\(\int (a+a \sin (e+f x)) (A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2} \, dx\) [82]

Optimal result

Integrand size = 36, antiderivative size = 157 \[ \int (a+a \sin (e+f x)) (A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2} \, dx=\frac {64 a (3 A-B) c^4 \cos ^3(e+f x)}{315 f (c-c \sin (e+f x))^{3/2}}+\frac {16 a (3 A-B) c^3 \cos ^3(e+f x)}{105 f \sqrt {c-c \sin (e+f x)}}+\frac {2 a (3 A-B) c^2 \cos ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{21 f}-\frac {2 a B c \cos ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{9 f} \]

[Out]

64/315*a*(3*A-B)*c^4*cos(f*x+e)^3/f/(c-c*sin(f*x+e))^(3/2)-2/9*a*B*c*cos(f*x+e)^3*(c-c*sin(f*x+e))^(3/2)/f+16/
105*a*(3*A-B)*c^3*cos(f*x+e)^3/f/(c-c*sin(f*x+e))^(1/2)+2/21*a*(3*A-B)*c^2*cos(f*x+e)^3*(c-c*sin(f*x+e))^(1/2)
/f

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {3046, 2935, 2753, 2752} \[ \int (a+a \sin (e+f x)) (A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2} \, dx=\frac {64 a c^4 (3 A-B) \cos ^3(e+f x)}{315 f (c-c \sin (e+f x))^{3/2}}+\frac {16 a c^3 (3 A-B) \cos ^3(e+f x)}{105 f \sqrt {c-c \sin (e+f x)}}+\frac {2 a c^2 (3 A-B) \cos ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{21 f}-\frac {2 a B c \cos ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{9 f} \]

[In]

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

[Out]

(64*a*(3*A - B)*c^4*Cos[e + f*x]^3)/(315*f*(c - c*Sin[e + f*x])^(3/2)) + (16*a*(3*A - B)*c^3*Cos[e + f*x]^3)/(
105*f*Sqrt[c - c*Sin[e + f*x]]) + (2*a*(3*A - B)*c^2*Cos[e + f*x]^3*Sqrt[c - c*Sin[e + f*x]])/(21*f) - (2*a*B*
c*Cos[e + f*x]^3*(c - c*Sin[e + f*x])^(3/2))/(9*f)

Rule 2752

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[b*(g*C
os[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^(m - 1)/(f*g*(m - 1))), x] /; FreeQ[{a, b, e, f, g, m, p}, x] && Eq
Q[a^2 - b^2, 0] && EqQ[2*m + p - 1, 0] && NeQ[m, 1]

Rule 2753

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(-b)*(
g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^(m - 1)/(f*g*(m + p))), x] + Dist[a*((2*m + p - 1)/(m + p)), Int
[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2,
0] && IGtQ[Simplify[(2*m + p - 1)/2], 0] && NeQ[m + p, 0]

Rule 2935

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Simp[(-d)*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^m/(f*g*(m + p + 1))), x
] + Dist[(a*d*m + b*c*(m + p + 1))/(b*(m + p + 1)), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^m, x], x] /; F
reeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && IGtQ[Simplify[(2*m + p + 1)/2], 0] && NeQ[m + p +
1, 0]

Rule 3046

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a^m*c^m, Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m)*(A + B
*Sin[e + f*x]), 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] && I
ntegerQ[m] &&  !(IntegerQ[n] && ((LtQ[m, 0] && GtQ[n, 0]) || LtQ[0, n, m] || LtQ[m, n, 0]))

