\(\int (a+a \sin (e+f x)) (A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2} \, dx\) [83]

Optimal result

Integrand size = 36, antiderivative size = 116 \[ \int (a+a \sin (e+f x)) (A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2} \, dx=\frac {8 a (7 A-B) c^3 \cos ^3(e+f x)}{105 f (c-c \sin (e+f x))^{3/2}}+\frac {2 a (7 A-B) c^2 \cos ^3(e+f x)}{35 f \sqrt {c-c \sin (e+f x)}}-\frac {2 a B c \cos ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{7 f} \] Output:

8/105*a*(7*A-B)*c^3*cos(f*x+e)^3/f/(c-c*sin(f*x+e))^(3/2)+2/35*a*(7*A-B)*c 
^2*cos(f*x+e)^3/f/(c-c*sin(f*x+e))^(1/2)-2/7*a*B*c*cos(f*x+e)^3*(c-c*sin(f 
*x+e))^(1/2)/f
 

Mathematica [A] (verified)

Time = 3.77 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.90 \[ \int (a+a \sin (e+f x)) (A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2} \, dx=\frac {a c \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3 (98 A-59 B+15 B \cos (2 (e+f x))+(-42 A+66 B) \sin (e+f x)) \sqrt {c-c \sin (e+f x)}}{105 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )} \] Input:

Integrate[(a + a*Sin[e + f*x])*(A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^( 
3/2),x]
 

Output:

(a*c*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3*(98*A - 59*B + 15*B*Cos[2*(e 
+ f*x)] + (-42*A + 66*B)*Sin[e + f*x])*Sqrt[c - c*Sin[e + f*x]])/(105*f*(C 
os[(e + f*x)/2] - Sin[(e + f*x)/2]))
 

Rubi [A] (verified)

Time = 0.69 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.96, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {3042, 3446, 3042, 3335, 3042, 3153, 3042, 3152}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int (a \sin (e+f x)+a) (c-c \sin (e+f x))^{3/2} (A+B \sin (e+f x)) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (a \sin (e+f x)+a) (c-c \sin (e+f x))^{3/2} (A+B \sin (e+f x))dx\)

\(\Big \downarrow \) 3446

\(\displaystyle a c \int \cos ^2(e+f x) (A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle a c \int \cos (e+f x)^2 (A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)}dx\)

\(\Big \downarrow \) 3335

\(\displaystyle a c \left (\frac {1}{7} (7 A-B) \int \cos ^2(e+f x) \sqrt {c-c \sin (e+f x)}dx-\frac {2 B \cos ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{7 f}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle a c \left (\frac {1}{7} (7 A-B) \int \cos (e+f x)^2 \sqrt {c-c \sin (e+f x)}dx-\frac {2 B \cos ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{7 f}\right )\)

\(\Big \downarrow \) 3153

\(\displaystyle a c \left (\frac {1}{7} (7 A-B) \left (\frac {4}{5} c \int \frac {\cos ^2(e+f x)}{\sqrt {c-c \sin (e+f x)}}dx+\frac {2 c \cos ^3(e+f x)}{5 f \sqrt {c-c \sin (e+f x)}}\right )-\frac {2 B \cos ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{7 f}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle a c \left (\frac {1}{7} (7 A-B) \left (\frac {4}{5} c \int \frac {\cos (e+f x)^2}{\sqrt {c-c \sin (e+f x)}}dx+\frac {2 c \cos ^3(e+f x)}{5 f \sqrt {c-c \sin (e+f x)}}\right )-\frac {2 B \cos ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{7 f}\right )\)

\(\Big \downarrow \) 3152

\(\displaystyle a c \left (\frac {1}{7} (7 A-B) \left (\frac {8 c^2 \cos ^3(e+f x)}{15 f (c-c \sin (e+f x))^{3/2}}+\frac {2 c \cos ^3(e+f x)}{5 f \sqrt {c-c \sin (e+f x)}}\right )-\frac {2 B \cos ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{7 f}\right )\)

Input:

Int[(a + a*Sin[e + f*x])*(A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^(3/2),x 
]
 

