3.17 \(\int \frac{\sec ^{-1}(a x^n)}{x} \, dx\)

Optimal. Leaf size=69 \[ \frac{i \text{PolyLog}\left (2,-e^{2 i \sec ^{-1}\left (a x^n\right )}\right )}{2 n}+\frac{i \sec ^{-1}\left (a x^n\right )^2}{2 n}-\frac{\sec ^{-1}\left (a x^n\right ) \log \left (1+e^{2 i \sec ^{-1}\left (a x^n\right )}\right )}{n} \]

[Out]

((I/2)*ArcSec[a*x^n]^2)/n - (ArcSec[a*x^n]*Log[1 + E^((2*I)*ArcSec[a*x^n])])/n + ((I/2)*PolyLog[2, -E^((2*I)*A
rcSec[a*x^n])])/n

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Rubi [A]  time = 0.0925888, antiderivative size = 69, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.6, Rules used = {5218, 4626, 3719, 2190, 2279, 2391} \[ \frac{i \text{PolyLog}\left (2,-e^{2 i \sec ^{-1}\left (a x^n\right )}\right )}{2 n}+\frac{i \sec ^{-1}\left (a x^n\right )^2}{2 n}-\frac{\sec ^{-1}\left (a x^n\right ) \log \left (1+e^{2 i \sec ^{-1}\left (a x^n\right )}\right )}{n} \]

Antiderivative was successfully verified.

[In]

Int[ArcSec[a*x^n]/x,x]

[Out]

((I/2)*ArcSec[a*x^n]^2)/n - (ArcSec[a*x^n]*Log[1 + E^((2*I)*ArcSec[a*x^n])])/n + ((I/2)*PolyLog[2, -E^((2*I)*A
rcSec[a*x^n])])/n

Rule 5218

Int[((a_.) + ArcSec[(c_.)*(x_)]*(b_.))/(x_), x_Symbol] :> -Subst[Int[(a + b*ArcCos[x/c])/x, x], x, 1/x] /; Fre
eQ[{a, b, c}, x]

Rule 4626

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)/(x_), x_Symbol] :> -Subst[Int[(a + b*x)^n/Cot[x], x], x, ArcCos[c
*x]] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0]

Rule 3719

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[(I*(c + d*x)^(m + 1))/(d*(m + 1)), x
] - Dist[2*I, Int[((c + d*x)^m*E^(2*I*(e + f*x)))/(1 + E^(2*I*(e + f*x))), x], x] /; FreeQ[{c, d, e, f}, x] &&
IGtQ[m, 0]

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rubi steps

\begin{align*} \int \frac{\sec ^{-1}\left (a x^n\right )}{x} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\sec ^{-1}(a x)}{x} \, dx,x,x^n\right )}{n}\\ &=-\frac{\operatorname{Subst}\left (\int \frac{\cos ^{-1}\left (\frac{x}{a}\right )}{x} \, dx,x,x^{-n}\right )}{n}\\ &=\frac{\operatorname{Subst}\left (\int x \tan (x) \, dx,x,\cos ^{-1}\left (\frac{x^{-n}}{a}\right )\right )}{n}\\ &=\frac{i \cos ^{-1}\left (\frac{x^{-n}}{a}\right )^2}{2 n}-\frac{(2 i) \operatorname{Subst}\left (\int \frac{e^{2 i x} x}{1+e^{2 i x}} \, dx,x,\cos ^{-1}\left (\frac{x^{-n}}{a}\right )\right )}{n}\\ &=\frac{i \cos ^{-1}\left (\frac{x^{-n}}{a}\right )^2}{2 n}-\frac{\cos ^{-1}\left (\frac{x^{-n}}{a}\right ) \log \left (1+e^{2 i \cos ^{-1}\left (\frac{x^{-n}}{a}\right )}\right )}{n}+\frac{\operatorname{Subst}\left (\int \log \left (1+e^{2 i x}\right ) \, dx,x,\cos ^{-1}\left (\frac{x^{-n}}{a}\right )\right )}{n}\\ &=\frac{i \cos ^{-1}\left (\frac{x^{-n}}{a}\right )^2}{2 n}-\frac{\cos ^{-1}\left (\frac{x^{-n}}{a}\right ) \log \left (1+e^{2 i \cos ^{-1}\left (\frac{x^{-n}}{a}\right )}\right )}{n}-\frac{i \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{2 i \cos ^{-1}\left (\frac{x^{-n}}{a}\right )}\right )}{2 n}\\ &=\frac{i \cos ^{-1}\left (\frac{x^{-n}}{a}\right )^2}{2 n}-\frac{\cos ^{-1}\left (\frac{x^{-n}}{a}\right ) \log \left (1+e^{2 i \cos ^{-1}\left (\frac{x^{-n}}{a}\right )}\right )}{n}+\frac{i \text{Li}_2\left (-e^{2 i \cos ^{-1}\left (\frac{x^{-n}}{a}\right )}\right )}{2 n}\\ \end{align*}

