\(\int (a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)} \, dx\) [152]

Optimal result

Integrand size = 40, antiderivative size = 96 \[ \int (a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)} \, dx=\frac {(A-B) c \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{3 f \sqrt {c-c \sin (e+f x)}}+\frac {B c \cos (e+f x) (a+a \sin (e+f x))^{7/2}}{4 a f \sqrt {c-c \sin (e+f x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 96, 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} \[ \int (a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)} \, dx=\frac {c (A-B) \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{3 f \sqrt {c-c \sin (e+f x)}}+\frac {B c \cos (e+f x) (a \sin (e+f x)+a)^{7/2}}{4 a f \sqrt {c-c \sin (e+f x)}} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 1.86 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.06 \[ \int (a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)} \, dx=\frac {a^2 \sec (e+f x) \sqrt {a (1+\sin (e+f x))} \sqrt {c-c \sin (e+f x)} (3 B \cos (4 (e+f x))+16 (7 A+2 B) \sin (e+f x)-4 \cos (2 (e+f x)) (12 A+9 B+4 (A+2 B) \sin (e+f x)))}{96 f} \]

[In]

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

[Out]

(a^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 = 3.40 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.99

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

[In]

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

[Out]

-1/12*a^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 = 117, normalized size of antiderivative = 1.22 \[ \int (a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)} \, dx=\frac {{\left (3 \, B a^{2} \cos \left (f x + e\right )^{4} - 12 \, {\left (A + B\right )} a^{2} \cos \left (f x + e\right )^{2} + 3 \, {\left (4 \, A + 3 \, B\right )} a^{2} - 4 \, {\left ({\left (A + 2 \, B\right )} a^{2} \cos \left (f x + e\right )^{2} - 2 \, {\left (2 \, A + B\right )} a^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{12 \, f \cos \left (f x + e\right )} \]

[In]

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

[Out]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.43 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.56 \[ \int (a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)} \, dx=-\frac {4 \, {\left (3 \, B a^{2} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{8} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 2 \, A a^{2} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 2 \, B a^{2} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sqrt {a} \sqrt {c}}{3 \, f} \]

[In]

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

[Out]

-4/3*(3*B*a^2*cos(-1/4*pi + 1/2*f*x + 1/2*e)^8*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x +
1/2*e)) + 2*A*a^2*cos(-1/4*pi + 1/2*f*x + 1/2*e)^6*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*
f*x + 1/2*e)) - 2*B*a^2*cos(-1/4*pi + 1/2*f*x + 1/2*e)^6*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi +
1/2*f*x + 1/2*e)))*sqrt(a)*sqrt(c)/f

Mupad [B] (verification not implemented)

Time = 3.03 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.55 \[ \int (a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)} \, dx=-\frac {a^2\,\sqrt {a\,\left (\sin \left (e+f\,x\right )+1\right )}\,\sqrt {-c\,\left (\sin \left (e+f\,x\right )-1\right )}\,\left (48\,A\,\cos \left (e+f\,x\right )+36\,B\,\cos \left (e+f\,x\right )+48\,A\,\cos \left (3\,e+3\,f\,x\right )+33\,B\,\cos \left (3\,e+3\,f\,x\right )-3\,B\,\cos \left (5\,e+5\,f\,x\right )-112\,A\,\sin \left (2\,e+2\,f\,x\right )+8\,A\,\sin \left (4\,e+4\,f\,x\right )-32\,B\,\sin \left (2\,e+2\,f\,x\right )+16\,B\,\sin \left (4\,e+4\,f\,x\right )\right )}{96\,f\,\left (\cos \left (2\,e+2\,f\,x\right )+1\right )} \]

[In]

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

[Out]

-(a^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))