\(\int \cos ^2(c+d x) (b \sec (c+d x))^n (A+C \sec ^2(c+d x)) \, dx\) [32]

Optimal result

Integrand size = 31, antiderivative size = 132 \[ \int \cos ^2(c+d x) (b \sec (c+d x))^n \left (A+C \sec ^2(c+d x)\right ) \, dx=-\frac {b^3 (A (1-n)+C (2-n)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3-n}{2},\frac {5-n}{2},\cos ^2(c+d x)\right ) (b \sec (c+d x))^{-3+n} \sin (c+d x)}{d (1-n) (3-n) \sqrt {\sin ^2(c+d x)}}-\frac {b^2 C (b \sec (c+d x))^{-2+n} \tan (c+d x)}{d (1-n)} \]

[Out]

-b^3*(A*(1-n)+C*(2-n))*hypergeom([1/2, 3/2-1/2*n],[5/2-1/2*n],cos(d*x+c)^2)*(b*sec(d*x+c))^(-3+n)*sin(d*x+c)/d
/(n^2-4*n+3)/(sin(d*x+c)^2)^(1/2)-b^2*C*(b*sec(d*x+c))^(-2+n)*tan(d*x+c)/d/(1-n)

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {16, 4131, 3857, 2722} \[ \int \cos ^2(c+d x) (b \sec (c+d x))^n \left (A+C \sec ^2(c+d x)\right ) \, dx=-\frac {b^3 (A (1-n)+C (2-n)) \sin (c+d x) (b \sec (c+d x))^{n-3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3-n}{2},\frac {5-n}{2},\cos ^2(c+d x)\right )}{d (1-n) (3-n) \sqrt {\sin ^2(c+d x)}}-\frac {b^2 C \tan (c+d x) (b \sec (c+d x))^{n-2}}{d (1-n)} \]

[In]

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

[Out]

-((b^3*(A*(1 - n) + C*(2 - n))*Hypergeometric2F1[1/2, (3 - n)/2, (5 - n)/2, Cos[c + d*x]^2]*(b*Sec[c + d*x])^(
-3 + n)*Sin[c + d*x])/(d*(1 - n)*(3 - n)*Sqrt[Sin[c + d*x]^2])) - (b^2*C*(b*Sec[c + d*x])^(-2 + n)*Tan[c + d*x
])/(d*(1 - n))

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 4131

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] + Dist[(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]

Rubi steps \begin{align*} \text {integral}& = b^2 \int (b \sec (c+d x))^{-2+n} \left (A+C \sec ^2(c+d x)\right ) \, dx \\ & = -\frac {b^2 C (b \sec (c+d x))^{-2+n} \tan (c+d x)}{d (1-n)}+\left (b^2 \left (A+\frac {C (2-n)}{1-n}\right )\right ) \int (b \sec (c+d x))^{-2+n} \, dx \\ & = -\frac {b^2 C (b \sec (c+d x))^{-2+n} \tan (c+d x)}{d (1-n)}+\left (b^2 \left (A+\frac {C (2-n)}{1-n}\right ) \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 )^{2-n} \, dx \\ & = -\frac {\left (A+\frac {C (2-n)}{1-n}\right ) \cos ^3(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3-n}{2},\frac {5-n}{2},\cos ^2(c+d x)\right ) (b \sec (c+d x))^n \sin (c+d x)}{d (3-n) \sqrt {\sin ^2(c+d x)}}-\frac {b^2 C (b \sec (c+d x))^{-2+n} \tan (c+d x)}{d (1-n)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.81 \[ \int \cos ^2(c+d x) (b \sec (c+d x))^n \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {\cot (c+d x) \left (A n \cos ^2(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-2+n),\frac {n}{2},\sec ^2(c+d x)\right )+C (-2+n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {n}{2},\frac {2+n}{2},\sec ^2(c+d x)\right )\right ) (b \sec (c+d x))^n \sqrt {-\tan ^2(c+d x)}}{d (-2+n) n} \]

[In]

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

[Out]

(Cot[c + d*x]*(A*n*Cos[c + d*x]^2*Hypergeometric2F1[1/2, (-2 + n)/2, n/2, Sec[c + d*x]^2] + C*(-2 + n)*Hyperge
ometric2F1[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*(-2 + n)*n)

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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