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

Optimal. Leaf size=133 \[ \frac {p \sqrt {\frac {1}{a x^p+1}} \sqrt {a x^p+1} x^{m-p+1} \, _2F_1\left (\frac {1}{2},\frac {m-p+1}{2 p};\frac {m+p+1}{2 p};a^2 x^{2 p}\right )}{a (m+1) (m-p+1)}+\frac {p x^{m-p+1}}{a (m+1) (m-p+1)}+\frac {x^{m+1} e^{\text {sech}^{-1}\left (a x^p\right )}}{m+1} \]

[Out]

(1/a/(x^p)+(1/a/(x^p)-1)^(1/2)*(1/a/(x^p)+1)^(1/2))*x^(1+m)/(1+m)+p*x^(1+m-p)/a/(1+m)/(1+m-p)+p*x^(1+m-p)*hype
rgeom([1/2, 1/2*(1+m-p)/p],[1/2*(1+m+p)/p],a^2*x^(2*p))*(1/(1+a*x^p))^(1/2)*(1+a*x^p)^(1/2)/a/(1+m)/(1+m-p)

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Rubi [A]  time = 0.08, antiderivative size = 133, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6335, 30, 259, 364} \[ \frac {p \sqrt {\frac {1}{a x^p+1}} \sqrt {a x^p+1} x^{m-p+1} \, _2F_1\left (\frac {1}{2},\frac {m-p+1}{2 p};\frac {m+p+1}{2 p};a^2 x^{2 p}\right )}{a (m+1) (m-p+1)}+\frac {p x^{m-p+1}}{a (m+1) (m-p+1)}+\frac {x^{m+1} e^{\text {sech}^{-1}\left (a x^p\right )}}{m+1} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(E^ArcSech[a*x^p]*x^(1 + m))/(1 + m) + (p*x^(1 + m - p))/(a*(1 + m)*(1 + m - p)) + (p*x^(1 + m - p)*Sqrt[(1 +
a*x^p)^(-1)]*Sqrt[1 + a*x^p]*Hypergeometric2F1[1/2, (1 + m - p)/(2*p), (1 + m + p)/(2*p), a^2*x^(2*p)])/(a*(1
+ m)*(1 + m - 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 6335

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

Rubi steps

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

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Mathematica [A]  time = 5.17, size = 186, normalized size = 1.40 \[ \frac {2^{\frac {m+1}{p}} x^{m+1} \left (a x^p\right )^{-\frac {m+1}{p}} e^{\text {sech}^{-1}\left (a x^p\right )} \left (\frac {e^{\text {sech}^{-1}\left (a x^p\right )}}{e^{2 \text {sech}^{-1}\left (a x^p\right )}+1}\right )^{\frac {m+1}{p}} \left ((m+3 p+1) \, _2F_1\left (1,1-\frac {m+p+1}{2 p};\frac {m+3 p+1}{2 p};-e^{2 \text {sech}^{-1}\left (a x^p\right )}\right )-(m+p+1) e^{2 \text {sech}^{-1}\left (a x^p\right )} \, _2F_1\left (1,-\frac {m-3 p+1}{2 p};\frac {m+5 p+1}{2 p};-e^{2 \text {sech}^{-1}\left (a x^p\right )}\right )\right )}{(m+p+1) (m+3 p+1)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2^((1 + m)/p)*E^ArcSech[a*x^p]*(E^ArcSech[a*x^p]/(1 + E^(2*ArcSech[a*x^p])))^((1 + m)/p)*x^(1 + m)*(-(E^(2*Ar
cSech[a*x^p])*(1 + m + p)*Hypergeometric2F1[1, -1/2*(1 + m - 3*p)/p, (1 + m + 5*p)/(2*p), -E^(2*ArcSech[a*x^p]
)]) + (1 + m + 3*p)*Hypergeometric2F1[1, 1 - (1 + m + p)/(2*p), (1 + m + 3*p)/(2*p), -E^(2*ArcSech[a*x^p])]))/
((1 + m + p)*(1 + m + 3*p)*(a*x^p)^((1 + m)/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^m,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 \[ \int x^{m} {\left (\sqrt {\frac {1}{a x^{p}} + 1} \sqrt {\frac {1}{a x^{p}} - 1} + \frac {1}{a x^{p}}\right )}\,{d x} \]

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^m,x, algorithm="giac")

[Out]

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

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maple [F]  time = 1.08, size = 0, normalized size = 0.00 \[ \int \left (\frac {x^{-p}}{a}+\sqrt {\frac {x^{-p}}{a}-1}\, \sqrt {\frac {x^{-p}}{a}+1}\right ) x^{m}\, dx \]

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^m,x)

[Out]

int((1/a/(x^p)+(1/a/(x^p)-1)^(1/2)*(1/a/(x^p)+1)^(1/2))*x^m,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^m,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(m-p>0)', see `assume?` for mor
e details)Is m-p equal to -1?

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x^m*((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 \[ \frac {\int x^{m} x^{- p}\, dx + \int a x^{m} \sqrt {-1 + \frac {x^{- p}}{a}} \sqrt {1 + \frac {x^{- p}}{a}}\, dx}{a} \]

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**m,x)

[Out]

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

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