Integrand size = 33, antiderivative size = 142 \[ \int \frac {(b \sec (c+d x))^n \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {2 C (b \sec (c+d x))^n \sin (c+d x)}{d (3-2 n) \sec ^{\frac {3}{2}}(c+d x)}-\frac {2 (A (3-2 n)+C (5-2 n)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (7-2 n),\frac {1}{4} (11-2 n),\cos ^2(c+d x)\right ) (b \sec (c+d x))^n \sin (c+d x)}{d (3-2 n) (7-2 n) \sec ^{\frac {7}{2}}(c+d x) \sqrt {\sin ^2(c+d x)}} \]
-2*C*(b*sec(d*x+c))^n*sin(d*x+c)/d/(3-2*n)/sec(d*x+c)^(3/2)-2*(A*(3-2*n)+C *(5-2*n))*hypergeom([1/2, 7/4-1/2*n],[11/4-1/2*n],cos(d*x+c)^2)*(b*sec(d*x +c))^n*sin(d*x+c)/d/(4*n^2-20*n+21)/sec(d*x+c)^(7/2)/(sin(d*x+c)^2)^(1/2)
Time = 0.37 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.00 \[ \int \frac {(b \sec (c+d x))^n \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\frac {2 \csc (c+d x) (b \sec (c+d x))^n \left (A (-1+2 n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (-5+2 n),\frac {1}{4} (-1+2 n),\sec ^2(c+d x)\right )+C (-5+2 n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (-1+2 n),\frac {1}{4} (3+2 n),\sec ^2(c+d x)\right ) \sec ^2(c+d x)\right ) \sqrt {-\tan ^2(c+d x)}}{d (-5+2 n) (-1+2 n) \sec ^{\frac {7}{2}}(c+d x)} \]
(2*Csc[c + d*x]*(b*Sec[c + d*x])^n*(A*(-1 + 2*n)*Hypergeometric2F1[1/2, (- 5 + 2*n)/4, (-1 + 2*n)/4, Sec[c + d*x]^2] + C*(-5 + 2*n)*Hypergeometric2F1 [1/2, (-1 + 2*n)/4, (3 + 2*n)/4, Sec[c + d*x]^2]*Sec[c + d*x]^2)*Sqrt[-Tan [c + d*x]^2])/(d*(-5 + 2*n)*(-1 + 2*n)*Sec[c + d*x]^(7/2))
Time = 0.57 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.04, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {2034, 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.
\(\displaystyle \int \frac {\left (A+C \sec ^2(c+d x)\right ) (b \sec (c+d x))^n}{\sec ^{\frac {5}{2}}(c+d x)} \, dx\) |
\(\Big \downarrow \) 2034 |
\(\displaystyle \sec ^{-n}(c+d x) (b \sec (c+d x))^n \int \sec ^{n-\frac {5}{2}}(c+d x) \left (C \sec ^2(c+d x)+A\right )dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sec ^{-n}(c+d x) (b \sec (c+d x))^n \int \csc \left (c+d x+\frac {\pi }{2}\right )^{n-\frac {5}{2}} \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A\right )dx\) |
\(\Big \downarrow \) 4534 |
\(\displaystyle \sec ^{-n}(c+d x) (b \sec (c+d x))^n \left (\frac {(A (3-2 n)+C (5-2 n)) \int \sec ^{n-\frac {5}{2}}(c+d x)dx}{3-2 n}-\frac {2 C \sin (c+d x) \sec ^{n-\frac {3}{2}}(c+d x)}{d (3-2 n)}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sec ^{-n}(c+d x) (b \sec (c+d x))^n \left (\frac {(A (3-2 n)+C (5-2 n)) \int \csc \left (c+d x+\frac {\pi }{2}\right )^{n-\frac {5}{2}}dx}{3-2 n}-\frac {2 C \sin (c+d x) \sec ^{n-\frac {3}{2}}(c+d x)}{d (3-2 n)}\right )\) |
\(\Big \downarrow \) 4259 |
\(\displaystyle \sec ^{-n}(c+d x) (b \sec (c+d x))^n \left (\frac {(A (3-2 n)+C (5-2 n)) \cos ^{n+\frac {1}{2}}(c+d x) \sec ^{n+\frac {1}{2}}(c+d x) \int \cos ^{\frac {5}{2}-n}(c+d x)dx}{3-2 n}-\frac {2 C \sin (c+d x) \sec ^{n-\frac {3}{2}}(c+d x)}{d (3-2 n)}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sec ^{-n}(c+d x) (b \sec (c+d x))^n \left (\frac {(A (3-2 n)+C (5-2 n)) \cos ^{n+\frac {1}{2}}(c+d x) \sec ^{n+\frac {1}{2}}(c+d x) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{\frac {5}{2}-n}dx}{3-2 n}-\frac {2 C \sin (c+d x) \sec ^{n-\frac {3}{2}}(c+d x)}{d (3-2 n)}\right )\) |
\(\Big \downarrow \) 3122 |
\(\displaystyle \sec ^{-n}(c+d x) (b \sec (c+d x))^n \left (-\frac {2 (A (3-2 n)+C (5-2 n)) \sin (c+d x) \sec ^{n-\frac {7}{2}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (7-2 n),\frac {1}{4} (11-2 n),\cos ^2(c+d x)\right )}{d (3-2 n) (7-2 n) \sqrt {\sin ^2(c+d x)}}-\frac {2 C \sin (c+d x) \sec ^{n-\frac {3}{2}}(c+d x)}{d (3-2 n)}\right )\) |
((b*Sec[c + d*x])^n*((-2*C*Sec[c + d*x]^(-3/2 + n)*Sin[c + d*x])/(d*(3 - 2 *n)) - (2*(A*(3 - 2*n) + C*(5 - 2*n))*Hypergeometric2F1[1/2, (7 - 2*n)/4, (11 - 2*n)/4, Cos[c + d*x]^2]*Sec[c + d*x]^(-7/2 + n)*Sin[c + d*x])/(d*(3 - 2*n)*(7 - 2*n)*Sqrt[Sin[c + d*x]^2])))/Sec[c + d*x]^n
3.1.39.3.1 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_)]*(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]
\[\int \frac {\left (b \sec \left (d x +c \right )\right )^{n} \left (A +C \sec \left (d x +c \right )^{2}\right )}{\sec \left (d x +c \right )^{\frac {5}{2}}}d x\]
\[ \int \frac {(b \sec (c+d x))^n \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} \left (b \sec \left (d x + c\right )\right )^{n}}{\sec \left (d x + c\right )^{\frac {5}{2}}} \,d x } \]
\[ \int \frac {(b \sec (c+d x))^n \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\int \frac {\left (b \sec {\left (c + d x \right )}\right )^{n} \left (A + C \sec ^{2}{\left (c + d x \right )}\right )}{\sec ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx \]
\[ \int \frac {(b \sec (c+d x))^n \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} \left (b \sec \left (d x + c\right )\right )^{n}}{\sec \left (d x + c\right )^{\frac {5}{2}}} \,d x } \]
\[ \int \frac {(b \sec (c+d x))^n \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} \left (b \sec \left (d x + c\right )\right )^{n}}{\sec \left (d x + c\right )^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {(b \sec (c+d x))^n \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\int \frac {\left (A+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )\,{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^n}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]