\(\int \sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2} \, dx\) [131]

Optimal result

Integrand size = 40, antiderivative size = 94 \[ \int \sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2} \, dx=-\frac {a (A+B) \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{4 f \sqrt {a+a \sin (e+f x)}}+\frac {a B \cos (e+f x) (c-c \sin (e+f x))^{9/2}}{5 c f \sqrt {a+a \sin (e+f x)}} \]

[Out]

-1/4*a*(A+B)*cos(f*x+e)*(c-c*sin(f*x+e))^(7/2)/f/(a+a*sin(f*x+e))^(1/2)+1/5*a*B*cos(f*x+e)*(c-c*sin(f*x+e))^(9
/2)/c/f/(a+a*sin(f*x+e))^(1/2)

Rubi [A] (verified)

Time = 0.23 (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} \[ \int \sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2} \, dx=\frac {a B \cos (e+f x) (c-c \sin (e+f x))^{9/2}}{5 c f \sqrt {a \sin (e+f x)+a}}-\frac {a (A+B) \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{4 f \sqrt {a \sin (e+f x)+a}} \]

[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]

-1/4*(a*(A + B)*Cos[e + f*x]*(c - c*Sin[e + f*x])^(7/2))/(f*Sqrt[a + a*Sin[e + f*x]]) + (a*B*Cos[e + f*x]*(c -
c*Sin[e + f*x])^(9/2))/(5*c*f*Sqrt[a + a*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}& = (A+B) \int \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{7/2} \, dx-\frac {B \int \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{9/2} \, dx}{c} \\ & = -\frac {a (A+B) \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{4 f \sqrt {a+a \sin (e+f x)}}+\frac {a B \cos (e+f x) (c-c \sin (e+f x))^{9/2}}{5 c f \sqrt {a+a \sin (e+f x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.39 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.26 \[ \int \sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2} \, dx=-\frac {c^3 \sec (e+f x) \sqrt {a (1+\sin (e+f x))} \sqrt {c-c \sin (e+f x)} (4 (-60 A+23 B) \sin (e+f x)+4 \cos (2 (e+f x)) (-35 A+25 B+4 (5 A-6 B) \sin (e+f x))+\cos (4 (e+f x)) (5 A-15 B+4 B \sin (e+f x)))}{160 f} \]

[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]

-1/160*(c^3*Sec[e + f*x]*Sqrt[a*(1 + Sin[e + f*x])]*Sqrt[c - c*Sin[e + f*x]]*(4*(-60*A + 23*B)*Sin[e + f*x] +
4*Cos[2*(e + f*x)]*(-35*A + 25*B + 4*(5*A - 6*B)*Sin[e + f*x]) + Cos[4*(e + f*x)]*(5*A - 15*B + 4*B*Sin[e + f*
x])))/f

Maple [A] (verified)

Time = 3.70 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.39

method	result	size
default	\(\frac {c^{3} \tan \left (f x +e \right ) \left (4 B \left (\sin ^{2}\left (f x +e \right )\right ) \left (\cos ^{2}\left (f x +e \right )\right )+5 A \sin \left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )\right )+15 B \left (\sin ^{3}\left (f x +e \right )\right )-20 A \left (\cos ^{2}\left (f x +e \right )\right )-24 B \left (\sin ^{2}\left (f x +e \right )\right )-35 A \sin \left (f x +e \right )+10 B \sin \left (f x +e \right )+40 A \right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+\sin \left (f x +e \right )\right )}}{20 f}\)	\(131\)
parts	\(-\frac {A \sqrt {a \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, c^{3} \left (\cos ^{3}\left (f x +e \right )+4 \cos \left (f x +e \right ) \sin \left (f x +e \right )-8 \cos \left (f x +e \right )-8 \tan \left (f x +e \right )+7 \sec \left (f x +e \right )\right )}{4 f}-\frac {B \sec \left (f x +e \right ) \left (4 \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )-15 \left (\cos ^{2}\left (f x +e \right )\right )-24 \sin \left (f x +e \right )+25\right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+\sin \left (f x +e \right )\right )}\, c^{3} \left (\cos ^{2}\left (f x +e \right )-1\right )}{20 f}\)	\(169\)

[In]

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

[Out]

