\(\int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2}}{(a+a \sin (e+f x))^3} \, dx\) [126]

Optimal result

Integrand size = 38, antiderivative size = 121 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2}}{(a+a \sin (e+f x))^3} \, dx=\frac {4 (A+9 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{15 a^3 f}-\frac {(A+9 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{5 a^3 c f}-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{9/2}}{5 a^3 c^3 f} \] Output:

4/15*(A+9*B)*sec(f*x+e)^3*(c-c*sin(f*x+e))^(3/2)/a^3/f-1/5*(A+9*B)*sec(f*x 
+e)^3*(c-c*sin(f*x+e))^(5/2)/a^3/c/f-1/5*(A-B)*sec(f*x+e)^5*(c-c*sin(f*x+e 
))^(9/2)/a^3/c^3/f
 

Mathematica [A] (verified)

Time = 8.97 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.93 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2}}{(a+a \sin (e+f x))^3} \, dx=\frac {c \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) (-2 A+27 B-15 B \cos (2 (e+f x))+10 (A+3 B) \sin (e+f x)) \sqrt {c-c \sin (e+f x)}}{15 a^3 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) (1+\sin (e+f x))^3} \] Input:

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

Output:

(c*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(-2*A + 27*B - 15*B*Cos[2*(e + f* 
x)] + 10*(A + 3*B)*Sin[e + f*x])*Sqrt[c - c*Sin[e + f*x]])/(15*a^3*f*(Cos[ 
(e + f*x)/2] - Sin[(e + f*x)/2])*(1 + Sin[e + f*x])^3)
 

Rubi [A] (verified)

Time = 0.78 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.96, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {3042, 3446, 3042, 3334, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3446

\(\displaystyle \frac {\int \sec ^6(e+f x) (A+B \sin (e+f x)) (c-c \sin (e+f x))^{9/2}dx}{a^3 c^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{9/2}}{\cos (e+f x)^6}dx}{a^3 c^3}\)

\(\Big \downarrow \) 3334

\(\displaystyle \frac {\frac {1}{10} c (A+9 B) \int \sec ^4(e+f x) (c-c \sin (e+f x))^{7/2}dx-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{9/2}}{5 f}}{a^3 c^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{10} c (A+9 B) \int \frac {(c-c \sin (e+f x))^{7/2}}{\cos (e+f x)^4}dx-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{9/2}}{5 f}}{a^3 c^3}\)

\(\Big \downarrow \) 3153

\(\displaystyle \frac {\frac {1}{10} c (A+9 B) \left (-4 c \int \sec ^4(e+f x) (c-c \sin (e+f x))^{5/2}dx-\frac {2 c \sec ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{f}\right )-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{9/2}}{5 f}}{a^3 c^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{10} c (A+9 B) \left (-4 c \int \frac {(c-c \sin (e+f x))^{5/2}}{\cos (e+f x)^4}dx-\frac {2 c \sec ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{f}\right )-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{9/2}}{5 f}}{a^3 c^3}\)

\(\Big \downarrow \) 3152

\(\displaystyle \frac {\frac {1}{10} c (A+9 B) \left (\frac {8 c^2 \sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{3 f}-\frac {2 c \sec ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{f}\right )-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{9/2}}{5 f}}{a^3 c^3}\)

Input:

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

Output:

(-1/5*((A - B)*Sec[e + f*x]^5*(c - c*Sin[e + f*x])^(9/2))/f + ((A + 9*B)*c 
*((8*c^2*Sec[e + f*x]^3*(c - c*Sin[e + f*x])^(3/2))/(3*f) - (2*c*Sec[e + f 
*x]^3*(c - c*Sin[e + f*x])^(5/2))/f))/10)/(a^3*c^3)
 

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 3334

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

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 = 0.38 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.69

method	result	size
default	\(-\frac {2 c^{2} \left (\sin \left (f x +e \right )-1\right ) \left (-15 B \cos \left (f x +e \right )^{2}+\sin \left (f x +e \right ) \left (5 A +15 B \right )-A +21 B \right )}{15 a^{3} \left (1+\sin \left (f x +e \right )\right )^{2} \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\)	\(83\)

