Integrand size = 28, antiderivative size = 38 \[ \int \frac {\sqrt {a-a \sin (e+f x)}}{\sqrt {-\sin (e+f x)}} \, dx=\frac {2 \sqrt {a} \arcsin \left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {a-a \sin (e+f x)}}\right )}{f} \] Output:
2*a^(1/2)*arcsin(a^(1/2)*cos(f*x+e)/(a-a*sin(f*x+e))^(1/2))/f
Result contains complex when optimal does not.
Time = 0.83 (sec) , antiderivative size = 119, normalized size of antiderivative = 3.13 \[ \int \frac {\sqrt {a-a \sin (e+f x)}}{\sqrt {-\sin (e+f x)}} \, dx=-\frac {\sqrt {-1+e^{2 i (e+f x)}} \left (\arctan \left (\sqrt {-1+e^{2 i (e+f x)}}\right )+i \text {arctanh}\left (\frac {e^{i (e+f x)}}{\sqrt {-1+e^{2 i (e+f x)}}}\right )\right ) \sqrt {a-a \sin (e+f x)}}{\left (-i+e^{i (e+f x)}\right ) f \sqrt {-\sin (e+f x)}} \] Input:
Integrate[Sqrt[a - a*Sin[e + f*x]]/Sqrt[-Sin[e + f*x]],x]
Output:
-((Sqrt[-1 + E^((2*I)*(e + f*x))]*(ArcTan[Sqrt[-1 + E^((2*I)*(e + f*x))]] + I*ArcTanh[E^(I*(e + f*x))/Sqrt[-1 + E^((2*I)*(e + f*x))]])*Sqrt[a - a*Si n[e + f*x]])/((-I + E^(I*(e + f*x)))*f*Sqrt[-Sin[e + f*x]]))
Time = 0.25 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {3042, 3253, 223}
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.
\(\displaystyle \int \frac {\sqrt {a-a \sin (e+f x)}}{\sqrt {-\sin (e+f x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sqrt {a-a \sin (e+f x)}}{\sqrt {-\sin (e+f x)}}dx\) |
\(\Big \downarrow \) 3253 |
\(\displaystyle -\frac {2 \int \frac {1}{\sqrt {1-\frac {a \cos ^2(e+f x)}{a-a \sin (e+f x)}}}d\left (-\frac {a \cos (e+f x)}{\sqrt {a-a \sin (e+f x)}}\right )}{f}\) |
\(\Big \downarrow \) 223 |
\(\displaystyle \frac {2 \sqrt {a} \arcsin \left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {a-a \sin (e+f x)}}\right )}{f}\) |
Input:
Int[Sqrt[a - a*Sin[e + f*x]]/Sqrt[-Sin[e + f*x]],x]
Output:
(2*Sqrt[a]*ArcSin[(Sqrt[a]*Cos[e + f*x])/Sqrt[a - a*Sin[e + f*x]]])/f
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]
Leaf count of result is larger than twice the leaf count of optimal. \(271\) vs. \(2(32)=64\).
Time = 9.28 (sec) , antiderivative size = 272, normalized size of antiderivative = 7.16
method | result | size |
default | \(\frac {\sqrt {2}\, \sqrt {-\frac {2 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )}{\left (\cos \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}}\, \sqrt {-\left (2 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right ) a}\, \left (\arctan \left (\frac {\sqrt {-\frac {2 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )}{\left (\cos \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}}\, \sin \left (\frac {f x}{2}+\frac {e}{2}\right )+\cos \left (\frac {f x}{2}+\frac {e}{2}\right )-1}{-1+\cos \left (\frac {f x}{2}+\frac {e}{2}\right )}\right )-\arctan \left (\frac {-\sqrt {-\frac {2 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )}{\left (\cos \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}}\, \sin \left (\frac {f x}{2}+\frac {e}{2}\right )+\cos \left (\frac {f x}{2}+\frac {e}{2}\right )-1}{-1+\cos \left (\frac {f x}{2}+\frac {e}{2}\right )}\right )\right ) \left (\cos \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{f \left (\cos \left (\frac {f x}{2}+\frac {e}{2}\right )-\sin \left (\frac {f x}{2}+\frac {e}{2}\right )\right ) \sqrt {-\sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )}}\) | \(272\) |
