\(\int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^{-m} \, dx\) [215]

Optimal result

Integrand size = 38, antiderivative size = 159 \[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^{-m} \, dx=-\frac {B \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-m}}{f}+\frac {2^{\frac {1}{2}-m} (A+2 B m) \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (1+2 m),\frac {1}{2} (1+2 m),\frac {1}{2} (3+2 m),\frac {1}{2} (1+\sin (e+f x))\right ) (1-\sin (e+f x))^{\frac {1}{2} (-1+2 m)} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-m}}{f (1+2 m)} \] Output:

-B*cos(f*x+e)*(a+a*sin(f*x+e))^m/f/((c-c*sin(f*x+e))^m)+2^(1/2-m)*(2*B*m+A 
)*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/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 = 9.30 (sec) , antiderivative size = 402, normalized size of antiderivative = 2.53 \[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^{-m} \, dx=-\frac {i 2^{-1-m} \left (a \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2\right )^m (c-c \sin (e+f x))^{-m} (\cosh (m \log (2))+\sinh (m \log (2))) \left (\frac {A \operatorname {Hypergeometric2F1}\left (1,-2 m,1-2 m,-\frac {\cos (e+f x)+i (-1+\sin (e+f x))}{\cos (e+f x)+i (1+\sin (e+f x))}\right )}{m}-\frac {2 B \operatorname {Hypergeometric2F1}\left (2,1-2 m,2-2 m,-\frac {\cos (e+f x)+i (-1+\sin (e+f x))}{\cos (e+f x)+i (1+\sin (e+f x))}\right ) (\cos (e+f x)+i (-1+\sin (e+f x)))}{(-1+2 m) (\cos (e+f x)+i (1+\sin (e+f x)))}+(1-i \cos (e+f x)+\sin (e+f x))^{-2 m} \left (\frac {B \operatorname {Hypergeometric2F1}\left (2 m,1+2 m,2+2 m,\frac {1}{2} (1-i \cos (e+f x)+\sin (e+f x))\right ) (\cos (e+f x) (\cos (e+f x)+i \sin (e+f x)))^{2 m} (1-i \cos (e+f x)+\sin (e+f x))}{1+2 m}-\frac {4^{-m} A \operatorname {Hypergeometric2F1}\left (-2 m,-2 m,1-2 m,\frac {1}{2} (1+i \cos (e+f x)-\sin (e+f x))\right ) (\cosh (m \log (16))+\sinh (m \log (16)))}{m}\right )\right )}{f} \] Input:

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

Output:

((-I)*2^(-1 - m)*(a*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2)^m*(Cosh[m*Log 
[2]] + Sinh[m*Log[2]])*((A*Hypergeometric2F1[1, -2*m, 1 - 2*m, -((Cos[e + 
f*x] + I*(-1 + Sin[e + f*x]))/(Cos[e + f*x] + I*(1 + Sin[e + f*x])))])/m - 
 (2*B*Hypergeometric2F1[2, 1 - 2*m, 2 - 2*m, -((Cos[e + f*x] + I*(-1 + Sin 
[e + f*x]))/(Cos[e + f*x] + I*(1 + Sin[e + f*x])))]*(Cos[e + f*x] + I*(-1 
+ Sin[e + f*x])))/((-1 + 2*m)*(Cos[e + f*x] + I*(1 + Sin[e + f*x]))) + ((B 
*Hypergeometric2F1[2*m, 1 + 2*m, 2 + 2*m, (1 - I*Cos[e + f*x] + Sin[e + f* 
x])/2]*(Cos[e + f*x]*(Cos[e + f*x] + I*Sin[e + f*x]))^(2*m)*(1 - I*Cos[e + 
 f*x] + Sin[e + f*x]))/(1 + 2*m) - (A*Hypergeometric2F1[-2*m, -2*m, 1 - 2* 
m, (1 + I*Cos[e + f*x] - Sin[e + f*x])/2]*(Cosh[m*Log[16]] + Sinh[m*Log[16 
]]))/(4^m*m))/(1 - I*Cos[e + f*x] + Sin[e + f*x])^(2*m)))/(f*(c - c*Sin[e 
+ f*x])^m)
 

Rubi [A] (verified)

Time = 0.64 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.23, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {3042, 3452, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3452

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3224

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3168

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

\(\Big \downarrow \) 80

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

\(\Big \downarrow \) 79

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

Input:

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

Output:

-((B*Cos[e + f*x]*(a + a*Sin[e + f*x])^m)/(f*(c - c*Sin[e + f*x])^m)) + (2 
^(1/2 - m)*c*(A + 2*B*m)*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 3452

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[(-B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^n/(f*(m + 
n + 1))), x] - Simp[(B*c*(m - n) - A*d*(m + n + 1))/(d*(m + n + 1))   Int[( 
a + b*Sin[e + f*x])^m*(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)] && NeQ[m + n + 1, 0]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

integrate((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e))/((c-c*sin(f*x+e))^m),x, algo 
rithm="fricas")
 

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

integrate((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e))/((c-c*sin(f*x+e))^m),x, algo 
rithm="maxima")
 

Output:

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

Giac [F]

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

integrate((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e))/((c-c*sin(f*x+e))^m),x, algo 
rithm="giac")
 

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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