Integrand size = 35, antiderivative size = 176 \[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=-\frac {(7 A-3 B) \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{2 \sqrt {2} a^{3/2} d}-\frac {(A-B) \sqrt {\sec (c+d x)} \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {(5 A-B) \sqrt {\sec (c+d x)} \sin (c+d x)}{2 a d \sqrt {a+a \cos (c+d x)}} \] Output:
-1/4*(7*A-3*B)*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)*sec(d*x+c)^(1/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(3/2)+1/2*(5*A-B)*s ec(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 = 5.16 (sec) , antiderivative size = 443, normalized size of antiderivative = 2.52 \[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\frac {\cos ^3\left (\frac {1}{2} (c+d x)\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (30 (A-B) \arctan \left (\frac {1-2 \sin \left (\frac {1}{2} (c+d x)\right )}{\sqrt {\cos (c+d x)}}\right )-30 (A-B) \arctan \left (\frac {1+2 \sin \left (\frac {1}{2} (c+d x)\right )}{\sqrt {\cos (c+d x)}}\right )-\frac {20 (A-B) \sqrt {\cos (c+d x)}}{-1+\sin \left (\frac {1}{2} (c+d x)\right )}-\frac {20 (A-B) \sqrt {\cos (c+d x)}}{1+\sin \left (\frac {1}{2} (c+d x)\right )}+\frac {5 (A-B) \left (-1+2 \sin \left (\frac {1}{2} (c+d x)\right )\right )}{\sqrt {\cos (c+d x)} \left (\cos \left (\frac {1}{4} (c+d x)\right )+\sin \left (\frac {1}{4} (c+d x)\right )\right )^2}-\frac {5 (A-B) \left (1+2 \sin \left (\frac {1}{2} (c+d x)\right )\right )}{\sqrt {\cos (c+d x)} \left (-1+\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {(A+3 B) \csc ^3\left (\frac {1}{2} (c+d x)\right ) \left (5 (1+4 \cos (c+d x)+\cos (2 (c+d x))) \left (1-\cos (c+d x)+\text {arctanh}\left (\sqrt {-\sec (c+d x) \sin ^2\left (\frac {1}{2} (c+d x)\right )}\right ) \cos (c+d x) \sqrt {2-2 \sec (c+d x)}\right )-2 \operatorname {Hypergeometric2F1}\left (2,\frac {5}{2},\frac {7}{2},-\sec (c+d x) \sin ^2\left (\frac {1}{2} (c+d x)\right )\right ) \sin ^4\left (\frac {1}{2} (c+d x)\right ) \sin (c+d x) \tan (c+d x)\right )}{2 \cos ^{\frac {3}{2}}(c+d x)}\right )}{10 d (a (1+\cos (c+d x)))^{3/2}} \] Input:
Integrate[((A + B*Cos[c + d*x])*Sec[c + d*x]^(3/2))/(a + a*Cos[c + d*x])^( 3/2),x]
Output:
(Cos[(c + d*x)/2]^3*Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(30*(A - B)*ArcT an[(1 - 2*Sin[(c + d*x)/2])/Sqrt[Cos[c + d*x]]] - 30*(A - B)*ArcTan[(1 + 2 *Sin[(c + d*x)/2])/Sqrt[Cos[c + d*x]]] - (20*(A - B)*Sqrt[Cos[c + d*x]])/( -1 + Sin[(c + d*x)/2]) - (20*(A - B)*Sqrt[Cos[c + d*x]])/(1 + Sin[(c + d*x )/2]) + (5*(A - B)*(-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)*(1 + 2*Sin[(c + d*x)/2]))/(S qrt[Cos[c + d*x]]*(-1 + Sin[(c + d*x)/2])) + ((A + 3*B)*Csc[(c + d*x)/2]^3 *(5*(1 + 4*Cos[c + d*x] + Cos[2*(c + d*x)])*(1 - Cos[c + d*x] + ArcTanh[Sq rt[-(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 = 0.94 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.02, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.314, Rules used = {3042, 3440, 3042, 3457, 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.
