Integrand size = 17, antiderivative size = 61 \[ \int \sec (c+d x) (b \sec (c+d x))^n \, dx=\frac {\operatorname {Hypergeometric2F1}\left (\frac {1}{2},-\frac {n}{2},\frac {2-n}{2},\cos ^2(c+d x)\right ) (b \sec (c+d x))^n \sin (c+d x)}{d n \sqrt {\sin ^2(c+d x)}} \] Output:
hypergeom([1/2, -1/2*n],[1-1/2*n],cos(d*x+c)^2)*(b*sec(d*x+c))^n*sin(d*x+c )/d/n/(sin(d*x+c)^2)^(1/2)
Time = 0.03 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.07 \[ \int \sec (c+d x) (b \sec (c+d x))^n \, dx=\frac {\csc (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+n}{2},\frac {3+n}{2},\sec ^2(c+d x)\right ) (b \sec (c+d x))^n \sqrt {-\tan ^2(c+d x)}}{d (1+n)} \] Input:
Integrate[Sec[c + d*x]*(b*Sec[c + d*x])^n,x]
Output:
(Csc[c + d*x]*Hypergeometric2F1[1/2, (1 + n)/2, (3 + n)/2, Sec[c + d*x]^2] *(b*Sec[c + d*x])^n*Sqrt[-Tan[c + d*x]^2])/(d*(1 + n))
Time = 0.30 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {2030, 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.
\(\displaystyle \int \sec (c+d x) (b \sec (c+d x))^n \, dx\) |
\(\Big \downarrow \) 2030 |
\(\displaystyle \frac {\int (b \sec (c+d x))^{n+1}dx}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{n+1}dx}{b}\) |
\(\Big \downarrow \) 4259 |
\(\displaystyle \frac {\left (\frac {\cos (c+d x)}{b}\right )^n (b \sec (c+d x))^n \int \left (\frac {\cos (c+d x)}{b}\right )^{-n-1}dx}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\left (\frac {\cos (c+d x)}{b}\right )^n (b \sec (c+d x))^n \int \left (\frac {\sin \left (c+d x+\frac {\pi }{2}\right )}{b}\right )^{-n-1}dx}{b}\) |
\(\Big \downarrow \) 3122 |
\(\displaystyle \frac {\sin (c+d x) (b \sec (c+d x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-\frac {n}{2},\frac {2-n}{2},\cos ^2(c+d x)\right )}{d n \sqrt {\sin ^2(c+d x)}}\) |
Input:
Int[Sec[c + d*x]*(b*Sec[c + d*x])^n,x]
Output:
(Hypergeometric2F1[1/2, -1/2*n, (2 - n)/2, Cos[c + d*x]^2]*(b*Sec[c + d*x] )^n*Sin[c + d*x])/(d*n*Sqrt[Sin[c + d*x]^2])
Int[(Fx_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Simp[1/b^m Int[(b*v) ^(m + n)*Fx, x], x] /; FreeQ[{b, n}, x] && IntegerQ[m]
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]
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]
\[\int \sec \left (d x +c \right ) \left (b \sec \left (d x +c \right )\right )^{n}d x\]
Input:
int(sec(d*x+c)*(b*sec(d*x+c))^n,x)
Output:
int(sec(d*x+c)*(b*sec(d*x+c))^n,x)
\[ \int \sec (c+d x) (b \sec (c+d x))^n \, dx=\int { \left (b \sec \left (d x + c\right )\right )^{n} \sec \left (d x + c\right ) \,d x } \] Input:
integrate(sec(d*x+c)*(b*sec(d*x+c))^n,x, algorithm="fricas")
Output:
integral((b*sec(d*x + c))^n*sec(d*x + c), x)
\[ \int \sec (c+d x) (b \sec (c+d x))^n \, dx=\int \left (b \sec {\left (c + d x \right )}\right )^{n} \sec {\left (c + d x \right )}\, dx \] Input:
integrate(sec(d*x+c)*(b*sec(d*x+c))**n,x)
Output:
Integral((b*sec(c + d*x))**n*sec(c + d*x), x)
\[ \int \sec (c+d x) (b \sec (c+d x))^n \, dx=\int { \left (b \sec \left (d x + c\right )\right )^{n} \sec \left (d x + c\right ) \,d x } \] Input:
integrate(sec(d*x+c)*(b*sec(d*x+c))^n,x, algorithm="maxima")
Output:
integrate((b*sec(d*x + c))^n*sec(d*x + c), x)
\[ \int \sec (c+d x) (b \sec (c+d x))^n \, dx=\int { \left (b \sec \left (d x + c\right )\right )^{n} \sec \left (d x + c\right ) \,d x } \] Input:
integrate(sec(d*x+c)*(b*sec(d*x+c))^n,x, algorithm="giac")
Output:
integrate((b*sec(d*x + c))^n*sec(d*x + c), x)
Timed out. \[ \int \sec (c+d x) (b \sec (c+d x))^n \, dx=\int \frac {{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^n}{\cos \left (c+d\,x\right )} \,d x \] Input:
int((b/cos(c + d*x))^n/cos(c + d*x),x)
Output:
int((b/cos(c + d*x))^n/cos(c + d*x), x)
\[ \int \sec (c+d x) (b \sec (c+d x))^n \, dx=b^{n} \left (\int \sec \left (d x +c \right )^{n} \sec \left (d x +c \right )d x \right ) \] Input:
int(sec(d*x+c)*(b*sec(d*x+c))^n,x)
Output:
b**n*int(sec(c + d*x)**n*sec(c + d*x),x)