\(\int \sec ^n(e+f x) (1+\sec (e+f x))^{5/2} \, dx\) [294]

Optimal result

Integrand size = 21, antiderivative size = 162 \[ \int \sec ^n(e+f x) (1+\sec (e+f x))^{5/2} \, dx=\frac {2 (7+4 n) \sec ^{1+n}(e+f x) \sin (e+f x)}{f (1+2 n) (3+2 n) \sqrt {1+\sec (e+f x)}}+\frac {2 \sec ^{1+n}(e+f x) \sqrt {1+\sec (e+f x)} \sin (e+f x)}{f (3+2 n)}+\frac {2 \left (3+24 n+16 n^2\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},1-n,\frac {3}{2},1-\sec (e+f x)\right ) \tan (e+f x)}{f (1+2 n) (3+2 n) \sqrt {1+\sec (e+f x)}} \]

[Out]

2*(7+4*n)*sec(f*x+e)^(1+n)*sin(f*x+e)/f/(4*n^2+8*n+3)/(1+sec(f*x+e))^(1/2)+2*sec(f*x+e)^(1+n)*sin(f*x+e)*(1+se
c(f*x+e))^(1/2)/f/(3+2*n)+2*(16*n^2+24*n+3)*hypergeom([1/2, 1-n],[3/2],1-sec(f*x+e))*tan(f*x+e)/f/(4*n^2+8*n+3
)/(1+sec(f*x+e))^(1/2)

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3899, 4101, 3891, 67} \[ \int \sec ^n(e+f x) (1+\sec (e+f x))^{5/2} \, dx=\frac {2 \left (16 n^2+24 n+3\right ) \tan (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},1-n,\frac {3}{2},1-\sec (e+f x)\right )}{f (2 n+1) (2 n+3) \sqrt {\sec (e+f x)+1}}+\frac {2 \sin (e+f x) \sqrt {\sec (e+f x)+1} \sec ^{n+1}(e+f x)}{f (2 n+3)}+\frac {2 (4 n+7) \sin (e+f x) \sec ^{n+1}(e+f x)}{f (2 n+1) (2 n+3) \sqrt {\sec (e+f x)+1}} \]

[In]

Int[Sec[e + f*x]^n*(1 + Sec[e + f*x])^(5/2),x]

[Out]

(2*(7 + 4*n)*Sec[e + f*x]^(1 + n)*Sin[e + f*x])/(f*(1 + 2*n)*(3 + 2*n)*Sqrt[1 + Sec[e + f*x]]) + (2*Sec[e + f*
x]^(1 + n)*Sqrt[1 + Sec[e + f*x]]*Sin[e + f*x])/(f*(3 + 2*n)) + (2*(3 + 24*n + 16*n^2)*Hypergeometric2F1[1/2,
1 - n, 3/2, 1 - Sec[e + f*x]]*Tan[e + f*x])/(f*(1 + 2*n)*(3 + 2*n)*Sqrt[1 + Sec[e + f*x]])

Rule 67

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((c + d*x)^(n + 1)/(d*(n + 1)*(-d/(b*c))^m))
*Hypergeometric2F1[-m, n + 1, n + 2, 1 + d*(x/c)], x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[n] && (Intege
rQ[m] || 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]

