3.38 \(\int (c \sec (a+b x))^n \, dx\)

Optimal. Leaf size=73 \[ -\frac{c \sin (a+b x) (c \sec (a+b x))^{n-1} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{1-n}{2},\frac{3-n}{2},\cos ^2(a+b x)\right )}{b (1-n) \sqrt{\sin ^2(a+b x)}} \]

[Out]

-((c*Hypergeometric2F1[1/2, (1 - n)/2, (3 - n)/2, Cos[a + b*x]^2]*(c*Sec[a + b*x])^(-1 + n)*Sin[a + b*x])/(b*(
1 - n)*Sqrt[Sin[a + b*x]^2]))

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Rubi [A]  time = 0.0311051, antiderivative size = 73, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3772, 2643} \[ -\frac{c \sin (a+b x) (c \sec (a+b x))^{n-1} \, _2F_1\left (\frac{1}{2},\frac{1-n}{2};\frac{3-n}{2};\cos ^2(a+b x)\right )}{b (1-n) \sqrt{\sin ^2(a+b x)}} \]

Antiderivative was successfully verified.

[In]

Int[(c*Sec[a + b*x])^n,x]

[Out]

-((c*Hypergeometric2F1[1/2, (1 - n)/2, (3 - n)/2, Cos[a + b*x]^2]*(c*Sec[a + b*x])^(-1 + n)*Sin[a + b*x])/(b*(
1 - n)*Sqrt[Sin[a + b*x]^2]))

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]

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]

Rubi steps

\begin{align*} \int (c \sec (a+b x))^n \, dx &=\left (\frac{\cos (a+b x)}{c}\right )^n (c \sec (a+b x))^n \int \left (\frac{\cos (a+b x)}{c}\right )^{-n} \, dx\\ &=-\frac{\cos (a+b x) \, _2F_1\left (\frac{1}{2},\frac{1-n}{2};\frac{3-n}{2};\cos ^2(a+b x)\right ) (c \sec (a+b x))^n \sin (a+b x)}{b (1-n) \sqrt{\sin ^2(a+b x)}}\\ \end{align*}

Mathematica [A]  time = 0.0469835, size = 61, normalized size = 0.84 \[ \frac{\sqrt{-\tan ^2(a+b x)} \cot (a+b x) (c \sec (a+b x))^n \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{n}{2},\frac{n+2}{2},\sec ^2(a+b x)\right )}{b n} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(c*Sec[a + b*x])^n,x]

[Out]

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

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Maple [F]  time = 0.265, size = 0, normalized size = 0. \begin{align*} \int \left ( c\sec \left ( bx+a \right ) \right ) ^{n}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (c \sec \left (b x + a\right )\right )^{n}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((c*sec(b*x + a))^n, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\left (c \sec \left (b x + a\right )\right )^{n}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((c*sec(b*x + a))^n, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (c \sec{\left (a + b x \right )}\right )^{n}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((c*sec(a + b*x))**n, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (c \sec \left (b x + a\right )\right )^{n}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((c*sec(b*x + a))^n, x)