1/20*c^3/f*tan(f*x+e)*(4*B*sin(f*x+e)^2*cos(f*x+e)^2+5*A*sin(f*x+e)*cos(f*x+e)^2+15*B*sin(f*x+e)^3-20*A*cos(f*
x+e)^2-24*B*sin(f*x+e)^2-35*A*sin(f*x+e)+10*B*sin(f*x+e)+40*A)*(-c*(sin(f*x+e)-1))^(1/2)*(a*(1+sin(f*x+e)))^(1
/2)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.49 \[ \int \sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2} \, dx=-\frac {{\left (5 \, {\left (A - 3 \, B\right )} c^{3} \cos \left (f x + e\right )^{4} - 40 \, {\left (A - B\right )} c^{3} \cos \left (f x + e\right )^{2} + 5 \, {\left (7 \, A - 5 \, B\right )} c^{3} + 4 \, {\left (B c^{3} \cos \left (f x + e\right )^{4} + {\left (5 \, A - 7 \, B\right )} c^{3} \cos \left (f x + e\right )^{2} - 2 \, {\left (5 \, A - 3 \, B\right )} c^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{20 \, f \cos \left (f x + e\right )} \]

[In]

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

[Out]

-1/20*(5*(A - 3*B)*c^3*cos(f*x + e)^4 - 40*(A - B)*c^3*cos(f*x + e)^2 + 5*(7*A - 5*B)*c^3 + 4*(B*c^3*cos(f*x +
e)^4 + (5*A - 7*B)*c^3*cos(f*x + e)^2 - 2*(5*A - 3*B)*c^3)*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin
(f*x + e) + c)/(f*cos(f*x + e))

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(7/2)*(a+a*sin(f*x+e))^(1/2),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 = 0.37 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.60 \[ \int \sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2} \, dx=-\frac {4 \, {\left (8 \, B c^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{10} - 5 \, A c^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{8} - 5 \, B c^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{8}\right )} \sqrt {a} \sqrt {c}}{5 \, f} \]

[In]

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

[Out]

-4/5*(8*B*c^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x +
1/2*e)^10 - 5*A*c^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*
f*x + 1/2*e)^8 - 5*B*c^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi +
1/2*f*x + 1/2*e)^8)*sqrt(a)*sqrt(c)/f

Mupad [B] (verification not implemented)

Time = 16.39 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.84 \[ \int \sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2} \, dx=-\frac {c^3\,\sqrt {a\,\left (\sin \left (e+f\,x\right )+1\right )}\,\sqrt {-c\,\left (\sin \left (e+f\,x\right )-1\right )}\,\left (100\,B\,\cos \left (e+f\,x\right )-140\,A\,\cos \left (e+f\,x\right )-135\,A\,\cos \left (3\,e+3\,f\,x\right )+5\,A\,\cos \left (5\,e+5\,f\,x\right )+85\,B\,\cos \left (3\,e+3\,f\,x\right )-15\,B\,\cos \left (5\,e+5\,f\,x\right )-240\,A\,\sin \left (2\,e+2\,f\,x\right )+40\,A\,\sin \left (4\,e+4\,f\,x\right )+90\,B\,\sin \left (2\,e+2\,f\,x\right )-48\,B\,\sin \left (4\,e+4\,f\,x\right )+2\,B\,\sin \left (6\,e+6\,f\,x\right )\right )}{160\,f\,\left (\cos \left (2\,e+2\,f\,x\right )+1\right )} \]

[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]

-(c^3*(a*(sin(e + f*x) + 1))^(1/2)*(-c*(sin(e + f*x) - 1))^(1/2)*(100*B*cos(e + f*x) - 140*A*cos(e + f*x) - 13
5*A*cos(3*e + 3*f*x) + 5*A*cos(5*e + 5*f*x) + 85*B*cos(3*e + 3*f*x) - 15*B*cos(5*e + 5*f*x) - 240*A*sin(2*e +
2*f*x) + 40*A*sin(4*e + 4*f*x) + 90*B*sin(2*e + 2*f*x) - 48*B*sin(4*e + 4*f*x) + 2*B*sin(6*e + 6*f*x)))/(160*f
*(cos(2*e + 2*f*x) + 1))