 4.78 (sec) , antiderivative size = 206, normalized size of antiderivative = 0.87 \[ \int \frac {(a+a \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=\frac {a^2 e^{-i d x} \sqrt {\sec (c+d x)} (\cos (d x)+i \sin (d x)) \left (240 (7 A+5 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-448 i (3 A+2 C) e^{i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )+\cos (c+d x) (4032 i A+2688 i C+60 (28 A+23 C) \sin (c+d x)+14 (18 A+37 C) \sin (2 (c+d x))+180 C \sin (3 (c+d x))+35 C \sin (4 (c+d x)))\right )}{1260 d} \]

[In]

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

[Out]

(a^2*Sqrt[Sec[c + d*x]]*(Cos[d*x] + I*Sin[d*x])*(240*(7*A + 5*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]
- (448*I)*(3*A + 2*C)*E^(I*(c + d*x))*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)
*(c + d*x))] + Cos[c + d*x]*((4032*I)*A + (2688*I)*C + 60*(28*A + 23*C)*Sin[c + d*x] + 14*(18*A + 37*C)*Sin[2*
(c + d*x)] + 180*C*Sin[3*(c + d*x)] + 35*C*Sin[4*(c + d*x)])))/(1260*d*E^(I*d*x))

Maple [A] (verified)

Time = 8.29 (sec) , antiderivative size = 408, normalized size of antiderivative = 1.72

method	result	size
default	\(-\frac {4 \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 )}\, a^{2} \left (-560 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1840 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-252 A -2368 C \right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (672 A +1568 C \right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-273 A -387 C \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+105 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{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 )-252 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{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 )+75 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{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 )-168 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{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 )\right )}{315 \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}\)	\(408\)
parts	\(\text {Expression too large to display}\)	\(949\)

[In]

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

[Out]

-4/315*((-1+2*cos(1/2*d*x+1/2*c)^2)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^2*(-560*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2
*c)^10+1840*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8+(-252*A-2368*C)*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+
(672*A+1568*C)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-273*A-387*C)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+
105*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-252*
A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+75*C*(si
n(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-168*C*(sin(1/
2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))/(-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 = 0.11 (sec) , antiderivative size = 227, normalized size of antiderivative = 0.96 \[ \int \frac {(a+a \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=-\frac {2 \, {\left (15 i \, \sqrt {2} {\left (7 \, A + 5 \, C\right )} a^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 15 i \, \sqrt {2} {\left (7 \, A + 5 \, C\right )} a^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 84 i \, \sqrt {2} {\left (3 \, A + 2 \, C\right )} a^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 84 i \, \sqrt {2} {\left (3 \, A + 2 \, C\right )} a^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {{\left (35 \, C a^{2} \cos \left (d x + c\right )^{4} + 90 \, C a^{2} \cos \left (d x + c\right )^{3} + 7 \, {\left (9 \, A + 16 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 30 \, {\left (7 \, A + 5 \, C\right )} a^{2} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}\right )}}{315 \, d} \]

[In]

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

[Out]

-2/315*(15*I*sqrt(2)*(7*A + 5*C)*a^2*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) - 15*I*sqrt(2)*
(7*A + 5*C)*a^2*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 84*I*sqrt(2)*(3*A + 2*C)*a^2*weier
strassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 84*I*sqrt(2)*(3*A + 2*C)*a^2*we
ierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - (35*C*a^2*cos(d*x + c)^4 + 9
0*C*a^2*cos(d*x + c)^3 + 7*(9*A + 16*C)*a^2*cos(d*x + c)^2 + 30*(7*A + 5*C)*a^2*cos(d*x + c))*sin(d*x + c)/sqr
t(cos(d*x + c)))/d

Sympy [F]

\[ \int \frac {(a+a \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=a^{2} \left (\int \frac {A}{\sqrt {\sec {\left (c + d x \right )}}}\, dx + \int \frac {2 A \cos {\left (c + d x \right )}}{\sqrt {\sec {\left (c + d x \right )}}}\, dx + \int \frac {A \cos ^{2}{\left (c + d x \right )}}{\sqrt {\sec {\left (c + d x \right )}}}\, dx + \int \frac {C \cos ^{2}{\left (c + d x \right )}}{\sqrt {\sec {\left (c + d x \right )}}}\, dx + \int \frac {2 C \cos ^{3}{\left (c + d x \right )}}{\sqrt {\sec {\left (c + d x \right )}}}\, dx + \int \frac {C \cos ^{4}{\left (c + d x \right )}}{\sqrt {\sec {\left (c + d x \right )}}}\, dx\right ) \]

[In]

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

[Out]

a**2*(Integral(A/sqrt(sec(c + d*x)), x) + Integral(2*A*cos(c + d*x)/sqrt(sec(c + d*x)), x) + Integral(A*cos(c
+ d*x)**2/sqrt(sec(c + d*x)), x) + Integral(C*cos(c + d*x)**2/sqrt(sec(c + d*x)), x) + Integral(2*C*cos(c + d*
x)**3/sqrt(sec(c + d*x)), x) + Integral(C*cos(c + d*x)**4/sqrt(sec(c + d*x)), x))

Maxima [F]

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

[In]

integrate((a+a*cos(d*x+c))^2*(A+C*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]

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

[In]

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

[In]

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

[Out]

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