\(\int e^{2 \text {csch}^{-1}(a x)} x^4 \, dx\) [49]

   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 [B] (verification not implemented)

Optimal result

Integrand size = 12, antiderivative size = 85 \[ \int e^{2 \text {csch}^{-1}(a x)} x^4 \, dx=\frac {\sqrt {1+\frac {1}{a^2 x^2}} x^2}{4 a^3}+\frac {2 x^3}{3 a^2}+\frac {\sqrt {1+\frac {1}{a^2 x^2}} x^4}{2 a}+\frac {x^5}{5}-\frac {\text {arctanh}\left (\sqrt {1+\frac {1}{a^2 x^2}}\right )}{4 a^5} \]

[Out]

2/3*x^3/a^2+1/5*x^5-1/4*arctanh((1+1/a^2/x^2)^(1/2))/a^5+1/4*x^2*(1+1/a^2/x^2)^(1/2)/a^3+1/2*x^4*(1+1/a^2/x^2)
^(1/2)/a

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.583, Rules used = {6473, 6874, 272, 43, 44, 65, 214} \[ \int e^{2 \text {csch}^{-1}(a x)} x^4 \, dx=\frac {2 x^3}{3 a^2}+\frac {x^4 \sqrt {\frac {1}{a^2 x^2}+1}}{2 a}-\frac {\text {arctanh}\left (\sqrt {\frac {1}{a^2 x^2}+1}\right )}{4 a^5}+\frac {x^2 \sqrt {\frac {1}{a^2 x^2}+1}}{4 a^3}+\frac {x^5}{5} \]

[In]

Int[E^(2*ArcCsch[a*x])*x^4,x]

[Out]

(Sqrt[1 + 1/(a^2*x^2)]*x^2)/(4*a^3) + (2*x^3)/(3*a^2) + (Sqrt[1 + 1/(a^2*x^2)]*x^4)/(2*a) + x^5/5 - ArcTanh[Sq
rt[1 + 1/(a^2*x^2)]]/(4*a^5)

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 44

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[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^4 \, dx \\ & = \int \left (\frac {2 x^2}{a^2}+\frac {2 \sqrt {1+\frac {1}{a^2 x^2}} x^3}{a}+x^4\right ) \, dx \\ & = \frac {2 x^3}{3 a^2}+\frac {x^5}{5}+\frac {2 \int \sqrt {1+\frac {1}{a^2 x^2}} x^3 \, dx}{a} \\ & = \frac {2 x^3}{3 a^2}+\frac {x^5}{5}-\frac {\text {Subst}\left (\int \frac {\sqrt {1+\frac {x}{a^2}}}{x^3} \, dx,x,\frac {1}{x^2}\right )}{a} \\ & = \frac {2 x^3}{3 a^2}+\frac {\sqrt {1+\frac {1}{a^2 x^2}} x^4}{2 a}+\frac {x^5}{5}-\frac {\text {Subst}\left (\int \frac {1}{x^2 \sqrt {1+\frac {x}{a^2}}} \, dx,x,\frac {1}{x^2}\right )}{4 a^3} \\ & = \frac {\sqrt {1+\frac {1}{a^2 x^2}} x^2}{4 a^3}+\frac {2 x^3}{3 a^2}+\frac {\sqrt {1+\frac {1}{a^2 x^2}} x^4}{2 a}+\frac {x^5}{5}+\frac {\text {Subst}\left (\int \frac {1}{x \sqrt {1+\frac {x}{a^2}}} \, dx,x,\frac {1}{x^2}\right )}{8 a^5} \\ & = \frac {\sqrt {1+\frac {1}{a^2 x^2}} x^2}{4 a^3}+\frac {2 x^3}{3 a^2}+\frac {\sqrt {1+\frac {1}{a^2 x^2}} x^4}{2 a}+\frac {x^5}{5}+\frac {\text {Subst}\left (\int \frac {1}{-a^2+a^2 x^2} \, dx,x,\sqrt {1+\frac {1}{a^2 x^2}}\right )}{4 a^3} \\ & = \frac {\sqrt {1+\frac {1}{a^2 x^2}} x^2}{4 a^3}+\frac {2 x^3}{3 a^2}+\frac {\sqrt {1+\frac {1}{a^2 x^2}} x^4}{2 a}+\frac {x^5}{5}-\frac {\text {arctanh}\left (\sqrt {1+\frac {1}{a^2 x^2}}\right )}{4 a^5} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.99 \[ \int e^{2 \text {csch}^{-1}(a x)} x^4 \, dx=\frac {a^2 x^2 \left (15 \sqrt {1+\frac {1}{a^2 x^2}}+40 a x+30 a^2 \sqrt {1+\frac {1}{a^2 x^2}} x^2+12 a^3 x^3\right )-15 \log \left (\left (1+\sqrt {1+\frac {1}{a^2 x^2}}\right ) x\right )}{60 a^5} \]

