\(\int (d \sec (e+f x))^m (a+b \sec ^2(e+f x))^p \, dx\) [298]

Optimal result

Integrand size = 25, antiderivative size = 112 \[ \int (d \sec (e+f x))^m \left (a+b \sec ^2(e+f x)\right )^p \, dx=\frac {\operatorname {AppellF1}\left (\frac {m}{2},\frac {1}{2},-p,\frac {2+m}{2},\sec ^2(e+f x),-\frac {b \sec ^2(e+f x)}{a}\right ) \cot (e+f x) (d \sec (e+f x))^m \left (a+b \sec ^2(e+f x)\right )^p \left (\frac {a+b \sec ^2(e+f x)}{a}\right )^{-p} \sqrt {-\tan ^2(e+f x)}}{f m} \] Output:

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

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(2195\) vs. \(2(112)=224\).

Time = 16.95 (sec) , antiderivative size = 2195, normalized size of antiderivative = 19.60 \[ \int (d \sec (e+f x))^m \left (a+b \sec ^2(e+f x)\right )^p \, dx=\text {Result too large to show} \] Input:

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

Output:

(3*(a + b)*AppellF1[1/2, 1 - m/2, -p, 3/2, -Tan[e + f*x]^2, -((b*Tan[e + f 
*x]^2)/(a + b))]*(a + 2*b + a*Cos[2*(e + f*x)])^p*(d*Sec[e + f*x])^m*(Sec[ 
e + f*x]^2)^(-1 + m/2 + p)*(a + b*Sec[e + f*x]^2)^p*Tan[e + f*x])/(f*(3*(a 
 + b)*AppellF1[1/2, 1 - m/2, -p, 3/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2 
)/(a + b))] + (2*b*p*AppellF1[3/2, 1 - m/2, 1 - p, 5/2, -Tan[e + f*x]^2, - 
((b*Tan[e + f*x]^2)/(a + b))] + (a + b)*(-2 + m)*AppellF1[3/2, 2 - m/2, -p 
, 5/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/(a + b))])*Tan[e + f*x]^2)*(( 
3*(a + b)*AppellF1[1/2, 1 - m/2, -p, 3/2, -Tan[e + f*x]^2, -((b*Tan[e + f* 
x]^2)/(a + b))]*(a + 2*b + a*Cos[2*(e + f*x)])^p*(Sec[e + f*x]^2)^(m/2 + p 
))/(3*(a + b)*AppellF1[1/2, 1 - m/2, -p, 3/2, -Tan[e + f*x]^2, -((b*Tan[e 
+ f*x]^2)/(a + b))] + (2*b*p*AppellF1[3/2, 1 - m/2, 1 - p, 5/2, -Tan[e + f 
*x]^2, -((b*Tan[e + f*x]^2)/(a + b))] + (a + b)*(-2 + m)*AppellF1[3/2, 2 - 
 m/2, -p, 5/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/(a + b))])*Tan[e + f* 
x]^2) - (6*a*(a + b)*p*AppellF1[1/2, 1 - m/2, -p, 3/2, -Tan[e + f*x]^2, -( 
(b*Tan[e + f*x]^2)/(a + b))]*(a + 2*b + a*Cos[2*(e + f*x)])^(-1 + p)*(Sec[ 
e + f*x]^2)^(-1 + m/2 + p)*Sin[2*(e + f*x)]*Tan[e + f*x])/(3*(a + b)*Appel 
lF1[1/2, 1 - m/2, -p, 3/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/(a + b))] 
 + (2*b*p*AppellF1[3/2, 1 - m/2, 1 - p, 5/2, -Tan[e + f*x]^2, -((b*Tan[e + 
 f*x]^2)/(a + b))] + (a + b)*(-2 + m)*AppellF1[3/2, 2 - m/2, -p, 5/2, -Tan 
[e + f*x]^2, -((b*Tan[e + f*x]^2)/(a + b))])*Tan[e + f*x]^2) + (6*(a + ...
 

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4638

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

Input:

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

Output:

$Aborted
 

Defintions of rubi rules used

rule 3042

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

rule 4638

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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