\(\int \frac {(A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)}}{(a+a \sin (e+f x))^{5/2}} \, dx\) [191]

Optimal result

Integrand size = 40, antiderivative size = 94 \[ \int \frac {(A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)}}{(a+a \sin (e+f x))^{5/2}} \, dx=-\frac {(A-B) c \cos (e+f x)}{2 f (a+a \sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)}}-\frac {B c \cos (e+f x)}{a f (a+a \sin (e+f x))^{3/2} \sqrt {c-c \sin (e+f x)}} \]

[Out]

-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)

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {3050, 2817} \[ \int \frac {(A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)}}{(a+a \sin (e+f x))^{5/2}} \, dx=-\frac {c (A-B) \cos (e+f x)}{2 f (a \sin (e+f x)+a)^{5/2} \sqrt {c-c \sin (e+f x)}}-\frac {B c \cos (e+f x)}{a f (a \sin (e+f x)+a)^{3/2} \sqrt {c-c \sin (e+f x)}} \]

[In]

Int[((A + B*Sin[e + f*x])*Sqrt[c - c*Sin[e + f*x]])/(a + a*Sin[e + f*x])^(5/2),x]

[Out]

-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]])

Rule 2817

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)]

Rule 3050

Int[Sqrt[(a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[B/d, Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n + 1), x], x
] - Dist[(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]

Rubi steps \begin{align*} \text {integral}& = \frac {B \int \frac {\sqrt {c-c \sin (e+f x)}}{(a+a \sin (e+f x))^{3/2}} \, dx}{a}-(-A+B) \int \frac {\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)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.14 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.05 \[ \int \frac {(A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)}}{(a+a \sin (e+f x))^{5/2}} \, dx=-\frac {\sqrt {a (1+\sin (e+f x))} (A+B+2 B \sin (e+f x)) \sqrt {c-c \sin (e+f x)}}{2 a^3 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^5} \]

[In]

Integrate[((A + B*Sin[e + f*x])*Sqrt[c - c*Sin[e + f*x]])/(a + a*Sin[e + f*x])^(5/2),x]

[Out]

-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)

Maple [A] (verified)

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	\(-\frac {A \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, \left (\cos \left (f x +e \right )-2 \tan \left (f x +e \right )-\sec \left (f x +e \right )\right )}{2 f \left (1+\sin \left (f x +e \right )\right ) \sqrt {a \left (1+\sin \left (f x +e \right )\right )}\, a^{2}}-\frac {B \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, \left (\cos \left (f x +e \right )-\sec \left (f x +e \right )\right )}{2 f \left (1+\sin \left (f x +e \right )\right ) \sqrt {a \left (1+\sin \left (f x +e \right )\right )}\, a^{2}}\)	\(128\)

[In]

int((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(1/2)/(a+a*sin(f*x+e))^(5/2),x,method=_RETURNVERBOSE)

[Out]

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)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.90 \[ \int \frac {(A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)}}{(a+a \sin (e+f x))^{5/2}} \, dx=\frac {{\left (2 \, B \sin \left (f x + e\right ) + A + B\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{2 \, {\left (a^{3} f \cos \left (f x + e\right )^{3} - 2 \, a^{3} f \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, a^{3} f \cos \left (f x + e\right )\right )}} \]

[In]

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")

[Out]

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))

Sympy [F]

\[ \int \frac {(A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)}}{(a+a \sin (e+f x))^{5/2}} \, dx=\int \frac {\sqrt {- c \left (\sin {\left (e + f x \right )} - 1\right )} \left (A + B \sin {\left (e + f x \right )}\right )}{\left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{\frac {5}{2}}}\, dx \]

[In]

integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))**(1/2)/(a+a*sin(f*x+e))**(5/2),x)

[Out]

Integral(sqrt(-c*(sin(e + f*x) - 1))*(A + B*sin(e + f*x))/(a*(sin(e + f*x) + 1))**(5/2), x)

Maxima [F]

\[ \int \frac {(A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)}}{(a+a \sin (e+f x))^{5/2}} \, dx=\int { \frac {{\left (B \sin \left (f x + e\right ) + A\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]

[In]

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")

[Out]

integrate((B*sin(f*x + e) + A)*sqrt(-c*sin(f*x + e) + c)/(a*sin(f*x + e) + a)^(5/2), x)

Giac [A] (verification not implemented)

none

Time = 0.47 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.18 \[ \int \frac {(A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)}}{(a+a \sin (e+f x))^{5/2}} \, dx=\frac {{\left (4 \, B \sqrt {a} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + A \sqrt {a} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - B \sqrt {a} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sqrt {c}}{8 \, a^{3} f \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} \]

[In]

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")

[Out]

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)*sgn(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)))

Mupad [B] (verification not implemented)

Time = 15.09 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.66 \[ \int \frac {(A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)}}{(a+a \sin (e+f x))^{5/2}} \, dx=-\frac {2\,\sqrt {-c\,\left (\sin \left (e+f\,x\right )-1\right )}\,\left (A\,\sin \left (2\,e+2\,f\,x\right )+3\,B\,\sin \left (2\,e+2\,f\,x\right )-2\,A\,\left (2\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-1\right )-3\,B\,\left (2\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-1\right )+B\,\left (2\,{\sin \left (\frac {3\,e}{2}+\frac {3\,f\,x}{2}\right )}^2-1\right )\right )}{a^2\,f\,\sqrt {a\,\left (\sin \left (e+f\,x\right )+1\right )}\,\left (-8\,{\sin \left (e+f\,x\right )}^2+4\,\sin \left (e+f\,x\right )+2\,{\sin \left (2\,e+2\,f\,x\right )}^2+4\,\sin \left (3\,e+3\,f\,x\right )+8\right )} \]

[In]

int(((A + B*sin(e + f*x))*(c - c*sin(e + f*x))^(1/2))/(a + a*sin(e + f*x))^(5/2),x)

[Out]

-(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))