\(\int (b \cos (c+d x))^n (A+B \cos (c+d x)) \sec ^2(c+d x) \, dx\) [916]

Optimal result

Integrand size = 29, antiderivative size = 131 \[ \int (b \cos (c+d x))^n (A+B \cos (c+d x)) \sec ^2(c+d x) \, dx=\frac {A b (b \cos (c+d x))^{-1+n} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-1+n),\frac {1+n}{2},\cos ^2(c+d x)\right ) \sin (c+d x)}{d (1-n) \sqrt {\sin ^2(c+d x)}}-\frac {B (b \cos (c+d x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {n}{2},\frac {2+n}{2},\cos ^2(c+d x)\right ) \sin (c+d x)}{d n \sqrt {\sin ^2(c+d x)}} \]

[Out]

A*b*(b*cos(d*x+c))^(-1+n)*hypergeom([1/2, -1/2+1/2*n],[1/2+1/2*n],cos(d*x+c)^2)*sin(d*x+c)/d/(1-n)/(sin(d*x+c)
^2)^(1/2)-B*(b*cos(d*x+c))^n*hypergeom([1/2, 1/2*n],[1+1/2*n],cos(d*x+c)^2)*sin(d*x+c)/d/n/(sin(d*x+c)^2)^(1/2
)

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {16, 2827, 2722} \[ \int (b \cos (c+d x))^n (A+B \cos (c+d x)) \sec ^2(c+d x) \, dx=\frac {A b \sin (c+d x) (b \cos (c+d x))^{n-1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {n-1}{2},\frac {n+1}{2},\cos ^2(c+d x)\right )}{d (1-n) \sqrt {\sin ^2(c+d x)}}-\frac {B \sin (c+d x) (b \cos (c+d x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {n}{2},\frac {n+2}{2},\cos ^2(c+d x)\right )}{d n \sqrt {\sin ^2(c+d x)}} \]

[In]

Int[(b*Cos[c + d*x])^n*(A + B*Cos[c + d*x])*Sec[c + d*x]^2,x]

[Out]

(A*b*(b*Cos[c + d*x])^(-1 + n)*Hypergeometric2F1[1/2, (-1 + n)/2, (1 + n)/2, Cos[c + d*x]^2]*Sin[c + d*x])/(d*
(1 - n)*Sqrt[Sin[c + d*x]^2]) - (B*(b*Cos[c + d*x])^n*Hypergeometric2F1[1/2, n/2, (2 + n)/2, Cos[c + d*x]^2]*S
in[c + d*x])/(d*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 2827

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rubi steps \begin{align*} \text {integral}& = b^2 \int (b \cos (c+d x))^{-2+n} (A+B \cos (c+d x)) \, dx \\ & = \left (A b^2\right ) \int (b \cos (c+d x))^{-2+n} \, dx+(b B) \int (b \cos (c+d x))^{-1+n} \, dx \\ & = \frac {A b (b \cos (c+d x))^{-1+n} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-1+n),\frac {1+n}{2},\cos ^2(c+d x)\right ) \sin (c+d x)}{d (1-n) \sqrt {\sin ^2(c+d x)}}-\frac {B (b \cos (c+d x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {n}{2},\frac {2+n}{2},\cos ^2(c+d x)\right ) \sin (c+d x)}{d n \sqrt {\sin ^2(c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.83 \[ \int (b \cos (c+d x))^n (A+B \cos (c+d x)) \sec ^2(c+d x) \, dx=-\frac {b (b \cos (c+d x))^{-1+n} \csc (c+d x) \left (A n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-1+n),\frac {1+n}{2},\cos ^2(c+d x)\right )+B (-1+n) \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {n}{2},\frac {2+n}{2},\cos ^2(c+d x)\right )\right ) \sqrt {\sin ^2(c+d x)}}{d (-1+n) n} \]

[In]

