Integrand size = 29, antiderivative size = 143 \[ \int \sec ^2(c+d x) (b \sec (c+d x))^n (A+B \sec (c+d x)) \, dx=\frac {A \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-1-n),\frac {1-n}{2},\cos ^2(c+d x)\right ) (b \sec (c+d x))^{1+n} \sin (c+d x)}{b d (1+n) \sqrt {\sin ^2(c+d x)}}+\frac {B \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-2-n),-\frac {n}{2},\cos ^2(c+d x)\right ) (b \sec (c+d x))^{2+n} \sin (c+d x)}{b^2 d (2+n) \sqrt {\sin ^2(c+d x)}} \]
A*hypergeom([1/2, -1/2-1/2*n],[1/2-1/2*n],cos(d*x+c)^2)*(b*sec(d*x+c))^(1+ n)*sin(d*x+c)/b/d/(1+n)/(sin(d*x+c)^2)^(1/2)+B*hypergeom([1/2, -1-1/2*n],[ -1/2*n],cos(d*x+c)^2)*(b*sec(d*x+c))^(2+n)*sin(d*x+c)/b^2/d/(2+n)/(sin(d*x +c)^2)^(1/2)
Time = 0.17 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.83 \[ \int \sec ^2(c+d x) (b \sec (c+d x))^n (A+B \sec (c+d x)) \, dx=\frac {\csc (c+d x) \sec (c+d x) (b \sec (c+d x))^n \left (A (3+n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+n}{2},\frac {4+n}{2},\sec ^2(c+d x)\right )+B (2+n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3+n}{2},\frac {5+n}{2},\sec ^2(c+d x)\right ) \sec (c+d x)\right ) \sqrt {-\tan ^2(c+d x)}}{d (2+n) (3+n)} \]
(Csc[c + d*x]*Sec[c + d*x]*(b*Sec[c + d*x])^n*(A*(3 + n)*Hypergeometric2F1 [1/2, (2 + n)/2, (4 + n)/2, Sec[c + d*x]^2] + B*(2 + n)*Hypergeometric2F1[ 1/2, (3 + n)/2, (5 + n)/2, Sec[c + d*x]^2]*Sec[c + d*x])*Sqrt[-Tan[c + d*x ]^2])/(d*(2 + n)*(3 + n))
Time = 0.52 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.99, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {2030, 3042, 4274, 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.
\(\displaystyle \int \sec ^2(c+d x) (A+B \sec (c+d x)) (b \sec (c+d x))^n \, dx\) |
\(\Big \downarrow \) 2030 |
\(\displaystyle \frac {\int (b \sec (c+d x))^{n+2} (A+B \sec (c+d x))dx}{b^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{n+2} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{b^2}\) |
\(\Big \downarrow \) 4274 |
\(\displaystyle \frac {A \int (b \sec (c+d x))^{n+2}dx+\frac {B \int (b \sec (c+d x))^{n+3}dx}{b}}{b^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {A \int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{n+2}dx+\frac {B \int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{n+3}dx}{b}}{b^2}\) |
\(\Big \downarrow \) 4259 |
\(\displaystyle \frac {A \left (\frac {\cos (c+d x)}{b}\right )^n (b \sec (c+d x))^n \int \left (\frac {\cos (c+d x)}{b}\right )^{-n-2}dx+\frac {B \left (\frac {\cos (c+d x)}{b}\right )^n (b \sec (c+d x))^n \int \left (\frac {\cos (c+d x)}{b}\right )^{-n-3}dx}{b}}{b^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {A \left (\frac {\cos (c+d x)}{b}\right )^n (b \sec (c+d x))^n \int \left (\frac {\sin \left (c+d x+\frac {\pi }{2}\right )}{b}\right )^{-n-2}dx+\frac {B \left (\frac {\cos (c+d x)}{b}\right )^n (b \sec (c+d x))^n \int \left (\frac {\sin \left (c+d x+\frac {\pi }{2}\right )}{b}\right )^{-n-3}dx}{b}}{b^2}\) |
\(\Big \downarrow \) 3122 |
\(\displaystyle \frac {\frac {A b \sin (c+d x) (b \sec (c+d x))^{n+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-n-1),\frac {1-n}{2},\cos ^2(c+d x)\right )}{d (n+1) \sqrt {\sin ^2(c+d x)}}+\frac {B \sin (c+d x) (b \sec (c+d x))^{n+2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-n-2),-\frac {n}{2},\cos ^2(c+d x)\right )}{d (n+2) \sqrt {\sin ^2(c+d x)}}}{b^2}\) |
((A*b*Hypergeometric2F1[1/2, (-1 - n)/2, (1 - n)/2, Cos[c + d*x]^2]*(b*Sec [c + d*x])^(1 + n)*Sin[c + d*x])/(d*(1 + n)*Sqrt[Sin[c + d*x]^2]) + (B*Hyp ergeometric2F1[1/2, (-2 - n)/2, -1/2*n, Cos[c + d*x]^2]*(b*Sec[c + d*x])^( 2 + n)*Sin[c + d*x])/(d*(2 + n)*Sqrt[Sin[c + d*x]^2]))/b^2
3.1.34.3.1 Defintions of rubi rules used
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[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[a Int[(d*Csc[e + f*x])^n, x], x] + Simp[b/d In t[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]
\[\int \sec \left (d x +c \right )^{2} \left (b \sec \left (d x +c \right )\right )^{n} \left (A +B \sec \left (d x +c \right )\right )d x\]
\[ \int \sec ^2(c+d x) (b \sec (c+d x))^n (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{n} \sec \left (d x + c\right )^{2} \,d x } \]
\[ \int \sec ^2(c+d x) (b \sec (c+d x))^n (A+B \sec (c+d x)) \, dx=\int \left (b \sec {\left (c + d x \right )}\right )^{n} \left (A + B \sec {\left (c + d x \right )}\right ) \sec ^{2}{\left (c + d x \right )}\, dx \]
\[ \int \sec ^2(c+d x) (b \sec (c+d x))^n (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{n} \sec \left (d x + c\right )^{2} \,d x } \]
\[ \int \sec ^2(c+d x) (b \sec (c+d x))^n (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{n} \sec \left (d x + c\right )^{2} \,d x } \]
Timed out. \[ \int \sec ^2(c+d x) (b \sec (c+d x))^n (A+B \sec (c+d x)) \, dx=\int \frac {\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^n}{{\cos \left (c+d\,x\right )}^2} \,d x \]