Integrand size = 37, antiderivative size = 191 \[ \int \cos (c+d x) (b \sec (c+d x))^n \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {b^2 (C (1-n)-A n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2-n}{2},\frac {4-n}{2},\cos ^2(c+d x)\right ) (b \sec (c+d x))^{-2+n} \sin (c+d x)}{d (2-n) n \sqrt {\sin ^2(c+d x)}}-\frac {b B \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1-n}{2},\frac {3-n}{2},\cos ^2(c+d x)\right ) (b \sec (c+d x))^{-1+n} \sin (c+d x)}{d (1-n) \sqrt {\sin ^2(c+d x)}}+\frac {b C (b \sec (c+d x))^{-1+n} \tan (c+d x)}{d n} \]
b^2*(C*(1-n)-A*n)*hypergeom([1/2, 1-1/2*n],[2-1/2*n],cos(d*x+c)^2)*(b*sec( d*x+c))^(-2+n)*sin(d*x+c)/d/(2-n)/n/(sin(d*x+c)^2)^(1/2)-b*B*hypergeom([1/ 2, 1/2-1/2*n],[3/2-1/2*n],cos(d*x+c)^2)*(b*sec(d*x+c))^(-1+n)*sin(d*x+c)/d /(1-n)/(sin(d*x+c)^2)^(1/2)+b*C*(b*sec(d*x+c))^(-1+n)*tan(d*x+c)/d/n
Time = 0.52 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.84 \[ \int \cos (c+d x) (b \sec (c+d x))^n \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {\left (A n (1+n) \cos (c+d x) \cot (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-1+n),\frac {1+n}{2},\sec ^2(c+d x)\right )+(-1+n) \csc (c+d x) \left (B (1+n) \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {n}{2},\frac {2+n}{2},\sec ^2(c+d x)\right )+C n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+n}{2},\frac {3+n}{2},\sec ^2(c+d x)\right )\right )\right ) (b \sec (c+d x))^n \sqrt {-\tan ^2(c+d x)}}{d (-1+n) n (1+n)} \]
((A*n*(1 + n)*Cos[c + d*x]*Cot[c + d*x]*Hypergeometric2F1[1/2, (-1 + n)/2, (1 + n)/2, Sec[c + d*x]^2] + (-1 + n)*Csc[c + d*x]*(B*(1 + n)*Cos[c + d*x ]*Hypergeometric2F1[1/2, n/2, (2 + n)/2, Sec[c + d*x]^2] + C*n*Hypergeomet ric2F1[1/2, (1 + n)/2, (3 + n)/2, Sec[c + d*x]^2]))*(b*Sec[c + d*x])^n*Sqr t[-Tan[c + d*x]^2])/(d*(-1 + n)*n*(1 + n))
Time = 0.86 (sec) , antiderivative size = 189, normalized size of antiderivative = 0.99, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.324, Rules used = {3042, 2030, 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.
\(\displaystyle \int \cos (c+d x) (b \sec (c+d x))^n \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^n \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx\) |
\(\Big \downarrow \) 2030 |
\(\displaystyle b \int \left (b \csc \left (\frac {1}{2} (2 c+\pi )+d x\right )\right )^{n-1} \left (C \csc \left (\frac {1}{2} (2 c+\pi )+d x\right )^2+B \csc \left (\frac {1}{2} (2 c+\pi )+d x\right )+A\right )dx\) |
\(\Big \downarrow \) 4535 |
\(\displaystyle b \left (\int \left (b \csc \left (\frac {1}{2} (2 c+\pi )+d x\right )\right )^{n-1} \left (C \csc ^2\left (\frac {1}{2} (2 c+\pi )+d x\right )+A\right )dx+\frac {B \int (b \sec (c+d x))^ndx}{b}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle b \left (\int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{n-1} \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A\right )dx+\frac {B \int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^ndx}{b}\right )\) |
\(\Big \downarrow \) 4259 |
\(\displaystyle b \left (\int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{n-1} \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A\right )dx+\frac {B \left (\frac {\cos (c+d x)}{b}\right )^n (b \sec (c+d x))^n \int \left (\frac {\cos (c+d x)}{b}\right )^{-n}dx}{b}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle b \left (\int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{n-1} \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A\right )dx+\frac {B \left (\frac {\cos (c+d x)}{b}\right )^n (b \sec (c+d x))^n \int \left (\frac {\sin \left (c+d x+\frac {\pi }{2}\right )}{b}\right )^{-n}dx}{b}\right )\) |
\(\Big \downarrow \) 3122 |
\(\displaystyle b \left (\int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{n-1} \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A\right )dx-\frac {B \sin (c+d x) (b \sec (c+d x))^{n-1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1-n}{2},\frac {3-n}{2},\cos ^2(c+d x)\right )}{d (1-n) \sqrt {\sin ^2(c+d x)}}\right )\) |
