\(\int \frac {x \text {arcsinh}(a x)^3}{\sqrt {1+a^2 x^2}} \, dx\) [345]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [F(-1)]

Optimal result

Integrand size = 21, antiderivative size = 64 \[ \int \frac {x \text {arcsinh}(a x)^3}{\sqrt {1+a^2 x^2}} \, dx=-\frac {6 x}{a}+\frac {6 \sqrt {1+a^2 x^2} \text {arcsinh}(a x)}{a^2}-\frac {3 x \text {arcsinh}(a x)^2}{a}+\frac {\sqrt {1+a^2 x^2} \text {arcsinh}(a x)^3}{a^2} \]

[Out]

-6*x/a-3*x*arcsinh(a*x)^2/a+6*arcsinh(a*x)*(a^2*x^2+1)^(1/2)/a^2+arcsinh(a*x)^3*(a^2*x^2+1)^(1/2)/a^2

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {5798, 5772, 8} \[ \int \frac {x \text {arcsinh}(a x)^3}{\sqrt {1+a^2 x^2}} \, dx=\frac {\sqrt {a^2 x^2+1} \text {arcsinh}(a x)^3}{a^2}+\frac {6 \sqrt {a^2 x^2+1} \text {arcsinh}(a x)}{a^2}-\frac {3 x \text {arcsinh}(a x)^2}{a}-\frac {6 x}{a} \]

[In]

Int[(x*ArcSinh[a*x]^3)/Sqrt[1 + a^2*x^2],x]

[Out]

(-6*x)/a + (6*Sqrt[1 + a^2*x^2]*ArcSinh[a*x])/a^2 - (3*x*ArcSinh[a*x]^2)/a + (Sqrt[1 + a^2*x^2]*ArcSinh[a*x]^3
)/a^2

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 5772

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.), x_Symbol] :> Simp[x*(a + b*ArcSinh[c*x])^n, x] - Dist[b*c*n, In
t[x*((a + b*ArcSinh[c*x])^(n - 1)/Sqrt[1 + c^2*x^2]), x], x] /; FreeQ[{a, b, c}, x] && GtQ[n, 0]

Rule 5798

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

Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {1+a^2 x^2} \text {arcsinh}(a x)^3}{a^2}-\frac {3 \int \text {arcsinh}(a x)^2 \, dx}{a} \\ & = -\frac {3 x \text {arcsinh}(a x)^2}{a}+\frac {\sqrt {1+a^2 x^2} \text {arcsinh}(a x)^3}{a^2}+6 \int \frac {x \text {arcsinh}(a x)}{\sqrt {1+a^2 x^2}} \, dx \\ & = \frac {6 \sqrt {1+a^2 x^2} \text {arcsinh}(a x)}{a^2}-\frac {3 x \text {arcsinh}(a x)^2}{a}+\frac {\sqrt {1+a^2 x^2} \text {arcsinh}(a x)^3}{a^2}-\frac {6 \int 1 \, dx}{a} \\ & = -\frac {6 x}{a}+\frac {6 \sqrt {1+a^2 x^2} \text {arcsinh}(a x)}{a^2}-\frac {3 x \text {arcsinh}(a x)^2}{a}+\frac {\sqrt {1+a^2 x^2} \text {arcsinh}(a x)^3}{a^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.91 \[ \int \frac {x \text {arcsinh}(a x)^3}{\sqrt {1+a^2 x^2}} \, dx=\frac {-6 a x+6 \sqrt {1+a^2 x^2} \text {arcsinh}(a x)-3 a x \text {arcsinh}(a x)^2+\sqrt {1+a^2 x^2} \text {arcsinh}(a x)^3}{a^2} \]

[In]

Integrate[(x*ArcSinh[a*x]^3)/Sqrt[1 + a^2*x^2],x]

[Out]

(-6*a*x + 6*Sqrt[1 + a^2*x^2]*ArcSinh[a*x] - 3*a*x*ArcSinh[a*x]^2 + Sqrt[1 + a^2*x^2]*ArcSinh[a*x]^3)/a^2

Maple [A] (verified)

Time = 0.29 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.41

method result size
default \(\frac {\operatorname {arcsinh}\left (a x \right )^{3} a^{2} x^{2}+\operatorname {arcsinh}\left (a x \right )^{3}-3 \operatorname {arcsinh}\left (a x \right )^{2} \sqrt {a^{2} x^{2}+1}\, a x +6 a^{2} x^{2} \operatorname {arcsinh}\left (a x \right )+6 \,\operatorname {arcsinh}\left (a x \right )-6 a x \sqrt {a^{2} x^{2}+1}}{a^{2} \sqrt {a^{2} x^{2}+1}}\) \(90\)

