\(\int \frac {e^{\text {sech}^{-1}(a x^p)}}{x} \, dx\) [63]

Optimal result

Integrand size = 12, antiderivative size = 87 \[ \int \frac {e^{\text {sech}^{-1}\left (a x^p\right )}}{x} \, dx=-\frac {x^{-p}}{a p}-\frac {x^{-p} \sqrt {1-a x^p}}{a p \sqrt {\frac {1}{1+a x^p}}}-\frac {\sqrt {\frac {1}{1+a x^p}} \sqrt {1+a x^p} \arcsin \left (a x^p\right )}{p} \]

[Out]

-1/a/p/(x^p)-(1-a*x^p)^(1/2)/a/p/(x^p)/(1/(1+a*x^p))^(1/2)-arcsin(a*x^p)*(1/(1+a*x^p))^(1/2)*(1+a*x^p)^(1/2)/p

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.22, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6469, 265, 352, 248, 283, 222} \[ \int \frac {e^{\text {sech}^{-1}\left (a x^p\right )}}{x} \, dx=-\frac {x^{-p} \sqrt {\frac {1}{a x^p+1}} \sqrt {a x^p+1} \sqrt {1-a^2 x^{2 p}}}{a p}-\frac {x^{-p}}{a p}-\frac {\sqrt {\frac {1}{a x^p+1}} \sqrt {a x^p+1} \csc ^{-1}\left (\frac {x^{-p}}{a}\right )}{p} \]

[In]

Int[E^ArcSech[a*x^p]/x,x]

[Out]

-(1/(a*p*x^p)) - (Sqrt[(1 + a*x^p)^(-1)]*Sqrt[1 + a*x^p]*Sqrt[1 - a^2*x^(2*p)])/(a*p*x^p) - (Sqrt[(1 + a*x^p)^
(-1)]*Sqrt[1 + a*x^p]*ArcCsc[1/(a*x^p)])/p

Rule 222

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt[a])]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rule 248

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^2, x], x, 1/x] /; FreeQ[{a, b, p},
x] && ILtQ[n, 0]

Rule 265

Int[((c_.)*(x_))^(m_.)*((a1_) + (b1_.)*(x_)^(n_))^(p_)*((a2_) + (b2_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[(c*x)
^m*(a1*a2 + b1*b2*x^(2*n))^p, x] /; FreeQ[{a1, b1, a2, b2, c, m, n, p}, x] && EqQ[a2*b1 + a1*b2, 0] && (Intege
rQ[p] || (GtQ[a1, 0] && GtQ[a2, 0]))

Rule 283

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^p/(c*(m + 1
))), x] - Dist[b*n*(p/(c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] &&
IGtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !ILtQ[(m + n*p + n + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 352

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/(m + 1), Subst[Int[(a + b*x^Simplify[n/(m +
1)])^p, x], x, x^(m + 1)], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[n/(m + 1)]] &&  !IntegerQ[n]

Rule 6469

Int[E^ArcSech[(a_.)*(x_)^(p_.)]/(x_), x_Symbol] :> -Simp[(a*p*x^p)^(-1), x] + Dist[(Sqrt[1 + a*x^p]/a)*Sqrt[1/
(1 + a*x^p)], Int[Sqrt[1 + a*x^p]*(Sqrt[1 - a*x^p]/x^(p + 1)), x], x] /; FreeQ[{a, p}, x]

