\(\int (-3+5 \sin (e+f x))^{-1-m} (a+a \sin (e+f x))^m \, dx\) [649]

Optimal result

Integrand size = 29, antiderivative size = 72 \[ \int (-3+5 \sin (e+f x))^{-1-m} (a+a \sin (e+f x))^m \, dx=-\frac {\cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},1+m,\frac {3}{2},\frac {4 (a-a \sin (e+f x))}{a+a \sin (e+f x)}\right ) (1+\sin (e+f x))^{-1-m} (a+a \sin (e+f x))^m}{f} \] Output:

-cos(f*x+e)*hypergeom([1/2, 1+m],[3/2],4*(a-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 [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(201\) vs. \(2(72)=144\).

Time = 9.73 (sec) , antiderivative size = 201, normalized size of antiderivative = 2.79 \[ \int (-3+5 \sin (e+f x))^{-1-m} (a+a \sin (e+f x))^m \, dx=-\frac {\operatorname {Hypergeometric2F1}\left (1+m,1+2 m,2 (1+m),-\frac {2 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )}{\cos \left (\frac {1}{2} (e+f x)\right )-3 \sin \left (\frac {1}{2} (e+f x)\right )}\right ) \left (\frac {3 \cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )}{\cos \left (\frac {1}{2} (e+f x)\right )-3 \sin \left (\frac {1}{2} (e+f x)\right )}\right )^m \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) (a (1+\sin (e+f x)))^m (-3+5 \sin (e+f x))^{-m}}{2 f (1+2 m) \left (\cos \left (\frac {1}{2} (e+f x)\right )-3 \sin \left (\frac {1}{2} (e+f x)\right )\right )} \] Input:

Integrate[(-3 + 5*Sin[e + f*x])^(-1 - m)*(a + a*Sin[e + f*x])^m,x]
 

Output:

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

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.60, 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.

\(\displaystyle \int (5 \sin (e+f x)-3)^{-m-1} (a \sin (e+f x)+a)^m \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (5 \sin (e+f x)-3)^{-m-1} (a \sin (e+f x)+a)^mdx\)

\(\Big \downarrow \) 3267

\(\displaystyle \frac {a^2 \cos (e+f x) \int \frac {(5 \sin (e+f x)-3)^{-m-1} (\sin (e+f x) a+a)^{m-\frac {1}{2}}}{\sqrt {a-a \sin (e+f x)}}d\sin (e+f x)}{f \sqrt {a-a \sin (e+f x)} \sqrt {a \sin (e+f x)+a}}\)

\(\Big \downarrow \) 142

\(\displaystyle -\frac {a \sqrt {\frac {1-\sin (e+f x)}{\sin (e+f x)+1}} \cos (e+f x) (5 \sin (e+f x)-3)^{-m} (a \sin (e+f x)+a)^m \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-m,1-m,-\frac {3-5 \sin (e+f x)}{\sin (e+f x)+1}\right )}{4 f m (a-a \sin (e+f x))}\)

Input:

Int[(-3 + 5*Sin[e + f*x])^(-1 - m)*(a + a*Sin[e + f*x])^m,x]
 

Output:

-1/4*(a*Cos[e + f*x]*Hypergeometric2F1[1/2, -m, 1 - m, -((3 - 5*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)/(f*m*(-3 + 5*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]

\[\int \left (-3+5 \sin \left (f x +e \right )\right )^{-1-m} \left (a +a \sin \left (f x +e \right )\right )^{m}d x\]

Input:

int((-3+5*sin(f*x+e))^(-1-m)*(a+a*sin(f*x+e))^m,x)
 

Output:

int((-3+5*sin(f*x+e))^(-1-m)*(a+a*sin(f*x+e))^m,x)
 

Fricas [F]

\[ \int (-3+5 \sin (e+f x))^{-1-m} (a+a \sin (e+f x))^m \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (5 \, \sin \left (f x + e\right ) - 3\right )}^{-m - 1} \,d x } \] Input:

integrate((-3+5*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*(5*sin(f*x + e) - 3)^(-m - 1), x)
 

Sympy [F]

\[ \int (-3+5 \sin (e+f x))^{-1-m} (a+a \sin (e+f x))^m \, dx=\int \left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{m} \left (5 \sin {\left (e + f x \right )} - 3\right )^{- m - 1}\, dx \] Input:

integrate((-3+5*sin(f*x+e))**(-1-m)*(a+a*sin(f*x+e))**m,x)
 

Output:

Integral((a*(sin(e + f*x) + 1))**m*(5*sin(e + f*x) - 3)**(-m - 1), x)
 

Maxima [F]

\[ \int (-3+5 \sin (e+f x))^{-1-m} (a+a \sin (e+f x))^m \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (5 \, \sin \left (f x + e\right ) - 3\right )}^{-m - 1} \,d x } \] Input:

integrate((-3+5*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*(5*sin(f*x + e) - 3)^(-m - 1), x)
 

Giac [F]

\[ \int (-3+5 \sin (e+f x))^{-1-m} (a+a \sin (e+f x))^m \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (5 \, \sin \left (f x + e\right ) - 3\right )}^{-m - 1} \,d x } \] Input:

integrate((-3+5*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*(5*sin(f*x + e) - 3)^(-m - 1), x)
 

Mupad [F(-1)]

Timed out. \[ \int (-3+5 \sin (e+f x))^{-1-m} (a+a \sin (e+f x))^m \, dx=\int \frac {{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m}{{\left (5\,\sin \left (e+f\,x\right )-3\right )}^{m+1}} \,d x \] Input:

int((a + a*sin(e + f*x))^m/(5*sin(e + f*x) - 3)^(m + 1),x)
 

Output:

int((a + a*sin(e + f*x))^m/(5*sin(e + f*x) - 3)^(m + 1), x)
 

Reduce [F]

\[ \int (-3+5 \sin (e+f x))^{-1-m} (a+a \sin (e+f x))^m \, dx=\int \frac {\left (a +a \sin \left (f x +e \right )\right )^{m}}{5 \left (5 \sin \left (f x +e \right )-3\right )^{m} \sin \left (f x +e \right )-3 \left (5 \sin \left (f x +e \right )-3\right )^{m}}d x \] Input:

int((-3+5*sin(f*x+e))^(-1-m)*(a+a*sin(f*x+e))^m,x)
 

Output:

int((sin(e + f*x)*a + a)**m/(5*(5*sin(e + f*x) - 3)**m*sin(e + f*x) - 3*(5 
*sin(e + f*x) - 3)**m),x)