\(\int (g \cos (e+f x))^{-5-2 m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n \, dx\) [180]

Optimal result

Integrand size = 40, antiderivative size = 88 \[ \int (g \cos (e+f x))^{-5-2 m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n \, dx=\frac {c^2 (g \cos (e+f x))^{-2 m} \operatorname {Hypergeometric2F1}\left (3,-2-m+n,-1-m+n,\frac {1}{2} (1-\sin (e+f x))\right ) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-2+n}}{8 f g^5 (2+m-n)} \] Output:

1/8*c^2*hypergeom([3, -2-m+n],[-1-m+n],1/2-1/2*sin(f*x+e))*(a+a*sin(f*x+e) 
)^m*(c-c*sin(f*x+e))^(-2+n)/f/g^5/(2+m-n)/((g*cos(f*x+e))^(2*m))
 

Mathematica [F]

\[ \int (g \cos (e+f x))^{-5-2 m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n \, dx=\int (g \cos (e+f x))^{-5-2 m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n \, dx \] Input:

Integrate[(g*Cos[e + f*x])^(-5 - 2*m)*(a + a*Sin[e + f*x])^m*(c - c*Sin[e 
+ f*x])^n,x]
 

Output:

Integrate[(g*Cos[e + f*x])^(-5 - 2*m)*(a + a*Sin[e + f*x])^m*(c - c*Sin[e 
+ f*x])^n, x]
 

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.05, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {3042, 3332, 27, 3042, 3146, 78}

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.

\(\displaystyle \int (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^n (g \cos (e+f x))^{-2 m-5} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^n (g \cos (e+f x))^{-2 m-5}dx\)

\(\Big \downarrow \) 3332

\(\displaystyle (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^m (g \cos (e+f x))^{-2 m} \int \frac {\sec ^5(e+f x) (c-c \sin (e+f x))^{n-m}}{g^5}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {(a \sin (e+f x)+a)^m (c-c \sin (e+f x))^m (g \cos (e+f x))^{-2 m} \int \sec ^5(e+f x) (c-c \sin (e+f x))^{n-m}dx}{g^5}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(a \sin (e+f x)+a)^m (c-c \sin (e+f x))^m (g \cos (e+f x))^{-2 m} \int \frac {(c-c \sin (e+f x))^{n-m}}{\cos (e+f x)^5}dx}{g^5}\)

\(\Big \downarrow \) 3146

\(\displaystyle -\frac {c^5 (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^m (g \cos (e+f x))^{-2 m} \int \frac {(c-c \sin (e+f x))^{-m+n-3}}{(\sin (e+f x) c+c)^3}d(-c \sin (e+f x))}{f g^5}\)

\(\Big \downarrow \) 78

\(\displaystyle \frac {c^2 (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{n-2} (g \cos (e+f x))^{-2 m} \operatorname {Hypergeometric2F1}\left (3,-m+n-2,-m+n-1,\frac {c-c \sin (e+f x)}{2 c}\right )}{8 f g^5 (m-n+2)}\)

Input:

Int[(g*Cos[e + f*x])^(-5 - 2*m)*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x] 
)^n,x]
 

Output:

(c^2*Hypergeometric2F1[3, -2 - m + n, -1 - m + n, (c - c*Sin[e + f*x])/(2* 
c)]*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^(-2 + n))/(8*f*g^5*(2 + m 
- n)*(g*Cos[e + f*x])^(2*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 78

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

rule 3042

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

rule 3146

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

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]

\[\int \left (g \cos \left (f x +e \right )\right )^{-5-2 m} \left (a +a \sin \left (f x +e \right )\right )^{m} \left (c -c \sin \left (f x +e \right )\right )^{n}d x\]

Input:

int((g*cos(f*x+e))^(-5-2*m)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^n,x)
 

Output:

int((g*cos(f*x+e))^(-5-2*m)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^n,x)
 

Fricas [F]

\[ \int (g \cos (e+f x))^{-5-2 m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n \, dx=\int { \left (g \cos \left (f x + e\right )\right )^{-2 \, m - 5} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{n} \,d x } \] Input:

integrate((g*cos(f*x+e))^(-5-2*m)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^n,x, 
 algorithm="fricas")
 

Output:

integral((g*cos(f*x + e))^(-2*m - 5)*(a*sin(f*x + e) + a)^m*(-c*sin(f*x + 
e) + c)^n, x)
 

Sympy [F(-1)]

Timed out. \[ \int (g \cos (e+f x))^{-5-2 m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n \, dx=\text {Timed out} \] Input:

integrate((g*cos(f*x+e))**(-5-2*m)*(a+a*sin(f*x+e))**m*(c-c*sin(f*x+e))**n 
,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int (g \cos (e+f x))^{-5-2 m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n \, dx=\int { \left (g \cos \left (f x + e\right )\right )^{-2 \, m - 5} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{n} \,d x } \] Input:

integrate((g*cos(f*x+e))^(-5-2*m)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^n,x, 
 algorithm="maxima")
 

Output:

integrate((g*cos(f*x + e))^(-2*m - 5)*(a*sin(f*x + e) + a)^m*(-c*sin(f*x + 
 e) + c)^n, x)
 

Giac [F]

\[ \int (g \cos (e+f x))^{-5-2 m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n \, dx=\int { \left (g \cos \left (f x + e\right )\right )^{-2 \, m - 5} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{n} \,d x } \] Input:

integrate((g*cos(f*x+e))^(-5-2*m)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^n,x, 
 algorithm="giac")
 

Output:

integrate((g*cos(f*x + e))^(-2*m - 5)*(a*sin(f*x + e) + a)^m*(-c*sin(f*x + 
 e) + c)^n, x)
 

Mupad [F(-1)]

Timed out. \[ \int (g \cos (e+f x))^{-5-2 m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n \, dx=\int \frac {{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^n}{{\left (g\,\cos \left (e+f\,x\right )\right )}^{2\,m+5}} \,d x \] Input:

int(((a + a*sin(e + f*x))^m*(c - c*sin(e + f*x))^n)/(g*cos(e + f*x))^(2*m 
+ 5),x)
 

Output:

int(((a + a*sin(e + f*x))^m*(c - c*sin(e + f*x))^n)/(g*cos(e + f*x))^(2*m 
+ 5), x)
 

Reduce [F]

\[ \int (g \cos (e+f x))^{-5-2 m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n \, dx=\frac {\int \frac {\left (a +a \sin \left (f x +e \right )\right )^{m} \left (-\sin \left (f x +e \right ) c +c \right )^{n}}{\cos \left (f x +e \right )^{2 m} \cos \left (f x +e \right )^{5}}d x}{g^{2 m} g^{5}} \] Input:

int((g*cos(f*x+e))^(-5-2*m)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^n,x)
 

Output:

int(((sin(e + f*x)*a + a)**m*( - sin(e + f*x)*c + c)**n)/(cos(e + f*x)**(2 
*m)*cos(e + f*x)**5),x)/(g**(2*m)*g**5)