\(\int \frac {(g \cos (e+f x))^p (c+d \sin (e+f x))^n}{(a+a \sin (e+f x))^4} \, dx\) [1044]

Optimal result

Integrand size = 35, antiderivative size = 149 \[ \int \frac {(g \cos (e+f x))^p (c+d \sin (e+f x))^n}{(a+a \sin (e+f x))^4} \, dx=-\frac {2^{-\frac {7}{2}+\frac {p}{2}} \operatorname {AppellF1}\left (\frac {1+p}{2},\frac {9-p}{2},-n,\frac {3+p}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {d (1-\sin (e+f x))}{c+d}\right ) (g \cos (e+f x))^{1+p} (1+\sin (e+f x))^{-4+\frac {7-p}{2}} (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c+d}\right )^{-n}}{a^4 f g (1+p)} \] Output:

-2^(-7/2+1/2*p)*AppellF1(1/2*p+1/2,-n,9/2-1/2*p,3/2+1/2*p,d*(1-sin(f*x+e)) 
/(c+d),1/2-1/2*sin(f*x+e))*(g*cos(f*x+e))^(p+1)*(1+sin(f*x+e))^(-1/2-1/2*p 
)*(c+d*sin(f*x+e))^n/a^4/f/g/(p+1)/(((c+d*sin(f*x+e))/(c+d))^n)
 

Mathematica [F]

\[ \int \frac {(g \cos (e+f x))^p (c+d \sin (e+f x))^n}{(a+a \sin (e+f x))^4} \, dx=\int \frac {(g \cos (e+f x))^p (c+d \sin (e+f x))^n}{(a+a \sin (e+f x))^4} \, dx \] Input:

Integrate[((g*Cos[e + f*x])^p*(c + d*Sin[e + f*x])^n)/(a + a*Sin[e + f*x]) 
^4,x]
 

Output:

Integrate[((g*Cos[e + f*x])^p*(c + d*Sin[e + f*x])^n)/(a + a*Sin[e + f*x]) 
^4, x]
 

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.15, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {3042, 3399, 156, 155}

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.

\(\displaystyle \int \frac {(g \cos (e+f x))^p (c+d \sin (e+f x))^n}{(a \sin (e+f x)+a)^4} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(g \cos (e+f x))^p (c+d \sin (e+f x))^n}{(a \sin (e+f x)+a)^4}dx\)

\(\Big \downarrow \) 3399

\(\displaystyle \frac {g (1-\sin (e+f x))^{\frac {1-p}{2}} (\sin (e+f x)+1)^{\frac {1-p}{2}} (g \cos (e+f x))^{p-1} \int (1-\sin (e+f x))^{\frac {p-1}{2}} (\sin (e+f x)+1)^{\frac {p-9}{2}} (c+d \sin (e+f x))^nd\sin (e+f x)}{a^4 f}\)

\(\Big \downarrow \) 156

\(\displaystyle \frac {g (1-\sin (e+f x))^{\frac {1-p}{2}} (\sin (e+f x)+1)^{\frac {1-p}{2}} (g \cos (e+f x))^{p-1} (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c+d}\right )^{-n} \int (1-\sin (e+f x))^{\frac {p-1}{2}} (\sin (e+f x)+1)^{\frac {p-9}{2}} \left (\frac {c}{c+d}+\frac {d \sin (e+f x)}{c+d}\right )^nd\sin (e+f x)}{a^4 f}\)

\(\Big \downarrow \) 155

\(\displaystyle -\frac {g 2^{\frac {p-7}{2}} (1-\sin (e+f x))^{\frac {1-p}{2}+\frac {p+1}{2}} (\sin (e+f x)+1)^{\frac {1-p}{2}} (g \cos (e+f x))^{p-1} (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c+d}\right )^{-n} \operatorname {AppellF1}\left (\frac {p+1}{2},\frac {9-p}{2},-n,\frac {p+3}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {d (1-\sin (e+f x))}{c+d}\right )}{a^4 f (p+1)}\)

Input:

Int[((g*Cos[e + f*x])^p*(c + d*Sin[e + f*x])^n)/(a + a*Sin[e + f*x])^4,x]
 

Output:

-((2^((-7 + p)/2)*g*AppellF1[(1 + p)/2, (9 - p)/2, -n, (3 + p)/2, (1 - Sin 
[e + f*x])/2, (d*(1 - Sin[e + f*x]))/(c + d)]*(g*Cos[e + f*x])^(-1 + p)*(1 
 - Sin[e + f*x])^((1 - p)/2 + (1 + p)/2)*(1 + Sin[e + f*x])^((1 - p)/2)*(c 
 + d*Sin[e + f*x])^n)/(a^4*f*(1 + p)*((c + d*Sin[e + f*x])/(c + d))^n))
 

Defintions of rubi rules used

rule 155

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) 
^(p_), x_] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*Simplify[b/(b*c - a*d)]^n* 
Simplify[b/(b*e - a*f)]^p))*AppellF1[m + 1, -n, -p, m + 2, (-d)*((a + b*x)/ 
(b*c - a*d)), (-f)*((a + b*x)/(b*e - a*f))], x] /; FreeQ[{a, b, c, d, e, f, 
 m, n, p}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] &&  !IntegerQ[p] && GtQ[Sim 
plify[b/(b*c - a*d)], 0] && GtQ[Simplify[b/(b*e - a*f)], 0] &&  !(GtQ[Simpl 
ify[d/(d*a - c*b)], 0] && GtQ[Simplify[d/(d*e - c*f)], 0] && SimplerQ[c + d 
*x, a + b*x]) &&  !(GtQ[Simplify[f/(f*a - e*b)], 0] && GtQ[Simplify[f/(f*c 
- e*d)], 0] && SimplerQ[e + f*x, a + b*x])
 

