\(\int \frac {(A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^{\frac {3}{2}}(c+d x)}{(a+a \cos (c+d x))^3} \, dx\) [1304]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [B] (warning: unable to verify)
   Fricas [C] (verification not implemented)
   Sympy [F(-1)]
   Maxima [F(-1)]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 43, antiderivative size = 277 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x)}{(a+a \cos (c+d x))^3} \, dx=-\frac {(49 A-9 B-C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{10 a^3 d}-\frac {(13 A-3 B-C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{6 a^3 d}+\frac {(49 A-9 B-C) \sqrt {\sec (c+d x)} \sin (c+d x)}{10 a^3 d}-\frac {(A-B+C) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(8 A-3 B-2 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {(13 A-3 B-C) \sqrt {\sec (c+d x)} \sin (c+d x)}{6 d \left (a^3+a^3 \cos (c+d x)\right )} \]

[Out]

1/10*(49*A-9*B-C)*sin(d*x+c)*sec(d*x+c)^(1/2)/a^3/d-1/5*(A-B+C)*sin(d*x+c)*sec(d*x+c)^(1/2)/d/(a+a*cos(d*x+c))
^3-1/15*(8*A-3*B-2*C)*sin(d*x+c)*sec(d*x+c)^(1/2)/a/d/(a+a*cos(d*x+c))^2-1/6*(13*A-3*B-C)*sin(d*x+c)*sec(d*x+c
)^(1/2)/d/(a^3+a^3*cos(d*x+c))-1/10*(49*A-9*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^3/d-1/6*(13*A-3*B-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^3/d

Rubi [A] (verified)

Time = 0.78 (sec) , antiderivative size = 277, 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, 2719, 2720} \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {(49 A-9 B-C) \sin (c+d x) \sqrt {\sec (c+d x)}}{10 a^3 d}-\frac {(13 A-3 B-C) \sin (c+d x) \sqrt {\sec (c+d x)}}{6 d \left (a^3 \cos (c+d x)+a^3\right )}-\frac {(13 A-3 B-C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{6 a^3 d}-\frac {(49 A-9 B-C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^3 d}-\frac {(8 A-3 B-2 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{15 a d (a \cos (c+d x)+a)^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{5 d (a \cos (c+d x)+a)^3} \]

[In]

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

[Out]

-1/10*((49*A - 9*B - C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(a^3*d) - ((13*A - 3*
B - C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(6*a^3*d) + ((49*A - 9*B - C)*Sqrt[Sec
[c + d*x]]*Sin[c + d*x])/(10*a^3*d) - ((A - B + C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(5*d*(a + a*Cos[c + d*x])^
3) - ((8*A - 3*B - 2*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(15*a*d*(a + a*Cos[c + d*x])^2) - ((13*A - 3*B - C)*S
qrt[Sec[c + d*x]]*Sin[c + d*x])/(6*d*(a^3 + a^3*Cos[c + d*x]))

