Integrand size = 41, antiderivative size = 221 \[ \int \frac {(b \sec (c+d x))^n \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=-\frac {2 C (b \sec (c+d x))^n \sin (c+d x)}{d (1-2 n) \sqrt {\sec (c+d x)}}-\frac {2 (A+C (3-2 n)-2 A n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (5-2 n),\frac {1}{4} (9-2 n),\cos ^2(c+d x)\right ) (b \sec (c+d x))^n \sin (c+d x)}{d (1-2 n) (5-2 n) \sec ^{\frac {5}{2}}(c+d x) \sqrt {\sin ^2(c+d x)}}-\frac {2 B \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (3-2 n),\frac {1}{4} (7-2 n),\cos ^2(c+d x)\right ) (b \sec (c+d x))^n \sin (c+d x)}{d (3-2 n) \sec ^{\frac {3}{2}}(c+d x) \sqrt {\sin ^2(c+d x)}} \]
-2*C*(b*sec(d*x+c))^n*sin(d*x+c)/d/(1-2*n)/sec(d*x+c)^(1/2)-2*(A+C*(3-2*n) -2*A*n)*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/(4*n^2-12*n+5)/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 = 1.77 (sec) , antiderivative size = 201, normalized size of antiderivative = 0.91 \[ \int \frac {(b \sec (c+d x))^n \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {2 \csc (c+d x) (b \sec (c+d x))^n \left (A \left (-1+4 n^2\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (-3+2 n),\frac {1}{4} (1+2 n),\sec ^2(c+d x)\right )+(-3+2 n) \left (B (1+2 n) \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (-1+2 n),\frac {1}{4} (3+2 n),\sec ^2(c+d x)\right )+C (-1+2 n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (1+2 n),\frac {1}{4} (5+2 n),\sec ^2(c+d x)\right )\right ) \sec ^2(c+d x)\right ) \sqrt {-\tan ^2(c+d x)}}{d (-3+2 n) (-1+2 n) (1+2 n) \sec ^{\frac {5}{2}}(c+d x)} \]
(2*Csc[c + d*x]*(b*Sec[c + d*x])^n*(A*(-1 + 4*n^2)*Hypergeometric2F1[1/2, (-3 + 2*n)/4, (1 + 2*n)/4, Sec[c + d*x]^2] + (-3 + 2*n)*(B*(1 + 2*n)*Cos[c + d*x]*Hypergeometric2F1[1/2, (-1 + 2*n)/4, (3 + 2*n)/4, Sec[c + d*x]^2] + C*(-1 + 2*n)*Hypergeometric2F1[1/2, (1 + 2*n)/4, (5 + 2*n)/4, Sec[c + d* x]^2])*Sec[c + d*x]^2)*Sqrt[-Tan[c + d*x]^2])/(d*(-3 + 2*n)*(-1 + 2*n)*(1 + 2*n)*Sec[c + d*x]^(5/2))
Time = 0.95 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.00, 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.
| \(\displaystyle \int \frac {(b \sec (c+d x))^n \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx\) |
| \(\Big \downarrow \) 2034 |
| \(\displaystyle \sec ^{-n}(c+d x) (b \sec (c+d x))^n \int \sec ^{n-\frac {3}{2}}(c+d x) \left (C \sec ^2(c+d x)+B \sec (c+d x)+A\right )dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sec ^{-n}(c+d x) (b \sec (c+d x))^n \int \csc \left (c+d x+\frac {\pi }{2}\right )^{n-\frac {3}{2}} \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+B \csc \left (c+d x+\frac {\pi }{2}\right )+A\right )dx\) |
| \(\Big \downarrow \) 4535 |
| \(\displaystyle \sec ^{-n}(c+d x) (b \sec (c+d x))^n \left (\int \sec ^{n-\frac {3}{2}}(c+d x) \left (C \sec ^2(c+d x)+A\right )dx+B \int \sec ^{n-\frac {1}{2}}(c+d x)dx\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sec ^{-n}(c+d x) (b \sec (c+d x))^n \left (\int \csc \left (c+d x+\frac {\pi }{2}\right )^{n-\frac {3}{2}} \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A\right )dx+B \int \csc \left (c+d x+\frac {\pi }{2}\right )^{n-\frac {1}{2}}dx\right )\) |
| \(\Big \downarrow \) 4259 |
| \(\displaystyle \sec ^{-n}(c+d x) (b \sec (c+d x))^n \left (\int \csc \left (c+d x+\frac {\pi }{2}\right )^{n-\frac {3}{2}} \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A\right )dx+B \cos ^{n+\frac {1}{2}}(c+d x) \sec ^{n+\frac {1}{2}}(c+d x) \int \cos ^{\frac {1}{2}-n}(c+d x)dx\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sec ^{-n}(c+d x) (b \sec (c+d x))^n \left (\int \csc \left (c+d x+\frac {\pi }{2}\right )^{n-\frac {3}{2}} \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A\right )dx+B \cos ^{n+\frac {1}{2}}(c+d x) \sec ^{n+\frac {1}{2}}(c+d x) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{\frac {1}{2}-n}dx\right )\) |
| \(\Big \downarrow \) 3122 |
| \(\displaystyle \sec ^{-n}(c+d x) (b \sec (c+d x))^n \left (\int \csc \left (c+d x+\frac {\pi }{2}\right )^{n-\frac {3}{2}} \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A\right )dx-\frac {2 B \sin (c+d x) \sec ^{n-\frac {3}{2}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (3-2 n),\frac {1}{4} (7-2 n),\cos ^2(c+d x)\right )}{d (3-2 n) \sqrt {\sin ^2(c+d x)}}\right )\) |
| \(\Big \downarrow \) 4534 |
| \(\displaystyle \sec ^{-n}(c+d x) (b \sec (c+d x))^n \left (\frac {(A (1-2 n)+C (3-2 n)) \int \sec ^{n-\frac {3}{2}}(c+d x)dx}{1-2 n}-\frac {2 B \sin (c+d x) \sec ^{n-\frac {3}{2}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (3-2 n),\frac {1}{4} (7-2 n),\cos ^2(c+d x)\right )}{d (3-2 n) \sqrt {\sin ^2(c+d x)}}-\frac {2 C \sin (c+d x) \sec ^{n-\frac {1}{2}}(c+d x)}{d (1-2 n)}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sec ^{-n}(c+d x) (b \sec (c+d x))^n \left (\frac {(A (1-2 n)+C (3-2 n)) \int \csc \left (c+d x+\frac {\pi }{2}\right )^{n-\frac {3}{2}}dx}{1-2 n}-\frac {2 B \sin (c+d x) \sec ^{n-\frac {3}{2}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (3-2 n),\frac {1}{4} (7-2 n),\cos ^2(c+d x)\right )}{d (3-2 n) \sqrt {\sin ^2(c+d x)}}-\frac {2 C \sin (c+d x) \sec ^{n-\frac {1}{2}}(c+d x)}{d (1-2 n)}\right )\) |
| \(\Big \downarrow \) 4259 |
| \(\displaystyle \sec ^{-n}(c+d x) (b \sec (c+d x))^n \left (\frac {(A (1-2 n)+C (3-2 n)) \cos ^{n+\frac {1}{2}}(c+d x) \sec ^{n+\frac {1}{2}}(c+d x) \int \cos ^{\frac {3}{2}-n}(c+d x)dx}{1-2 n}-\frac {2 B \sin (c+d x) \sec ^{n-\frac {3}{2}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (3-2 n),\frac {1}{4} (7-2 n),\cos ^2(c+d x)\right )}{d (3-2 n) \sqrt {\sin ^2(c+d x)}}-\frac {2 C \sin (c+d x) \sec ^{n-\frac {1}{2}}(c+d x)}{d (1-2 n)}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sec ^{-n}(c+d x) (b \sec (c+d x))^n \left (\frac {(A (1-2 n)+C (3-2 n)) \cos ^{n+\frac {1}{2}}(c+d x) \sec ^{n+\frac {1}{2}}(c+d x) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{\frac {3}{2}-n}dx}{1-2 n}-\frac {2 B \sin (c+d x) \sec ^{n-\frac {3}{2}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (3-2 n),\frac {1}{4} (7-2 n),\cos ^2(c+d x)\right )}{d (3-2 n) \sqrt {\sin ^2(c+d x)}}-\frac {2 C \sin (c+d x) \sec ^{n-\frac {1}{2}}(c+d x)}{d (1-2 n)}\right )\) |
| \(\Big \downarrow \) 3122 |
| \(\displaystyle \sec ^{-n}(c+d x) (b \sec (c+d x))^n \left (-\frac {2 (A (1-2 n)+C (3-2 n)) \sin (c+d x) \sec ^{n-\frac {5}{2}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (5-2 n),\frac {1}{4} (9-2 n),\cos ^2(c+d x)\right )}{d (1-2 n) (5-2 n) \sqrt {\sin ^2(c+d x)}}-\frac {2 B \sin (c+d x) \sec ^{n-\frac {3}{2}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (3-2 n),\frac {1}{4} (7-2 n),\cos ^2(c+d x)\right )}{d (3-2 n) \sqrt {\sin ^2(c+d x)}}-\frac {2 C \sin (c+d x) \sec ^{n-\frac {1}{2}}(c+d x)}{d (1-2 n)}\right )\) |
((b*Sec[c + d*x])^n*((-2*C*Sec[c + d*x]^(-1/2 + n)*Sin[c + d*x])/(d*(1 - 2 *n)) - (2*(A*(1 - 2*n) + C*(3 - 2*n))*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*(1 - 2*n)*(5 - 2*n)*Sqrt[Sin[c + d*x]^2]) - (2*B*Hypergeometric2F1[1/2, (3 - 2 *n)/4, (7 - 2*n)/4, Cos[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
3.1.82.3.1 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]
\[\int \frac {\left (b \sec \left (d x +c \right )\right )^{n} \left (A +B \sec \left (d x +c \right )+C \sec \left (d x +c \right )^{2}\right )}{\sec \left (d x +c \right )^{\frac {3}{2}}}d x\]
\[ \int \frac {(b \sec (c+d x))^n \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{n}}{\sec \left (d x + c\right )^{\frac {3}{2}}} \,d x } \]
integrate((b*sec(d*x+c))^n*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(3/2 ),x, algorithm="fricas")
\[ \int \frac {(b \sec (c+d x))^n \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {\left (b \sec {\left (c + d x \right )}\right )^{n} \left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right )}{\sec ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx \]
Integral((b*sec(c + d*x))**n*(A + B*sec(c + d*x) + C*sec(c + d*x)**2)/sec( c + d*x)**(3/2), x)
\[ \int \frac {(b \sec (c+d x))^n \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{n}}{\sec \left (d x + c\right )^{\frac {3}{2}}} \,d x } \]
integrate((b*sec(d*x+c))^n*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(3/2 ),x, algorithm="maxima")
\[ \int \frac {(b \sec (c+d x))^n \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{n}}{\sec \left (d x + c\right )^{\frac {3}{2}}} \,d x } \]
integrate((b*sec(d*x+c))^n*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(3/2 ),x, algorithm="giac")
Timed out. \[ \int \frac {(b \sec (c+d x))^n \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^n\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \]