\(\int (-\sec (e+f x))^n (1+\sec (e+f x))^{3/2} \, dx\) [299]

Optimal result

Integrand size = 23, antiderivative size = 117 \[ \int (-\sec (e+f x))^n (1+\sec (e+f x))^{3/2} \, dx=\frac {2 (-\sec (e+f x))^n \tan (e+f x)}{f (1+2 n) \sqrt {1+\sec (e+f x)}}-\frac {(1+4 n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},n,1+n,\sec (e+f x)\right ) (-\sec (e+f x))^n \tan (e+f x)}{f n (1+2 n) \sqrt {1-\sec (e+f x)} \sqrt {1+\sec (e+f x)}} \]

[Out]

2*(-sec(f*x+e))^n*tan(f*x+e)/f/(1+2*n)/(1+sec(f*x+e))^(1/2)-(1+4*n)*hypergeom([1/2, n],[1+n],sec(f*x+e))*(-sec
(f*x+e))^n*tan(f*x+e)/f/n/(1+2*n)/(1-sec(f*x+e))^(1/2)/(1+sec(f*x+e))^(1/2)

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3899, 21, 3891, 66} \[ \int (-\sec (e+f x))^n (1+\sec (e+f x))^{3/2} \, dx=\frac {2 \tan (e+f x) (-\sec (e+f x))^n}{f (2 n+1) \sqrt {\sec (e+f x)+1}}-\frac {(4 n+1) \tan (e+f x) (-\sec (e+f x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},n,n+1,\sec (e+f x)\right )}{f n (2 n+1) \sqrt {1-\sec (e+f x)} \sqrt {\sec (e+f x)+1}} \]

[In]

Int[(-Sec[e + f*x])^n*(1 + Sec[e + f*x])^(3/2),x]

[Out]

(2*(-Sec[e + f*x])^n*Tan[e + f*x])/(f*(1 + 2*n)*Sqrt[1 + Sec[e + f*x]]) - ((1 + 4*n)*Hypergeometric2F1[1/2, n,
1 + n, Sec[e + f*x]]*(-Sec[e + f*x])^n*Tan[e + f*x])/(f*n*(1 + 2*n)*Sqrt[1 - Sec[e + f*x]]*Sqrt[1 + Sec[e + f
*x]])

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 66

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[c^n*((b*x)^(m + 1)/(b*(m + 1)))*Hypergeometr
ic2F1[-n, m + 1, m + 2, (-d)*(x/c)], x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[m] && (IntegerQ[n] || (GtQ[
c, 0] &&  !(EqQ[n, -2^(-1)] && EqQ[c^2 - d^2, 0] && GtQ[-d/(b*c), 0])))

Rule 3891

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

Rule 3899

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-b^2)
*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 2)*((d*Csc[e + f*x])^n/(f*(m + n - 1))), x] + Dist[b/(m + n - 1), Int[
(a + b*Csc[e + f*x])^(m - 2)*(d*Csc[e + f*x])^n*(b*(m + 2*n - 1) + a*(3*m + 2*n - 4)*Csc[e + f*x]), x], x] /;
FreeQ[{a, b, d, e, f, n}, x] && EqQ[a^2 - b^2, 0] && GtQ[m, 1] && NeQ[m + n - 1, 0] && IntegerQ[2*m]

