Integrand size = 35, antiderivative size = 230 \[ \int \frac {A+B \cos (c+d x)}{\sqrt {a+a \cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x)} \, dx=-\frac {(4 A-7 B) \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{4 \sqrt {a} d}+\frac {\sqrt {2} (A-B) \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{\sqrt {a} d}+\frac {B \sin (c+d x)}{2 d \sqrt {a+a \cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x)}+\frac {(4 A-B) \sin (c+d x)}{4 d \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}} \] Output:
-1/4*(4*A-7*B)*arcsin(a^(1/2)*sin(d*x+c)/(a+a*cos(d*x+c))^(1/2))*cos(d*x+c )^(1/2)*sec(d*x+c)^(1/2)/a^(1/2)/d+2^(1/2)*(A-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)/a^(1/2)/d+1/2*B*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)/sec(d*x+ c)^(3/2)+1/4*(4*A-B)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)/sec(d*x+c)^(1/2)
Result contains complex when optimal does not.
Time = 2.63 (sec) , antiderivative size = 412, normalized size of antiderivative = 1.79 \[ \int \frac {A+B \cos (c+d x)}{\sqrt {a+a \cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x)} \, dx=-\frac {i e^{-3 i (c+d x)} \left (1+e^{i (c+d x)}\right ) \left (-B-4 A e^{i (c+d x)}+2 B e^{i (c+d x)}+4 A e^{2 i (c+d x)}-3 B e^{2 i (c+d x)}-4 A e^{3 i (c+d x)}+3 B e^{3 i (c+d x)}+4 A e^{4 i (c+d x)}-2 B e^{4 i (c+d x)}+B e^{5 i (c+d x)}-(4 A-7 B) e^{2 i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \text {arcsinh}\left (e^{i (c+d x)}\right )-8 \sqrt {2} (A-B) e^{2 i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \text {arctanh}\left (\frac {1-e^{i (c+d x)}}{\sqrt {2} \sqrt {1+e^{2 i (c+d x)}}}\right )+4 A e^{2 i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \text {arctanh}\left (\sqrt {1+e^{2 i (c+d x)}}\right )-7 B e^{2 i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \text {arctanh}\left (\sqrt {1+e^{2 i (c+d x)}}\right )\right ) \sqrt {\sec (c+d x)}}{16 d \sqrt {a (1+\cos (c+d x))}} \] Input:
Integrate[(A + B*Cos[c + d*x])/(Sqrt[a + a*Cos[c + d*x]]*Sec[c + d*x]^(3/2 )),x]
Output:
((-1/16*I)*(1 + E^(I*(c + d*x)))*(-B - 4*A*E^(I*(c + d*x)) + 2*B*E^(I*(c + d*x)) + 4*A*E^((2*I)*(c + d*x)) - 3*B*E^((2*I)*(c + d*x)) - 4*A*E^((3*I)* (c + d*x)) + 3*B*E^((3*I)*(c + d*x)) + 4*A*E^((4*I)*(c + d*x)) - 2*B*E^((4 *I)*(c + d*x)) + B*E^((5*I)*(c + d*x)) - (4*A - 7*B)*E^((2*I)*(c + d*x))*S qrt[1 + E^((2*I)*(c + d*x))]*ArcSinh[E^(I*(c + d*x))] - 8*Sqrt[2]*(A - B)* E^((2*I)*(c + d*x))*Sqrt[1 + E^((2*I)*(c + d*x))]*ArcTanh[(1 - E^(I*(c + d *x)))/(Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))])] + 4*A*E^((2*I)*(c + d*x))*S qrt[1 + E^((2*I)*(c + d*x))]*ArcTanh[Sqrt[1 + E^((2*I)*(c + d*x))]] - 7*B* E^((2*I)*(c + d*x))*Sqrt[1 + E^((2*I)*(c + d*x))]*ArcTanh[Sqrt[1 + E^((2*I )*(c + d*x))]])*Sqrt[Sec[c + d*x]])/(d*E^((3*I)*(c + d*x))*Sqrt[a*(1 + Cos [c + d*x])])
Time = 1.34 (sec) , antiderivative size = 224, normalized size of antiderivative = 0.97, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3042, 3440, 3042, 3462, 27, 3042, 3462, 27, 3042, 3461, 3042, 3253, 223, 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.
