\(\int (3-2 \sin (e+f x))^{-1-m} (a+a \sin (e+f x))^m \, dx\) [645]

Optimal result

Integrand size = 29, antiderivative size = 77 \[ \int (3-2 \sin (e+f x))^{-1-m} (a+a \sin (e+f x))^m \, dx=-\frac {2^{1+m} \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},1+m,\frac {3}{2},-\frac {5 (a-a \sin (e+f x))}{a+a \sin (e+f x)}\right ) (1+\sin (e+f x))^{-1-m} (a+a \sin (e+f x))^m}{f} \] Output:

-2^(1+m)*cos(f*x+e)*hypergeom([1/2, 1+m],[3/2],(-5*a+5*a*sin(f*x+e))/(a+a* 
sin(f*x+e)))*(1+sin(f*x+e))^(-1-m)*(a+a*sin(f*x+e))^m/f
 

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 11.54 (sec) , antiderivative size = 272, normalized size of antiderivative = 3.53 \[ \int (3-2 \sin (e+f x))^{-1-m} (a+a \sin (e+f x))^m \, dx=-\frac {\operatorname {Hypergeometric2F1}\left (1+m,1+2 m,2 (1+m),\frac {4 \sqrt {5} (1-i \cos (e+f x)+\sin (e+f x))}{\left (5+\sqrt {5}\right ) \left (-3+\sqrt {5}-2 i \cos (e+f x)+2 \sin (e+f x)\right )}\right ) (3-2 \sin (e+f x))^{-1-m} \left (3+\sqrt {5}+2 i \cos (e+f x)-2 \sin (e+f x)\right ) (a (1+\sin (e+f x)))^m (i \cos (e+f x)+\sin (e+f x)) \left (\frac {\left (-5+\sqrt {5}\right ) \left (3+\sqrt {5}+2 i \cos (e+f x)-2 \sin (e+f x)\right )}{-3+\sqrt {5}-2 i \cos (e+f x)+2 \sin (e+f x)}\right )^m (\cos (e+f x)+i (1+\sin (e+f x))) \left (\cosh \left (m \log \left (5+\sqrt {5}\right )\right )-\sinh \left (m \log \left (5+\sqrt {5}\right )\right )\right )}{\left (5+\sqrt {5}\right ) f (1+2 m)} \] Input:

Integrate[(3 - 2*Sin[e + f*x])^(-1 - m)*(a + a*Sin[e + f*x])^m,x]
 

Output:

-((Hypergeometric2F1[1 + m, 1 + 2*m, 2*(1 + m), (4*Sqrt[5]*(1 - I*Cos[e + 
f*x] + Sin[e + f*x]))/((5 + Sqrt[5])*(-3 + Sqrt[5] - (2*I)*Cos[e + f*x] + 
2*Sin[e + f*x]))]*(3 - 2*Sin[e + f*x])^(-1 - m)*(3 + Sqrt[5] + (2*I)*Cos[e 
 + f*x] - 2*Sin[e + f*x])*(a*(1 + Sin[e + f*x]))^m*(I*Cos[e + f*x] + Sin[e 
 + f*x])*(((-5 + Sqrt[5])*(3 + Sqrt[5] + (2*I)*Cos[e + f*x] - 2*Sin[e + f* 
x]))/(-3 + Sqrt[5] - (2*I)*Cos[e + f*x] + 2*Sin[e + f*x]))^m*(Cos[e + f*x] 
 + I*(1 + Sin[e + f*x]))*(Cosh[m*Log[5 + Sqrt[5]]] - Sinh[m*Log[5 + Sqrt[5 
]]]))/((5 + Sqrt[5])*f*(1 + 2*m)))
 

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.53, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {3042, 3267, 142}

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.

\(\displaystyle \int (3-2 \sin (e+f x))^{-m-1} (a \sin (e+f x)+a)^m \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (3-2 \sin (e+f x))^{-m-1} (a \sin (e+f x)+a)^mdx\)

\(\Big \downarrow \) 3267

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

\(\Big \downarrow \) 142

\(\displaystyle \frac {a \sqrt {-\frac {1-\sin (e+f x)}{\sin (e+f x)+1}} \cos (e+f x) (3-2 \sin (e+f x))^{-m} (a \sin (e+f x)+a)^m \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-m,1-m,\frac {2 (3-2 \sin (e+f x))}{\sin (e+f x)+1}\right )}{\sqrt {5} f m (a-a \sin (e+f x))}\)

Input:

