\(\int (a+b \sec ^2(e+f x))^p (d \sin (e+f x))^m \, dx\) [132]

Optimal result

Integrand size = 25, antiderivative size = 123 \[ \int \left (a+b \sec ^2(e+f x)\right )^p (d \sin (e+f x))^m \, dx=\frac {\operatorname {AppellF1}\left (\frac {1+m}{2},\frac {1}{2}+p,-p,\frac {3+m}{2},\sin ^2(e+f x),\frac {a \sin ^2(e+f x)}{a+b}\right ) \cos ^2(e+f x)^{\frac {1}{2}+p} \left (a+b \sec ^2(e+f x)\right )^p (d \sin (e+f x))^m \left (\frac {a+b-a \sin ^2(e+f x)}{a+b}\right )^{-p} \tan (e+f x)}{f (1+m)} \] Output:

AppellF1(1/2+1/2*m,1/2+p,-p,3/2+1/2*m,sin(f*x+e)^2,a*sin(f*x+e)^2/(a+b))*( 
cos(f*x+e)^2)^(1/2+p)*(a+b*sec(f*x+e)^2)^p*(d*sin(f*x+e))^m*tan(f*x+e)/f/( 
1+m)/(((a+b-a*sin(f*x+e)^2)/(a+b))^p)
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(286\) vs. \(2(123)=246\).

Time = 3.93 (sec) , antiderivative size = 286, normalized size of antiderivative = 2.33 \[ \int \left (a+b \sec ^2(e+f x)\right )^p (d \sin (e+f x))^m \, dx=\frac {\operatorname {AppellF1}\left (\frac {1+m}{2},\frac {2+m}{2},-p,\frac {3+m}{2},-\tan ^2(e+f x),-\frac {b \tan ^2(e+f x)}{a+b}\right ) \cos (e+f x) \left (a+b \sec ^2(e+f x)\right )^p \sin (e+f x) (d \sin (e+f x))^m}{f (1+m) \left (\operatorname {AppellF1}\left (\frac {1+m}{2},\frac {2+m}{2},-p,\frac {3+m}{2},-\tan ^2(e+f x),-\frac {b \tan ^2(e+f x)}{a+b}\right )-\frac {\left (-2 b p \operatorname {AppellF1}\left (\frac {3+m}{2},\frac {2+m}{2},1-p,\frac {5+m}{2},-\tan ^2(e+f x),-\frac {b \tan ^2(e+f x)}{a+b}\right )+(a+b) (2+m) \operatorname {AppellF1}\left (\frac {3+m}{2},\frac {4+m}{2},-p,\frac {5+m}{2},-\tan ^2(e+f x),-\frac {b \tan ^2(e+f x)}{a+b}\right )\right ) \tan ^2(e+f x)}{(a+b) (3+m)}\right )} \] Input:

Integrate[(a + b*Sec[e + f*x]^2)^p*(d*Sin[e + f*x])^m,x]
 

Output:

(AppellF1[(1 + m)/2, (2 + m)/2, -p, (3 + m)/2, -Tan[e + f*x]^2, -((b*Tan[e 
 + f*x]^2)/(a + b))]*Cos[e + f*x]*(a + b*Sec[e + f*x]^2)^p*Sin[e + f*x]*(d 
*Sin[e + f*x])^m)/(f*(1 + m)*(AppellF1[(1 + m)/2, (2 + m)/2, -p, (3 + m)/2 
, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/(a + b))] - ((-2*b*p*AppellF1[(3 + 
 m)/2, (2 + m)/2, 1 - p, (5 + m)/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/ 
(a + b))] + (a + b)*(2 + m)*AppellF1[(3 + m)/2, (4 + m)/2, -p, (5 + m)/2, 
-Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/(a + b))])*Tan[e + f*x]^2)/((a + b)* 
(3 + m))))
 

Rubi [F]

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.

