\(\int (a+a \sin (e+f x))^3 (c-c \sin (e+f x))^{3/2} \, dx\) [317]

Optimal result

Integrand size = 28, antiderivative size = 73 \[ \int (a+a \sin (e+f x))^3 (c-c \sin (e+f x))^{3/2} \, dx=\frac {8 a^3 c^5 \cos ^7(e+f x)}{63 f (c-c \sin (e+f x))^{7/2}}+\frac {2 a^3 c^4 \cos ^7(e+f x)}{9 f (c-c \sin (e+f x))^{5/2}} \] Output:

8/63*a^3*c^5*cos(f*x+e)^7/f/(c-c*sin(f*x+e))^(7/2)+2/9*a^3*c^4*cos(f*x+e)^ 
7/f/(c-c*sin(f*x+e))^(5/2)
 

Mathematica [A] (verified)

Time = 9.71 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.15 \[ \int (a+a \sin (e+f x))^3 (c-c \sin (e+f x))^{3/2} \, dx=-\frac {2 a^3 c \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^7 (-11+7 \sin (e+f x)) \sqrt {c-c \sin (e+f x)}}{63 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )} \] Input:

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

Output:

(-2*a^3*c*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^7*(-11 + 7*Sin[e + f*x])*S 
qrt[c - c*Sin[e + f*x]])/(63*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2]))
 

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.99, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3042, 3215, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3215

\(\displaystyle a^3 c^3 \int \frac {\cos ^6(e+f x)}{(c-c \sin (e+f x))^{3/2}}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle a^3 c^3 \int \frac {\cos (e+f x)^6}{(c-c \sin (e+f x))^{3/2}}dx\)

\(\Big \downarrow \) 3153

\(\displaystyle a^3 c^3 \left (\frac {4}{9} c \int \frac {\cos ^6(e+f x)}{(c-c \sin (e+f x))^{5/2}}dx+\frac {2 c \cos ^7(e+f x)}{9 f (c-c \sin (e+f x))^{5/2}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle a^3 c^3 \left (\frac {4}{9} c \int \frac {\cos (e+f x)^6}{(c-c \sin (e+f x))^{5/2}}dx+\frac {2 c \cos ^7(e+f x)}{9 f (c-c \sin (e+f x))^{5/2}}\right )\)

\(\Big \downarrow \) 3152

\(\displaystyle a^3 c^3 \left (\frac {8 c^2 \cos ^7(e+f x)}{63 f (c-c \sin (e+f x))^{7/2}}+\frac {2 c \cos ^7(e+f x)}{9 f (c-c \sin (e+f x))^{5/2}}\right )\)

Input:

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

Output:

a^3*c^3*((8*c^2*Cos[e + f*x]^7)/(63*f*(c - c*Sin[e + f*x])^(7/2)) + (2*c*C 
os[e + f*x]^7)/(9*f*(c - c*Sin[e + f*x])^(5/2)))
 

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 3215

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

Maple [A] (verified)

Time = 2.20 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.84

method	result	size
default	\(\frac {2 \left (-1+\sin \left (f x +e \right )\right ) c^{2} \left (1+\sin \left (f x +e \right )\right )^{4} a^{3} \left (7 \sin \left (f x +e \right )-11\right )}{63 \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\)	\(61\)
parts	\(\frac {2 a^{3} \left (-1+\sin \left (f x +e \right )\right ) c^{2} \left (1+\sin \left (f x +e \right )\right ) \left (\sin \left (f x +e \right )-5\right )}{3 \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}+\frac {2 a^{3} \left (-1+\sin \left (f x +e \right )\right ) c^{2} \left (1+\sin \left (f x +e \right )\right ) \left (35 \sin \left (f x +e \right )^{4}-85 \sin \left (f x +e \right )^{3}+102 \sin \left (f x +e \right )^{2}-136 \sin \left (f x +e \right )+272\right )}{315 \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}+\frac {6 a^{3} \left (-1+\sin \left (f x +e \right )\right ) c^{2} \left (1+\sin \left (f x +e \right )\right ) \left (\sin \left (f x +e \right )^{2}-3 \sin \left (f x +e \right )+6\right )}{5 \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}+\frac {2 a^{3} \left (-1+\sin \left (f x +e \right )\right ) c^{2} \left (1+\sin \left (f x +e \right )\right ) \left (15 \sin \left (f x +e \right )^{3}-39 \sin \left (f x +e \right )^{2}+52 \sin \left (f x +e \right )-104\right )}{35 \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\)	\(290\)

Input:

int((a+sin(f*x+e)*a)^3*(c-c*sin(f*x+e))^(3/2),x,method=_RETURNVERBOSE)
 

Output:

2/63*(-1+sin(f*x+e))*c^2*(1+sin(f*x+e))^4*a^3*(7*sin(f*x+e)-11)/cos(f*x+e) 
/(c-c*sin(f*x+e))^(1/2)/f
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 179 vs. \(2 (65) = 130\).