Rubi steps \begin{align*} \text {integral}& = -\frac {x^{-p}}{a p}+\frac {\left (\sqrt {\frac {1}{1+a x^p}} \sqrt {1+a x^p}\right ) \int x^{-1-p} \sqrt {1-a x^p} \sqrt {1+a x^p} \, dx}{a} \\ & = -\frac {x^{-p}}{a p}+\frac {\left (\sqrt {\frac {1}{1+a x^p}} \sqrt {1+a x^p}\right ) \int x^{-1-p} \sqrt {1-a^2 x^{2 p}} \, dx}{a} \\ & = -\frac {x^{-p}}{a p}-\frac {\left (\sqrt {\frac {1}{1+a x^p}} \sqrt {1+a x^p}\right ) \text {Subst}\left (\int \sqrt {1-\frac {a^2}{x^2}} \, dx,x,x^{-p}\right )}{a p} \\ & = -\frac {x^{-p}}{a p}+\frac {\left (\sqrt {\frac {1}{1+a x^p}} \sqrt {1+a x^p}\right ) \text {Subst}\left (\int \frac {\sqrt {1-a^2 x^2}}{x^2} \, dx,x,x^p\right )}{a p} \\ & = -\frac {x^{-p}}{a p}-\frac {x^{-p} \sqrt {\frac {1}{1+a x^p}} \sqrt {1+a x^p} \sqrt {1-a^2 x^{2 p}}}{a p}-\frac {\left (a \sqrt {\frac {1}{1+a x^p}} \sqrt {1+a x^p}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-a^2 x^2}} \, dx,x,x^p\right )}{p} \\ & = -\frac {x^{-p}}{a p}-\frac {x^{-p} \sqrt {\frac {1}{1+a x^p}} \sqrt {1+a x^p} \sqrt {1-a^2 x^{2 p}}}{a p}-\frac {\sqrt {\frac {1}{1+a x^p}} \sqrt {1+a x^p} \arcsin \left (a x^p\right )}{p} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.15 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.10 \[ \int \frac {e^{\text {sech}^{-1}\left (a x^p\right )}}{x} \, dx=-\frac {i \left (-i x^{-p}-i \left (a+x^{-p}\right ) \sqrt {\frac {1-a x^p}{1+a x^p}}+a \log \left (-2 i a x^p+2 \sqrt {\frac {1-a x^p}{1+a x^p}} \left (1+a x^p\right )\right )\right )}{a p} \]

[In]

Integrate[E^ArcSech[a*x^p]/x,x]

[Out]

((-I)*((-I)/x^p - I*(a + x^(-p))*Sqrt[(1 - a*x^p)/(1 + a*x^p)] + a*Log[(-2*I)*a*x^p + 2*Sqrt[(1 - a*x^p)/(1 +
a*x^p)]*(1 + a*x^p)]))/(a*p)

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 1.17 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.33

method	result	size
derivativedivides	\(\frac {-\frac {\sqrt {-\frac {\left (a \,x^{p}-1\right ) x^{-p}}{a}}\, \sqrt {\frac {\left (1+a \,x^{p}\right ) x^{-p}}{a}}\, \left (\arctan \left (\frac {\operatorname {csgn}\left (a \right ) a \,x^{p}}{\sqrt {-a^{2} x^{2 p}+1}}\right ) a \,x^{p}+\sqrt {-a^{2} x^{2 p}+1}\, \operatorname {csgn}\left (a \right )\right ) \operatorname {csgn}\left (a \right )}{\sqrt {-a^{2} x^{2 p}+1}}-\frac {x^{-p}}{a}}{p}\)	\(116\)
default	\(\frac {-\frac {\sqrt {-\frac {\left (a \,x^{p}-1\right ) x^{-p}}{a}}\, \sqrt {\frac {\left (1+a \,x^{p}\right ) x^{-p}}{a}}\, \left (\arctan \left (\frac {\operatorname {csgn}\left (a \right ) a \,x^{p}}{\sqrt {-a^{2} x^{2 p}+1}}\right ) a \,x^{p}+\sqrt {-a^{2} x^{2 p}+1}\, \operatorname {csgn}\left (a \right )\right ) \operatorname {csgn}\left (a \right )}{\sqrt {-a^{2} x^{2 p}+1}}-\frac {x^{-p}}{a}}{p}\)	\(116\)

[In]

int((1/a/(x^p)+(1/a/(x^p)-1)^(1/2)*(1/a/(x^p)+1)^(1/2))/x,x,method=_RETURNVERBOSE)

[Out]

