Integrand size = 31, antiderivative size = 163 \[ \int \frac {(b \sec (c+d x))^n (A+B \sec (c+d x))}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=-\frac {2 A \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (5-2 n),\frac {1}{4} (9-2 n),\cos ^2(c+d x)\right ) (b \sec (c+d x))^n \sin (c+d x)}{d (5-2 n) \sec ^{\frac {5}{2}}(c+d x) \sqrt {\sin ^2(c+d x)}}-\frac {2 B \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (3-2 n),\frac {1}{4} (7-2 n),\cos ^2(c+d x)\right ) (b \sec (c+d x))^n \sin (c+d x)}{d (3-2 n) \sec ^{\frac {3}{2}}(c+d x) \sqrt {\sin ^2(c+d x)}} \]
-2*A*hypergeom([1/2, 5/4-1/2*n],[9/4-1/2*n],cos(d*x+c)^2)*(b*sec(d*x+c))^n *sin(d*x+c)/d/(5-2*n)/sec(d*x+c)^(5/2)/(sin(d*x+c)^2)^(1/2)-2*B*hypergeom( [1/2, 3/4-1/2*n],[7/4-1/2*n],cos(d*x+c)^2)*(b*sec(d*x+c))^n*sin(d*x+c)/d/( 3-2*n)/sec(d*x+c)^(3/2)/(sin(d*x+c)^2)^(1/2)
Time = 0.25 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.86 \[ \int \frac {(b \sec (c+d x))^n (A+B \sec (c+d x))}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {2 \csc (c+d x) (b \sec (c+d x))^n \left (A (-1+2 n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (-3+2 n),\frac {1}{4} (1+2 n),\sec ^2(c+d x)\right )+B (-3+2 n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (-1+2 n),\frac {1}{4} (3+2 n),\sec ^2(c+d x)\right ) \sec (c+d x)\right ) \sqrt {-\tan ^2(c+d x)}}{d (-3+2 n) (-1+2 n) \sec ^{\frac {5}{2}}(c+d x)} \]
(2*Csc[c + d*x]*(b*Sec[c + d*x])^n*(A*(-1 + 2*n)*Hypergeometric2F1[1/2, (- 3 + 2*n)/4, (1 + 2*n)/4, Sec[c + d*x]^2] + B*(-3 + 2*n)*Hypergeometric2F1[ 1/2, (-1 + 2*n)/4, (3 + 2*n)/4, Sec[c + d*x]^2]*Sec[c + d*x])*Sqrt[-Tan[c + d*x]^2])/(d*(-3 + 2*n)*(-1 + 2*n)*Sec[c + d*x]^(5/2))
Time = 0.55 (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.
| \(\displaystyle \int \frac {(A+B \sec (c+d x)) (b \sec (c+d x))^n}{\sec ^{\frac {3}{2}}(c+d x)} \, dx\) |
| \(\Big \downarrow \) 2034 |
| \(\displaystyle \sec ^{-n}(c+d x) (b \sec (c+d x))^n \int \sec ^{n-\frac {3}{2}}(c+d x) (A+B \sec (c+d x))dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sec ^{-n}(c+d x) (b \sec (c+d x))^n \int \csc \left (c+d x+\frac {\pi }{2}\right )^{n-\frac {3}{2}} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx\) |
| \(\Big \downarrow \) 4274 |
| \(\displaystyle \sec ^{-n}(c+d x) (b \sec (c+d x))^n \left (A \int \sec ^{n-\frac {3}{2}}(c+d x)dx+B \int \sec ^{n-\frac {1}{2}}(c+d x)dx\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sec ^{-n}(c+d x) (b \sec (c+d x))^n \left (A \int \csc \left (c+d x+\frac {\pi }{2}\right )^{n-\frac {3}{2}}dx+B \int \csc \left (c+d x+\frac {\pi }{2}\right )^{n-\frac {1}{2}}dx\right )\) |
| \(\Big \downarrow \) 4259 |
| \(\displaystyle \sec ^{-n}(c+d x) (b \sec (c+d x))^n \left (A \cos ^{n+\frac {1}{2}}(c+d x) \sec ^{n+\frac {1}{2}}(c+d x) \int \cos ^{\frac {3}{2}-n}(c+d x)dx+B \cos ^{n+\frac {1}{2}}(c+d x) \sec ^{n+\frac {1}{2}}(c+d x) \int \cos ^{\frac {1}{2}-n}(c+d x)dx\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sec ^{-n}(c+d x) (b \sec (c+d x))^n \left (A \cos ^{n+\frac {1}{2}}(c+d x) \sec ^{n+\frac {1}{2}}(c+d x) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{\frac {3}{2}-n}dx+B \cos ^{n+\frac {1}{2}}(c+d x) \sec ^{n+\frac {1}{2}}(c+d x) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{\frac {1}{2}-n}dx\right )\) |
| \(\Big \downarrow \) 3122 |
| \(\displaystyle \sec ^{-n}(c+d x) (b \sec (c+d x))^n \left (-\frac {2 A \sin (c+d x) \sec ^{n-\frac {5}{2}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (5-2 n),\frac {1}{4} (9-2 n),\cos ^2(c+d x)\right )}{d (5-2 n) \sqrt {\sin ^2(c+d x)}}-\frac {2 B \sin (c+d x) \sec ^{n-\frac {3}{2}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (3-2 n),\frac {1}{4} (7-2 n),\cos ^2(c+d x)\right )}{d (3-2 n) \sqrt {\sin ^2(c+d x)}}\right )\) |
((b*Sec[c + d*x])^n*((-2*A*Hypergeometric2F1[1/2, (5 - 2*n)/4, (9 - 2*n)/4 , Cos[c + d*x]^2]*Sec[c + d*x]^(-5/2 + n)*Sin[c + d*x])/(d*(5 - 2*n)*Sqrt[ Sin[c + d*x]^2]) - (2*B*Hypergeometric2F1[1/2, (3 - 2*n)/4, (7 - 2*n)/4, C os[c + d*x]^2]*Sec[c + d*x]^(-3/2 + n)*Sin[c + d*x])/(d*(3 - 2*n)*Sqrt[Sin [c + d*x]^2])))/Sec[c + d*x]^n
3.1.42.3.1 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]
\[\int \frac {\left (b \sec \left (d x +c \right )\right )^{n} \left (A +B \sec \left (d x +c \right )\right )}{\sec \left (d x +c \right )^{\frac {3}{2}}}d x\]
\[ \int \frac {(b \sec (c+d x))^n (A+B \sec (c+d x))}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\int { \frac {{\left (B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{n}}{\sec \left (d x + c\right )^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {(b \sec (c+d x))^n (A+B \sec (c+d x))}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {\left (b \sec {\left (c + d x \right )}\right )^{n} \left (A + B \sec {\left (c + d x \right )}\right )}{\sec ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx \]
\[ \int \frac {(b \sec (c+d x))^n (A+B \sec (c+d x))}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\int { \frac {{\left (B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{n}}{\sec \left (d x + c\right )^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {(b \sec (c+d x))^n (A+B \sec (c+d x))}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\int { \frac {{\left (B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{n}}{\sec \left (d x + c\right )^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {(b \sec (c+d x))^n (A+B \sec (c+d x))}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^n}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \]