\(\int \frac {(A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^{\frac {5}{2}}(c+d x)}{(a+a \cos (c+d x))^2} \, dx\) [1297]

Optimal result

Integrand size = 43, antiderivative size = 251 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {(7 A-4 B+C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{a^2 d}+\frac {(10 A-5 B+2 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 a^2 d}-\frac {(7 A-4 B+C) \sqrt {\sec (c+d x)} \sin (c+d x)}{a^2 d}+\frac {(10 A-5 B+2 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a^2 d}-\frac {(7 A-4 B+C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac {(A-B+C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2} \]

[Out]

1/3*(10*A-5*B+2*C)*sec(d*x+c)^(3/2)*sin(d*x+c)/a^2/d-1/3*(7*A-4*B+C)*sec(d*x+c)^(3/2)*sin(d*x+c)/a^2/d/(1+cos(
d*x+c))-1/3*(A-B+C)*sec(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^2-(7*A-4*B+C)*sin(d*x+c)*sec(d*x+c)^(1/2)/a
^2/d+(7*A-4*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))*cos(d*x
+c)^(1/2)*sec(d*x+c)^(1/2)/a^2/d+1/3*(10*A-5*B+2*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 = 0.57 (sec) , antiderivative size = 251, 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, 3057, 2827, 2716, 2720, 2719} \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {(10 A-5 B+2 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 a^2 d}-\frac {(7 A-4 B+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{a^2 d}-\frac {(7 A-4 B+C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 a^2 d (\cos (c+d x)+1)}+\frac {(10 A-5 B+2 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 a^2 d}+\frac {(7 A-4 B+C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2} \]

[In]

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

[Out]

((7*A - 4*B + C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(a^2*d) + ((10*A - 5*B + 2*C
)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*a^2*d) - ((7*A - 4*B + C)*Sqrt[Sec[c + d
*x]]*Sin[c + d*x])/(a^2*d) + ((10*A - 5*B + 2*C)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(3*a^2*d) - ((7*A - 4*B + C)
*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(3*a^2*d*(1 + Cos[c + d*x])) - ((A - B + C)*Sec[c + d*x]^(3/2)*Sin[c + d*x])
/(3*d*(a + a*Cos[c + d*x])^2)

Rule 2716

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

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

Mathematica [A] (verified)

Time = 5.01 (sec) , antiderivative size = 212, normalized size of antiderivative = 0.84 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {2 \cos ^4\left (\frac {1}{2} (c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x) \left (6 (7 A-4 B+C) \cos ^{\frac {3}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+2 (10 A-5 B+2 C) \cos ^{\frac {3}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-\frac {1}{4} (56 A-38 B+8 C+(95 A-60 B+9 C) \cos (c+d x)+(64 A-38 B+8 C) \cos (2 (c+d x))+21 A \cos (3 (c+d x))-12 B \cos (3 (c+d x))+3 C \cos (3 (c+d x))) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{3 a^2 d (1+\cos (c+d x))^2} \]

[In]

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

[Out]

(2*Cos[(c + d*x)/2]^4*Sec[c + d*x]^(3/2)*(6*(7*A - 4*B + C)*Cos[c + d*x]^(3/2)*EllipticE[(c + d*x)/2, 2] + 2*(
10*A - 5*B + 2*C)*Cos[c + d*x]^(3/2)*EllipticF[(c + d*x)/2, 2] - ((56*A - 38*B + 8*C + (95*A - 60*B + 9*C)*Cos
[c + d*x] + (64*A - 38*B + 8*C)*Cos[2*(c + d*x)] + 21*A*Cos[3*(c + d*x)] - 12*B*Cos[3*(c + d*x)] + 3*C*Cos[3*(
c + d*x)])*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2])/4))/(3*a^2*d*(1 + Cos[c + d*x])^2)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(723\) vs. \(2(279)=558\).

