\(\int \frac {\sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{5/2}} \, dx\) [137]

Optimal result

Integrand size = 40, antiderivative size = 92 \[ \int \frac {\sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{5/2}} \, dx=\frac {a (A+B) \cos (e+f x)}{2 f \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2}}-\frac {a B \cos (e+f x)}{c f \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}} \]

[Out]

1/2*a*(A+B)*cos(f*x+e)/f/(c-c*sin(f*x+e))^(5/2)/(a+a*sin(f*x+e))^(1/2)-a*B*cos(f*x+e)/c/f/(c-c*sin(f*x+e))^(3/
2)/(a+a*sin(f*x+e))^(1/2)

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 92, 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} \[ \int \frac {\sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{5/2}} \, dx=\frac {a (A+B) \cos (e+f x)}{2 f \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}}-\frac {a B \cos (e+f x)}{c f \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{3/2}} \]

[In]

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

[Out]

(a*(A + B)*Cos[e + f*x])/(2*f*Sqrt[a + a*Sin[e + f*x]]*(c - c*Sin[e + f*x])^(5/2)) - (a*B*Cos[e + f*x])/(c*f*S
qrt[a + a*Sin[e + f*x]]*(c - c*Sin[e + f*x])^(3/2))

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}& = (A+B) \int \frac {\sqrt {a+a \sin (e+f x)}}{(c-c \sin (e+f x))^{5/2}} \, dx-\frac {B \int \frac {\sqrt {a+a \sin (e+f x)}}{(c-c \sin (e+f x))^{3/2}} \, dx}{c} \\ & = \frac {a (A+B) \cos (e+f x)}{2 f \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2}}-\frac {a B \cos (e+f x)}{c f \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.38 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.10 \[ \int \frac {\sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{5/2}} \, dx=\frac {\sqrt {a (1+\sin (e+f x))} (A-B+2 B \sin (e+f x)) \sqrt {c-c \sin (e+f x)}}{2 c^3 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^5 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )} \]

[In]

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

[Out]

(Sqrt[a*(1 + Sin[e + f*x])]*(A - B + 2*B*Sin[e + f*x])*Sqrt[c - c*Sin[e + f*x]])/(2*c^3*f*(Cos[(e + f*x)/2] -
Sin[(e + f*x)/2])^5*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]))

Maple [A] (verified)

Time = 3.41 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.77

method	result	size
default	\(\frac {\tan \left (f x +e \right ) \left (A \sin \left (f x +e \right )-B \sin \left (f x +e \right )-2 A \right ) \sqrt {a \left (1+\sin \left (f x +e \right )\right )}}{2 c^{2} f \left (\sin \left (f x +e \right )-1\right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}}\)	\(71\)
parts	\(-\frac {A \sqrt {a \left (1+\sin \left (f x +e \right )\right )}\, \left (\cos \left (f x +e \right )+2 \tan \left (f x +e \right )-\sec \left (f x +e \right )\right )}{2 f \left (\sin \left (f x +e \right )-1\right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, c^{2}}+\frac {B \sqrt {a \left (1+\sin \left (f x +e \right )\right )}\, \left (\cos \left (f x +e \right )-\sec \left (f x +e \right )\right )}{2 f \left (\sin \left (f x +e \right )-1\right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, c^{2}}\)	\(128\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.95 \[ \int \frac {\sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{5/2}} \, dx=-\frac {{\left (2 \, B \sin \left (f x + e\right ) + A - B\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{2 \, {\left (c^{3} f \cos \left (f x + e\right )^{3} + 2 \, c^{3} f \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, c^{3} f \cos \left (f x + e\right )\right )}} \]

[In]

integrate((A+B*sin(f*x+e))*(a+a*sin(f*x+e))^(1/2)/(c-c*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)/(c^3*f*cos(f*x + e)^3 + 2*c
^3*f*cos(f*x + e)*sin(f*x + e) - 2*c^3*f*cos(f*x + e))

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

integrate((A+B*sin(f*x+e))*(a+a*sin(f*x+e))^(1/2)/(c-c*sin(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.45 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.22 \[ \int \frac {\sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{5/2}} \, dx=\frac {{\left (4 \, B \sqrt {c} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - A \sqrt {c} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - B \sqrt {c} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sqrt {a}}{8 \, c^{3} f \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4}} \]

[In]

integrate((A+B*sin(f*x+e))*(a+a*sin(f*x+e))^(1/2)/(c-c*sin(f*x+e))^(5/2),x, algorithm="giac")

[Out]

1/8*(4*B*sqrt(c)*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e)^2 - A*sqrt(c)*sgn(cos(-1/4
*pi + 1/2*f*x + 1/2*e)) - B*sqrt(c)*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)))*sqrt(a)/(c^3*f*sgn(sin(-1/4*pi + 1/2*
f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e)^4)

Mupad [F(-1)]

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

[In]

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

[Out]

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