Output:

a*c*((-2*B*Cos[e + f*x]^3*Sqrt[c - c*Sin[e + f*x]])/(7*f) + ((7*A - B)*((8 
*c^2*Cos[e + f*x]^3)/(15*f*(c - c*Sin[e + f*x])^(3/2)) + (2*c*Cos[e + f*x] 
^3)/(5*f*Sqrt[c - c*Sin[e + f*x]])))/7)
 

Defintions of rubi rules used

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3152

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 - 1))), x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 
 - b^2, 0] && EqQ[2*m + p - 1, 0] && NeQ[m, 1]
 

rule 3153

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] + Simp[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 3335

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] + S 
imp[(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] /; FreeQ[{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 3446

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + 
 (f_.)*(x_)])*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Si 
mp[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] && IntegerQ[m] &&  !(IntegerQ[n] && ((LtQ[m, 0] 
&& GtQ[n, 0]) || LtQ[0, n, m] || LtQ[m, n, 0]))
 

Maple [A] (verified)

Time = 0.63 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.70

method	result	size
default	\(\frac {2 \left (\sin \left (f x +e \right )-1\right ) c^{2} \left (1+\sin \left (f x +e \right )\right )^{2} a \left (-15 B \cos \left (f x +e \right )^{2}+\sin \left (f x +e \right ) \left (21 A -33 B \right )-49 A +37 B \right )}{105 \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\)	\(81\)
parts	\(\frac {2 A a \left (\sin \left (f x +e \right )-1\right ) c^{2} \left (1+\sin \left (f x +e \right )\right ) \left (\sin \left (f x +e \right )-5\right )}{3 \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}+\frac {2 B a \left (\sin \left (f x +e \right )-1\right ) c^{2} \left (1+\sin \left (f x +e \right )\right ) \left (15 \sin \left (f x +e \right )^{3}-39 \sin \left (f x +e \right )^{2}+52 \sin \left (f x +e \right )-104\right )}{105 \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}+\frac {2 a \left (A +B \right ) \left (\sin \left (f x +e \right )-1\right ) c^{2} \left (1+\sin \left (f x +e \right )\right ) \left (\sin \left (f x +e \right )^{2}-3 \sin \left (f x +e \right )+6\right )}{5 \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\)	\(201\)

Input:

int((a+a*sin(f*x+e))*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(3/2),x,method=_RET 
URNVERBOSE)
 

Output:

2/105*(sin(f*x+e)-1)*c^2*(1+sin(f*x+e))^2*a*(-15*B*cos(f*x+e)^2+sin(f*x+e) 
*(21*A-33*B)-49*A+37*B)/cos(f*x+e)/(c-c*sin(f*x+e))^(1/2)/f
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.59 \[ \int (a+a \sin (e+f x)) (A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2} \, dx=-\frac {2 \, {\left (15 \, B a c \cos \left (f x + e\right )^{4} - 3 \, {\left (7 \, A - 6 \, B\right )} a c \cos \left (f x + e\right )^{3} + {\left (7 \, A - B\right )} a c \cos \left (f x + e\right )^{2} - 4 \, {\left (7 \, A - B\right )} a c \cos \left (f x + e\right ) - 8 \, {\left (7 \, A - B\right )} a c - {\left (15 \, B a c \cos \left (f x + e\right )^{3} + 3 \, {\left (7 \, A - B\right )} a c \cos \left (f x + e\right )^{2} + 4 \, {\left (7 \, A - B\right )} a c \cos \left (f x + e\right ) + 8 \, {\left (7 \, A - B\right )} a c\right )} \sin \left (f x + e\right )\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{105 \, {\left (f \cos \left (f x + e\right ) - f \sin \left (f x + e\right ) + f\right )}} \] Input:

integrate((a+a*sin(f*x+e))*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(3/2),x, algo 
rithm="fricas")
 

Output:

-2/105*(15*B*a*c*cos(f*x + e)^4 - 3*(7*A - 6*B)*a*c*cos(f*x + e)^3 + (7*A 
- B)*a*c*cos(f*x + e)^2 - 4*(7*A - B)*a*c*cos(f*x + e) - 8*(7*A - B)*a*c - 
 (15*B*a*c*cos(f*x + e)^3 + 3*(7*A - B)*a*c*cos(f*x + e)^2 + 4*(7*A - B)*a 
*c*cos(f*x + e) + 8*(7*A - B)*a*c)*sin(f*x + e))*sqrt(-c*sin(f*x + e) + c) 
/(f*cos(f*x + e) - f*sin(f*x + e) + f)
 

