\(\int (a+b \sec ^2(e+f x))^p (d \tan (e+f x))^m \, dx\) [441]

Optimal result

Integrand size = 25, antiderivative size = 107 \[ \int \left (a+b \sec ^2(e+f x)\right )^p (d \tan (e+f x))^m \, dx=\frac {\operatorname {AppellF1}\left (\frac {1+m}{2},1,-p,\frac {3+m}{2},-\tan ^2(e+f x),-\frac {b \tan ^2(e+f x)}{a+b}\right ) (d \tan (e+f x))^{1+m} \left (a+b+b \tan ^2(e+f x)\right )^p \left (\frac {a+b+b \tan ^2(e+f x)}{a+b}\right )^{-p}}{d f (1+m)} \] Output:

AppellF1(1/2+1/2*m,1,-p,3/2+1/2*m,-tan(f*x+e)^2,-b*tan(f*x+e)^2/(a+b))*(d* 
tan(f*x+e))^(1+m)*(a+b+b*tan(f*x+e)^2)^p/d/f/(1+m)/(((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. \(259\) vs. \(2(107)=214\).

Time = 2.97 (sec) , antiderivative size = 259, normalized size of antiderivative = 2.42 \[ \int \left (a+b \sec ^2(e+f x)\right )^p (d \tan (e+f x))^m \, dx=\frac {\operatorname {AppellF1}\left (\frac {1+m}{2},-p,1,\frac {3+m}{2},-\frac {b \tan ^2(e+f x)}{a+b},-\tan ^2(e+f x)\right ) \cos (e+f x) \left (a+b \sec ^2(e+f x)\right )^p \sin (e+f x) (d \tan (e+f x))^m}{f (1+m) \left (\operatorname {AppellF1}\left (\frac {1+m}{2},-p,1,\frac {3+m}{2},-\frac {b \tan ^2(e+f x)}{a+b},-\tan ^2(e+f x)\right )+\frac {2 \left (b p \operatorname {AppellF1}\left (\frac {3+m}{2},1-p,1,\frac {5+m}{2},-\frac {b \tan ^2(e+f x)}{a+b},-\tan ^2(e+f x)\right )-(a+b) \operatorname {AppellF1}\left (\frac {3+m}{2},-p,2,\frac {5+m}{2},-\frac {b \tan ^2(e+f x)}{a+b},-\tan ^2(e+f x)\right )\right ) \tan ^2(e+f x)}{(a+b) (3+m)}\right )} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.98, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 4629, 2075, 395, 394}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4629

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

\(\Big \downarrow \) 2075

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

\(\Big \downarrow \) 395

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

\(\Big \downarrow \) 394

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 394

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ 
), x_Symbol] :> Simp[a^p*c^q*((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m + 1)/2 
, -p, -q, 1 + (m + 1)/2, (-b)*(x^2/a), (-d)*(x^2/c)], x] /; FreeQ[{a, b, c, 
 d, e, m, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, 1] && (Int 
egerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
 

rule 395

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

rule 2075

Int[(u_)^(p_.)*(v_)^(q_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Int[(e*x)^m*Expa 
ndToSum[u, x]^p*ExpandToSum[v, x]^q, x] /; FreeQ[{e, m, p, q}, x] && Binomi 
alQ[{u, v}, x] && EqQ[BinomialDegree[u, x] - BinomialDegree[v, x], 0] &&  ! 
BinomialMatchQ[{u, v}, x]
 

rule 3042

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

rule 4629

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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