3.120 \(\int x^m \sinh ^{-1}(a x) \, dx\)

Optimal. Leaf size=60 \[ \frac{x^{m+1} \sinh ^{-1}(a x)}{m+1}-\frac{a x^{m+2} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m+2}{2},\frac{m+4}{2},-a^2 x^2\right )}{m^2+3 m+2} \]

[Out]

(x^(1 + m)*ArcSinh[a*x])/(1 + m) - (a*x^(2 + m)*Hypergeometric2F1[1/2, (2 + m)/2, (4 + m)/2, -(a^2*x^2)])/(2 +
3*m + m^2)

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Rubi [A]  time = 0.0195273, antiderivative size = 60, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {5661, 364} \[ \frac{x^{m+1} \sinh ^{-1}(a x)}{m+1}-\frac{a x^{m+2} \, _2F_1\left (\frac{1}{2},\frac{m+2}{2};\frac{m+4}{2};-a^2 x^2\right )}{m^2+3 m+2} \]

Antiderivative was successfully verified.

[In]

Int[x^m*ArcSinh[a*x],x]

[Out]

(x^(1 + m)*ArcSinh[a*x])/(1 + m) - (a*x^(2 + m)*Hypergeometric2F1[1/2, (2 + m)/2, (4 + m)/2, -(a^2*x^2)])/(2 +
3*m + m^2)

Rule 5661

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcS
inh[c*x])^n)/(d*(m + 1)), x] - Dist[(b*c*n)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcSinh[c*x])^(n - 1))/Sqrt
[1 + c^2*x^2], x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[m, -1]

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])

Rubi steps

\begin{align*} \int x^m \sinh ^{-1}(a x) \, dx &=\frac{x^{1+m} \sinh ^{-1}(a x)}{1+m}-\frac{a \int \frac{x^{1+m}}{\sqrt{1+a^2 x^2}} \, dx}{1+m}\\ &=\frac{x^{1+m} \sinh ^{-1}(a x)}{1+m}-\frac{a x^{2+m} \, _2F_1\left (\frac{1}{2},\frac{2+m}{2};\frac{4+m}{2};-a^2 x^2\right )}{2+3 m+m^2}\\ \end{align*}

Mathematica [A]  time = 0.0222289, size = 55, normalized size = 0.92 \[ \frac{x^{m+1} \left ((m+2) \sinh ^{-1}(a x)-a x \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m+2}{2},\frac{m+4}{2},-a^2 x^2\right )\right )}{(m+1) (m+2)} \]

Antiderivative was successfully verified.

[In]

Integrate[x^m*ArcSinh[a*x],x]

[Out]

(x^(1 + m)*((2 + m)*ArcSinh[a*x] - a*x*Hypergeometric2F1[1/2, (2 + m)/2, (4 + m)/2, -(a^2*x^2)]))/((1 + m)*(2
+ m))

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Maple [F]  time = 0.546, size = 0, normalized size = 0. \begin{align*} \int{x}^{m}{\it Arcsinh} \left ( ax \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m*arcsinh(a*x),x)

[Out]

int(x^m*arcsinh(a*x),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m*arcsinh(a*x),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (x^{m} \operatorname{arsinh}\left (a x\right ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m*arcsinh(a*x),x, algorithm="fricas")

[Out]

integral(x^m*arcsinh(a*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**m*asinh(a*x),x)

[Out]

Integral(x**m*asinh(a*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m*arcsinh(a*x),x, algorithm="giac")

[Out]

integrate(x^m*arcsinh(a*x), x)