Time = 0.09 (sec) , antiderivative size = 179, normalized size of antiderivative = 2.45 \[ \int (a+a \sin (e+f x))^3 (c-c \sin (e+f x))^{3/2} \, dx=-\frac {2 \, {\left (7 \, a^{3} c \cos \left (f x + e\right )^{5} + 17 \, a^{3} c \cos \left (f x + e\right )^{4} - 2 \, a^{3} c \cos \left (f x + e\right )^{3} + 4 \, a^{3} c \cos \left (f x + e\right )^{2} - 16 \, a^{3} c \cos \left (f x + e\right ) - 32 \, a^{3} c + {\left (7 \, a^{3} c \cos \left (f x + e\right )^{4} - 10 \, a^{3} c \cos \left (f x + e\right )^{3} - 12 \, a^{3} c \cos \left (f x + e\right )^{2} - 16 \, a^{3} c \cos \left (f x + e\right ) - 32 \, a^{3} c\right )} \sin \left (f x + e\right )\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{63 \, {\left (f \cos \left (f x + e\right ) - f \sin \left (f x + e\right ) + f\right )}} \] Input:

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

Output:

-2/63*(7*a^3*c*cos(f*x + e)^5 + 17*a^3*c*cos(f*x + e)^4 - 2*a^3*c*cos(f*x 
+ e)^3 + 4*a^3*c*cos(f*x + e)^2 - 16*a^3*c*cos(f*x + e) - 32*a^3*c + (7*a^ 
3*c*cos(f*x + e)^4 - 10*a^3*c*cos(f*x + e)^3 - 12*a^3*c*cos(f*x + e)^2 - 1 
6*a^3*c*cos(f*x + e) - 32*a^3*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]

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

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

Output:

a**3*(Integral(c*sqrt(-c*sin(e + f*x) + c), x) + Integral(2*c*sqrt(-c*sin( 
e + f*x) + c)*sin(e + f*x), x) + Integral(-2*c*sqrt(-c*sin(e + f*x) + c)*s 
in(e + f*x)**3, x) + Integral(-c*sqrt(-c*sin(e + f*x) + c)*sin(e + f*x)**4 
, x))
 

Maxima [F]

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

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 136 vs. \(2 (65) = 130\).

Time = 0.19 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.86 \[ \int (a+a \sin (e+f x))^3 (c-c \sin (e+f x))^{3/2} \, dx=-\frac {\sqrt {2} {\left (378 \, a^{3} c \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 168 \, a^{3} c \cos \left (-\frac {3}{4} \, \pi + \frac {3}{2} \, f x + \frac {3}{2} \, e\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 27 \, a^{3} c \cos \left (-\frac {7}{4} \, \pi + \frac {7}{2} \, f x + \frac {7}{2} \, e\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 7 \, a^{3} c \cos \left (-\frac {9}{4} \, \pi + \frac {9}{2} \, f x + \frac {9}{2} \, e\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sqrt {c}}{504 \, f} \] Input:

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

Output:

-1/504*sqrt(2)*(378*a^3*c*cos(-1/4*pi + 1/2*f*x + 1/2*e)*sgn(sin(-1/4*pi + 
 1/2*f*x + 1/2*e)) + 168*a^3*c*cos(-3/4*pi + 3/2*f*x + 3/2*e)*sgn(sin(-1/4 
*pi + 1/2*f*x + 1/2*e)) - 27*a^3*c*cos(-7/4*pi + 7/2*f*x + 7/2*e)*sgn(sin( 
-1/4*pi + 1/2*f*x + 1/2*e)) - 7*a^3*c*cos(-9/4*pi + 9/2*f*x + 9/2*e)*sgn(s 
in(-1/4*pi + 1/2*f*x + 1/2*e)))*sqrt(c)/f
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int (a+a \sin (e+f x))^3 (c-c \sin (e+f x))^{3/2} \, dx=\sqrt {c}\, a^{3} c \left (\int \sqrt {-\sin \left (f x +e \right )+1}d x -\left (\int \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{4}d x \right )-2 \left (\int \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{3}d x \right )+2 \left (\int \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )d x \right )\right ) \] Input:

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

Output:

sqrt(c)*a**3*c*(int(sqrt( - sin(e + f*x) + 1),x) - int(sqrt( - sin(e + f*x 
) + 1)*sin(e + f*x)**4,x) - 2*int(sqrt( - sin(e + f*x) + 1)*sin(e + f*x)** 
3,x) + 2*int(sqrt( - sin(e + f*x) + 1)*sin(e + f*x),x))