Input:

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

Output:

-2/15*c^2/a^3*(sin(f*x+e)-1)/(1+sin(f*x+e))^2*(-15*B*cos(f*x+e)^2+sin(f*x+ 
e)*(5*A+15*B)-A+21*B)/cos(f*x+e)/(c-c*sin(f*x+e))^(1/2)/f
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.79 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2}}{(a+a \sin (e+f x))^3} \, dx=\frac {2 \, {\left (15 \, B c \cos \left (f x + e\right )^{2} - 5 \, {\left (A + 3 \, B\right )} c \sin \left (f x + e\right ) + {\left (A - 21 \, B\right )} c\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{15 \, {\left (a^{3} f \cos \left (f x + e\right )^{3} - 2 \, a^{3} f \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, a^{3} f \cos \left (f x + e\right )\right )}} \] Input:

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

Output:

2/15*(15*B*c*cos(f*x + e)^2 - 5*(A + 3*B)*c*sin(f*x + e) + (A - 21*B)*c)*s 
qrt(-c*sin(f*x + e) + c)/(a^3*f*cos(f*x + e)^3 - 2*a^3*f*cos(f*x + e)*sin( 
f*x + e) - 2*a^3*f*cos(f*x + e))
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 663 vs. \(2 (109) = 218\).

Time = 0.15 (sec) , antiderivative size = 663, normalized size of antiderivative = 5.48 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2}}{(a+a \sin (e+f x))^3} \, dx=\text {Too large to display} \] Input:

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

Output:

2/15*((c^(3/2) - 10*c^(3/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 4*c^(3/2)*si 
n(f*x + e)^2/(cos(f*x + e) + 1)^2 - 30*c^(3/2)*sin(f*x + e)^3/(cos(f*x + e 
) + 1)^3 + 6*c^(3/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 30*c^(3/2)*sin( 
f*x + e)^5/(cos(f*x + e) + 1)^5 + 4*c^(3/2)*sin(f*x + e)^6/(cos(f*x + e) + 
 1)^6 - 10*c^(3/2)*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 + c^(3/2)*sin(f*x + 
 e)^8/(cos(f*x + e) + 1)^8)*A/((a^3 + 5*a^3*sin(f*x + e)/(cos(f*x + e) + 1 
) + 10*a^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 10*a^3*sin(f*x + e)^3/(co 
s(f*x + e) + 1)^3 + 5*a^3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + a^3*sin(f* 
x + e)^5/(cos(f*x + e) + 1)^5)*(sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1)^( 
3/2)) - 6*(c^(3/2) + 5*c^(3/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 14*c^(3/2 
)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 15*c^(3/2)*sin(f*x + e)^3/(cos(f*x 
 + e) + 1)^3 + 26*c^(3/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 15*c^(3/2) 
*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 14*c^(3/2)*sin(f*x + e)^6/(cos(f*x 
+ e) + 1)^6 + 5*c^(3/2)*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 + c^(3/2)*sin( 
f*x + e)^8/(cos(f*x + e) + 1)^8)*B/((a^3 + 5*a^3*sin(f*x + e)/(cos(f*x + e 
) + 1) + 10*a^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 10*a^3*sin(f*x + e)^ 
3/(cos(f*x + e) + 1)^3 + 5*a^3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + a^3*s 
in(f*x + e)^5/(cos(f*x + e) + 1)^5)*(sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 
 1)^(3/2)))/f
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 373 vs. \(2 (109) = 218\).

