3.212 \(\int \sec (c+d x) (b \sec (c+d x))^n \, dx\)

Optimal. Leaf size=61 \[ \frac {\sin (c+d x) (b \sec (c+d x))^n \, _2F_1\left (\frac {1}{2},-\frac {n}{2};\frac {2-n}{2};\cos ^2(c+d x)\right )}{d n \sqrt {\sin ^2(c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {16, 3772, 2643} \[ \frac {\sin (c+d x) (b \sec (c+d x))^n \, _2F_1\left (\frac {1}{2},-\frac {n}{2};\frac {2-n}{2};\cos ^2(c+d x)\right )}{d n \sqrt {\sin ^2(c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Hypergeometric2F1[1/2, -n/2, (2 - n)/2, Cos[c + d*x]^2]*(b*Sec[c + d*x])^n*Sin[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 2643

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Sin[c + d*x])^(n + 1)*Hypergeomet
ric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2])/(b*d*(n + 1)*Sqrt[Cos[c + d*x]^2]), x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rule 3772

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]

Rubi steps

\begin {align*} \int \sec (c+d x) (b \sec (c+d x))^n \, dx &=\frac {\int (b \sec (c+d x))^{1+n} \, dx}{b}\\ &=\frac {\left (\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}{b}\\ &=\frac {\, _2F_1\left (\frac {1}{2},-\frac {n}{2};\frac {2-n}{2};\cos ^2(c+d x)\right ) (b \sec (c+d x))^n \sin (c+d x)}{d n \sqrt {\sin ^2(c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 65, normalized size = 1.07 \[ \frac {\sqrt {-\tan ^2(c+d x)} \csc (c+d x) (b \sec (c+d x))^n \, _2F_1\left (\frac {1}{2},\frac {n+1}{2};\frac {n+3}{2};\sec ^2(c+d x)\right )}{d (n+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Csc[c + d*x]*Hypergeometric2F1[1/2, (1 + n)/2, (3 + n)/2, Sec[c + d*x]^2]*(b*Sec[c + d*x])^n*Sqrt[-Tan[c + d*
x]^2])/(d*(1 + n))

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fricas [F]  time = 0.75, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\left (b \sec \left (d x + c\right )\right )^{n} \sec \left (d x + c\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (b \sec \left (d x + c\right )\right )^{n} \sec \left (d x + c\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 1.69, size = 0, normalized size = 0.00 \[ \int \sec \left (d x +c \right ) \left (b \sec \left (d x +c \right )\right )^{n}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (b \sec \left (d x + c\right )\right )^{n} \sec \left (d x + c\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^n}{\cos \left (c+d\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (b \sec {\left (c + d x \right )}\right )^{n} \sec {\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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