Integrand size = 40, antiderivative size = 94 \[ \int \frac {(A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)}}{(a+a \sin (e+f x))^{5/2}} \, dx=-\frac {(A-B) c \cos (e+f x)}{2 f (a+a \sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)}}-\frac {B c \cos (e+f x)}{a f (a+a \sin (e+f x))^{3/2} \sqrt {c-c \sin (e+f x)}} \]
-1/2*(A-B)*c*cos(f*x+e)/f/(a+a*sin(f*x+e))^(5/2)/(c-c*sin(f*x+e))^(1/2)-B* c*cos(f*x+e)/a/f/(a+a*sin(f*x+e))^(3/2)/(c-c*sin(f*x+e))^(1/2)
Time = 2.14 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.05 \[ \int \frac {(A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)}}{(a+a \sin (e+f x))^{5/2}} \, dx=-\frac {\sqrt {a (1+\sin (e+f x))} (A+B+2 B \sin (e+f x)) \sqrt {c-c \sin (e+f x)}}{2 a^3 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^5} \]
-1/2*(Sqrt[a*(1 + Sin[e + f*x])]*(A + B + 2*B*Sin[e + f*x])*Sqrt[c - c*Sin [e + f*x]])/(a^3*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^5)
Time = 0.62 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {3042, 3450, 3042, 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.
\(\displaystyle \int \frac {\sqrt {c-c \sin (e+f x)} (A+B \sin (e+f x))}{(a \sin (e+f x)+a)^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sqrt {c-c \sin (e+f x)} (A+B \sin (e+f x))}{(a \sin (e+f x)+a)^{5/2}}dx\) |
\(\Big \downarrow \) 3450 |
\(\displaystyle (A-B) \int \frac {\sqrt {c-c \sin (e+f x)}}{(\sin (e+f x) a+a)^{5/2}}dx+\frac {B \int \frac {\sqrt {c-c \sin (e+f x)}}{(\sin (e+f x) a+a)^{3/2}}dx}{a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle (A-B) \int \frac {\sqrt {c-c \sin (e+f x)}}{(\sin (e+f x) a+a)^{5/2}}dx+\frac {B \int \frac {\sqrt {c-c \sin (e+f x)}}{(\sin (e+f x) a+a)^{3/2}}dx}{a}\) |
\(\Big \downarrow \) 3217 |
\(\displaystyle -\frac {c (A-B) \cos (e+f x)}{2 f (a \sin (e+f x)+a)^{5/2} \sqrt {c-c \sin (e+f x)}}-\frac {B c \cos (e+f x)}{a f (a \sin (e+f x)+a)^{3/2} \sqrt {c-c \sin (e+f x)}}\) |
-1/2*((A - B)*c*Cos[e + f*x])/(f*(a + a*Sin[e + f*x])^(5/2)*Sqrt[c - c*Sin [e + f*x]]) - (B*c*Cos[e + f*x])/(a*f*(a + a*Sin[e + f*x])^(3/2)*Sqrt[c - c*Sin[e + f*x]])
3.2.91.3.1 Defintions of rubi rules used
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 = 2.55 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.74
method | result | size |
default | \(\frac {\tan \left (f x +e \right ) \left (A \sin \left (f x +e \right )+B \sin \left (f x +e \right )+2 A \right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}}{2 a^{2} f \left (1+\sin \left (f x +e \right )\right ) \sqrt {a \left (1+\sin \left (f x +e \right )\right )}}\) | \(70\) |
parts | \(-\frac {A \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, \left (\cos \left (f x +e \right )-2 \tan \left (f x +e \right )-\sec \left (f x +e \right )\right )}{2 f \left (1+\sin \left (f x +e \right )\right ) \sqrt {a \left (1+\sin \left (f x +e \right )\right )}\, a^{2}}-\frac {B \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, \left (\cos \left (f x +e \right )-\sec \left (f x +e \right )\right )}{2 f \left (1+\sin \left (f x +e \right )\right ) \sqrt {a \left (1+\sin \left (f x +e \right )\right )}\, a^{2}}\) | \(128\) |
1/2/a^2/f*tan(f*x+e)*(A*sin(f*x+e)+B*sin(f*x+e)+2*A)*(-c*(sin(f*x+e)-1))^( 1/2)/(1+sin(f*x+e))/(a*(1+sin(f*x+e)))^(1/2)