Rule 4101

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

Rubi steps \begin{align*} \text {integral}& = \frac {2 \sec ^{1+n}(e+f x) \sqrt {1+\sec (e+f x)} \sin (e+f x)}{f (3+2 n)}+\frac {2 \int \sec ^n(e+f x) \sqrt {1+\sec (e+f x)} \left (\frac {3}{2}+2 n+\left (\frac {7}{2}+2 n\right ) \sec (e+f x)\right ) \, dx}{3+2 n} \\ & = \frac {2 (7+4 n) \sec ^{1+n}(e+f x) \sin (e+f x)}{f (1+2 n) (3+2 n) \sqrt {1+\sec (e+f x)}}+\frac {2 \sec ^{1+n}(e+f x) \sqrt {1+\sec (e+f x)} \sin (e+f x)}{f (3+2 n)}+\frac {\left (3+24 n+16 n^2\right ) \int \sec ^n(e+f x) \sqrt {1+\sec (e+f x)} \, dx}{3+8 n+4 n^2} \\ & = \frac {2 (7+4 n) \sec ^{1+n}(e+f x) \sin (e+f x)}{f (1+2 n) (3+2 n) \sqrt {1+\sec (e+f x)}}+\frac {2 \sec ^{1+n}(e+f x) \sqrt {1+\sec (e+f x)} \sin (e+f x)}{f (3+2 n)}-\frac {\left (\left (3+24 n+16 n^2\right ) \tan (e+f x)\right ) \text {Subst}\left (\int \frac {x^{-1+n}}{\sqrt {1-x}} \, dx,x,\sec (e+f x)\right )}{f \left (3+8 n+4 n^2\right ) \sqrt {1-\sec (e+f x)} \sqrt {1+\sec (e+f x)}} \\ & = \frac {2 (7+4 n) \sec ^{1+n}(e+f x) \sin (e+f x)}{f (1+2 n) (3+2 n) \sqrt {1+\sec (e+f x)}}+\frac {2 \sec ^{1+n}(e+f x) \sqrt {1+\sec (e+f x)} \sin (e+f x)}{f (3+2 n)}+\frac {2 \left (3+24 n+16 n^2\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},1-n,\frac {3}{2},1-\sec (e+f x)\right ) \tan (e+f x)}{f \left (3+8 n+4 n^2\right ) \sqrt {1+\sec (e+f x)}} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 31.83 (sec) , antiderivative size = 398, normalized size of antiderivative = 2.46 \[ \int \sec ^n(e+f x) (1+\sec (e+f x))^{5/2} \, dx=-\frac {i 2^{-\frac {5}{2}+n} e^{-\frac {1}{2} i (3+2 n) (e+f x)} \left (\frac {e^{i (e+f x)}}{1+e^{2 i (e+f x)}}\right )^{\frac {3}{2}+n} \left (\frac {10 e^{i (2+n) (e+f x)} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} (-1-n),\frac {4+n}{2},-e^{2 i (e+f x)}\right )}{2+n}+\frac {5 e^{i (4+n) (e+f x)} \operatorname {Hypergeometric2F1}\left (1,\frac {1-n}{2},\frac {6+n}{2},-e^{2 i (e+f x)}\right )}{4+n}+\frac {e^{i n (e+f x)} \operatorname {Hypergeometric2F1}\left (1,-\frac {3}{2}-\frac {n}{2},1+\frac {n}{2},-e^{2 i (e+f x)}\right )}{n}+\frac {5 e^{i (1+n) (e+f x)} \operatorname {Hypergeometric2F1}\left (1,-1-\frac {n}{2},\frac {3+n}{2},-e^{2 i (e+f x)}\right )}{1+n}+\frac {e^{i (5+n) (e+f x)} \operatorname {Hypergeometric2F1}\left (1,1-\frac {n}{2},\frac {7+n}{2},-e^{2 i (e+f x)}\right )}{5+n}+\frac {10 e^{i (3+n) (e+f x)} \operatorname {Hypergeometric2F1}\left (1,-\frac {n}{2},\frac {5+n}{2},-e^{2 i (e+f x)}\right )}{3+n}\right ) \sec ^5\left (\frac {1}{2} (e+f x)\right ) (1+\sec (e+f x))^{5/2}}{f \sec ^{\frac {5}{2}}(e+f x)} \]

[In]

Integrate[Sec[e + f*x]^n*(1 + Sec[e + f*x])^(5/2),x]

[Out]