Mathematica [C]  time = 0.0843848, size = 60, normalized size = 0.87 \[ \frac{x^{-n} \text{HypergeometricPFQ}\left (\left \{\frac{1}{2},\frac{1}{2},\frac{1}{2}\right \},\left \{\frac{3}{2},\frac{3}{2}\right \},\frac{x^{-2 n}}{a^2}\right )}{a n}+\log (x) \left (\sin ^{-1}\left (\frac{x^{-n}}{a}\right )+\sec ^{-1}\left (a x^n\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[ArcSec[a*x^n]/x,x]

[Out]

HypergeometricPFQ[{1/2, 1/2, 1/2}, {3/2, 3/2}, 1/(a^2*x^(2*n))]/(a*n*x^n) + (ArcSec[a*x^n] + ArcSin[1/(a*x^n)]
)*Log[x]

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Maple [A]  time = 0.336, size = 98, normalized size = 1.4 \begin{align*}{\frac{{\frac{i}{2}} \left ({\rm arcsec} \left (a{x}^{n}\right ) \right ) ^{2}}{n}}-{\frac{{\rm arcsec} \left (a{x}^{n}\right )}{n}\ln \left ( 1+ \left ({\frac{1}{a{x}^{n}}}+i\sqrt{1-{\frac{1}{{a}^{2} \left ({x}^{n} \right ) ^{2}}}} \right ) ^{2} \right ) }+{\frac{{\frac{i}{2}}}{n}{\it polylog} \left ( 2,- \left ({\frac{1}{a{x}^{n}}}+i\sqrt{1-{\frac{1}{{a}^{2} \left ({x}^{n} \right ) ^{2}}}} \right ) ^{2} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsec(a*x^n)/x,x)

[Out]

1/2*I*arcsec(a*x^n)^2/n-arcsec(a*x^n)*ln(1+(1/a/(x^n)+I*(1-1/a^2/(x^n)^2)^(1/2))^2)/n+1/2*I*polylog(2,-(1/a/(x
^n)+I*(1-1/a^2/(x^n)^2)^(1/2))^2)/n

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -a^{2} n \int \frac{\sqrt{a x^{n} + 1} \sqrt{a x^{n} - 1} \log \left (x\right )}{a^{4} x x^{2 \, n} - a^{2} x}\,{d x} + \arctan \left (\sqrt{a x^{n} + 1} \sqrt{a x^{n} - 1}\right ) \log \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsec(a*x^n)/x,x, algorithm="maxima")

[Out]

-a^2*n*integrate(sqrt(a*x^n + 1)*sqrt(a*x^n - 1)*log(x)/(a^4*x*x^(2*n) - a^2*x), x) + arctan(sqrt(a*x^n + 1)*s
qrt(a*x^n - 1))*log(x)

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Fricas [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: UnboundLocalError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsec(a*x^n)/x,x, algorithm="fricas")

[Out]

Exception raised: UnboundLocalError

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(asec(a*x**n)/x,x)

[Out]

Integral(asec(a*x**n)/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsec(a*x^n)/x,x, algorithm="giac")

[Out]

integrate(arcsec(a*x^n)/x, x)