\(\int \frac {(d \sec (e+f x))^n}{(1+\sec (e+f x))^{3/2}} \, dx\) [306]

Optimal result

Integrand size = 23, antiderivative size = 84 \[ \int \frac {(d \sec (e+f x))^n}{(1+\sec (e+f x))^{3/2}} \, dx=\frac {\operatorname {AppellF1}\left (\frac {1}{2},1-n,2,\frac {3}{2},1-\sec (e+f x),\frac {1}{2} (1-\sec (e+f x))\right ) \sec ^{1-n}(e+f x) (d \sec (e+f x))^n \sin (e+f x)}{2 f \sqrt {1+\sec (e+f x)}} \] Output:

1/2*AppellF1(1/2,1-n,2,3/2,1-sec(f*x+e),1/2-1/2*sec(f*x+e))*sec(f*x+e)^(1- 
n)*(d*sec(f*x+e))^n*sin(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. \(3003\) vs. \(2(84)=168\).

Time = 6.14 (sec) , antiderivative size = 3003, normalized size of antiderivative = 35.75 \[ \int \frac {(d \sec (e+f x))^n}{(1+\sec (e+f x))^{3/2}} \, dx=\text {Result too large to show} \] Input:

Integrate[(d*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 - n + (-3 + 2*n)/2)*(d*Sec[ 
e + f*x])^n*(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, 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)*((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, T 
an[(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)*Appe 
llF1[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*Ap 
pellF1[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, -Ta 
n[(e + f*x)/2]^2] + (-3 + 2*n)*AppellF1[3/2, -1/2 + n, 1 - n, 5/2, Tan[...
 

Rubi [A] (warning: unable to verify)

Time = 0.28 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.87, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3042, 4314, 150}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4314

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

\(\Big \downarrow \) 150

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 150

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_ 
] :> Simp[c^n*e^p*((b*x)^(m + 1)/(b*(m + 1)))*AppellF1[m + 1, -n, -p, m + 2 
, (-d)*(x/c), (-f)*(x/e)], x] /; FreeQ[{b, c, d, e, f, m, n, p}, x] &&  !In 
tegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])
 

rule 3042

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

rule 4314

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
(a_))^(m_), 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)*((a + b*x)^(m - 1/2 
)/Sqrt[a - b*x]), x], x, 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]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral((d*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]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

d**n*int((sec(e + f*x)**n*sqrt(sec(e + f*x) + 1))/(sec(e + f*x)**2 + 2*sec 
(e + f*x) + 1),x)