\(\int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{1-m} \, dx\) [431]

Optimal result

Integrand size = 30, antiderivative size = 115 \[ \int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{1-m} \, dx=\frac {2^{\frac {3}{2}-m} \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-1+2 m),\frac {1}{2} (1+2 m),\frac {1}{2} (3+2 m),\frac {1}{2} (1+\sin (e+f x))\right ) (1-\sin (e+f x))^{\frac {1}{2} (-3+2 m)} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{1-m}}{f (1+2 m)} \] Output:

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

Mathematica [A] (verified)

Time = 0.47 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.99 \[ \int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{1-m} \, dx=\frac {2^{\frac {1}{2}+m} c \operatorname {Hypergeometric2F1}\left (\frac {1}{2}-m,\frac {3}{2}-m,\frac {5}{2}-m,\frac {1}{2} (1-\sin (e+f x))\right ) \sec (e+f x) (-1+\sin (e+f x))^2 (1+\sin (e+f x))^{\frac {1}{2}-m} (a (1+\sin (e+f x)))^m (c-c \sin (e+f x))^{-m}}{f (-3+2 m)} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.33, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3224

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3168

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

\(\Big \downarrow \) 80

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

\(\Big \downarrow \) 79

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

Input:

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

Output:

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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