\(\int (d \sec (e+f x))^m (a+b \tan ^2(e+f x))^p \, dx\) [477]

Optimal result

Integrand size = 25, antiderivative size = 109 \[ \int (d \sec (e+f x))^m \left (a+b \tan ^2(e+f x)\right )^p \, dx=\frac {\operatorname {AppellF1}\left (\frac {1}{2},1-\frac {m}{2},-p,\frac {3}{2},-\tan ^2(e+f x),-\frac {b \tan ^2(e+f x)}{a}\right ) (d \sec (e+f x))^m \sec ^2(e+f x)^{-m/2} \tan (e+f x) \left (a+b \tan ^2(e+f x)\right )^p \left (\frac {a+b \tan ^2(e+f x)}{a}\right )^{-p}}{f} \] Output:

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

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(2033\) vs. \(2(109)=218\).

Time = 15.24 (sec) , antiderivative size = 2033, normalized size of antiderivative = 18.65 \[ \int (d \sec (e+f x))^m \left (a+b \tan ^2(e+f x)\right )^p \, dx=\text {Result too large to show} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.01, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {3042, 4162, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4162

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

\(\Big \downarrow \) 334

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

\(\Big \downarrow \) 333

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

Input:

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

Output:

(AppellF1[1/2, (2 - m)/2, -p, 3/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/a 
)]*(d*Sec[e + f*x])^m*Tan[e + f*x]*(a + b*Tan[e + f*x]^2)^p)/(f*(Sec[e + f 
*x]^2)^(m/2)*(1 + (b*Tan[e + f*x]^2)/a)^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 4162

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x 
_)]^2)^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff 
*((d*Sec[e + f*x])^m/(f*(Sec[e + f*x]^2)^(m/2)))   Subst[Int[(1 + ff^2*x^2) 
^(m/2 - 1)*(a + b*ff^2*x^2)^p, x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, 
 d, e, f, m, p}, x] &&  !IntegerQ[m]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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