Integrand size = 45, antiderivative size = 181 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=-\frac {(7 A-3 B-C) \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{2 \sqrt {2} a^{3/2} d}-\frac {(A-B+C) \sqrt {\sec (c+d x)} \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {(5 A-B+C) \sqrt {\sec (c+d x)} \sin (c+d x)}{2 a d \sqrt {a+a \cos (c+d x)}} \] Output:
-1/4*(7*A-3*B-C)*arctan(1/2*a^(1/2)*sin(d*x+c)*2^(1/2)/cos(d*x+c)^(1/2)/(a +a*cos(d*x+c))^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)*2^(1/2)/a^(3/2)/d- 1/2*(A-B+C)*sec(d*x+c)^(1/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(3/2)+1/2*(5*A- B+C)*sec(d*x+c)^(1/2)*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 6.39 (sec) , antiderivative size = 791, normalized size of antiderivative = 4.37 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx =\text {Too large to display} \] Input:
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])^(3/2),x]
Output:
(2*Cos[c/2 + (d*x)/2]^3*Sqrt[(1 - 2*Sin[c/2 + (d*x)/2]^2)^(-1)]*Sqrt[1 - 2 *Sin[c/2 + (d*x)/2]^2]*((3*(A - B + C)*ArcTan[(1 - 2*Sin[c/2 + (d*x)/2])/S qrt[1 - 2*Sin[c/2 + (d*x)/2]^2]])/2 - (3*(A - B + C)*ArcTan[(1 + 2*Sin[c/2 + (d*x)/2])/Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2]])/2 + (4*C*Sin[c/2 + (d*x)/2 ])/Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2] - ((A - B + C)*(1 - 2*Sin[c/2 + (d*x)/ 2]))/(4*(1 + Sin[c/2 + (d*x)/2])*Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2]) + ((A - B + C)*(1 + 2*Sin[c/2 + (d*x)/2]))/(4*(1 - Sin[c/2 + (d*x)/2])*Sqrt[1 - 2 *Sin[c/2 + (d*x)/2]^2]) + ((A - B + C)*Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2])/( 1 - Sin[c/2 + (d*x)/2]) - ((A - B + C)*Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2])/( 1 + Sin[c/2 + (d*x)/2]) - ((A + 3*B - 7*C)*Sin[c/2 + (d*x)/2]*((2*Cos[(c + d*x)/2]^2*Hypergeometric2F1[2, 5/2, 7/2, -(Sin[c/2 + (d*x)/2]^2/(1 - 2*Si n[c/2 + (d*x)/2]^2))]*Sin[c/2 + (d*x)/2]^2)/(1 - 2*Sin[c/2 + (d*x)/2]^2) + 5*Csc[c/2 + (d*x)/2]^4*Sqrt[-(Sin[c/2 + (d*x)/2]^2/(1 - 2*Sin[c/2 + (d*x) /2]^2))]*(1 - 2*Sin[c/2 + (d*x)/2]^2)^2*(3 - 2*Sin[c/2 + (d*x)/2]^2)*(-Arc Tanh[Sqrt[-(Sin[c/2 + (d*x)/2]^2/(1 - 2*Sin[c/2 + (d*x)/2]^2))]] + Sqrt[-( Sin[c/2 + (d*x)/2]^2/(1 - 2*Sin[c/2 + (d*x)/2]^2))])))/(10*(1 - 2*Sin[c/2 + (d*x)/2]^2)^(3/2))))/(d*(a*(1 + Cos[c + d*x]))^(3/2))
Time = 1.03 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.02, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.244, Rules used = {3042, 4709, 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.
