\(\int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^2 \sec ^{\frac {5}{2}}(c+d x)} \, dx\) [1302]

Optimal result

Integrand size = 43, antiderivative size = 254 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^2 \sec ^{\frac {5}{2}}(c+d x)} \, dx=\frac {(20 A-35 B+56 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 a^2 d}-\frac {5 (A-2 B+3 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 a^2 d}-\frac {(A-B+C) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2 \sec ^{\frac {7}{2}}(c+d x)}-\frac {(A-2 B+3 C) \sin (c+d x)}{a^2 d (1+\cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x)}+\frac {(20 A-35 B+56 C) \sin (c+d x)}{15 a^2 d \sec ^{\frac {3}{2}}(c+d x)}-\frac {5 (A-2 B+3 C) \sin (c+d x)}{3 a^2 d \sqrt {\sec (c+d x)}} \]

[Out]

-1/3*(A-B+C)*sin(d*x+c)/d/(a+a*cos(d*x+c))^2/sec(d*x+c)^(7/2)-(A-2*B+3*C)*sin(d*x+c)/a^2/d/(1+cos(d*x+c))/sec(
d*x+c)^(5/2)+1/15*(20*A-35*B+56*C)*sin(d*x+c)/a^2/d/sec(d*x+c)^(3/2)-5/3*(A-2*B+3*C)*sin(d*x+c)/a^2/d/sec(d*x+
c)^(1/2)+1/5*(20*A-35*B+56*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-5/3*(A-2*B+3*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 = 0.68 (sec) , antiderivative size = 254, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.163, Rules used = {4306, 3120, 3056, 2827, 2715, 2720, 2719} \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^2 \sec ^{\frac {5}{2}}(c+d x)} \, dx=\frac {(20 A-35 B+56 C) \sin (c+d x)}{15 a^2 d \sec ^{\frac {3}{2}}(c+d x)}-\frac {5 (A-2 B+3 C) \sin (c+d x)}{3 a^2 d \sqrt {\sec (c+d x)}}-\frac {(A-2 B+3 C) \sin (c+d x)}{a^2 d (\cos (c+d x)+1) \sec ^{\frac {5}{2}}(c+d x)}-\frac {5 (A-2 B+3 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 a^2 d}+\frac {(20 A-35 B+56 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a^2 d}-\frac {(A-B+C) \sin (c+d x)}{3 d \sec ^{\frac {7}{2}}(c+d x) (a \cos (c+d x)+a)^2} \]

[In]

Int[(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)/((a + a*Cos[c + d*x])^2*Sec[c + d*x]^(5/2)),x]

[Out]

((20*A - 35*B + 56*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*a^2*d) - (5*(A - 2*B
+ 3*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*a^2*d) - ((A - B + C)*Sin[c + d*x]
)/(3*d*(a + a*Cos[c + d*x])^2*Sec[c + d*x]^(7/2)) - ((A - 2*B + 3*C)*Sin[c + d*x])/(a^2*d*(1 + Cos[c + d*x])*S
ec[c + d*x]^(5/2)) + ((20*A - 35*B + 56*C)*Sin[c + d*x])/(15*a^2*d*Sec[c + d*x]^(3/2)) - (5*(A - 2*B + 3*C)*Si
n[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 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)]