\(\Big \downarrow \) 4534 |
\(\displaystyle b \left (-\frac {(C (1-n)-A n) \int (b \sec (c+d x))^{n-1}dx}{n}-\frac {B \sin (c+d x) (b \sec (c+d x))^{n-1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1-n}{2},\frac {3-n}{2},\cos ^2(c+d x)\right )}{d (1-n) \sqrt {\sin ^2(c+d x)}}+\frac {C \tan (c+d x) (b \sec (c+d x))^{n-1}}{d n}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle b \left (-\frac {(C (1-n)-A n) \int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{n-1}dx}{n}-\frac {B \sin (c+d x) (b \sec (c+d x))^{n-1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1-n}{2},\frac {3-n}{2},\cos ^2(c+d x)\right )}{d (1-n) \sqrt {\sin ^2(c+d x)}}+\frac {C \tan (c+d x) (b \sec (c+d x))^{n-1}}{d n}\right )\) |
\(\Big \downarrow \) 4259 |
\(\displaystyle b \left (-\frac {(C (1-n)-A n) \left (\frac {\cos (c+d x)}{b}\right )^n (b \sec (c+d x))^n \int \left (\frac {\cos (c+d x)}{b}\right )^{1-n}dx}{n}-\frac {B \sin (c+d x) (b \sec (c+d x))^{n-1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1-n}{2},\frac {3-n}{2},\cos ^2(c+d x)\right )}{d (1-n) \sqrt {\sin ^2(c+d x)}}+\frac {C \tan (c+d x) (b \sec (c+d x))^{n-1}}{d n}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle b \left (-\frac {(C (1-n)-A n) \left (\frac {\cos (c+d x)}{b}\right )^n (b \sec (c+d x))^n \int \left (\frac {\sin \left (c+d x+\frac {\pi }{2}\right )}{b}\right )^{1-n}dx}{n}-\frac {B \sin (c+d x) (b \sec (c+d x))^{n-1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1-n}{2},\frac {3-n}{2},\cos ^2(c+d x)\right )}{d (1-n) \sqrt {\sin ^2(c+d x)}}+\frac {C \tan (c+d x) (b \sec (c+d x))^{n-1}}{d n}\right )\) |
\(\Big \downarrow \) 3122 |
\(\displaystyle b \left (\frac {b (C (1-n)-A n) \sin (c+d x) (b \sec (c+d x))^{n-2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2-n}{2},\frac {4-n}{2},\cos ^2(c+d x)\right )}{d (2-n) n \sqrt {\sin ^2(c+d x)}}-\frac {B \sin (c+d x) (b \sec (c+d x))^{n-1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1-n}{2},\frac {3-n}{2},\cos ^2(c+d x)\right )}{d (1-n) \sqrt {\sin ^2(c+d x)}}+\frac {C \tan (c+d x) (b \sec (c+d x))^{n-1}}{d n}\right )\) |
b*((b*(C*(1 - n) - A*n)*Hypergeometric2F1[1/2, (2 - n)/2, (4 - n)/2, Cos[c + d*x]^2]*(b*Sec[c + d*x])^(-2 + n)*Sin[c + d*x])/(d*(2 - n)*n*Sqrt[Sin[c + d*x]^2]) - (B*Hypergeometric2F1[1/2, (1 - n)/2, (3 - n)/2, Cos[c + d*x] ^2]*(b*Sec[c + d*x])^(-1 + n)*Sin[c + d*x])/(d*(1 - n)*Sqrt[Sin[c + d*x]^2 ]) + (C*(b*Sec[c + d*x])^(-1 + n)*Tan[c + d*x])/(d*n))
3.1.75.3.1 Defintions of rubi rules used
Int[(Fx_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Simp[1/b^m Int[(b*v) ^(m + n)*Fx, x], x] /; FreeQ[{b, n}, x] && IntegerQ[m]
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]
\[\int \cos \left (d x +c \right ) \left (b \sec \left (d x +c \right )\right )^{n} \left (A +B \sec \left (d x +c \right )+C \sec \left (d x +c \right )^{2}\right )d x\]
\[ \int \cos (c+d x) (b \sec (c+d x))^n \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{n} \cos \left (d x + c\right ) \,d x } \]
integral((C*cos(d*x + c)*sec(d*x + c)^2 + B*cos(d*x + c)*sec(d*x + c) + A* cos(d*x + c))*(b*sec(d*x + c))^n, x)
\[ \int \cos (c+d x) (b \sec (c+d x))^n \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int \left (b \sec {\left (c + d x \right )}\right )^{n} \left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \cos {\left (c + d x \right )}\, dx \]
\[ \int \cos (c+d x) (b \sec (c+d x))^n \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{n} \cos \left (d x + c\right ) \,d x } \]
\[ \int \cos (c+d x) (b \sec (c+d x))^n \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{n} \cos \left (d x + c\right ) \,d x } \]
Timed out. \[ \int \cos (c+d x) (b \sec (c+d x))^n \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int \cos \left (c+d\,x\right )\,{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^n\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right ) \,d x \]