\(\int (1-\sec (e+f x))^m \sec ^n(e+f x) \, dx\) [330]

Optimal result

Integrand size = 21, antiderivative size = 89 \[ \int (1-\sec (e+f x))^m \sec ^n(e+f x) \, dx=\frac {\sqrt {2} \operatorname {AppellF1}\left (\frac {1}{2}+m,1-n,\frac {1}{2},\frac {3}{2}+m,1-\sec (e+f x),\frac {1}{2} (1-\sec (e+f x))\right ) (1-\sec (e+f x))^m \tan (e+f x)}{f (1+2 m) \sqrt {1+\sec (e+f x)}} \]

[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))*(1-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 = 0.09 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {3911, 138} \[ \int (1-\sec (e+f x))^m \sec ^n(e+f x) \, dx=\frac {\sqrt {2} \tan (e+f x) (1-\sec (e+f x))^m \operatorname {AppellF1}\left (m+\frac {1}{2},1-n,\frac {1}{2},m+\frac {3}{2},1-\sec (e+f x),\frac {1}{2} (1-\sec (e+f x))\right )}{f (2 m+1) \sqrt {\sec (e+f x)+1}} \]

[In]

Int[(1 - Sec[e + f*x])^m*Sec[e + f*x]^n,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]*(1 - Sec[e + f*x])^m*T
an[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]

Rubi steps \begin{align*} \text {integral}& = \frac {\tan (e+f x) \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)} \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 ) (1-\sec (e+f x))^m \tan (e+f x)}{f (1+2 m) \sqrt {1+\sec (e+f x)}} \\ \end{align*}

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(255\) vs. \(2(89)=178\).

Time = 3.22 (sec) , antiderivative size = 255, normalized size of antiderivative = 2.87 \[ \int (1-\sec (e+f x))^m \sec ^n(e+f x) \, dx=\frac {(3+2 m) \operatorname {AppellF1}\left (\frac {1}{2}+m,m+n,1-n,\frac {3}{2}+m,\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) (1-\sec (e+f x))^m \sec ^n(e+f x) \sin (e+f x)}{f (1+2 m) \left ((3+2 m) \operatorname {AppellF1}\left (\frac {1}{2}+m,m+n,1-n,\frac {3}{2}+m,\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )+2 \left ((-1+n) \operatorname {AppellF1}\left (\frac {3}{2}+m,m+n,2-n,\frac {5}{2}+m,\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )+(m+n) \operatorname {AppellF1}\left (\frac {3}{2}+m,1+m+n,1-n,\frac {5}{2}+m,\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )\right ) \tan ^2\left (\frac {1}{2} (e+f x)\right )\right )} \]

[In]

Integrate[(1 - Sec[e + f*x])^m*Sec[e + f*x]^n,x]

[Out]

((3 + 2*m)*AppellF1[1/2 + m, m + n, 1 - n, 3/2 + m, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*(1 - Sec[e + f*x]
)^m*Sec[e + f*x]^n*Sin[e + f*x])/(f*(1 + 2*m)*((3 + 2*m)*AppellF1[1/2 + m, m + n, 1 - n, 3/2 + m, Tan[(e + f*x
)/2]^2, -Tan[(e + f*x)/2]^2] + 2*((-1 + n)*AppellF1[3/2 + m, m + n, 2 - n, 5/2 + m, Tan[(e + f*x)/2]^2, -Tan[(
e + f*x)/2]^2] + (m + n)*AppellF1[3/2 + m, 1 + m + n, 1 - n, 5/2 + m, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]
)*Tan[(e + f*x)/2]^2))

Maple [F]

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

[In]

int((1-sec(f*x+e))^m*sec(f*x+e)^n,x)

[Out]

int((1-sec(f*x+e))^m*sec(f*x+e)^n,x)

Fricas [F]

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

[In]

integrate((1-sec(f*x+e))^m*sec(f*x+e)^n,x, algorithm="fricas")

[Out]

integral((-sec(f*x + e) + 1)^m*sec(f*x + e)^n, x)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

integrate((1-sec(f*x+e))^m*sec(f*x+e)^n,x, algorithm="maxima")

[Out]

integrate((-sec(f*x + e) + 1)^m*sec(f*x + e)^n, x)

Giac [F]

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

[In]

integrate((1-sec(f*x+e))^m*sec(f*x+e)^n,x, algorithm="giac")

[Out]

integrate((-sec(f*x + e) + 1)^m*sec(f*x + e)^n, x)

Mupad [F(-1)]

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

[In]

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

[Out]

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