\(\int \sec ^n(e+f x) (1+\sec (e+f x))^{3/2} \, dx\) [295]

Optimal result

Integrand size = 21, antiderivative size = 98 \[ \int \sec ^n(e+f x) (1+\sec (e+f x))^{3/2} \, dx=\frac {2 \sec ^{1+n}(e+f x) \sin (e+f x)}{f (1+2 n) \sqrt {1+\sec (e+f x)}}+\frac {2 (1+4 n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},1-n,\frac {3}{2},1-\sec (e+f x)\right ) \tan (e+f x)}{f (1+2 n) \sqrt {1+\sec (e+f x)}} \] Output:

2*sec(f*x+e)^(1+n)*sin(f*x+e)/f/(1+2*n)/(1+sec(f*x+e))^(1/2)+2*(1+4*n)*hyp 
ergeom([1/2, 1-n],[3/2],1-sec(f*x+e))*tan(f*x+e)/f/(1+2*n)/(1+sec(f*x+e))^ 
(1/2)
 

Mathematica [A] (verified)

Time = 0.31 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.85 \[ \int \sec ^n(e+f x) (1+\sec (e+f x))^{3/2} \, dx=\frac {\left (-1+(1+4 n) \cos ^{\frac {1}{2}+n}(e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{2}+n,\frac {3}{2},2 \sin ^2\left (\frac {1}{2} (e+f x)\right )\right )\right ) \sec ^n(e+f x) \sqrt {1+\sec (e+f x)} \tan \left (\frac {1}{2} (e+f x)\right )}{f n} \] Input:

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

Output:

((-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)
 

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 4301, 27, 2011, 3042, 4293, 75}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int (\sec (e+f x)+1)^{3/2} \sec ^n(e+f x) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \left (\csc \left (e+f x+\frac {\pi }{2}\right )+1\right )^{3/2} \csc \left (e+f x+\frac {\pi }{2}\right )^ndx\)

\(\Big \downarrow \) 4301

\(\displaystyle \frac {2 \int \frac {\sec ^n(e+f x) (4 n+(4 n+1) \sec (e+f x)+1)}{2 \sqrt {\sec (e+f x)+1}}dx}{2 n+1}+\frac {2 \sin (e+f x) \sec ^{n+1}(e+f x)}{f (2 n+1) \sqrt {\sec (e+f x)+1}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {\sec ^n(e+f x) (4 n+(4 n+1) \sec (e+f x)+1)}{\sqrt {\sec (e+f x)+1}}dx}{2 n+1}+\frac {2 \sin (e+f x) \sec ^{n+1}(e+f x)}{f (2 n+1) \sqrt {\sec (e+f x)+1}}\)

\(\Big \downarrow \) 2011

\(\displaystyle \frac {(4 n+1) \int \sec ^n(e+f x) \sqrt {\sec (e+f x)+1}dx}{2 n+1}+\frac {2 \sin (e+f x) \sec ^{n+1}(e+f x)}{f (2 n+1) \sqrt {\sec (e+f x)+1}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(4 n+1) \int \csc \left (e+f x+\frac {\pi }{2}\right )^n \sqrt {\csc \left (e+f x+\frac {\pi }{2}\right )+1}dx}{2 n+1}+\frac {2 \sin (e+f x) \sec ^{n+1}(e+f x)}{f (2 n+1) \sqrt {\sec (e+f x)+1}}\)

\(\Big \downarrow \) 4293

\(\displaystyle \frac {2 \sin (e+f x) \sec ^{n+1}(e+f x)}{f (2 n+1) \sqrt {\sec (e+f x)+1}}-\frac {(4 n+1) \tan (e+f x) \int \frac {\sec ^{n-1}(e+f x)}{\sqrt {1-\sec (e+f x)}}d\sec (e+f x)}{f (2 n+1) \sqrt {1-\sec (e+f x)} \sqrt {\sec (e+f x)+1}}\)

\(\Big \downarrow \) 75

\(\displaystyle \frac {2 (4 n+1) \tan (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},1-n,\frac {3}{2},1-\sec (e+f x)\right )}{f (2 n+1) \sqrt {\sec (e+f x)+1}}+\frac {2 \sin (e+f x) \sec ^{n+1}(e+f x)}{f (2 n+1) \sqrt {\sec (e+f x)+1}}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 75

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] && (IntegerQ[m] 
 || GtQ[-d/(b*c), 0])
 

rule 2011

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> 
 Simp[(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 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4293

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) 
 + (a_)], x_Symbol] :> Simp[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 4301

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] + Simp[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]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int(sec(e + f*x)**n*sqrt(sec(e + f*x) + 1)*sec(e + f*x),x) + int(sec(e + f 
*x)**n*sqrt(sec(e + f*x) + 1),x)