3.6.30 \(\int \frac {x \sinh ^{-1}(a x)^n}{\sqrt {1+a^2 x^2}} \, dx\) [530]

Optimal. Leaf size=49 \[ \frac {\left (-\sinh ^{-1}(a x)\right )^{-n} \sinh ^{-1}(a x)^n \Gamma \left (1+n,-\sinh ^{-1}(a x)\right )}{2 a^2}+\frac {\Gamma \left (1+n,\sinh ^{-1}(a x)\right )}{2 a^2} \]

[Out]

1/2*arcsinh(a*x)^n*GAMMA(1+n,-arcsinh(a*x))/a^2/((-arcsinh(a*x))^n)+1/2*GAMMA(1+n,arcsinh(a*x))/a^2

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Rubi [A]
time = 0.07, antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {5819, 3389, 2212} \begin {gather*} \frac {\sinh ^{-1}(a x)^n \left (-\sinh ^{-1}(a x)\right )^{-n} \text {Gamma}\left (n+1,-\sinh ^{-1}(a x)\right )}{2 a^2}+\frac {\text {Gamma}\left (n+1,\sinh ^{-1}(a x)\right )}{2 a^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x*ArcSinh[a*x]^n)/Sqrt[1 + a^2*x^2],x]

[Out]

(ArcSinh[a*x]^n*Gamma[1 + n, -ArcSinh[a*x]])/(2*a^2*(-ArcSinh[a*x])^n) + Gamma[1 + n, ArcSinh[a*x]]/(2*a^2)

Rule 2212

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[(-F^(g*(e - c*(f/d))))*((c
+ d*x)^FracPart[m]/(d*((-f)*g*(Log[F]/d))^(IntPart[m] + 1)*((-f)*g*Log[F]*((c + d*x)/d))^FracPart[m]))*Gamma[m
+ 1, ((-f)*g*(Log[F]/d))*(c + d*x)], x] /; FreeQ[{F, c, d, e, f, g, m}, x] &&  !IntegerQ[m]

Rule 3389

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

Rule 5819

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[(1/(b*
c^(m + 1)))*Simp[(d + e*x^2)^p/(1 + c^2*x^2)^p], Subst[Int[x^n*Sinh[-a/b + x/b]^m*Cosh[-a/b + x/b]^(2*p + 1),
x], x, a + b*ArcSinh[c*x]], x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[e, c^2*d] && IGtQ[2*p + 2, 0] && IGtQ[m,
0]

Rubi steps

\begin {align*} \int \frac {x \sinh ^{-1}(a x)^n}{\sqrt {1+a^2 x^2}} \, dx &=\frac {\text {Subst}\left (\int x^n \sinh (x) \, dx,x,\sinh ^{-1}(a x)\right )}{a^2}\\ &=-\frac {\text {Subst}\left (\int e^{-x} x^n \, dx,x,\sinh ^{-1}(a x)\right )}{2 a^2}+\frac {\text {Subst}\left (\int e^x x^n \, dx,x,\sinh ^{-1}(a x)\right )}{2 a^2}\\ &=\frac {\left (-\sinh ^{-1}(a x)\right )^{-n} \sinh ^{-1}(a x)^n \Gamma \left (1+n,-\sinh ^{-1}(a x)\right )}{2 a^2}+\frac {\Gamma \left (1+n,\sinh ^{-1}(a x)\right )}{2 a^2}\\ \end {align*}

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Mathematica [A]
time = 0.06, size = 43, normalized size = 0.88 \begin {gather*} \frac {\left (-\sinh ^{-1}(a x)\right )^{-n} \sinh ^{-1}(a x)^n \Gamma \left (1+n,-\sinh ^{-1}(a x)\right )+\Gamma \left (1+n,\sinh ^{-1}(a x)\right )}{2 a^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x*ArcSinh[a*x]^n)/Sqrt[1 + a^2*x^2],x]

[Out]

((ArcSinh[a*x]^n*Gamma[1 + n, -ArcSinh[a*x]])/(-ArcSinh[a*x])^n + Gamma[1 + n, ArcSinh[a*x]])/(2*a^2)

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Maple [F]
time = 180.00, size = 0, normalized size = 0.00 \[\int \frac {x \arcsinh \left (a x \right )^{n}}{\sqrt {a^{2} x^{2}+1}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*arcsinh(a*x)^n/(a^2*x^2+1)^(1/2),x)

[Out]

int(x*arcsinh(a*x)^n/(a^2*x^2+1)^(1/2),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(x*arcsinh(a*x)^n/(a^2*x^2+1)^(1/2),x, algorithm="maxima")

[Out]

integrate(x*arcsinh(a*x)^n/sqrt(a^2*x^2 + 1), 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(x*arcsinh(a*x)^n/(a^2*x^2+1)^(1/2),x, algorithm="fricas")

[Out]

integral(x*arcsinh(a*x)^n/sqrt(a^2*x^2 + 1), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x \operatorname {asinh}^{n}{\left (a x \right )}}{\sqrt {a^{2} x^{2} + 1}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*asinh(a*x)**n/(a**2*x**2+1)**(1/2),x)

[Out]

Integral(x*asinh(a*x)**n/sqrt(a**2*x**2 + 1), 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(x*arcsinh(a*x)^n/(a^2*x^2+1)^(1/2),x, algorithm="giac")

[Out]

integrate(x*arcsinh(a*x)^n/sqrt(a^2*x^2 + 1), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {x\,{\mathrm {asinh}\left (a\,x\right )}^n}{\sqrt {a^2\,x^2+1}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*asinh(a*x)^n)/(a^2*x^2 + 1)^(1/2),x)

[Out]

int((x*asinh(a*x)^n)/(a^2*x^2 + 1)^(1/2), x)

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