Time = 0.27 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.90 \[ \int \frac {(A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)}}{(a+a \sin (e+f x))^{5/2}} \, dx=\frac {{\left (2 \, B \sin \left (f x + e\right ) + A + B\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{2 \, {\left (a^{3} f \cos \left (f x + e\right )^{3} - 2 \, a^{3} f \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, a^{3} f \cos \left (f x + e\right )\right )}} \]
integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(1/2)/(a+a*sin(f*x+e))^(5/2),x , algorithm="fricas")
1/2*(2*B*sin(f*x + e) + A + B)*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)/(a^3*f*cos(f*x + e)^3 - 2*a^3*f*cos(f*x + e)*sin(f*x + e) - 2*a^3* f*cos(f*x + e))
\[ \int \frac {(A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)}}{(a+a \sin (e+f x))^{5/2}} \, dx=\int \frac {\sqrt {- c \left (\sin {\left (e + f x \right )} - 1\right )} \left (A + B \sin {\left (e + f x \right )}\right )}{\left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{\frac {5}{2}}}\, dx \]
\[ \int \frac {(A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)}}{(a+a \sin (e+f x))^{5/2}} \, dx=\int { \frac {{\left (B \sin \left (f x + e\right ) + A\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(1/2)/(a+a*sin(f*x+e))^(5/2),x , algorithm="maxima")
Time = 0.47 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.18 \[ \int \frac {(A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)}}{(a+a \sin (e+f x))^{5/2}} \, dx=\frac {{\left (4 \, B \sqrt {a} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + A \sqrt {a} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - B \sqrt {a} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sqrt {c}}{8 \, a^{3} f \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} \]
integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(1/2)/(a+a*sin(f*x+e))^(5/2),x , algorithm="giac")
1/8*(4*B*sqrt(a)*cos(-1/4*pi + 1/2*f*x + 1/2*e)^2*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)*sg n(sin(-1/4*pi + 1/2*f*x + 1/2*e)))*sqrt(c)/(a^3*f*cos(-1/4*pi + 1/2*f*x + 1/2*e)^4*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)))
Time = 15.09 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.66 \[ \int \frac {(A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)}}{(a+a \sin (e+f x))^{5/2}} \, dx=-\frac {2\,\sqrt {-c\,\left (\sin \left (e+f\,x\right )-1\right )}\,\left (A\,\sin \left (2\,e+2\,f\,x\right )+3\,B\,\sin \left (2\,e+2\,f\,x\right )-2\,A\,\left (2\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-1\right )-3\,B\,\left (2\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-1\right )+B\,\left (2\,{\sin \left (\frac {3\,e}{2}+\frac {3\,f\,x}{2}\right )}^2-1\right )\right )}{a^2\,f\,\sqrt {a\,\left (\sin \left (e+f\,x\right )+1\right )}\,\left (-8\,{\sin \left (e+f\,x\right )}^2+4\,\sin \left (e+f\,x\right )+2\,{\sin \left (2\,e+2\,f\,x\right )}^2+4\,\sin \left (3\,e+3\,f\,x\right )+8\right )} \]
-(2*(-c*(sin(e + f*x) - 1))^(1/2)*(A*sin(2*e + 2*f*x) + 3*B*sin(2*e + 2*f* x) - 2*A*(2*sin(e/2 + (f*x)/2)^2 - 1) - 3*B*(2*sin(e/2 + (f*x)/2)^2 - 1) + B*(2*sin((3*e)/2 + (3*f*x)/2)^2 - 1)))/(a^2*f*(a*(sin(e + f*x) + 1))^(1/2 )*(4*sin(e + f*x) + 4*sin(3*e + 3*f*x) + 2*sin(2*e + 2*f*x)^2 - 8*sin(e + f*x)^2 + 8))