Integrand size = 14, antiderivative size = 47 \[ \int \frac {1}{\sqrt {a+a \sin (c+d x)}} \, dx=-\frac {\sqrt {2} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a+a \sin (c+d x)}}\right )}{\sqrt {a} d} \] Output:
-2^(1/2)*arctanh(1/2*a^(1/2)*cos(d*x+c)*2^(1/2)/(a+a*sin(d*x+c))^(1/2))/a^ (1/2)/d
Result contains complex when optimal does not.
Time = 0.02 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.55 \[ \int \frac {1}{\sqrt {a+a \sin (c+d x)}} \, dx=\frac {(2+2 i) (-1)^{3/4} \text {arctanh}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \left (-1+\tan \left (\frac {1}{4} (c+d x)\right )\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d \sqrt {a (1+\sin (c+d x))}} \] Input:
Integrate[1/Sqrt[a + a*Sin[c + d*x]],x]
Output:
((2 + 2*I)*(-1)^(3/4)*ArcTanh[(1/2 + I/2)*(-1)^(3/4)*(-1 + Tan[(c + d*x)/4 ])]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))/(d*Sqrt[a*(1 + Sin[c + d*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.
\(\displaystyle \int \frac {1}{\sqrt {a \sin (c+d x)+a}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sqrt {a \sin (c+d x)+a}}dx\) |
\(\Big \downarrow \) 3128 |
\(\displaystyle -\frac {2 \int \frac {1}{2 a-\frac {a^2 \cos ^2(c+d x)}{\sin (c+d x) a+a}}d\frac {a \cos (c+d x)}{\sqrt {\sin (c+d x) a+a}}}{d}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {\sqrt {2} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a \sin (c+d x)+a}}\right )}{\sqrt {a} d}\) |
Input:
Int[1/Sqrt[a + a*Sin[c + d*x]],x]
Output:
-((Sqrt[2]*ArcTanh[(Sqrt[a]*Cos[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sin[c + d*x] ])])/(Sqrt[a]*d))
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 = 0.27 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.60
method | result | size |
default | \(-\frac {\left (1+\sin \left (d x +c \right )\right ) \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{\sqrt {a}\, \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\) | \(75\) |
risch | \(\frac {2 i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) \sqrt {2}\, {\mathrm e}^{-i \left (d x +c \right )}}{d \sqrt {-a \left (i {\mathrm e}^{2 i \left (d x +c \right )}-2 \,{\mathrm e}^{i \left (d x +c \right )}-i\right ) {\mathrm e}^{-i \left (d x +c \right )}}}-\frac {2 i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) \left (a^{\frac {3}{2}}+\arctan \left (\frac {\sqrt {-i {\mathrm e}^{i \left (d x +c \right )} a}}{\sqrt {a}}\right ) a \sqrt {-i {\mathrm e}^{i \left (d x +c \right )} a}\right ) \sqrt {2}\, {\mathrm e}^{-i \left (d x +c \right )}}{d \,a^{\frac {3}{2}} \sqrt {-a \left (i {\mathrm e}^{2 i \left (d x +c \right )}-2 \,{\mathrm e}^{i \left (d x +c \right )}-i\right ) {\mathrm e}^{-i \left (d x +c \right )}}}\) | \(198\) |
Input:
int(1/(a+a*sin(d*x+c))^(1/2),x,method=_RETURNVERBOSE)
Output:
-(1+sin(d*x+c))*(-a*(sin(d*x+c)-1))^(1/2)*2^(1/2)/a^(1/2)*arctanh(1/2*(-a* (sin(d*x+c)-1))^(1/2)*2^(1/2)/a^(1/2))/cos(d*x+c)/(a+a*sin(d*x+c))^(1/2)/d
Time = 0.15 (sec) , antiderivative size = 193, normalized size of antiderivative = 4.11 \[ \int \frac {1}{\sqrt {a+a \sin (c+d x)}} \, dx=\left [\frac {\sqrt {2} \log \left (-\frac {\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) - \frac {2 \, \sqrt {2} \sqrt {a \sin \left (d x + c\right ) + a} {\left (\cos \left (d x + c\right ) - \sin \left (d x + c\right ) + 1\right )}}{\sqrt {a}} + 3 \, \cos \left (d x + c\right ) + 2}{\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 2}\right )}{2 \, \sqrt {a} d}, -\frac {\sqrt {2} \sqrt {-\frac {1}{a}} \arctan \left (-\frac {\sqrt {2} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {-\frac {1}{a}} {\left (\cos \left (d x + c\right ) - \sin \left (d x + c\right ) + 1\right )}}{2 \, {\left (\cos \left (d x + c\right ) + \sin \left (d x + c\right ) + 1\right )}}\right )}{d}\right ] \] Input:
integrate(1/(a+a*sin(d*x+c))^(1/2),x, algorithm="fricas")
Output:
[1/2*sqrt(2)*log(-(cos(d*x + c)^2 - (cos(d*x + c) - 2)*sin(d*x + c) - 2*sq rt(2)*sqrt(a*sin(d*x + c) + a)*(cos(d*x + c) - sin(d*x + c) + 1)/sqrt(a) + 3*cos(d*x + c) + 2)/(cos(d*x + c)^2 - (cos(d*x + c) + 2)*sin(d*x + c) - c os(d*x + c) - 2))/(sqrt(a)*d), -sqrt(2)*sqrt(-1/a)*arctan(-1/2*sqrt(2)*sqr t(a*sin(d*x + c) + a)*sqrt(-1/a)*(cos(d*x + c) - sin(d*x + c) + 1)/(cos(d* x + c) + sin(d*x + c) + 1))/d]
\[ \int \frac {1}{\sqrt {a+a \sin (c+d x)}} \, dx=\int \frac {1}{\sqrt {a \sin {\left (c + d x \right )} + a}}\, dx \] Input:
integrate(1/(a+a*sin(d*x+c))**(1/2),x)
Output:
Integral(1/sqrt(a*sin(c + d*x) + a), x)
\[ \int \frac {1}{\sqrt {a+a \sin (c+d x)}} \, dx=\int { \frac {1}{\sqrt {a \sin \left (d x + c\right ) + a}} \,d x } \] Input:
integrate(1/(a+a*sin(d*x+c))^(1/2),x, algorithm="maxima")
Output:
integrate(1/sqrt(a*sin(d*x + c) + a), x)
Time = 0.14 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.30 \[ \int \frac {1}{\sqrt {a+a \sin (c+d x)}} \, dx=\frac {\sqrt {2} {\left (\log \left ({\left | \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - \log \left ({\left | \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )\right )}}{2 \, \sqrt {a} d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \] Input:
integrate(1/(a+a*sin(d*x+c))^(1/2),x, algorithm="giac")
Output:
1/2*sqrt(2)*(log(abs(sin(-1/4*pi + 1/2*d*x + 1/2*c) + 1)) - log(abs(sin(-1 /4*pi + 1/2*d*x + 1/2*c) - 1)))/(sqrt(a)*d*sgn(cos(-1/4*pi + 1/2*d*x + 1/2 *c)))
Time = 0.13 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.04 \[ \int \frac {1}{\sqrt {a+a \sin (c+d x)}} \, dx=-\frac {\mathrm {F}\left (\frac {\pi }{4}-\frac {c}{2}-\frac {d\,x}{2}\middle |1\right )\,\sqrt {\frac {2\,\left (a+a\,\sin \left (c+d\,x\right )\right )}{a}}}{d\,\sqrt {a+a\,\sin \left (c+d\,x\right )}} \] Input:
int(1/(a + a*sin(c + d*x))^(1/2),x)
Output:
-(ellipticF(pi/4 - c/2 - (d*x)/2, 1)*((2*(a + a*sin(c + d*x)))/a)^(1/2))/( d*(a + a*sin(c + d*x))^(1/2))
\[ \int \frac {1}{\sqrt {a+a \sin (c+d x)}} \, dx=\frac {\sqrt {a}\, \left (\int \frac {\sqrt {\sin \left (d x +c \right )+1}}{\sin \left (d x +c \right )+1}d x \right )}{a} \] Input:
int(1/(a+a*sin(d*x+c))^(1/2),x)
Output:
(sqrt(a)*int(sqrt(sin(c + d*x) + 1)/(sin(c + d*x) + 1),x))/a