Integrand size = 40, antiderivative size = 100 \[ \int \frac {(A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)}}{(a+a \sin (e+f x))^{3/2}} \, dx=-\frac {(A-B) c \cos (e+f x)}{f (a+a \sin (e+f x))^{3/2} \sqrt {c-c \sin (e+f x)}}+\frac {B c \cos (e+f x) \log (1+\sin (e+f x))}{a f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}} \]
-(A-B)*c*cos(f*x+e)/f/(a+a*sin(f*x+e))^(3/2)/(c-c*sin(f*x+e))^(1/2)+B*c*co s(f*x+e)*ln(1+sin(f*x+e))/a/f/(a+a*sin(f*x+e))^(1/2)/(c-c*sin(f*x+e))^(1/2 )
Result contains complex when optimal does not.
Time = 2.37 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.43 \[ \int \frac {(A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)}}{(a+a \sin (e+f x))^{3/2}} \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {c-c \sin (e+f x)} \left (-A+B-i B f x+2 B \log \left (i+e^{i (e+f x)}\right )+B \left (-i f x+2 \log \left (i+e^{i (e+f x)}\right )\right ) \sin (e+f x)\right )}{f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) (a (1+\sin (e+f x)))^{3/2}} \]
((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*Sqrt[c - c*Sin[e + f*x]]*(-A + B - I*B*f*x + 2*B*Log[I + E^(I*(e + f*x))] + B*((-I)*f*x + 2*Log[I + E^(I*(e + f*x))])*Sin[e + f*x]))/(f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(a*(1 + S in[e + f*x]))^(3/2))
Time = 0.88 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.02, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 3450, 3042, 3216, 3042, 3146, 16, 3217}
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.
| \(\displaystyle \int \frac {\sqrt {c-c \sin (e+f x)} (A+B \sin (e+f x))}{(a \sin (e+f x)+a)^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sqrt {c-c \sin (e+f x)} (A+B \sin (e+f x))}{(a \sin (e+f x)+a)^{3/2}}dx\) |
| \(\Big \downarrow \) 3450 |
| \(\displaystyle (A-B) \int \frac {\sqrt {c-c \sin (e+f x)}}{(\sin (e+f x) a+a)^{3/2}}dx+\frac {B \int \frac {\sqrt {c-c \sin (e+f x)}}{\sqrt {\sin (e+f x) a+a}}dx}{a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle (A-B) \int \frac {\sqrt {c-c \sin (e+f x)}}{(\sin (e+f x) a+a)^{3/2}}dx+\frac {B \int \frac {\sqrt {c-c \sin (e+f x)}}{\sqrt {\sin (e+f x) a+a}}dx}{a}\) |
| \(\Big \downarrow \) 3216 |
| \(\displaystyle (A-B) \int \frac {\sqrt {c-c \sin (e+f x)}}{(\sin (e+f x) a+a)^{3/2}}dx+\frac {B c \cos (e+f x) \int \frac {\cos (e+f x)}{\sin (e+f x) a+a}dx}{\sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle (A-B) \int \frac {\sqrt {c-c \sin (e+f x)}}{(\sin (e+f x) a+a)^{3/2}}dx+\frac {B c \cos (e+f x) \int \frac {\cos (e+f x)}{\sin (e+f x) a+a}dx}{\sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}\) |
| \(\Big \downarrow \) 3146 |
| \(\displaystyle (A-B) \int \frac {\sqrt {c-c \sin (e+f x)}}{(\sin (e+f x) a+a)^{3/2}}dx+\frac {B c \cos (e+f x) \int \frac {1}{\sin (e+f x) a+a}d(a \sin (e+f x))}{a f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}\) |
| \(\Big \downarrow \) 16 |
| \(\displaystyle (A-B) \int \frac {\sqrt {c-c \sin (e+f x)}}{(\sin (e+f x) a+a)^{3/2}}dx+\frac {B c \cos (e+f x) \log (a \sin (e+f x)+a)}{a f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}\) |
| \(\Big \downarrow \) 3217 |
| \(\displaystyle \frac {B c \cos (e+f x) \log (a \sin (e+f x)+a)}{a f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}-\frac {c (A-B) \cos (e+f x)}{f (a \sin (e+f x)+a)^{3/2} \sqrt {c-c \sin (e+f x)}}\) |
-(((A - B)*c*Cos[e + f*x])/(f*(a + a*Sin[e + f*x])^(3/2)*Sqrt[c - c*Sin[e + f*x]])) + (B*c*Cos[e + f*x]*Log[a + a*Sin[e + f*x]])/(a*f*Sqrt[a + a*Sin [e + f*x]]*Sqrt[c - c*Sin[e + f*x]])
3.2.83.3.1 Defintions of rubi rules used
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m _.), x_Symbol] :> Simp[1/(b^p*f) Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x )^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && I ntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] || !IntegerQ[m + 1/ 2])
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Simp[a*c*(Cos[e + f*x]/(Sqrt[a + b*Sin[e + f*x ]]*Sqrt[c + d*Sin[e + f*x]])) Int[Cos[e + f*x]/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0 ]
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_) + (d_.)*sin[(e_.) + (f _.)*(x_)])^(n_), x_Symbol] :> Simp[-2*b*Cos[e + f*x]*((c + d*Sin[e + f*x])^ n/(f*(2*n + 1)*Sqrt[a + b*Sin[e + f*x]])), x] /; FreeQ[{a, b, c, d, e, f, n }, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[n, -2^(-1)]
