3.62 \(\int e^{\text {sech}^{-1}(a x^p)} \, dx\)

Optimal. Leaf size=105 \[ \frac {p x^{1-p} \sqrt {\frac {1}{a x^p+1}} \sqrt {a x^p+1} \, _2F_1\left (\frac {1}{2},\frac {1}{2} \left (\frac {1}{p}-1\right );\frac {p+1}{2 p};a^2 x^{2 p}\right )}{a (1-p)}+\frac {p x^{1-p}}{a (1-p)}+x e^{\text {sech}^{-1}\left (a x^p\right )} \]

[Out]

(1/a/(x^p)+(1/a/(x^p)-1)^(1/2)*(1/a/(x^p)+1)^(1/2))*x+p*x^(1-p)/a/(1-p)+p*x^(1-p)*hypergeom([1/2, -1/2+1/2/p],
[1/2*(1+p)/p],a^2*x^(2*p))*(1/(1+a*x^p))^(1/2)*(1+a*x^p)^(1/2)/a/(1-p)

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Rubi [A]  time = 0.05, antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6330, 30, 259, 364} \[ \frac {p x^{1-p} \sqrt {\frac {1}{a x^p+1}} \sqrt {a x^p+1} \, _2F_1\left (\frac {1}{2},\frac {1}{2} \left (\frac {1}{p}-1\right );\frac {p+1}{2 p};a^2 x^{2 p}\right )}{a (1-p)}+\frac {p x^{1-p}}{a (1-p)}+x e^{\text {sech}^{-1}\left (a x^p\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

E^ArcSech[a*x^p]*x + (p*x^(1 - p))/(a*(1 - p)) + (p*x^(1 - p)*Sqrt[(1 + a*x^p)^(-1)]*Sqrt[1 + a*x^p]*Hypergeom
etric2F1[1/2, (-1 + p^(-1))/2, (1 + p)/(2*p), a^2*x^(2*p)])/(a*(1 - p))

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 259

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 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 6330

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

Rubi steps

\begin {align*} \int e^{\text {sech}^{-1}\left (a x^p\right )} \, dx &=e^{\text {sech}^{-1}\left (a x^p\right )} x+\frac {p \int x^{-p} \, dx}{a}+\frac {\left (p \sqrt {\frac {1}{1+a x^p}} \sqrt {1+a x^p}\right ) \int \frac {x^{-p}}{\sqrt {1-a x^p} \sqrt {1+a x^p}} \, dx}{a}\\ &=e^{\text {sech}^{-1}\left (a x^p\right )} x+\frac {p x^{1-p}}{a (1-p)}+\frac {\left (p \sqrt {\frac {1}{1+a x^p}} \sqrt {1+a x^p}\right ) \int \frac {x^{-p}}{\sqrt {1-a^2 x^{2 p}}} \, dx}{a}\\ &=e^{\text {sech}^{-1}\left (a x^p\right )} x+\frac {p x^{1-p}}{a (1-p)}+\frac {p x^{1-p} \sqrt {\frac {1}{1+a x^p}} \sqrt {1+a x^p} \, _2F_1\left (\frac {1}{2},\frac {1}{2} \left (-1+\frac {1}{p}\right );\frac {1+p}{2 p};a^2 x^{2 p}\right )}{a (1-p)}\\ \end {align*}

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Mathematica [A]  time = 0.41, size = 139, normalized size = 1.32 \[ \frac {x \left (-\frac {a^2 p x^p \sqrt {\frac {1-a x^p}{a x^p+1}} \sqrt {1-a^2 x^{2 p}} \, _2F_1\left (\frac {1}{2},\frac {p+1}{2 p};\frac {1}{2} \left (3+\frac {1}{p}\right );a^2 x^{2 p}\right )}{(p+1) \left (a x^p-1\right )}+\left (a+x^{-p}\right ) \sqrt {\frac {1-a x^p}{a x^p+1}}+x^{-p}\right )}{a-a p} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(x*(x^(-p) + (a + x^(-p))*Sqrt[(1 - a*x^p)/(1 + a*x^p)] - (a^2*p*x^p*Sqrt[(1 - a*x^p)/(1 + a*x^p)]*Sqrt[1 - a^
2*x^(2*p)]*Hypergeometric2F1[1/2, (1 + p)/(2*p), (3 + p^(-1))/2, a^2*x^(2*p)])/((1 + p)*(-1 + a*x^p))))/(a - a
*p)

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fricas [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (ha
s polynomial part)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {\frac {1}{a x^{p}} + 1} \sqrt {\frac {1}{a x^{p}} - 1} + \frac {1}{a x^{p}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 1.07, size = 0, normalized size = 0.00 \[ \int \frac {x^{-p}}{a}+\sqrt {\frac {x^{-p}}{a}-1}\, \sqrt {\frac {x^{-p}}{a}+1}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(-p>0)', see `assume?` for more
details)Is -p equal to -1?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \sqrt {\frac {1}{a\,x^p}-1}\,\sqrt {\frac {1}{a\,x^p}+1}+\frac {1}{a\,x^p} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int x^{- p}\, dx + \int a \sqrt {-1 + \frac {x^{- p}}{a}} \sqrt {1 + \frac {x^{- p}}{a}}\, dx}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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