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 {5}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx \\ & = -\frac {(A-B+C) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2 \sec ^{\frac {7}{2}}(c+d x)}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (-\frac {1}{2} a (A-7 B+7 C)+\frac {1}{2} a (5 A-5 B+11 C) \cos (c+d x)\right )}{a+a \cos (c+d x)} \, dx}{3 a^2} \\ & = -\frac {(A-B+C) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2 \sec ^{\frac {7}{2}}(c+d x)}-\frac {(A-2 B+3 C) \sin (c+d x)}{a^2 d (1+\cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x)}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \cos ^{\frac {3}{2}}(c+d x) \left (-\frac {15}{2} a^2 (A-2 B+3 C)+\frac {1}{2} a^2 (20 A-35 B+56 C) \cos (c+d x)\right ) \, dx}{3 a^4} \\ & = -\frac {(A-B+C) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2 \sec ^{\frac {7}{2}}(c+d x)}-\frac {(A-2 B+3 C) \sin (c+d x)}{a^2 d (1+\cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x)}-\frac {\left (5 (A-2 B+3 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \cos ^{\frac {3}{2}}(c+d x) \, dx}{2 a^2}+\frac {\left ((20 A-35 B+56 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \cos ^{\frac {5}{2}}(c+d x) \, dx}{6 a^2} \\ & = -\frac {(A-B+C) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2 \sec ^{\frac {7}{2}}(c+d x)}-\frac {(A-2 B+3 C) \sin (c+d x)}{a^2 d (1+\cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x)}+\frac {(20 A-35 B+56 C) \sin (c+d x)}{15 a^2 d \sec ^{\frac {3}{2}}(c+d x)}-\frac {5 (A-2 B+3 C) \sin (c+d x)}{3 a^2 d \sqrt {\sec (c+d x)}}-\frac {\left (5 (A-2 B+3 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 ((20 A-35 B+56 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{10 a^2} \\ & = \frac {(20 A-35 B+56 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 a^2 d}-\frac {5 (A-2 B+3 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-B+C) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2 \sec ^{\frac {7}{2}}(c+d x)}-\frac {(A-2 B+3 C) \sin (c+d x)}{a^2 d (1+\cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x)}+\frac {(20 A-35 B+56 C) \sin (c+d x)}{15 a^2 d \sec ^{\frac {3}{2}}(c+d x)}-\frac {5 (A-2 B+3 C) \sin (c+d x)}{3 a^2 d \sqrt {\sec (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 5.37 (sec) , antiderivative size = 200, normalized size of antiderivative = 0.79 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^2 \sec ^{\frac {5}{2}}(c+d x)} \, dx=\frac {2 \cos ^4\left (\frac {1}{2} (c+d x)\right ) \sqrt {\sec (c+d x)} \left (6 (20 A-35 B+56 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )-50 (A-2 B+3 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+\frac {1}{4} (50 A-110 B+158 C+(60 A-130 B+179 C) \cos (c+d x)+(-10 B+8 C) \cos (2 (c+d x))-3 C \cos (3 (c+d x))) \sec ^3\left (\frac {1}{2} (c+d x)\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {3}{2} (c+d x)\right )\right )\right )}{15 a^2 d (1+\cos (c+d x))^2} \]

[In]

Integrate[(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)/((a + a*Cos[c + d*x])^2*Sec[c + d*x]^(5/2)),x]

[Out]

(2*Cos[(c + d*x)/2]^4*Sqrt[Sec[c + d*x]]*(6*(20*A - 35*B + 56*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]
- 50*(A - 2*B + 3*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] + ((50*A - 110*B + 158*C + (60*A - 130*B + 1
79*C)*Cos[c + d*x] + (-10*B + 8*C)*Cos[2*(c + d*x)] - 3*C*Cos[3*(c + d*x)])*Sec[(c + d*x)/2]^3*(Sin[(c + d*x)/
2] - Sin[(3*(c + d*x))/2]))/4))/(15*a^2*d*(1 + Cos[c + d*x])^2)

Maple [A] (verified)

Time = 4.31 (sec) , antiderivative size = 491, normalized size of antiderivative = 1.93

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 \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}\, \left (25 A F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+60 A E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-50 B F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-105 B E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+75 C F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+168 C E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \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}\, \left (25 A F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+60 A E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-50 B F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-105 B E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+75 C F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+168 C E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+96 C \left (\sin ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-80 B -128 C \right ) \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-120 A +380 B -328 C \right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (170 A -420 B +526 C \right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-55 A +125 B -171 C \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}{30 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}\)	\(491\)

[In]

