\(\int \sec ^n(e+f x) (a-a \sec (e+f x))^m \, dx\) [332]

   Optimal result
   Rubi [A] (verified)
   Mathematica [F]
   Maple [F]
   Fricas [F]
   Sympy [F]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 22, antiderivative size = 90 \[ \int \sec ^n(e+f x) (a-a \sec (e+f x))^m \, dx=\frac {\sqrt {2} \operatorname {AppellF1}\left (\frac {1}{2}+m,1-n,\frac {1}{2},\frac {3}{2}+m,1-\sec (e+f x),\frac {1}{2} (1-\sec (e+f x))\right ) (a-a \sec (e+f x))^m \tan (e+f x)}{f (1+2 m) \sqrt {1+\sec (e+f x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {3913, 3911, 138} \[ \int \sec ^n(e+f x) (a-a \sec (e+f x))^m \, dx=\frac {\sqrt {2} \tan (e+f x) (a-a \sec (e+f x))^m \operatorname {AppellF1}\left (m+\frac {1}{2},1-n,\frac {1}{2},m+\frac {3}{2},1-\sec (e+f x),\frac {1}{2} (1-\sec (e+f x))\right )}{f (2 m+1) \sqrt {\sec (e+f x)+1}} \]

[In]

Int[Sec[e + f*x]^n*(a - a*Sec[e + f*x])^m,x]

[Out]

(Sqrt[2]*AppellF1[1/2 + m, 1 - n, 1/2, 3/2 + m, 1 - Sec[e + f*x], (1 - Sec[e + f*x])/2]*(a - a*Sec[e + f*x])^m
*Tan[e + f*x])/(f*(1 + 2*m)*Sqrt[1 + Sec[e + f*x]])

Rule 138

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[c^n*e^p*((b*x)^(m +
 1)/(b*(m + 1)))*AppellF1[m + 1, -n, -p, m + 2, (-d)*(x/c), (-f)*(x/e)], x] /; FreeQ[{b, c, d, e, f, m, n, p},
 x] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])

Rule 3911

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Dist[(-((-a
)*(d/b))^n)*(Cot[e + f*x]/(a^(n - 1)*f*Sqrt[a + b*Csc[e + f*x]]*Sqrt[a - b*Csc[e + f*x]])), Subst[Int[x^(m - 1
/2)*((a - x)^(n - 1)/Sqrt[2*a - x]), x], x, a + b*Csc[e + f*x]], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[
a^2 - b^2, 0] &&  !IntegerQ[m] && GtQ[a, 0] &&  !IntegerQ[n] && LtQ[a*(d/b), 0]

Rule 3913

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Dist[a^Int
Part[m]*((a + b*Csc[e + f*x])^FracPart[m]/(1 + (b/a)*Csc[e + f*x])^FracPart[m]), Int[(1 + (b/a)*Csc[e + f*x])^
m*(d*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[m] &&  !GtQ
[a, 0]

Rubi steps \begin{align*} \text {integral}& = \left ((1-\sec (e+f x))^{-m} (a-a \sec (e+f x))^m\right ) \int (1-\sec (e+f x))^m \sec ^n(e+f x) \, dx \\ & = \frac {\left ((1-\sec (e+f x))^{-\frac {1}{2}-m} (a-a \sec (e+f x))^m \tan (e+f x)\right ) \text {Subst}\left (\int \frac {(1-x)^{-1+n} x^{-\frac {1}{2}+m}}{\sqrt {2-x}} \, dx,x,1-\sec (e+f x)\right )}{f \sqrt {1+\sec (e+f x)}} \\ & = \frac {\sqrt {2} \operatorname {AppellF1}\left (\frac {1}{2}+m,1-n,\frac {1}{2},\frac {3}{2}+m,1-\sec (e+f x),\frac {1}{2} (1-\sec (e+f x))\right ) (a-a \sec (e+f x))^m \tan (e+f x)}{f (1+2 m) \sqrt {1+\sec (e+f x)}} \\ \end{align*}

Mathematica [F]

\[ \int \sec ^n(e+f x) (a-a \sec (e+f x))^m \, dx=\int \sec ^n(e+f x) (a-a \sec (e+f x))^m \, dx \]

[In]

Integrate[Sec[e + f*x]^n*(a - a*Sec[e + f*x])^m,x]

[Out]

Integrate[Sec[e + f*x]^n*(a - a*Sec[e + f*x])^m, x]

Maple [F]

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

[In]

int(sec(f*x+e)^n*(a-a*sec(f*x+e))^m,x)

[Out]

int(sec(f*x+e)^n*(a-a*sec(f*x+e))^m,x)

Fricas [F]

\[ \int \sec ^n(e+f x) (a-a \sec (e+f x))^m \, dx=\int { {\left (-a \sec \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right )^{n} \,d x } \]

[In]

integrate(sec(f*x+e)^n*(a-a*sec(f*x+e))^m,x, algorithm="fricas")

[Out]

integral((-a*sec(f*x + e) + a)^m*sec(f*x + e)^n, x)

Sympy [F]

\[ \int \sec ^n(e+f x) (a-a \sec (e+f x))^m \, dx=\int \left (- a \left (\sec {\left (e + f x \right )} - 1\right )\right )^{m} \sec ^{n}{\left (e + f x \right )}\, dx \]

[In]

integrate(sec(f*x+e)**n*(a-a*sec(f*x+e))**m,x)

[Out]

Integral((-a*(sec(e + f*x) - 1))**m*sec(e + f*x)**n, x)

Maxima [F]

\[ \int \sec ^n(e+f x) (a-a \sec (e+f x))^m \, dx=\int { {\left (-a \sec \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right )^{n} \,d x } \]

[In]

integrate(sec(f*x+e)^n*(a-a*sec(f*x+e))^m,x, algorithm="maxima")

[Out]

integrate((-a*sec(f*x + e) + a)^m*sec(f*x + e)^n, x)

Giac [F]

\[ \int \sec ^n(e+f x) (a-a \sec (e+f x))^m \, dx=\int { {\left (-a \sec \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right )^{n} \,d x } \]

[In]

integrate(sec(f*x+e)^n*(a-a*sec(f*x+e))^m,x, algorithm="giac")

[Out]

integrate((-a*sec(f*x + e) + a)^m*sec(f*x + e)^n, x)

Mupad [F(-1)]

Timed out. \[ \int \sec ^n(e+f x) (a-a \sec (e+f x))^m \, dx=\int {\left (a-\frac {a}{\cos \left (e+f\,x\right )}\right )}^m\,{\left (\frac {1}{\cos \left (e+f\,x\right )}\right )}^n \,d x \]

[In]

int((a - a/cos(e + f*x))^m*(1/cos(e + f*x))^n,x)

[Out]

int((a - a/cos(e + f*x))^m*(1/cos(e + f*x))^n, x)