\(\int (d \sec (e+f x))^m (a+b \tan (e+f x)) \, dx\) [650]

Optimal result

Integrand size = 21, antiderivative size = 93 \[ \int (d \sec (e+f x))^m (a+b \tan (e+f x)) \, dx=\frac {b (d \sec (e+f x))^m}{f m}-\frac {a d \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1-m}{2},\frac {3-m}{2},\cos ^2(e+f x)\right ) (d \sec (e+f x))^{-1+m} \sin (e+f x)}{f (1-m) \sqrt {\sin ^2(e+f x)}} \] Output:

b*(d*sec(f*x+e))^m/f/m-a*d*hypergeom([1/2, 1/2-1/2*m],[3/2-1/2*m],cos(f*x+ 
e)^2)*(d*sec(f*x+e))^(-1+m)*sin(f*x+e)/f/(1-m)/(sin(f*x+e)^2)^(1/2)
 

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.70 \[ \int (d \sec (e+f x))^m (a+b \tan (e+f x)) \, dx=\frac {(d \sec (e+f x))^m \left (b+a \cot (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m}{2},\frac {2+m}{2},\sec ^2(e+f x)\right ) \sqrt {-\tan ^2(e+f x)}\right )}{f m} \] Input:

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

Output:

((d*Sec[e + f*x])^m*(b + a*Cot[e + f*x]*Hypergeometric2F1[1/2, m/2, (2 + m 
)/2, Sec[e + f*x]^2]*Sqrt[-Tan[e + f*x]^2]))/(f*m)
 

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 3967, 3042, 4259, 3042, 3122}

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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int (a+b \tan (e+f x)) (d \sec (e+f x))^mdx\)

\(\Big \downarrow \) 3967

\(\displaystyle a \int (d \sec (e+f x))^mdx+\frac {b (d \sec (e+f x))^m}{f m}\)

\(\Big \downarrow \) 3042

\(\displaystyle a \int \left (d \csc \left (e+f x+\frac {\pi }{2}\right )\right )^mdx+\frac {b (d \sec (e+f x))^m}{f m}\)

\(\Big \downarrow \) 4259

\(\displaystyle a \left (\frac {\cos (e+f x)}{d}\right )^m (d \sec (e+f x))^m \int \left (\frac {\cos (e+f x)}{d}\right )^{-m}dx+\frac {b (d \sec (e+f x))^m}{f m}\)

\(\Big \downarrow \) 3042

\(\displaystyle a \left (\frac {\cos (e+f x)}{d}\right )^m (d \sec (e+f x))^m \int \left (\frac {\sin \left (e+f x+\frac {\pi }{2}\right )}{d}\right )^{-m}dx+\frac {b (d \sec (e+f x))^m}{f m}\)

\(\Big \downarrow \) 3122

\(\displaystyle \frac {b (d \sec (e+f x))^m}{f m}-\frac {a d \sin (e+f x) (d \sec (e+f x))^{m-1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1-m}{2},\frac {3-m}{2},\cos ^2(e+f x)\right )}{f (1-m) \sqrt {\sin ^2(e+f x)}}\)

Input:

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

Output:

(b*(d*Sec[e + f*x])^m)/(f*m) - (a*d*Hypergeometric2F1[1/2, (1 - m)/2, (3 - 
 m)/2, Cos[e + f*x]^2]*(d*Sec[e + f*x])^(-1 + m)*Sin[e + f*x])/(f*(1 - m)* 
Sqrt[Sin[e + f*x]^2])
 

Defintions of rubi rules used

rule 3042

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

rule 3122

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( 
b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1)*Sqrt[Cos[c + d*x]^2]))*Hypergeometric2 
F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n}, x] 
 &&  !IntegerQ[2*n]
 

rule 3967

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*( 
x_)]), x_Symbol] :> Simp[b*((d*Sec[e + f*x])^m/(f*m)), x] + Simp[a   Int[(d 
*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2*m] 
|| NeQ[a^2 + b^2, 0])
 

rule 4259

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x] 
)^(n - 1)*((Sin[c + d*x]/b)^(n - 1)   Int[1/(Sin[c + d*x]/b)^n, x]), x] /; 
FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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