\(\int (a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2} \, dx\) [150]

Optimal result

Integrand size = 40, antiderivative size = 180 \[ \int (a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2} \, dx=-\frac {2 a^3 A \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{15 f \sqrt {a+a \sin (e+f x)}}-\frac {a^2 A \cos (e+f x) \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2}}{5 f}-\frac {a A \cos (e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2}}{5 f}-\frac {B \cos (e+f x) (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{5/2}}{6 f} \]

[Out]

-1/5*a*A*cos(f*x+e)*(a+a*sin(f*x+e))^(3/2)*(c-c*sin(f*x+e))^(5/2)/f-1/6*B*cos(f*x+e)*(a+a*sin(f*x+e))^(5/2)*(c
-c*sin(f*x+e))^(5/2)/f-2/15*a^3*A*cos(f*x+e)*(c-c*sin(f*x+e))^(5/2)/f/(a+a*sin(f*x+e))^(1/2)-1/5*a^2*A*cos(f*x
+e)*(c-c*sin(f*x+e))^(5/2)*(a+a*sin(f*x+e))^(1/2)/f

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.075, Rules used = {3052, 2819, 2817} \[ \int (a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2} \, dx=-\frac {2 a^3 A \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{15 f \sqrt {a \sin (e+f x)+a}}-\frac {a^2 A \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}}{5 f}-\frac {a A \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{5/2}}{5 f}-\frac {B \cos (e+f x) (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{5/2}}{6 f} \]

[In]

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

[Out]

(-2*a^3*A*Cos[e + f*x]*(c - c*Sin[e + f*x])^(5/2))/(15*f*Sqrt[a + a*Sin[e + f*x]]) - (a^2*A*Cos[e + f*x]*Sqrt[
a + a*Sin[e + f*x]]*(c - c*Sin[e + f*x])^(5/2))/(5*f) - (a*A*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2)*(c - c*Si
n[e + f*x])^(5/2))/(5*f) - (B*Cos[e + f*x]*(a + a*Sin[e + f*x])^(5/2)*(c - c*Sin[e + f*x])^(5/2))/(6*f)

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 2819

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp
[(-b)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[e + f*x])^n/(f*(m + n))), x] + Dist[a*((2*m - 1)/(
m + n)), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&
EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IGtQ[m - 1/2, 0] &&  !LtQ[n, -1] &&  !(IGtQ[n - 1/2, 0] && LtQ[n, m
]) &&  !(ILtQ[m + n, 0] && GtQ[2*m + n + 1, 0])

Rule 3052

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^n/(f*
(m + n + 1))), x] - Dist[(B*c*(m - n) - A*d*(m + n + 1))/(d*(m + n + 1)), Int[(a + b*Sin[e + f*x])^m*(c + d*Si
n[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] &&
!LtQ[m, -2^(-1)] && NeQ[m + n + 1, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {B \cos (e+f x) (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{5/2}}{6 f}+A \int (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{5/2} \, dx \\ & = -\frac {a A \cos (e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2}}{5 f}-\frac {B \cos (e+f x) (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{5/2}}{6 f}+\frac {1}{5} (4 a A) \int (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2} \, dx \\ & = -\frac {a^2 A \cos (e+f x) \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2}}{5 f}-\frac {a A \cos (e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2}}{5 f}-\frac {B \cos (e+f x) (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{5/2}}{6 f}+\frac {1}{5} \left (2 a^2 A\right ) \int \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2} \, dx \\ & = -\frac {2 a^3 A \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{15 f \sqrt {a+a \sin (e+f x)}}-\frac {a^2 A \cos (e+f x) \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2}}{5 f}-\frac {a A \cos (e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2}}{5 f}-\frac {B \cos (e+f x) (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{5/2}}{6 f} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.56 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.64 \[ \int (a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2} \, dx=-\frac {a^2 c^2 \sec (e+f x) \sqrt {a (1+\sin (e+f x))} \sqrt {c-c \sin (e+f x)} (50 B+75 B \cos (2 (e+f x))+30 B \cos (4 (e+f x))+5 B \cos (6 (e+f x))-600 A \sin (e+f x)-100 A \sin (3 (e+f x))-12 A \sin (5 (e+f x)))}{960 f} \]

[In]

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

[Out]

-1/960*(a^2*c^2*Sec[e + f*x]*Sqrt[a*(1 + Sin[e + f*x])]*Sqrt[c - c*Sin[e + f*x]]*(50*B + 75*B*Cos[2*(e + f*x)]
+ 30*B*Cos[4*(e + f*x)] + 5*B*Cos[6*(e + f*x)] - 600*A*Sin[e + f*x] - 100*A*Sin[3*(e + f*x)] - 12*A*Sin[5*(e
+ f*x)]))/f

Maple [A] (verified)

Time = 72.76 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.62

method	result	size
default	\(\frac {a^{2} c^{2} \tan \left (f x +e \right ) \left (5 B \left (\cos ^{4}\left (f x +e \right )\right ) \sin \left (f x +e \right )+6 A \left (\cos ^{4}\left (f x +e \right )\right )+5 B \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )+8 A \left (\cos ^{2}\left (f x +e \right )\right )+5 B \sin \left (f x +e \right )+16 A \right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+\sin \left (f x +e \right )\right )}}{30 f}\)	\(112\)
parts	\(\frac {A \tan \left (f x +e \right ) a^{2} c^{2} \left (3 \left (\cos ^{4}\left (f x +e \right )\right )+4 \left (\cos ^{2}\left (f x +e \right )\right )+8\right ) \sqrt {a \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}}{15 f}-\frac {B \sec \left (f x +e \right ) a^{2} c^{2} \left (\cos ^{4}\left (f x +e \right )+\cos ^{2}\left (f x +e \right )+1\right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+\sin \left (f x +e \right )\right )}\, \left (\cos ^{2}\left (f x +e \right )-1\right )}{6 f}\)	\(138\)

[In]

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

[Out]

1/30*a^2*c^2/f*tan(f*x+e)*(5*B*cos(f*x+e)^4*sin(f*x+e)+6*A*cos(f*x+e)^4+5*B*cos(f*x+e)^2*sin(f*x+e)+8*A*cos(f*
x+e)^2+5*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 = 0.65 \[ \int (a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2} \, dx=-\frac {{\left (5 \, B a^{2} c^{2} \cos \left (f x + e\right )^{6} - 5 \, B a^{2} c^{2} - 2 \, {\left (3 \, A a^{2} c^{2} \cos \left (f x + e\right )^{4} + 4 \, A a^{2} c^{2} \cos \left (f x + e\right )^{2} + 8 \, A a^{2} c^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{30 \, f \cos \left (f x + e\right )} \]

[In]

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

[Out]

-1/30*(5*B*a^2*c^2*cos(f*x + e)^6 - 5*B*a^2*c^2 - 2*(3*A*a^2*c^2*cos(f*x + e)^4 + 4*A*a^2*c^2*cos(f*x + e)^2 +
8*A*a^2*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. \[ \int (a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 355 vs. \(2 (156) = 312\).

Time = 0.44 (sec) , antiderivative size = 355, normalized size of antiderivative = 1.97 \[ \int (a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2} \, dx=-\frac {16 \, {\left (10 \, B a^{2} 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 )^{12} - 6 \, A a^{2} 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 )^{10} - 30 \, B a^{2} 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 )^{10} + 15 \, A a^{2} 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} + 30 \, B a^{2} 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} - 10 \, A a^{2} 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} - 10 \, B a^{2} 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}}{15 \, f} \]

[In]

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

[Out]

-16/15*(10*B*a^2*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)^12 - 6*A*a^2*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)^10 - 30*B*a^2*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)^10 + 15*A*a^2*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 + 30*B*a^2*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 - 10*A*a^2*c^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*s
gn(sin(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e)^6 - 10*B*a^2*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 = 16.76 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.73 \[ \int (a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2} \, dx=-\frac {a^2\,c^2\,\sqrt {a\,\left (\sin \left (e+f\,x\right )+1\right )}\,\sqrt {-c\,\left (\sin \left (e+f\,x\right )-1\right )}\,\left (75\,B\,\cos \left (e+f\,x\right )+105\,B\,\cos \left (3\,e+3\,f\,x\right )+35\,B\,\cos \left (5\,e+5\,f\,x\right )+5\,B\,\cos \left (7\,e+7\,f\,x\right )-700\,A\,\sin \left (2\,e+2\,f\,x\right )-112\,A\,\sin \left (4\,e+4\,f\,x\right )-12\,A\,\sin \left (6\,e+6\,f\,x\right )\right )}{960\,f\,\left (\cos \left (2\,e+2\,f\,x\right )+1\right )} \]

[In]

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

[Out]

-(a^2*c^2*(a*(sin(e + f*x) + 1))^(1/2)*(-c*(sin(e + f*x) - 1))^(1/2)*(75*B*cos(e + f*x) + 105*B*cos(3*e + 3*f*
x) + 35*B*cos(5*e + 5*f*x) + 5*B*cos(7*e + 7*f*x) - 700*A*sin(2*e + 2*f*x) - 112*A*sin(4*e + 4*f*x) - 12*A*sin
(6*e + 6*f*x)))/(960*f*(cos(2*e + 2*f*x) + 1))