\(\int \frac {\sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{7/2}} \, dx\) [138]

Optimal result

Integrand size = 40, antiderivative size = 94 \[ \int \frac {\sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{7/2}} \, dx=\frac {a (A+B) \cos (e+f x)}{3 f \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{7/2}}-\frac {a B \cos (e+f x)}{2 c f \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2}} \]

[Out]

1/3*a*(A+B)*cos(f*x+e)/f/(c-c*sin(f*x+e))^(7/2)/(a+a*sin(f*x+e))^(1/2)-1/2*a*B*cos(f*x+e)/c/f/(c-c*sin(f*x+e))
^(5/2)/(a+a*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} \[ \int \frac {\sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{7/2}} \, dx=\frac {a (A+B) \cos (e+f x)}{3 f \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{7/2}}-\frac {a B \cos (e+f x)}{2 c f \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}} \]

[In]

Int[(Sqrt[a + a*Sin[e + f*x]]*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x])^(7/2),x]

[Out]

(a*(A + B)*Cos[e + f*x])/(3*f*Sqrt[a + a*Sin[e + f*x]]*(c - c*Sin[e + f*x])^(7/2)) - (a*B*Cos[e + f*x])/(2*c*f
*Sqrt[a + a*Sin[e + f*x]]*(c - c*Sin[e + f*x])^(5/2))

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))^{7/2}} \, dx-\frac {B \int \frac {\sqrt {a+a \sin (e+f x)}}{(c-c \sin (e+f x))^{5/2}} \, dx}{c} \\ & = \frac {a (A+B) \cos (e+f x)}{3 f \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{7/2}}-\frac {a B \cos (e+f x)}{2 c f \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 3.23 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.10 \[ \int \frac {\sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{7/2}} \, dx=\frac {\sqrt {a (1+\sin (e+f x))} (2 A-B+3 B \sin (e+f x)) \sqrt {c-c \sin (e+f x)}}{6 c^4 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^7 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )} \]

[In]

Integrate[(Sqrt[a + a*Sin[e + f*x]]*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x])^(7/2),x]

[Out]

(Sqrt[a*(1 + Sin[e + f*x])]*(2*A - B + 3*B*Sin[e + f*x])*Sqrt[c - c*Sin[e + f*x]])/(6*c^4*f*(Cos[(e + f*x)/2]
- Sin[(e + f*x)/2])^7*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]))

Maple [A] (verified)

Time = 3.56 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.10

method	result	size
default	\(\frac {\tan \left (f x +e \right ) \left (2 A \left (\cos ^{2}\left (f x +e \right )\right )+B \left (\sin ^{2}\left (f x +e \right )\right )+6 A \sin \left (f x +e \right )-3 B \sin \left (f x +e \right )-8 A \right ) \sqrt {a \left (1+\sin \left (f x +e \right )\right )}}{6 c^{3} f \left (\cos ^{2}\left (f x +e \right )+2 \sin \left (f x +e \right )-2\right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}}\)	\(103\)
parts	\(\frac {A \sqrt {a \left (1+\sin \left (f x +e \right )\right )}\, \left (\cos \left (f x +e \right ) \sin \left (f x +e \right )-3 \cos \left (f x +e \right )-4 \tan \left (f x +e \right )+3 \sec \left (f x +e \right )\right )}{3 f \left (\cos ^{2}\left (f x +e \right )+2 \sin \left (f x +e \right )-2\right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, c^{3}}-\frac {B \sec \left (f x +e \right ) \left (\cos \left (f x +e \right )-1\right ) \left (1+\cos \left (f x +e \right )\right ) \left (\sin \left (f x +e \right )-3\right ) \sqrt {a \left (1+\sin \left (f x +e \right )\right )}}{6 f \left (\cos ^{2}\left (f x +e \right )+2 \sin \left (f x +e \right )-2\right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, c^{3}}\)	\(178\)

[In]

int((A+B*sin(f*x+e))*(a+a*sin(f*x+e))^(1/2)/(c-c*sin(f*x+e))^(7/2),x,method=_RETURNVERBOSE)

[Out]