Rubi steps \begin{align*} \text {integral}& = (a c) \int \cos ^2(e+f x) (A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2} \, dx \\ & = -\frac {2 a B c \cos ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{9 f}+\frac {1}{3} (a (3 A-B) c) \int \cos ^2(e+f x) (c-c \sin (e+f x))^{3/2} \, dx \\ & = \frac {2 a (3 A-B) c^2 \cos ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{21 f}-\frac {2 a B c \cos ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{9 f}+\frac {1}{21} \left (8 a (3 A-B) c^2\right ) \int \cos ^2(e+f x) \sqrt {c-c \sin (e+f x)} \, dx \\ & = \frac {16 a (3 A-B) c^3 \cos ^3(e+f x)}{105 f \sqrt {c-c \sin (e+f x)}}+\frac {2 a (3 A-B) c^2 \cos ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{21 f}-\frac {2 a B c \cos ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{9 f}+\frac {1}{105} \left (32 a (3 A-B) c^3\right ) \int \frac {\cos ^2(e+f x)}{\sqrt {c-c \sin (e+f x)}} \, dx \\ & = \frac {64 a (3 A-B) c^4 \cos ^3(e+f x)}{315 f (c-c \sin (e+f x))^{3/2}}+\frac {16 a (3 A-B) c^3 \cos ^3(e+f x)}{105 f \sqrt {c-c \sin (e+f x)}}+\frac {2 a (3 A-B) c^2 \cos ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{21 f}-\frac {2 a B c \cos ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{9 f} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.97 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.78 \[ \int (a+a \sin (e+f x)) (A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2} \, dx=-\frac {a c^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3 \sqrt {c-c \sin (e+f x)} (-942 A+664 B+30 (3 A-8 B) \cos (2 (e+f x))+(648 A-741 B) \sin (e+f x)+35 B \sin (3 (e+f x)))}{630 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )} \]

[In]

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

[Out]

-1/630*(a*c^2*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3*Sqrt[c - c*Sin[e + f*x]]*(-942*A + 664*B + 30*(3*A - 8*B
)*Cos[2*(e + f*x)] + (648*A - 741*B)*Sin[e + f*x] + 35*B*Sin[3*(e + f*x)]))/(f*(Cos[(e + f*x)/2] - Sin[(e + f*
x)/2]))

Maple [A] (verified)

Time = 7.36 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.66

method	result	size
default	\(-\frac {2 \left (\sin \left (f x +e \right )-1\right ) c^{3} \left (1+\sin \left (f x +e \right )\right )^{2} a \left (-35 B \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )+\left (-45 A +120 B \right ) \left (\cos ^{2}\left (f x +e \right )\right )+\left (-162 A +194 B \right ) \sin \left (f x +e \right )+258 A -226 B \right )}{315 \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\)	\(103\)
parts	\(-\frac {2 a A \left (\sin \left (f x +e \right )-1\right ) c^{3} \left (1+\sin \left (f x +e \right )\right ) \left (3 \left (\sin ^{2}\left (f x +e \right )\right )-14 \sin \left (f x +e \right )+43\right )}{15 \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}-\frac {2 B a \left (\sin \left (f x +e \right )-1\right ) c^{3} \left (1+\sin \left (f x +e \right )\right ) \left (35 \left (\sin ^{4}\left (f x +e \right )\right )-130 \left (\sin ^{3}\left (f x +e \right )\right )+219 \left (\sin ^{2}\left (f x +e \right )\right )-292 \sin \left (f x +e \right )+584\right )}{315 \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}-\frac {2 a \left (A +B \right ) \left (\sin \left (f x +e \right )-1\right ) c^{3} \left (1+\sin \left (f x +e \right )\right ) \left (3 \left (\sin ^{3}\left (f x +e \right )\right )-12 \left (\sin ^{2}\left (f x +e \right )\right )+23 \sin \left (f x +e \right )-46\right )}{21 \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\)	\(235\)

[In]

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

[Out]

-2/315*(sin(f*x+e)-1)*c^3*(1+sin(f*x+e))^2*a*(-35*B*cos(f*x+e)^2*sin(f*x+e)+(-45*A+120*B)*cos(f*x+e)^2+(-162*A
+194*B)*sin(f*x+e)+258*A-226*B)/cos(f*x+e)/(c-c*sin(f*x+e))^(1/2)/f

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.55 \[ \int (a+a \sin (e+f x)) (A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2} \, dx=\frac {2 \, {\left (35 \, B a c^{2} \cos \left (f x + e\right )^{5} + 5 \, {\left (9 \, A - 10 \, B\right )} a c^{2} \cos \left (f x + e\right )^{4} + {\left (117 \, A - 109 \, B\right )} a c^{2} \cos \left (f x + e\right )^{3} - 8 \, {\left (3 \, A - B\right )} a c^{2} \cos \left (f x + e\right )^{2} + 32 \, {\left (3 \, A - B\right )} a c^{2} \cos \left (f x + e\right ) + 64 \, {\left (3 \, A - B\right )} a c^{2} + {\left (35 \, B a c^{2} \cos \left (f x + e\right )^{4} - 5 \, {\left (9 \, A - 17 \, B\right )} a c^{2} \cos \left (f x + e\right )^{3} + 24 \, {\left (3 \, A - B\right )} a c^{2} \cos \left (f x + e\right )^{2} + 32 \, {\left (3 \, A - B\right )} a c^{2} \cos \left (f x + e\right ) + 64 \, {\left (3 \, A - B\right )} a c^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{315 \, {\left (f \cos \left (f x + e\right ) - f \sin \left (f x + e\right ) + f\right )}} \]

