Integrand size = 31, antiderivative size = 261 \[ \int \sec ^m(c+d x) (a+b \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=\frac {b (A b (2+m)+a B (3+m)) \sec ^{1+m}(c+d x) \sin (c+d x)}{d (1+m) (2+m)}+\frac {b B \sec ^{1+m}(c+d x) (a+b \sec (c+d x)) \sin (c+d x)}{d (2+m)}-\frac {\left (A b^2 m+2 a b B m+a^2 A (1+m)\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1-m}{2},\frac {3-m}{2},\cos ^2(c+d x)\right ) \sec ^{-1+m}(c+d x) \sin (c+d x)}{d \left (1-m^2\right ) \sqrt {\sin ^2(c+d x)}}+\frac {\left (b^2 B (1+m)+a (2 A b+a B) (2+m)\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-\frac {m}{2},\frac {2-m}{2},\cos ^2(c+d x)\right ) \sec ^m(c+d x) \sin (c+d x)}{d m (2+m) \sqrt {\sin ^2(c+d x)}} \]
b*(A*b*(2+m)+a*B*(3+m))*sec(d*x+c)^(1+m)*sin(d*x+c)/d/(1+m)/(2+m)+b*B*sec( d*x+c)^(1+m)*(a+b*sec(d*x+c))*sin(d*x+c)/d/(2+m)-(A*b^2*m+2*a*b*B*m+a^2*A* (1+m))*hypergeom([1/2, -1/2*m+1/2],[3/2-1/2*m],cos(d*x+c)^2)*sec(d*x+c)^(- 1+m)*sin(d*x+c)/d/(-m^2+1)/(sin(d*x+c)^2)^(1/2)+(b^2*B*(1+m)+a*(2*A*b+B*a) *(2+m))*hypergeom([1/2, -1/2*m],[1-1/2*m],cos(d*x+c)^2)*sec(d*x+c)^m*sin(d *x+c)/d/m/(2+m)/(sin(d*x+c)^2)^(1/2)
Time = 1.09 (sec) , antiderivative size = 239, normalized size of antiderivative = 0.92 \[ \int \sec ^m(c+d x) (a+b \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=\frac {\csc (c+d x) \left (a^2 A \left (6+11 m+6 m^2+m^3\right ) \cos ^3(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m}{2},\frac {2+m}{2},\sec ^2(c+d x)\right )+a (2 A b+a B) m \left (6+5 m+m^2\right ) \cos ^2(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},\sec ^2(c+d x)\right )+b m (1+m) \left ((A b+2 a B) (3+m) \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},\sec ^2(c+d x)\right )+b B (2+m) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3+m}{2},\frac {5+m}{2},\sec ^2(c+d x)\right )\right )\right ) \sec ^{2+m}(c+d x) \sqrt {-\tan ^2(c+d x)}}{d m (1+m) (2+m) (3+m)} \]
(Csc[c + d*x]*(a^2*A*(6 + 11*m + 6*m^2 + m^3)*Cos[c + d*x]^3*Hypergeometri c2F1[1/2, m/2, (2 + m)/2, Sec[c + d*x]^2] + a*(2*A*b + a*B)*m*(6 + 5*m + m ^2)*Cos[c + d*x]^2*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, Sec[c + d* x]^2] + b*m*(1 + m)*((A*b + 2*a*B)*(3 + m)*Cos[c + d*x]*Hypergeometric2F1[ 1/2, (2 + m)/2, (4 + m)/2, Sec[c + d*x]^2] + b*B*(2 + m)*Hypergeometric2F1 [1/2, (3 + m)/2, (5 + m)/2, Sec[c + d*x]^2]))*Sec[c + d*x]^(2 + m)*Sqrt[-T an[c + d*x]^2])/(d*m*(1 + m)*(2 + m)*(3 + m))
Time = 1.28 (sec) , antiderivative size = 264, normalized size of antiderivative = 1.01, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.419, Rules used = {3042, 4514, 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.
