Integrand size = 21, antiderivative size = 72 \[ \int \frac {\sin (c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\frac {\sqrt {2} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a+a \sin (c+d x)}}\right )}{\sqrt {a} d}-\frac {2 \cos (c+d x)}{d \sqrt {a+a \sin (c+d x)}} \]
arctanh(1/2*cos(d*x+c)*a^(1/2)*2^(1/2)/(a+a*sin(d*x+c))^(1/2))*2^(1/2)/d/a ^(1/2)-2*cos(d*x+c)/d/(a+a*sin(d*x+c))^(1/2)
Result contains complex when optimal does not.
Time = 0.20 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.36 \[ \int \frac {\sin (c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=-\frac {2 \left ((1+i) (-1)^{3/4} \text {arctanh}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \left (-1+\tan \left (\frac {1}{4} (c+d x)\right )\right )\right )+\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d \sqrt {a (1+\sin (c+d x))}} \]
(-2*((1 + 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])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))/(d*Sqrt[a*(1 + Sin[c + d*x])])
Time = 0.29 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3042, 3230, 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.
\(\displaystyle \int \frac {\sin (c+d x)}{\sqrt {a \sin (c+d x)+a}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (c+d x)}{\sqrt {a \sin (c+d x)+a}}dx\) |
\(\Big \downarrow \) 3230 |
\(\displaystyle -\int \frac {1}{\sqrt {\sin (c+d x) a+a}}dx-\frac {2 \cos (c+d x)}{d \sqrt {a \sin (c+d x)+a}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\int \frac {1}{\sqrt {\sin (c+d x) a+a}}dx-\frac {2 \cos (c+d x)}{d \sqrt {a \sin (c+d x)+a}}\) |
\(\Big \downarrow \) 3128 |
\(\displaystyle \frac {2 \int \frac {1}{2 a-\frac {a^2 \cos ^2(c+d x)}{\sin (c+d x) a+a}}d\frac {a \cos (c+d x)}{\sqrt {\sin (c+d x) a+a}}}{d}-\frac {2 \cos (c+d x)}{d \sqrt {a \sin (c+d x)+a}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\sqrt {2} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a \sin (c+d x)+a}}\right )}{\sqrt {a} d}-\frac {2 \cos (c+d x)}{d \sqrt {a \sin (c+d x)+a}}\) |
(Sqrt[2]*ArcTanh[(Sqrt[a]*Cos[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sin[c + d*x]]) ])/(Sqrt[a]*d) - (2*Cos[c + d*x])/(d*Sqrt[a + a*Sin[c + d*x]])
3.1.63.3.1 Defintions of rubi rules used
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]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-d)*Cos[e + f*x]*((a + b*Sin[e + f*x])^m/( f*(m + 1))), x] + Simp[(a*d*m + b*c*(m + 1))/(b*(m + 1)) Int[(a + b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && !LtQ[m, -2^(-1)]
Time = 0.83 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.33
method | result | size |
default | \(-\frac {\left (1+\sin \left (d x +c \right )\right ) \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, \left (-\sqrt {a}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )+2 \sqrt {a -a \sin \left (d x +c \right )}\right )}{a \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\) | \(96\) |
risch | \(-\frac {\left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{2} \sqrt {2}\, {\mathrm e}^{-i \left (d x +c \right )}}{d \sqrt {-a \left (i {\mathrm e}^{2 i \left (d x +c \right )}-i-2 \,{\mathrm e}^{i \left (d x +c \right )}\right ) {\mathrm e}^{-i \left (d x +c \right )}}}+\frac {2 i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) \left (\arctan \left (\frac {\sqrt {-i a \,{\mathrm e}^{i \left (d x +c \right )}}}{\sqrt {a}}\right ) a \sqrt {-i a \,{\mathrm e}^{i \left (d x +c \right )}}+a^{\frac {3}{2}}\right ) \sqrt {2}\, {\mathrm e}^{-i \left (d x +c \right )}}{d \,a^{\frac {3}{2}} \sqrt {-a \left (i {\mathrm e}^{2 i \left (d x +c \right )}-i-2 \,{\mathrm e}^{i \left (d x +c \right )}\right ) {\mathrm e}^{-i \left (d x +c \right )}}}\) | \(199\) |
-(1+sin(d*x+c))*(-a*(sin(d*x+c)-1))^(1/2)*(-a^(1/2)*2^(1/2)*arctanh(1/2*(a -a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))+2*(a-a*sin(d*x+c))^(1/2))/a/cos(d*x+ c)/(a+a*sin(d*x+c))^(1/2)/d
Leaf count of result is larger than twice the leaf count of optimal. 191 vs. \(2 (61) = 122\).
