\(\int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^{-1-m} \, dx\) [214]

Optimal result

Integrand size = 40, antiderivative size = 168 \[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^{-1-m} \, dx=\frac {(A+B) \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1-m}}{f (1+2 m)}-\frac {2^{\frac {1}{2}-m} B \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (1+2 m),\frac {1}{2} (1+2 m),\frac {1}{2} (3+2 m),\frac {1}{2} (1+\sin (e+f x))\right ) (1-\sin (e+f x))^{\frac {1}{2} (-1+2 m)} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-m}}{c f (1+2 m)} \] Output:

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

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 12.30 (sec) , antiderivative size = 310, normalized size of antiderivative = 1.85 \[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^{-1-m} \, dx=\frac {2^{-1-m} \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^{-2 (-1-m)-2 m} (a (1+\sin (e+f x)))^m (c-c \sin (e+f x))^{-1-m} \left (\frac {1-\tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {\frac {1}{1+\cos (e+f x)}}}\right )^{2 m} \left (\frac {1-\tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {\sec ^2\left (\frac {1}{2} (e+f x)\right )}}\right )^{-2 m} \left (i B (1+2 m) \operatorname {Hypergeometric2F1}\left (1,-2 m,1-2 m,-\frac {i \left (-1+\tan \left (\frac {1}{2} (e+f x)\right )\right )}{1+\tan \left (\frac {1}{2} (e+f x)\right )}\right ) \left (-1+\tan \left (\frac {1}{2} (e+f x)\right )\right )-i B (1+2 m) \operatorname {Hypergeometric2F1}\left (1,-2 m,1-2 m,\frac {i \left (-1+\tan \left (\frac {1}{2} (e+f x)\right )\right )}{1+\tan \left (\frac {1}{2} (e+f x)\right )}\right ) \left (-1+\tan \left (\frac {1}{2} (e+f x)\right )\right )-2 (A+B) m \left (1+\tan \left (\frac {1}{2} (e+f x)\right )\right )\right )}{f m (1+2 m) \left (-1+\tan \left (\frac {1}{2} (e+f x)\right )\right )} \] Input:

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

Output:

(2^(-1 - m)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^(-2*(-1 - m) - 2*m)*(a*( 
1 + Sin[e + f*x]))^m*(c - c*Sin[e + f*x])^(-1 - m)*((1 - Tan[(e + f*x)/2]) 
/Sqrt[(1 + Cos[e + f*x])^(-1)])^(2*m)*(I*B*(1 + 2*m)*Hypergeometric2F1[1, 
-2*m, 1 - 2*m, ((-I)*(-1 + Tan[(e + f*x)/2]))/(1 + Tan[(e + f*x)/2])]*(-1 
+ Tan[(e + f*x)/2]) - I*B*(1 + 2*m)*Hypergeometric2F1[1, -2*m, 1 - 2*m, (I 
*(-1 + Tan[(e + f*x)/2]))/(1 + Tan[(e + f*x)/2])]*(-1 + Tan[(e + f*x)/2]) 
- 2*(A + B)*m*(1 + Tan[(e + f*x)/2])))/(f*m*(1 + 2*m)*((1 - Tan[(e + f*x)/ 
2])/Sqrt[Sec[(e + f*x)/2]^2])^(2*m)*(-1 + Tan[(e + f*x)/2]))
 

Rubi [A] (verified)

Time = 0.73 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.20, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 3451, 3042, 3224, 3042, 3168, 80, 79}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3451

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3224

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3168

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

\(\Big \downarrow \) 80

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

\(\Big \downarrow \) 79

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 79

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(( 
a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c - a*d))^n))*Hypergeometric2F1[-n, m + 1 
, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}, x] 
&&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] 
 ||  !(RationalQ[n] && GtQ[-d/(b*c - a*d), 0]))
 

rule 80

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(c 
 + d*x)^FracPart[n]/((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, x], x] /; FreeQ[{a, b, c, d, m, n}, x] &&  !IntegerQ[m] &&  !Integ 
erQ[n] && (RationalQ[m] ||  !SimplerQ[n + 1, m + 1])
 

rule 3042

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

rule 3168

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_.), x_Symbol] :> Simp[a^2*((g*Cos[e + f*x])^(p + 1)/(f*g*(a + b*Sin 
[e + f*x])^((p + 1)/2)*(a - b*Sin[e + f*x])^((p + 1)/2)))   Subst[Int[(a + 
b*x)^(m + (p - 1)/2)*(a - b*x)^((p - 1)/2), x], x, Sin[e + f*x]], x] /; Fre 
eQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[m]
 

rule 3224

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + ( 
f_.)*(x_)])^(n_), 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]/Cos[e + f*x]^(2*FracP 
art[m]))   Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m), x], x] /; F 
reeQ[{a, b, c, d, e, f, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] 
 && (FractionQ[m] ||  !FractionQ[n])
 

rule 3451

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + 
(f_.)*(x_)])*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Sim 
p[(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^n/( 
a*f*(2*m + 1))), x] + Simp[(a*B*(m - n) + A*b*(m + n + 1))/(a*b*(2*m + 1)) 
  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] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0 
] && (LtQ[m, -2^(-1)] || (ILtQ[m + n, 0] &&  !SumSimplerQ[n, 1])) && NeQ[2* 
m + 1, 0]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Giac [F]

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

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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