[In]

Integrate[E^(2*ArcCsch[a*x])*x^4,x]

[Out]

(a^2*x^2*(15*Sqrt[1 + 1/(a^2*x^2)] + 40*a*x + 30*a^2*Sqrt[1 + 1/(a^2*x^2)]*x^2 + 12*a^3*x^3) - 15*Log[(1 + Sqr
t[1 + 1/(a^2*x^2)])*x])/(60*a^5)

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.49

method result size
default \(\frac {\frac {1}{5} a^{2} x^{5}+\frac {1}{3} x^{3}}{a^{2}}-\frac {\sqrt {\frac {a^{2} x^{2}+1}{a^{2} x^{2}}}\, x \left (-2 x \left (\frac {a^{2} x^{2}+1}{a^{2}}\right )^{\frac {3}{2}} a^{4}+x \sqrt {\frac {a^{2} x^{2}+1}{a^{2}}}\, a^{2}+\ln \left (x +\sqrt {\frac {a^{2} x^{2}+1}{a^{2}}}\right )\right )}{4 a^{5} \sqrt {\frac {a^{2} x^{2}+1}{a^{2}}}}+\frac {x^{3}}{3 a^{2}}\) \(127\)

[In]

int((1/a/x+(1+1/a^2/x^2)^(1/2))^2*x^4,x,method=_RETURNVERBOSE)

[Out]

1/a^2*(1/5*a^2*x^5+1/3*x^3)-1/4/a^5*((a^2*x^2+1)/a^2/x^2)^(1/2)*x*(-2*x*((a^2*x^2+1)/a^2)^(3/2)*a^4+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)+1/3*x^3/a^2

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.02 \[ \int e^{2 \text {csch}^{-1}(a x)} x^4 \, dx=\frac {12 \, a^{5} x^{5} + 40 \, a^{3} x^{3} + 15 \, {\left (2 \, a^{4} x^{4} + a^{2} x^{2}\right )} \sqrt {\frac {a^{2} x^{2} + 1}{a^{2} x^{2}}} + 15 \, \log \left (a x \sqrt {\frac {a^{2} x^{2} + 1}{a^{2} x^{2}}} - a x\right )}{60 \, a^{5}} \]

[In]

integrate((1/a/x+(1+1/a^2/x^2)^(1/2))^2*x^4,x, algorithm="fricas")

[Out]

1/60*(12*a^5*x^5 + 40*a^3*x^3 + 15*(2*a^4*x^4 + a^2*x^2)*sqrt((a^2*x^2 + 1)/(a^2*x^2)) + 15*log(a*x*sqrt((a^2*
x^2 + 1)/(a^2*x^2)) - a*x))/a^5

Sympy [A] (verification not implemented)

Time = 2.61 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.96 \[ \int e^{2 \text {csch}^{-1}(a x)} x^4 \, dx=\frac {x^{5}}{5} + \frac {x^{5}}{2 \sqrt {a^{2} x^{2} + 1}} + \frac {2 x^{3}}{3 a^{2}} + \frac {3 x^{3}}{4 a^{2} \sqrt {a^{2} x^{2} + 1}} + \frac {x}{4 a^{4} \sqrt {a^{2} x^{2} + 1}} - \frac {\operatorname {asinh}{\left (a x \right )}}{4 a^{5}} \]

[In]

integrate((1/a/x+(1+1/a**2/x**2)**(1/2))**2*x**4,x)

