Integrand size = 41, antiderivative size = 223 \[ \int \sec ^{\frac {5}{2}}(c+d x) (b \sec (c+d x))^n \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 C \sec ^{\frac {7}{2}}(c+d x) (b \sec (c+d x))^n \sin (c+d x)}{d (7+2 n)}+\frac {2 (C (5+2 n)+A (7+2 n)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (-3-2 n),\frac {1}{4} (1-2 n),\cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x) (b \sec (c+d x))^n \sin (c+d x)}{d (3+2 n) (7+2 n) \sqrt {\sin ^2(c+d x)}}+\frac {2 B \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (-5-2 n),\frac {1}{4} (-1-2 n),\cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) (b \sec (c+d x))^n \sin (c+d x)}{d (5+2 n) \sqrt {\sin ^2(c+d x)}} \]
2*C*sec(d*x+c)^(7/2)*(b*sec(d*x+c))^n*sin(d*x+c)/d/(7+2*n)+2*(C*(5+2*n)+A* (7+2*n))*hypergeom([1/2, -3/4-1/2*n],[1/4-1/2*n],cos(d*x+c)^2)*sec(d*x+c)^ (3/2)*(b*sec(d*x+c))^n*sin(d*x+c)/d/(4*n^2+20*n+21)/(sin(d*x+c)^2)^(1/2)+2 *B*hypergeom([1/2, -5/4-1/2*n],[-1/4-1/2*n],cos(d*x+c)^2)*sec(d*x+c)^(5/2) *(b*sec(d*x+c))^n*sin(d*x+c)/d/(5+2*n)/(sin(d*x+c)^2)^(1/2)
Time = 2.01 (sec) , antiderivative size = 204, normalized size of antiderivative = 0.91 \[ \int \sec ^{\frac {5}{2}}(c+d x) (b \sec (c+d x))^n \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 \csc (c+d x) \sec ^{\frac {3}{2}}(c+d x) (b \sec (c+d x))^n \left (A \left (63+32 n+4 n^2\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (5+2 n),\frac {1}{4} (9+2 n),\sec ^2(c+d x)\right )+(5+2 n) \left (B (9+2 n) \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (7+2 n),\frac {1}{4} (11+2 n),\sec ^2(c+d x)\right )+C (7+2 n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (9+2 n),\frac {1}{4} (13+2 n),\sec ^2(c+d x)\right )\right ) \sec ^2(c+d x)\right ) \sqrt {-\tan ^2(c+d x)}}{d (5+2 n) (7+2 n) (9+2 n)} \]
(2*Csc[c + d*x]*Sec[c + d*x]^(3/2)*(b*Sec[c + d*x])^n*(A*(63 + 32*n + 4*n^ 2)*Hypergeometric2F1[1/2, (5 + 2*n)/4, (9 + 2*n)/4, Sec[c + d*x]^2] + (5 + 2*n)*(B*(9 + 2*n)*Cos[c + d*x]*Hypergeometric2F1[1/2, (7 + 2*n)/4, (11 + 2*n)/4, Sec[c + d*x]^2] + C*(7 + 2*n)*Hypergeometric2F1[1/2, (9 + 2*n)/4, (13 + 2*n)/4, Sec[c + d*x]^2])*Sec[c + d*x]^2)*Sqrt[-Tan[c + d*x]^2])/(d*( 5 + 2*n)*(7 + 2*n)*(9 + 2*n))
Time = 0.97 (sec) , antiderivative size = 220, normalized size of antiderivative = 0.99, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.293, Rules used = {2034, 3042, 4535, 3042, 4259, 3042, 3122, 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.
