Integrand size = 28, antiderivative size = 83 \[ \int \frac {1}{(a+a \sin (e+f x)) \sqrt {c-c \sin (e+f x)}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{\sqrt {2} a \sqrt {c} f}-\frac {\sec (e+f x) \sqrt {c-c \sin (e+f x)}}{a c f} \] Output:
1/2*arctanh(1/2*c^(1/2)*cos(f*x+e)*2^(1/2)/(c-c*sin(f*x+e))^(1/2))*2^(1/2) /a/c^(1/2)/f-sec(f*x+e)*(c-c*sin(f*x+e))^(1/2)/a/c/f
Result contains complex when optimal does not.
Time = 2.60 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.17 \[ \int \frac {1}{(a+a \sin (e+f x)) \sqrt {c-c \sin (e+f x)}} \, dx=-\frac {\cos (e+f x) \left (1+(1+i) \sqrt [4]{-1} \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt [4]{-1} \left (1+\tan \left (\frac {1}{4} (e+f x)\right )\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )\right )}{a f (1+\sin (e+f x)) \sqrt {c-c \sin (e+f x)}} \] Input:
Integrate[1/((a + a*Sin[e + f*x])*Sqrt[c - c*Sin[e + f*x]]),x]
Output:
-((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])))/(a*f*(1 + Sin[ e + f*x])*Sqrt[c - c*Sin[e + f*x]]))
Time = 0.48 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.98, 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 |
\(\displaystyle \frac {\frac {\sqrt {c} \text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{\sqrt {2} f}-\frac {\sec (e+f x) \sqrt {c-c \sin (e+f x)}}{f}}{a c}\) |
Input:
Int[1/((a + a*Sin[e + f*x])*Sqrt[c - c*Sin[e + f*x]]),x]
Output:
((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)
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 = 0.46 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.02
method | result | size |
default | \(-\frac {\left (-1+\sin \left (f x +e \right )\right ) \left (\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c \sqrt {c \left (1+\sin \left (f x +e \right )\right )}-2 c^{\frac {3}{2}}\right )}{2 a \,c^{\frac {3}{2}} \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\) | \(85\) |
Input:
int(1/(a+sin(f*x+e)*a)/(c-c*sin(f*x+e))^(1/2),x,method=_RETURNVERBOSE)
Output:
-1/2/a*(-1+sin(f*x+e))*(2^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/ 2)/c^(1/2))*c*(c*(1+sin(f*x+e)))^(1/2)-2*c^(3/2))/c^(3/2)/cos(f*x+e)/(c-c* sin(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 = 0.09 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.86 \[ \int \frac {1}{(a+a \sin (e+f x)) \sqrt {c-c \sin (e+f x)}} \, dx=\frac {\sqrt {2} \sqrt {c} \cos \left (f x + e\right ) \log \left (-\frac {\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) - 2\right )} \sin \left (f x + e\right ) + \frac {2 \, \sqrt {2} \sqrt {-c \sin \left (f x + e\right ) + c} {\left (\cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right )}}{\sqrt {c}} + 3 \, \cos \left (f x + e\right ) + 2}{\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right ) - 4 \, \sqrt {-c \sin \left (f x + e\right ) + c}}{4 \, a c f \cos \left (f x + e\right )} \] Input:
integrate(1/(a+a*sin(f*x+e))/(c-c*sin(f*x+e))^(1/2),x, algorithm="fricas")
Output:
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))
\[ \int \frac {1}{(a+a \sin (e+f x)) \sqrt {c-c \sin (e+f x)}} \, dx=\frac {\int \frac {1}{\sqrt {- c \sin {\left (e + f x \right )} + c} \sin {\left (e + f x \right )} + \sqrt {- c \sin {\left (e + f x \right )} + c}}\, dx}{a} \] Input:
integrate(1/(a+a*sin(f*x+e))/(c-c*sin(f*x+e))**(1/2),x)
Output:
Integral(1/(sqrt(-c*sin(e + f*x) + c)*sin(e + f*x) + sqrt(-c*sin(e + f*x) + c)), x)/a
\[ \int \frac {1}{(a+a \sin (e+f x)) \sqrt {c-c \sin (e+f x)}} \, dx=\int { \frac {1}{{\left (a \sin \left (f x + e\right ) + a\right )} \sqrt {-c \sin \left (f x + e\right ) + c}} \,d x } \] Input:
integrate(1/(a+a*sin(f*x+e))/(c-c*sin(f*x+e))^(1/2),x, algorithm="maxima")
Output:
integrate(1/((a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)), x)
Exception generated. \[ \int \frac {1}{(a+a \sin (e+f x)) \sqrt {c-c \sin (e+f x)}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(1/(a+a*sin(f*x+e))/(c-c*sin(f*x+e))^(1/2),x, algorithm="giac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value
Timed out. \[ \int \frac {1}{(a+a \sin (e+f x)) \sqrt {c-c \sin (e+f x)}} \, dx=\int \frac {1}{\left (a+a\,\sin \left (e+f\,x\right )\right )\,\sqrt {c-c\,\sin \left (e+f\,x\right )}} \,d x \] Input:
int(1/((a + a*sin(e + f*x))*(c - c*sin(e + f*x))^(1/2)),x)
Output:
int(1/((a + a*sin(e + f*x))*(c - c*sin(e + f*x))^(1/2)), x)
\[ \int \frac {1}{(a+a \sin (e+f x)) \sqrt {c-c \sin (e+f x)}} \, dx=-\frac {\sqrt {c}\, \left (\int \frac {\sqrt {-\sin \left (f x +e \right )+1}}{\sin \left (f x +e \right )^{2}-1}d x \right )}{a c} \] Input:
int(1/(a+a*sin(f*x+e))/(c-c*sin(f*x+e))^(1/2),x)
Output:
( - sqrt(c)*int(sqrt( - sin(e + f*x) + 1)/(sin(e + f*x)**2 - 1),x))/(a*c)