\(\int \sec ^m(c+d x) (b \sec (c+d x))^{2/3} \, dx\) [205]

Optimal result

Integrand size = 21, antiderivative size = 82 \[ \int \sec ^m(c+d x) (b \sec (c+d x))^{2/3} \, dx=-\frac {3 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (1-3 m),\frac {1}{6} (7-3 m),\cos ^2(c+d x)\right ) \sec ^{-1+m}(c+d x) (b \sec (c+d x))^{2/3} \sin (c+d x)}{d (1-3 m) \sqrt {\sin ^2(c+d x)}} \] Output:

-3*hypergeom([1/2, 1/6-1/2*m],[7/6-1/2*m],cos(d*x+c)^2)*sec(d*x+c)^(-1+m)* 
(b*sec(d*x+c))^(2/3)*sin(d*x+c)/d/(1-3*m)/(sin(d*x+c)^2)^(1/2)
 

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.01 \[ \int \sec ^m(c+d x) (b \sec (c+d x))^{2/3} \, dx=\frac {\csc (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} \left (\frac {2}{3}+m\right ),\frac {1}{2} \left (\frac {8}{3}+m\right ),\sec ^2(c+d x)\right ) \sec ^{-1+m}(c+d x) (b \sec (c+d x))^{2/3} \sqrt {-\tan ^2(c+d x)}}{d \left (\frac {2}{3}+m\right )} \] Input:

Integrate[Sec[c + d*x]^m*(b*Sec[c + d*x])^(2/3),x]
 

Output:

(Csc[c + d*x]*Hypergeometric2F1[1/2, (2/3 + m)/2, (8/3 + m)/2, Sec[c + d*x 
]^2]*Sec[c + d*x]^(-1 + m)*(b*Sec[c + d*x])^(2/3)*Sqrt[-Tan[c + d*x]^2])/( 
d*(2/3 + m))
 

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2034, 3042, 4259, 3042, 3122}

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.

\(\displaystyle \int (b \sec (c+d x))^{2/3} \sec ^m(c+d x) \, dx\)

\(\Big \downarrow \) 2034

\(\displaystyle \frac {(b \sec (c+d x))^{2/3} \int \sec ^{m+\frac {2}{3}}(c+d x)dx}{\sec ^{\frac {2}{3}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(b \sec (c+d x))^{2/3} \int \csc \left (c+d x+\frac {\pi }{2}\right )^{m+\frac {2}{3}}dx}{\sec ^{\frac {2}{3}}(c+d x)}\)

\(\Big \downarrow \) 4259

\(\displaystyle (b \sec (c+d x))^{2/3} \cos ^{m+\frac {2}{3}}(c+d x) \sec ^m(c+d x) \int \cos ^{-m-\frac {2}{3}}(c+d x)dx\)

\(\Big \downarrow \) 3042

\(\displaystyle (b \sec (c+d x))^{2/3} \cos ^{m+\frac {2}{3}}(c+d x) \sec ^m(c+d x) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{-m-\frac {2}{3}}dx\)

\(\Big \downarrow \) 3122

\(\displaystyle -\frac {3 \sin (c+d x) (b \sec (c+d x))^{2/3} \sec ^{m-1}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (1-3 m),\frac {1}{6} (7-3 m),\cos ^2(c+d x)\right )}{d (1-3 m) \sqrt {\sin ^2(c+d x)}}\)

Input:

Int[Sec[c + d*x]^m*(b*Sec[c + d*x])^(2/3),x]
 

Output:

(-3*Hypergeometric2F1[1/2, (1 - 3*m)/6, (7 - 3*m)/6, Cos[c + d*x]^2]*Sec[c 
 + d*x]^(-1 + m)*(b*Sec[c + d*x])^(2/3)*Sin[c + d*x])/(d*(1 - 3*m)*Sqrt[Si 
n[c + d*x]^2])
 

Defintions of rubi rules used

rule 2034

Int[(Fx_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Simp[b^IntPart 
[n]*((b*v)^FracPart[n]/(a^IntPart[n]*(a*v)^FracPart[n]))   Int[(a*v)^(m + n 
)*Fx, x], x] /; FreeQ[{a, b, m, n}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] && 
  !IntegerQ[m + n]
 

rule 3042

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

rule 3122

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( 
b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1)*Sqrt[Cos[c + d*x]^2]))*Hypergeometric2 
F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n}, x] 
 &&  !IntegerQ[2*n]
 

rule 4259

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x] 
)^(n - 1)*((Sin[c + d*x]/b)^(n - 1)   Int[1/(Sin[c + d*x]/b)^n, x]), x] /; 
FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]
 

Maple [F]

\[\int \sec \left (d x +c \right )^{m} \left (b \sec \left (d x +c \right )\right )^{\frac {2}{3}}d x\]

Input:

int(sec(d*x+c)^m*(b*sec(d*x+c))^(2/3),x)
 

Output:

int(sec(d*x+c)^m*(b*sec(d*x+c))^(2/3),x)
 

Fricas [F]

\[ \int \sec ^m(c+d x) (b \sec (c+d x))^{2/3} \, dx=\int { \left (b \sec \left (d x + c\right )\right )^{\frac {2}{3}} \sec \left (d x + c\right )^{m} \,d x } \] Input:

integrate(sec(d*x+c)^m*(b*sec(d*x+c))^(2/3),x, algorithm="fricas")
 

Output:

integral((b*sec(d*x + c))^(2/3)*sec(d*x + c)^m, x)
 

Sympy [F]

\[ \int \sec ^m(c+d x) (b \sec (c+d x))^{2/3} \, dx=\int \left (b \sec {\left (c + d x \right )}\right )^{\frac {2}{3}} \sec ^{m}{\left (c + d x \right )}\, dx \] Input:

integrate(sec(d*x+c)**m*(b*sec(d*x+c))**(2/3),x)
 

Output:

Integral((b*sec(c + d*x))**(2/3)*sec(c + d*x)**m, x)
 

Maxima [F]

\[ \int \sec ^m(c+d x) (b \sec (c+d x))^{2/3} \, dx=\int { \left (b \sec \left (d x + c\right )\right )^{\frac {2}{3}} \sec \left (d x + c\right )^{m} \,d x } \] Input:

integrate(sec(d*x+c)^m*(b*sec(d*x+c))^(2/3),x, algorithm="maxima")
 

Output:

integrate((b*sec(d*x + c))^(2/3)*sec(d*x + c)^m, x)
 

Giac [F]

\[ \int \sec ^m(c+d x) (b \sec (c+d x))^{2/3} \, dx=\int { \left (b \sec \left (d x + c\right )\right )^{\frac {2}{3}} \sec \left (d x + c\right )^{m} \,d x } \] Input:

integrate(sec(d*x+c)^m*(b*sec(d*x+c))^(2/3),x, algorithm="giac")
 

Output:

integrate((b*sec(d*x + c))^(2/3)*sec(d*x + c)^m, x)
 

Mupad [F(-1)]

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

int((b/cos(c + d*x))^(2/3)*(1/cos(c + d*x))^m,x)
 

Output:

int((b/cos(c + d*x))^(2/3)*(1/cos(c + d*x))^m, x)
 

Reduce [F]

\[ \int \sec ^m(c+d x) (b \sec (c+d x))^{2/3} \, dx=b^{\frac {2}{3}} \left (\int \sec \left (d x +c \right )^{m +\frac {2}{3}}d x \right ) \] Input:

int(sec(d*x+c)^m*(b*sec(d*x+c))^(2/3),x)
 

Output:

b**(2/3)*int(sec(c + d*x)**((3*m + 2)/3),x)