\(\int x^m \text {arcsinh}(a x) \, dx\) [120]

Optimal result

Integrand size = 8, antiderivative size = 60 \[ \int x^m \text {arcsinh}(a x) \, dx=\frac {x^{1+m} \text {arcsinh}(a x)}{1+m}-\frac {a x^{2+m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},-a^2 x^2\right )}{2+3 m+m^2} \]

[Out]

x^(1+m)*arcsinh(a*x)/(1+m)-a*x^(2+m)*hypergeom([1/2, 1+1/2*m],[2+1/2*m],-a^2*x^2)/(m^2+3*m+2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {5776, 371} \[ \int x^m \text {arcsinh}(a x) \, dx=\frac {x^{m+1} \text {arcsinh}(a x)}{m+1}-\frac {a x^{m+2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+2}{2},\frac {m+4}{2},-a^2 x^2\right )}{m^2+3 m+2} \]

[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 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 5776

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]

Rubi steps \begin{align*} \text {integral}& = \frac {x^{1+m} \text {arcsinh}(a x)}{1+m}-\frac {a \int \frac {x^{1+m}}{\sqrt {1+a^2 x^2}} \, dx}{1+m} \\ & = \frac {x^{1+m} \text {arcsinh}(a x)}{1+m}-\frac {a x^{2+m} \operatorname {Hypergeometric2F1}\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] (verified)

Time = 0.02 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.92 \[ \int x^m \text {arcsinh}(a x) \, dx=\frac {x^{1+m} \left ((2+m) \text {arcsinh}(a x)-a x \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},-a^2 x^2\right )\right )}{(1+m) (2+m)} \]

[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))

Maple [F]

\[\int x^{m} \operatorname {arcsinh}\left (a x \right )d x\]

[In]

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

[Out]

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

Fricas [F]

\[ \int x^m \text {arcsinh}(a x) \, dx=\int { x^{m} \operatorname {arsinh}\left (a x\right ) \,d x } \]

[In]

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

[Out]

integral(x^m*arcsinh(a*x), x)

Sympy [F]

\[ \int x^m \text {arcsinh}(a x) \, dx=\int x^{m} \operatorname {asinh}{\left (a x \right )}\, dx \]

[In]

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

[Out]

Integral(x**m*asinh(a*x), x)

Maxima [F]

\[ \int x^m \text {arcsinh}(a x) \, dx=\int { x^{m} \operatorname {arsinh}\left (a x\right ) \,d x } \]

[In]

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

[Out]

-a^2*integrate(x^2*x^m/(a^2*(m + 1)*x^2 + m + 1), x) - a*integrate(x*x^m/(a^3*(m + 1)*x^3 + a*(m + 1)*x + (a^2
*(m + 1)*x^2 + m + 1)*sqrt(a^2*x^2 + 1)), x) + x*x^m*log(a*x + sqrt(a^2*x^2 + 1))/(m + 1)

Giac [F]

\[ \int x^m \text {arcsinh}(a x) \, dx=\int { x^{m} \operatorname {arsinh}\left (a x\right ) \,d x } \]

[In]

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

[Out]

integrate(x^m*arcsinh(a*x), x)

Mupad [F(-1)]

Timed out. \[ \int x^m \text {arcsinh}(a x) \, dx=\int x^m\,\mathrm {asinh}\left (a\,x\right ) \,d x \]

[In]

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

[Out]

int(x^m*asinh(a*x), x)