Integrand size = 45, antiderivative size = 161 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}} \, dx=-\frac {(7 A-3 B-C) \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {(A-B+C) \sin (c+d x)}{2 d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{3/2}}+\frac {(5 A-B+C) \sin (c+d x)}{2 a d \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \]
-1/4*(7*A-3*B-C)*arctan(1/2*sin(d*x+c)*a^(1/2)*2^(1/2)/cos(d*x+c)^(1/2)/(a +a*cos(d*x+c))^(1/2))/a^(3/2)/d*2^(1/2)-1/2*(A-B+C)*sin(d*x+c)/d/(a+a*cos( d*x+c))^(3/2)/cos(d*x+c)^(1/2)+1/2*(5*A-B+C)*sin(d*x+c)/a/d/cos(d*x+c)^(1/ 2)/(a+a*cos(d*x+c))^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 5.00 (sec) , antiderivative size = 455, normalized size of antiderivative = 2.83 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}} \, dx=\frac {\cos ^3\left (\frac {1}{2} (c+d x)\right ) \left (30 (A-B+C) \arctan \left (\frac {1-2 \sin \left (\frac {1}{2} (c+d x)\right )}{\sqrt {\cos (c+d x)}}\right )-30 (A-B+C) \arctan \left (\frac {1+2 \sin \left (\frac {1}{2} (c+d x)\right )}{\sqrt {\cos (c+d x)}}\right )-\frac {20 (A-B+C) \sqrt {\cos (c+d x)}}{-1+\sin \left (\frac {1}{2} (c+d x)\right )}+\frac {80 C \sin \left (\frac {1}{2} (c+d x)\right )}{\sqrt {\cos (c+d x)}}-\frac {20 (A-B+C) \sqrt {\cos (c+d x)}}{1+\sin \left (\frac {1}{2} (c+d x)\right )}+\frac {5 (A-B+C) \left (-1+2 \sin \left (\frac {1}{2} (c+d x)\right )\right )}{\sqrt {\cos (c+d x)} \left (\cos \left (\frac {1}{4} (c+d x)\right )+\sin \left (\frac {1}{4} (c+d x)\right )\right )^2}-\frac {5 (A-B+C) \left (1+2 \sin \left (\frac {1}{2} (c+d x)\right )\right )}{\sqrt {\cos (c+d x)} \left (-1+\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {(A+3 B-7 C) \csc ^3\left (\frac {1}{2} (c+d x)\right ) \left (5 (1+4 \cos (c+d x)+\cos (2 (c+d x))) \left (1-\cos (c+d x)+\text {arctanh}\left (\sqrt {-\sec (c+d x) \sin ^2\left (\frac {1}{2} (c+d x)\right )}\right ) \cos (c+d x) \sqrt {2-2 \sec (c+d x)}\right )-2 \operatorname {Hypergeometric2F1}\left (2,\frac {5}{2},\frac {7}{2},-\sec (c+d x) \sin ^2\left (\frac {1}{2} (c+d x)\right )\right ) \sin ^4\left (\frac {1}{2} (c+d x)\right ) \sin (c+d x) \tan (c+d x)\right )}{2 \cos ^{\frac {3}{2}}(c+d x)}\right )}{10 d (a (1+\cos (c+d x)))^{3/2}} \]
Integrate[(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)/(Cos[c + d*x]^(3/2)*(a + a*Cos[c + d*x])^(3/2)),x]
(Cos[(c + d*x)/2]^3*(30*(A - B + C)*ArcTan[(1 - 2*Sin[(c + d*x)/2])/Sqrt[C os[c + d*x]]] - 30*(A - B + C)*ArcTan[(1 + 2*Sin[(c + d*x)/2])/Sqrt[Cos[c + d*x]]] - (20*(A - B + C)*Sqrt[Cos[c + d*x]])/(-1 + Sin[(c + d*x)/2]) + ( 80*C*Sin[(c + d*x)/2])/Sqrt[Cos[c + d*x]] - (20*(A - B + C)*Sqrt[Cos[c + d *x]])/(1 + Sin[(c + d*x)/2]) + (5*(A - B + C)*(-1 + 2*Sin[(c + d*x)/2]))/( Sqrt[Cos[c + d*x]]*(Cos[(c + d*x)/4] + Sin[(c + d*x)/4])^2) - (5*(A - B + C)*(1 + 2*Sin[(c + d*x)/2]))/(Sqrt[Cos[c + d*x]]*(-1 + Sin[(c + d*x)/2])) + ((A + 3*B - 7*C)*Csc[(c + d*x)/2]^3*(5*(1 + 4*Cos[c + d*x] + Cos[2*(c + d*x)])*(1 - Cos[c + d*x] + ArcTanh[Sqrt[-(Sec[c + d*x]*Sin[(c + d*x)/2]^2) ]]*Cos[c + d*x]*Sqrt[2 - 2*Sec[c + d*x]]) - 2*Hypergeometric2F1[2, 5/2, 7/ 2, -(Sec[c + d*x]*Sin[(c + d*x)/2]^2)]*Sin[(c + d*x)/2]^4*Sin[c + d*x]*Tan [c + d*x]))/(2*Cos[c + d*x]^(3/2))))/(10*d*(a*(1 + Cos[c + d*x]))^(3/2))
Time = 0.76 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.01, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 3520, 27, 3042, 3463, 27, 3042, 3261, 218}
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.
