\(\int (g \sec (e+f x))^p (d \sin (e+f x))^n (a+a \sin (e+f x))^m \, dx\) [908]

Optimal result

Integrand size = 33, antiderivative size = 127 \[ \int (g \sec (e+f x))^p (d \sin (e+f x))^n (a+a \sin (e+f x))^m \, dx=\frac {\operatorname {AppellF1}\left (1+n,\frac {1+p}{2},\frac {1}{2} (1-2 m+p),2+n,\sin (e+f x),-\sin (e+f x)\right ) \sec (e+f x) (g \sec (e+f x))^p (1-\sin (e+f x))^{\frac {1+p}{2}} (d \sin (e+f x))^{1+n} (1+\sin (e+f x))^{\frac {1}{2} (1-2 m+p)} (a+a \sin (e+f x))^m}{d f (1+n)} \] Output:

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

Mathematica [F]

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

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

Output:

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

Rubi [A] (verified)

Time = 0.65 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.35, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {3042, 3405, 3042, 3365, 152, 152, 150}

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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int (a \sin (e+f x)+a)^m (d \sin (e+f x))^n (g \sec (e+f x))^pdx\)

\(\Big \downarrow \) 3405

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3365

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

\(\Big \downarrow \) 152

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

\(\Big \downarrow \) 152

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

\(\Big \downarrow \) 150

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

Input:

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

Output:

(AppellF1[1 + n, (1 + p)/2, (1 - 2*m + p)/2, 2 + n, Sin[e + f*x], -Sin[e + 
 f*x]]*Sec[e + f*x]*(g*Sec[e + f*x])^p*(1 - Sin[e + f*x])^((1 + p)/2)*(d*S 
in[e + f*x])^(1 + n)*(1 + Sin[e + f*x])^((1 - 2*m + 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))/(d*f*(1 + n))
 

Defintions of rubi rules used

rule 150

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

rule 152

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_ 
] :> Simp[c^IntPart[n]*((c + d*x)^FracPart[n]/(1 + d*(x/c))^FracPart[n]) 
Int[(b*x)^m*(1 + d*(x/c))^n*(e + f*x)^p, x], x] /; FreeQ[{b, c, d, e, f, m, 
 n, p}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] &&  !GtQ[c, 0]
 

rule 3042

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

rule 3365

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n 
_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), 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[(d*x)^n*(a + b*x)^(m + (p - 1)/2)*(a - b*x)^((p 
- 1)/2), x], x, Sin[e + f*x]], x] /; FreeQ[{a, b, 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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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