\(\int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2}}{a+a \sin (e+f x)} \, dx\) [110]

Optimal result

Integrand size = 38, antiderivative size = 118 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2}}{a+a \sin (e+f x)} \, dx=-\frac {4 (3 A-5 B) c^2 \cos (e+f x)}{3 a f \sqrt {c-c \sin (e+f x)}}-\frac {(3 A-5 B) c \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{3 a f}-\frac {(A-B) \sec (e+f x) (c-c \sin (e+f x))^{5/2}}{a c f} \] Output:

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

Mathematica [A] (verified)

Time = 11.43 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.96 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2}}{a+a \sin (e+f x)} \, dx=\frac {c \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) (-18 A+27 B+B \cos (2 (e+f x))+(-6 A+14 B) \sin (e+f x)) \sqrt {c-c \sin (e+f x)}}{3 a f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) (1+\sin (e+f x))} \] Input:

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

Output:

(c*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(-18*A + 27*B + B*Cos[2*(e + f*x) 
] + (-6*A + 14*B)*Sin[e + f*x])*Sqrt[c - c*Sin[e + f*x]])/(3*a*f*(Cos[(e + 
 f*x)/2] - Sin[(e + f*x)/2])*(1 + Sin[e + f*x]))
 

Rubi [A] (verified)

Time = 0.71 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.95, 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, 3126, 3042, 3125}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3446

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

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2}}{\cos (e+f x)^2}dx}{a c}\)

\(\Big \downarrow \) 3334

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3126

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3125

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 3042

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

rule 3125

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2*b*(Cos 
[c + d*x]/(d*Sqrt[a + b*Sin[c + d*x]])), x] /; FreeQ[{a, b, c, d}, x] && Eq 
Q[a^2 - b^2, 0]
 

rule 3126

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos 
[c + d*x]*((a + b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[a*((2*n - 1)/n) 
 Int[(a + b*Sin[c + d*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[ 
a^2 - b^2, 0] && IGtQ[n - 1/2, 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 = 0.35 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.62

method	result	size
default	\(\frac {2 c^{2} \left (\sin \left (f x +e \right )-1\right ) \left (-B \cos \left (f x +e \right )^{2}+\sin \left (f x +e \right ) \left (3 A -7 B \right )+9 A -13 B \right )}{3 a \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\)	\(73\)

Input:

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

Output:

2/3*c^2/a*(sin(f*x+e)-1)*(-B*cos(f*x+e)^2+sin(f*x+e)*(3*A-7*B)+9*A-13*B)/c 
os(f*x+e)/(c-c*sin(f*x+e))^(1/2)/f
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.57 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2}}{a+a \sin (e+f x)} \, dx=\frac {2 \, {\left (B c \cos \left (f x + e\right )^{2} - {\left (3 \, A - 7 \, B\right )} c \sin \left (f x + e\right ) - {\left (9 \, A - 13 \, B\right )} c\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{3 \, a f \cos \left (f x + e\right )} \] Input:

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

Output:

2/3*(B*c*cos(f*x + e)^2 - (3*A - 7*B)*c*sin(f*x + e) - (9*A - 13*B)*c)*sqr 
t(-c*sin(f*x + e) + c)/(a*f*cos(f*x + e))
 

Sympy [F]

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

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

Output:

(Integral(A*c*sqrt(-c*sin(e + f*x) + c)/(sin(e + f*x) + 1), x) + Integral( 
-A*c*sqrt(-c*sin(e + f*x) + c)*sin(e + f*x)/(sin(e + f*x) + 1), x) + Integ 
ral(B*c*sqrt(-c*sin(e + f*x) + c)*sin(e + f*x)/(sin(e + f*x) + 1), x) + In 
tegral(-B*c*sqrt(-c*sin(e + f*x) + c)*sin(e + f*x)**2/(sin(e + f*x) + 1), 
x))/a
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 294 vs. \(2 (108) = 216\).

Time = 0.13 (sec) , antiderivative size = 294, normalized size of antiderivative = 2.49 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2}}{a+a \sin (e+f x)} \, dx=\frac {2 \, {\left (\frac {3 \, {\left (3 \, c^{\frac {3}{2}} + \frac {2 \, c^{\frac {3}{2}} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {6 \, c^{\frac {3}{2}} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {2 \, c^{\frac {3}{2}} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {3 \, c^{\frac {3}{2}} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}}\right )} A}{{\left (a + \frac {a \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )} {\left (\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )}^{\frac {3}{2}}} - \frac {2 \, {\left (7 \, c^{\frac {3}{2}} + \frac {7 \, c^{\frac {3}{2}} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {12 \, c^{\frac {3}{2}} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {7 \, c^{\frac {3}{2}} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {7 \, c^{\frac {3}{2}} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}}\right )} B}{{\left (a + \frac {a \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )} {\left (\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )}^{\frac {3}{2}}}\right )}}{3 \, f} \] Input:

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

Output:

2/3*(3*(3*c^(3/2) + 2*c^(3/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 6*c^(3/2)* 
sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 2*c^(3/2)*sin(f*x + e)^3/(cos(f*x + 
e) + 1)^3 + 3*c^(3/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4)*A/((a + a*sin(f 
*x + e)/(cos(f*x + e) + 1))*(sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1)^(3/2 
)) - 2*(7*c^(3/2) + 7*c^(3/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 12*c^(3/2) 
*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 7*c^(3/2)*sin(f*x + e)^3/(cos(f*x + 
 e) + 1)^3 + 7*c^(3/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4)*B/((a + a*sin( 
f*x + e)/(cos(f*x + e) + 1))*(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. 353 vs. \(2 (108) = 216\).

Time = 0.29 (sec) , antiderivative size = 353, normalized size of antiderivative = 2.99 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2}}{a+a \sin (e+f x)} \, dx =\text {Too large to display} \] Input:

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

Output:

4/3*sqrt(2)*sqrt(c)*(3*(A*c*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) - B*c*sgn( 
sin(-1/4*pi + 1/2*f*x + 1/2*e)))/(a*((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)) - (3*A*c*sgn(sin(-1/4*pi + 1/2* 
f*x + 1/2*e)) - 7*B*c*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) - 6*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) + 18*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) + 
 3*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 - 3*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)/(a*((cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)/(cos(-1/4*p 
i + 1/2*f*x + 1/2*e) + 1) - 1)^3))/f
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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