\(\int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)} \, dx\) [207]

Optimal result

Integrand size = 38, antiderivative size = 104 \[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)} \, dx=\frac {2 (A-B) c \cos (e+f x) (a+a \sin (e+f x))^m}{f (1+2 m) \sqrt {c-c \sin (e+f x)}}+\frac {2 B c \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{a f (3+2 m) \sqrt {c-c \sin (e+f x)}} \] Output:

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

Mathematica [A] (verified)

Time = 2.12 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.12 \[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)} \, dx=\frac {2 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) (a (1+\sin (e+f x)))^m \sqrt {c-c \sin (e+f x)} (-2 B+A (3+2 m)+B (1+2 m) \sin (e+f x))}{f (1+2 m) (3+2 m) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )} \] Input:

Integrate[(a + a*Sin[e + f*x])^m*(A + B*Sin[e + f*x])*Sqrt[c - c*Sin[e + f 
*x]],x]
 

Output:

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

Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {3042, 3450, 3042, 3217}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3450

\(\displaystyle (A-B) \int (\sin (e+f x) a+a)^m \sqrt {c-c \sin (e+f x)}dx+\frac {B \int (\sin (e+f x) a+a)^{m+1} \sqrt {c-c \sin (e+f x)}dx}{a}\)

\(\Big \downarrow \) 3042

\(\displaystyle (A-B) \int (\sin (e+f x) a+a)^m \sqrt {c-c \sin (e+f x)}dx+\frac {B \int (\sin (e+f x) a+a)^{m+1} \sqrt {c-c \sin (e+f x)}dx}{a}\)

\(\Big \downarrow \) 3217

\(\displaystyle \frac {2 c (A-B) \cos (e+f x) (a \sin (e+f x)+a)^m}{f (2 m+1) \sqrt {c-c \sin (e+f x)}}+\frac {2 B c \cos (e+f x) (a \sin (e+f x)+a)^{m+1}}{a f (2 m+3) \sqrt {c-c \sin (e+f x)}}\)

Input:

Int[(a + a*Sin[e + f*x])^m*(A + B*Sin[e + f*x])*Sqrt[c - c*Sin[e + f*x]],x 
]
 

Output:

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

Defintions of rubi rules used

rule 3042

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

rule 3217

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_) + (d_.)*sin[(e_.) + (f 
_.)*(x_)])^(n_), x_Symbol] :> Simp[-2*b*Cos[e + f*x]*((c + d*Sin[e + f*x])^ 
n/(f*(2*n + 1)*Sqrt[a + b*Sin[e + f*x]])), x] /; FreeQ[{a, b, c, d, e, f, n 
}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[n, -2^(-1)]
 

rule 3450

Int[Sqrt[(a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + 
(f_.)*(x_)])*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp 
[B/d   Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n + 1), x], x] - 
Simp[(B*c - A*d)/d   Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^n, 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]
 

Maple [F]

\[\int \left (a +a \sin \left (f x +e \right )\right )^{m} \left (A +B \sin \left (f x +e \right )\right ) \sqrt {c -c \sin \left (f x +e \right )}d x\]

Input:

int((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(1/2),x)
 

Output:

int((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(1/2),x)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.59 \[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)} \, dx=-\frac {2 \, {\left ({\left (2 \, B m + B\right )} \cos \left (f x + e\right )^{2} - 2 \, {\left (A + B\right )} m - {\left (2 \, A m + 3 \, A - 2 \, B\right )} \cos \left (f x + e\right ) - {\left (2 \, {\left (A + B\right )} m + {\left (2 \, B m + B\right )} \cos \left (f x + e\right ) + 3 \, A - B\right )} \sin \left (f x + e\right ) - 3 \, A + B\right )} \sqrt {-c \sin \left (f x + e\right ) + c} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{4 \, f m^{2} + 8 \, f m + {\left (4 \, f m^{2} + 8 \, f m + 3 \, f\right )} \cos \left (f x + e\right ) - {\left (4 \, f m^{2} + 8 \, f m + 3 \, f\right )} \sin \left (f x + e\right ) + 3 \, f} \] Input:

integrate((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(1/2),x, al 
gorithm="fricas")
 

Output:

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

Sympy [F]

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

integrate((a+a*sin(f*x+e))**m*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))**(1/2),x)
 

Output:

Integral((a*(sin(e + f*x) + 1))**m*sqrt(-c*(sin(e + f*x) - 1))*(A + B*sin( 
e + f*x)), x)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 323 vs. \(2 (100) = 200\).