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 {3}{2}}(c+d x) (a+a \cos (c+d x))^3} \, dx \\ & = -\frac {(A-B+C) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {1}{2} a (11 A-B+C)-\frac {5}{2} a (A-B-C) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^2} \, dx}{5 a^2} \\ & = -\frac {(A-B+C) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(8 A-3 B-2 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {1}{2} a^2 (41 A-6 B+C)-\frac {3}{2} a^2 (8 A-3 B-2 C) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))} \, dx}{15 a^4} \\ & = -\frac {(A-B+C) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(8 A-3 B-2 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {(13 A-3 B-C) \sqrt {\sec (c+d x)} \sin (c+d x)}{6 d \left (a^3+a^3 \cos (c+d x)\right )}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {3}{4} a^3 (49 A-9 B-C)-\frac {5}{4} a^3 (13 A-3 B-C) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x)} \, dx}{15 a^6} \\ & = -\frac {(A-B+C) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(8 A-3 B-2 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {(13 A-3 B-C) \sqrt {\sec (c+d x)} \sin (c+d x)}{6 d \left (a^3+a^3 \cos (c+d x)\right )}+\frac {\left ((49 A-9 B-C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x)} \, dx}{20 a^3}-\frac {\left ((13 A-3 B-C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{12 a^3} \\ & = -\frac {(13 A-3 B-C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{6 a^3 d}+\frac {(49 A-9 B-C) \sqrt {\sec (c+d x)} \sin (c+d x)}{10 a^3 d}-\frac {(A-B+C) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(8 A-3 B-2 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {(13 A-3 B-C) \sqrt {\sec (c+d x)} \sin (c+d x)}{6 d \left (a^3+a^3 \cos (c+d x)\right )}-\frac {\left ((49 A-9 B-C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{20 a^3} \\ & = -\frac {(49 A-9 B-C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{10 a^3 d}-\frac {(13 A-3 B-C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{6 a^3 d}+\frac {(49 A-9 B-C) \sqrt {\sec (c+d x)} \sin (c+d x)}{10 a^3 d}-\frac {(A-B+C) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(8 A-3 B-2 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {(13 A-3 B-C) \sqrt {\sec (c+d x)} \sin (c+d x)}{6 d \left (a^3+a^3 \cos (c+d x)\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 3.89 (sec) , antiderivative size = 215, normalized size of antiderivative = 0.78 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x)}{(a+a \cos (c+d x))^3} \, dx=-\frac {2 \cos ^6\left (\frac {1}{2} (c+d x)\right ) \sqrt {\sec (c+d x)} \left (6 (49 A-9 B-C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+10 (13 A-3 B-C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-\frac {1}{8} (992 A-132 B-8 C+(1621 A-261 B+11 C) \cos (c+d x)+4 (188 A-33 B-2 C) \cos (2 (c+d x))+147 A \cos (3 (c+d x))-27 B \cos (3 (c+d x))-3 C \cos (3 (c+d x))) \sec ^4\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{15 a^3 d (1+\cos (c+d x))^3} \]

[In]

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

[Out]

(-2*Cos[(c + d*x)/2]^6*Sqrt[Sec[c + d*x]]*(6*(49*A - 9*B - C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2] + 1
0*(13*A - 3*B - C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] - ((992*A - 132*B - 8*C + (1621*A - 261*B + 11
*C)*Cos[c + d*x] + 4*(188*A - 33*B - 2*C)*Cos[2*(c + d*x)] + 147*A*Cos[3*(c + d*x)] - 27*B*Cos[3*(c + d*x)] -
3*C*Cos[3*(c + d*x)])*Sec[(c + d*x)/2]^4*Tan[(c + d*x)/2])/8))/(15*a^3*d*(1 + Cos[c + d*x])^3)

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(792\) vs. \(2(301)=602\).

Time = 3.29 (sec) , antiderivative size = 793, normalized size of antiderivative = 2.86

method result size
default \(\text {Expression too large to display}\) \(793\)

[In]

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

[Out]

