\(\int (a-a \sin (e+f x)) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{-1-m} \, dx\) [349]

Optimal result

Integrand size = 40, antiderivative size = 149 \[ \int (a-a \sin (e+f x)) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{-1-m} \, dx=-\frac {2^{\frac {1}{2}+m} \operatorname {AppellF1}\left (\frac {3}{2},\frac {1}{2}-m,1+m,\frac {5}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {d (1-\sin (e+f x))}{c+d}\right ) \sec (e+f x) (1+\sin (e+f x))^{\frac {1}{2}-m} (a-a \sin (e+f x))^2 (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{-m} \left (\frac {c+d \sin (e+f x)}{c+d}\right )^m}{3 a (c+d) f} \] Output:

-1/3*2^(1/2+m)*AppellF1(3/2,1+m,1/2-m,5/2,d*(1-sin(f*x+e))/(c+d),1/2-1/2*s 
in(f*x+e))*sec(f*x+e)*(1+sin(f*x+e))^(1/2-m)*(a-a*sin(f*x+e))^2*(a+a*sin(f 
*x+e))^m*((c+d*sin(f*x+e))/(c+d))^m/a/(c+d)/f/((c+d*sin(f*x+e))^m)
                                                                                    
                                                                                    
 

Mathematica [F]

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

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

Output:

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

Rubi [A] (warning: unable to verify)

Time = 0.44 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.03, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {3042, 3487, 157, 27, 156, 155}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3487

\(\displaystyle \frac {\sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {a \sin (e+f x)+a} \int \sqrt {a-a \sin (e+f x)} (\sin (e+f x) a+a)^{m-\frac {1}{2}} (c+d \sin (e+f x))^{-m-1}d\sin (e+f x)}{f}\)

\(\Big \downarrow \) 157

\(\displaystyle \frac {\sqrt {2} \sec (e+f x) (a-a \sin (e+f x)) \sqrt {a \sin (e+f x)+a} \int \frac {\sqrt {1-\sin (e+f x)} (\sin (e+f x) a+a)^{m-\frac {1}{2}} (c+d \sin (e+f x))^{-m-1}}{\sqrt {2}}d\sin (e+f x)}{f \sqrt {1-\sin (e+f x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\sec (e+f x) (a-a \sin (e+f x)) \sqrt {a \sin (e+f x)+a} \int \sqrt {1-\sin (e+f x)} (\sin (e+f x) a+a)^{m-\frac {1}{2}} (c+d \sin (e+f x))^{-m-1}d\sin (e+f x)}{f \sqrt {1-\sin (e+f x)}}\)

\(\Big \downarrow \) 156

\(\displaystyle \frac {\sec (e+f x) (a-a \sin (e+f x)) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^{-m} \left (\frac {c+d \sin (e+f x)}{c-d}\right )^m \int \sqrt {1-\sin (e+f x)} (\sin (e+f x) a+a)^{m-\frac {1}{2}} \left (\frac {c}{c-d}+\frac {d \sin (e+f x)}{c-d}\right )^{-m-1}d\sin (e+f x)}{f (c-d) \sqrt {1-\sin (e+f x)}}\)

\(\Big \downarrow \) 155

\(\displaystyle \frac {2 \sqrt {2} \sec (e+f x) (a-a \sin (e+f x)) (a \sin (e+f x)+a)^{m+1} (c+d \sin (e+f x))^{-m} \left (\frac {c+d \sin (e+f x)}{c-d}\right )^m \operatorname {AppellF1}\left (m+\frac {1}{2},-\frac {1}{2},m+1,m+\frac {3}{2},\frac {1}{2} (\sin (e+f x)+1),-\frac {d (\sin (e+f x)+1)}{c-d}\right )}{a f (2 m+1) (c-d) \sqrt {1-\sin (e+f x)}}\)

Input:

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

Output:

(2*Sqrt[2]*AppellF1[1/2 + m, -1/2, 1 + m, 3/2 + m, (1 + Sin[e + f*x])/2, - 
((d*(1 + Sin[e + f*x]))/(c - d))]*Sec[e + f*x]*(a - a*Sin[e + f*x])*(a + a 
*Sin[e + f*x])^(1 + m)*((c + d*Sin[e + f*x])/(c - d))^m)/(a*(c - d)*f*(1 + 
 2*m)*Sqrt[1 - Sin[e + f*x]]*(c + d*Sin[e + f*x])^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 155