((-I)*2^(-5/2 + n)*(E^(I*(e + f*x))/(1 + E^((2*I)*(e + f*x))))^(3/2 + n)*((10*E^(I*(2 + n)*(e + f*x))*Hypergeo
metric2F1[1, (-1 - n)/2, (4 + n)/2, -E^((2*I)*(e + f*x))])/(2 + n) + (5*E^(I*(4 + n)*(e + f*x))*Hypergeometric
2F1[1, (1 - n)/2, (6 + n)/2, -E^((2*I)*(e + f*x))])/(4 + n) + (E^(I*n*(e + f*x))*Hypergeometric2F1[1, -3/2 - n
/2, 1 + n/2, -E^((2*I)*(e + f*x))])/n + (5*E^(I*(1 + n)*(e + f*x))*Hypergeometric2F1[1, -1 - n/2, (3 + n)/2, -
E^((2*I)*(e + f*x))])/(1 + n) + (E^(I*(5 + n)*(e + f*x))*Hypergeometric2F1[1, 1 - n/2, (7 + n)/2, -E^((2*I)*(e
+ f*x))])/(5 + n) + (10*E^(I*(3 + n)*(e + f*x))*Hypergeometric2F1[1, -1/2*n, (5 + n)/2, -E^((2*I)*(e + f*x))]
)/(3 + n))*Sec[(e + f*x)/2]^5*(1 + Sec[e + f*x])^(5/2))/(E^((I/2)*(3 + 2*n)*(e + f*x))*f*Sec[e + f*x]^(5/2))

Maple [F]

\[\int \sec \left (f x +e \right )^{n} \left (\sec \left (f x +e \right )+1\right )^{\frac {5}{2}}d x\]

[In]

int(sec(f*x+e)^n*(sec(f*x+e)+1)^(5/2),x)

[Out]

int(sec(f*x+e)^n*(sec(f*x+e)+1)^(5/2),x)

Fricas [F]

\[ \int \sec ^n(e+f x) (1+\sec (e+f x))^{5/2} \, dx=\int { \sec \left (f x + e\right )^{n} {\left (\sec \left (f x + e\right ) + 1\right )}^{\frac {5}{2}} \,d x } \]

[In]

integrate(sec(f*x+e)^n*(1+sec(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

integral((sec(f*x + e)^2 + 2*sec(f*x + e) + 1)*sec(f*x + e)^n*sqrt(sec(f*x + e) + 1), x)

Sympy [F(-1)]

Timed out. \[ \int \sec ^n(e+f x) (1+\sec (e+f x))^{5/2} \, dx=\text {Timed out} \]

[In]

integrate(sec(f*x+e)**n*(1+sec(f*x+e))**(5/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \sec ^n(e+f x) (1+\sec (e+f x))^{5/2} \, dx=\int { \sec \left (f x + e\right )^{n} {\left (\sec \left (f x + e\right ) + 1\right )}^{\frac {5}{2}} \,d x } \]

[In]

integrate(sec(f*x+e)^n*(1+sec(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

integrate(sec(f*x + e)^n*(sec(f*x + e) + 1)^(5/2), x)

Giac [F]

\[ \int \sec ^n(e+f x) (1+\sec (e+f x))^{5/2} \, dx=\int { \sec \left (f x + e\right )^{n} {\left (\sec \left (f x + e\right ) + 1\right )}^{\frac {5}{2}} \,d x } \]

[In]

integrate(sec(f*x+e)^n*(1+sec(f*x+e))^(5/2),x, algorithm="giac")

[Out]

integrate(sec(f*x + e)^n*(sec(f*x + e) + 1)^(5/2), x)

Mupad [F(-1)]

Timed out. \[ \int \sec ^n(e+f x) (1+\sec (e+f x))^{5/2} \, dx=\int {\left (\frac {1}{\cos \left (e+f\,x\right )}+1\right )}^{5/2}\,{\left (\frac {1}{\cos \left (e+f\,x\right )}\right )}^n \,d x \]

[In]

int((1/cos(e + f*x) + 1)^(5/2)*(1/cos(e + f*x))^n,x)

[Out]

int((1/cos(e + f*x) + 1)^(5/2)*(1/cos(e + f*x))^n, x)