Integrate[(b*Cos[c + d*x])^n*(A + B*Cos[c + d*x])*Sec[c + d*x]^2,x]

[Out]

-((b*(b*Cos[c + d*x])^(-1 + n)*Csc[c + d*x]*(A*n*Hypergeometric2F1[1/2, (-1 + n)/2, (1 + n)/2, Cos[c + d*x]^2]
+ B*(-1 + n)*Cos[c + d*x]*Hypergeometric2F1[1/2, n/2, (2 + n)/2, Cos[c + d*x]^2])*Sqrt[Sin[c + d*x]^2])/(d*(-
1 + n)*n))

Maple [F]

\[\int \left (\cos \left (d x +c \right ) b \right )^{n} \left (A +B \cos \left (d x +c \right )\right ) \left (\sec ^{2}\left (d x +c \right )\right )d x\]

[In]

int((cos(d*x+c)*b)^n*(A+B*cos(d*x+c))*sec(d*x+c)^2,x)

[Out]

int((cos(d*x+c)*b)^n*(A+B*cos(d*x+c))*sec(d*x+c)^2,x)

Fricas [F]

\[ \int (b \cos (c+d x))^n (A+B \cos (c+d x)) \sec ^2(c+d x) \, dx=\int { {\left (B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{n} \sec \left (d x + c\right )^{2} \,d x } \]

[In]

integrate((b*cos(d*x+c))^n*(A+B*cos(d*x+c))*sec(d*x+c)^2,x, algorithm="fricas")

[Out]

integral((B*cos(d*x + c) + A)*(b*cos(d*x + c))^n*sec(d*x + c)^2, x)

Sympy [F]

\[ \int (b \cos (c+d x))^n (A+B \cos (c+d x)) \sec ^2(c+d x) \, dx=\int \left (b \cos {\left (c + d x \right )}\right )^{n} \left (A + B \cos {\left (c + d x \right )}\right ) \sec ^{2}{\left (c + d x \right )}\, dx \]

[In]

integrate((b*cos(d*x+c))**n*(A+B*cos(d*x+c))*sec(d*x+c)**2,x)

[Out]

Integral((b*cos(c + d*x))**n*(A + B*cos(c + d*x))*sec(c + d*x)**2, x)

Maxima [F]

\[ \int (b \cos (c+d x))^n (A+B \cos (c+d x)) \sec ^2(c+d x) \, dx=\int { {\left (B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{n} \sec \left (d x + c\right )^{2} \,d x } \]

[In]

integrate((b*cos(d*x+c))^n*(A+B*cos(d*x+c))*sec(d*x+c)^2,x, algorithm="maxima")

[Out]

integrate((B*cos(d*x + c) + A)*(b*cos(d*x + c))^n*sec(d*x + c)^2, x)

Giac [F]

\[ \int (b \cos (c+d x))^n (A+B \cos (c+d x)) \sec ^2(c+d x) \, dx=\int { {\left (B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{n} \sec \left (d x + c\right )^{2} \,d x } \]

[In]

integrate((b*cos(d*x+c))^n*(A+B*cos(d*x+c))*sec(d*x+c)^2,x, algorithm="giac")

[Out]

integrate((B*cos(d*x + c) + A)*(b*cos(d*x + c))^n*sec(d*x + c)^2, x)

Mupad [F(-1)]

Timed out. \[ \int (b \cos (c+d x))^n (A+B \cos (c+d x)) \sec ^2(c+d x) \, dx=\int \frac {{\left (b\,\cos \left (c+d\,x\right )\right )}^n\,\left (A+B\,\cos \left (c+d\,x\right )\right )}{{\cos \left (c+d\,x\right )}^2} \,d x \]

[In]

int(((b*cos(c + d*x))^n*(A + B*cos(c + d*x)))/cos(c + d*x)^2,x)

[Out]

int(((b*cos(c + d*x))^n*(A + B*cos(c + d*x)))/cos(c + d*x)^2, x)