\(\int (g \cos (e+f x))^{-1-m-n} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{1+n} \, dx\) [184]

Optimal result

Integrand size = 45, antiderivative size = 133 \[ \int (g \cos (e+f x))^{-1-m-n} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{1+n} \, dx=\frac {2^{1-\frac {m}{2}+\frac {n}{2}} (g \cos (e+f x))^{-m-n} \operatorname {Hypergeometric2F1}\left (\frac {m-n}{2},\frac {m-n}{2},\frac {1}{2} (2+m-n),\frac {1}{2} (1+\sin (e+f x))\right ) (1-\sin (e+f x))^{\frac {1}{2} (-2+m-n)} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{1+n}}{f g (m-n)} \] Output:

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 14.64 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.56 \[ \int (g \cos (e+f x))^{-1-m-n} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{1+n} \, dx=\frac {i c (g \cos (e+f x))^{-m-n} \left (\operatorname {Hypergeometric2F1}\left (1,1-m+n,2-m+n,-\frac {i \left (-1+\tan \left (\frac {1}{2} (e+f x)\right )\right )}{1+\tan \left (\frac {1}{2} (e+f x)\right )}\right )-\operatorname {Hypergeometric2F1}\left (1,1-m+n,2-m+n,\frac {i \left (-1+\tan \left (\frac {1}{2} (e+f x)\right )\right )}{1+\tan \left (\frac {1}{2} (e+f x)\right )}\right )\right ) (-1+\sin (e+f x)) (a (1+\sin (e+f x)))^m (c-c \sin (e+f x))^n}{f g (-1+m-n) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )} \] Input:

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

Output:

(I*c*(g*Cos[e + f*x])^(-m - n)*(Hypergeometric2F1[1, 1 - m + n, 2 - m + n, 
 ((-I)*(-1 + Tan[(e + f*x)/2]))/(1 + Tan[(e + f*x)/2])] - Hypergeometric2F 
1[1, 1 - m + n, 2 - m + n, (I*(-1 + Tan[(e + f*x)/2]))/(1 + Tan[(e + f*x)/ 
2])])*(-1 + Sin[e + f*x])*(a*(1 + Sin[e + f*x]))^m*(c - c*Sin[e + f*x])^n) 
/(f*g*(-1 + m - n)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(Cos[(e + f*x)/2] 
 + Sin[(e + f*x)/2]))
 

Rubi [A] (warning: unable to verify)

Time = 0.68 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.38, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {3042, 3332, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3332

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3168

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

\(\Big \downarrow \) 80

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

\(\Big \downarrow \) 79

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

Input:

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

Output:

-((2^(1 + (-2 + m - n)/2)*(g*Cos[e + f*x])^(-m - n)*Hypergeometric2F1[(2 - 
 m + n)/2, (2 - m + n)/2, (4 - m + n)/2, (1 - Sin[e + f*x])/2]*(1 + Sin[e 
+ f*x])^((-m + n)/2)*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^(m + (-m 
+ n)/2 + (2 - m + n)/2)*(c + c*Sin[e + f*x])^((m - n)/2 + (-m + n)/2))/(f* 
g*(2 - m + n)))
 

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 3332

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((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]/(g^(2*IntPart[m])*(g*Cos[e + f*x])^(2*FracPart[m])))   Int[ 
(g*Cos[e + f*x])^(2*m + p)*(c + d*Sin[e + f*x])^(n - m), x], x] /; FreeQ[{a 
, b, c, d, e, f, g, m, n, p}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] 
&& (FractionQ[m] ||  !FractionQ[n])
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

Timed out. \[ \int (g \cos (e+f x))^{-1-m-n} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{1+n} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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