Integrand size = 25, antiderivative size = 105 \[ \int \frac {1}{\sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}} \, dx=\frac {2 \arctan \left (\frac {\sqrt {a} \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{\sqrt {a} d}-\frac {\sqrt {2} \arctan \left (\frac {\sqrt {a} \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \cos (c+d x)}}\right )}{\sqrt {a} d} \]
2*arctan(sin(d*x+c)*a^(1/2)*sec(d*x+c)^(1/2)/(a+a*cos(d*x+c))^(1/2))/d/a^( 1/2)-arctan(1/2*sin(d*x+c)*a^(1/2)*sec(d*x+c)^(1/2)*2^(1/2)/(a+a*cos(d*x+c ))^(1/2))*2^(1/2)/d/a^(1/2)
Result contains complex when optimal does not.
Time = 0.34 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.65 \[ \int \frac {1}{\sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}} \, dx=\frac {i \sqrt {2} e^{-\frac {1}{2} i (c+d x)} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \left (-\text {arcsinh}\left (e^{i (c+d x)}\right )+\sqrt {2} \text {arctanh}\left (\frac {-1+e^{i (c+d x)}}{\sqrt {2} \sqrt {1+e^{2 i (c+d x)}}}\right )+\text {arctanh}\left (\sqrt {1+e^{2 i (c+d x)}}\right )\right ) \cos \left (\frac {1}{2} (c+d x)\right )}{d \sqrt {a (1+\cos (c+d x))}} \]
(I*Sqrt[2]*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2* I)*(c + d*x))]*(-ArcSinh[E^(I*(c + d*x))] + Sqrt[2]*ArcTanh[(-1 + E^(I*(c + d*x)))/(Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))])] + ArcTanh[Sqrt[1 + E^((2 *I)*(c + d*x))]])*Cos[(c + d*x)/2])/(d*E^((I/2)*(c + d*x))*Sqrt[a*(1 + Cos [c + d*x])])
Time = 0.64 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.10, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {3042, 4710, 3042, 3256, 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.
| \(\displaystyle \int \frac {1}{\sqrt {\sec (c+d x)} \sqrt {a \cos (c+d x)+a}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a \sin \left (c+d x+\frac {\pi }{2}\right )+a}}dx\) |
| \(\Big \downarrow \) 4710 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\sqrt {\cos (c+d x)}}{\sqrt {\cos (c+d x) a+a}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx\) |
| \(\Big \downarrow \) 3256 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {\sqrt {\cos (c+d x) a+a}}{\sqrt {\cos (c+d x)}}dx}{a}-\int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}dx\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}-\int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx\right )\) |
| \(\Big \downarrow \) 3253 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-\frac {2 \int \frac {1}{\sqrt {1-\frac {a \sin ^2(c+d x)}{\cos (c+d x) a+a}}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{a d}\right )\) |
| \(\Big \downarrow \) 223 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{\sqrt {a} d}-\int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx\right )\) |
| \(\Big \downarrow \) 3261 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 a \int \frac {1}{\frac {\sin (c+d x) \tan (c+d x) a^3}{\cos (c+d x) a+a}+2 a^2}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}\right )}{d}+\frac {2 \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{\sqrt {a} d}\right )\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{\sqrt {a} d}-\frac {\sqrt {2} \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{\sqrt {a} d}\right )\) |
((2*ArcSin[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/(Sqrt[a]*d) - (Sqrt[2]*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])])/(Sqrt[a]*d))*Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]
3.4.71.3.1 Defintions of rubi rules used