1/p*(-(-(a*x^p-1)/a/(x^p))^(1/2)*((1+a*x^p)/a/(x^p))^(1/2)*(arctan(csgn(a)*a*x^p/(-(x^p)^2*a^2+1)^(1/2))*a*x^p
+(-(x^p)^2*a^2+1)^(1/2)*csgn(a))*csgn(a)/(-(x^p)^2*a^2+1)^(1/2)-1/a/(x^p))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.17 \[ \int \frac {e^{\text {sech}^{-1}\left (a x^p\right )}}{x} \, dx=-\frac {a x^{p} \sqrt {\frac {a x^{p} + 1}{a x^{p}}} \sqrt {-\frac {a x^{p} - 1}{a x^{p}}} - a x^{p} \arctan \left (\sqrt {\frac {a x^{p} + 1}{a x^{p}}} \sqrt {-\frac {a x^{p} - 1}{a x^{p}}}\right ) + 1}{a p x^{p}} \]

[In]

integrate((1/a/(x^p)+(1/a/(x^p)-1)^(1/2)*(1/a/(x^p)+1)^(1/2))/x,x, algorithm="fricas")

[Out]

-(a*x^p*sqrt((a*x^p + 1)/(a*x^p))*sqrt(-(a*x^p - 1)/(a*x^p)) - a*x^p*arctan(sqrt((a*x^p + 1)/(a*x^p))*sqrt(-(a
*x^p - 1)/(a*x^p))) + 1)/(a*p*x^p)

Sympy [F]

\[ \int \frac {e^{\text {sech}^{-1}\left (a x^p\right )}}{x} \, dx=\frac {\int \frac {x^{- p}}{x}\, dx + \int \frac {a \sqrt {-1 + \frac {x^{- p}}{a}} \sqrt {1 + \frac {x^{- p}}{a}}}{x}\, dx}{a} \]

[In]

integrate((1/a/(x**p)+(1/a/(x**p)-1)**(1/2)*(1/a/(x**p)+1)**(1/2))/x,x)

[Out]

(Integral(1/(x*x**p), x) + Integral(a*sqrt(-1 + 1/(a*x**p))*sqrt(1 + 1/(a*x**p))/x, x))/a

Maxima [F]

\[ \int \frac {e^{\text {sech}^{-1}\left (a x^p\right )}}{x} \, dx=\int { \frac {\sqrt {\frac {1}{a x^{p}} + 1} \sqrt {\frac {1}{a x^{p}} - 1} + \frac {1}{a x^{p}}}{x} \,d x } \]

[In]

integrate((1/a/(x^p)+(1/a/(x^p)-1)^(1/2)*(1/a/(x^p)+1)^(1/2))/x,x, algorithm="maxima")

[Out]

integrate(sqrt(a*x^p + 1)*sqrt(-a*x^p + 1)/(x*x^p), x)/a - 1/(a*p*x^p)

Giac [F]

\[ \int \frac {e^{\text {sech}^{-1}\left (a x^p\right )}}{x} \, dx=\int { \frac {\sqrt {\frac {1}{a x^{p}} + 1} \sqrt {\frac {1}{a x^{p}} - 1} + \frac {1}{a x^{p}}}{x} \,d x } \]

[In]

integrate((1/a/(x^p)+(1/a/(x^p)-1)^(1/2)*(1/a/(x^p)+1)^(1/2))/x,x, algorithm="giac")

[Out]

integrate((sqrt(1/(a*x^p) + 1)*sqrt(1/(a*x^p) - 1) + 1/(a*x^p))/x, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {e^{\text {sech}^{-1}\left (a x^p\right )}}{x} \, dx=\int \frac {\sqrt {\frac {1}{a\,x^p}-1}\,\sqrt {\frac {1}{a\,x^p}+1}+\frac {1}{a\,x^p}}{x} \,d x \]

[In]

int(((1/(a*x^p) - 1)^(1/2)*(1/(a*x^p) + 1)^(1/2) + 1/(a*x^p))/x,x)

[Out]

int(((1/(a*x^p) - 1)^(1/2)*(1/(a*x^p) + 1)^(1/2) + 1/(a*x^p))/x, x)