\(\int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c+d \sin (e+f x))^{-1-m} \, dx\) [347]

Optimal result

Integrand size = 39, antiderivative size = 267 \[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c+d \sin (e+f x))^{-1-m} \, dx=-\frac {2^{\frac {1}{2}+m} a (A-B) \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2}-m,\frac {3}{2},\frac {(c-d) (1-\sin (e+f x))}{2 (c+d \sin (e+f x))}\right ) (a+a \sin (e+f x))^{-1+m} \left (\frac {(c+d) (1+\sin (e+f x))}{c+d \sin (e+f x)}\right )^{\frac {1}{2}-m} (c+d \sin (e+f x))^{-m}}{(c+d) f}-\frac {2^{\frac {3}{2}+m} B \operatorname {AppellF1}\left (\frac {1}{2},-\frac {1}{2}-m,1+m,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {d (1-\sin (e+f x))}{c+d}\right ) \cos (e+f x) (1+\sin (e+f x))^{-\frac {1}{2}-m} (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{-m} \left (\frac {c+d \sin (e+f x)}{c+d}\right )^m}{(c+d) f} \] Output:

-2^(1/2+m)*a*(A-B)*cos(f*x+e)*hypergeom([1/2, 1/2-m],[3/2],(c-d)*(1-sin(f* 
x+e))/(2*c+2*d*sin(f*x+e)))*(a+a*sin(f*x+e))^(-1+m)*((c+d)*(1+sin(f*x+e))/ 
(c+d*sin(f*x+e)))^(1/2-m)/(c+d)/f/((c+d*sin(f*x+e))^m)-2^(3/2+m)*B*AppellF 
1(1/2,1+m,-1/2-m,3/2,d*(1-sin(f*x+e))/(c+d),1/2-1/2*sin(f*x+e))*cos(f*x+e) 
*(1+sin(f*x+e))^(-1/2-m)*(a+a*sin(f*x+e))^m*((c+d*sin(f*x+e))/(c+d))^m/(c+ 
d)/f/((c+d*sin(f*x+e))^m)
                                                                                    
                                                                                    
 

Mathematica [F]

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

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

Output:

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

Rubi [A] (warning: unable to verify)

Time = 0.74 (sec) , antiderivative size = 287, normalized size of antiderivative = 1.07, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {3042, 3466, 3042, 3267, 142, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3466

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3267

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

\(\Big \downarrow \) 142

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

\(\Big \downarrow \) 157

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 156

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

\(\Big \downarrow \) 155

\(\displaystyle \frac {\sqrt {2} B \sqrt {1-\sin (e+f x)} \cos (e+f x) (a \sin (e+f x)+a)^{m+1} (c+d \sin (e+f x))^{-m} \left (\frac {c+d \sin (e+f x)}{c-d}\right )^m \operatorname {AppellF1}\left (m+\frac {3}{2},\frac {1}{2},m+1,m+\frac {5}{2},\frac {1}{2} (\sin (e+f x)+1),-\frac {d (\sin (e+f x)+1)}{c-d}\right )}{f (2 m+3) (c-d) (a-a \sin (e+f x))}-\frac {a 2^{m+\frac {1}{2}} (A-B) \cos (e+f x) (a \sin (e+f x)+a)^{m-1} \left (\frac {(c+d) (\sin (e+f x)+1)}{c+d \sin (e+f x)}\right )^{\frac {1}{2}-m} (c+d \sin (e+f x))^{-m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2}-m,\frac {3}{2},\frac {(c-d) (1-\sin (e+f x))}{2 (c+d \sin (e+f x))}\right )}{f (c+d)}\)

Input:

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

Output:

-((2^(1/2 + m)*a*(A - B)*Cos[e + f*x]*Hypergeometric2F1[1/2, 1/2 - m, 3/2, 
 ((c - d)*(1 - Sin[e + f*x]))/(2*(c + d*Sin[e + f*x]))]*(a + a*Sin[e + f*x 
])^(-1 + m)*(((c + d)*(1 + Sin[e + f*x]))/(c + d*Sin[e + f*x]))^(1/2 - m)) 
/((c + d)*f*(c + d*Sin[e + f*x])^m)) + (Sqrt[2]*B*AppellF1[3/2 + m, 1/2, 1 
 + m, 5/2 + m, (1 + Sin[e + f*x])/2, -((d*(1 + Sin[e + f*x]))/(c - d))]*Co 
s[e + f*x]*Sqrt[1 - Sin[e + f*x]]*(a + a*Sin[e + f*x])^(1 + m)*((c + d*Sin 
[e + f*x])/(c - d))^m)/((c - d)*f*(3 + 2*m)*(a - a*Sin[e + f*x])*(c + d*Si 
n[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 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 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 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]
 

rule 3466

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

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

Output:

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

Reduce [F]

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

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

Output:

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)*a + 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)*b