Sympy [F]

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

integrate((a+a*sin(f*x+e))*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))**(3/2),x)
 

Output:

a*(Integral(A*c*sqrt(-c*sin(e + f*x) + c), x) + Integral(-A*c*sqrt(-c*sin( 
e + f*x) + c)*sin(e + f*x)**2, x) + Integral(B*c*sqrt(-c*sin(e + f*x) + c) 
*sin(e + f*x), x) + Integral(-B*c*sqrt(-c*sin(e + f*x) + c)*sin(e + f*x)** 
3, x))
 

Maxima [F]

\[ \int (a+a \sin (e+f x)) (A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2} \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )} {\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}} \,d x } \] Input:

integrate((a+a*sin(f*x+e))*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(3/2),x, algo 
rithm="maxima")
 

Output:

integrate((B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)*(-c*sin(f*x + e) + c)^ 
(3/2), x)
 

Giac [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.67 \[ \int (a+a \sin (e+f x)) (A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2} \, dx=\frac {\sqrt {2} {\left (15 \, B a c \cos \left (-\frac {7}{4} \, \pi + \frac {7}{2} \, f x + \frac {7}{2} \, e\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 105 \, {\left (4 \, A a c \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - B a c \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 35 \, {\left (2 \, A a c \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + B a c \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \cos \left (-\frac {3}{4} \, \pi + \frac {3}{2} \, f x + \frac {3}{2} \, e\right ) + 21 \, {\left (2 \, A a c \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - B a c \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \cos \left (-\frac {5}{4} \, \pi + \frac {5}{2} \, f x + \frac {5}{2} \, e\right )\right )} \sqrt {c}}{420 \, f} \] Input:

integrate((a+a*sin(f*x+e))*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(3/2),x, algo 
rithm="giac")
 

Output:

1/420*sqrt(2)*(15*B*a*c*cos(-7/4*pi + 7/2*f*x + 7/2*e)*sgn(sin(-1/4*pi + 1 
/2*f*x + 1/2*e)) - 105*(4*A*a*c*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) - B*a* 
c*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)))*cos(-1/4*pi + 1/2*f*x + 1/2*e) - 35 
*(2*A*a*c*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) + B*a*c*sgn(sin(-1/4*pi + 1/ 
2*f*x + 1/2*e)))*cos(-3/4*pi + 3/2*f*x + 3/2*e) + 21*(2*A*a*c*sgn(sin(-1/4 
*pi + 1/2*f*x + 1/2*e)) - B*a*c*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)))*cos(- 
5/4*pi + 5/2*f*x + 5/2*e))*sqrt(c)/f
 

Mupad [F(-1)]

Timed out. \[ \int (a+a \sin (e+f x)) (A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2} \, dx=\int \left (A+B\,\sin \left (e+f\,x\right )\right )\,\left (a+a\,\sin \left (e+f\,x\right )\right )\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{3/2} \,d x \] Input:

int((A + B*sin(e + f*x))*(a + a*sin(e + f*x))*(c - c*sin(e + f*x))^(3/2),x 
)
 

Output:

int((A + B*sin(e + f*x))*(a + a*sin(e + f*x))*(c - c*sin(e + f*x))^(3/2), 
x)
 

Reduce [F]

\[ \int (a+a \sin (e+f x)) (A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2} \, dx=\sqrt {c}\, a c \left (\left (\int \sqrt {-\sin \left (f x +e \right )+1}d x \right ) a -\left (\int \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{3}d x \right ) b -\left (\int \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{2}d x \right ) a +\left (\int \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )d x \right ) b \right ) \] Input:

int((a+a*sin(f*x+e))*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(3/2),x)
 

Output:

sqrt(c)*a*c*(int(sqrt( - sin(e + f*x) + 1),x)*a - int(sqrt( - sin(e + f*x) 
 + 1)*sin(e + f*x)**3,x)*b - int(sqrt( - sin(e + f*x) + 1)*sin(e + f*x)**2 
,x)*a + int(sqrt( - sin(e + f*x) + 1)*sin(e + f*x),x)*b)