int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^2/sec(d*x+c)^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/30*((-1+2*cos(1/2*d*x+1/2*c)^2)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+
1/2*c)^2-1)^(1/2)*(25*A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+60*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-50*B*
EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-105*B*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+75*C*EllipticF(cos(1/2*d*x+1
/2*c),2^(1/2))+168*C*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2+2*(sin(1/2
*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(25*A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+60*A*Ellipti
cE(cos(1/2*d*x+1/2*c),2^(1/2))-50*B*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-105*B*EllipticE(cos(1/2*d*x+1/2*c),2
^(1/2))+75*C*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+168*C*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))*cos(1/2*d*x+1/
2*c)+96*C*sin(1/2*d*x+1/2*c)^10+(-80*B-128*C)*sin(1/2*d*x+1/2*c)^8+(-120*A+380*B-328*C)*sin(1/2*d*x+1/2*c)^6+(
170*A-420*B+526*C)*sin(1/2*d*x+1/2*c)^4+(-55*A+125*B-171*C)*sin(1/2*d*x+1/2*c)^2)/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 = 0.13 (sec) , antiderivative size = 435, normalized size of antiderivative = 1.71 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^2 \sec ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {25 \, {\left (\sqrt {2} {\left (-i \, A + 2 i \, B - 3 i \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} {\left (-i \, A + 2 i \, B - 3 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-i \, A + 2 i \, B - 3 i \, C\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 25 \, {\left (\sqrt {2} {\left (i \, A - 2 i \, B + 3 i \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} {\left (i \, A - 2 i \, B + 3 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (i \, A - 2 i \, B + 3 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 (-20 i \, A + 35 i \, B - 56 i \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} {\left (-20 i \, A + 35 i \, B - 56 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-20 i \, A + 35 i \, B - 56 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 (20 i \, A - 35 i \, B + 56 i \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} {\left (20 i \, A - 35 i \, B + 56 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (20 i \, A - 35 i \, B + 56 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 (6 \, C \cos \left (d x + c\right )^{4} + 2 \, {\left (5 \, B - 4 \, C\right )} \cos \left (d x + c\right )^{3} - {\left (30 \, A - 65 \, B + 94 \, C\right )} \cos \left (d x + c\right )^{2} - 25 \, {\left (A - 2 \, B + 3 \, C\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{30 \, {\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+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^2/sec(d*x+c)^(5/2),x, algorithm="fricas")

[Out]

-1/30*(25*(sqrt(2)*(-I*A + 2*I*B - 3*I*C)*cos(d*x + c)^2 + 2*sqrt(2)*(-I*A + 2*I*B - 3*I*C)*cos(d*x + c) + sqr
t(2)*(-I*A + 2*I*B - 3*I*C))*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 25*(sqrt(2)*(I*A - 2*
I*B + 3*I*C)*cos(d*x + c)^2 + 2*sqrt(2)*(I*A - 2*I*B + 3*I*C)*cos(d*x + c) + sqrt(2)*(I*A - 2*I*B + 3*I*C))*we
ierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 3*(sqrt(2)*(-20*I*A + 35*I*B - 56*I*C)*cos(d*x + c)^
2 + 2*sqrt(2)*(-20*I*A + 35*I*B - 56*I*C)*cos(d*x + c) + sqrt(2)*(-20*I*A + 35*I*B - 56*I*C))*weierstrassZeta(
-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*(sqrt(2)*(20*I*A - 35*I*B + 56*I*C)*cos(
d*x + c)^2 + 2*sqrt(2)*(20*I*A - 35*I*B + 56*I*C)*cos(d*x + c) + sqrt(2)*(20*I*A - 35*I*B + 56*I*C))*weierstra
ssZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - 2*(6*C*cos(d*x + c)^4 + 2*(5*B - 4*
C)*cos(d*x + c)^3 - (30*A - 65*B + 94*C)*cos(d*x + c)^2 - 25*(A - 2*B + 3*C)*cos(d*x + c))*sin(d*x + c)/sqrt(c
os(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. \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^2 \sec ^{\frac {5}{2}}(c+d x)} \, dx=\text {Timed out} \]

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)**2)/(a+a*cos(d*x+c))**2/sec(d*x+c)**(5/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^2 \sec ^{\frac {5}{2}}(c+d x)} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A}{{\left (a \cos \left (d x + c\right ) + a\right )}^{2} \sec \left (d x + c\right )^{\frac {5}{2}}} \,d x } \]

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^2/sec(d*x+c)^(5/2),x, algorithm="maxima")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)/((a*cos(d*x + c) + a)^2*sec(d*x + c)^(5/2)), x)

Giac [F]

\[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^2 \sec ^{\frac {5}{2}}(c+d x)} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A}{{\left (a \cos \left (d x + c\right ) + a\right )}^{2} \sec \left (d x + c\right )^{\frac {5}{2}}} \,d x } \]

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^2/sec(d*x+c)^(5/2),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)/((a*cos(d*x + c) + a)^2*sec(d*x + c)^(5/2)), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^2 \sec ^{\frac {5}{2}}(c+d x)} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^2} \,d x \]

[In]

int((A + B*cos(c + d*x) + C*cos(c + d*x)^2)/((1/cos(c + d*x))^(5/2)*(a + a*cos(c + d*x))^2),x)

[Out]

int((A + B*cos(c + d*x) + C*cos(c + d*x)^2)/((1/cos(c + d*x))^(5/2)*(a + a*cos(c + d*x))^2), x)