Integrand size = 29, antiderivative size = 109 \[ \int \sec (c+d x) (b \sec (c+d x))^n \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {(C (1+n)+A (2+n)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-\frac {n}{2},\frac {2-n}{2},\cos ^2(c+d x)\right ) (b \sec (c+d x))^n \sin (c+d x)}{d n (2+n) \sqrt {\sin ^2(c+d x)}}+\frac {C (b \sec (c+d x))^{1+n} \tan (c+d x)}{b d (2+n)} \]
(C*(1+n)+A*(2+n))*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/(2+n)/(sin(d*x+c)^2)^(1/2)+C*(b*sec(d*x+c))^(1+n)* tan(d*x+c)/b/d/(2+n)
Time = 0.34 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.06 \[ \int \sec (c+d x) (b \sec (c+d x))^n \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {\csc (c+d x) (b \sec (c+d x))^n \left (A (3+n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+n}{2},\frac {3+n}{2},\sec ^2(c+d x)\right )+C (1+n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3+n}{2},\frac {5+n}{2},\sec ^2(c+d x)\right ) \sec ^2(c+d x)\right ) \sqrt {-\tan ^2(c+d x)}}{d (1+n) (3+n)} \]
(Csc[c + d*x]*(b*Sec[c + d*x])^n*(A*(3 + n)*Hypergeometric2F1[1/2, (1 + n) /2, (3 + n)/2, Sec[c + d*x]^2] + C*(1 + n)*Hypergeometric2F1[1/2, (3 + n)/ 2, (5 + n)/2, Sec[c + d*x]^2]*Sec[c + d*x]^2)*Sqrt[-Tan[c + d*x]^2])/(d*(1 + n)*(3 + n))
Time = 0.45 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.98, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {2030, 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.
\(\displaystyle \int \sec (c+d x) \left (A+C \sec ^2(c+d x)\right ) (b \sec (c+d x))^n \, dx\) |
\(\Big \downarrow \) 2030 |
\(\displaystyle \frac {\int (b \sec (c+d x))^{n+1} \left (C \sec ^2(c+d x)+A\right )dx}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{n+1} \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A\right )dx}{b}\) |
\(\Big \downarrow \) 4534 |
\(\displaystyle \frac {\left (A+\frac {C (n+1)}{n+2}\right ) \int (b \sec (c+d x))^{n+1}dx+\frac {C \tan (c+d x) (b \sec (c+d x))^{n+1}}{d (n+2)}}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\left (A+\frac {C (n+1)}{n+2}\right ) \int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{n+1}dx+\frac {C \tan (c+d x) (b \sec (c+d x))^{n+1}}{d (n+2)}}{b}\) |
\(\Big \downarrow \) 4259 |
\(\displaystyle \frac {\left (A+\frac {C (n+1)}{n+2}\right ) \left (\frac {\cos (c+d x)}{b}\right )^n (b \sec (c+d x))^n \int \left (\frac {\cos (c+d x)}{b}\right )^{-n-1}dx+\frac {C \tan (c+d x) (b \sec (c+d x))^{n+1}}{d (n+2)}}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\left (A+\frac {C (n+1)}{n+2}\right ) \left (\frac {\cos (c+d x)}{b}\right )^n (b \sec (c+d x))^n \int \left (\frac {\sin \left (c+d x+\frac {\pi }{2}\right )}{b}\right )^{-n-1}dx+\frac {C \tan (c+d x) (b \sec (c+d x))^{n+1}}{d (n+2)}}{b}\) |
\(\Big \downarrow \) 3122 |
\(\displaystyle \frac {\frac {b \left (A+\frac {C (n+1)}{n+2}\right ) \sin (c+d x) (b \sec (c+d x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-\frac {n}{2},\frac {2-n}{2},\cos ^2(c+d x)\right )}{d n \sqrt {\sin ^2(c+d x)}}+\frac {C \tan (c+d x) (b \sec (c+d x))^{n+1}}{d (n+2)}}{b}\) |
((b*(A + (C*(1 + n))/(2 + n))*Hypergeometric2F1[1/2, -1/2*n, (2 - n)/2, Co s[c + d*x]^2]*(b*Sec[c + d*x])^n*Sin[c + d*x])/(d*n*Sqrt[Sin[c + d*x]^2]) + (C*(b*Sec[c + d*x])^(1 + n)*Tan[c + d*x])/(d*(2 + n)))/b
3.1.29.3.1 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_)]*(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]
\[\int \sec \left (d x +c \right ) \left (b \sec \left (d x +c \right )\right )^{n} \left (A +C \sec \left (d x +c \right )^{2}\right )d x\]
\[ \int \sec (c+d x) (b \sec (c+d x))^n \left (A+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + A\right )} \left (b \sec \left (d x + c\right )\right )^{n} \sec \left (d x + c\right ) \,d x } \]
\[ \int \sec (c+d x) (b \sec (c+d x))^n \left (A+C \sec ^2(c+d x)\right ) \, dx=\int \left (b \sec {\left (c + d x \right )}\right )^{n} \left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \sec {\left (c + d x \right )}\, dx \]
\[ \int \sec (c+d x) (b \sec (c+d x))^n \left (A+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + A\right )} \left (b \sec \left (d x + c\right )\right )^{n} \sec \left (d x + c\right ) \,d x } \]
\[ \int \sec (c+d x) (b \sec (c+d x))^n \left (A+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + A\right )} \left (b \sec \left (d x + c\right )\right )^{n} \sec \left (d x + c\right ) \,d x } \]
Timed out. \[ \int \sec (c+d x) (b \sec (c+d x))^n \left (A+C \sec ^2(c+d x)\right ) \, dx=\int \frac {\left (A+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )\,{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^n}{\cos \left (c+d\,x\right )} \,d x \]