\(\int \frac {\sec ^n(e+f x)}{(1+\sec (e+f x))^{3/2}} \, dx\) [298]

Optimal result

Integrand size = 21, antiderivative size = 62 \[ \int \frac {\sec ^n(e+f x)}{(1+\sec (e+f x))^{3/2}} \, dx=\frac {\operatorname {AppellF1}\left (\frac {1}{2},1-n,2,\frac {3}{2},1-\sec (e+f x),\frac {1}{2} (1-\sec (e+f x))\right ) \tan (e+f x)}{2 f \sqrt {1+\sec (e+f x)}} \] Output:

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)/f/( 
1+sec(f*x+e))^(1/2)
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(2990\) vs. \(2(62)=124\).

Time = 14.71 (sec) , antiderivative size = 2990, normalized size of antiderivative = 48.23 \[ \int \frac {\sec ^n(e+f x)}{(1+\sec (e+f x))^{3/2}} \, dx=\text {Result too large to show} \] Input:

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

Output:

(6*AppellF1[1/2, -3/2 + n, 1 - n, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/ 
2]^2]*(Sec[(e + f*x)/2]^2)^n*Sec[e + f*x]^(1/2 + (-3 + 2*n)/2)*(Cos[(e + f 
*x)/2]^2*Sec[e + f*x])^(3/2 + n)*Tan[(e + f*x)/2]*(-1 + Tan[(e + f*x)/2]^2 
)^2)/(f*(1 + Sec[e + f*x])^(3/2)*(3*AppellF1[1/2, -3/2 + n, 1 - n, 3/2, Ta 
n[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (2*(-1 + n)*AppellF1[3/2, -3/2 + 
n, 2 - n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (-3 + 2*n)*Appel 
lF1[3/2, -1/2 + n, 1 - n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*T 
an[(e + f*x)/2]^2)*((12*AppellF1[1/2, -3/2 + n, 1 - n, 3/2, Tan[(e + f*x)/ 
2]^2, -Tan[(e + f*x)/2]^2]*Cos[e + f*x]*(Sec[(e + f*x)/2]^2)^(1 + n)*(Cos[ 
(e + f*x)/2]^2*Sec[e + f*x])^(3/2 + n)*Tan[(e + f*x)/2]^2*(-1 + Tan[(e + f 
*x)/2]^2))/(3*AppellF1[1/2, -3/2 + n, 1 - n, 3/2, Tan[(e + f*x)/2]^2, -Tan 
[(e + f*x)/2]^2] + (2*(-1 + n)*AppellF1[3/2, -3/2 + n, 2 - n, 5/2, Tan[(e 
+ f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (-3 + 2*n)*AppellF1[3/2, -1/2 + n, 1 - 
 n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Tan[(e + f*x)/2]^2) + ( 
3*AppellF1[1/2, -3/2 + n, 1 - n, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2 
]^2]*Cos[e + f*x]*(Sec[(e + f*x)/2]^2)^(1 + n)*(Cos[(e + f*x)/2]^2*Sec[e + 
 f*x])^(3/2 + n)*(-1 + Tan[(e + f*x)/2]^2)^2)/(3*AppellF1[1/2, -3/2 + n, 1 
 - n, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (2*(-1 + n)*AppellF1 
[3/2, -3/2 + n, 2 - n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (-3 
 + 2*n)*AppellF1[3/2, -1/2 + n, 1 - n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e...
 

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3042, 4312, 148, 333}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4312

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

\(\Big \downarrow \) 148

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

\(\Big \downarrow \) 333

\(\displaystyle \frac {\tan (e+f x) \operatorname {AppellF1}\left (\frac {1}{2},1-n,2,\frac {3}{2},1-\sec (e+f x),\frac {1}{2} (1-\sec (e+f x))\right )}{2 f \sqrt {\sec (e+f x)+1}}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 148

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))^(p_.), 
x_] :> With[{k = Denominator[m]}, Simp[k/b   Subst[Int[x^(k*(m + 1) - 1)*(c 
 + d*(x^k/b))^n*(e + f*(x^k/b))^p, x], x, (b*x)^(1/k)], x]] /; FreeQ[{b, c, 
 d, e, f, n, p}, x] && FractionQ[m] && IntegerQ[p]
 

rule 333

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[a^p*c^q*x*AppellF1[1/2, -p, -q, 3/2, (-b)*(x^2/a), (-d)*(x^2/c)], x] /; F 
reeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[p] || GtQ[a, 
0]) && (IntegerQ[q] || GtQ[c, 0])
 

rule 3042

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

rule 4312

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

Maple [F]

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

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]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

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

Output:

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

Reduce [F]

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