Integrand size = 31, antiderivative size = 163 \[ \int \sqrt {\sec (c+d x)} (b \sec (c+d x))^n (A+B \sec (c+d x)) \, dx=-\frac {2 A \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (1-2 n),\frac {1}{4} (5-2 n),\cos ^2(c+d x)\right ) (b \sec (c+d x))^n \sin (c+d x)}{d (1-2 n) \sqrt {\sec (c+d x)} \sqrt {\sin ^2(c+d x)}}+\frac {2 B \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (-1-2 n),\frac {1}{4} (3-2 n),\cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)} (b \sec (c+d x))^n \sin (c+d x)}{d (1+2 n) \sqrt {\sin ^2(c+d x)}} \]
-2*A*hypergeom([1/2, 1/4-1/2*n],[5/4-1/2*n],cos(d*x+c)^2)*(b*sec(d*x+c))^n *sin(d*x+c)/d/(1-2*n)/sec(d*x+c)^(1/2)/(sin(d*x+c)^2)^(1/2)+2*B*hypergeom( [1/2, -1/4-1/2*n],[3/4-1/2*n],cos(d*x+c)^2)*(b*sec(d*x+c))^n*sin(d*x+c)*se c(d*x+c)^(1/2)/d/(1+2*n)/(sin(d*x+c)^2)^(1/2)
Time = 0.18 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.86 \[ \int \sqrt {\sec (c+d x)} (b \sec (c+d x))^n (A+B \sec (c+d x)) \, dx=\frac {2 \csc (c+d x) (b \sec (c+d x))^n \left (A (3+2 n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (1+2 n),\frac {1}{4} (5+2 n),\sec ^2(c+d x)\right )+B (1+2 n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (3+2 n),\frac {1}{4} (7+2 n),\sec ^2(c+d x)\right ) \sec (c+d x)\right ) \sqrt {-\tan ^2(c+d x)}}{d (1+2 n) (3+2 n) \sqrt {\sec (c+d x)}} \]
(2*Csc[c + d*x]*(b*Sec[c + d*x])^n*(A*(3 + 2*n)*Hypergeometric2F1[1/2, (1 + 2*n)/4, (5 + 2*n)/4, Sec[c + d*x]^2] + B*(1 + 2*n)*Hypergeometric2F1[1/2 , (3 + 2*n)/4, (7 + 2*n)/4, Sec[c + d*x]^2]*Sec[c + d*x])*Sqrt[-Tan[c + d* x]^2])/(d*(1 + 2*n)*(3 + 2*n)*Sqrt[Sec[c + d*x]])
Time = 0.52 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.03, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.226, Rules used = {2034, 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.
| \(\displaystyle \int \sqrt {\sec (c+d x)} (A+B \sec (c+d x)) (b \sec (c+d x))^n \, dx\) |
| \(\Big \downarrow \) 2034 |
| \(\displaystyle \sec ^{-n}(c+d x) (b \sec (c+d x))^n \int \sec ^{n+\frac {1}{2}}(c+d x) (A+B \sec (c+d x))dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sec ^{-n}(c+d x) (b \sec (c+d x))^n \int \csc \left (c+d x+\frac {\pi }{2}\right )^{n+\frac {1}{2}} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx\) |
| \(\Big \downarrow \) 4274 |
| \(\displaystyle \sec ^{-n}(c+d x) (b \sec (c+d x))^n \left (A \int \sec ^{n+\frac {1}{2}}(c+d x)dx+B \int \sec ^{n+\frac {3}{2}}(c+d x)dx\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sec ^{-n}(c+d x) (b \sec (c+d x))^n \left (A \int \csc \left (c+d x+\frac {\pi }{2}\right )^{n+\frac {1}{2}}dx+B \int \csc \left (c+d x+\frac {\pi }{2}\right )^{n+\frac {3}{2}}dx\right )\) |
| \(\Big \downarrow \) 4259 |
| \(\displaystyle \sec ^{-n}(c+d x) (b \sec (c+d x))^n \left (A \cos ^{n+\frac {1}{2}}(c+d x) \sec ^{n+\frac {1}{2}}(c+d x) \int \cos ^{-n-\frac {1}{2}}(c+d x)dx+B \cos ^{n+\frac {1}{2}}(c+d x) \sec ^{n+\frac {1}{2}}(c+d x) \int \cos ^{-n-\frac {3}{2}}(c+d x)dx\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sec ^{-n}(c+d x) (b \sec (c+d x))^n \left (A \cos ^{n+\frac {1}{2}}(c+d x) \sec ^{n+\frac {1}{2}}(c+d x) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{-n-\frac {1}{2}}dx+B \cos ^{n+\frac {1}{2}}(c+d x) \sec ^{n+\frac {1}{2}}(c+d x) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{-n-\frac {3}{2}}dx\right )\) |
| \(\Big \downarrow \) 3122 |
| \(\displaystyle \sec ^{-n}(c+d x) (b \sec (c+d x))^n \left (\frac {2 B \sin (c+d x) \sec ^{n+\frac {1}{2}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (-2 n-1),\frac {1}{4} (3-2 n),\cos ^2(c+d x)\right )}{d (2 n+1) \sqrt {\sin ^2(c+d x)}}-\frac {2 A \sin (c+d x) \sec ^{n-\frac {1}{2}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (1-2 n),\frac {1}{4} (5-2 n),\cos ^2(c+d x)\right )}{d (1-2 n) \sqrt {\sin ^2(c+d x)}}\right )\) |
((b*Sec[c + d*x])^n*((-2*A*Hypergeometric2F1[1/2, (1 - 2*n)/4, (5 - 2*n)/4 , Cos[c + d*x]^2]*Sec[c + d*x]^(-1/2 + n)*Sin[c + d*x])/(d*(1 - 2*n)*Sqrt[ Sin[c + d*x]^2]) + (2*B*Hypergeometric2F1[1/2, (-1 - 2*n)/4, (3 - 2*n)/4, Cos[c + d*x]^2]*Sec[c + d*x]^(1/2 + n)*Sin[c + d*x])/(d*(1 + 2*n)*Sqrt[Sin [c + d*x]^2])))/Sec[c + d*x]^n
3.1.40.3.1 Defintions of rubi rules used
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]
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]
\[\int \left (b \sec \left (d x +c \right )\right )^{n} \left (A +B \sec \left (d x +c \right )\right ) \sqrt {\sec \left (d x +c \right )}d x\]
\[ \int \sqrt {\sec (c+d x)} (b \sec (c+d x))^n (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{n} \sqrt {\sec \left (d x + c\right )} \,d x } \]
\[ \int \sqrt {\sec (c+d x)} (b \sec (c+d x))^n (A+B \sec (c+d x)) \, dx=\int \left (b \sec {\left (c + d x \right )}\right )^{n} \left (A + B \sec {\left (c + d x \right )}\right ) \sqrt {\sec {\left (c + d x \right )}}\, dx \]
\[ \int \sqrt {\sec (c+d x)} (b \sec (c+d x))^n (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{n} \sqrt {\sec \left (d x + c\right )} \,d x } \]
\[ \int \sqrt {\sec (c+d x)} (b \sec (c+d x))^n (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{n} \sqrt {\sec \left (d x + c\right )} \,d x } \]
Timed out. \[ \int \sqrt {\sec (c+d x)} (b \sec (c+d x))^n (A+B \sec (c+d x)) \, dx=\int \left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^n\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}} \,d x \]