\(\int e^{2 \text {csch}^{-1}(a x)} x^2 \, dx\) [51]

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

Optimal result

Integrand size = 12, antiderivative size = 52 \[ \int e^{2 \text {csch}^{-1}(a x)} x^2 \, dx=\frac {2 x}{a^2}+\frac {\sqrt {1+\frac {1}{a^2 x^2}} x^2}{a}+\frac {x^3}{3}+\frac {\text {arctanh}\left (\sqrt {1+\frac {1}{a^2 x^2}}\right )}{a^3} \]

[Out]

2*x/a^2+1/3*x^3+arctanh((1+1/a^2/x^2)^(1/2))/a^3+x^2*(1+1/a^2/x^2)^(1/2)/a

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6473, 6874, 272, 43, 65, 214} \[ \int e^{2 \text {csch}^{-1}(a x)} x^2 \, dx=\frac {x^2 \sqrt {\frac {1}{a^2 x^2}+1}}{a}+\frac {2 x}{a^2}+\frac {\text {arctanh}\left (\sqrt {\frac {1}{a^2 x^2}+1}\right )}{a^3}+\frac {x^3}{3} \]

[In]

Int[E^(2*ArcCsch[a*x])*x^2,x]

[Out]

(2*x)/a^2 + (Sqrt[1 + 1/(a^2*x^2)]*x^2)/a + x^3/3 + ArcTanh[Sqrt[1 + 1/(a^2*x^2)]]/a^3

Rule 43

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

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 6473

Int[E^(ArcCsch[u_]*(n_.))*(x_)^(m_.), x_Symbol] :> Int[x^m*(1/u + Sqrt[1 + 1/u^2])^n, x] /; FreeQ[m, x] && Int
egerQ[n]

Rule 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.10 \[ \int e^{2 \text {csch}^{-1}(a x)} x^2 \, dx=\frac {a x \left (6+3 a \sqrt {1+\frac {1}{a^2 x^2}} x+a^2 x^2\right )+3 \log \left (\left (1+\sqrt {1+\frac {1}{a^2 x^2}}\right ) x\right )}{3 a^3} \]

[In]

Integrate[E^(2*ArcCsch[a*x])*x^2,x]

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(97\) vs. \(2(46)=92\).

Time = 0.05 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.88

method result size
default \(\frac {\frac {1}{3} a^{2} x^{3}+x}{a^{2}}+\frac {\sqrt {\frac {a^{2} x^{2}+1}{a^{2} x^{2}}}\, x \left (x \sqrt {\frac {a^{2} x^{2}+1}{a^{2}}}\, a^{2}+\ln \left (x +\sqrt {\frac {a^{2} x^{2}+1}{a^{2}}}\right )\right )}{a^{3} \sqrt {\frac {a^{2} x^{2}+1}{a^{2}}}}+\frac {x}{a^{2}}\) \(98\)

[In]

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

[Out]

1/a^2*(1/3*a^2*x^3+x)+1/a^3*((a^2*x^2+1)/a^2/x^2)^(1/2)*x*(x*((a^2*x^2+1)/a^2)^(1/2)*a^2+ln(x+((a^2*x^2+1)/a^2
)^(1/2)))/((a^2*x^2+1)/a^2)^(1/2)+x/a^2

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.38 \[ \int e^{2 \text {csch}^{-1}(a x)} x^2 \, dx=\frac {a^{3} x^{3} + 3 \, a^{2} x^{2} \sqrt {\frac {a^{2} x^{2} + 1}{a^{2} x^{2}}} + 6 \, a x - 3 \, \log \left (a x \sqrt {\frac {a^{2} x^{2} + 1}{a^{2} x^{2}}} - a x\right )}{3 \, a^{3}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 1.61 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.69 \[ \int e^{2 \text {csch}^{-1}(a x)} x^2 \, dx=\frac {x^{3}}{3} + \frac {x \sqrt {a^{2} x^{2} + 1}}{a^{2}} + \frac {2 x}{a^{2}} + \frac {\operatorname {asinh}{\left (a x \right )}}{a^{3}} \]

[In]

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

[Out]

x**3/3 + x*sqrt(a**2*x**2 + 1)/a**2 + 2*x/a**2 + asinh(a*x)/a**3

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.71 \[ \int e^{2 \text {csch}^{-1}(a x)} x^2 \, dx=\frac {1}{3} \, x^{3} + \frac {\frac {2 \, \sqrt {\frac {1}{a^{2} x^{2}} + 1}}{a^{2} {\left (\frac {1}{a^{2} x^{2}} + 1\right )} - a^{2}} + \frac {\log \left (\sqrt {\frac {1}{a^{2} x^{2}} + 1} + 1\right )}{a^{2}} - \frac {\log \left (\sqrt {\frac {1}{a^{2} x^{2}} + 1} - 1\right )}{a^{2}}}{2 \, a} + \frac {2 \, x}{a^{2}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.19 \[ \int e^{2 \text {csch}^{-1}(a x)} x^2 \, dx=\frac {\sqrt {a^{2} x^{2} + 1} x {\left | a \right |} \mathrm {sgn}\left (x\right )}{a^{3}} + \frac {a^{2} x^{3} + 6 \, x}{3 \, a^{2}} - \frac {\log \left (-x {\left | a \right |} + \sqrt {a^{2} x^{2} + 1}\right ) \mathrm {sgn}\left (x\right )}{a^{3}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.06 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.98 \[ \int e^{2 \text {csch}^{-1}(a x)} x^2 \, dx=\frac {2\,x}{a^2}+\frac {x^3}{3}+\frac {x^2\,\sqrt {\frac {1}{a^2\,x^2}+1}}{a}-\frac {\mathrm {atan}\left (\sqrt {\frac {1}{a^2\,x^2}+1}\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{a^3} \]

[In]

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

[Out]

(2*x)/a^2 - (atan((1/(a^2*x^2) + 1)^(1/2)*1i)*1i)/a^3 + x^3/3 + (x^2*(1/(a^2*x^2) + 1)^(1/2))/a