Time = 0.15 (sec) , antiderivative size = 323, normalized size of antiderivative = 3.11 \[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)} \, dx=-\frac {2 \, {\left (\frac {2 \, {\left (\frac {2 \, a^{m} \sqrt {c} m \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {2 \, a^{m} \sqrt {c} m \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - a^{m} \sqrt {c} - \frac {a^{m} \sqrt {c} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}\right )} B e^{\left (2 \, m \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right ) - m \log \left (\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )\right )}}{{\left (4 \, m^{2} + 8 \, m + \frac {{\left (4 \, m^{2} + 8 \, m + 3\right )} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 3\right )} \sqrt {\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1}} + \frac {{\left (a^{m} \sqrt {c} + \frac {a^{m} \sqrt {c} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )} A e^{\left (2 \, m \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right ) - m \log \left (\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )\right )}}{{\left (2 \, m + 1\right )} \sqrt {\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1}}\right )}}{f} \] Input:

integrate((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(1/2),x, al 
gorithm="maxima")
 

Output:

-2*(2*(2*a^m*sqrt(c)*m*sin(f*x + e)/(cos(f*x + e) + 1) + 2*a^m*sqrt(c)*m*s 
in(f*x + e)^2/(cos(f*x + e) + 1)^2 - a^m*sqrt(c) - a^m*sqrt(c)*sin(f*x + e 
)^3/(cos(f*x + e) + 1)^3)*B*e^(2*m*log(sin(f*x + e)/(cos(f*x + e) + 1) + 1 
) - m*log(sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1))/((4*m^2 + 8*m + (4*m^2 
 + 8*m + 3)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 3)*sqrt(sin(f*x + e)^2/( 
cos(f*x + e) + 1)^2 + 1)) + (a^m*sqrt(c) + a^m*sqrt(c)*sin(f*x + e)/(cos(f 
*x + e) + 1))*A*e^(2*m*log(sin(f*x + e)/(cos(f*x + e) + 1) + 1) - m*log(si 
n(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1))/((2*m + 1)*sqrt(sin(f*x + e)^2/(co 
s(f*x + e) + 1)^2 + 1)))/f
 

Giac [F]

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

integrate((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(1/2),x, al 
gorithm="giac")
 

Output:

integrate((B*sin(f*x + e) + A)*sqrt(-c*sin(f*x + e) + c)*(a*sin(f*x + e) + 
 a)^m, x)
 

Mupad [B] (verification not implemented)

Time = 1.60 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.01 \[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)} \, dx=-\frac {{\left (a\,\left (\sin \left (e+f\,x\right )+1\right )\right )}^m\,\sqrt {-c\,\left (\sin \left (e+f\,x\right )-1\right )}\,\left (6\,A\,\cos \left (e+f\,x\right )-4\,B\,\cos \left (e+f\,x\right )+B\,\sin \left (2\,e+2\,f\,x\right )+4\,A\,m\,\cos \left (e+f\,x\right )+2\,B\,m\,\sin \left (2\,e+2\,f\,x\right )\right )}{f\,\left (\sin \left (e+f\,x\right )-1\right )\,\left (4\,m^2+8\,m+3\right )} \] Input:

int((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^m*(c - c*sin(e + f*x))^(1/2) 
,x)
 

Output:

-((a*(sin(e + f*x) + 1))^m*(-c*(sin(e + f*x) - 1))^(1/2)*(6*A*cos(e + f*x) 
 - 4*B*cos(e + f*x) + B*sin(2*e + 2*f*x) + 4*A*m*cos(e + f*x) + 2*B*m*sin( 
2*e + 2*f*x)))/(f*(sin(e + f*x) - 1)*(8*m + 4*m^2 + 3))
 

Reduce [F]

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

int((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(1/2),x)
 

Output:

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