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

Optimal. Leaf size=52 \[ \frac {x^2 \sqrt {\frac {1}{a^2 x^2}+1}}{a}+\frac {2 x}{a^2}+\frac {\tanh ^{-1}\left (\sqrt {\frac {1}{a^2 x^2}+1}\right )}{a^3}+\frac {x^3}{3} \]

[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

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Rubi [A]  time = 0.24, antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6338, 6742, 266, 47, 63, 208} \[ \frac {x^2 \sqrt {\frac {1}{a^2 x^2}+1}}{a}+\frac {\tanh ^{-1}\left (\sqrt {\frac {1}{a^2 x^2}+1}\right )}{a^3}+\frac {2 x}{a^2}+\frac {x^3}{3} \]

Antiderivative was successfully verified.

[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 47

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},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

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 208

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

Rule 266

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 6338

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 6742

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

Rubi steps

\begin {align*} \int e^{2 \text {csch}^{-1}(a x)} x^2 \, dx &=\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 {\operatorname {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 {\operatorname {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 {\operatorname {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 {\tanh ^{-1}\left (\sqrt {1+\frac {1}{a^2 x^2}}\right )}{a^3}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 57, normalized size = 1.10 \[ \frac {a x \left (a^2 x^2+3 a x \sqrt {\frac {1}{a^2 x^2}+1}+6\right )+3 \log \left (x \left (\sqrt {\frac {1}{a^2 x^2}+1}+1\right )\right )}{3 a^3} \]

Warning: Unable to verify antiderivative.

[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)

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fricas [A]  time = 0.77, size = 72, normalized size = 1.38 \[ \frac {a^{3} x^{3} + 3 \, a^{2} x^{2} \sqrt {\frac {a^{2} x^{2} + 1}{a^{2} x^{2}}} + 6 \, a x - 3 \, \log \left (a x \sqrt {\frac {a^{2} x^{2} + 1}{a^{2} x^{2}}} - a x\right )}{3 \, a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(t_nostep)]Error: Bad Argument Type

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maple [A]  time = 0.05, size = 90, normalized size = 1.73 \[ \frac {x^{3}}{3}+\frac {2 x}{a^{2}}+\frac {\sqrt {\frac {a^{2} x^{2}+1}{a^{2} x^{2}}}\, x \left (x \sqrt {\frac {a^{2} x^{2}+1}{a^{2}}}\, a^{2}+\ln \left (x +\sqrt {\frac {a^{2} x^{2}+1}{a^{2}}}\right )\right )}{a^{3} \sqrt {\frac {a^{2} x^{2}+1}{a^{2}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.32, size = 89, normalized size = 1.71 \[ \frac {1}{3} \, x^{3} + \frac {\frac {2 \, \sqrt {\frac {1}{a^{2} x^{2}} + 1}}{a^{2} {\left (\frac {1}{a^{2} x^{2}} + 1\right )} - a^{2}} + \frac {\log \left (\sqrt {\frac {1}{a^{2} x^{2}} + 1} + 1\right )}{a^{2}} - \frac {\log \left (\sqrt {\frac {1}{a^{2} x^{2}} + 1} - 1\right )}{a^{2}}}{2 \, a} + \frac {2 \, x}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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mupad [B]  time = 2.13, size = 51, normalized size = 0.98 \[ \frac {2\,x}{a^2}+\frac {x^3}{3}+\frac {x^2\,\sqrt {\frac {1}{a^2\,x^2}+1}}{a}-\frac {\mathrm {atan}\left (\sqrt {\frac {1}{a^2\,x^2}+1}\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{a^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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sympy [A]  time = 3.33, size = 36, normalized size = 0.69 \[ \frac {x^{3}}{3} + \frac {x \sqrt {a^{2} x^{2} + 1}}{a^{2}} + \frac {2 x}{a^{2}} + \frac {\operatorname {asinh}{\left (a x \right )}}{a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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