Integrand size = 38, antiderivative size = 73 \[ \int \frac {(A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)}}{a+a \sin (e+f x)} \, dx=-\frac {(A-3 B) c \cos (e+f x)}{a f \sqrt {c-c \sin (e+f x)}}-\frac {(A-B) \sec (e+f x) (c-c \sin (e+f x))^{3/2}}{a c f} \] Output:
-(A-3*B)*c*cos(f*x+e)/a/f/(c-c*sin(f*x+e))^(1/2)-(A-B)*sec(f*x+e)*(c-c*sin (f*x+e))^(3/2)/a/c/f
Time = 3.19 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.60 \[ \int \frac {(A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)}}{a+a \sin (e+f x)} \, dx=\frac {2 \sec (e+f x) (-A+2 B+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)}}{a f} \] Input:
Integrate[((A + B*Sin[e + f*x])*Sqrt[c - c*Sin[e + f*x]])/(a + a*Sin[e + f *x]),x]
Output:
(2*Sec[e + f*x]*(-A + 2*B + B*Sin[e + f*x])*Sqrt[c - c*Sin[e + f*x]])/(a*f )
Time = 0.58 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {3042, 3446, 3042, 3334, 3042, 3125}
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.
\(\displaystyle \int \frac {\sqrt {c-c \sin (e+f x)} (A+B \sin (e+f x))}{a \sin (e+f x)+a} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sqrt {c-c \sin (e+f x)} (A+B \sin (e+f x))}{a \sin (e+f x)+a}dx\) |
\(\Big \downarrow \) 3446 |
\(\displaystyle \frac {\int \sec ^2(e+f x) (A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2}dx}{a c}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2}}{\cos (e+f x)^2}dx}{a c}\) |
\(\Big \downarrow \) 3334 |
\(\displaystyle \frac {-\frac {1}{2} c (A-3 B) \int \sqrt {c-c \sin (e+f x)}dx-\frac {(A-B) \sec (e+f x) (c-c \sin (e+f x))^{3/2}}{f}}{a c}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {-\frac {1}{2} c (A-3 B) \int \sqrt {c-c \sin (e+f x)}dx-\frac {(A-B) \sec (e+f x) (c-c \sin (e+f x))^{3/2}}{f}}{a c}\) |
\(\Big \downarrow \) 3125 |
\(\displaystyle \frac {-\frac {c^2 (A-3 B) \cos (e+f x)}{f \sqrt {c-c \sin (e+f x)}}-\frac {(A-B) \sec (e+f x) (c-c \sin (e+f x))^{3/2}}{f}}{a c}\) |
Input:
Int[((A + B*Sin[e + f*x])*Sqrt[c - c*Sin[e + f*x]])/(a + a*Sin[e + f*x]),x ]
Output:
(-(((A - 3*B)*c^2*Cos[e + f*x])/(f*Sqrt[c - c*Sin[e + f*x]])) - ((A - B)*S ec[e + f*x]*(c - c*Sin[e + f*x])^(3/2))/f)/(a*c)
Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2*b*(Cos [c + d*x]/(d*Sqrt[a + b*Sin[c + d*x]])), x] /; FreeQ[{a, b, c, d}, x] && Eq Q[a^2 - b^2, 0]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-(b* c + a*d))*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^m/(a*f*g*(p + 1))) , x] + Simp[b*((a*d*m + b*c*(m + p + 1))/(a*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, c, d, e, f, g}, x] && EqQ[a^2 - b^2, 0] && GtQ[m, -1] && LtQ[p, -1]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Si mp[a^m*c^m Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m)*(A + B*Sin [e + f*x]), 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] && IntegerQ[m] && !(IntegerQ[n] && ((LtQ[m, 0] && GtQ[n, 0]) || LtQ[0, n, m] || LtQ[m, n, 0]))
Time = 0.35 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.73
method | result | size |
default | \(\frac {2 c \left (\sin \left (f x +e \right )-1\right ) \left (-B \sin \left (f x +e \right )+A -2 B \right )}{a \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\) | \(53\) |
Input:
int((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(1/2)/(a+a*sin(f*x+e)),x,method=_RET URNVERBOSE)
Output:
2*c/a*(sin(f*x+e)-1)*(-B*sin(f*x+e)+A-2*B)/cos(f*x+e)/(c-c*sin(f*x+e))^(1/ 2)/f
Time = 0.08 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.60 \[ \int \frac {(A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)}}{a+a \sin (e+f x)} \, dx=\frac {2 \, {\left (B \sin \left (f x + e\right ) - A + 2 \, B\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{a f \cos \left (f x + e\right )} \] Input:
integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(1/2)/(a+a*sin(f*x+e)),x, algo rithm="fricas")
Output:
2*(B*sin(f*x + e) - A + 2*B)*sqrt(-c*sin(f*x + e) + c)/(a*f*cos(f*x + e))
\[ \int \frac {(A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)}}{a+a \sin (e+f x)} \, dx=\frac {\int \frac {A \sqrt {- c \sin {\left (e + f x \right )} + c}}{\sin {\left (e + f x \right )} + 1}\, dx + \int \frac {B \sqrt {- c \sin {\left (e + f x \right )} + c} \sin {\left (e + f x \right )}}{\sin {\left (e + f x \right )} + 1}\, dx}{a} \] Input:
integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))**(1/2)/(a+a*sin(f*x+e)),x)
Output:
(Integral(A*sqrt(-c*sin(e + f*x) + c)/(sin(e + f*x) + 1), x) + Integral(B* sqrt(-c*sin(e + f*x) + c)*sin(e + f*x)/(sin(e + f*x) + 1), x))/a
Leaf count of result is larger than twice the leaf count of optimal. 174 vs. \(2 (69) = 138\).