Input:
int((a-sin(f*x+e)*a)^(1/2)/(-sin(f*x+e))^(1/2),x,method=_RETURNVERBOSE)
Output:
1/f*2^(1/2)*(-2*sin(1/2*f*x+1/2*e)*cos(1/2*f*x+1/2*e)/(cos(1/2*f*x+1/2*e)+ 1)^2)^(1/2)*(-(2*sin(1/2*f*x+1/2*e)*cos(1/2*f*x+1/2*e)-1)*a)^(1/2)*(arctan (((-2*sin(1/2*f*x+1/2*e)*cos(1/2*f*x+1/2*e)/(cos(1/2*f*x+1/2*e)+1)^2)^(1/2 )*sin(1/2*f*x+1/2*e)+cos(1/2*f*x+1/2*e)-1)/(-1+cos(1/2*f*x+1/2*e)))-arctan ((-(-2*sin(1/2*f*x+1/2*e)*cos(1/2*f*x+1/2*e)/(cos(1/2*f*x+1/2*e)+1)^2)^(1/ 2)*sin(1/2*f*x+1/2*e)+cos(1/2*f*x+1/2*e)-1)/(-1+cos(1/2*f*x+1/2*e))))*(cos (1/2*f*x+1/2*e)+1)/(cos(1/2*f*x+1/2*e)-sin(1/2*f*x+1/2*e))/(-sin(1/2*f*x+1 /2*e)*cos(1/2*f*x+1/2*e))^(1/2)
Leaf count of result is larger than twice the leaf count of optimal. 95 vs. \(2 (32) = 64\).
Time = 0.19 (sec) , antiderivative size = 341, normalized size of antiderivative = 8.97 \[ \int \frac {\sqrt {a-a \sin (e+f x)}}{\sqrt {-\sin (e+f x)}} \, dx=\left [\frac {\sqrt {-a} \log \left (\frac {128 \, a \cos \left (f x + e\right )^{5} - 128 \, a \cos \left (f x + e\right )^{4} - 416 \, a \cos \left (f x + e\right )^{3} + 128 \, a \cos \left (f x + e\right )^{2} + 8 \, {\left (16 \, \cos \left (f x + e\right )^{4} - 24 \, \cos \left (f x + e\right )^{3} - 66 \, \cos \left (f x + e\right )^{2} - {\left (16 \, \cos \left (f x + e\right )^{3} + 40 \, \cos \left (f x + e\right )^{2} - 26 \, \cos \left (f x + e\right ) - 51\right )} \sin \left (f x + e\right ) + 25 \, \cos \left (f x + e\right ) + 51\right )} \sqrt {-a \sin \left (f x + e\right ) + a} \sqrt {-a} \sqrt {-\sin \left (f x + e\right )} + 289 \, a \cos \left (f x + e\right ) - {\left (128 \, a \cos \left (f x + e\right )^{4} + 256 \, a \cos \left (f x + e\right )^{3} - 160 \, a \cos \left (f x + e\right )^{2} - 288 \, a \cos \left (f x + e\right ) + a\right )} \sin \left (f x + e\right ) + a}{\cos \left (f x + e\right ) - \sin \left (f x + e\right ) + 1}\right )}{4 \, f}, -\frac {\sqrt {a} \arctan \left (\frac {{\left (8 \, \cos \left (f x + e\right )^{2} - 8 \, \sin \left (f x + e\right ) - 9\right )} \sqrt {-a \sin \left (f x + e\right ) + a} \sqrt {a} \sqrt {-\sin \left (f x + e\right )}}{4 \, {\left (2 \, a \cos \left (f x + e\right )^{3} - a \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, a \cos \left (f x + e\right )\right )}}\right )}{2 \, f}\right ] \] Input:
integrate((a-a*sin(f*x+e))^(1/2)/(-sin(f*x+e))^(1/2),x, algorithm="fricas" )
Output:
[1/4*sqrt(-a)*log((128*a*cos(f*x + e)^5 - 128*a*cos(f*x + e)^4 - 416*a*cos (f*x + e)^3 + 128*a*cos(f*x + e)^2 + 8*(16*cos(f*x + e)^4 - 24*cos(f*x + e )^3 - 66*cos(f*x + e)^2 - (16*cos(f*x + e)^3 + 40*cos(f*x + e)^2 - 26*cos( f*x + e) - 51)*sin(f*x + e) + 25*cos(f*x + e) + 51)*sqrt(-a*sin(f*x + e) + a)*sqrt(-a)*sqrt(-sin(f*x + e)) + 289*a*cos(f*x + e) - (128*a*cos(f*x + e )^4 + 256*a*cos(f*x + e)^3 - 160*a*cos(f*x + e)^2 - 288*a*cos(f*x + e) + a )*sin(f*x + e) + a)/(cos(f*x + e) - sin(f*x + e) + 1))/f, -1/2*sqrt(a)*arc tan(1/4*(8*cos(f*x + e)^2 - 8*sin(f*x + e) - 9)*sqrt(-a*sin(f*x + e) + a)* sqrt(a)*sqrt(-sin(f*x + e))/(2*a*cos(f*x + e)^3 - a*cos(f*x + e)*sin(f*x + e) - 2*a*cos(f*x + e)))/f]
\[ \int \frac {\sqrt {a-a \sin (e+f x)}}{\sqrt {-\sin (e+f x)}} \, dx=\int \frac {\sqrt {- a \left (\sin {\left (e + f x \right )} - 1\right )}}{\sqrt {- \sin {\left (e + f x \right )}}}\, dx \] Input:
integrate((a-a*sin(f*x+e))**(1/2)/(-sin(f*x+e))**(1/2),x)
Output:
Integral(sqrt(-a*(sin(e + f*x) - 1))/sqrt(-sin(e + f*x)), x)
\[ \int \frac {\sqrt {a-a \sin (e+f x)}}{\sqrt {-\sin (e+f x)}} \, dx=\int { \frac {\sqrt {-a \sin \left (f x + e\right ) + a}}{\sqrt {-\sin \left (f x + e\right )}} \,d x } \] Input:
integrate((a-a*sin(f*x+e))^(1/2)/(-sin(f*x+e))^(1/2),x, algorithm="maxima" )
Output:
integrate(sqrt(-a*sin(f*x + e) + a)/sqrt(-sin(f*x + e)), x)
Leaf count of result is larger than twice the leaf count of optimal. 95 vs. \(2 (32) = 64\).
Time = 0.17 (sec) , antiderivative size = 95, normalized size of antiderivative = 2.50 \[ \int \frac {\sqrt {a-a \sin (e+f x)}}{\sqrt {-\sin (e+f x)}} \, dx=-\frac {4 \, \sqrt {a} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + \frac {2 \, {\left (2 \, \sqrt {2} - \sqrt {-\tan \left (-\frac {1}{8} \, \pi + \frac {1}{4} \, f x + \frac {1}{4} \, e\right )^{4} + 6 \, \tan \left (-\frac {1}{8} \, \pi + \frac {1}{4} \, f x + \frac {1}{4} \, e\right )^{2} - 1}\right )}}{\tan \left (-\frac {1}{8} \, \pi + \frac {1}{4} \, f x + \frac {1}{4} \, e\right )^{2} - 3}\right )}\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{f} \] Input:
integrate((a-a*sin(f*x+e))^(1/2)/(-sin(f*x+e))^(1/2),x, algorithm="giac")
Output:
-4*sqrt(a)*arctan(-1/2*sqrt(2)*(sqrt(2) + 2*(2*sqrt(2) - sqrt(-tan(-1/8*pi + 1/4*f*x + 1/4*e)^4 + 6*tan(-1/8*pi + 1/4*f*x + 1/4*e)^2 - 1))/(tan(-1/8 *pi + 1/4*f*x + 1/4*e)^2 - 3)))*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))/f
Timed out. \[ \int \frac {\sqrt {a-a \sin (e+f x)}}{\sqrt {-\sin (e+f x)}} \, dx=\int \frac {\sqrt {a-a\,\sin \left (e+f\,x\right )}}{\sqrt {-\sin \left (e+f\,x\right )}} \,d x \] Input:
int((a - a*sin(e + f*x))^(1/2)/(-sin(e + f*x))^(1/2),x)
Output:
int((a - a*sin(e + f*x))^(1/2)/(-sin(e + f*x))^(1/2), x)
\[ \int \frac {\sqrt {a-a \sin (e+f x)}}{\sqrt {-\sin (e+f x)}} \, dx=-\sqrt {a}\, \left (\int \frac {\sqrt {\sin \left (f x +e \right )}\, \sqrt {-\sin \left (f x +e \right )+1}}{\sin \left (f x +e \right )}d x \right ) i \] Input:
int((a-a*sin(f*x+e))^(1/2)/(-sin(f*x+e))^(1/2),x)
Output:
- sqrt(a)*int((sqrt(sin(e + f*x))*sqrt( - sin(e + f*x) + 1))/sin(e + f*x) ,x)*i