Integrand size = 23, antiderivative size = 113 \[ \int (b \sec (c+d x))^n \left (A+C \sec ^2(c+d x)\right ) \, dx=-\frac {b (A+A n+C n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1-n}{2},\frac {3-n}{2},\cos ^2(c+d x)\right ) (b \sec (c+d x))^{-1+n} \sin (c+d x)}{d (1-n) (1+n) \sqrt {\sin ^2(c+d x)}}+\frac {C (b \sec (c+d x))^n \tan (c+d x)}{d (1+n)} \]
-b*(A*n+C*n+A)*hypergeom([1/2, 1/2-1/2*n],[3/2-1/2*n],cos(d*x+c)^2)*(b*sec (d*x+c))^(-1+n)*sin(d*x+c)/d/(-n^2+1)/(sin(d*x+c)^2)^(1/2)+C*(b*sec(d*x+c) )^n*tan(d*x+c)/d/(1+n)
Time = 0.44 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.77 \[ \int (b \sec (c+d x))^n \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {\cot (c+d x) (b \sec (c+d x))^n \left (C n \tan ^2(c+d x)+(A+A n+C n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {n}{2},\frac {2+n}{2},\sec ^2(c+d x)\right ) \sqrt {-\tan ^2(c+d x)}\right )}{d n (1+n)} \]
(Cot[c + d*x]*(b*Sec[c + d*x])^n*(C*n*Tan[c + d*x]^2 + (A + A*n + C*n)*Hyp ergeometric2F1[1/2, n/2, (2 + n)/2, Sec[c + d*x]^2]*Sqrt[-Tan[c + d*x]^2]) )/(d*n*(1 + n))
Time = 0.43 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3042, 4534, 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.
\(\displaystyle \int \left (A+C \sec ^2(c+d x)\right ) (b \sec (c+d x))^n \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \left (A+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right ) \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^ndx\) |
\(\Big \downarrow \) 4534 |
\(\displaystyle \frac {(A n+A+C n) \int (b \sec (c+d x))^ndx}{n+1}+\frac {C \tan (c+d x) (b \sec (c+d x))^n}{d (n+1)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {(A n+A+C n) \int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^ndx}{n+1}+\frac {C \tan (c+d x) (b \sec (c+d x))^n}{d (n+1)}\) |
\(\Big \downarrow \) 4259 |
\(\displaystyle \frac {(A n+A+C n) \left (\frac {\cos (c+d x)}{b}\right )^n (b \sec (c+d x))^n \int \left (\frac {\cos (c+d x)}{b}\right )^{-n}dx}{n+1}+\frac {C \tan (c+d x) (b \sec (c+d x))^n}{d (n+1)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {(A n+A+C n) \left (\frac {\cos (c+d x)}{b}\right )^n (b \sec (c+d x))^n \int \left (\frac {\sin \left (c+d x+\frac {\pi }{2}\right )}{b}\right )^{-n}dx}{n+1}+\frac {C \tan (c+d x) (b \sec (c+d x))^n}{d (n+1)}\) |
\(\Big \downarrow \) 3122 |
\(\displaystyle \frac {C \tan (c+d x) (b \sec (c+d x))^n}{d (n+1)}-\frac {b (A n+A+C n) \sin (c+d x) (b \sec (c+d x))^{n-1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1-n}{2},\frac {3-n}{2},\cos ^2(c+d x)\right )}{d (1-n) (n+1) \sqrt {\sin ^2(c+d x)}}\) |
-((b*(A + A*n + C*n)*Hypergeometric2F1[1/2, (1 - n)/2, (3 - n)/2, Cos[c + d*x]^2]*(b*Sec[c + d*x])^(-1 + n)*Sin[c + d*x])/(d*(1 - n)*(1 + n)*Sqrt[Si n[c + d*x]^2])) + (C*(b*Sec[c + d*x])^n*Tan[c + d*x])/(d*(1 + n))
3.1.30.3.1 Defintions of rubi rules used
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_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[(-C)*Cot[e + f*x]*((b*Csc[e + f*x])^m/(f*(m + 1) )), x] + Simp[(C*m + A*(m + 1))/(m + 1) Int[(b*Csc[e + f*x])^m, x], x] /; FreeQ[{b, e, f, A, C, m}, x] && NeQ[C*m + A*(m + 1), 0] && !LeQ[m, -1]
\[\int \left (b \sec \left (d x +c \right )\right )^{n} \left (A +C \sec \left (d x +c \right )^{2}\right )d x\]
\[ \int (b \sec (c+d x))^n \left (A+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + A\right )} \left (b \sec \left (d x + c\right )\right )^{n} \,d x } \]
\[ \int (b \sec (c+d x))^n \left (A+C \sec ^2(c+d x)\right ) \, dx=\int \left (b \sec {\left (c + d x \right )}\right )^{n} \left (A + C \sec ^{2}{\left (c + d x \right )}\right )\, dx \]
\[ \int (b \sec (c+d x))^n \left (A+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + A\right )} \left (b \sec \left (d x + c\right )\right )^{n} \,d x } \]
\[ \int (b \sec (c+d x))^n \left (A+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + A\right )} \left (b \sec \left (d x + c\right )\right )^{n} \,d x } \]
Timed out. \[ \int (b \sec (c+d x))^n \left (A+C \sec ^2(c+d x)\right ) \, dx=\int \left (A+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )\,{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^n \,d x \]