3.324 \(\int \frac{(-\sec (e+f x))^n}{(a-a \sec (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=64 \[ \frac{\tan (e+f x) F_1\left (\frac{1}{2};1-n,2;\frac{3}{2};\sec (e+f x)+1,\frac{1}{2} (\sec (e+f x)+1)\right )}{2 a f \sqrt{a-a \sec (e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.167159, antiderivative size = 64, normalized size of antiderivative = 1., 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} \[ \frac{\tan (e+f x) F_1\left (\frac{1}{2};1-n,2;\frac{3}{2};\sec (e+f x)+1,\frac{1}{2} (\sec (e+f x)+1)\right )}{2 a f \sqrt{a-a \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[(-Sec[e + f*x])^n/(a - a*Sec[e + f*x])^(3/2),x]

[Out]

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

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]

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 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])

Rubi steps

\begin{align*} \int \frac{(-\sec (e+f x))^n}{(a-a \sec (e+f x))^{3/2}} \, dx &=\frac{\sqrt{1-\sec (e+f x)} \int \frac{(-\sec (e+f x))^n}{(1-\sec (e+f x))^{3/2}} \, dx}{a \sqrt{a-a \sec (e+f x)}}\\ &=\frac{\tan (e+f x) \operatorname{Subst}\left (\int \frac{(1-x)^{-1+n}}{(2-x)^2 \sqrt{x}} \, dx,x,1+\sec (e+f x)\right )}{a 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}}{\left (2-x^2\right )^2} \, dx,x,\sqrt{1+\sec (e+f x)}\right )}{a f \sqrt{1+\sec (e+f x)} \sqrt{a-a \sec (e+f x)}}\\ &=\frac{F_1\left (\frac{1}{2};1-n,2;\frac{3}{2};1+\sec (e+f x),\frac{1}{2} (1+\sec (e+f x))\right ) \tan (e+f x)}{2 a f \sqrt{a-a \sec (e+f x)}}\\ \end{align*}

Mathematica [F]  time = 1.8857, size = 0, normalized size = 0. \[ \int \frac{(-\sec (e+f x))^n}{(a-a \sec (e+f x))^{3/2}} \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[(-Sec[e + f*x])^n/(a - a*Sec[e + f*x])^(3/2),x]

[Out]

Integrate[(-Sec[e + f*x])^n/(a - a*Sec[e + f*x])^(3/2), x]

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Maple [F]  time = 0.171, size = 0, normalized size = 0. \begin{align*} \int{ \left ( -\sec \left ( fx+e \right ) \right ) ^{n} \left ( a-a\sec \left ( fx+e \right ) \right ) ^{-{\frac{3}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (-\sec \left (f x + e\right )\right )^{n}}{{\left (-a \sec \left (f x + e\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-sec(f*x+e))^n/(a-a*sec(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{-a \sec \left (f x + e\right ) + a} \left (-\sec \left (f x + e\right )\right )^{n}}{a^{2} \sec \left (f x + e\right )^{2} - 2 \, a^{2} \sec \left (f x + e\right ) + a^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-sec(f*x+e))^n/(a-a*sec(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(-a*sec(f*x + e) + a)*(-sec(f*x + e))^n/(a^2*sec(f*x + e)^2 - 2*a^2*sec(f*x + e) + a^2), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (- \sec{\left (e + f x \right )}\right )^{n}}{\left (- a \left (\sec{\left (e + f x \right )} - 1\right )\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-sec(f*x+e))**n/(a-a*sec(f*x+e))**(3/2),x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (-\sec \left (f x + e\right )\right )^{n}}{{\left (-a \sec \left (f x + e\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-sec(f*x+e))^n/(a-a*sec(f*x+e))^(3/2),x, algorithm="giac")

[Out]

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