\(\int \cos (c+d x) (b \sec (c+d x))^n (A+B \sec (c+d x)) \, dx\) [37]

Optimal result

Integrand size = 27, antiderivative size = 151 \[ \int \cos (c+d x) (b \sec (c+d x))^n (A+B \sec (c+d x)) \, dx=-\frac {A b^2 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2-n}{2},\frac {4-n}{2},\cos ^2(c+d x)\right ) (b \sec (c+d x))^{-2+n} \sin (c+d x)}{d (2-n) \sqrt {\sin ^2(c+d x)}}-\frac {b B \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1-n}{2},\frac {3-n}{2},\cos ^2(c+d x)\right ) (b \sec (c+d x))^{-1+n} \sin (c+d x)}{d (1-n) \sqrt {\sin ^2(c+d x)}} \]

[Out]

-A*b^2*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)/(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)

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {16, 3872, 3857, 2722} \[ \int \cos (c+d x) (b \sec (c+d x))^n (A+B \sec (c+d x)) \, dx=-\frac {A b^2 \sin (c+d x) (b \sec (c+d x))^{n-2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2-n}{2},\frac {4-n}{2},\cos ^2(c+d x)\right )}{d (2-n) \sqrt {\sin ^2(c+d x)}}-\frac {b B \sin (c+d x) (b \sec (c+d x))^{n-1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1-n}{2},\frac {3-n}{2},\cos ^2(c+d x)\right )}{d (1-n) \sqrt {\sin ^2(c+d x)}} \]

[In]

Int[Cos[c + d*x]*(b*Sec[c + d*x])^n*(A + B*Sec[c + d*x]),x]

[Out]

-((A*b^2*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)*Sqrt[Sin[c + d*x]^2])) - (b*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])

Rule 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

Rule 2722

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]))*Hypergeometric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rule 3857

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]

Rule 3872

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rubi steps \begin{align*} \text {integral}& = b \int (b \sec (c+d x))^{-1+n} (A+B \sec (c+d x)) \, dx \\ & = (A b) \int (b \sec (c+d x))^{-1+n} \, dx+B \int (b \sec (c+d x))^n \, dx \\ & = \left (A b \left (\frac {\cos (c+d x)}{b}\right )^n (b \sec (c+d x))^n\right ) \int \left (\frac {\cos (c+d x)}{b}\right )^{1-n} \, dx+\left (B \left (\frac {\cos (c+d x)}{b}\right )^n (b \sec (c+d x))^n\right ) \int \left (\frac {\cos (c+d x)}{b}\right )^{-n} \, dx \\ & = -\frac {B \cos (c+d x) \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))^n \sin (c+d x)}{d (1-n) \sqrt {\sin ^2(c+d x)}}-\frac {A \cos ^2(c+d x) \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))^n \sin (c+d x)}{d (2-n) \sqrt {\sin ^2(c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.71 \[ \int \cos (c+d x) (b \sec (c+d x))^n (A+B \sec (c+d x)) \, dx=\frac {\cot (c+d x) \left (A n \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-1+n),\frac {1+n}{2},\sec ^2(c+d x)\right )+B (-1+n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {n}{2},\frac {2+n}{2},\sec ^2(c+d x)\right )\right ) (b \sec (c+d x))^n \sqrt {-\tan ^2(c+d x)}}{d (-1+n) n} \]

[In]

Integrate[Cos[c + d*x]*(b*Sec[c + d*x])^n*(A + B*Sec[c + d*x]),x]

[Out]

(Cot[c + d*x]*(A*n*Cos[c + d*x]*Hypergeometric2F1[1/2, (-1 + n)/2, (1 + n)/2, Sec[c + d*x]^2] + B*(-1 + n)*Hyp
ergeometric2F1[1/2, n/2, (2 + n)/2, Sec[c + d*x]^2])*(b*Sec[c + d*x])^n*Sqrt[-Tan[c + d*x]^2])/(d*(-1 + n)*n)

Maple [F]

\[\int \cos \left (d x +c \right ) \left (b \sec \left (d x +c \right )\right )^{n} \left (A +B \sec \left (d x +c \right )\right )d x\]

[In]

int(cos(d*x+c)*(b*sec(d*x+c))^n*(A+B*sec(d*x+c)),x)

[Out]

int(cos(d*x+c)*(b*sec(d*x+c))^n*(A+B*sec(d*x+c)),x)

Fricas [F]

\[ \int \cos (c+d x) (b \sec (c+d x))^n (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{n} \cos \left (d x + c\right ) \,d x } \]

[In]

integrate(cos(d*x+c)*(b*sec(d*x+c))^n*(A+B*sec(d*x+c)),x, algorithm="fricas")

[Out]

integral((B*cos(d*x + c)*sec(d*x + c) + A*cos(d*x + c))*(b*sec(d*x + c))^n, x)

Sympy [F]

\[ \int \cos (c+d x) (b \sec (c+d x))^n (A+B \sec (c+d x)) \, dx=\int \left (b \sec {\left (c + d x \right )}\right )^{n} \left (A + B \sec {\left (c + d x \right )}\right ) \cos {\left (c + d x \right )}\, dx \]

[In]

integrate(cos(d*x+c)*(b*sec(d*x+c))**n*(A+B*sec(d*x+c)),x)

[Out]

Integral((b*sec(c + d*x))**n*(A + B*sec(c + d*x))*cos(c + d*x), x)

Maxima [F]

\[ \int \cos (c+d x) (b \sec (c+d x))^n (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{n} \cos \left (d x + c\right ) \,d x } \]

[In]

integrate(cos(d*x+c)*(b*sec(d*x+c))^n*(A+B*sec(d*x+c)),x, algorithm="maxima")

[Out]

integrate((B*sec(d*x + c) + A)*(b*sec(d*x + c))^n*cos(d*x + c), x)

Giac [F]

\[ \int \cos (c+d x) (b \sec (c+d x))^n (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{n} \cos \left (d x + c\right ) \,d x } \]

[In]

integrate(cos(d*x+c)*(b*sec(d*x+c))^n*(A+B*sec(d*x+c)),x, algorithm="giac")

[Out]

integrate((B*sec(d*x + c) + A)*(b*sec(d*x + c))^n*cos(d*x + c), x)

Mupad [F(-1)]

Timed out. \[ \int \cos (c+d x) (b \sec (c+d x))^n (A+B \sec (c+d x)) \, dx=\int \cos \left (c+d\,x\right )\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^n \,d x \]

[In]

int(cos(c + d*x)*(A + B/cos(c + d*x))*(b/cos(c + d*x))^n,x)

[Out]

int(cos(c + d*x)*(A + B/cos(c + d*x))*(b/cos(c + d*x))^n, x)