Time = 0.29 (sec) , antiderivative size = 373, normalized size of antiderivative = 3.08 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2}}{(a+a \sin (e+f x))^3} \, dx =\text {Too large to display} \] Input:

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

Output:

-2/15*sqrt(2)*(A*c*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) + 9*B*c*sgn(sin(-1/ 
4*pi + 1/2*f*x + 1/2*e)) + 5*A*c*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)*sgn( 
sin(-1/4*pi + 1/2*f*x + 1/2*e))/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1) + 45* 
B*c*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e 
))/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1) - 5*A*c*(cos(-1/4*pi + 1/2*f*x + 1 
/2*e) - 1)^2*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))/(cos(-1/4*pi + 1/2*f*x + 
1/2*e) + 1)^2 + 75*B*c*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^2*sgn(sin(-1/4 
*pi + 1/2*f*x + 1/2*e))/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^2 + 15*A*c*(c 
os(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^3*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))/( 
cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^3 + 15*B*c*(cos(-1/4*pi + 1/2*f*x + 1/ 
2*e) - 1)^3*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))/(cos(-1/4*pi + 1/2*f*x + 1 
/2*e) + 1)^3)*sqrt(c)/(a^3*f*((cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)/(cos(-1 
/4*pi + 1/2*f*x + 1/2*e) + 1) + 1)^5)
 

Mupad [B] (verification not implemented)

Time = 42.48 (sec) , antiderivative size = 683, normalized size of antiderivative = 5.64 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2}}{(a+a \sin (e+f x))^3} \, dx=\text {Too large to display} \] Input:

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

Output:

(exp(e*1i + f*x*1i)*(c - c*((exp(- e*1i - f*x*1i)*1i)/2 - (exp(e*1i + f*x* 
1i)*1i)/2))^(1/2)*((4*B*c)/(5*a^3*f) - (2*c*(2*A - 3*B))/(5*a^3*f) - (4*c* 
(3*A - 2*B))/(5*a^3*f) + (c*(A*2i - B*3i)*2i)/(5*a^3*f) + (c*(A*3i - B*2i) 
*4i)/(5*a^3*f)))/((exp(e*1i + f*x*1i) - 1i)*(exp(e*1i + f*x*1i) + 1i)^5) - 
 (exp(e*1i + f*x*1i)*(c - c*((exp(- e*1i - f*x*1i)*1i)/2 - (exp(e*1i + f*x 
*1i)*1i)/2))^(1/2)*((2*B*c)/(3*a^3*f) + (c*(A*2i - B*5i)*2i)/(3*a^3*f) - ( 
2*c*(10*A - 13*B))/(3*a^3*f) + (c*(A*8i - B*13i)*2i)/(15*a^3*f)))/((exp(e* 
1i + f*x*1i) - 1i)*(exp(e*1i + f*x*1i) + 1i)^3) + (exp(e*1i + f*x*1i)*(c - 
 c*((exp(- e*1i - f*x*1i)*1i)/2 - (exp(e*1i + f*x*1i)*1i)/2))^(1/2)*((c*(2 
*A - 3*B)*1i)/(3*a^3*f) - (B*c*1i)/(a^3*f) + (2*c*(A*1i - B*3i))/(a^3*f))) 
/((exp(e*1i + f*x*1i) - 1i)*(exp(e*1i + f*x*1i) + 1i)^2) - (exp(e*1i + f*x 
*1i)*(c - c*((exp(- e*1i - f*x*1i)*1i)/2 - (exp(e*1i + f*x*1i)*1i)/2))^(1/ 
2)*((c*(A - B)*4i)/(a^3*f) - (B*c*1i)/(2*a^3*f) + (c*(8*A - 3*B)*1i)/(10*a 
^3*f) + (c*(A*1i - B*2i))/(a^3*f) + (c*(A*7i - B*6i))/(a^3*f)))/((exp(e*1i 
 + f*x*1i) - 1i)*(exp(e*1i + f*x*1i) + 1i)^4) + (4*B*c*exp(e*1i + f*x*1i)* 
(c - c*((exp(- e*1i - f*x*1i)*1i)/2 - (exp(e*1i + f*x*1i)*1i)/2))^(1/2))/( 
a^3*f*(exp(e*1i + f*x*1i) - 1i)*(exp(e*1i + f*x*1i) + 1i))
 

Reduce [F]

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

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

Output:

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