3.4.24 \(\int \frac {(-\sec (e+f x))^n}{(a-a \sec (e+f x))^{3/2}} \, dx\) [324]

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

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.12, antiderivative size = 64, 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 = {3913, 3910, 129, 440} \begin {gather*} \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)}} \end {gather*}

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 129

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 440

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 3910

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 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*} \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) \text {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)) \text {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 = 74.99, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(-\sec (e+f x))^n}{(a-a \sec (e+f x))^{3/2}} \, dx \end {gather*}

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]

________________________________________________________________________________________

Maple [F]
time = 0.08, size = 0, normalized size = 0.00 \[\int \frac {\left (-\sec \left (f x +e \right )\right )^{n}}{\left (a -a \sec \left (f x +e \right )\right )^{\frac {3}{2}}}\, dx\]

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)

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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)

________________________________________________________________________________________

Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 {gather*}

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)

________________________________________________________________________________________

Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

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]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Ch
eck [abs(co

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {{\left (-\frac {1}{\cos \left (e+f\,x\right )}\right )}^n}{{\left (a-\frac {a}{\cos \left (e+f\,x\right )}\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-1/cos(e + f*x))^n/(a - a/cos(e + f*x))^(3/2),x)

[Out]

int((-1/cos(e + f*x))^n/(a - a/cos(e + f*x))^(3/2), x)

________________________________________________________________________________________