[In]

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

[Out]

2/315*(35*B*a*c^2*cos(f*x + e)^5 + 5*(9*A - 10*B)*a*c^2*cos(f*x + e)^4 + (117*A - 109*B)*a*c^2*cos(f*x + e)^3
- 8*(3*A - B)*a*c^2*cos(f*x + e)^2 + 32*(3*A - B)*a*c^2*cos(f*x + e) + 64*(3*A - B)*a*c^2 + (35*B*a*c^2*cos(f*
x + e)^4 - 5*(9*A - 17*B)*a*c^2*cos(f*x + e)^3 + 24*(3*A - B)*a*c^2*cos(f*x + e)^2 + 32*(3*A - B)*a*c^2*cos(f*
x + e) + 64*(3*A - B)*a*c^2)*sin(f*x + e))*sqrt(-c*sin(f*x + e) + c)/(f*cos(f*x + e) - f*sin(f*x + e) + f)

Sympy [F]

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

[In]

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

[Out]

a*(Integral(A*c**2*sqrt(-c*sin(e + f*x) + c), x) + Integral(-A*c**2*sqrt(-c*sin(e + f*x) + c)*sin(e + f*x), x)
+ Integral(-A*c**2*sqrt(-c*sin(e + f*x) + c)*sin(e + f*x)**2, x) + Integral(A*c**2*sqrt(-c*sin(e + f*x) + c)*
sin(e + f*x)**3, x) + Integral(B*c**2*sqrt(-c*sin(e + f*x) + c)*sin(e + f*x), x) + Integral(-B*c**2*sqrt(-c*si
n(e + f*x) + c)*sin(e + f*x)**2, x) + Integral(-B*c**2*sqrt(-c*sin(e + f*x) + c)*sin(e + f*x)**3, x) + Integra
l(B*c**2*sqrt(-c*sin(e + f*x) + c)*sin(e + f*x)**4, x))

Maxima [F]

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

[In]

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

Giac [A] (verification not implemented)

none

Time = 0.50 (sec) , antiderivative size = 262, normalized size of antiderivative = 1.67 \[ \int (a+a \sin (e+f x)) (A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2} \, dx=-\frac {\sqrt {2} {\left (35 \, B a c^{2} \cos \left (-\frac {9}{4} \, \pi + \frac {9}{2} \, f x + \frac {9}{2} \, e\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 630 \, {\left (5 \, A a c^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 2 \, B a c^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 210 \, {\left (A a c^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + B a c^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \cos \left (-\frac {3}{4} \, \pi + \frac {3}{2} \, f x + \frac {3}{2} \, e\right ) - 126 \, {\left (3 \, A a c^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - B a c^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \cos \left (-\frac {5}{4} \, \pi + \frac {5}{2} \, f x + \frac {5}{2} \, e\right ) + 45 \, {\left (2 \, A a c^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 3 \, B a c^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \cos \left (-\frac {7}{4} \, \pi + \frac {7}{2} \, f x + \frac {7}{2} \, e\right )\right )} \sqrt {c}}{2520 \, f} \]

[In]

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

[Out]

-1/2520*sqrt(2)*(35*B*a*c^2*cos(-9/4*pi + 9/2*f*x + 9/2*e)*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) + 630*(5*A*a*c^
2*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) - 2*B*a*c^2*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)))*cos(-1/4*pi + 1/2*f*x +
1/2*e) + 210*(A*a*c^2*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) + B*a*c^2*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)))*cos(
-3/4*pi + 3/2*f*x + 3/2*e) - 126*(3*A*a*c^2*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) - B*a*c^2*sgn(sin(-1/4*pi + 1/
2*f*x + 1/2*e)))*cos(-5/4*pi + 5/2*f*x + 5/2*e) + 45*(2*A*a*c^2*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) - 3*B*a*c^
2*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)))*cos(-7/4*pi + 7/2*f*x + 7/2*e))*sqrt(c)/f

Mupad [F(-1)]

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

[In]

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

[Out]

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