\(\int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2}}{(a+a \sin (e+f x))^2} \, dx\) [118]

Optimal result

Integrand size = 38, antiderivative size = 115 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2}}{(a+a \sin (e+f x))^2} \, dx=\frac {4 (A-7 B) c \sec (e+f x) \sqrt {c-c \sin (e+f x)}}{3 a^2 f}-\frac {(A-7 B) \sec (e+f x) (c-c \sin (e+f x))^{3/2}}{3 a^2 f}-\frac {(A-B) \sec ^3(e+f x) (c-c \sin (e+f x))^{7/2}}{3 a^2 c^2 f} \] Output:

4/3*(A-7*B)*c*sec(f*x+e)*(c-c*sin(f*x+e))^(1/2)/a^2/f-1/3*(A-7*B)*sec(f*x+ 
e)*(c-c*sin(f*x+e))^(3/2)/a^2/f-1/3*(A-B)*sec(f*x+e)^3*(c-c*sin(f*x+e))^(7 
/2)/a^2/c^2/f
 

Mathematica [A] (verified)

Time = 8.47 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.98 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2}}{(a+a \sin (e+f x))^2} \, dx=\frac {c \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) (2 A-23 B+3 B \cos (2 (e+f x))+6 (A-5 B) \sin (e+f x)) \sqrt {c-c \sin (e+f x)}}{3 a^2 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) (1+\sin (e+f x))^2} \] Input:

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

Output:

(c*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(2*A - 23*B + 3*B*Cos[2*(e + f*x) 
] + 6*(A - 5*B)*Sin[e + f*x])*Sqrt[c - c*Sin[e + f*x]])/(3*a^2*f*(Cos[(e + 
 f*x)/2] - Sin[(e + f*x)/2])*(1 + Sin[e + f*x])^2)
 

Rubi [A] (verified)

Time = 0.79 (sec) , antiderivative size = 110, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3446

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3334

\(\displaystyle \frac {-\frac {1}{6} c (A-7 B) \int \sec ^2(e+f x) (c-c \sin (e+f x))^{5/2}dx-\frac {(A-B) \sec ^3(e+f x) (c-c \sin (e+f x))^{7/2}}{3 f}}{a^2 c^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {1}{6} c (A-7 B) \int \frac {(c-c \sin (e+f x))^{5/2}}{\cos (e+f x)^2}dx-\frac {(A-B) \sec ^3(e+f x) (c-c \sin (e+f x))^{7/2}}{3 f}}{a^2 c^2}\)

\(\Big \downarrow \) 3153

\(\displaystyle \frac {-\frac {1}{6} c (A-7 B) \left (4 c \int \sec ^2(e+f x) (c-c \sin (e+f x))^{3/2}dx+\frac {2 c \sec (e+f x) (c-c \sin (e+f x))^{3/2}}{f}\right )-\frac {(A-B) \sec ^3(e+f x) (c-c \sin (e+f x))^{7/2}}{3 f}}{a^2 c^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {1}{6} c (A-7 B) \left (4 c \int \frac {(c-c \sin (e+f x))^{3/2}}{\cos (e+f x)^2}dx+\frac {2 c \sec (e+f x) (c-c \sin (e+f x))^{3/2}}{f}\right )-\frac {(A-B) \sec ^3(e+f x) (c-c \sin (e+f x))^{7/2}}{3 f}}{a^2 c^2}\)

\(\Big \downarrow \) 3152

\(\displaystyle \frac {-\frac {1}{6} c (A-7 B) \left (\frac {2 c \sec (e+f x) (c-c \sin (e+f x))^{3/2}}{f}-\frac {8 c^2 \sec (e+f x) \sqrt {c-c \sin (e+f x)}}{f}\right )-\frac {(A-B) \sec ^3(e+f x) (c-c \sin (e+f x))^{7/2}}{3 f}}{a^2 c^2}\)

Input:

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

Output:

(-1/3*((A - B)*Sec[e + f*x]^3*(c - c*Sin[e + f*x])^(7/2))/f - ((A - 7*B)*c 
*((-8*c^2*Sec[e + f*x]*Sqrt[c - c*Sin[e + f*x]])/f + (2*c*Sec[e + f*x]*(c 
- c*Sin[e + f*x])^(3/2))/f))/6)/(a^2*c^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 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 = 0.39 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.70

method	result	size
default	\(-\frac {2 c^{2} \left (\sin \left (f x +e \right )-1\right ) \left (3 B \cos \left (f x +e \right )^{2}+\sin \left (f x +e \right ) \left (3 A -15 B \right )+A -13 B \right )}{3 a^{2} \left (1+\sin \left (f x +e \right )\right ) \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\)	\(81\)

Input:

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

Output:

-2/3*c^2/a^2*(sin(f*x+e)-1)/(1+sin(f*x+e))*(3*B*cos(f*x+e)^2+sin(f*x+e)*(3 
*A-15*B)+A-13*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 = 80, normalized size of antiderivative = 0.70 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2}}{(a+a \sin (e+f x))^2} \, dx=\frac {2 \, {\left (3 \, B c \cos \left (f x + e\right )^{2} + 3 \, {\left (A - 5 \, B\right )} c \sin \left (f x + e\right ) + {\left (A - 13 \, B\right )} c\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{3 \, {\left (a^{2} f \cos \left (f x + e\right ) \sin \left (f x + e\right ) + a^{2} f \cos \left (f x + e\right )\right )}} \] Input:

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

Output:

2/3*(3*B*c*cos(f*x + e)^2 + 3*(A - 5*B)*c*sin(f*x + e) + (A - 13*B)*c)*sqr 
t(-c*sin(f*x + e) + c)/(a^2*f*cos(f*x + e)*sin(f*x + e) + a^2*f*cos(f*x + 
e))
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 482 vs. \(2 (103) = 206\).

Time = 0.16 (sec) , antiderivative size = 482, normalized size of antiderivative = 4.19 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2}}{(a+a \sin (e+f x))^2} \, dx =\text {Too large to display} \] Input:

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

Output:

-2/3*((c^(3/2) + 6*c^(3/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 3*c^(3/2)*sin 
(f*x + e)^2/(cos(f*x + e) + 1)^2 + 12*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 + 6*c^(3/2)*sin(f* 
x + e)^5/(cos(f*x + e) + 1)^5 + c^(3/2)*sin(f*x + e)^6/(cos(f*x + e) + 1)^ 
6)*A/((a^2 + 3*a^2*sin(f*x + e)/(cos(f*x + e) + 1) + 3*a^2*sin(f*x + e)^2/ 
(cos(f*x + e) + 1)^2 + a^2*sin(f*x + e)^3/(cos(f*x + e) + 1)^3)*(sin(f*x + 
 e)^2/(cos(f*x + e) + 1)^2 + 1)^(3/2)) - 2*(5*c^(3/2) + 15*c^(3/2)*sin(f*x 
 + e)/(cos(f*x + e) + 1) + 21*c^(3/2)*sin(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 + 21*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 + 
 5*c^(3/2)*sin(f*x + e)^6/(cos(f*x + e) + 1)^6)*B/((a^2 + 3*a^2*sin(f*x + 
e)/(cos(f*x + e) + 1) + 3*a^2*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + a^2*si 
n(f*x + e)^3/(cos(f*x + e) + 1)^3)*(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. 283 vs. \(2 (103) = 206\).

Time = 0.31 (sec) , antiderivative size = 283, normalized size of antiderivative = 2.46 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2}}{(a+a \sin (e+f x))^2} \, dx=-\frac {4 \, \sqrt {2} \sqrt {c} {\left (\frac {3 \, B c \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{a^{2} {\left (\frac {\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1} - 1\right )}} + \frac {A c \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 4 \, B c \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + \frac {3 \, A c {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1} - \frac {9 \, B c {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1} - \frac {3 \, B c {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{2}}}{a^{2} {\left (\frac {\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1} + 1\right )}^{3}}\right )}}{3 \, f} \] Input:

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

Output:

-4/3*sqrt(2)*sqrt(c)*(3*B*c*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))/(a^2*((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) 
) + (A*c*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) - 4*B*c*sgn(sin(-1/4*pi + 1/2 
*f*x + 1/2*e)) + 3*A*c*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)*sgn(sin(-1/4*p 
i + 1/2*f*x + 1/2*e))/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1) - 9*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*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^2*((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))/f
 

Mupad [B] (verification not implemented)

Time = 40.64 (sec) , antiderivative size = 492, normalized size of antiderivative = 4.28 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2}}{(a+a \sin (e+f x))^2} \, dx =\text {Too large to display} \] Input:

int(((A + B*sin(e + f*x))*(c - c*sin(e + f*x))^(3/2))/(a + a*sin(e + f*x)) 
^2,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)*((2*B*c)/(3*a^2*f) - (c*(2*A - 3*B))/(3*a^2*f) - (2*c*(3 
*A - 2*B))/(3*a^2*f) + (c*(A*2i - B*3i)*1i)/(3*a^2*f) + (c*(A*3i - B*2i)*2 
i)/(3*a^2*f)))/((exp(e*1i + f*x*1i) - 1i)*(exp(e*1i + f*x*1i) + 1i)^3) - ( 
(c - c*((exp(- e*1i - f*x*1i)*1i)/2 - (exp(e*1i + f*x*1i)*1i)/2))^(1/2)*(( 
2*B*c)/(a^2*f) - (B*c*exp(e*1i + f*x*1i)*2i)/(a^2*f)))/(exp(e*1i + f*x*1i) 
 - 1i) - (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^2*f) + (c*(A*1i - B*2i))/(a^ 
2*f) + (c*(A*1i + B*2i))/(3*a^2*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)*((4*B*c)/(a^2*f) + (c*(A*1i - B*2i) 
*4i)/(a^2*f)))/((exp(e*1i + f*x*1i) - 1i)*(exp(e*1i + f*x*1i) + 1i))
 

Reduce [F]

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

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

Output:

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