| \(\displaystyle \int \frac {A+B \cos (c+d x)}{\sec ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \sin \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a \sin \left (c+d x+\frac {\pi }{2}\right )+a}}dx\) |
| \(\Big \downarrow \) 3440 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \cos (c+d x))}{\sqrt {\cos (c+d x) a+a}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx\) |
| \(\Big \downarrow \) 3462 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {\sqrt {\cos (c+d x)} (3 a B+a (4 A-B) \cos (c+d x))}{2 \sqrt {\cos (c+d x) a+a}}dx}{2 a}+\frac {B \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {\sqrt {\cos (c+d x)} (3 a B+a (4 A-B) \cos (c+d x))}{\sqrt {\cos (c+d x) a+a}}dx}{4 a}+\frac {B \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (3 a B+a (4 A-B) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a}+\frac {B \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )\) |
| \(\Big \downarrow \) 3462 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\int \frac {a^2 (4 A-B)-a^2 (4 A-7 B) \cos (c+d x)}{2 \sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}dx}{a}+\frac {a (4 A-B) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}}{4 a}+\frac {B \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\int \frac {a^2 (4 A-B)-a^2 (4 A-7 B) \cos (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}dx}{2 a}+\frac {a (4 A-B) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}}{4 a}+\frac {B \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\int \frac {a^2 (4 A-B)-a^2 (4 A-7 B) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{2 a}+\frac {a (4 A-B) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}}{4 a}+\frac {B \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )\) |
| \(\Big \downarrow \) 3461 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {8 a^2 (A-B) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}dx-a (4 A-7 B) \int \frac {\sqrt {\cos (c+d x) a+a}}{\sqrt {\cos (c+d x)}}dx}{2 a}+\frac {a (4 A-B) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}}{4 a}+\frac {B \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {8 a^2 (A-B) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-a (4 A-7 B) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 a}+\frac {a (4 A-B) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}}{4 a}+\frac {B \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )\) |
| \(\Big \downarrow \) 3253 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {8 a^2 (A-B) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx+\frac {2 a (4 A-7 B) \int \frac {1}{\sqrt {1-\frac {a \sin ^2(c+d x)}{\cos (c+d x) a+a}}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}}{2 a}+\frac {a (4 A-B) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}}{4 a}+\frac {B \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )\) |
| \(\Big \downarrow \) 223 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {8 a^2 (A-B) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-\frac {2 a^{3/2} (4 A-7 B) \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}}{2 a}+\frac {a (4 A-B) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}}{4 a}+\frac {B \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )\) |
| \(\Big \downarrow \) 3261 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {-\frac {16 a^3 (A-B) \int \frac {1}{\frac {\sin (c+d x) \tan (c+d x) a^3}{\cos (c+d x) a+a}+2 a^2}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}\right )}{d}-\frac {2 a^{3/2} (4 A-7 B) \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}}{2 a}+\frac {a (4 A-B) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}}{4 a}+\frac {B \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {8 \sqrt {2} a^{3/2} (A-B) \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{d}-\frac {2 a^{3/2} (4 A-7 B) \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}}{2 a}+\frac {a (4 A-B) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}}{4 a}+\frac {B \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )\) |