1/6/c^3/f*tan(f*x+e)*(2*A*cos(f*x+e)^2+B*sin(f*x+e)^2+6*A*sin(f*x+e)-3*B*sin(f*x+e)-8*A)*(a*(1+sin(f*x+e)))^(1
/2)/(cos(f*x+e)^2+2*sin(f*x+e)-2)/(-c*(sin(f*x+e)-1))^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.13 \[ \int \frac {\sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{7/2}} \, dx=-\frac {{\left (3 \, B \sin \left (f x + e\right ) + 2 \, A - B\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{6 \, {\left (3 \, c^{4} f \cos \left (f x + e\right )^{3} - 4 \, c^{4} f \cos \left (f x + e\right ) - {\left (c^{4} f \cos \left (f x + e\right )^{3} - 4 \, c^{4} f \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )}} \]

[In]

integrate((A+B*sin(f*x+e))*(a+a*sin(f*x+e))^(1/2)/(c-c*sin(f*x+e))^(7/2),x, algorithm="fricas")

[Out]

-1/6*(3*B*sin(f*x + e) + 2*A - B)*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)/(3*c^4*f*cos(f*x + e)^3 -
4*c^4*f*cos(f*x + e) - (c^4*f*cos(f*x + e)^3 - 4*c^4*f*cos(f*x + e))*sin(f*x + e))

Sympy [F(-1)]

Timed out. \[ \int \frac {\sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{7/2}} \, dx=\text {Timed out} \]

[In]

integrate((A+B*sin(f*x+e))*(a+a*sin(f*x+e))**(1/2)/(c-c*sin(f*x+e))**(7/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {\sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{7/2}} \, dx=\int { \frac {{\left (B \sin \left (f x + e\right ) + A\right )} \sqrt {a \sin \left (f x + e\right ) + a}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {7}{2}}} \,d x } \]

[In]

integrate((A+B*sin(f*x+e))*(a+a*sin(f*x+e))^(1/2)/(c-c*sin(f*x+e))^(7/2),x, algorithm="maxima")

[Out]

integrate((B*sin(f*x + e) + A)*sqrt(a*sin(f*x + e) + a)/(-c*sin(f*x + e) + c)^(7/2), x)

Giac [A] (verification not implemented)

none

Time = 0.43 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.19 \[ \int \frac {\sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{7/2}} \, dx=\frac {{\left (3 \, B \sqrt {c} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - A \sqrt {c} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - B \sqrt {c} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sqrt {a}}{24 \, c^{4} f \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6}} \]

[In]

integrate((A+B*sin(f*x+e))*(a+a*sin(f*x+e))^(1/2)/(c-c*sin(f*x+e))^(7/2),x, algorithm="giac")

[Out]

1/24*(3*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^4*f*sgn(sin(-1/4*pi + 1/2
*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e)^6)

Mupad [B] (verification not implemented)

Time = 18.33 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.63 \[ \int \frac {\sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{7/2}} \, dx=-\frac {2\,A\,\sqrt {a+a\,\sin \left (e+f\,x\right )}\,\sqrt {c-c\,\sin \left (e+f\,x\right )}-B\,\sqrt {a+a\,\sin \left (e+f\,x\right )}\,\sqrt {c-c\,\sin \left (e+f\,x\right )}+3\,B\,\sin \left (e+f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}\,\sqrt {c-c\,\sin \left (e+f\,x\right )}}{\frac {9\,c^4\,f\,\cos \left (3\,e+3\,f\,x\right )}{2}+\frac {21\,c^4\,f\,\sin \left (2\,e+2\,f\,x\right )}{2}-\frac {3\,c^4\,f\,\sin \left (4\,e+4\,f\,x\right )}{4}-\frac {21\,c^4\,f\,\cos \left (e+f\,x\right )}{2}} \]

[In]

int(((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^(1/2))/(c - c*sin(e + f*x))^(7/2),x)

[Out]

-(2*A*(a + a*sin(e + f*x))^(1/2)*(c - c*sin(e + f*x))^(1/2) - B*(a + a*sin(e + f*x))^(1/2)*(c - c*sin(e + f*x)
)^(1/2) + 3*B*sin(e + f*x)*(a + a*sin(e + f*x))^(1/2)*(c - c*sin(e + f*x))^(1/2))/((9*c^4*f*cos(3*e + 3*f*x))/
2 + (21*c^4*f*sin(2*e + 2*f*x))/2 - (3*c^4*f*sin(4*e + 4*f*x))/4 - (21*c^4*f*cos(e + f*x))/2)