Rubi steps \begin{align*} \text {integral}& = \frac {2 (-\sec (e+f x))^n \tan (e+f x)}{f (1+2 n) \sqrt {1+\sec (e+f x)}}+\frac {2 \int \frac {(-\sec (e+f x))^n \left (\frac {1}{2}+2 n+\left (\frac {1}{2}+2 n\right ) \sec (e+f x)\right )}{\sqrt {1+\sec (e+f x)}} \, dx}{1+2 n} \\ & = \frac {2 (-\sec (e+f x))^n \tan (e+f x)}{f (1+2 n) \sqrt {1+\sec (e+f x)}}+\frac {(1+4 n) \int (-\sec (e+f x))^n \sqrt {1+\sec (e+f x)} \, dx}{1+2 n} \\ & = \frac {2 (-\sec (e+f x))^n \tan (e+f x)}{f (1+2 n) \sqrt {1+\sec (e+f x)}}+\frac {((1+4 n) \tan (e+f x)) \text {Subst}\left (\int \frac {(-x)^{-1+n}}{\sqrt {1-x}} \, dx,x,\sec (e+f x)\right )}{f (1+2 n) \sqrt {1-\sec (e+f x)} \sqrt {1+\sec (e+f x)}} \\ & = \frac {2 (-\sec (e+f x))^n \tan (e+f x)}{f (1+2 n) \sqrt {1+\sec (e+f x)}}-\frac {(1+4 n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},n,1+n,\sec (e+f x)\right ) (-\sec (e+f x))^n \tan (e+f x)}{f n (1+2 n) \sqrt {1-\sec (e+f x)} \sqrt {1+\sec (e+f x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.50 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.73 \[ \int (-\sec (e+f x))^n (1+\sec (e+f x))^{3/2} \, dx=\frac {\left (-1+(1+4 n) \cos ^{\frac {1}{2}+n}(e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{2}+n,\frac {3}{2},2 \sin ^2\left (\frac {1}{2} (e+f x)\right )\right )\right ) (-\sec (e+f x))^n \sqrt {1+\sec (e+f x)} \tan \left (\frac {1}{2} (e+f x)\right )}{f n} \]

[In]

Integrate[(-Sec[e + f*x])^n*(1 + Sec[e + f*x])^(3/2),x]

[Out]

((-1 + (1 + 4*n)*Cos[e + f*x]^(1/2 + n)*Hypergeometric2F1[1/2, 3/2 + n, 3/2, 2*Sin[(e + f*x)/2]^2])*(-Sec[e +
f*x])^n*Sqrt[1 + Sec[e + f*x]]*Tan[(e + f*x)/2])/(f*n)

Maple [F]

\[\int \left (-\sec \left (f x +e \right )\right )^{n} \left (\sec \left (f x +e \right )+1\right )^{\frac {3}{2}}d x\]

[In]

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

[Out]

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

Fricas [F]

\[ \int (-\sec (e+f x))^n (1+\sec (e+f x))^{3/2} \, dx=\int { \left (-\sec \left (f x + e\right )\right )^{n} {\left (\sec \left (f x + e\right ) + 1\right )}^{\frac {3}{2}} \,d x } \]

[In]

integrate((-sec(f*x+e))^n*(1+sec(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int (-\sec (e+f x))^n (1+\sec (e+f x))^{3/2} \, dx=\int \left (- \sec {\left (e + f x \right )}\right )^{n} \left (\sec {\left (e + f x \right )} + 1\right )^{\frac {3}{2}}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int (-\sec (e+f x))^n (1+\sec (e+f x))^{3/2} \, dx=\int { \left (-\sec \left (f x + e\right )\right )^{n} {\left (\sec \left (f x + e\right ) + 1\right )}^{\frac {3}{2}} \,d x } \]

[In]

integrate((-sec(f*x+e))^n*(1+sec(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int (-\sec (e+f x))^n (1+\sec (e+f x))^{3/2} \, dx=\int { \left (-\sec \left (f x + e\right )\right )^{n} {\left (\sec \left (f x + e\right ) + 1\right )}^{\frac {3}{2}} \,d x } \]

[In]

integrate((-sec(f*x+e))^n*(1+sec(f*x+e))^(3/2),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int (-\sec (e+f x))^n (1+\sec (e+f x))^{3/2} \, dx=\int {\left (\frac {1}{\cos \left (e+f\,x\right )}+1\right )}^{3/2}\,{\left (-\frac {1}{\cos \left (e+f\,x\right )}\right )}^n \,d x \]

[In]

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

[Out]

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