| \(\displaystyle \int \sec ^m(c+d x) (a+b \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^m \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^2 \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx\) |
| \(\Big \downarrow \) 4514 |
| \(\displaystyle \frac {\int \sec ^m(c+d x) \left (b (A b (m+2)+a B (m+3)) \sec ^2(c+d x)+\left (B (m+1) b^2+a (2 A b+a B) (m+2)\right ) \sec (c+d x)+a (b B m+a A (m+2))\right )dx}{m+2}+\frac {b B \sin (c+d x) \sec ^{m+1}(c+d x) (a+b \sec (c+d x))}{d (m+2)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \csc \left (c+d x+\frac {\pi }{2}\right )^m \left (b (A b (m+2)+a B (m+3)) \csc \left (c+d x+\frac {\pi }{2}\right )^2+\left (B (m+1) b^2+a (2 A b+a B) (m+2)\right ) \csc \left (c+d x+\frac {\pi }{2}\right )+a (b B m+a A (m+2))\right )dx}{m+2}+\frac {b B \sin (c+d x) \sec ^{m+1}(c+d x) (a+b \sec (c+d x))}{d (m+2)}\) |
| \(\Big \downarrow \) 4535 |
| \(\displaystyle \frac {\left (a (m+2) (a B+2 A b)+b^2 B (m+1)\right ) \int \sec ^{m+1}(c+d x)dx+\int \sec ^m(c+d x) \left (b (A b (m+2)+a B (m+3)) \sec ^2(c+d x)+a (b B m+a A (m+2))\right )dx}{m+2}+\frac {b B \sin (c+d x) \sec ^{m+1}(c+d x) (a+b \sec (c+d x))}{d (m+2)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\left (a (m+2) (a B+2 A b)+b^2 B (m+1)\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )^{m+1}dx+\int \csc \left (c+d x+\frac {\pi }{2}\right )^m \left (b (A b (m+2)+a B (m+3)) \csc \left (c+d x+\frac {\pi }{2}\right )^2+a (b B m+a A (m+2))\right )dx}{m+2}+\frac {b B \sin (c+d x) \sec ^{m+1}(c+d x) (a+b \sec (c+d x))}{d (m+2)}\) |
| \(\Big \downarrow \) 4259 |
| \(\displaystyle \frac {\left (a (m+2) (a B+2 A b)+b^2 B (m+1)\right ) \cos ^m(c+d x) \sec ^m(c+d x) \int \cos ^{-m-1}(c+d x)dx+\int \csc \left (c+d x+\frac {\pi }{2}\right )^m \left (b (A b (m+2)+a B (m+3)) \csc \left (c+d x+\frac {\pi }{2}\right )^2+a (b B m+a A (m+2))\right )dx}{m+2}+\frac {b B \sin (c+d x) \sec ^{m+1}(c+d x) (a+b \sec (c+d x))}{d (m+2)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\left (a (m+2) (a B+2 A b)+b^2 B (m+1)\right ) \cos ^m(c+d x) \sec ^m(c+d x) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{-m-1}dx+\int \csc \left (c+d x+\frac {\pi }{2}\right )^m \left (b (A b (m+2)+a B (m+3)) \csc \left (c+d x+\frac {\pi }{2}\right )^2+a (b B m+a A (m+2))\right )dx}{m+2}+\frac {b B \sin (c+d x) \sec ^{m+1}(c+d x) (a+b \sec (c+d x))}{d (m+2)}\) |
| \(\Big \downarrow \) 3122 |
| \(\displaystyle \frac {\int \csc \left (c+d x+\frac {\pi }{2}\right )^m \left (b (A b (m+2)+a B (m+3)) \csc \left (c+d x+\frac {\pi }{2}\right )^2+a (b B m+a A (m+2))\right )dx+\frac {\sin (c+d x) \left (a (m+2) (a B+2 A b)+b^2 B (m+1)\right ) \sec ^m(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-\frac {m}{2},\frac {2-m}{2},\cos ^2(c+d x)\right )}{d m \sqrt {\sin ^2(c+d x)}}}{m+2}+\frac {b B \sin (c+d x) \sec ^{m+1}(c+d x) (a+b \sec (c+d x))}{d (m+2)}\) |
| \(\Big \downarrow \) 4534 |
| \(\displaystyle \frac {\frac {(m+2) \left (a^2 A (m+1)+2 a b B m+A b^2 m\right ) \int \sec ^m(c+d x)dx}{m+1}+\frac {\sin (c+d x) \left (a (m+2) (a B+2 A b)+b^2 B (m+1)\right ) \sec ^m(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-\frac {m}{2},\frac {2-m}{2},\cos ^2(c+d x)\right )}{d m \sqrt {\sin ^2(c+d x)}}+\frac {b \sin (c+d x) (a B (m+3)+A b (m+2)) \sec ^{m+1}(c+d x)}{d (m+1)}}{m+2}+\frac {b B \sin (c+d x) \sec ^{m+1}(c+d x) (a+b \sec (c+d x))}{d (m+2)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {(m+2) \left (a^2 A (m+1)+2 a b B m+A b^2 m\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )^mdx}{m+1}+\frac {\sin (c+d x) \left (a (m+2) (a B+2 A b)+b^2 B (m+1)\right ) \sec ^m(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-\frac {m}{2},\frac {2-m}{2},\cos ^2(c+d x)\right )}{d m \sqrt {\sin ^2(c+d x)}}+\frac {b \sin (c+d x) (a B (m+3)+A b (m+2)) \sec ^{m+1}(c+d x)}{d (m+1)}}{m+2}+\frac {b B \sin (c+d x) \sec ^{m+1}(c+d x) (a+b \sec (c+d x))}{d (m+2)}\) |
| \(\Big \downarrow \) 4259 |