\(\displaystyle \int (d \sin (e+f x))^m \left (a+b \sec ^2(e+f x)\right )^p \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (d \sin (e+f x))^m \left (a+b \sec (e+f x)^2\right )^pdx\)

\(\Big \downarrow \) 4623

\(\displaystyle \int (d \sin (e+f x))^m \left (a+b \sec ^2(e+f x)\right )^pdx\)

Input:

Int[(a + b*Sec[e + f*x]^2)^p*(d*Sin[e + f*x])^m,x]
 

Output:

$Aborted
 

Defintions of rubi rules used

rule 3042

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

rule 4623

Int[((a_) + (b_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*((d_.)*sin[(e 
_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Unintegrable[(a + b*(c*Sec[e + f*x]) 
^n)^p*(d*Sin[e + f*x])^m, x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x]
 

Maple [F]

\[\int \left (a +b \sec \left (f x +e \right )^{2}\right )^{p} \left (d \sin \left (f x +e \right )\right )^{m}d x\]

Input:

int((a+b*sec(f*x+e)^2)^p*(d*sin(f*x+e))^m,x)
 

Output:

int((a+b*sec(f*x+e)^2)^p*(d*sin(f*x+e))^m,x)
 

Fricas [F]

\[ \int \left (a+b \sec ^2(e+f x)\right )^p (d \sin (e+f x))^m \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \left (d \sin \left (f x + e\right )\right )^{m} \,d x } \] Input:

integrate((a+b*sec(f*x+e)^2)^p*(d*sin(f*x+e))^m,x, algorithm="fricas")
 

Output:

integral((b*sec(f*x + e)^2 + a)^p*(d*sin(f*x + e))^m, x)
 

Sympy [F(-1)]

Timed out. \[ \int \left (a+b \sec ^2(e+f x)\right )^p (d \sin (e+f x))^m \, dx=\text {Timed out} \] Input:

integrate((a+b*sec(f*x+e)**2)**p*(d*sin(f*x+e))**m,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \left (a+b \sec ^2(e+f x)\right )^p (d \sin (e+f x))^m \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \left (d \sin \left (f x + e\right )\right )^{m} \,d x } \] Input:

integrate((a+b*sec(f*x+e)^2)^p*(d*sin(f*x+e))^m,x, algorithm="maxima")
 

Output:

integrate((b*sec(f*x + e)^2 + a)^p*(d*sin(f*x + e))^m, x)
 

Giac [F]

\[ \int \left (a+b \sec ^2(e+f x)\right )^p (d \sin (e+f x))^m \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \left (d \sin \left (f x + e\right )\right )^{m} \,d x } \] Input:

integrate((a+b*sec(f*x+e)^2)^p*(d*sin(f*x+e))^m,x, algorithm="giac")
 

Output:

integrate((b*sec(f*x + e)^2 + a)^p*(d*sin(f*x + e))^m, x)
 

Mupad [F(-1)]

Timed out. \[ \int \left (a+b \sec ^2(e+f x)\right )^p (d \sin (e+f x))^m \, dx=\int {\left (d\,\sin \left (e+f\,x\right )\right )}^m\,{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^p \,d x \] Input:

int((d*sin(e + f*x))^m*(a + b/cos(e + f*x)^2)^p,x)
 

Output:

int((d*sin(e + f*x))^m*(a + b/cos(e + f*x)^2)^p, x)
 

Reduce [F]

\[ \int \left (a+b \sec ^2(e+f x)\right )^p (d \sin (e+f x))^m \, dx=d^{m} \left (\int \sin \left (f x +e \right )^{m} \left (\sec \left (f x +e \right )^{2} b +a \right )^{p}d x \right ) \] Input:

int((a+b*sec(f*x+e)^2)^p*(d*sin(f*x+e))^m,x)
 

Output:

d**m*int(sin(e + f*x)**m*(sec(e + f*x)**2*b + a)**p,x)