Time = 0.13 (sec) , antiderivative size = 174, normalized size of antiderivative = 2.38 \[ \int \frac {(A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)}}{a+a \sin (e+f x)} \, dx=-\frac {2 \, {\left (\frac {2 \, B {\left (\sqrt {c} + \frac {\sqrt {c} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {\sqrt {c} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )}}{{\left (a + \frac {a \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )} \sqrt {\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1}} - \frac {A {\left (\sqrt {c} + \frac {\sqrt {c} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )}}{{\left (a + \frac {a \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )} \sqrt {\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1}}\right )}}{f} \] Input:
integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(1/2)/(a+a*sin(f*x+e)),x, algo rithm="maxima")
Output:
-2*(2*B*(sqrt(c) + sqrt(c)*sin(f*x + e)/(cos(f*x + e) + 1) + sqrt(c)*sin(f *x + e)^2/(cos(f*x + e) + 1)^2)/((a + a*sin(f*x + e)/(cos(f*x + e) + 1))*s qrt(sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1)) - A*(sqrt(c) + sqrt(c)*sin(f *x + e)^2/(cos(f*x + e) + 1)^2)/((a + a*sin(f*x + e)/(cos(f*x + e) + 1))*s qrt(sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1)))/f
Leaf count of result is larger than twice the leaf count of optimal. 175 vs. \(2 (69) = 138\).
Time = 0.26 (sec) , antiderivative size = 175, normalized size of antiderivative = 2.40 \[ \int \frac {(A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)}}{a+a \sin (e+f x)} \, dx=-\frac {2 \, \sqrt {2} {\left (A \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 3 \, B \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - \frac {A {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1} - \frac {B {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1}\right )} \sqrt {c}}{a f {\left (\frac {{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{2}}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{2}} - 1\right )}} \] Input:
integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(1/2)/(a+a*sin(f*x+e)),x, algo rithm="giac")
Output:
-2*sqrt(2)*(A*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) - 3*B*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) - A*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1) - B*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1))*sqrt(c)/(a*f*((cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1 )^2/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^2 - 1))
Time = 38.59 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.75 \[ \int \frac {(A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)}}{a+a \sin (e+f x)} \, dx=\frac {2\,\sqrt {-c\,\left (\sin \left (e+f\,x\right )-1\right )}\,\left (2\,B\,\sin \left (2\,e+2\,f\,x\right )-2\,A\,\sin \left (2\,e+2\,f\,x\right )-4\,A\,\left (2\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-1\right )+7\,B\,\left (2\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-1\right )+B\,\left (2\,{\sin \left (\frac {3\,e}{2}+\frac {3\,f\,x}{2}\right )}^2-1\right )\right )}{a\,f\,\left (4\,{\sin \left (e+f\,x\right )}^2+\sin \left (e+f\,x\right )+\sin \left (3\,e+3\,f\,x\right )-4\right )} \] Input:
int(((A + B*sin(e + f*x))*(c - c*sin(e + f*x))^(1/2))/(a + a*sin(e + f*x)) ,x)
Output:
(2*(-c*(sin(e + f*x) - 1))^(1/2)*(2*B*sin(2*e + 2*f*x) - 2*A*sin(2*e + 2*f *x) - 4*A*(2*sin(e/2 + (f*x)/2)^2 - 1) + 7*B*(2*sin(e/2 + (f*x)/2)^2 - 1) + B*(2*sin((3*e)/2 + (3*f*x)/2)^2 - 1)))/(a*f*(sin(e + f*x) + sin(3*e + 3* f*x) + 4*sin(e + f*x)^2 - 4))
\[ \int \frac {(A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)}}{a+a \sin (e+f x)} \, dx=\frac {\sqrt {c}\, \left (\left (\int \frac {\sqrt {-\sin \left (f x +e \right )+1}}{\sin \left (f x +e \right )+1}d x \right ) a +\left (\int \frac {\sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )}{\sin \left (f x +e \right )+1}d x \right ) b \right )}{a} \] Input:
int((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(1/2)/(a+a*sin(f*x+e)),x)
Output:
(sqrt(c)*(int(sqrt( - sin(e + f*x) + 1)/(sin(e + f*x) + 1),x)*a + int((sqr t( - sin(e + f*x) + 1)*sin(e + f*x))/(sin(e + f*x) + 1),x)*b))/a