\(\displaystyle \int \frac {\sec ^{\frac {3}{2}}(c+d x) (A+B \cos (c+d x))}{(a \cos (c+d x)+a)^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^{3/2}}dx\) |
\(\Big \downarrow \) 3440 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (\cos (c+d x) a+a)^{3/2}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {A+B \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}}dx\) |
\(\Big \downarrow \) 3457 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {a (5 A-B)-2 a (A-B) \cos (c+d x)}{2 \cos ^{\frac {3}{2}}(c+d x) \sqrt {\cos (c+d x) a+a}}dx}{2 a^2}-\frac {(A-B) \sin (c+d x)}{2 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {a (5 A-B)-2 a (A-B) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {\cos (c+d x) a+a}}dx}{4 a^2}-\frac {(A-B) \sin (c+d x)}{2 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {a (5 A-B)-2 a (A-B) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {(A-B) \sin (c+d x)}{2 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}\right )\) |
\(\Big \downarrow \) 3463 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 \int -\frac {a^2 (7 A-3 B)}{2 \sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}dx}{a}+\frac {2 a (5 A-B) \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}}{4 a^2}-\frac {(A-B) \sin (c+d x)}{2 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 a (5 A-B) \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}-a (7 A-3 B) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}dx}{4 a^2}-\frac {(A-B) \sin (c+d x)}{2 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 a (5 A-B) \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}-a (7 A-3 B) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {(A-B) \sin (c+d x)}{2 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}\right )\) |
\(\Big \downarrow \) 3261 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 a^2 (7 A-3 B) \int \frac {1}{\frac {\sin (c+d x) \tan (c+d x) a^3}{\cos (c+d x) a+a}+2 a^2}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}\right )}{d}+\frac {2 a (5 A-B) \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}}{4 a^2}-\frac {(A-B) \sin (c+d x)}{2 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}\right )\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 a (5 A-B) \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}-\frac {\sqrt {2} \sqrt {a} (7 A-3 B) \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{d}}{4 a^2}-\frac {(A-B) \sin (c+d x)}{2 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}\right )\) |
Input:
Int[((A + B*Cos[c + d*x])*Sec[c + d*x]^(3/2))/(a + a*Cos[c + d*x])^(3/2),x ]
Output:
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(-1/2*((A - B)*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)*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Co s[c + d*x]])])/d) + (2*a*(5*A - B)*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]*Sqr t[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[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*((a_.) + (b_.)*sin[(e_.) + (f_.)* (x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Sim p[(g*Csc[e + f*x])^p*(g*Sin[e + f*x])^p Int[(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^n/(g*Sin[e + f*x])^p), x], x] /; FreeQ[{a, b, c, d, e, f, g , m, n, p}, x] && NeQ[b*c - a*d, 0] && !IntegerQ[p] && !(IntegerQ[m] && I ntegerQ[n])
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*(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] + Simp[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])
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])
Time = 14.27 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.13
method | result | size |
default | \(\frac {\sqrt {2}\, \sqrt {a \left (\cos \left (d x +c \right )+1\right )}\, \cos \left (d x +c \right ) \left (\frac {7 \left (3+\cos \left (2 d x +2 c \right )+4 \cos \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, A \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )}{2}-\frac {3 \left (3+\cos \left (2 d x +2 c \right )+4 \cos \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, B \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )}{2}+\sin \left (d x +c \right ) \left (5 \cos \left (d x +c \right )+4\right ) \sqrt {2}\, A -\frac {\sqrt {2}\, \sin \left (2 d x +2 c \right ) B}{2}\right ) \sec \left (d x +c \right )^{\frac {3}{2}}}{4 d \left (\cos \left (d x +c \right )+1\right )^{2} a^{2}}\) | \(199\) |
