\(\int (a+b \sec ^2(e+f x))^p \, dx\) [140]

Optimal result

Integrand size = 14, antiderivative size = 85 \[ \int \left (a+b \sec ^2(e+f x)\right )^p \, dx=\frac {\operatorname {AppellF1}\left (\frac {1}{2},1,-p,\frac {3}{2},-\tan ^2(e+f x),-\frac {b \tan ^2(e+f x)}{a+b}\right ) \tan (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^p \left (\frac {a+b+b \tan ^2(e+f x)}{a+b}\right )^{-p}}{f} \] Output:

AppellF1(1/2,1,-p,3/2,-tan(f*x+e)^2,-b*tan(f*x+e)^2/(a+b))*tan(f*x+e)*(a+b 
+b*tan(f*x+e)^2)^p/f/(((a+b+b*tan(f*x+e)^2)/(a+b))^p)
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(2137\) vs. \(2(85)=170\).

Time = 14.23 (sec) , antiderivative size = 2137, normalized size of antiderivative = 25.14 \[ \int \left (a+b \sec ^2(e+f x)\right )^p \, dx=\text {Result too large to show} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.98, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 4616, 334, 333}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4616

\(\displaystyle \frac {\int \frac {\left (b \tan ^2(e+f x)+a+b\right )^p}{\tan ^2(e+f x)+1}d\tan (e+f x)}{f}\)

\(\Big \downarrow \) 334

\(\displaystyle \frac {\left (a+b \tan ^2(e+f x)+b\right )^p \left (\frac {b \tan ^2(e+f x)}{a+b}+1\right )^{-p} \int \frac {\left (\frac {b \tan ^2(e+f x)}{a+b}+1\right )^p}{\tan ^2(e+f x)+1}d\tan (e+f x)}{f}\)

\(\Big \downarrow \) 333

\(\displaystyle \frac {\tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^p \left (\frac {b \tan ^2(e+f x)}{a+b}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},1,-p,\frac {3}{2},-\tan ^2(e+f x),-\frac {b \tan ^2(e+f x)}{a+b}\right )}{f}\)

Input:

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

Output:

(AppellF1[1/2, 1, -p, 3/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/(a + b))] 
*Tan[e + f*x]*(a + b + b*Tan[e + f*x]^2)^p)/(f*(1 + (b*Tan[e + f*x]^2)/(a 
+ b))^p)
 

Defintions of rubi rules used

rule 333

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[a^p*c^q*x*AppellF1[1/2, -p, -q, 3/2, (-b)*(x^2/a), (-d)*(x^2/c)], x] /; F 
reeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[p] || GtQ[a, 
0]) && (IntegerQ[q] || GtQ[c, 0])
 

rule 334

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[a^IntPart[p]*((a + b*x^2)^FracPart[p]/(1 + b*(x^2/a))^FracPart[p])   Int[ 
(1 + b*(x^2/a))^p*(c + d*x^2)^q, x], x] /; FreeQ[{a, b, c, d, p, q}, x] && 
NeQ[b*c - a*d, 0] &&  !(IntegerQ[p] || GtQ[a, 0])
 

rule 3042

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

rule 4616

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> With[{ff = 
FreeFactors[Tan[e + f*x], x]}, Simp[ff/f   Subst[Int[(a + b + b*ff^2*x^2)^p 
/(1 + ff^2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] 
&& NeQ[a + b, 0] && NeQ[p, -1]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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