| \(\displaystyle \frac {\frac {(m+2) \left (a^2 A (m+1)+2 a b B m+A b^2 m\right ) \cos ^m(c+d x) \sec ^m(c+d x) \int \cos ^{-m}(c+d x)dx}{m+1}+\frac {\sin (c+d x) \left (a (m+2) (a B+2 A b)+b^2 B (m+1)\right ) \sec ^m(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-\frac {m}{2},\frac {2-m}{2},\cos ^2(c+d x)\right )}{d m \sqrt {\sin ^2(c+d x)}}+\frac {b \sin (c+d x) (a B (m+3)+A b (m+2)) \sec ^{m+1}(c+d x)}{d (m+1)}}{m+2}+\frac {b B \sin (c+d x) \sec ^{m+1}(c+d x) (a+b \sec (c+d x))}{d (m+2)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {(m+2) \left (a^2 A (m+1)+2 a b B m+A b^2 m\right ) \cos ^m(c+d x) \sec ^m(c+d x) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{-m}dx}{m+1}+\frac {\sin (c+d x) \left (a (m+2) (a B+2 A b)+b^2 B (m+1)\right ) \sec ^m(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-\frac {m}{2},\frac {2-m}{2},\cos ^2(c+d x)\right )}{d m \sqrt {\sin ^2(c+d x)}}+\frac {b \sin (c+d x) (a B (m+3)+A b (m+2)) \sec ^{m+1}(c+d x)}{d (m+1)}}{m+2}+\frac {b B \sin (c+d x) \sec ^{m+1}(c+d x) (a+b \sec (c+d x))}{d (m+2)}\) |
| \(\Big \downarrow \) 3122 |
| \(\displaystyle \frac {-\frac {(m+2) \sin (c+d x) \left (a^2 A (m+1)+2 a b B m+A b^2 m\right ) \sec ^{m-1}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1-m}{2},\frac {3-m}{2},\cos ^2(c+d x)\right )}{d (1-m) (m+1) \sqrt {\sin ^2(c+d x)}}+\frac {\sin (c+d x) \left (a (m+2) (a B+2 A b)+b^2 B (m+1)\right ) \sec ^m(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-\frac {m}{2},\frac {2-m}{2},\cos ^2(c+d x)\right )}{d m \sqrt {\sin ^2(c+d x)}}+\frac {b \sin (c+d x) (a B (m+3)+A b (m+2)) \sec ^{m+1}(c+d x)}{d (m+1)}}{m+2}+\frac {b B \sin (c+d x) \sec ^{m+1}(c+d x) (a+b \sec (c+d x))}{d (m+2)}\) |
(b*B*Sec[c + d*x]^(1 + m)*(a + b*Sec[c + d*x])*Sin[c + d*x])/(d*(2 + m)) + ((b*(A*b*(2 + m) + a*B*(3 + m))*Sec[c + d*x]^(1 + m)*Sin[c + d*x])/(d*(1 + m)) - ((2 + m)*(A*b^2*m + 2*a*b*B*m + a^2*A*(1 + m))*Hypergeometric2F1[1 /2, (1 - m)/2, (3 - m)/2, Cos[c + d*x]^2]*Sec[c + d*x]^(-1 + m)*Sin[c + d* x])/(d*(1 - m)*(1 + m)*Sqrt[Sin[c + d*x]^2]) + ((b^2*B*(1 + m) + a*(2*A*b + a*B)*(2 + m))*Hypergeometric2F1[1/2, -1/2*m, (2 - m)/2, Cos[c + d*x]^2]* Sec[c + d*x]^m*Sin[c + d*x])/(d*m*Sqrt[Sin[c + d*x]^2]))/(2 + m)
3.5.81.3.1 Defintions of rubi rules used
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_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-b)*B* Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*((d*Csc[e + f*x])^n/(f*(m + n))), x] + Simp[1/(m + n) Int[(a + b*Csc[e + f*x])^(m - 2)*(d*Csc[e + f*x])^n* Simp[a^2*A*(m + n) + a*b*B*n + (a*(2*A*b + a*B)*(m + n) + b^2*B*(m + n - 1) )*Csc[e + f*x] + b*(A*b*(m + n) + a*B*(2*m + n - 1))*Csc[e + f*x]^2, x], x] , x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 1] && !(IGtQ[n, 1] && !IntegerQ[m])
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]
\[\int \sec \left (d x +c \right )^{m} \left (a +b \sec \left (d x +c \right )\right )^{2} \left (A +B \sec \left (d x +c \right )\right )d x\]
\[ \int \sec ^m(c+d x) (a+b \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{2} \sec \left (d x + c\right )^{m} \,d x } \]
integral((B*b^2*sec(d*x + c)^3 + A*a^2 + (2*B*a*b + A*b^2)*sec(d*x + c)^2 + (B*a^2 + 2*A*a*b)*sec(d*x + c))*sec(d*x + c)^m, x)
\[ \int \sec ^m(c+d x) (a+b \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=\int \left (A + B \sec {\left (c + d x \right )}\right ) \left (a + b \sec {\left (c + d x \right )}\right )^{2} \sec ^{m}{\left (c + d x \right )}\, dx \]
\[ \int \sec ^m(c+d x) (a+b \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{2} \sec \left (d x + c\right )^{m} \,d x } \]
\[ \int \sec ^m(c+d x) (a+b \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{2} \sec \left (d x + c\right )^{m} \,d x } \]
Timed out. \[ \int \sec ^m(c+d x) (a+b \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=\int \left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^2\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^m \,d x \]