\(\int \frac {(d \tan (e+f x))^n}{a+b \sec (e+f x)} \, dx\) [344]

Optimal result

Integrand size = 23, antiderivative size = 266 \[ \int \frac {(d \tan (e+f x))^n}{a+b \sec (e+f x)} \, dx=\frac {d \operatorname {AppellF1}\left (1-n,\frac {1-n}{2},\frac {1-n}{2},2-n,\frac {a+b}{a+b \sec (e+f x)},\frac {a-b}{a+b \sec (e+f x)}\right ) \left (-\frac {b (1-\sec (e+f x))}{a+b \sec (e+f x)}\right )^{\frac {1-n}{2}} \left (\frac {b (1+\sec (e+f x))}{a+b \sec (e+f x)}\right )^{\frac {1-n}{2}} (d \tan (e+f x))^{-1+n} \left (-\tan ^2(e+f x)\right )^{\frac {1-n}{2}+\frac {1}{2} (-1+n)}}{a f (1-n)}-\frac {d \operatorname {Hypergeometric2F1}\left (1,\frac {1+n}{2},\frac {3+n}{2},-\tan ^2(e+f x)\right ) (d \tan (e+f x))^{-1+n} \left (-\tan ^2(e+f x)\right )^{\frac {1-n}{2}+\frac {1+n}{2}}}{a f (1+n)} \] Output:

d*AppellF1(1-n,1/2-1/2*n,1/2-1/2*n,2-n,(a-b)/(a+b*sec(f*x+e)),(a+b)/(a+b*s 
ec(f*x+e)))*(-b*(1-sec(f*x+e))/(a+b*sec(f*x+e)))^(1/2-1/2*n)*(b*(1+sec(f*x 
+e))/(a+b*sec(f*x+e)))^(1/2-1/2*n)*(d*tan(f*x+e))^(-1+n)/a/f/(1-n)+d*hyper 
geom([1, 1/2+1/2*n],[3/2+1/2*n],-tan(f*x+e)^2)*(d*tan(f*x+e))^(-1+n)*tan(f 
*x+e)^2/a/f/(1+n)
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(786\) vs. \(2(266)=532\).

Time = 3.73 (sec) , antiderivative size = 786, normalized size of antiderivative = 2.95 \[ \int \frac {(d \tan (e+f x))^n}{a+b \sec (e+f x)} \, dx =\text {Too large to display} \] Input:

Integrate[(d*Tan[e + f*x])^n/(a + b*Sec[e + f*x]),x]
 

Output:

(2*((a + b)*AppellF1[(1 + n)/2, n, 1, (3 + n)/2, Tan[(e + f*x)/2]^2, -Tan[ 
(e + f*x)/2]^2] - b*AppellF1[(1 + n)/2, n, 1, (3 + n)/2, Tan[(e + f*x)/2]^ 
2, ((a - b)*Tan[(e + f*x)/2]^2)/(a + b)])*Tan[(e + f*x)/2]*(d*Tan[e + f*x] 
)^n)/(f*(a + b*Sec[e + f*x])*(((a + b)*AppellF1[(1 + n)/2, n, 1, (3 + n)/2 
, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - b*AppellF1[(1 + n)/2, n, 1, ( 
3 + n)/2, Tan[(e + f*x)/2]^2, ((a - b)*Tan[(e + f*x)/2]^2)/(a + b)])*Sec[( 
e + f*x)/2]^2 - 16*n*((a + b)*AppellF1[(1 + n)/2, n, 1, (3 + n)/2, Tan[(e 
+ f*x)/2]^2, -Tan[(e + f*x)/2]^2] - b*AppellF1[(1 + n)/2, n, 1, (3 + n)/2, 
 Tan[(e + f*x)/2]^2, ((a - b)*Tan[(e + f*x)/2]^2)/(a + b)])*Cos[(e + f*x)/ 
2]*Csc[e + f*x]^3*Sec[e + f*x]*Sin[(e + f*x)/2]^5 + 2*n*((a + b)*AppellF1[ 
(1 + n)/2, n, 1, (3 + n)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - b*A 
ppellF1[(1 + n)/2, n, 1, (3 + n)/2, Tan[(e + f*x)/2]^2, ((a - b)*Tan[(e + 
f*x)/2]^2)/(a + b)])*Csc[e + f*x]*Sec[e + f*x]*Tan[(e + f*x)/2] - (2*(1 + 
n)*((a - b)*b*AppellF1[(3 + n)/2, n, 2, (5 + n)/2, Tan[(e + f*x)/2]^2, ((a 
 - b)*Tan[(e + f*x)/2]^2)/(a + b)] + (a + b)^2*(AppellF1[(3 + n)/2, n, 2, 
(5 + n)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - n*AppellF1[(3 + n)/2 
, 1 + n, 1, (5 + n)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]) + b*(a + 
b)*n*AppellF1[(3 + n)/2, 1 + n, 1, (5 + n)/2, Tan[(e + f*x)/2]^2, ((a - b) 
*Tan[(e + f*x)/2]^2)/(a + b)])*Sec[(e + f*x)/2]^2*Tan[(e + f*x)/2]^2)/((a 
+ b)*(3 + n))))
 

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.

\(\displaystyle \int \frac {(d \tan (e+f x))^n}{a+b \sec (e+f x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (-d \cot \left (e+f x+\frac {\pi }{2}\right )\right )^n}{a+b \csc \left (e+f x+\frac {\pi }{2}\right )}dx\)

\(\Big \downarrow \) 4387

\(\displaystyle \int \frac {(d \tan (e+f x))^n}{a+b \sec (e+f x)}dx\)

Input:

Int[(d*Tan[e + f*x])^n/(a + b*Sec[e + f*x]),x]
 

Output:

$Aborted
 

Defintions of rubi rules used

rule 3042

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

rule 4387

Int[((a_.) + csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_.)*(cot[(c_.) + (d_.)*(x_)]* 
(e_.))^(m_.), x_Symbol] :> Unintegrable[(e*Cot[c + d*x])^m*(a + b*Csc[c + d 
*x])^n, x] /; FreeQ[{a, b, c, d, e, m, n}, x]
 

Maple [F]

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

Input:

int((d*tan(f*x+e))^n/(a+b*sec(f*x+e)),x)
 

Output:

int((d*tan(f*x+e))^n/(a+b*sec(f*x+e)),x)
 

Fricas [F]

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

integrate((d*tan(f*x+e))^n/(a+b*sec(f*x+e)),x, algorithm="fricas")
 

Output:

integral((d*tan(f*x + e))^n/(b*sec(f*x + e) + a), x)
 

Sympy [F]

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

integrate((d*tan(f*x+e))**n/(a+b*sec(f*x+e)),x)
 

Output:

Integral((d*tan(e + f*x))**n/(a + b*sec(e + f*x)), x)
 

Maxima [F]

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

integrate((d*tan(f*x+e))^n/(a+b*sec(f*x+e)),x, algorithm="maxima")
 

Output:

integrate((d*tan(f*x + e))^n/(b*sec(f*x + e) + a), x)
 

Giac [F]

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

integrate((d*tan(f*x+e))^n/(a+b*sec(f*x+e)),x, algorithm="giac")
 

Output:

integrate((d*tan(f*x + e))^n/(b*sec(f*x + e) + a), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

int((d*tan(f*x+e))^n/(a+b*sec(f*x+e)),x)
 

Output:

d**n*int(tan(e + f*x)**n/(sec(e + f*x)*b + a),x)