\(\int (g \sec (e+f x))^p (a+a \sin (e+f x))^m (c+d \sin (e+f x))^n \, dx\) [1045]

Optimal result

Integrand size = 35, antiderivative size = 176 \[ \int (g \sec (e+f x))^p (a+a \sin (e+f x))^m (c+d \sin (e+f x))^n \, dx=-\frac {2^{\frac {1}{2}+m-\frac {p}{2}} \operatorname {AppellF1}\left (\frac {1-p}{2},\frac {1}{2} (1-2 m+p),-n,\frac {3-p}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {d (1-\sin (e+f x))}{c+d}\right ) \sec (e+f x) (g \sec (e+f x))^p (1-\sin (e+f x)) (1+\sin (e+f x))^{\frac {1}{2} (1-2 m+p)} (a+a \sin (e+f x))^m (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c+d}\right )^{-n}}{f (1-p)} \] Output:

-2^(1/2+m-1/2*p)*AppellF1(1/2-1/2*p,-n,1/2-m+1/2*p,3/2-1/2*p,d*(1-sin(f*x+ 
e))/(c+d),1/2-1/2*sin(f*x+e))*sec(f*x+e)*(g*sec(f*x+e))^p*(1-sin(f*x+e))*( 
1+sin(f*x+e))^(1/2-m+1/2*p)*(a+a*sin(f*x+e))^m*(c+d*sin(f*x+e))^n/f/(1-p)/ 
(((c+d*sin(f*x+e))/(c+d))^n)
 

Mathematica [F]

\[ \int (g \sec (e+f x))^p (a+a \sin (e+f x))^m (c+d \sin (e+f x))^n \, dx=\int (g \sec (e+f x))^p (a+a \sin (e+f x))^m (c+d \sin (e+f x))^n \, dx \] Input:

Integrate[(g*Sec[e + f*x])^p*(a + a*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^n 
,x]
 

Output:

Integrate[(g*Sec[e + f*x])^p*(a + a*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^n 
, x]
 

Rubi [A] (warning: unable to verify)

Time = 0.72 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.26, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 3405, 3042, 3400, 157, 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.

\(\displaystyle \int (a \sin (e+f x)+a)^m (g \sec (e+f x))^p (c+d \sin (e+f x))^n \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (a \sin (e+f x)+a)^m (g \sec (e+f x))^p (c+d \sin (e+f x))^ndx\)

\(\Big \downarrow \) 3405

\(\displaystyle (g \cos (e+f x))^p (g \sec (e+f x))^p \int (g \cos (e+f x))^{-p} (\sin (e+f x) a+a)^m (c+d \sin (e+f x))^ndx\)

\(\Big \downarrow \) 3042

\(\displaystyle (g \cos (e+f x))^p (g \sec (e+f x))^p \int (g \cos (e+f x))^{-p} (\sin (e+f x) a+a)^m (c+d \sin (e+f x))^ndx\)

\(\Big \downarrow \) 3400

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

\(\Big \downarrow \) 157

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

\(\Big \downarrow \) 156

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

\(\Big \downarrow \) 155

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

Input:

Int[(g*Sec[e + f*x])^p*(a + a*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^n,x]
 

Output:

(2^(1/2 - p/2)*AppellF1[(1 + 2*m - p)/2, (1 + p)/2, -n, (3 + 2*m - p)/2, ( 
1 + Sin[e + f*x])/2, -((d*(1 + Sin[e + f*x]))/(c - d))]*Sec[e + f*x]*(g*Se 
c[e + f*x])^p*(1 - Sin[e + f*x])^((1 + p)/2)*(a - a*Sin[e + f*x])^((-1 - p 
)/2 + (1 + p)/2)*(a + a*Sin[e + f*x])^((1 + 2*m - p)/2 + (1 + p)/2)*(c + d 
*Sin[e + f*x])^n)/(a*f*(1 + 2*m - 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 157

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) 
^(p_), x_] :> Simp[(c + d*x)^FracPart[n]/(Simplify[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*(e + 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] &&  !SimplerQ[c + d*x, a + b*x] &&  !Si 
mplerQ[e + f*x, a + b*x]
 

rule 3042

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

rule 3400

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[g* 
((g*Cos[e + f*x])^(p - 1)/(f*(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)*(c + d*x)^n, x], x, Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, 
 n, p}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[m]
 

rule 3405

Int[((g_.)*sec[(e_.) + (f_.)*(x_)])^(p_)*((a_.) + (b_.)*sin[(e_.) + (f_.)*( 
x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Sim 
p[g^(2*IntPart[p])*(g*Cos[e + f*x])^FracPart[p]*(g*Sec[e + f*x])^FracPart[p 
]   Int[(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^n/(g*Cos[e + f*x])^p), 
 x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p}, x] &&  !IntegerQ[p]
 

Maple [F]

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

Input:

int((g*sec(f*x+e))^p*(a+a*sin(f*x+e))^m*(c+d*sin(f*x+e))^n,x)
 

Output:

int((g*sec(f*x+e))^p*(a+a*sin(f*x+e))^m*(c+d*sin(f*x+e))^n,x)
 

Fricas [F]

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

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

Output:

integral((g*sec(f*x + e))^p*(a*sin(f*x + e) + a)^m*(d*sin(f*x + e) + c)^n, 
 x)
 

Sympy [F(-1)]

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

integrate((g*sec(f*x+e))**p*(a+a*sin(f*x+e))**m*(c+d*sin(f*x+e))**n,x)
 

Output:

Timed out
 

Maxima [F]

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

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

Output:

integrate((g*sec(f*x + e))^p*(a*sin(f*x + e) + a)^m*(d*sin(f*x + e) + c)^n 
, x)
 

Giac [F]

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

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

Output:

integrate((g*sec(f*x + e))^p*(a*sin(f*x + e) + a)^m*(d*sin(f*x + e) + c)^n 
, x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

int((g*sec(f*x+e))^p*(a+a*sin(f*x+e))^m*(c+d*sin(f*x+e))^n,x)
                                                                                    
                                                                                    
 

Output:

g**p*int(sec(e + f*x)**p*(sin(e + f*x)*d + c)**n*(sin(e + f*x)*a + a)**m,x 
)