Integrand size = 43, antiderivative size = 214 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+a \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {3 (5 A-5 B+7 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 a d}-\frac {(3 A-5 B+5 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 a d}-\frac {(A-B+C) \sin (c+d x)}{d (a+a \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x)}+\frac {(5 A-5 B+7 C) \sin (c+d x)}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {(3 A-5 B+5 C) \sin (c+d x)}{3 a d \sqrt {\sec (c+d x)}} \]
-(A-B+C)*sin(d*x+c)/d/(a+a*cos(d*x+c))/sec(d*x+c)^(5/2)+1/5*(5*A-5*B+7*C)* sin(d*x+c)/a/d/sec(d*x+c)^(3/2)-1/3*(3*A-5*B+5*C)*sin(d*x+c)/a/d/sec(d*x+c )^(1/2)+3/5*(5*A-5*B+7*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/ d-1/3*(3*A-5*B+5*C)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Ellipt icF(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/a/d
Time = 3.22 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.83 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+a \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {\sec (c+d x)} \left (18 (5 A-5 B+7 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )-10 (3 A-5 B+5 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+(15 A-25 B+22 C+(-10 B+4 C) \cos (c+d x)-3 C \cos (2 (c+d x))) \sec \left (\frac {1}{2} (c+d x)\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {3}{2} (c+d x)\right )\right )\right )}{15 a d (1+\cos (c+d x))} \]
Integrate[(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)/((a + a*Cos[c + d*x])*Se c[c + d*x]^(3/2)),x]
(Cos[(c + d*x)/2]^2*Sqrt[Sec[c + d*x]]*(18*(5*A - 5*B + 7*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2] - 10*(3*A - 5*B + 5*C)*Sqrt[Cos[c + d*x]]* EllipticF[(c + d*x)/2, 2] + (15*A - 25*B + 22*C + (-10*B + 4*C)*Cos[c + d* x] - 3*C*Cos[2*(c + d*x)])*Sec[(c + d*x)/2]*(Sin[(c + d*x)/2] - Sin[(3*(c + d*x))/2])))/(15*a*d*(1 + Cos[c + d*x]))
Time = 0.91 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.83, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.279, Rules used = {3042, 4709, 3042, 3520, 27, 3042, 3227, 3042, 3115, 3042, 3119, 3120}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \cos (c+d x)+C \cos (c+d x)^2}{\sec (c+d x)^{3/2} (a \cos (c+d x)+a)}dx\) |
| \(\Big \downarrow \) 4709 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (C \cos ^2(c+d x)+B \cos (c+d x)+A\right )}{\cos (c+d x) a+a}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (C \sin \left (c+d x+\frac {\pi }{2}\right )^2+B \sin \left (c+d x+\frac {\pi }{2}\right )+A\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx\) |
| \(\Big \downarrow \) 3520 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int -\frac {1}{2} \cos ^{\frac {3}{2}}(c+d x) (a (3 A-5 B+5 C)-a (5 A-5 B+7 C) \cos (c+d x))dx}{a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{d (a \cos (c+d x)+a)}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {\int \cos ^{\frac {3}{2}}(c+d x) (a (3 A-5 B+5 C)-a (5 A-5 B+7 C) \cos (c+d x))dx}{2 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{d (a \cos (c+d x)+a)}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {\int \sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (a (3 A-5 B+5 C)-a (5 A-5 B+7 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx}{2 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{d (a \cos (c+d x)+a)}\right )\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {a (3 A-5 B+5 C) \int \cos ^{\frac {3}{2}}(c+d x)dx-a (5 A-5 B+7 C) \int \cos ^{\frac {5}{2}}(c+d x)dx}{2 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{d (a \cos (c+d x)+a)}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {a (3 A-5 B+5 C) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}dx-a (5 A-5 B+7 C) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}dx}{2 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{d (a \cos (c+d x)+a)}\right )\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {a (3 A-5 B+5 C) \left (\frac {1}{3} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )-a (5 A-5 B+7 C) \left (\frac {3}{5} \int \sqrt {\cos (c+d x)}dx+\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )}{2 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{d (a \cos (c+d x)+a)}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {a (3 A-5 B+5 C) \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )-a (5 A-5 B+7 C) \left (\frac {3}{5} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )}{2 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{d (a \cos (c+d x)+a)}\right )\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {a (3 A-5 B+5 C) \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )-a (5 A-5 B+7 C) \left (\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )}{2 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{d (a \cos (c+d x)+a)}\right )\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {a (3 A-5 B+5 C) \left (\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )-a (5 A-5 B+7 C) \left (\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )}{2 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{d (a \cos (c+d x)+a)}\right )\) |
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(-(((A - B + C)*Cos[c + d*x]^(5/2)*S in[c + d*x])/(d*(a + a*Cos[c + d*x]))) - (a*(3*A - 5*B + 5*C)*((2*Elliptic F[(c + d*x)/2, 2])/(3*d) + (2*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*d)) - a* (5*A - 5*B + 7*C)*((6*EllipticE[(c + d*x)/2, 2])/(5*d) + (2*Cos[c + d*x]^( 3/2)*Sin[c + d*x])/(5*d)))/(2*a^2))
3.13.95.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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] + Simp[b^2*((n - 1)/n) Int[(b*Sin [c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 2*n]
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]
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]
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[c Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b Int [(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)*sin[(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 ] + Simp[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)]