Time = 0.30 (sec) , antiderivative size = 191, normalized size of antiderivative = 2.65 \[ \int \frac {\sin (c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\frac {\frac {\sqrt {2} {\left (a \cos \left (d x + c\right ) + a \sin \left (d x + c\right ) + a\right )} \log \left (-\frac {\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + \frac {2 \, \sqrt {2} \sqrt {a \sin \left (d x + c\right ) + a} {\left (\cos \left (d x + c\right ) - \sin \left (d x + c\right ) + 1\right )}}{\sqrt {a}} + 3 \, \cos \left (d x + c\right ) + 2}{\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 2}\right )}{\sqrt {a}} - 4 \, \sqrt {a \sin \left (d x + c\right ) + a} {\left (\cos \left (d x + c\right ) - \sin \left (d x + c\right ) + 1\right )}}{2 \, {\left (a d \cos \left (d x + c\right ) + a d \sin \left (d x + c\right ) + a d\right )}} \]
1/2*(sqrt(2)*(a*cos(d*x + c) + a*sin(d*x + c) + a)*log(-(cos(d*x + c)^2 - (cos(d*x + c) - 2)*sin(d*x + c) + 2*sqrt(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) - cos(d*x + c) - 2))/sqrt(a) - 4*sqrt(a *sin(d*x + c) + a)*(cos(d*x + c) - sin(d*x + c) + 1))/(a*d*cos(d*x + c) + a*d*sin(d*x + c) + a*d)
\[ \int \frac {\sin (c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\int \frac {\sin {\left (c + d x \right )}}{\sqrt {a \left (\sin {\left (c + d x \right )} + 1\right )}}\, dx \]
\[ \int \frac {\sin (c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\int { \frac {\sin \left (d x + c\right )}{\sqrt {a \sin \left (d x + c\right ) + a}} \,d x } \]
Time = 0.34 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.64 \[ \int \frac {\sin (c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=-\frac {\frac {\sqrt {2} \log \left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{\sqrt {a} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {\sqrt {2} \log \left (-\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{\sqrt {a} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {4 \, \sqrt {2} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{2 \, d} \]
-1/2*(sqrt(2)*log(sin(-1/4*pi + 1/2*d*x + 1/2*c) + 1)/(sqrt(a)*sgn(cos(-1/ 4*pi + 1/2*d*x + 1/2*c))) - sqrt(2)*log(-sin(-1/4*pi + 1/2*d*x + 1/2*c) + 1)/(sqrt(a)*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))) - 4*sqrt(2)*sin(-1/4*pi + 1/2*d*x + 1/2*c)/(sqrt(a)*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))))/d
Time = 0.80 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.38 \[ \int \frac {\sin (c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=-\frac {\left (4\,\mathrm {E}\left (\mathrm {asin}\left (\frac {\sqrt {2}\,\sqrt {1-\sin \left (c+d\,x\right )}}{2}\right )\middle |1\right )-2\,\mathrm {F}\left (\mathrm {asin}\left (\frac {\sqrt {2}\,\sqrt {1-\sin \left (c+d\,x\right )}}{2}\right )\middle |1\right )\right )\,\sqrt {{\cos \left (c+d\,x\right )}^2}\,\sqrt {\frac {a+a\,\sin \left (c+d\,x\right )}{2\,a}}}{d\,\cos \left (c+d\,x\right )\,\sqrt {a+a\,\sin \left (c+d\,x\right )}} \]