\(\int \frac {x^2 \text {arcsinh}(a x)^2}{\sqrt {1+a^2 x^2}} \, dx\) [284]

Optimal result

Integrand size = 23, antiderivative size = 87 \[ \int \frac {x^2 \text {arcsinh}(a x)^2}{\sqrt {1+a^2 x^2}} \, dx=\frac {x \sqrt {1+a^2 x^2}}{4 a^2}-\frac {\text {arcsinh}(a x)}{4 a^3}-\frac {x^2 \text {arcsinh}(a x)}{2 a}+\frac {x \sqrt {1+a^2 x^2} \text {arcsinh}(a x)^2}{2 a^2}-\frac {\text {arcsinh}(a x)^3}{6 a^3} \]

[Out]

-1/4*arcsinh(a*x)/a^3-1/2*x^2*arcsinh(a*x)/a-1/6*arcsinh(a*x)^3/a^3+1/4*x*(a^2*x^2+1)^(1/2)/a^2+1/2*x*arcsinh(
a*x)^2*(a^2*x^2+1)^(1/2)/a^2

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {5812, 5783, 5776, 327, 221} \[ \int \frac {x^2 \text {arcsinh}(a x)^2}{\sqrt {1+a^2 x^2}} \, dx=-\frac {\text {arcsinh}(a x)^3}{6 a^3}-\frac {\text {arcsinh}(a x)}{4 a^3}+\frac {x \sqrt {a^2 x^2+1} \text {arcsinh}(a x)^2}{2 a^2}+\frac {x \sqrt {a^2 x^2+1}}{4 a^2}-\frac {x^2 \text {arcsinh}(a x)}{2 a} \]

[In]

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

[Out]

(x*Sqrt[1 + a^2*x^2])/(4*a^2) - ArcSinh[a*x]/(4*a^3) - (x^2*ArcSinh[a*x])/(2*a) + (x*Sqrt[1 + a^2*x^2]*ArcSinh
[a*x]^2)/(2*a^2) - ArcSinh[a*x]^3/(6*a^3)

Rule 221

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt[a])]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

Rule 327

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

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]

Rule 5783

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

Rule 5812

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

Rubi steps \begin{align*} \text {integral}& = \frac {x \sqrt {1+a^2 x^2} \text {arcsinh}(a x)^2}{2 a^2}-\frac {\int \frac {\text {arcsinh}(a x)^2}{\sqrt {1+a^2 x^2}} \, dx}{2 a^2}-\frac {\int x \text {arcsinh}(a x) \, dx}{a} \\ & = -\frac {x^2 \text {arcsinh}(a x)}{2 a}+\frac {x \sqrt {1+a^2 x^2} \text {arcsinh}(a x)^2}{2 a^2}-\frac {\text {arcsinh}(a x)^3}{6 a^3}+\frac {1}{2} \int \frac {x^2}{\sqrt {1+a^2 x^2}} \, dx \\ & = \frac {x \sqrt {1+a^2 x^2}}{4 a^2}-\frac {x^2 \text {arcsinh}(a x)}{2 a}+\frac {x \sqrt {1+a^2 x^2} \text {arcsinh}(a x)^2}{2 a^2}-\frac {\text {arcsinh}(a x)^3}{6 a^3}-\frac {\int \frac {1}{\sqrt {1+a^2 x^2}} \, dx}{4 a^2} \\ & = \frac {x \sqrt {1+a^2 x^2}}{4 a^2}-\frac {\text {arcsinh}(a x)}{4 a^3}-\frac {x^2 \text {arcsinh}(a x)}{2 a}+\frac {x \sqrt {1+a^2 x^2} \text {arcsinh}(a x)^2}{2 a^2}-\frac {\text {arcsinh}(a x)^3}{6 a^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.83 \[ \int \frac {x^2 \text {arcsinh}(a x)^2}{\sqrt {1+a^2 x^2}} \, dx=\frac {3 a x \sqrt {1+a^2 x^2}-3 \left (1+2 a^2 x^2\right ) \text {arcsinh}(a x)+6 a x \sqrt {1+a^2 x^2} \text {arcsinh}(a x)^2-2 \text {arcsinh}(a x)^3}{12 a^3} \]

[In]

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

[Out]

(3*a*x*Sqrt[1 + a^2*x^2] - 3*(1 + 2*a^2*x^2)*ArcSinh[a*x] + 6*a*x*Sqrt[1 + a^2*x^2]*ArcSinh[a*x]^2 - 2*ArcSinh
[a*x]^3)/(12*a^3)

Maple [A] (verified)

Time = 0.19 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.79

method	result	size
default	\(-\frac {-6 \operatorname {arcsinh}\left (a x \right )^{2} \sqrt {a^{2} x^{2}+1}\, a x +6 a^{2} x^{2} \operatorname {arcsinh}\left (a x \right )+2 \operatorname {arcsinh}\left (a x \right )^{3}-3 a x \sqrt {a^{2} x^{2}+1}+3 \,\operatorname {arcsinh}\left (a x \right )}{12 a^{3}}\)	\(69\)

[In]

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

[Out]

-1/12*(-6*arcsinh(a*x)^2*(a^2*x^2+1)^(1/2)*a*x+6*a^2*x^2*arcsinh(a*x)+2*arcsinh(a*x)^3-3*a*x*(a^2*x^2+1)^(1/2)
+3*arcsinh(a*x))/a^3

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.17 \[ \int \frac {x^2 \text {arcsinh}(a x)^2}{\sqrt {1+a^2 x^2}} \, dx=\frac {6 \, \sqrt {a^{2} x^{2} + 1} a x \log \left (a x + \sqrt {a^{2} x^{2} + 1}\right )^{2} - 2 \, \log \left (a x + \sqrt {a^{2} x^{2} + 1}\right )^{3} + 3 \, \sqrt {a^{2} x^{2} + 1} a x - 3 \, {\left (2 \, a^{2} x^{2} + 1\right )} \log \left (a x + \sqrt {a^{2} x^{2} + 1}\right )}{12 \, a^{3}} \]

[In]

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

[Out]

1/12*(6*sqrt(a^2*x^2 + 1)*a*x*log(a*x + sqrt(a^2*x^2 + 1))^2 - 2*log(a*x + sqrt(a^2*x^2 + 1))^3 + 3*sqrt(a^2*x
^2 + 1)*a*x - 3*(2*a^2*x^2 + 1)*log(a*x + sqrt(a^2*x^2 + 1)))/a^3

Sympy [A] (verification not implemented)

Time = 0.39 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.90 \[ \int \frac {x^2 \text {arcsinh}(a x)^2}{\sqrt {1+a^2 x^2}} \, dx=\begin {cases} - \frac {x^{2} \operatorname {asinh}{\left (a x \right )}}{2 a} + \frac {x \sqrt {a^{2} x^{2} + 1} \operatorname {asinh}^{2}{\left (a x \right )}}{2 a^{2}} + \frac {x \sqrt {a^{2} x^{2} + 1}}{4 a^{2}} - \frac {\operatorname {asinh}^{3}{\left (a x \right )}}{6 a^{3}} - \frac {\operatorname {asinh}{\left (a x \right )}}{4 a^{3}} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((-x**2*asinh(a*x)/(2*a) + x*sqrt(a**2*x**2 + 1)*asinh(a*x)**2/(2*a**2) + x*sqrt(a**2*x**2 + 1)/(4*a*
*2) - asinh(a*x)**3/(6*a**3) - asinh(a*x)/(4*a**3), Ne(a, 0)), (0, True))

Maxima [F]

\[ \int \frac {x^2 \text {arcsinh}(a x)^2}{\sqrt {1+a^2 x^2}} \, dx=\int { \frac {x^{2} \operatorname {arsinh}\left (a x\right )^{2}}{\sqrt {a^{2} x^{2} + 1}} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {x^2 \text {arcsinh}(a x)^2}{\sqrt {1+a^2 x^2}} \, dx=\int { \frac {x^{2} \operatorname {arsinh}\left (a x\right )^{2}}{\sqrt {a^{2} x^{2} + 1}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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