Integrand size = 25, antiderivative size = 56 \[ \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {a+a \cos (c+d x)}} \, dx=\frac {\sqrt {2} \arctan \left (\frac {\sqrt {a} \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \cos (c+d x)}}\right )}{\sqrt {a} d} \] Output:
2^(1/2)*arctan(1/2*a^(1/2)*sec(d*x+c)^(1/2)*sin(d*x+c)*2^(1/2)/(a+a*cos(d* x+c))^(1/2))/a^(1/2)/d
Time = 0.11 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.27 \[ \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {a+a \cos (c+d x)}} \, dx=\frac {2 \arctan \left (\frac {\sin \left (\frac {1}{2} (c+d x)\right )}{\sqrt {\cos (c+d x)}}\right ) \cos \left (\frac {1}{2} (c+d x)\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{d \sqrt {a (1+\cos (c+d x))}} \] Input:
Integrate[Sqrt[Sec[c + d*x]]/Sqrt[a + a*Cos[c + d*x]],x]
Output:
(2*ArcTan[Sin[(c + d*x)/2]/Sqrt[Cos[c + d*x]]]*Cos[(c + d*x)/2]*Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]])/(d*Sqrt[a*(1 + Cos[c + d*x])])
Time = 0.36 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.36, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 4710, 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.
\(\displaystyle \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {a \cos (c+d x)+a}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {a \sin \left (c+d x+\frac {\pi }{2}\right )+a}}dx\) |
\(\Big \downarrow \) 4710 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx\) |
\(\Big \downarrow \) 3261 |
\(\displaystyle -\frac {2 a \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\frac {\sin (c+d x) \tan (c+d x) a^3}{\cos (c+d x) a+a}+2 a^2}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}\right )}{d}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{\sqrt {a} d}\) |
Input:
Int[Sqrt[Sec[c + d*x]]/Sqrt[a + a*Cos[c + d*x]],x]
Output:
(Sqrt[2]*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])]*Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]])/(Sqrt[a]*d)
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[(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 = 7.03 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.48
method | result | size |
default | \(-\frac {\sqrt {2}\, \sqrt {a \left (\cos \left (d x +c \right )+1\right )}\, \sqrt {\sec \left (d x +c \right )}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \cos \left (d x +c \right )}{d \left (\cos \left (d x +c \right )+1\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, a}\) | \(83\) |
Input:
int(sec(d*x+c)^(1/2)/(a+a*cos(d*x+c))^(1/2),x,method=_RETURNVERBOSE)
Output:
-1/d*2^(1/2)*(a*(cos(d*x+c)+1))^(1/2)*sec(d*x+c)^(1/2)*arcsin(cot(d*x+c)-c sc(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)/(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)/a
Time = 0.11 (sec) , antiderivative size = 144, normalized size of antiderivative = 2.57 \[ \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {a+a \cos (c+d x)}} \, dx=\left [\frac {\sqrt {2} \sqrt {-\frac {1}{a}} \log \left (-\frac {2 \, \sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {-\frac {1}{a}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 3 \, \cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right ) + 1}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right )}{2 \, d}, -\frac {\sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right )}{\sqrt {a} d}\right ] \] Input:
integrate(sec(d*x+c)^(1/2)/(a+a*cos(d*x+c))^(1/2),x, algorithm="fricas")
Output:
[1/2*sqrt(2)*sqrt(-1/a)*log(-(2*sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(-1/a )*sqrt(cos(d*x + c))*sin(d*x + c) - 3*cos(d*x + c)^2 - 2*cos(d*x + c) + 1) /(cos(d*x + c)^2 + 2*cos(d*x + c) + 1))/d, -sqrt(2)*arctan(sqrt(2)*sqrt(a* cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c)))/(sqrt(a)*d)]
\[ \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {a+a \cos (c+d x)}} \, dx=\int \frac {\sqrt {\sec {\left (c + d x \right )}}}{\sqrt {a \left (\cos {\left (c + d x \right )} + 1\right )}}\, dx \] Input:
integrate(sec(d*x+c)**(1/2)/(a+a*cos(d*x+c))**(1/2),x)
Output:
Integral(sqrt(sec(c + d*x))/sqrt(a*(cos(c + d*x) + 1)), x)
Result contains complex when optimal does not.