| \(\displaystyle \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 3520 |
| \(\displaystyle \frac {\int \frac {a (5 A-B+C)-2 a (A-B-C) \cos (c+d x)}{2 \cos ^{\frac {3}{2}}(c+d x) \sqrt {\cos (c+d x) a+a}}dx}{2 a^2}-\frac {(A-B+C) \sin (c+d x)}{2 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {a (5 A-B+C)-2 a (A-B-C) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {\cos (c+d x) a+a}}dx}{4 a^2}-\frac {(A-B+C) \sin (c+d x)}{2 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {a (5 A-B+C)-2 a (A-B-C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {(A-B+C) \sin (c+d x)}{2 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3463 |
| \(\displaystyle \frac {\frac {2 \int -\frac {a^2 (7 A-3 B-C)}{2 \sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}dx}{a}+\frac {2 a (5 A-B+C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}}{4 a^2}-\frac {(A-B+C) \sin (c+d x)}{2 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {2 a (5 A-B+C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}-a (7 A-3 B-C) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}dx}{4 a^2}-\frac {(A-B+C) \sin (c+d x)}{2 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {2 a (5 A-B+C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}-a (7 A-3 B-C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {(A-B+C) \sin (c+d x)}{2 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3261 |
| \(\displaystyle \frac {\frac {2 a^2 (7 A-3 B-C) \int \frac {1}{\frac {\sin (c+d x) \tan (c+d x) a^3}{\cos (c+d x) a+a}+2 a^2}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}\right )}{d}+\frac {2 a (5 A-B+C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}}{4 a^2}-\frac {(A-B+C) \sin (c+d x)}{2 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {2 a (5 A-B+C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}-\frac {\sqrt {2} \sqrt {a} (7 A-3 B-C) \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{d}}{4 a^2}-\frac {(A-B+C) \sin (c+d x)}{2 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}\) |
Int[(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)/(Cos[c + d*x]^(3/2)*(a + a*Cos [c + d*x])^(3/2)),x]
-1/2*((A - B + C)*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x]) ^(3/2)) + (-((Sqrt[2]*Sqrt[a]*(7*A - 3*B - C)*ArcTan[(Sqrt[a]*Sin[c + d*x] )/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])])/d) + (2*a*(5*A - B + C)*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]]))/(4* a^2)
3.6.14.3.1 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[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[1/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin[(e _.) + (f_.)*(x_)]]), x_Symbol] :> Simp[-2*(a/f) Subst[Int[1/(2*b^2 - (a*c - b*d)*x^2), x], x, b*(Cos[e + f*x]/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*S in[e + f*x]]))], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(f*(n + 1)*(c^2 - d^2))), x] + Simp[1/(b*(n + 1)*(c^2 - d^2)) Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*(a*d*m + b*c*(n + 1)) - B*(a*c*m + b*d*(n + 1)) + b*(B*c - A*d)*(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && Eq Q[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[n, -1] && (IntegerQ[n] || EqQ[m + 1/2, 0])
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)]
Leaf count of result is larger than twice the leaf count of optimal. \(362\) vs. \(2(136)=272\).