[In]

int(x*arcsinh(a*x)^3/(a^2*x^2+1)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/a^2/(a^2*x^2+1)^(1/2)*(arcsinh(a*x)^3*a^2*x^2+arcsinh(a*x)^3-3*arcsinh(a*x)^2*(a^2*x^2+1)^(1/2)*a*x+6*a^2*x^
2*arcsinh(a*x)+6*arcsinh(a*x)-6*a*x*(a^2*x^2+1)^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.44 \[ \int \frac {x \text {arcsinh}(a x)^3}{\sqrt {1+a^2 x^2}} \, dx=-\frac {3 \, a x \log \left (a x + \sqrt {a^{2} x^{2} + 1}\right )^{2} - \sqrt {a^{2} x^{2} + 1} \log \left (a x + \sqrt {a^{2} x^{2} + 1}\right )^{3} + 6 \, a x - 6 \, \sqrt {a^{2} x^{2} + 1} \log \left (a x + \sqrt {a^{2} x^{2} + 1}\right )}{a^{2}} \]

[In]

integrate(x*arcsinh(a*x)^3/(a^2*x^2+1)^(1/2),x, algorithm="fricas")

[Out]

-(3*a*x*log(a*x + sqrt(a^2*x^2 + 1))^2 - sqrt(a^2*x^2 + 1)*log(a*x + sqrt(a^2*x^2 + 1))^3 + 6*a*x - 6*sqrt(a^2
*x^2 + 1)*log(a*x + sqrt(a^2*x^2 + 1)))/a^2

Sympy [A] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.95 \[ \int \frac {x \text {arcsinh}(a x)^3}{\sqrt {1+a^2 x^2}} \, dx=\begin {cases} - \frac {3 x \operatorname {asinh}^{2}{\left (a x \right )}}{a} - \frac {6 x}{a} + \frac {\sqrt {a^{2} x^{2} + 1} \operatorname {asinh}^{3}{\left (a x \right )}}{a^{2}} + \frac {6 \sqrt {a^{2} x^{2} + 1} \operatorname {asinh}{\left (a x \right )}}{a^{2}} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \]

[In]

integrate(x*asinh(a*x)**3/(a**2*x**2+1)**(1/2),x)

[Out]

Piecewise((-3*x*asinh(a*x)**2/a - 6*x/a + sqrt(a**2*x**2 + 1)*asinh(a*x)**3/a**2 + 6*sqrt(a**2*x**2 + 1)*asinh
(a*x)/a**2, Ne(a, 0)), (0, True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.95 \[ \int \frac {x \text {arcsinh}(a x)^3}{\sqrt {1+a^2 x^2}} \, dx=-\frac {3 \, x \operatorname {arsinh}\left (a x\right )^{2}}{a} + \frac {\sqrt {a^{2} x^{2} + 1} \operatorname {arsinh}\left (a x\right )^{3}}{a^{2}} - \frac {6 \, {\left (x - \frac {\sqrt {a^{2} x^{2} + 1} \operatorname {arsinh}\left (a x\right )}{a}\right )}}{a} \]

[In]

integrate(x*arcsinh(a*x)^3/(a^2*x^2+1)^(1/2),x, algorithm="maxima")

[Out]

-3*x*arcsinh(a*x)^2/a + sqrt(a^2*x^2 + 1)*arcsinh(a*x)^3/a^2 - 6*(x - sqrt(a^2*x^2 + 1)*arcsinh(a*x)/a)/a

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.58 \[ \int \frac {x \text {arcsinh}(a x)^3}{\sqrt {1+a^2 x^2}} \, dx=\frac {\sqrt {a^{2} x^{2} + 1} \log \left (a x + \sqrt {a^{2} x^{2} + 1}\right )^{3}}{a^{2}} - \frac {3 \, {\left (x \log \left (a x + \sqrt {a^{2} x^{2} + 1}\right )^{2} + 2 \, a {\left (\frac {x}{a} - \frac {\sqrt {a^{2} x^{2} + 1} \log \left (a x + \sqrt {a^{2} x^{2} + 1}\right )}{a^{2}}\right )}\right )}}{a} \]

[In]

integrate(x*arcsinh(a*x)^3/(a^2*x^2+1)^(1/2),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {x \text {arcsinh}(a x)^3}{\sqrt {1+a^2 x^2}} \, dx=\int \frac {x\,{\mathrm {asinh}\left (a\,x\right )}^3}{\sqrt {a^2\,x^2+1}} \,d x \]

[In]

int((x*asinh(a*x)^3)/(a^2*x^2 + 1)^(1/2),x)

[Out]

int((x*asinh(a*x)^3)/(a^2*x^2 + 1)^(1/2), x)