rule 156

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) 
^(p_), x_] :> Simp[(e + f*x)^FracPart[p]/(Simplify[b/(b*e - a*f)]^IntPart[p 
]*(b*((e + f*x)/(b*e - a*f)))^FracPart[p])   Int[(a + b*x)^m*(c + d*x)^n*Si 
mp[b*(e/(b*e - a*f)) + b*f*(x/(b*e - a*f)), x]^p, x], x] /; FreeQ[{a, b, c, 
 d, e, f, m, n, p}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] &&  !IntegerQ[p] & 
& GtQ[Simplify[b/(b*c - a*d)], 0] &&  !GtQ[Simplify[b/(b*e - a*f)], 0]
 

rule 3042

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

rule 3399

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^ 
m*g*((g*Cos[e + f*x])^(p - 1)/(f*(1 + Sin[e + f*x])^((p - 1)/2)*(1 - Sin[e 
+ f*x])^((p - 1)/2)))   Subst[Int[(1 + (b/a)*x)^(m + (p - 1)/2)*(1 - (b/a)* 
x)^((p - 1)/2)*(c + d*x)^n, x], x, Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, 
e, f, n, p}, x] && EqQ[a^2 - b^2, 0] && IntegerQ[m]
 

Maple [F]

\[\int \frac {\left (g \cos \left (f x +e \right )\right )^{p} \left (c +d \sin \left (f x +e \right )\right )^{n}}{\left (a +a \sin \left (f x +e \right )\right )^{4}}d x\]

Input:

int((g*cos(f*x+e))^p*(c+d*sin(f*x+e))^n/(a+a*sin(f*x+e))^4,x)
 

Output:

int((g*cos(f*x+e))^p*(c+d*sin(f*x+e))^n/(a+a*sin(f*x+e))^4,x)
 

Fricas [F]

\[ \int \frac {(g \cos (e+f x))^p (c+d \sin (e+f x))^n}{(a+a \sin (e+f x))^4} \, dx=\int { \frac {\left (g \cos \left (f x + e\right )\right )^{p} {\left (d \sin \left (f x + e\right ) + c\right )}^{n}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{4}} \,d x } \] Input:

integrate((g*cos(f*x+e))^p*(c+d*sin(f*x+e))^n/(a+a*sin(f*x+e))^4,x, algori 
thm="fricas")
 

Output:

integral((g*cos(f*x + e))^p*(d*sin(f*x + e) + c)^n/(a^4*cos(f*x + e)^4 - 8 
*a^4*cos(f*x + e)^2 + 8*a^4 - 4*(a^4*cos(f*x + e)^2 - 2*a^4)*sin(f*x + e)) 
, x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(g \cos (e+f x))^p (c+d \sin (e+f x))^n}{(a+a \sin (e+f x))^4} \, dx=\text {Timed out} \] Input:

integrate((g*cos(f*x+e))**p*(c+d*sin(f*x+e))**n/(a+a*sin(f*x+e))**4,x)
 

Output:

Timed out
 

Maxima [F(-1)]

Timed out. \[ \int \frac {(g \cos (e+f x))^p (c+d \sin (e+f x))^n}{(a+a \sin (e+f x))^4} \, dx=\text {Timed out} \] Input:

integrate((g*cos(f*x+e))^p*(c+d*sin(f*x+e))^n/(a+a*sin(f*x+e))^4,x, algori 
thm="maxima")
 

Output:

Timed out
 

Giac [F]

\[ \int \frac {(g \cos (e+f x))^p (c+d \sin (e+f x))^n}{(a+a \sin (e+f x))^4} \, dx=\int { \frac {\left (g \cos \left (f x + e\right )\right )^{p} {\left (d \sin \left (f x + e\right ) + c\right )}^{n}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{4}} \,d x } \] Input:

integrate((g*cos(f*x+e))^p*(c+d*sin(f*x+e))^n/(a+a*sin(f*x+e))^4,x, algori 
thm="giac")
 

Output:

integrate((g*cos(f*x + e))^p*(d*sin(f*x + e) + c)^n/(a*sin(f*x + e) + a)^4 
, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(g \cos (e+f x))^p (c+d \sin (e+f x))^n}{(a+a \sin (e+f x))^4} \, dx=\int \frac {{\left (g\,\cos \left (e+f\,x\right )\right )}^p\,{\left (c+d\,\sin \left (e+f\,x\right )\right )}^n}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^4} \,d x \] Input:

int(((g*cos(e + f*x))^p*(c + d*sin(e + f*x))^n)/(a + a*sin(e + f*x))^4,x)
 

Output:

int(((g*cos(e + f*x))^p*(c + d*sin(e + f*x))^n)/(a + a*sin(e + f*x))^4, x)
 

Reduce [F]

\[ \int \frac {(g \cos (e+f x))^p (c+d \sin (e+f x))^n}{(a+a \sin (e+f x))^4} \, dx=\frac {g^{p} \left (\int \frac {\left (\sin \left (f x +e \right ) d +c \right )^{n} \cos \left (f x +e \right )^{p}}{\sin \left (f x +e \right )^{4}+4 \sin \left (f x +e \right )^{3}+6 \sin \left (f x +e \right )^{2}+4 \sin \left (f x +e \right )+1}d x \right )}{a^{4}} \] Input:

int((g*cos(f*x+e))^p*(c+d*sin(f*x+e))^n/(a+a*sin(f*x+e))^4,x)
 

Output:

(g**p*int(((sin(e + f*x)*d + c)**n*cos(e + f*x)**p)/(sin(e + f*x)**4 + 4*s 
in(e + f*x)**3 + 6*sin(e + f*x)**2 + 4*sin(e + f*x) + 1),x))/a**4