Integrand size = 33, antiderivative size = 140 \[ \int \sqrt {\sec (c+d x)} (b \sec (c+d x))^n \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {2 C \sec ^{\frac {3}{2}}(c+d x) (b \sec (c+d x))^n \sin (c+d x)}{d (3+2 n)}-\frac {2 (C+2 C n+A (3+2 n)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (1-2 n),\frac {1}{4} (5-2 n),\cos ^2(c+d x)\right ) (b \sec (c+d x))^n \sin (c+d x)}{d (1-2 n) (3+2 n) \sqrt {\sec (c+d x)} \sqrt {\sin ^2(c+d x)}} \]
2*C*sec(d*x+c)^(3/2)*(b*sec(d*x+c))^n*sin(d*x+c)/d/(3+2*n)-2*(C+2*C*n+A*(3 +2*n))*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/(-4*n^2-4*n+3)/sec(d*x+c)^(1/2)/(sin(d*x+c)^2)^(1/2)
Time = 0.43 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.01 \[ \int \sqrt {\sec (c+d x)} (b \sec (c+d x))^n \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {2 \csc (c+d x) (b \sec (c+d x))^n \left (A (5+2 n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (1+2 n),\frac {1}{4} (5+2 n),\sec ^2(c+d x)\right )+C (1+2 n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (5+2 n),\frac {1}{4} (9+2 n),\sec ^2(c+d x)\right ) \sec ^2(c+d x)\right ) \sqrt {-\tan ^2(c+d x)}}{d (1+2 n) (5+2 n) \sqrt {\sec (c+d x)}} \]
(2*Csc[c + d*x]*(b*Sec[c + d*x])^n*(A*(5 + 2*n)*Hypergeometric2F1[1/2, (1 + 2*n)/4, (5 + 2*n)/4, Sec[c + d*x]^2] + C*(1 + 2*n)*Hypergeometric2F1[1/2 , (5 + 2*n)/4, (9 + 2*n)/4, Sec[c + d*x]^2]*Sec[c + d*x]^2)*Sqrt[-Tan[c + d*x]^2])/(d*(1 + 2*n)*(5 + 2*n)*Sqrt[Sec[c + d*x]])
Time = 0.53 (sec) , antiderivative size = 145, 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.
\(\displaystyle \int \sqrt {\sec (c+d x)} \left (A+C \sec ^2(c+d x)\right ) (b \sec (c+d x))^n \, dx\) |
\(\Big \downarrow \) 2034 |
\(\displaystyle \sec ^{-n}(c+d x) (b \sec (c+d x))^n \int \sec ^{n+\frac {1}{2}}(c+d x) \left (C \sec ^2(c+d x)+A\right )dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sec ^{-n}(c+d x) (b \sec (c+d x))^n \int \csc \left (c+d x+\frac {\pi }{2}\right )^{n+\frac {1}{2}} \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A\right )dx\) |
\(\Big \downarrow \) 4534 |
\(\displaystyle \sec ^{-n}(c+d x) (b \sec (c+d x))^n \left (\frac {(A (2 n+3)+2 C n+C) \int \sec ^{n+\frac {1}{2}}(c+d x)dx}{2 n+3}+\frac {2 C \sin (c+d x) \sec ^{n+\frac {3}{2}}(c+d x)}{d (2 n+3)}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sec ^{-n}(c+d x) (b \sec (c+d x))^n \left (\frac {(A (2 n+3)+2 C n+C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^{n+\frac {1}{2}}dx}{2 n+3}+\frac {2 C \sin (c+d x) \sec ^{n+\frac {3}{2}}(c+d x)}{d (2 n+3)}\right )\) |
\(\Big \downarrow \) 4259 |
\(\displaystyle \sec ^{-n}(c+d x) (b \sec (c+d x))^n \left (\frac {(A (2 n+3)+2 C n+C) \cos ^{n+\frac {1}{2}}(c+d x) \sec ^{n+\frac {1}{2}}(c+d x) \int \cos ^{-n-\frac {1}{2}}(c+d x)dx}{2 n+3}+\frac {2 C \sin (c+d x) \sec ^{n+\frac {3}{2}}(c+d x)}{d (2 n+3)}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sec ^{-n}(c+d x) (b \sec (c+d x))^n \left (\frac {(A (2 n+3)+2 C n+C) \cos ^{n+\frac {1}{2}}(c+d x) \sec ^{n+\frac {1}{2}}(c+d x) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{-n-\frac {1}{2}}dx}{2 n+3}+\frac {2 C \sin (c+d x) \sec ^{n+\frac {3}{2}}(c+d x)}{d (2 n+3)}\right )\) |
\(\Big \downarrow \) 3122 |
\(\displaystyle \sec ^{-n}(c+d x) (b \sec (c+d x))^n \left (\frac {2 C \sin (c+d x) \sec ^{n+\frac {3}{2}}(c+d x)}{d (2 n+3)}-\frac {2 (A (2 n+3)+2 C n+C) \sin (c+d x) \sec ^{n-\frac {1}{2}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (1-2 n),\frac {1}{4} (5-2 n),\cos ^2(c+d x)\right )}{d (1-2 n) (2 n+3) \sqrt {\sin ^2(c+d x)}}\right )\) |
((b*Sec[c + d*x])^n*((2*C*Sec[c + d*x]^(3/2 + n)*Sin[c + d*x])/(d*(3 + 2*n )) - (2*(C + 2*C*n + A*(3 + 2*n))*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 )*(3 + 2*n)*Sqrt[Sin[c + d*x]^2])))/Sec[c + d*x]^n
3.1.36.3.1 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]
\[\int \left (b \sec \left (d x +c \right )\right )^{n} \left (A +C \sec \left (d x +c \right )^{2}\right ) \sqrt {\sec \left (d x +c \right )}d x\]
\[ \int \sqrt {\sec (c+d x)} (b \sec (c+d x))^n \left (A+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + A\right )} \left (b \sec \left (d x + c\right )\right )^{n} \sqrt {\sec \left (d x + c\right )} \,d x } \]
Timed out. \[ \int \sqrt {\sec (c+d x)} (b \sec (c+d x))^n \left (A+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]
\[ \int \sqrt {\sec (c+d x)} (b \sec (c+d x))^n \left (A+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + A\right )} \left (b \sec \left (d x + c\right )\right )^{n} \sqrt {\sec \left (d x + c\right )} \,d x } \]
\[ \int \sqrt {\sec (c+d x)} (b \sec (c+d x))^n \left (A+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + A\right )} \left (b \sec \left (d x + c\right )\right )^{n} \sqrt {\sec \left (d x + c\right )} \,d x } \]
Timed out. \[ \int \sqrt {\sec (c+d x)} (b \sec (c+d x))^n \left (A+C \sec ^2(c+d x)\right ) \, dx=\int \left (A+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )\,{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^n\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}} \,d x \]