Integrand size = 14, antiderivative size = 47 \[ \int \frac {1}{\sqrt {a+a \sin (e+f x)}} \, dx=-\frac {\sqrt {2} \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)}}\right )}{\sqrt {a} f} \] Output:
-arctanh(1/2*a^(1/2)*cos(f*x+e)*2^(1/2)/(a+a*sin(f*x+e))^(1/2))*2^(1/2)/a^ (1/2)/f
Result contains complex when optimal does not.
Time = 0.09 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.55 \[ \int \frac {1}{\sqrt {a+a \sin (e+f x)}} \, dx=\frac {(2+2 i) (-1)^{3/4} \text {arctanh}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \left (-1+\tan \left (\frac {1}{4} (e+f x)\right )\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )}{f \sqrt {a (1+\sin (e+f x))}} \] Input:
Integrate[1/Sqrt[a + a*Sin[e + f*x]],x]
Output:
((2 + 2*I)*(-1)^(3/4)*ArcTanh[(1/2 + I/2)*(-1)^(3/4)*(-1 + Tan[(e + f*x)/4 ])]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]))/(f*Sqrt[a*(1 + Sin[e + f*x])])
Time = 0.21 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3042, 3128, 219}
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.
\(\displaystyle \int \frac {1}{\sqrt {a \sin (e+f x)+a}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sqrt {a \sin (e+f x)+a}}dx\) |
\(\Big \downarrow \) 3128 |
\(\displaystyle -\frac {2 \int \frac {1}{2 a-\frac {a^2 \cos ^2(e+f x)}{\sin (e+f x) a+a}}d\frac {a \cos (e+f x)}{\sqrt {\sin (e+f x) a+a}}}{f}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {\sqrt {2} \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a}}\right )}{\sqrt {a} f}\) |
Input:
Int[1/Sqrt[a + a*Sin[e + f*x]],x]
Output:
-((Sqrt[2]*ArcTanh[(Sqrt[a]*Cos[e + f*x])/(Sqrt[2]*Sqrt[a + a*Sin[e + f*x] ])])/(Sqrt[a]*f))
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2/d Subst[Int[1/(2*a - x^2), x], x, b*(Cos[c + d*x]/Sqrt[a + b*Sin[c + d*x]])], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]
Time = 0.78 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.60
method | result | size |
default | \(-\frac {\left (1+\sin \left (f x +e \right )\right ) \sqrt {-a \left (-1+\sin \left (f x +e \right )\right )}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-a \left (-1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{\sqrt {a}\, \cos \left (f x +e \right ) \sqrt {a +\sin \left (f x +e \right ) a}\, f}\) | \(75\) |
risch | \(\frac {2 i \left ({\mathrm e}^{i \left (f x +e \right )}+i\right ) \sqrt {2}\, {\mathrm e}^{-i \left (f x +e \right )}}{f \sqrt {-a \left (i {\mathrm e}^{2 i \left (f x +e \right )}-i-2 \,{\mathrm e}^{i \left (f x +e \right )}\right ) {\mathrm e}^{-i \left (f x +e \right )}}}-\frac {2 i \left ({\mathrm e}^{i \left (f x +e \right )}+i\right ) \left (\arctan \left (\frac {\sqrt {-i a \,{\mathrm e}^{i \left (f x +e \right )}}}{\sqrt {a}}\right ) a \sqrt {-i a \,{\mathrm e}^{i \left (f x +e \right )}}+a^{\frac {3}{2}}\right ) \sqrt {2}\, {\mathrm e}^{-i \left (f x +e \right )}}{f \,a^{\frac {3}{2}} \sqrt {-a \left (i {\mathrm e}^{2 i \left (f x +e \right )}-i-2 \,{\mathrm e}^{i \left (f x +e \right )}\right ) {\mathrm e}^{-i \left (f x +e \right )}}}\) | \(198\) |
Input:
int(1/(a+sin(f*x+e)*a)^(1/2),x,method=_RETURNVERBOSE)
Output:
-(1+sin(f*x+e))*(-a*(-1+sin(f*x+e)))^(1/2)*2^(1/2)/a^(1/2)*arctanh(1/2*(-a *(-1+sin(f*x+e)))^(1/2)*2^(1/2)/a^(1/2))/cos(f*x+e)/(a+sin(f*x+e)*a)^(1/2) /f
