\(\int \sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2} \, dx\) [132]

Optimal result

Integrand size = 40, antiderivative size = 94 \[ \int \sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2} \, dx=-\frac {a (A+B) \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{3 f \sqrt {a+a \sin (e+f x)}}+\frac {a B \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{4 c f \sqrt {a+a \sin (e+f x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.24 (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} \[ \int \sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2} \, dx=\frac {a B \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{4 c f \sqrt {a \sin (e+f x)+a}}-\frac {a (A+B) \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{3 f \sqrt {a \sin (e+f x)+a}} \]

[In]

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

[Out]

-1/3*(a*(A + B)*Cos[e + f*x]*(c - c*Sin[e + f*x])^(5/2))/(f*Sqrt[a + a*Sin[e + f*x]]) + (a*B*Cos[e + f*x]*(c -
c*Sin[e + f*x])^(7/2))/(4*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))^{5/2} \, dx-\frac {B \int \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{7/2} \, dx}{c} \\ & = -\frac {a (A+B) \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{3 f \sqrt {a+a \sin (e+f x)}}+\frac {a B \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{4 c f \sqrt {a+a \sin (e+f x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.11 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.09 \[ \int \sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2} \, dx=\frac {c^2 \sec (e+f x) \sqrt {a (1+\sin (e+f x))} \sqrt {c-c \sin (e+f x)} (3 B \cos (4 (e+f x))+16 (7 A-2 B) \sin (e+f x)-4 \cos (2 (e+f x)) (-12 A+9 B+4 (A-2 B) \sin (e+f x)))}{96 f} \]

[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]

(c^2*Sec[e + f*x]*Sqrt[a*(1 + Sin[e + f*x])]*Sqrt[c - c*Sin[e + f*x]]*(3*B*Cos[4*(e + f*x)] + 16*(7*A - 2*B)*S
in[e + f*x] - 4*Cos[2*(e + f*x)]*(-12*A + 9*B + 4*(A - 2*B)*Sin[e + f*x])))/(96*f)

Maple [A] (verified)

Time = 3.59 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.01

method	result	size
default	\(-\frac {c^{2} \tan \left (f x +e \right ) \left (-3 B \left (\sin ^{3}\left (f x +e \right )\right )+4 A \left (\cos ^{2}\left (f x +e \right )\right )+8 B \left (\sin ^{2}\left (f x +e \right )\right )+12 A \sin \left (f x +e \right )-6 B \sin \left (f x +e \right )-16 A \right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+\sin \left (f x +e \right )\right )}}{12 f}\)	\(95\)
parts	\(-\frac {A \,c^{2} \sqrt {a \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, \left (\cos \left (f x +e \right ) \sin \left (f x +e \right )-3 \cos \left (f x +e \right )-4 \tan \left (f x +e \right )+3 \sec \left (f x +e \right )\right )}{3 f}+\frac {B \sec \left (f x +e \right ) \left (3 \left (\cos ^{2}\left (f x +e \right )\right )-9+8 \sin \left (f x +e \right )\right ) c^{2} \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+\sin \left (f x +e \right )\right )}\, \left (\cos ^{2}\left (f x +e \right )-1\right )}{12 f}\)	\(144\)

[In]

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

[Out]

-1/12*c^2/f*tan(f*x+e)*(-3*B*sin(f*x+e)^3+4*A*cos(f*x+e)^2+8*B*sin(f*x+e)^2+12*A*sin(f*x+e)-6*B*sin(f*x+e)-16*
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 = 121, normalized size of antiderivative = 1.29 \[ \int \sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2} \, dx=\frac {{\left (3 \, B c^{2} \cos \left (f x + e\right )^{4} + 12 \, {\left (A - B\right )} c^{2} \cos \left (f x + e\right )^{2} - 3 \, {\left (4 \, A - 3 \, B\right )} c^{2} - 4 \, {\left ({\left (A - 2 \, B\right )} c^{2} \cos \left (f x + e\right )^{2} - 2 \, {\left (2 \, A - B\right )} c^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{12 \, f \cos \left (f x + e\right )} \]

[In]

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

[Out]

1/12*(3*B*c^2*cos(f*x + e)^4 + 12*(A - B)*c^2*cos(f*x + e)^2 - 3*(4*A - 3*B)*c^2 - 4*((A - 2*B)*c^2*cos(f*x +
e)^2 - 2*(2*A - B)*c^2)*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. \[ \int \sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

Giac [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.60 \[ \int \sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2} \, dx=-\frac {4 \, {\left (3 \, B c^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{8} - 2 \, A c^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} - 2 \, B c^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6}\right )} \sqrt {a} \sqrt {c}}{3 \, f} \]

[In]

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

[Out]

-4/3*(3*B*c^2*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 - 2*A*c^2*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)^6 - 2*B*c^2*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)^6)*sqrt(a)*sqrt(c)/f

Mupad [B] (verification not implemented)

Time = 15.01 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.59 \[ \int \sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2} \, dx=\frac {c^2\,\sqrt {a\,\left (\sin \left (e+f\,x\right )+1\right )}\,\sqrt {-c\,\left (\sin \left (e+f\,x\right )-1\right )}\,\left (48\,A\,\cos \left (e+f\,x\right )-36\,B\,\cos \left (e+f\,x\right )+48\,A\,\cos \left (3\,e+3\,f\,x\right )-33\,B\,\cos \left (3\,e+3\,f\,x\right )+3\,B\,\cos \left (5\,e+5\,f\,x\right )+112\,A\,\sin \left (2\,e+2\,f\,x\right )-8\,A\,\sin \left (4\,e+4\,f\,x\right )-32\,B\,\sin \left (2\,e+2\,f\,x\right )+16\,B\,\sin \left (4\,e+4\,f\,x\right )\right )}{96\,f\,\left (\cos \left (2\,e+2\,f\,x\right )+1\right )} \]

[In]

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

[Out]

(c^2*(a*(sin(e + f*x) + 1))^(1/2)*(-c*(sin(e + f*x) - 1))^(1/2)*(48*A*cos(e + f*x) - 36*B*cos(e + f*x) + 48*A*
cos(3*e + 3*f*x) - 33*B*cos(3*e + 3*f*x) + 3*B*cos(5*e + 5*f*x) + 112*A*sin(2*e + 2*f*x) - 8*A*sin(4*e + 4*f*x
) - 32*B*sin(2*e + 2*f*x) + 16*B*sin(4*e + 4*f*x)))/(96*f*(cos(2*e + 2*f*x) + 1))