Time = 13.17 (sec) , antiderivative size = 724, normalized size of antiderivative = 2.88

method	result	size
default	\(\text {Expression too large to display}\)	\(724\)

[In]

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

[Out]

-1/2*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/a^2*((4*A-2*B)*(cos(1/2*d*x+1/2*c)*(2*sin(1/2*d
*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(EllipticF(cos(1/2*d*x+1/2*c),2^(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)/cos(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^4+sin
(1/2*d*x+1/2*c)^2)^(1/2)+1/3*(A-B+C)*(2*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*Ellip
ticF(cos(1/2*d*x+1/2*c),2^(1/2))-3*EllipticE(cos(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*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-3
*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))*cos(1/2*d*x+1/2*c)-12*sin(1/2*d*x+1/2*c)^6+20*sin(1/2*d*x+1/2*c)^4-7*s
in(1/2*d*x+1/2*c)^2)/cos(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(-1+sin(1/2*d*x+1
/2*c)^2)+4*A*(-1/6*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)
^2-1/2)^2+1/3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*
d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)))+(-8*A+4*B)/sin(1/2*d*x+1/2*c)^2/(2*sin(1/2*d*x+1/2*
c)^2-1)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*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))))/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 = 464, normalized size of antiderivative = 1.85 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {{\left (\sqrt {2} {\left (-10 i \, A + 5 i \, B - 2 i \, C\right )} \cos \left (d x + c\right )^{3} - 2 \, \sqrt {2} {\left (10 i \, A - 5 i \, B + 2 i \, C\right )} \cos \left (d x + c\right )^{2} + \sqrt {2} {\left (-10 i \, A + 5 i \, B - 2 i \, C\right )} \cos \left (d x + 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 (10 i \, A - 5 i \, B + 2 i \, C\right )} \cos \left (d x + c\right )^{3} - 2 \, \sqrt {2} {\left (-10 i \, A + 5 i \, B - 2 i \, C\right )} \cos \left (d x + c\right )^{2} + \sqrt {2} {\left (10 i \, A - 5 i \, B + 2 i \, C\right )} \cos \left (d x + 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 (-7 i \, A + 4 i \, B - i \, C\right )} \cos \left (d x + c\right )^{3} + 2 \, \sqrt {2} {\left (-7 i \, A + 4 i \, B - i \, C\right )} \cos \left (d x + c\right )^{2} + \sqrt {2} {\left (-7 i \, A + 4 i \, B - i \, C\right )} \cos \left (d x + 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 (7 i \, A - 4 i \, B + i \, C\right )} \cos \left (d x + c\right )^{3} + 2 \, \sqrt {2} {\left (7 i \, A - 4 i \, B + i \, C\right )} \cos \left (d x + c\right )^{2} + \sqrt {2} {\left (7 i \, A - 4 i \, B + i \, C\right )} \cos \left (d x + 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 (3 \, {\left (7 \, A - 4 \, B + C\right )} \cos \left (d x + c\right )^{3} + {\left (32 \, A - 19 \, B + 4 \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (4 \, A - 3 \, B\right )} \cos \left (d x + c\right ) - 2 \, A\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 )^{3} + 2 \, a^{2} d \cos \left (d x + c\right )^{2} + a^{2} d \cos \left (d x + c\right )\right )}} \]

[In]

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

[Out]

1/6*((sqrt(2)*(-10*I*A + 5*I*B - 2*I*C)*cos(d*x + c)^3 - 2*sqrt(2)*(10*I*A - 5*I*B + 2*I*C)*cos(d*x + c)^2 + s
qrt(2)*(-10*I*A + 5*I*B - 2*I*C)*cos(d*x + c))*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + (sq
rt(2)*(10*I*A - 5*I*B + 2*I*C)*cos(d*x + c)^3 - 2*sqrt(2)*(-10*I*A + 5*I*B - 2*I*C)*cos(d*x + c)^2 + sqrt(2)*(
10*I*A - 5*I*B + 2*I*C)*cos(d*x + c))*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 3*(sqrt(2)*(
-7*I*A + 4*I*B - I*C)*cos(d*x + c)^3 + 2*sqrt(2)*(-7*I*A + 4*I*B - I*C)*cos(d*x + c)^2 + sqrt(2)*(-7*I*A + 4*I
*B - I*C)*cos(d*x + c))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 3*
(sqrt(2)*(7*I*A - 4*I*B + I*C)*cos(d*x + c)^3 + 2*sqrt(2)*(7*I*A - 4*I*B + I*C)*cos(d*x + c)^2 + sqrt(2)*(7*I*
A - 4*I*B + I*C)*cos(d*x + c))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)
)) - 2*(3*(7*A - 4*B + C)*cos(d*x + c)^3 + (32*A - 19*B + 4*C)*cos(d*x + c)^2 + 2*(4*A - 3*B)*cos(d*x + c) - 2
*A)*sin(d*x + c)/sqrt(cos(d*x + c)))/(a^2*d*cos(d*x + c)^3 + 2*a^2*d*cos(d*x + c)^2 + a^2*d*cos(d*x + c))

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F(-1)]

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

[In]

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

[Out]

Timed out

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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