Time = 0.63 (sec) , antiderivative size = 522, normalized size of antiderivative = 9.32 \[ \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {a+a \cos (c+d x)}} \, dx =\text {Too large to display} \] Input:
integrate(sec(d*x+c)^(1/2)/(a+a*cos(d*x+c))^(1/2),x, algorithm="maxima")
Output:
sqrt(2)*arctan2(((abs(e^(I*d*x + I*c) + 1)^4 + cos(d*x + c)^4 + sin(d*x + c)^4 + 2*(cos(d*x + c)^2 - sin(d*x + c)^2 - 2*cos(d*x + c) + 1)*abs(e^(I*d *x + I*c) + 1)^2 - 4*cos(d*x + c)^3 + 2*(cos(d*x + c)^2 - 2*cos(d*x + c) + 1)*sin(d*x + c)^2 + 6*cos(d*x + c)^2 - 4*cos(d*x + c) + 1)^(1/4)*sin(1/2* arctan2(2*(cos(d*x + c) - 1)*sin(d*x + c)/abs(e^(I*d*x + I*c) + 1)^2, (abs (e^(I*d*x + I*c) + 1)^2 + cos(d*x + c)^2 - sin(d*x + c)^2 - 2*cos(d*x + c) + 1)/abs(e^(I*d*x + I*c) + 1)^2)) + sin(d*x + c))/abs(e^(I*d*x + I*c) + 1 ), ((abs(e^(I*d*x + I*c) + 1)^4 + cos(d*x + c)^4 + sin(d*x + c)^4 + 2*(cos (d*x + c)^2 - sin(d*x + c)^2 - 2*cos(d*x + c) + 1)*abs(e^(I*d*x + I*c) + 1 )^2 - 4*cos(d*x + c)^3 + 2*(cos(d*x + c)^2 - 2*cos(d*x + c) + 1)*sin(d*x + c)^2 + 6*cos(d*x + c)^2 - 4*cos(d*x + c) + 1)^(1/4)*sqrt(a)*cos(1/2*arcta n2(2*(cos(d*x + c) - 1)*sin(d*x + c)/abs(e^(I*d*x + I*c) + 1)^2, (abs(e^(I *d*x + I*c) + 1)^2 + cos(d*x + c)^2 - sin(d*x + c)^2 - 2*cos(d*x + c) + 1) /abs(e^(I*d*x + I*c) + 1)^2)) + sqrt(a)*cos(d*x + c) - sqrt(a))/(sqrt(a)*a bs(e^(I*d*x + I*c) + 1)))/(sqrt(a)*d)
\[ \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {a+a \cos (c+d x)}} \, dx=\int { \frac {\sqrt {\sec \left (d x + c\right )}}{\sqrt {a \cos \left (d x + c\right ) + a}} \,d x } \] Input:
integrate(sec(d*x+c)^(1/2)/(a+a*cos(d*x+c))^(1/2),x, algorithm="giac")
Output:
integrate(sqrt(sec(d*x + c))/sqrt(a*cos(d*x + c) + a), x)
Timed out. \[ \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {a+a \cos (c+d x)}} \, dx=\int \frac {\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}}{\sqrt {a+a\,\cos \left (c+d\,x\right )}} \,d x \] Input:
int((1/cos(c + d*x))^(1/2)/(a + a*cos(c + d*x))^(1/2),x)
Output:
int((1/cos(c + d*x))^(1/2)/(a + a*cos(c + d*x))^(1/2), x)
\[ \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {a+a \cos (c+d x)}} \, dx=\frac {\sqrt {a}\, \left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )+1}}{\cos \left (d x +c \right )+1}d x \right )}{a} \] Input:
int(sec(d*x+c)^(1/2)/(a+a*cos(d*x+c))^(1/2),x)
Output:
(sqrt(a)*int((sqrt(sec(c + d*x))*sqrt(cos(c + d*x) + 1))/(cos(c + d*x) + 1 ),x))/a