Int[(3 - 2*Sin[e + f*x])^(-1 - m)*(a + a*Sin[e + f*x])^m,x]
 

Output:

(a*Cos[e + f*x]*Hypergeometric2F1[1/2, -m, 1 - m, (2*(3 - 2*Sin[e + f*x])) 
/(1 + Sin[e + f*x])]*Sqrt[-((1 - Sin[e + f*x])/(1 + Sin[e + f*x]))]*(a + a 
*Sin[e + f*x])^m)/(Sqrt[5]*f*m*(3 - 2*Sin[e + f*x])^m*(a - a*Sin[e + f*x]) 
)
 

Defintions of rubi rules used

rule 142

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n*((e + f*x)^(p + 1)/((b*e 
 - a*f)*(m + 1)))*Hypergeometric2F1[m + 1, -n, m + 2, (-(d*e - c*f))*((a + 
b*x)/((b*c - a*d)*(e + f*x)))])/((b*e - a*f)*((c + d*x)/((b*c - a*d)*(e + f 
*x))))^n, x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[m + n + p + 2, 
 0] &&  !IntegerQ[n]
 

rule 3042

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

rule 3267

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

Maple [F]

\[\int \left (3-2 \sin \left (f x +e \right )\right )^{-1-m} \left (a +a \sin \left (f x +e \right )\right )^{m}d x\]

Input:

int((3-2*sin(f*x+e))^(-1-m)*(a+a*sin(f*x+e))^m,x)
 

Output:

int((3-2*sin(f*x+e))^(-1-m)*(a+a*sin(f*x+e))^m,x)
 

Fricas [F]

\[ \int (3-2 \sin (e+f x))^{-1-m} (a+a \sin (e+f x))^m \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-2 \, \sin \left (f x + e\right ) + 3\right )}^{-m - 1} \,d x } \] Input:

integrate((3-2*sin(f*x+e))^(-1-m)*(a+a*sin(f*x+e))^m,x, algorithm="fricas" 
)
 

Output:

integral((a*sin(f*x + e) + a)^m*(-2*sin(f*x + e) + 3)^(-m - 1), x)
 

Sympy [F]

\[ \int (3-2 \sin (e+f x))^{-1-m} (a+a \sin (e+f x))^m \, dx=\int \left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{m} \left (3 - 2 \sin {\left (e + f x \right )}\right )^{- m - 1}\, dx \] Input:

integrate((3-2*sin(f*x+e))**(-1-m)*(a+a*sin(f*x+e))**m,x)
 

Output:

Integral((a*(sin(e + f*x) + 1))**m*(3 - 2*sin(e + f*x))**(-m - 1), x)
 

Maxima [F]

\[ \int (3-2 \sin (e+f x))^{-1-m} (a+a \sin (e+f x))^m \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-2 \, \sin \left (f x + e\right ) + 3\right )}^{-m - 1} \,d x } \] Input:

integrate((3-2*sin(f*x+e))^(-1-m)*(a+a*sin(f*x+e))^m,x, algorithm="maxima" 
)
 

Output:

integrate((a*sin(f*x + e) + a)^m*(-2*sin(f*x + e) + 3)^(-m - 1), x)
 

Giac [F]

\[ \int (3-2 \sin (e+f x))^{-1-m} (a+a \sin (e+f x))^m \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-2 \, \sin \left (f x + e\right ) + 3\right )}^{-m - 1} \,d x } \] Input:

integrate((3-2*sin(f*x+e))^(-1-m)*(a+a*sin(f*x+e))^m,x, algorithm="giac")
 

Output:

integrate((a*sin(f*x + e) + a)^m*(-2*sin(f*x + e) + 3)^(-m - 1), x)
 

Mupad [F(-1)]

Timed out. \[ \int (3-2 \sin (e+f x))^{-1-m} (a+a \sin (e+f x))^m \, dx=\int \frac {{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m}{{\left (3-2\,\sin \left (e+f\,x\right )\right )}^{m+1}} \,d x \] Input:

int((a + a*sin(e + f*x))^m/(3 - 2*sin(e + f*x))^(m + 1),x)
 

Output:

int((a + a*sin(e + f*x))^m/(3 - 2*sin(e + f*x))^(m + 1), x)
 

Reduce [F]

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

int((3-2*sin(f*x+e))^(-1-m)*(a+a*sin(f*x+e))^m,x)
 

Output:

 - int((sin(e + f*x)*a + a)**m/(2*( - 2*sin(e + f*x) + 3)**m*sin(e + f*x) 
- 3*( - 2*sin(e + f*x) + 3)**m),x)