[Out]

x**5/5 + x**5/(2*sqrt(a**2*x**2 + 1)) + 2*x**3/(3*a**2) + 3*x**3/(4*a**2*sqrt(a**2*x**2 + 1)) + x/(4*a**4*sqrt
(a**2*x**2 + 1)) - asinh(a*x)/(4*a**5)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.38 \[ \int e^{2 \text {csch}^{-1}(a x)} x^4 \, dx=\frac {1}{5} \, x^{5} + \frac {2 \, x^{3}}{3 \, a^{2}} + \frac {\frac {2 \, {\left ({\left (\frac {1}{a^{2} x^{2}} + 1\right )}^{\frac {3}{2}} + \sqrt {\frac {1}{a^{2} x^{2}} + 1}\right )}}{a^{4} {\left (\frac {1}{a^{2} x^{2}} + 1\right )}^{2} - 2 \, a^{4} {\left (\frac {1}{a^{2} x^{2}} + 1\right )} + a^{4}} - \frac {\log \left (\sqrt {\frac {1}{a^{2} x^{2}} + 1} + 1\right )}{a^{4}} + \frac {\log \left (\sqrt {\frac {1}{a^{2} x^{2}} + 1} - 1\right )}{a^{4}}}{8 \, a} \]

[In]

integrate((1/a/x+(1+1/a^2/x^2)^(1/2))^2*x^4,x, algorithm="maxima")

[Out]

1/5*x^5 + 2/3*x^3/a^2 + 1/8*(2*((1/(a^2*x^2) + 1)^(3/2) + sqrt(1/(a^2*x^2) + 1))/(a^4*(1/(a^2*x^2) + 1)^2 - 2*
a^4*(1/(a^2*x^2) + 1) + a^4) - log(sqrt(1/(a^2*x^2) + 1) + 1)/a^4 + log(sqrt(1/(a^2*x^2) + 1) - 1)/a^4)/a

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.94 \[ \int e^{2 \text {csch}^{-1}(a x)} x^4 \, dx=\frac {1}{4} \, \sqrt {a^{2} x^{2} + 1} x {\left (\frac {2 \, x^{2} {\left | a \right |} \mathrm {sgn}\left (x\right )}{a^{3}} + \frac {{\left | a \right |} \mathrm {sgn}\left (x\right )}{a^{5}}\right )} + \frac {3 \, a^{2} x^{5} + 10 \, x^{3}}{15 \, a^{2}} + \frac {\log \left (-x {\left | a \right |} + \sqrt {a^{2} x^{2} + 1}\right ) \mathrm {sgn}\left (x\right )}{4 \, a^{5}} \]

[In]

integrate((1/a/x+(1+1/a^2/x^2)^(1/2))^2*x^4,x, algorithm="giac")

[Out]

1/4*sqrt(a^2*x^2 + 1)*x*(2*x^2*abs(a)*sgn(x)/a^3 + abs(a)*sgn(x)/a^5) + 1/15*(3*a^2*x^5 + 10*x^3)/a^2 + 1/4*lo
g(-x*abs(a) + sqrt(a^2*x^2 + 1))*sgn(x)/a^5

Mupad [B] (verification not implemented)

Time = 4.84 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.86 \[ \int e^{2 \text {csch}^{-1}(a x)} x^4 \, dx=\frac {x^5}{5}+\frac {2\,x^3}{3\,a^2}+\frac {x^4\,\sqrt {\frac {1}{a^2\,x^2}+1}}{2\,a}+\frac {x^2\,\sqrt {\frac {1}{a^2\,x^2}+1}}{4\,a^3}+\frac {\mathrm {atan}\left (\sqrt {\frac {1}{a^2\,x^2}+1}\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{4\,a^5} \]

[In]

int(x^4*((1/(a^2*x^2) + 1)^(1/2) + 1/(a*x))^2,x)

[Out]

(atan((1/(a^2*x^2) + 1)^(1/2)*1i)*1i)/(4*a^5) + x^5/5 + (2*x^3)/(3*a^2) + (x^4*(1/(a^2*x^2) + 1)^(1/2))/(2*a)
+ (x^2*(1/(a^2*x^2) + 1)^(1/2))/(4*a^3)