3.323 \(\int \frac {(-\sec (e+f x))^n}{\sqrt {a-a \sec (e+f x)}} \, dx\)

Optimal. Leaf size=58 \[ \frac {\tan (e+f x) F_1\left (\frac {1}{2};1-n,1;\frac {3}{2};\sec (e+f x)+1,\frac {1}{2} (\sec (e+f x)+1)\right )}{f \sqrt {a-a \sec (e+f x)}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.15, antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {3828, 3825, 130, 429} \[ \frac {\tan (e+f x) F_1\left (\frac {1}{2};1-n,1;\frac {3}{2};\sec (e+f x)+1,\frac {1}{2} (\sec (e+f x)+1)\right )}{f \sqrt {a-a \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 130

Int[((e_.)*(x_))^(p_)*((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> With[{k = Denominator[p]
}, Dist[k/e, Subst[Int[x^(k*(p + 1) - 1)*(a + (b*x^k)/e)^m*(c + (d*x^k)/e)^n, x], x, (e*x)^(1/k)], x]] /; Free
Q[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] && FractionQ[p] && IntegerQ[m]

Rule 429

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

Rule 3825

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 - 2)*f*Sqrt[a + b*Csc[e + f*x]]*Sqrt[a - b*Csc[e + f*x]]), Subst[Int[((a - x)^(n -
1)*(2*a - x)^(m - 1/2))/Sqrt[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] && GtQ[(a*d)/b, 0]

Rule 3828

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Dist[(a^In
tPart[m]*(a + b*Csc[e + f*x])^FracPart[m])/(1 + (b*Csc[e + f*x])/a)^FracPart[m], Int[(1 + (b*Csc[e + f*x])/a)^
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*} \int \frac {(-\sec (e+f x))^n}{\sqrt {a-a \sec (e+f x)}} \, dx &=\frac {\sqrt {1-\sec (e+f x)} \int \frac {(-\sec (e+f x))^n}{\sqrt {1-\sec (e+f x)}} \, dx}{\sqrt {a-a \sec (e+f x)}}\\ &=\frac {\tan (e+f x) \operatorname {Subst}\left (\int \frac {(1-x)^{-1+n}}{(2-x) \sqrt {x}} \, dx,x,1+\sec (e+f x)\right )}{f \sqrt {1+\sec (e+f x)} \sqrt {a-a \sec (e+f x)}}\\ &=\frac {(2 \tan (e+f x)) \operatorname {Subst}\left (\int \frac {\left (1-x^2\right )^{-1+n}}{2-x^2} \, dx,x,\sqrt {1+\sec (e+f x)}\right )}{f \sqrt {1+\sec (e+f x)} \sqrt {a-a \sec (e+f x)}}\\ &=\frac {F_1\left (\frac {1}{2};1-n,1;\frac {3}{2};1+\sec (e+f x),\frac {1}{2} (1+\sec (e+f x))\right ) \tan (e+f x)}{f \sqrt {a-a \sec (e+f x)}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [F]  time = 1.33, size = 0, normalized size = 0.00 \[ \int \frac {(-\sec (e+f x))^n}{\sqrt {a-a \sec (e+f x)}} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [F]  time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-a \sec \left (f x + e\right ) + a} \left (-\sec \left (f x + e\right )\right )^{n}}{a \sec \left (f x + e\right ) - a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-sec(f*x+e))^n/(a-a*sec(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (-\sec \left (f x + e\right )\right )^{n}}{\sqrt {-a \sec \left (f x + e\right ) + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-sec(f*x+e))^n/(a-a*sec(f*x+e))^(1/2),x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [F]  time = 1.02, size = 0, normalized size = 0.00 \[ \int \frac {\left (-\sec \left (f x +e \right )\right )^{n}}{\sqrt {a -a \sec \left (f x +e \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-sec(f*x+e))^n/(a-a*sec(f*x+e))^(1/2),x)

[Out]

int((-sec(f*x+e))^n/(a-a*sec(f*x+e))^(1/2),x)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (-\sec \left (f x + e\right )\right )^{n}}{\sqrt {-a \sec \left (f x + e\right ) + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-sec(f*x+e))^n/(a-a*sec(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {{\left (-\frac {1}{\cos \left (e+f\,x\right )}\right )}^n}{\sqrt {a-\frac {a}{\cos \left (e+f\,x\right )}}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-1/cos(e + f*x))^n/(a - a/cos(e + f*x))^(1/2),x)

[Out]

int((-1/cos(e + f*x))^n/(a - a/cos(e + f*x))^(1/2), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (- \sec {\left (e + f x \right )}\right )^{n}}{\sqrt {- a \left (\sec {\left (e + f x \right )} - 1\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-sec(f*x+e))**n/(a-a*sec(f*x+e))**(1/2),x)

[Out]

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

________________________________________________________________________________________