\(\int (g \cos (e+f x))^{-1-2 m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^m \, dx\) [181]

Optimal result

Integrand size = 40, antiderivative size = 51 \[ \int (g \cos (e+f x))^{-1-2 m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^m \, dx=\frac {\text {arctanh}(\sin (e+f x)) (g \cos (e+f x))^{-2 m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^m}{f g} \] Output:

arctanh(sin(f*x+e))*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^m/f/g/((g*cos(f*x+ 
e))^(2*m))
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 2.26 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.84 \[ \int (g \cos (e+f x))^{-1-2 m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^m \, dx=\frac {e^{m (-2 \log (\cos (e+f x))+\log (a (1+\sin (e+f x)))+\log (c-c \sin (e+f x)))} \arcsin (\sec (e+f x)) \cos ^{2 (1+m)}(e+f x) (g \cos (e+f x))^{-1-2 m} \csc (e+f x) \sqrt {-\tan ^2(e+f x)}}{f} \] Input:

Integrate[(g*Cos[e + f*x])^(-1 - 2*m)*(a + a*Sin[e + f*x])^m*(c - c*Sin[e 
+ f*x])^m,x]
 

Output:

(E^(m*(-2*Log[Cos[e + f*x]] + Log[a*(1 + Sin[e + f*x])] + Log[c - c*Sin[e 
+ f*x]]))*ArcSin[Sec[e + f*x]]*Cos[e + f*x]^(2*(1 + m))*(g*Cos[e + f*x])^( 
-1 - 2*m)*Csc[e + f*x]*Sqrt[-Tan[e + f*x]^2])/f
 

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3042, 3326, 27, 3042, 4257}

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.

\(\displaystyle \int (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^m (g \cos (e+f x))^{-2 m-1} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^m (g \cos (e+f x))^{-2 m-1}dx\)

\(\Big \downarrow \) 3326

\(\displaystyle (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^m (g \cos (e+f x))^{-2 m} \int \frac {\sec (e+f x)}{g}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {(a \sin (e+f x)+a)^m (c-c \sin (e+f x))^m (g \cos (e+f x))^{-2 m} \int \sec (e+f x)dx}{g}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(a \sin (e+f x)+a)^m (c-c \sin (e+f x))^m (g \cos (e+f x))^{-2 m} \int \csc \left (e+f x+\frac {\pi }{2}\right )dx}{g}\)

\(\Big \downarrow \) 4257

\(\displaystyle \frac {\text {arctanh}(\sin (e+f x)) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^m (g \cos (e+f x))^{-2 m}}{f g}\)

Input:

Int[(g*Cos[e + f*x])^(-1 - 2*m)*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x] 
)^m,x]
 

Output:

(ArcTanh[Sin[e + f*x]]*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^m)/(f*g 
*(g*Cos[e + f*x])^(2*m))
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 3042

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

rule 3326

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

rule 4257

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] 
 /; FreeQ[{c, d}, x]
 

Maple [F]

\[\int \left (g \cos \left (f x +e \right )\right )^{-1-2 m} \left (a +a \sin \left (f x +e \right )\right )^{m} \left (c -c \sin \left (f x +e \right )\right )^{m}d x\]

Input:

int((g*cos(f*x+e))^(-1-2*m)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^m,x)
 

Output:

int((g*cos(f*x+e))^(-1-2*m)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^m,x)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.94 \[ \int (g \cos (e+f x))^{-1-2 m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^m \, dx=\frac {\left (\frac {a c}{g^{2}}\right )^{m} \log \left (\sin \left (f x + e\right ) + 1\right ) - \left (\frac {a c}{g^{2}}\right )^{m} \log \left (-\sin \left (f x + e\right ) + 1\right )}{2 \, f g} \] Input:

integrate((g*cos(f*x+e))^(-1-2*m)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^m,x, 
 algorithm="fricas")
 

Output:

1/2*((a*c/g^2)^m*log(sin(f*x + e) + 1) - (a*c/g^2)^m*log(-sin(f*x + e) + 1 
))/(f*g)
 

Sympy [F(-1)]

Timed out. \[ \int (g \cos (e+f x))^{-1-2 m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^m \, dx=\text {Timed out} \] Input:

integrate((g*cos(f*x+e))**(-1-2*m)*(a+a*sin(f*x+e))**m*(c-c*sin(f*x+e))**m 
,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int (g \cos (e+f x))^{-1-2 m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^m \, dx=\int { \left (g \cos \left (f x + e\right )\right )^{-2 \, m - 1} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{m} \,d x } \] Input:

integrate((g*cos(f*x+e))^(-1-2*m)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^m,x, 
 algorithm="maxima")
 

Output:

integrate((g*cos(f*x + e))^(-2*m - 1)*(a*sin(f*x + e) + a)^m*(-c*sin(f*x + 
 e) + c)^m, x)
 

Giac [F]

\[ \int (g \cos (e+f x))^{-1-2 m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^m \, dx=\int { \left (g \cos \left (f x + e\right )\right )^{-2 \, m - 1} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{m} \,d x } \] Input:

integrate((g*cos(f*x+e))^(-1-2*m)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^m,x, 
 algorithm="giac")
 

Output:

integrate((g*cos(f*x + e))^(-2*m - 1)*(a*sin(f*x + e) + a)^m*(-c*sin(f*x + 
 e) + c)^m, x)
                                                                                    
                                                                                    
 

Mupad [F(-1)]

Timed out. \[ \int (g \cos (e+f x))^{-1-2 m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^m \, dx=\int \frac {{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^m}{{\left (g\,\cos \left (e+f\,x\right )\right )}^{2\,m+1}} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int (g \cos (e+f x))^{-1-2 m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^m \, dx=\frac {\int \frac {\left (a +a \sin \left (f x +e \right )\right )^{m} \left (-\sin \left (f x +e \right ) c +c \right )^{m}}{\cos \left (f x +e \right )^{2 m} \cos \left (f x +e \right )}d x}{g^{2 m} g} \] Input:

int((g*cos(f*x+e))^(-1-2*m)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^m,x)
 

Output:

int(((sin(e + f*x)*a + a)**m*( - sin(e + f*x)*c + c)**m)/(cos(e + f*x)**(2 
*m)*cos(e + f*x)),x)/(g**(2*m)*g)