3.5.100 \(\int (b \sec (e+f x))^n \, dx\) [500]

Optimal. Leaf size=73 \[ -\frac {b \, _2F_1\left (\frac {1}{2},\frac {1-n}{2};\frac {3-n}{2};\cos ^2(e+f x)\right ) (b \sec (e+f x))^{-1+n} \sin (e+f x)}{f (1-n) \sqrt {\sin ^2(e+f x)}} \]

[Out]

-b*hypergeom([1/2, 1/2-1/2*n],[3/2-1/2*n],cos(f*x+e)^2)*(b*sec(f*x+e))^(-1+n)*sin(f*x+e)/f/(1-n)/(sin(f*x+e)^2
)^(1/2)

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Rubi [A]
time = 0.02, antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3857, 2722} \begin {gather*} -\frac {b \sin (e+f x) (b \sec (e+f x))^{n-1} \, _2F_1\left (\frac {1}{2},\frac {1-n}{2};\frac {3-n}{2};\cos ^2(e+f x)\right )}{f (1-n) \sqrt {\sin ^2(e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*Sec[e + f*x])^n,x]

[Out]

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

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]

Rubi steps

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

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Mathematica [A]
time = 0.05, size = 61, normalized size = 0.84 \begin {gather*} \frac {\cot (e+f x) \, _2F_1\left (\frac {1}{2},\frac {n}{2};\frac {2+n}{2};\sec ^2(e+f x)\right ) (b \sec (e+f x))^n \sqrt {-\tan ^2(e+f x)}}{f n} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b*Sec[e + f*x])^n,x]

[Out]

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

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \left (b \sec \left (f x +e \right )\right )^{n}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*sec(f*x+e))^n,x)

[Out]

int((b*sec(f*x+e))^n,x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(f*x+e))^n,x, algorithm="maxima")

[Out]

integrate((b*sec(f*x + e))^n, x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(f*x+e))^n,x, algorithm="fricas")

[Out]

integral((b*sec(f*x + e))^n, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (b \sec {\left (e + f x \right )}\right )^{n}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(f*x+e))**n,x)

[Out]

Integral((b*sec(e + f*x))**n, x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(f*x+e))^n,x, algorithm="giac")

[Out]

integrate((b*sec(f*x + e))^n, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (\frac {b}{\cos \left (e+f\,x\right )}\right )}^n \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b/cos(e + f*x))^n,x)

[Out]

int((b/cos(e + f*x))^n, x)

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