| \(\displaystyle \int \sec ^{\frac {5}{2}}(c+d x) (b \sec (c+d x))^n \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 2034 |
| \(\displaystyle \sec ^{-n}(c+d x) (b \sec (c+d x))^n \int \sec ^{n+\frac {5}{2}}(c+d x) \left (C \sec ^2(c+d x)+B \sec (c+d x)+A\right )dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sec ^{-n}(c+d x) (b \sec (c+d x))^n \int \csc \left (c+d x+\frac {\pi }{2}\right )^{n+\frac {5}{2}} \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+B \csc \left (c+d x+\frac {\pi }{2}\right )+A\right )dx\) |
| \(\Big \downarrow \) 4535 |
| \(\displaystyle \sec ^{-n}(c+d x) (b \sec (c+d x))^n \left (\int \sec ^{n+\frac {5}{2}}(c+d x) \left (C \sec ^2(c+d x)+A\right )dx+B \int \sec ^{n+\frac {7}{2}}(c+d x)dx\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sec ^{-n}(c+d x) (b \sec (c+d x))^n \left (\int \csc \left (c+d x+\frac {\pi }{2}\right )^{n+\frac {5}{2}} \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A\right )dx+B \int \csc \left (c+d x+\frac {\pi }{2}\right )^{n+\frac {7}{2}}dx\right )\) |
| \(\Big \downarrow \) 4259 |
| \(\displaystyle \sec ^{-n}(c+d x) (b \sec (c+d x))^n \left (\int \csc \left (c+d x+\frac {\pi }{2}\right )^{n+\frac {5}{2}} \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A\right )dx+B \cos ^{n+\frac {1}{2}}(c+d x) \sec ^{n+\frac {1}{2}}(c+d x) \int \cos ^{-n-\frac {7}{2}}(c+d x)dx\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sec ^{-n}(c+d x) (b \sec (c+d x))^n \left (\int \csc \left (c+d x+\frac {\pi }{2}\right )^{n+\frac {5}{2}} \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A\right )dx+B \cos ^{n+\frac {1}{2}}(c+d x) \sec ^{n+\frac {1}{2}}(c+d x) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{-n-\frac {7}{2}}dx\right )\) |
| \(\Big \downarrow \) 3122 |
| \(\displaystyle \sec ^{-n}(c+d x) (b \sec (c+d x))^n \left (\int \csc \left (c+d x+\frac {\pi }{2}\right )^{n+\frac {5}{2}} \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A\right )dx+\frac {2 B \sin (c+d x) \sec ^{n+\frac {5}{2}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (-2 n-5),\frac {1}{4} (-2 n-1),\cos ^2(c+d x)\right )}{d (2 n+5) \sqrt {\sin ^2(c+d x)}}\right )\) |
| \(\Big \downarrow \) 4534 |
| \(\displaystyle \sec ^{-n}(c+d x) (b \sec (c+d x))^n \left (\frac {(A (2 n+7)+C (2 n+5)) \int \sec ^{n+\frac {5}{2}}(c+d x)dx}{2 n+7}+\frac {2 B \sin (c+d x) \sec ^{n+\frac {5}{2}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (-2 n-5),\frac {1}{4} (-2 n-1),\cos ^2(c+d x)\right )}{d (2 n+5) \sqrt {\sin ^2(c+d x)}}+\frac {2 C \sin (c+d x) \sec ^{n+\frac {7}{2}}(c+d x)}{d (2 n+7)}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sec ^{-n}(c+d x) (b \sec (c+d x))^n \left (\frac {(A (2 n+7)+C (2 n+5)) \int \csc \left (c+d x+\frac {\pi }{2}\right )^{n+\frac {5}{2}}dx}{2 n+7}+\frac {2 B \sin (c+d x) \sec ^{n+\frac {5}{2}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (-2 n-5),\frac {1}{4} (-2 n-1),\cos ^2(c+d x)\right )}{d (2 n+5) \sqrt {\sin ^2(c+d x)}}+\frac {2 C \sin (c+d x) \sec ^{n+\frac {7}{2}}(c+d x)}{d (2 n+7)}\right )\) |
| \(\Big \downarrow \) 4259 |
| \(\displaystyle \sec ^{-n}(c+d x) (b \sec (c+d x))^n \left (\frac {(A (2 n+7)+C (2 n+5)) \cos ^{n+\frac {1}{2}}(c+d x) \sec ^{n+\frac {1}{2}}(c+d x) \int \cos ^{-n-\frac {5}{2}}(c+d x)dx}{2 n+7}+\frac {2 B \sin (c+d x) \sec ^{n+\frac {5}{2}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (-2 n-5),\frac {1}{4} (-2 n-1),\cos ^2(c+d x)\right )}{d (2 n+5) \sqrt {\sin ^2(c+d x)}}+\frac {2 C \sin (c+d x) \sec ^{n+\frac {7}{2}}(c+d x)}{d (2 n+7)}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sec ^{-n}(c+d x) (b \sec (c+d x))^n \left (\frac {(A (2 n+7)+C (2 n+5)) \cos ^{n+\frac {1}{2}}(c+d x) \sec ^{n+\frac {1}{2}}(c+d x) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{-n-\frac {5}{2}}dx}{2 n+7}+\frac {2 B \sin (c+d x) \sec ^{n+\frac {5}{2}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (-2 n-5),\frac {1}{4} (-2 n-1),\cos ^2(c+d x)\right )}{d (2 n+5) \sqrt {\sin ^2(c+d x)}}+\frac {2 C \sin (c+d x) \sec ^{n+\frac {7}{2}}(c+d x)}{d (2 n+7)}\right )\) |
| \(\Big \downarrow \) 3122 |
| \(\displaystyle \sec ^{-n}(c+d x) (b \sec (c+d x))^n \left (\frac {2 (A (2 n+7)+C (2 n+5)) \sin (c+d x) \sec ^{n+\frac {3}{2}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (-2 n-3),\frac {1}{4} (1-2 n),\cos ^2(c+d x)\right )}{d (2 n+3) (2 n+7) \sqrt {\sin ^2(c+d x)}}+\frac {2 B \sin (c+d x) \sec ^{n+\frac {5}{2}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (-2 n-5),\frac {1}{4} (-2 n-1),\cos ^2(c+d x)\right )}{d (2 n+5) \sqrt {\sin ^2(c+d x)}}+\frac {2 C \sin (c+d x) \sec ^{n+\frac {7}{2}}(c+d x)}{d (2 n+7)}\right )\) |
((b*Sec[c + d*x])^n*((2*C*Sec[c + d*x]^(7/2 + n)*Sin[c + d*x])/(d*(7 + 2*n )) + (2*(C*(5 + 2*n) + A*(7 + 2*n))*Hypergeometric2F1[1/2, (-3 - 2*n)/4, ( 1 - 2*n)/4, Cos[c + d*x]^2]*Sec[c + d*x]^(3/2 + n)*Sin[c + d*x])/(d*(3 + 2 *n)*(7 + 2*n)*Sqrt[Sin[c + d*x]^2]) + (2*B*Hypergeometric2F1[1/2, (-5 - 2* n)/4, (-1 - 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])))/Sec[c + d*x]^n
3.1.78.3.1 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[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]* (B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.)), x_Symbol] :> Simp[B/b Int[(b*Cs c[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x]^2) , x] /; FreeQ[{b, e, f, A, B, C, m}, x]
\[\int \sec \left (d x +c \right )^{\frac {5}{2}} \left (b \sec \left (d x +c \right )\right )^{n} \left (A +B \sec \left (d x +c \right )+C \sec \left (d x +c \right )^{2}\right )d x\]
\[ \int \sec ^{\frac {5}{2}}(c+d x) (b \sec (c+d x))^n \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{n} \sec \left (d x + c\right )^{\frac {5}{2}} \,d x } \]
integrate(sec(d*x+c)^(5/2)*(b*sec(d*x+c))^n*(A+B*sec(d*x+c)+C*sec(d*x+c)^2 ),x, algorithm="fricas")
integral((C*sec(d*x + c)^4 + B*sec(d*x + c)^3 + A*sec(d*x + c)^2)*(b*sec(d *x + c))^n*sqrt(sec(d*x + c)), x)
Timed out. \[ \int \sec ^{\frac {5}{2}}(c+d x) (b \sec (c+d x))^n \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]
\[ \int \sec ^{\frac {5}{2}}(c+d x) (b \sec (c+d x))^n \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{n} \sec \left (d x + c\right )^{\frac {5}{2}} \,d x } \]
integrate(sec(d*x+c)^(5/2)*(b*sec(d*x+c))^n*(A+B*sec(d*x+c)+C*sec(d*x+c)^2 ),x, algorithm="maxima")
\[ \int \sec ^{\frac {5}{2}}(c+d x) (b \sec (c+d x))^n \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{n} \sec \left (d x + c\right )^{\frac {5}{2}} \,d x } \]
integrate(sec(d*x+c)^(5/2)*(b*sec(d*x+c))^n*(A+B*sec(d*x+c)+C*sec(d*x+c)^2 ),x, algorithm="giac")
Timed out. \[ \int \sec ^{\frac {5}{2}}(c+d x) (b \sec (c+d x))^n \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int {\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^n\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right ) \,d x \]