Integrand size = 29, antiderivative size = 172 \[ \int \sec ^m(c+d x) (b \sec (c+d x))^n (A+B \sec (c+d x)) \, dx=-\frac {A \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (1-m-n),\frac {1}{2} (3-m-n),\cos ^2(c+d x)\right ) \sec ^{-1+m}(c+d x) (b \sec (c+d x))^n \sin (c+d x)}{d (1-m-n) \sqrt {\sin ^2(c+d x)}}+\frac {B \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-m-n),\frac {1}{2} (2-m-n),\cos ^2(c+d x)\right ) \sec ^m(c+d x) (b \sec (c+d x))^n \sin (c+d x)}{d (m+n) \sqrt {\sin ^2(c+d x)}} \]
-A*hypergeom([1/2, 1/2-1/2*m-1/2*n],[3/2-1/2*m-1/2*n],cos(d*x+c)^2)*sec(d* x+c)^(-1+m)*(b*sec(d*x+c))^n*sin(d*x+c)/d/(1-m-n)/(sin(d*x+c)^2)^(1/2)+B*h ypergeom([1/2, -1/2*m-1/2*n],[1-1/2*m-1/2*n],cos(d*x+c)^2)*sec(d*x+c)^m*(b *sec(d*x+c))^n*sin(d*x+c)/d/(m+n)/(sin(d*x+c)^2)^(1/2)
Time = 0.17 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.73 \[ \int \sec ^m(c+d x) (b \sec (c+d x))^n (A+B \sec (c+d x)) \, dx=\frac {\csc (c+d x) \left (A (1+m+n) \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+n}{2},\frac {1}{2} (2+m+n),\sec ^2(c+d x)\right )+B (m+n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (1+m+n),\frac {1}{2} (3+m+n),\sec ^2(c+d x)\right )\right ) \sec ^m(c+d x) (b \sec (c+d x))^n \sqrt {-\tan ^2(c+d x)}}{d (m+n) (1+m+n)} \]
(Csc[c + d*x]*(A*(1 + m + n)*Cos[c + d*x]*Hypergeometric2F1[1/2, (m + n)/2 , (2 + m + n)/2, Sec[c + d*x]^2] + B*(m + n)*Hypergeometric2F1[1/2, (1 + m + n)/2, (3 + m + n)/2, Sec[c + d*x]^2])*Sec[c + d*x]^m*(b*Sec[c + d*x])^n *Sqrt[-Tan[c + d*x]^2])/(d*(m + n)*(1 + m + n))
Time = 0.54 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.02, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {2034, 3042, 4274, 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.
\(\displaystyle \int \sec ^m(c+d x) (A+B \sec (c+d x)) (b \sec (c+d x))^n \, dx\) |
\(\Big \downarrow \) 2034 |
\(\displaystyle \sec ^{-n}(c+d x) (b \sec (c+d x))^n \int \sec ^{m+n}(c+d x) (A+B \sec (c+d x))dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sec ^{-n}(c+d x) (b \sec (c+d x))^n \int \csc \left (c+d x+\frac {\pi }{2}\right )^{m+n} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx\) |
\(\Big \downarrow \) 4274 |
\(\displaystyle \sec ^{-n}(c+d x) (b \sec (c+d x))^n \left (A \int \sec ^{m+n}(c+d x)dx+B \int \sec ^{m+n+1}(c+d x)dx\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sec ^{-n}(c+d x) (b \sec (c+d x))^n \left (A \int \csc \left (c+d x+\frac {\pi }{2}\right )^{m+n}dx+B \int \csc \left (c+d x+\frac {\pi }{2}\right )^{m+n+1}dx\right )\) |
\(\Big \downarrow \) 4259 |
\(\displaystyle \sec ^{-n}(c+d x) (b \sec (c+d x))^n \left (A \cos ^{m+n}(c+d x) \sec ^{m+n}(c+d x) \int \cos ^{-m-n}(c+d x)dx+B \cos ^{m+n}(c+d x) \sec ^{m+n}(c+d x) \int \cos ^{-m-n-1}(c+d x)dx\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sec ^{-n}(c+d x) (b \sec (c+d x))^n \left (A \cos ^{m+n}(c+d x) \sec ^{m+n}(c+d x) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{-m-n}dx+B \cos ^{m+n}(c+d x) \sec ^{m+n}(c+d x) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{-m-n-1}dx\right )\) |
\(\Big \downarrow \) 3122 |
\(\displaystyle \sec ^{-n}(c+d x) (b \sec (c+d x))^n \left (\frac {B \sin (c+d x) \sec ^{m+n}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-m-n),\frac {1}{2} (-m-n+2),\cos ^2(c+d x)\right )}{d (m+n) \sqrt {\sin ^2(c+d x)}}-\frac {A \sin (c+d x) \sec ^{m+n-1}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-m-n+1),\frac {1}{2} (-m-n+3),\cos ^2(c+d x)\right )}{d (-m-n+1) \sqrt {\sin ^2(c+d x)}}\right )\) |
((b*Sec[c + d*x])^n*(-((A*Hypergeometric2F1[1/2, (1 - m - n)/2, (3 - m - n )/2, Cos[c + d*x]^2]*Sec[c + d*x]^(-1 + m + n)*Sin[c + d*x])/(d*(1 - m - n )*Sqrt[Sin[c + d*x]^2])) + (B*Hypergeometric2F1[1/2, (-m - n)/2, (2 - m - n)/2, Cos[c + d*x]^2]*Sec[c + d*x]^(m + n)*Sin[c + d*x])/(d*(m + n)*Sqrt[S in[c + d*x]^2])))/Sec[c + d*x]^n
3.1.33.3.1 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_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[a Int[(d*Csc[e + f*x])^n, x], x] + Simp[b/d In t[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]
\[\int \sec \left (d x +c \right )^{m} \left (b \sec \left (d x +c \right )\right )^{n} \left (A +B \sec \left (d x +c \right )\right )d x\]
\[ \int \sec ^m(c+d x) (b \sec (c+d x))^n (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{n} \sec \left (d x + c\right )^{m} \,d x } \]
\[ \int \sec ^m(c+d x) (b \sec (c+d x))^n (A+B \sec (c+d x)) \, dx=\int \left (b \sec {\left (c + d x \right )}\right )^{n} \left (A + B \sec {\left (c + d x \right )}\right ) \sec ^{m}{\left (c + d x \right )}\, dx \]
\[ \int \sec ^m(c+d x) (b \sec (c+d x))^n (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{n} \sec \left (d x + c\right )^{m} \,d x } \]
\[ \int \sec ^m(c+d x) (b \sec (c+d x))^n (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{n} \sec \left (d x + c\right )^{m} \,d x } \]
Timed out. \[ \int \sec ^m(c+d x) (b \sec (c+d x))^n (A+B \sec (c+d x)) \, dx=\int \left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^n\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^m \,d x \]