Integrand size = 28, antiderivative size = 82 \[ \int \frac {1}{(3+3 \sin (e+f x)) \sqrt {c-c \sin (e+f x)}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{3 \sqrt {2} \sqrt {c} f}-\frac {\sec (e+f x) \sqrt {c-c \sin (e+f x)}}{3 c f} \]
1/2*arctanh(1/2*cos(f*x+e)*c^(1/2)*2^(1/2)/(c-c*sin(f*x+e))^(1/2))/a/f*2^( 1/2)/c^(1/2)-sec(f*x+e)*(c-c*sin(f*x+e))^(1/2)/a/c/f
Result contains complex when optimal does not.
Time = 1.22 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.17 \[ \int \frac {1}{(3+3 \sin (e+f x)) \sqrt {c-c \sin (e+f x)}} \, dx=-\frac {\cos (e+f x) \left (1+(1+i) \sqrt [4]{-1} \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt [4]{-1} \left (1+\tan \left (\frac {1}{4} (e+f x)\right )\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )\right )}{3 f (1+\sin (e+f x)) \sqrt {c-c \sin (e+f x)}} \]
-1/3*(Cos[e + f*x]*(1 + (1 + I)*(-1)^(1/4)*ArcTan[(1/2 + I/2)*(-1)^(1/4)*( 1 + Tan[(e + f*x)/4])]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])))/(f*(1 + Sin [e + f*x])*Sqrt[c - c*Sin[e + f*x]])
Time = 0.43 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.99, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3042, 3215, 3042, 3154, 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.
\(\displaystyle \int \frac {1}{(a \sin (e+f x)+a) \sqrt {c-c \sin (e+f x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{(a \sin (e+f x)+a) \sqrt {c-c \sin (e+f x)}}dx\) |
\(\Big \downarrow \) 3215 |
\(\displaystyle \frac {\int \sec ^2(e+f x) \sqrt {c-c \sin (e+f x)}dx}{a c}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {\sqrt {c-c \sin (e+f x)}}{\cos (e+f x)^2}dx}{a c}\) |
\(\Big \downarrow \) 3154 |
\(\displaystyle \frac {\frac {1}{2} c \int \frac {1}{\sqrt {c-c \sin (e+f x)}}dx-\frac {\sec (e+f x) \sqrt {c-c \sin (e+f x)}}{f}}{a c}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {1}{2} c \int \frac {1}{\sqrt {c-c \sin (e+f x)}}dx-\frac {\sec (e+f x) \sqrt {c-c \sin (e+f x)}}{f}}{a c}\) |
\(\Big \downarrow \) 3128 |
\(\displaystyle \frac {-\frac {c \int \frac {1}{2 c-\frac {c^2 \cos ^2(e+f x)}{c-c \sin (e+f x)}}d\left (-\frac {c \cos (e+f x)}{\sqrt {c-c \sin (e+f x)}}\right )}{f}-\frac {\sec (e+f x) \sqrt {c-c \sin (e+f x)}}{f}}{a c}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\frac {\sqrt {c} \text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{\sqrt {2} f}-\frac {\sec (e+f x) \sqrt {c-c \sin (e+f x)}}{f}}{a c}\) |
((Sqrt[c]*ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]] )])/(Sqrt[2]*f) - (Sec[e + f*x]*Sqrt[c - c*Sin[e + f*x]])/f)/(a*c)
3.4.21.3.1 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[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_), x_Symbol] :> Simp[(-b)*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^m/(a*f*g*(p + 1))), x] + Simp[a*((m + p + 1)/(g^2*(p + 1))) Int[(g* Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2, 0] && GtQ[m, 0] && LeQ[p, -2*m] && IntegersQ [m + 1/2, 2*p]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[a^m*c^m Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && EqQ[ b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] && !(IntegerQ[n] && ((Lt Q[m, 0] && GtQ[n, 0]) || LtQ[0, n, m] || LtQ[m, n, 0]))
Time = 0.91 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.05
method | result | size |
default | \(\frac {\left (\sin \left (f x +e \right )-1\right ) \left (-\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (\sin \left (f x +e \right )+1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c \sqrt {c \left (\sin \left (f x +e \right )+1\right )}+2 c^{\frac {3}{2}}\right )}{2 a \,c^{\frac {3}{2}} \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\) | \(86\) |
1/2*(sin(f*x+e)-1)*(-2^(1/2)*arctanh(1/2*(c*(sin(f*x+e)+1))^(1/2)*2^(1/2)/ c^(1/2))*c*(c*(sin(f*x+e)+1))^(1/2)+2*c^(3/2))/a/c^(3/2)/cos(f*x+e)/(c-c*s in(f*x+e))^(1/2)/f
Leaf count of result is larger than twice the leaf count of optimal. 154 vs. \(2 (73) = 146\).