-1/60*(-2*(-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)^2-1)^(1/2)*(sin(1/2*d*x+1
/2*c)^2)^(1/2)*(65*A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-147*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-15*B*El
lipticF(cos(1/2*d*x+1/2*c),2^(1/2))+27*B*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-5*C*EllipticF(cos(1/2*d*x+1/2*c
),2^(1/2))+3*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)^4+4*(-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)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(65*A*Ell
ipticF(cos(1/2*d*x+1/2*c),2^(1/2))-147*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-15*B*EllipticF(cos(1/2*d*x+1/2*
c),2^(1/2))+27*B*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-5*C*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+3*C*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)^4+sin(1/2*d*x+1
/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(65*A*EllipticF(cos(1/2*d*x+1/2*c
),2^(1/2))-147*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-15*B*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+27*B*Ellipti
cE(cos(1/2*d*x+1/2*c),2^(1/2))-5*C*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+3*C*EllipticE(cos(1/2*d*x+1/2*c),2^(1
/2)))*cos(1/2*d*x+1/2*c)+12*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(49*A-9*B-C)*sin(1/2*d*x+1/2*
c)^8-2*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(817*A-147*B-13*C)*sin(1/2*d*x+1/2*c)^6+6*(-2*sin(
1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(248*A-43*B-2*C)*sin(1/2*d*x+1/2*c)^4-(-2*sin(1/2*d*x+1/2*c)^4+si
n(1/2*d*x+1/2*c)^2)^(1/2)*(439*A-69*B-C)*sin(1/2*d*x+1/2*c)^2)/a^3/cos(1/2*d*x+1/2*c)^5/(-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 = 536, normalized size of antiderivative = 1.94 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x)}{(a+a \cos (c+d x))^3} \, dx=-\frac {5 \, {\left (\sqrt {2} {\left (-13 i \, A + 3 i \, B + i \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, \sqrt {2} {\left (-13 i \, A + 3 i \, B + i \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, \sqrt {2} {\left (-13 i \, A + 3 i \, B + i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-13 i \, A + 3 i \, B + i \, C\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, {\left (\sqrt {2} {\left (13 i \, A - 3 i \, B - i \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, \sqrt {2} {\left (13 i \, A - 3 i \, B - i \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, \sqrt {2} {\left (13 i \, A - 3 i \, B - i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (13 i \, A - 3 i \, B - 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 (49 i \, A - 9 i \, B - i \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, \sqrt {2} {\left (49 i \, A - 9 i \, B - i \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, \sqrt {2} {\left (49 i \, A - 9 i \, B - i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (49 i \, A - 9 i \, B - 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 (-49 i \, A + 9 i \, B + i \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, \sqrt {2} {\left (-49 i \, A + 9 i \, B + i \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, \sqrt {2} {\left (-49 i \, A + 9 i \, B + i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-49 i \, A + 9 i \, B + 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 (3 \, {\left (49 \, A - 9 \, B - C\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left (188 \, A - 33 \, B - 2 \, C\right )} \cos \left (d x + c\right )^{2} + 5 \, {\left (59 \, A - 9 \, B + C\right )} \cos \left (d x + c\right ) + 60 \, A\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{60 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \]

[In]

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

[Out]

-1/60*(5*(sqrt(2)*(-13*I*A + 3*I*B + I*C)*cos(d*x + c)^3 + 3*sqrt(2)*(-13*I*A + 3*I*B + I*C)*cos(d*x + c)^2 +
3*sqrt(2)*(-13*I*A + 3*I*B + I*C)*cos(d*x + c) + sqrt(2)*(-13*I*A + 3*I*B + I*C))*weierstrassPInverse(-4, 0, c
os(d*x + c) + I*sin(d*x + c)) + 5*(sqrt(2)*(13*I*A - 3*I*B - I*C)*cos(d*x + c)^3 + 3*sqrt(2)*(13*I*A - 3*I*B -
 I*C)*cos(d*x + c)^2 + 3*sqrt(2)*(13*I*A - 3*I*B - I*C)*cos(d*x + c) + sqrt(2)*(13*I*A - 3*I*B - I*C))*weierst
rassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 3*(sqrt(2)*(49*I*A - 9*I*B - I*C)*cos(d*x + c)^3 + 3*sqrt
(2)*(49*I*A - 9*I*B - I*C)*cos(d*x + c)^2 + 3*sqrt(2)*(49*I*A - 9*I*B - I*C)*cos(d*x + c) + sqrt(2)*(49*I*A -
9*I*B - I*C))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*(sqrt(2)*(
-49*I*A + 9*I*B + I*C)*cos(d*x + c)^3 + 3*sqrt(2)*(-49*I*A + 9*I*B + I*C)*cos(d*x + c)^2 + 3*sqrt(2)*(-49*I*A
+ 9*I*B + I*C)*cos(d*x + c) + sqrt(2)*(-49*I*A + 9*I*B + I*C))*weierstrassZeta(-4, 0, weierstrassPInverse(-4,
0, cos(d*x + c) - I*sin(d*x + c))) - 2*(3*(49*A - 9*B - C)*cos(d*x + c)^3 + 2*(188*A - 33*B - 2*C)*cos(d*x + c
)^2 + 5*(59*A - 9*B + C)*cos(d*x + c) + 60*A)*sin(d*x + c)/sqrt(cos(d*x + c)))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d
*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F(-1)]

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

[In]

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

[Out]

Timed out

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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