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) 
^(p_), x_] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*Simplify[b/(b*c - a*d)]^n* 
Simplify[b/(b*e - a*f)]^p))*AppellF1[m + 1, -n, -p, m + 2, (-d)*((a + b*x)/ 
(b*c - a*d)), (-f)*((a + b*x)/(b*e - a*f))], x] /; FreeQ[{a, b, c, d, e, f, 
 m, n, p}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] &&  !IntegerQ[p] && GtQ[Sim 
plify[b/(b*c - a*d)], 0] && GtQ[Simplify[b/(b*e - a*f)], 0] &&  !(GtQ[Simpl 
ify[d/(d*a - c*b)], 0] && GtQ[Simplify[d/(d*e - c*f)], 0] && SimplerQ[c + d 
*x, a + b*x]) &&  !(GtQ[Simplify[f/(f*a - e*b)], 0] && GtQ[Simplify[f/(f*c 
- e*d)], 0] && SimplerQ[e + f*x, a + b*x])
 

rule 156

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) 
^(p_), x_] :> Simp[(e + f*x)^FracPart[p]/(Simplify[b/(b*e - a*f)]^IntPart[p 
]*(b*((e + f*x)/(b*e - a*f)))^FracPart[p])   Int[(a + b*x)^m*(c + d*x)^n*Si 
mp[b*(e/(b*e - a*f)) + b*f*(x/(b*e - a*f)), x]^p, x], x] /; FreeQ[{a, b, c, 
 d, e, f, m, n, p}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] &&  !IntegerQ[p] & 
& GtQ[Simplify[b/(b*c - a*d)], 0] &&  !GtQ[Simplify[b/(b*e - a*f)], 0]
 

rule 157

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) 
^(p_), x_] :> Simp[(c + d*x)^FracPart[n]/(Simplify[b/(b*c - a*d)]^IntPart[n 
]*(b*((c + d*x)/(b*c - a*d)))^FracPart[n])   Int[(a + b*x)^m*Simp[b*(c/(b*c 
 - a*d)) + b*d*(x/(b*c - a*d)), x]^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, 
 d, e, f, m, n, p}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] &&  !IntegerQ[p] & 
&  !GtQ[Simplify[b/(b*c - a*d)], 0] &&  !SimplerQ[c + d*x, a + b*x] &&  !Si 
mplerQ[e + f*x, a + b*x]
 

rule 3042

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

rule 3487

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + 
 (f_.)*(x_)])^(p_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] 
:> Simp[Sqrt[a + b*Sin[e + f*x]]*(Sqrt[c + d*Sin[e + f*x]]/(f*Cos[e + f*x]) 
)   Subst[Int[(a + b*x)^(m - 1/2)*(c + d*x)^(n - 1/2)*(A + B*x)^p, x], x, S 
in[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n, p}, x] && EqQ[b*c 
+ a*d, 0] && EqQ[a^2 - b^2, 0]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

integrate((a-a*sin(f*x+e))*(a+a*sin(f*x+e))^m*(c+d*sin(f*x+e))^(-1-m),x, a 
lgorithm="fricas")
 

Output:

integral(-(a*sin(f*x + e) - a)*(a*sin(f*x + e) + a)^m*(d*sin(f*x + e) + c) 
^(-m - 1), x)
 

Sympy [F(-1)]

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

integrate((a-a*sin(f*x+e))*(a+a*sin(f*x+e))**m*(c+d*sin(f*x+e))**(-1-m),x)
 

Output:

Timed out
 

Maxima [F]

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

integrate((a-a*sin(f*x+e))*(a+a*sin(f*x+e))^m*(c+d*sin(f*x+e))^(-1-m),x, a 
lgorithm="maxima")
 

Output:

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

Giac [F]

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

integrate((a-a*sin(f*x+e))*(a+a*sin(f*x+e))^m*(c+d*sin(f*x+e))^(-1-m),x, a 
lgorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

a*(int((sin(e + f*x)*a + a)**m/((sin(e + f*x)*d + c)**m*sin(e + f*x)*d + ( 
sin(e + f*x)*d + c)**m*c),x) - int(((sin(e + f*x)*a + a)**m*sin(e + f*x))/ 
((sin(e + f*x)*d + c)**m*sin(e + f*x)*d + (sin(e + f*x)*d + c)**m*c),x))