Time = 13.28 (sec) , antiderivative size = 363, normalized size of antiderivative = 2.25
| method | result | size |
| default | \(\frac {\left (7 A \sqrt {2}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \left (\cos ^{2}\left (d x +c \right )\right )-3 B \sqrt {2}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \left (\cos ^{2}\left (d x +c \right )\right )-C \sqrt {2}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \left (\cos ^{2}\left (d x +c \right )\right )+7 A \sqrt {2}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \cos \left (d x +c \right )+10 A \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}-3 B \sqrt {2}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \cos \left (d x +c \right )-2 B \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}-C \cos \left (d x +c \right ) \sqrt {2}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )+2 C \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+8 A \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )\right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}}{4 a^{2} d \left (1+\cos \left (d x +c \right )\right )^{2} \sqrt {\cos \left (d x +c \right )}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\) | \(363\) |
| parts | \(\frac {A \left (7 \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (\cos ^{2}\left (d x +c \right )\right ) \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )+5 \sin \left (d x +c \right ) \cos \left (d x +c \right ) \sqrt {2}+14 \cos \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )+4 \sqrt {2}\, \sin \left (d x +c \right )+7 \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )\right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \sqrt {2}}{4 d \left (1+\cos \left (d x +c \right )\right )^{2} \sqrt {\cos \left (d x +c \right )}\, a^{2}}-\frac {B \sqrt {-\left (\csc ^{2}\left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{2}+1}\, \sqrt {\frac {a}{\left (\csc ^{2}\left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{2}+1}}\, \left (\sqrt {-\left (\csc ^{2}\left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{2}+1}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )+3 \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )\right ) \sqrt {2}}{4 d \sqrt {-\frac {\left (\csc ^{2}\left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{2}-1}{\left (\csc ^{2}\left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{2}+1}}\, a^{2}}+\frac {C \left (\sin \left (d x +c \right ) \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}-\arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \cos \left (d x +c \right )-\arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )\right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) \sqrt {2}}{4 d \left (1+\cos \left (d x +c \right )\right )^{2} \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, a^{2}}\) | \(511\) |
int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2)/(a+cos(d*x+c)*a)^(3/2 ),x,method=_RETURNVERBOSE)
1/4/a^2/d*(7*A*2^(1/2)*arcsin(cot(d*x+c)-csc(d*x+c))*cos(d*x+c)^2-3*B*2^(1 /2)*arcsin(cot(d*x+c)-csc(d*x+c))*cos(d*x+c)^2-C*2^(1/2)*arcsin(cot(d*x+c) -csc(d*x+c))*cos(d*x+c)^2+7*A*2^(1/2)*arcsin(cot(d*x+c)-csc(d*x+c))*cos(d* x+c)+10*A*cos(d*x+c)*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)-3*B*2^(1 /2)*arcsin(cot(d*x+c)-csc(d*x+c))*cos(d*x+c)-2*B*cos(d*x+c)*sin(d*x+c)*(co s(d*x+c)/(1+cos(d*x+c)))^(1/2)-C*cos(d*x+c)*2^(1/2)*arcsin(cot(d*x+c)-csc( d*x+c))+2*C*cos(d*x+c)*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+8*A*(c os(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c))*(a*(1+cos(d*x+c)))^(1/2)/(1+co s(d*x+c))^2/cos(d*x+c)^(1/2)/(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)
Time = 0.33 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.31 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}} \, dx=-\frac {\sqrt {2} {\left ({\left (7 \, A - 3 \, B - C\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left (7 \, A - 3 \, B - C\right )} \cos \left (d x + c\right )^{2} + {\left (7 \, A - 3 \, B - C\right )} \cos \left (d x + c\right )\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, {\left (a \cos \left (d x + c\right )^{2} + a \cos \left (d x + c\right )\right )}}\right ) - 2 \, {\left ({\left (5 \, A - B + C\right )} \cos \left (d x + c\right ) + 4 \, A\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{4 \, {\left (a^{2} d \cos \left (d x + c\right )^{3} + 2 \, a^{2} d \cos \left (d x + c\right )^{2} + a^{2} d \cos \left (d x + c\right )\right )}} \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2)/(a+a*cos(d*x+c) )^(3/2),x, algorithm="fricas")
-1/4*(sqrt(2)*((7*A - 3*B - C)*cos(d*x + c)^3 + 2*(7*A - 3*B - C)*cos(d*x + c)^2 + (7*A - 3*B - C)*cos(d*x + c))*sqrt(a)*arctan(1/2*sqrt(2)*sqrt(a*c os(d*x + c) + a)*sqrt(a)*sqrt(cos(d*x + c))*sin(d*x + c)/(a*cos(d*x + c)^2 + a*cos(d*x + c))) - 2*((5*A - B + C)*cos(d*x + c) + 4*A)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))*sin(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))
\[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}} \, dx=\int \frac {A + B \cos {\left (c + d x \right )} + C \cos ^{2}{\left (c + d x \right )}}{\left (a \left (\cos {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}} \cos ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx \]
Integral((A + B*cos(c + d*x) + C*cos(c + d*x)**2)/((a*(cos(c + d*x) + 1))* *(3/2)*cos(c + d*x)**(3/2)), x)
\[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \cos \left (d x + c\right )^{\frac {3}{2}}} \,d x } \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2)/(a+a*cos(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)^(3 /2)*cos(d*x + c)^(3/2)), x)
\[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \cos \left (d x + c\right )^{\frac {3}{2}}} \,d x } \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2)/(a+a*cos(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)^(3 /2)*cos(d*x + c)^(3/2)), x)
Timed out. \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A}{{\cos \left (c+d\,x\right )}^{3/2}\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{3/2}} \,d x \]
int((A + B*cos(c + d*x) + C*cos(c + d*x)^2)/(cos(c + d*x)^(3/2)*(a + a*cos (c + d*x))^(3/2)),x)