3.4.12 \(\int (-\sec (e+f x))^n (a+a \sec (e+f x))^{3/2} \, dx\) [312]

3.4.12.1 Optimal result

Integrand size = 25, antiderivative size = 130 \[ \int (-\sec (e+f x))^n (a+a \sec (e+f x))^{3/2} \, dx=\frac {2 a^2 (1+4 n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},1-n,\frac {3}{2},1-\sec (e+f x)\right ) (-\sec (e+f x))^n \sec ^{1-n}(e+f x) \sin (e+f x)}{f (1+2 n) \sqrt {a+a \sec (e+f x)}}+\frac {2 a^2 (-\sec (e+f x))^n \tan (e+f x)}{f (1+2 n) \sqrt {a+a \sec (e+f x)}} \]

output

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

3.4.12.2 Mathematica [A] (verified)

Time = 0.50 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.68 \[ \int (-\sec (e+f x))^n (a+a \sec (e+f x))^{3/2} \, dx=\frac {a \left (-1+(1+4 n) \cos ^{\frac {1}{2}+n}(e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{2}+n,\frac {3}{2},2 \sin ^2\left (\frac {1}{2} (e+f x)\right )\right )\right ) (-\sec (e+f x))^n \sqrt {a (1+\sec (e+f x))} \tan \left (\frac {1}{2} (e+f x)\right )}{f n} \]

input

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

output

(a*(-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[a*(1 + Sec[e + f*x])]*T 
an[(e + f*x)/2])/(f*n)
 

3.4.12.3 Rubi [A] (verified)

Time = 0.51 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4301

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2011

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4293

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

\(\Big \downarrow \) 77

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

\(\Big \downarrow \) 75

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

input

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

output

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

3.4.12.3.1 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]
 

3.4.12.4 Maple [F]

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

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)
 

3.4.12.5 Fricas [F]

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

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

3.4.12.6 Sympy [F]

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

input

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

output

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

3.4.12.7 Maxima [F]

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

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)
 

3.4.12.8 Giac [F]

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

input

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

output

sage0*x
 

3.4.12.9 Mupad [F(-1)]

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

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)