parts | \(\frac {A \cos \left (d x +c \right ) \sqrt {2}\, \sqrt {a \left (\cos \left (d x +c \right )+1\right )}\, \left (\sin \left (d x +c \right ) \left (5 \cos \left (d x +c \right )+4\right ) \sqrt {2}+\frac {7 \left (3+\cos \left (2 d x +2 c \right )+4 \cos \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )}{2}\right ) \sec \left (d x +c \right )^{\frac {3}{2}}}{4 d \left (\cos \left (d x +c \right )+1\right )^{2} a^{2}}-\frac {B \sqrt {2}\, \sqrt {a \left (\cos \left (d x +c \right )+1\right )}\, \sec \left (d x +c \right )^{\frac {3}{2}} \left (\sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \left (3 \cos \left (d x +c \right )^{3}+6 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )\right )+\sin \left (d x +c \right ) \cos \left (d x +c \right )^{2} \sqrt {2}\right )}{4 d \left (\cos \left (d x +c \right )+1\right )^{2} a^{2}}\) | \(253\) |
Input:
int((A+B*cos(d*x+c))*sec(d*x+c)^(3/2)/(a+a*cos(d*x+c))^(3/2),x,method=_RET URNVERBOSE)
Output:
1/4/d*2^(1/2)*(a*(cos(d*x+c)+1))^(1/2)*cos(d*x+c)*(7/2*(3+cos(2*d*x+2*c)+4 *cos(d*x+c))*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*A*arcsin(cot(d*x+c)-csc(d*x +c))-3/2*(3+cos(2*d*x+2*c)+4*cos(d*x+c))*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2) *B*arcsin(cot(d*x+c)-csc(d*x+c))+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)*sec(d*x+c)^(3/2)/(cos(d*x+c)+1)^2/a^2
Time = 0.11 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.93 \[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\frac {\sqrt {2} {\left ({\left (7 \, A - 3 \, B\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (7 \, A - 3 \, B\right )} \cos \left (d x + c\right ) + 7 \, A - 3 \, B\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) + \frac {2 \, {\left ({\left (5 \, A - B\right )} \cos \left (d x + c\right ) + 4 \, A\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{4 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \] Input:
integrate((A+B*cos(d*x+c))*sec(d*x+c)^(3/2)/(a+a*cos(d*x+c))^(3/2),x, algo rithm="fricas")
Output:
1/4*(sqrt(2)*((7*A - 3*B)*cos(d*x + c)^2 + 2*(7*A - 3*B)*cos(d*x + c) + 7* A - 3*B)*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)*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. \[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\text {Timed out} \] Input:
integrate((A+B*cos(d*x+c))*sec(d*x+c)**(3/2)/(a+a*cos(d*x+c))**(3/2),x)
Output:
Timed out
\[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\int { \frac {{\left (B \cos \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{\frac {3}{2}}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate((A+B*cos(d*x+c))*sec(d*x+c)^(3/2)/(a+a*cos(d*x+c))^(3/2),x, algo rithm="maxima")
Output:
integrate((B*cos(d*x + c) + A)*sec(d*x + c)^(3/2)/(a*cos(d*x + c) + a)^(3/ 2), x)
\[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\int { \frac {{\left (B \cos \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{\frac {3}{2}}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate((A+B*cos(d*x+c))*sec(d*x+c)^(3/2)/(a+a*cos(d*x+c))^(3/2),x, algo rithm="giac")
Output:
integrate((B*cos(d*x + c) + A)*sec(d*x + c)^(3/2)/(a*cos(d*x + c) + a)^(3/ 2), x)
Timed out. \[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\int \frac {\left (A+B\,\cos \left (c+d\,x\right )\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}}{{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{3/2}} \,d x \] Input:
int(((A + B*cos(c + d*x))*(1/cos(c + d*x))^(3/2))/(a + a*cos(c + d*x))^(3/ 2),x)
Output:
int(((A + B*cos(c + d*x))*(1/cos(c + d*x))^(3/2))/(a + a*cos(c + d*x))^(3/ 2), x)
\[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\frac {\sqrt {a}\, \left (\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right ) \sec \left (d x +c \right )}{\cos \left (d x +c \right )^{2}+2 \cos \left (d x +c \right )+1}d x \right ) b +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )+1}\, \sec \left (d x +c \right )}{\cos \left (d x +c \right )^{2}+2 \cos \left (d x +c \right )+1}d x \right ) a \right )}{a^{2}} \] Input:
int((A+B*cos(d*x+c))*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)*sec(c + d*x))/(cos(c + d*x)**2 + 2*cos(c + d* x) + 1),x)*a))/a**2