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[Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Simp[d/b Int[Sqrt[a + b*Sin[e + f*x]]/Sqrt[ c + d*Sin[e + f*x]], x], x] + Simp[(b*c - a*d)/b Int[1/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]]), 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[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[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Simp[(c*Csc[a + b*x])^m*(c*Sin[a + b*x])^m Int[ActivateTrig[u]/(c*Sin[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] && !IntegerQ[m] && KnownSineIntegrandQ[u, x]
Time = 3.98 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.03
| method | result | size |
| default | \(\frac {\sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \left (\sqrt {2}\, \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )+\arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )\right ) \sqrt {2}}{d \left (1+\cos \left (d x +c \right )\right ) \sqrt {\sec \left (d x +c \right )}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, a}\) | \(108\) |
1/d*(a*(1+cos(d*x+c)))^(1/2)*(2^(1/2)*arctan(tan(d*x+c)*(cos(d*x+c)/(1+cos (d*x+c)))^(1/2))+arcsin(cot(d*x+c)-csc(d*x+c)))/(1+cos(d*x+c))/sec(d*x+c)^ (1/2)/(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2)/a
Time = 0.29 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.85 \[ \int \frac {1}{\sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}} \, dx=\frac {\sqrt {2} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) - 2 \, \sqrt {a} \arctan \left (\frac {\sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right )}{a d} \]
(sqrt(2)*sqrt(a)*arctan(sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c) )/(sqrt(a)*sin(d*x + c))) - 2*sqrt(a)*arctan(sqrt(a*cos(d*x + c) + a)*sqrt (cos(d*x + c))/(sqrt(a)*sin(d*x + c))))/(a*d)
\[ \int \frac {1}{\sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}} \, dx=\int \frac {1}{\sqrt {a \left (\cos {\left (c + d x \right )} + 1\right )} \sqrt {\sec {\left (c + d x \right )}}}\, dx \]
Result contains complex when optimal does not.
Time = 0.60 (sec) , antiderivative size = 698, normalized size of antiderivative = 6.65 \[ \int \frac {1}{\sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}} \, dx=-\frac {\sqrt {2} \sqrt {a} \arctan \left (\frac {{\left ({\left | 2 \, e^{\left (i \, d x + i \, c\right )} + 2 \right |}^{4} + 16 \, \cos \left (d x + c\right )^{4} + 16 \, \sin \left (d x + c\right )^{4} + 8 \, {\left (\cos \left (d x + c\right )^{2} - \sin \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right ) + 1\right )} {\left | 2 \, e^{\left (i \, d x + i \, c\right )} + 2 \right |}^{2} - 64 \, \cos \left (d x + c\right )^{3} + 32 \, {\left (\cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right ) + 1\right )} \sin \left (d x + c\right )^{2} + 96 \, \cos \left (d x + c\right )^{2} - 64 \, \cos \left (d x + c\right ) + 16\right )}^{\frac {1}{4}} \sin \left (\frac {1}{2} \, \arctan \left (\frac {8 \, {\left (\cos \left (d x + c\right ) - 1\right )} \sin \left (d x + c\right )}{{\left | 2 \, e^{\left (i \, d x + i \, c\right )} + 2 \right |}^{2}}, \frac {{\left | 2 \, e^{\left (i \, d x + i \, c\right )} + 2 \right |}^{2} + 4 \, \cos \left (d x + c\right )^{2} - 4 \, \sin \left (d x + c\right )^{2} - 8 \, \cos \left (d x + c\right ) + 4}{{\left | 2 \, e^{\left (i \, d x + i \, c\right )} + 2 \right |}^{2}}\right )\right ) + 2 \, \sin \left (d x + c\right )}{{\left | 2 \, e^{\left (i \, d x + i \, c\right )} + 2 \right |}}, \frac {{\left ({\left | 2 \, e^{\left (i \, d x + i \, c\right )} + 2 \right |}^{4} + 16 \, \cos \left (d x + c\right )^{4} + 16 \, \sin \left (d x + c\right )^{4} + 8 \, {\left (\cos \left (d x + c\right )^{2} - \sin \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right ) + 1\right )} {\left | 2 \, e^{\left (i \, d x + i \, c\right )} + 2 \right |}^{2} - 64 \, \cos \left (d x + c\right )^{3} + 32 \, {\left (\cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right ) + 1\right )} \sin \left (d x + c\right )^{2} + 96 \, \cos \left (d x + c\right )^{2} - 64 \, \cos \left (d x + c\right ) + 16\right )}^{\frac {1}{4}} \cos \left (\frac {1}{2} \, \arctan \left (\frac {8 \, {\left (\cos \left (d x + c\right ) - 1\right )} \sin \left (d x + c\right )}{{\left | 2 \, e^{\left (i \, d x + i \, c\right )} + 2 \right |}^{2}}, \frac {{\left | 2 \, e^{\left (i \, d x + i \, c\right )} + 2 \right |}^{2} + 4 \, \cos \left (d x + c\right )^{2} - 4 \, \sin \left (d x + c\right )^{2} - 8 \, \cos \left (d x + c\right ) + 4}{{\left | 2 \, e^{\left (i \, d x + i \, c\right )} + 2 \right |}^{2}}\right )\right ) + 2 \, \cos \left (d x + c\right ) - 2}{{\left | 2 \, e^{\left (i \, d x + i \, c\right )} + 2 \right |}}\right ) - \sqrt {a} \arctan \left ({\left (\cos \left (2 \, d x + 2 \, c\right )^{2} + \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )}^{\frac {1}{4}} \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right ) + 1\right )\right ) + \sin \left (d x + c\right ), {\left (\cos \left (2 \, d x + 2 \, c\right )^{2} + \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )}^{\frac {1}{4}} \cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right ) + 1\right )\right ) + \cos \left (d x + c\right )\right )}{a d} \]
-(sqrt(2)*sqrt(a)*arctan2(((abs(2*e^(I*d*x + I*c) + 2)^4 + 16*cos(d*x + c) ^4 + 16*sin(d*x + c)^4 + 8*(cos(d*x + c)^2 - sin(d*x + c)^2 - 2*cos(d*x + c) + 1)*abs(2*e^(I*d*x + I*c) + 2)^2 - 64*cos(d*x + c)^3 + 32*(cos(d*x + c )^2 - 2*cos(d*x + c) + 1)*sin(d*x + c)^2 + 96*cos(d*x + c)^2 - 64*cos(d*x + c) + 16)^(1/4)*sin(1/2*arctan2(8*(cos(d*x + c) - 1)*sin(d*x + c)/abs(2*e ^(I*d*x + I*c) + 2)^2, (abs(2*e^(I*d*x + I*c) + 2)^2 + 4*cos(d*x + c)^2 - 4*sin(d*x + c)^2 - 8*cos(d*x + c) + 4)/abs(2*e^(I*d*x + I*c) + 2)^2)) + 2* sin(d*x + c))/abs(2*e^(I*d*x + I*c) + 2), ((abs(2*e^(I*d*x + I*c) + 2)^4 + 16*cos(d*x + c)^4 + 16*sin(d*x + c)^4 + 8*(cos(d*x + c)^2 - sin(d*x + c)^ 2 - 2*cos(d*x + c) + 1)*abs(2*e^(I*d*x + I*c) + 2)^2 - 64*cos(d*x + c)^3 + 32*(cos(d*x + c)^2 - 2*cos(d*x + c) + 1)*sin(d*x + c)^2 + 96*cos(d*x + c) ^2 - 64*cos(d*x + c) + 16)^(1/4)*cos(1/2*arctan2(8*(cos(d*x + c) - 1)*sin( d*x + c)/abs(2*e^(I*d*x + I*c) + 2)^2, (abs(2*e^(I*d*x + I*c) + 2)^2 + 4*c os(d*x + c)^2 - 4*sin(d*x + c)^2 - 8*cos(d*x + c) + 4)/abs(2*e^(I*d*x + I* c) + 2)^2)) + 2*cos(d*x + c) - 2)/abs(2*e^(I*d*x + I*c) + 2)) - sqrt(a)*ar ctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^( 1/4)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + sin(d*x + c), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/ 4)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + cos(d*x + c) ))/(a*d)
\[ \int \frac {1}{\sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}} \, dx=\int { \frac {1}{\sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\sec \left (d x + c\right )}} \,d x } \]
Timed out. \[ \int \frac {1}{\sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}} \, dx=\int \frac {1}{\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}\,\sqrt {a+a\,\cos \left (c+d\,x\right )}} \,d x \]