Int[(u_)*((c_.)*sec[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Simp[(c*Sec[a + b*x])^m*(c*Cos[a + b*x])^m Int[ActivateTrig[u]/(c*Cos[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] && !IntegerQ[m] && KnownSineIntegrandQ[u, x]
Time = 4.04 (sec) , antiderivative size = 319, normalized size of antiderivative = 1.49
| method | result | size |
| default | \(\frac {\sqrt {\left (-1+2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \left (15 A F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+45 A E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-25 B F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-45 B E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+25 C F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+63 C E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )-48 C \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (40 B +56 C \right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (30 A -90 B +30 C \right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-15 A +35 B -23 C \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}{15 a \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-1+2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}\) | \(319\) |
int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))/sec(d*x+c)^(3/2),x,me thod=_RETURNVERBOSE)
1/15*((-1+2*cos(1/2*d*x+1/2*c)^2)*sin(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)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(15* A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+45*A*EllipticE(cos(1/2*d*x+1/2*c), 2^(1/2))-25*B*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-45*B*EllipticE(cos(1/2 *d*x+1/2*c),2^(1/2))+25*C*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+63*C*Ellip ticE(cos(1/2*d*x+1/2*c),2^(1/2)))-48*C*sin(1/2*d*x+1/2*c)^8+(40*B+56*C)*si n(1/2*d*x+1/2*c)^6+(30*A-90*B+30*C)*sin(1/2*d*x+1/2*c)^4+(-15*A+35*B-23*C) *sin(1/2*d*x+1/2*c)^2)/a/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)/sin(1/2*d*x+1/2*c)/(-1+2*cos(1/2*d*x+1/2*c)^2)^(1/2 )/d
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.11 (sec) , antiderivative size = 307, normalized size of antiderivative = 1.43 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+a \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x)} \, dx=-\frac {5 \, {\left (\sqrt {2} {\left (-3 i \, A + 5 i \, B - 5 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-3 i \, A + 5 i \, B - 5 i \, C\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, {\left (\sqrt {2} {\left (3 i \, A - 5 i \, B + 5 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (3 i \, A - 5 i \, B + 5 i \, C\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 9 \, {\left (\sqrt {2} {\left (-5 i \, A + 5 i \, B - 7 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-5 i \, A + 5 i \, B - 7 i \, C\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 9 \, {\left (\sqrt {2} {\left (5 i \, A - 5 i \, B + 7 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (5 i \, A - 5 i \, B + 7 i \, C\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {2 \, {\left (6 \, C \cos \left (d x + c\right )^{3} + 2 \, {\left (5 \, B - 2 \, C\right )} \cos \left (d x + c\right )^{2} - 5 \, {\left (3 \, A - 5 \, B + 5 \, C\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{30 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}} \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))/sec(d*x+c)^(3/2 ),x, algorithm="fricas")
-1/30*(5*(sqrt(2)*(-3*I*A + 5*I*B - 5*I*C)*cos(d*x + c) + sqrt(2)*(-3*I*A + 5*I*B - 5*I*C))*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c) ) + 5*(sqrt(2)*(3*I*A - 5*I*B + 5*I*C)*cos(d*x + c) + sqrt(2)*(3*I*A - 5*I *B + 5*I*C))*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 9 *(sqrt(2)*(-5*I*A + 5*I*B - 7*I*C)*cos(d*x + c) + sqrt(2)*(-5*I*A + 5*I*B - 7*I*C))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 9*(sqrt(2)*(5*I*A - 5*I*B + 7*I*C)*cos(d*x + c) + sqrt (2)*(5*I*A - 5*I*B + 7*I*C))*weierstrassZeta(-4, 0, weierstrassPInverse(-4 , 0, cos(d*x + c) - I*sin(d*x + c))) - 2*(6*C*cos(d*x + c)^3 + 2*(5*B - 2* C)*cos(d*x + c)^2 - 5*(3*A - 5*B + 5*C)*cos(d*x + c))*sin(d*x + c)/sqrt(co s(d*x + c)))/(a*d*cos(d*x + c) + a*d)
\[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+a \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {\int \frac {A}{\cos {\left (c + d x \right )} \sec ^{\frac {3}{2}}{\left (c + d x \right )} + \sec ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx + \int \frac {B \cos {\left (c + d x \right )}}{\cos {\left (c + d x \right )} \sec ^{\frac {3}{2}}{\left (c + d x \right )} + \sec ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx + \int \frac {C \cos ^{2}{\left (c + d x \right )}}{\cos {\left (c + d x \right )} \sec ^{\frac {3}{2}}{\left (c + d x \right )} + \sec ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx}{a} \]
(Integral(A/(cos(c + d*x)*sec(c + d*x)**(3/2) + sec(c + d*x)**(3/2)), x) + Integral(B*cos(c + d*x)/(cos(c + d*x)*sec(c + d*x)**(3/2) + sec(c + d*x)* *(3/2)), x) + Integral(C*cos(c + d*x)**2/(cos(c + d*x)*sec(c + d*x)**(3/2) + sec(c + d*x)**(3/2)), x))/a
\[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+a \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x)} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A}{{\left (a \cos \left (d x + c\right ) + a\right )} \sec \left (d x + c\right )^{\frac {3}{2}}} \,d x } \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))/sec(d*x+c)^(3/2 ),x, algorithm="maxima")
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)/((a*cos(d*x + c) + a)*se c(d*x + c)^(3/2)), x)
\[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+a \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x)} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A}{{\left (a \cos \left (d x + c\right ) + a\right )} \sec \left (d x + c\right )^{\frac {3}{2}}} \,d x } \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))/sec(d*x+c)^(3/2 ),x, algorithm="giac")
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)/((a*cos(d*x + c) + a)*se c(d*x + c)^(3/2)), x)
Timed out. \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+a \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}\,\left (a+a\,\cos \left (c+d\,x\right )\right )} \,d x \]