| \(\displaystyle \int \frac {\sec ^{\frac {3}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a \cos (c+d x)+a)^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sec (c+d x)^{3/2} \left (A+B \cos (c+d x)+C \cos (c+d x)^2\right )}{(a \cos (c+d x)+a)^{3/2}}dx\) |
| \(\Big \downarrow \) 4709 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {C \cos ^2(c+d x)+B \cos (c+d x)+A}{\cos ^{\frac {3}{2}}(c+d x) (\cos (c+d x) a+a)^{3/2}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {C \sin \left (c+d x+\frac {\pi }{2}\right )^2+B \sin \left (c+d x+\frac {\pi }{2}\right )+A}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 3520 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {a (5 A-B+C)-2 a (A-B-C) \cos (c+d x)}{2 \cos ^{\frac {3}{2}}(c+d x) \sqrt {\cos (c+d x) a+a}}dx}{2 a^2}-\frac {(A-B+C) \sin (c+d x)}{2 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {a (5 A-B+C)-2 a (A-B-C) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {\cos (c+d x) a+a}}dx}{4 a^2}-\frac {(A-B+C) \sin (c+d x)}{2 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {a (5 A-B+C)-2 a (A-B-C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {(A-B+C) \sin (c+d x)}{2 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}\right )\) |
| \(\Big \downarrow \) 3463 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 \int -\frac {a^2 (7 A-3 B-C)}{2 \sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}dx}{a}+\frac {2 a (5 A-B+C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}}{4 a^2}-\frac {(A-B+C) \sin (c+d x)}{2 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 a (5 A-B+C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}-a (7 A-3 B-C) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}dx}{4 a^2}-\frac {(A-B+C) \sin (c+d x)}{2 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 a (5 A-B+C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}-a (7 A-3 B-C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {(A-B+C) \sin (c+d x)}{2 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}\right )\) |
| \(\Big \downarrow \) 3261 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 a^2 (7 A-3 B-C) \int \frac {1}{\frac {\sin (c+d x) \tan (c+d x) a^3}{\cos (c+d x) a+a}+2 a^2}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}\right )}{d}+\frac {2 a (5 A-B+C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}}{4 a^2}-\frac {(A-B+C) \sin (c+d x)}{2 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}\right )\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 a (5 A-B+C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}-\frac {\sqrt {2} \sqrt {a} (7 A-3 B-C) \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{d}}{4 a^2}-\frac {(A-B+C) \sin (c+d x)}{2 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}\right )\) |
Input:
Int[((A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^(3/2))/(a + a*Co s[c + d*x])^(3/2),x]
Output:
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*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))
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)]
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 = 2.58 (sec) , antiderivative size = 271, normalized size of antiderivative = 1.50
| method | result | size |
| default | \(\frac {\cos \left (d x +c \right ) \sqrt {2}\, \sqrt {\left (1+\cos \left (d x +c \right )\right ) a}\, \left (\frac {7 \left (\cos \left (2 d x +2 c \right )+3+4 \cos \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, A \arcsin \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{2}-\frac {3 \left (\cos \left (2 d x +2 c \right )+3+4 \cos \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, B \arcsin \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{2}+\left (-\cos \left (d x +c \right )^{2}-2 \cos \left (d x +c \right )-1\right ) C \arcsin \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+\sin \left (d x +c \right ) \left (5 \cos \left (d x +c \right )+4\right ) \sqrt {2}\, A -\frac {\sqrt {2}\, \sin \left (2 d x +2 c \right ) B}{2}+\frac {\sqrt {2}\, \sin \left (2 d x +2 c \right ) C}{2}\right ) \sec \left (d x +c \right )^{\frac {3}{2}}}{4 d \,a^{2} \left (1+\cos \left (d x +c \right )\right )^{2}}\) | \(271\) |