Time = 0.09 (sec) , antiderivative size = 193, normalized size of antiderivative = 4.11 \[ \int \frac {1}{\sqrt {a+a \sin (e+f x)}} \, dx=\left [\frac {\sqrt {2} \log \left (-\frac {\cos \left (f x + e\right )^{2} - {\left (\cos \left (f x + e\right ) - 2\right )} \sin \left (f x + e\right ) - \frac {2 \, \sqrt {2} \sqrt {a \sin \left (f x + e\right ) + a} {\left (\cos \left (f x + e\right ) - \sin \left (f x + e\right ) + 1\right )}}{\sqrt {a}} + 3 \, \cos \left (f x + e\right ) + 2}{\cos \left (f x + e\right )^{2} - {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right )}{2 \, \sqrt {a} f}, -\frac {\sqrt {2} \sqrt {-\frac {1}{a}} \arctan \left (-\frac {\sqrt {2} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-\frac {1}{a}} {\left (\cos \left (f x + e\right ) - \sin \left (f x + e\right ) + 1\right )}}{2 \, {\left (\cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right )}}\right )}{f}\right ] \] Input:
integrate(1/(a+a*sin(f*x+e))^(1/2),x, algorithm="fricas")
Output:
[1/2*sqrt(2)*log(-(cos(f*x + e)^2 - (cos(f*x + e) - 2)*sin(f*x + e) - 2*sq rt(2)*sqrt(a*sin(f*x + e) + a)*(cos(f*x + e) - sin(f*x + e) + 1)/sqrt(a) + 3*cos(f*x + e) + 2)/(cos(f*x + e)^2 - (cos(f*x + e) + 2)*sin(f*x + e) - c os(f*x + e) - 2))/(sqrt(a)*f), -sqrt(2)*sqrt(-1/a)*arctan(-1/2*sqrt(2)*sqr t(a*sin(f*x + e) + a)*sqrt(-1/a)*(cos(f*x + e) - sin(f*x + e) + 1)/(cos(f* x + e) + sin(f*x + e) + 1))/f]
\[ \int \frac {1}{\sqrt {a+a \sin (e+f x)}} \, dx=\int \frac {1}{\sqrt {a \sin {\left (e + f x \right )} + a}}\, dx \] Input:
integrate(1/(a+a*sin(f*x+e))**(1/2),x)
Output:
Integral(1/sqrt(a*sin(e + f*x) + a), x)
\[ \int \frac {1}{\sqrt {a+a \sin (e+f x)}} \, dx=\int { \frac {1}{\sqrt {a \sin \left (f x + e\right ) + a}} \,d x } \] Input:
integrate(1/(a+a*sin(f*x+e))^(1/2),x, algorithm="maxima")
Output:
integrate(1/sqrt(a*sin(f*x + e) + a), x)
Time = 0.15 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.30 \[ \int \frac {1}{\sqrt {a+a \sin (e+f x)}} \, dx=\frac {\sqrt {2} {\left (\log \left ({\left | \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1 \right |}\right ) - \log \left ({\left | \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1 \right |}\right )\right )}}{2 \, \sqrt {a} f \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} \] Input:
integrate(1/(a+a*sin(f*x+e))^(1/2),x, algorithm="giac")
Output:
1/2*sqrt(2)*(log(abs(sin(-1/4*pi + 1/2*f*x + 1/2*e) + 1)) - log(abs(sin(-1 /4*pi + 1/2*f*x + 1/2*e) - 1)))/(sqrt(a)*f*sgn(cos(-1/4*pi + 1/2*f*x + 1/2 *e)))
Time = 16.95 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.04 \[ \int \frac {1}{\sqrt {a+a \sin (e+f x)}} \, dx=-\frac {\mathrm {F}\left (\frac {\pi }{4}-\frac {e}{2}-\frac {f\,x}{2}\middle |1\right )\,\sqrt {\frac {2\,\left (a+a\,\sin \left (e+f\,x\right )\right )}{a}}}{f\,\sqrt {a+a\,\sin \left (e+f\,x\right )}} \] Input:
int(1/(a + a*sin(e + f*x))^(1/2),x)
Output:
-(ellipticF(pi/4 - e/2 - (f*x)/2, 1)*((2*(a + a*sin(e + f*x)))/a)^(1/2))/( f*(a + a*sin(e + f*x))^(1/2))
\[ \int \frac {1}{\sqrt {a+a \sin (e+f x)}} \, dx=\frac {\sqrt {a}\, \left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}}{\sin \left (f x +e \right )+1}d x \right )}{a} \] Input:
int(1/(a+a*sin(f*x+e))^(1/2),x)
Output:
(sqrt(a)*int(sqrt(sin(e + f*x) + 1)/(sin(e + f*x) + 1),x))/a