Input:
Int[(A + B*Cos[c + d*x])/(Sqrt[a + a*Cos[c + d*x]]*Sec[c + d*x]^(3/2)),x]
Output:
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((B*Cos[c + d*x]^(3/2)*Sin[c + d*x]) /(2*d*Sqrt[a + a*Cos[c + d*x]]) + (((-2*a^(3/2)*(4*A - 7*B)*ArcSin[(Sqrt[a ]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/d + (8*Sqrt[2]*a^(3/2)*(A - B)* ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])])/d)/(2*a) + (a*(4*A - B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(d*Sq rt[a + a*Cos[c + d*x]]))/(4*a))
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_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt [a])]/Rt[-b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b]
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(d_.)*sin[(e_.) + (f_.) *(x_)]], x_Symbol] :> Simp[-2/f Subst[Int[1/Sqrt[1 - x^2/a], x], x, b*(Co s[e + f*x]/Sqrt[a + b*Sin[e + f*x]])], x] /; FreeQ[{a, b, d, e, f}, x] && E qQ[a^2 - b^2, 0] && EqQ[d, 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_)])/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Sim p[(A*b - a*B)/b Int[1/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]]) , x], x] + Simp[B/b Int[Sqrt[a + b*Sin[e + f*x]]/Sqrt[c + d*Sin[e + f*x]] , x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, 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)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^n/(f*(m + n + 1))), x] + Simp[1/(b*(m + n + 1)) Int[(a + b*Sin[e + f*x])^m*(c + d*S in[e + f*x])^(n - 1)*Simp[A*b*c*(m + n + 1) + B*(a*c*m + b*d*n) + (A*b*d*(m + n + 1) + B*(a*d*m + b*c*n))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[n, 0] && (IntegerQ[n] || EqQ[m + 1/2, 0])
Time = 17.96 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.04
| method | result | size |
| default | \(\frac {\left (-4 A \sqrt {2}\, \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right )+7 B \sqrt {2}\, \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right )+4 A \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sin \left (d x +c \right )+\sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (\sin \left (2 d x +2 c \right )-\sin \left (d x +c \right )\right ) B -8 A \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )+8 B \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )\right ) \sqrt {2}\, \sqrt {a \left (\cos \left (d x +c \right )+1\right )}}{8 d \sqrt {\sec \left (d x +c \right )}\, \left (\cos \left (d x +c \right )+1\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, a}\) | \(239\) |
| parts | \(\frac {A \sqrt {2}\, \left (\sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sin \left (d x +c \right )-\sqrt {2}\, \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right )-2 \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )\right ) \sqrt {a \left (\cos \left (d x +c \right )+1\right )}}{2 d \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (\cos \left (d x +c \right )+1\right ) \sqrt {\sec \left (d x +c \right )}\, a}+\frac {B \left (7 \sqrt {2}\, \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right )+\sin \left (d x +c \right ) \left (2 \cos \left (d x +c \right )-1\right ) \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+8 \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )\right ) \sqrt {2}\, \sqrt {a \left (\cos \left (d x +c \right )+1\right )}}{8 d \sqrt {\sec \left (d x +c \right )}\, \left (\cos \left (d x +c \right )+1\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, a}\) | \(294\) |
Input:
int((A+B*cos(d*x+c))/(a+a*cos(d*x+c))^(1/2)/sec(d*x+c)^(3/2),x,method=_RET URNVERBOSE)
Output:
1/8/d/sec(d*x+c)^(1/2)*(-4*A*2^(1/2)*arctan(tan(d*x+c)*(cos(d*x+c)/(cos(d* x+c)+1))^(1/2))+7*B*2^(1/2)*arctan(tan(d*x+c)*(cos(d*x+c)/(cos(d*x+c)+1))^ (1/2))+4*A*2^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+2^(1/2)*(c os(d*x+c)/(cos(d*x+c)+1))^(1/2)*(sin(2*d*x+2*c)-sin(d*x+c))*B-8*A*arcsin(c ot(d*x+c)-csc(d*x+c))+8*B*arcsin(cot(d*x+c)-csc(d*x+c)))*2^(1/2)*(a*(cos(d *x+c)+1))^(1/2)/(cos(d*x+c)+1)/(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)/a
Time = 2.08 (sec) , antiderivative size = 194, normalized size of antiderivative = 0.84 \[ \int \frac {A+B \cos (c+d x)}{\sqrt {a+a \cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {{\left ({\left (4 \, A - 7 \, B\right )} \cos \left (d x + c\right ) + 4 \, A - 7 \, B\right )} \sqrt {a} \arctan \left (\frac {\sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) - \frac {4 \, \sqrt {2} {\left ({\left (A - B\right )} a \cos \left (d x + c\right ) + {\left (A - B\right )} a\right )} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right )}{\sqrt {a}} + \frac {{\left (2 \, B \cos \left (d x + c\right )^{2} + {\left (4 \, A - B\right )} \cos \left (d x + c\right )\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{4 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}} \] Input:
integrate((A+B*cos(d*x+c))/(a+a*cos(d*x+c))^(1/2)/sec(d*x+c)^(3/2),x, algo rithm="fricas")
Output:
1/4*(((4*A - 7*B)*cos(d*x + c) + 4*A - 7*B)*sqrt(a)*arctan(sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c))) - 4*sqrt(2)*((A - B)* a*cos(d*x + c) + (A - B)*a)*arctan(sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(c os(d*x + c))/(sqrt(a)*sin(d*x + c)))/sqrt(a) + (2*B*cos(d*x + c)^2 + (4*A - B)*cos(d*x + c))*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/sqrt(cos(d*x + c) ))/(a*d*cos(d*x + c) + a*d)
\[ \int \frac {A+B \cos (c+d x)}{\sqrt {a+a \cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {A + B \cos {\left (c + d x \right )}}{\sqrt {a \left (\cos {\left (c + d x \right )} + 1\right )} \sec ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx \] Input:
integrate((A+B*cos(d*x+c))/(a+a*cos(d*x+c))**(1/2)/sec(d*x+c)**(3/2),x)
Output:
Integral((A + B*cos(c + d*x))/(sqrt(a*(cos(c + d*x) + 1))*sec(c + d*x)**(3 /2)), x)
\[ \int \frac {A+B \cos (c+d x)}{\sqrt {a+a \cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x)} \, dx=\int { \frac {B \cos \left (d x + c\right ) + A}{\sqrt {a \cos \left (d x + c\right ) + a} \sec \left (d x + c\right )^{\frac {3}{2}}} \,d x } \] Input:
integrate((A+B*cos(d*x+c))/(a+a*cos(d*x+c))^(1/2)/sec(d*x+c)^(3/2),x, algo rithm="maxima")
Output:
integrate((B*cos(d*x + c) + A)/(sqrt(a*cos(d*x + c) + a)*sec(d*x + c)^(3/2 )), x)
Timed out. \[ \int \frac {A+B \cos (c+d x)}{\sqrt {a+a \cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x)} \, dx=\text {Timed out} \] Input:
integrate((A+B*cos(d*x+c))/(a+a*cos(d*x+c))^(1/2)/sec(d*x+c)^(3/2),x, algo rithm="giac")
Output:
Timed out
Timed out. \[ \int \frac {A+B \cos (c+d x)}{\sqrt {a+a \cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {A+B\,\cos \left (c+d\,x\right )}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}\,\sqrt {a+a\,\cos \left (c+d\,x\right )}} \,d x \] Input:
int((A + B*cos(c + d*x))/((1/cos(c + d*x))^(3/2)*(a + a*cos(c + d*x))^(1/2 )),x)
Output:
int((A + B*cos(c + d*x))/((1/cos(c + d*x))^(3/2)*(a + a*cos(c + d*x))^(1/2 )), x)
\[ \int \frac {A+B \cos (c+d x)}{\sqrt {a+a \cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {\sqrt {a}\, \left (\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right )}{\cos \left (d x +c \right ) \sec \left (d x +c \right )^{2}+\sec \left (d x +c \right )^{2}}d x \right ) b +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )+1}}{\cos \left (d x +c \right ) \sec \left (d x +c \right )^{2}+\sec \left (d x +c \right )^{2}}d x \right ) a \right )}{a} \] Input:
int((A+B*cos(d*x+c))/(a+a*cos(d*x+c))^(1/2)/sec(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))/(co s(c + d*x)*sec(c + d*x)**2 + sec(c + d*x)**2),x)*b + int((sqrt(sec(c + d*x ))*sqrt(cos(c + d*x) + 1))/(cos(c + d*x)*sec(c + d*x)**2 + sec(c + d*x)**2 ),x)*a))/a