\(\int \sec ^2(c+d x) (b \sec (c+d x))^n (A+C \sec ^2(c+d x)) \, dx\) [28]

Optimal result

Integrand size = 31, antiderivative size = 120 \[ \int \sec ^2(c+d x) (b \sec (c+d x))^n \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {(C (2+n)+A (3+n)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-1-n),\frac {1-n}{2},\cos ^2(c+d x)\right ) (b \sec (c+d x))^{1+n} \sin (c+d x)}{b d (1+n) (3+n) \sqrt {\sin ^2(c+d x)}}+\frac {C (b \sec (c+d x))^{2+n} \tan (c+d x)}{b^2 d (3+n)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 120, 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} \[ \int \sec ^2(c+d x) (b \sec (c+d x))^n \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {(A (n+3)+C (n+2)) \sin (c+d x) (b \sec (c+d x))^{n+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-n-1),\frac {1-n}{2},\cos ^2(c+d x)\right )}{b d (n+1) (n+3) \sqrt {\sin ^2(c+d x)}}+\frac {C \tan (c+d x) (b \sec (c+d x))^{n+2}}{b^2 d (n+3)} \]

[In]

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

[Out]

((C*(2 + n) + A*(3 + n))*Hypergeometric2F1[1/2, (-1 - n)/2, (1 - n)/2, Cos[c + d*x]^2]*(b*Sec[c + d*x])^(1 + n
)*Sin[c + d*x])/(b*d*(1 + n)*(3 + n)*Sqrt[Sin[c + d*x]^2]) + (C*(b*Sec[c + d*x])^(2 + n)*Tan[c + d*x])/(b^2*d*
(3 + 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}& = \frac {\int (b \sec (c+d x))^{2+n} \left (A+C \sec ^2(c+d x)\right ) \, dx}{b^2} \\ & = \frac {C (b \sec (c+d x))^{2+n} \tan (c+d x)}{b^2 d (3+n)}+\frac {\left (A+\frac {C (2+n)}{3+n}\right ) \int (b \sec (c+d x))^{2+n} \, dx}{b^2} \\ & = \frac {C (b \sec (c+d x))^{2+n} \tan (c+d x)}{b^2 d (3+n)}+\frac {\left (\left (A+\frac {C (2+n)}{3+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}{b^2} \\ & = \frac {\left (A+\frac {C (2+n)}{3+n}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-1-n),\frac {1-n}{2},\cos ^2(c+d x)\right ) (b \sec (c+d x))^{1+n} \sin (c+d x)}{b d (1+n) \sqrt {\sin ^2(c+d x)}}+\frac {C (b \sec (c+d x))^{2+n} \tan (c+d x)}{b^2 d (3+n)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.37 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.01 \[ \int \sec ^2(c+d x) (b \sec (c+d x))^n \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {\csc (c+d x) \sec (c+d x) (b \sec (c+d x))^n \left (A (4+n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+n}{2},\frac {4+n}{2},\sec ^2(c+d x)\right )+C (2+n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {4+n}{2},\frac {6+n}{2},\sec ^2(c+d x)\right ) \sec ^2(c+d x)\right ) \sqrt {-\tan ^2(c+d x)}}{d (2+n) (4+n)} \]

[In]

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

[Out]

(Csc[c + d*x]*Sec[c + d*x]*(b*Sec[c + d*x])^n*(A*(4 + n)*Hypergeometric2F1[1/2, (2 + n)/2, (4 + n)/2, Sec[c +
d*x]^2] + C*(2 + n)*Hypergeometric2F1[1/2, (4 + n)/2, (6 + n)/2, Sec[c + d*x]^2]*Sec[c + d*x]^2)*Sqrt[-Tan[c +
d*x]^2])/(d*(2 + n)*(4 + n))

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

integral((C*sec(d*x + c)^4 + A*sec(d*x + c)^2)*(b*sec(d*x + c))^n, x)

Sympy [F]

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

[In]

integrate(sec(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)*sec(c + d*x)**2, x)

Maxima [F]

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

[In]

integrate(sec(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*sec(d*x + c)^2, x)

Giac [F]

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

[In]

integrate(sec(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*sec(d*x + c)^2, x)

Mupad [F(-1)]

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

[In]

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

[Out]

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