| parts | \(\frac {A \cos \left (d x +c \right ) \sqrt {2}\, \sqrt {\left (1+\cos \left (d x +c \right )\right ) a}\, \left (\sin \left (d x +c \right ) \left (5 \cos \left (d x +c \right )+4\right ) \sqrt {2}+\frac {7 \left (\cos \left (2 d x +2 c \right )+3+4 \cos \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arcsin \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{2}\right ) \sec \left (d x +c \right )^{\frac {3}{2}}}{4 d \,a^{2} \left (1+\cos \left (d x +c \right )\right )^{2}}-\frac {B \sqrt {2}\, \sqrt {\left (1+\cos \left (d x +c \right )\right ) a}\, \sec \left (d x +c \right )^{\frac {3}{2}} \left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arcsin \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ) \left (3 \cos \left (d x +c \right )^{3}+6 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )\right )+\sqrt {2}\, \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )\right )}{4 d \,a^{2} \left (1+\cos \left (d x +c \right )\right )^{2}}+\frac {C \sqrt {2}\, \sqrt {\left (1+\cos \left (d x +c \right )\right ) a}\, \sec \left (d x +c \right )^{\frac {3}{2}} \left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arcsin \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ) \left (-\cos \left (d x +c \right )^{3}-2 \cos \left (d x +c \right )^{2}-\cos \left (d x +c \right )\right )+\sqrt {2}\, \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )\right )}{4 d \,a^{2} \left (1+\cos \left (d x +c \right )\right )^{2}}\) | \(379\) |
Input:
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))^(3/2 ),x,method=_RETURNVERBOSE)
Output:
1/4/d/a^2*cos(d*x+c)*2^(1/2)*((1+cos(d*x+c))*a)^(1/2)*(7/2*(cos(2*d*x+2*c) +3+4*cos(d*x+c))*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*A*arcsin(-csc(d*x+c)+co t(d*x+c))-3/2*(cos(2*d*x+2*c)+3+4*cos(d*x+c))*(cos(d*x+c)/(1+cos(d*x+c)))^ (1/2)*B*arcsin(-csc(d*x+c)+cot(d*x+c))+(-cos(d*x+c)^2-2*cos(d*x+c)-1)*C*ar csin(-csc(d*x+c)+cot(d*x+c))*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+sin(d*x+c)* (5*cos(d*x+c)+4)*2^(1/2)*A-1/2*2^(1/2)*sin(2*d*x+2*c)*B+1/2*2^(1/2)*sin(2* d*x+2*c)*C)*sec(d*x+c)^(3/2)/(1+cos(d*x+c))^2
Time = 0.12 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.96 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\frac {\sqrt {2} {\left ({\left (7 \, A - 3 \, B - C\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (7 \, A - 3 \, B - C\right )} \cos \left (d x + c\right ) + 7 \, A - 3 \, B - C\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) + \frac {2 \, {\left ({\left (5 \, A - B + C\right )} \cos \left (d x + c\right ) + 4 \, A\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{4 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \] Input:
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) )^(3/2),x, algorithm="fricas")
Output:
1/4*(sqrt(2)*((7*A - 3*B - C)*cos(d*x + c)^2 + 2*(7*A - 3*B - C)*cos(d*x + c) + 7*A - 3*B - C)*sqrt(a)*arctan(sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt( cos(d*x + c))/(sqrt(a)*sin(d*x + c))) + 2*((5*A - B + C)*cos(d*x + c) + 4* A)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/sqrt(cos(d*x + c)))/(a^2*d*cos(d* x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)
Timed out. \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\text {Timed out} \] Input:
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))**(3/2),x)
Output:
Timed out
\[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{\frac {3}{2}}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \] Input:
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) )^(3/2),x, algorithm="maxima")
Output:
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*sec(d*x + c)^(3/2)/(a*co s(d*x + c) + a)^(3/2), x)
\[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{\frac {3}{2}}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \] Input:
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) )^(3/2),x, algorithm="giac")
Output:
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*sec(d*x + c)^(3/2)/(a*co s(d*x + c) + a)^(3/2), x)
Timed out. \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\int \frac {{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right )}{{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{3/2}} \,d x \] Input:
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))^(3/2),x)
Output:
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))^(3/2), x)
\[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\frac {\sqrt {a}\, \left (\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right ) \sec \left (d x +c \right )}{\cos \left (d x +c \right )^{2}+2 \cos \left (d x +c \right )+1}d x \right ) b +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )}{\cos \left (d x +c \right )^{2}+2 \cos \left (d x +c \right )+1}d x \right ) c +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )+1}\, \sec \left (d x +c \right )}{\cos \left (d x +c \right )^{2}+2 \cos \left (d x +c \right )+1}d x \right ) a \right )}{a^{2}} \] Input:
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))^(3/2 ),x)
Output:
(sqrt(a)*(int((sqrt(sec(c + d*x))*sqrt(cos(c + d*x) + 1)*cos(c + d*x)*sec( c + d*x))/(cos(c + d*x)**2 + 2*cos(c + d*x) + 1),x)*b + int((sqrt(sec(c + d*x))*sqrt(cos(c + d*x) + 1)*cos(c + d*x)**2*sec(c + d*x))/(cos(c + d*x)** 2 + 2*cos(c + d*x) + 1),x)*c + int((sqrt(sec(c + d*x))*sqrt(cos(c + d*x) + 1)*sec(c + d*x))/(cos(c + d*x)**2 + 2*cos(c + d*x) + 1),x)*a))/a**2