Time = 0.29 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.88 \[ \int \frac {1}{(3+3 \sin (e+f x)) \sqrt {c-c \sin (e+f x)}} \, dx=\frac {\sqrt {2} \sqrt {c} \cos \left (f x + e\right ) \log \left (-\frac {\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) - 2\right )} \sin \left (f x + e\right ) + \frac {2 \, \sqrt {2} \sqrt {-c \sin \left (f x + e\right ) + c} {\left (\cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right )}}{\sqrt {c}} + 3 \, \cos \left (f x + e\right ) + 2}{\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right ) - 4 \, \sqrt {-c \sin \left (f x + e\right ) + c}}{4 \, a c f \cos \left (f x + e\right )} \]
1/4*(sqrt(2)*sqrt(c)*cos(f*x + e)*log(-(cos(f*x + e)^2 + (cos(f*x + e) - 2 )*sin(f*x + e) + 2*sqrt(2)*sqrt(-c*sin(f*x + e) + c)*(cos(f*x + e) + sin(f *x + e) + 1)/sqrt(c) + 3*cos(f*x + e) + 2)/(cos(f*x + e)^2 + (cos(f*x + e) + 2)*sin(f*x + e) - cos(f*x + e) - 2)) - 4*sqrt(-c*sin(f*x + e) + c))/(a* c*f*cos(f*x + e))
\[ \int \frac {1}{(3+3 \sin (e+f x)) \sqrt {c-c \sin (e+f x)}} \, dx=\frac {\int \frac {1}{\sqrt {- c \sin {\left (e + f x \right )} + c} \sin {\left (e + f x \right )} + \sqrt {- c \sin {\left (e + f x \right )} + c}}\, dx}{a} \]
\[ \int \frac {1}{(3+3 \sin (e+f x)) \sqrt {c-c \sin (e+f x)}} \, dx=\int { \frac {1}{{\left (a \sin \left (f x + e\right ) + a\right )} \sqrt {-c \sin \left (f x + e\right ) + c}} \,d x } \]
Time = 0.33 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.52 \[ \int \frac {1}{(3+3 \sin (e+f x)) \sqrt {c-c \sin (e+f x)}} \, dx=\frac {\frac {\sqrt {2} \log \left (-\frac {\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1}\right )}{a \sqrt {c} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} + \frac {4 \, \sqrt {2}}{a \sqrt {c} {\left (\frac {\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1} + 1\right )} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{4 \, f} \]
1/4*(sqrt(2)*log(-(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)/(cos(-1/4*pi + 1/2* f*x + 1/2*e) + 1))/(a*sqrt(c)*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))) + 4*sqr t(2)/(a*sqrt(c)*((cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)/(cos(-1/4*pi + 1/2*f *x + 1/2*e) + 1) + 1)*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))))/f
Timed out. \[ \int \frac {1}{(3+3 \sin (e+f x)) \sqrt {c-c \sin (e+f x)}} \, dx=\int \frac {1}{\left (a+a\,\sin \left (e+f\,x\right )\right )\,\sqrt {c-c\,\sin \left (e+f\,x\right )}} \,d x \]