Int[Sqrt[(a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp [B/d Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n + 1), x], x] - Simp[(B*c - A*d)/d Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^n, x ], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && EqQ[b*c + a*d, 0] && EqQ[ a^2 - b^2, 0]
Time = 2.54 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.41
| method | result | size |
| default | \(\frac {\sec \left (f x +e \right ) \left (2 B \sin \left (f x +e \right ) \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1\right )-B \sin \left (f x +e \right ) \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )+A \sin \left (f x +e \right )-B \sin \left (f x +e \right )+2 B \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1\right )-B \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )\right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}}{a f \sqrt {a \left (1+\sin \left (f x +e \right )\right )}}\) | \(141\) |
| parts | \(\frac {A \tan \left (f x +e \right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}}{f a \sqrt {a \left (1+\sin \left (f x +e \right )\right )}}+\frac {B \sec \left (f x +e \right ) \left (2 \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1\right ) \sin \left (f x +e \right )-\ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right ) \sin \left (f x +e \right )+2 \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1\right )-\ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )-\sin \left (f x +e \right )\right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}}{f a \sqrt {a \left (1+\sin \left (f x +e \right )\right )}}\) | \(169\) |
1/a/f*sec(f*x+e)*(2*B*sin(f*x+e)*ln(-cot(f*x+e)+csc(f*x+e)+1)-B*sin(f*x+e) *ln(2/(1+cos(f*x+e)))+A*sin(f*x+e)-B*sin(f*x+e)+2*B*ln(-cot(f*x+e)+csc(f*x +e)+1)-B*ln(2/(1+cos(f*x+e))))*(-c*(sin(f*x+e)-1))^(1/2)/(a*(1+sin(f*x+e)) )^(1/2)
\[ \int \frac {(A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)}}{(a+a \sin (e+f x))^{3/2}} \, dx=\int { \frac {{\left (B \sin \left (f x + e\right ) + A\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]
integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(1/2)/(a+a*sin(f*x+e))^(3/2),x , algorithm="fricas")
integral(-(B*sin(f*x + e) + A)*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)/(a^2*cos(f*x + e)^2 - 2*a^2*sin(f*x + e) - 2*a^2), x)
\[ \int \frac {(A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)}}{(a+a \sin (e+f x))^{3/2}} \, dx=\int \frac {\sqrt {- c \left (\sin {\left (e + f x \right )} - 1\right )} \left (A + B \sin {\left (e + f x \right )}\right )}{\left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {(A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)}}{(a+a \sin (e+f x))^{3/2}} \, dx=\int { \frac {{\left (B \sin \left (f x + e\right ) + A\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]
integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(1/2)/(a+a*sin(f*x+e))^(3/2),x , algorithm="maxima")
Time = 0.35 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.14 \[ \int \frac {(A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)}}{(a+a \sin (e+f x))^{3/2}} \, dx=-\frac {{\left (4 \, B \sqrt {a} \log \left ({\left | \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) \right |}\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - \frac {A \sqrt {a} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - B \sqrt {a} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2}}\right )} \sqrt {c}}{2 \, a^{2} f \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} \]
integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(1/2)/(a+a*sin(f*x+e))^(3/2),x , algorithm="giac")
-1/2*(4*B*sqrt(a)*log(abs(cos(-1/4*pi + 1/2*f*x + 1/2*e)))*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) - (A*sqrt(a)*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) - B* sqrt(a)*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)))/cos(-1/4*pi + 1/2*f*x + 1/2*e )^2)*sqrt(c)/(a^2*f*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)))
Timed out. \[ \int \frac {(A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)}}{(a+a \sin (e+f x))^{3/2}} \, dx=\int \frac {\left (A+B\,\sin \left (e+f\,x\right )\right )\,\sqrt {c-c\,\sin \left (e+f\,x\right )}}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{3/2}} \,d x \]