3.4.25 \(\int \sec ^n(e+f x) (a-a \sec (e+f x))^{3/2} \, dx\) [325]

3.4.25.1 Optimal result

Integrand size = 24, antiderivative size = 130 \[ \int \sec ^n(e+f x) (a-a \sec (e+f x))^{3/2} \, dx=\frac {2 a^2 \sec ^{1+n}(e+f x) \sin (e+f x)}{f (1+2 n) \sqrt {a-a \sec (e+f x)}}+\frac {2 a^2 (1+4 n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},1-n,\frac {3}{2},1+\sec (e+f x)\right ) (-\sec (e+f x))^{-n} \sec ^{1+n}(e+f x) \sin (e+f x)}{f (1+2 n) \sqrt {a-a \sec (e+f x)}} \]

output

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

3.4.25.2 Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 2.73 (sec) , antiderivative size = 302, normalized size of antiderivative = 2.32 \[ \int \sec ^n(e+f x) (a-a \sec (e+f x))^{3/2} \, dx=\frac {2^{-\frac {3}{2}+n} e^{-\frac {1}{2} i (e+f x)} \left (\frac {e^{i (e+f x)}}{1+e^{2 i (e+f x)}}\right )^{\frac {1}{2}+n} \left (1+e^{2 i (e+f x)}\right )^{\frac {1}{2}+n} \csc ^3\left (\frac {1}{2} (e+f x)\right ) \left (-\frac {\operatorname {Hypergeometric2F1}\left (\frac {n}{2},\frac {3}{2}+n,\frac {2+n}{2},-e^{2 i (e+f x)}\right )}{n}+\frac {3 e^{i (e+f x)} \operatorname {Hypergeometric2F1}\left (\frac {1+n}{2},\frac {3}{2}+n,\frac {3+n}{2},-e^{2 i (e+f x)}\right )}{1+n}-\frac {3 e^{2 i (e+f x)} \operatorname {Hypergeometric2F1}\left (\frac {3}{2}+n,\frac {2+n}{2},\frac {4+n}{2},-e^{2 i (e+f x)}\right )}{2+n}+\frac {e^{3 i (e+f x)} \operatorname {Hypergeometric2F1}\left (\frac {3}{2}+n,\frac {3+n}{2},\frac {5+n}{2},-e^{2 i (e+f x)}\right )}{3+n}\right ) (a-a \sec (e+f x))^{3/2}}{f \sec ^{\frac {3}{2}}(e+f x)} \]

input

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

output

(2^(-3/2 + n)*(E^(I*(e + f*x))/(1 + E^((2*I)*(e + f*x))))^(1/2 + n)*(1 + E 
^((2*I)*(e + f*x)))^(1/2 + n)*Csc[(e + f*x)/2]^3*(-(Hypergeometric2F1[n/2, 
 3/2 + n, (2 + n)/2, -E^((2*I)*(e + f*x))]/n) + (3*E^(I*(e + f*x))*Hyperge 
ometric2F1[(1 + n)/2, 3/2 + n, (3 + n)/2, -E^((2*I)*(e + f*x))])/(1 + n) - 
 (3*E^((2*I)*(e + f*x))*Hypergeometric2F1[3/2 + n, (2 + n)/2, (4 + n)/2, - 
E^((2*I)*(e + f*x))])/(2 + n) + (E^((3*I)*(e + f*x))*Hypergeometric2F1[3/2 
 + n, (3 + n)/2, (5 + n)/2, -E^((2*I)*(e + f*x))])/(3 + n))*(a - a*Sec[e + 
 f*x])^(3/2))/(E^((I/2)*(e + f*x))*f*Sec[e + f*x]^(3/2))
 

3.4.25.3 Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 4301, 27, 2011, 3042, 4293, 77, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4301

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2011

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4293

\(\displaystyle \frac {2 a^2 \sin (e+f x) \sec ^{n+1}(e+f x)}{f (2 n+1) \sqrt {a-a \sec (e+f x)}}-\frac {a^3 (4 n+1) \tan (e+f x) \int \frac {\sec ^{n-1}(e+f x)}{\sqrt {\sec (e+f x) a+a}}d\sec (e+f x)}{f (2 n+1) \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\)

\(\Big \downarrow \) 77

\(\displaystyle \frac {a^3 (4 n+1) \sin (e+f x) \sec ^{n+1}(e+f x) (-\sec (e+f x))^{-n} \int \frac {(-\sec (e+f x))^{n-1}}{\sqrt {\sec (e+f x) a+a}}d\sec (e+f x)}{f (2 n+1) \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}+\frac {2 a^2 \sin (e+f x) \sec ^{n+1}(e+f x)}{f (2 n+1) \sqrt {a-a \sec (e+f x)}}\)

\(\Big \downarrow \) 75

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

input

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

output

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

3.4.25.3.1 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 77

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

3.4.25.4 Maple [F]

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

input

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

output

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

3.4.25.5 Fricas [F]

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

input

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

output

integral(-(a*sec(f*x + e) - a)*sqrt(-a*sec(f*x + e) + a)*sec(f*x + e)^n, x 
)
 

3.4.25.6 Sympy [F]

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

input

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

output

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

3.4.25.7 Maxima [F]

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

input

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

output

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

3.4.25.8 Giac [F]

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

input

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

output

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

3.4.25.9 Mupad [F(-1)]

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

input

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

output

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