### 3.39 $$\int e^{\text{csch}^{-1}(a x^2)} x^3 \, dx$$

Optimal. Leaf size=52 $\frac{1}{4} x^4 \sqrt{\frac{1}{a^2 x^4}+1}+\frac{\tanh ^{-1}\left (\sqrt{\frac{1}{a^2 x^4}+1}\right )}{4 a^2}+\frac{x^2}{2 a}$

[Out]

x^2/(2*a) + (Sqrt[1 + 1/(a^2*x^4)]*x^4)/4 + ArcTanh[Sqrt[1 + 1/(a^2*x^4)]]/(4*a^2)

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Rubi [A]  time = 0.0383185, antiderivative size = 52, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 12, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.5, Rules used = {6336, 30, 266, 47, 63, 208} $\frac{1}{4} x^4 \sqrt{\frac{1}{a^2 x^4}+1}+\frac{\tanh ^{-1}\left (\sqrt{\frac{1}{a^2 x^4}+1}\right )}{4 a^2}+\frac{x^2}{2 a}$

Antiderivative was successfully veriﬁed.

[In]

Int[E^ArcCsch[a*x^2]*x^3,x]

[Out]

x^2/(2*a) + (Sqrt[1 + 1/(a^2*x^4)]*x^4)/4 + ArcTanh[Sqrt[1 + 1/(a^2*x^4)]]/(4*a^2)

Rule 6336

Int[E^ArcCsch[(a_.)*(x_)^(p_.)]*(x_)^(m_.), x_Symbol] :> Dist[1/a, Int[x^(m - p), x], x] + Int[x^m*Sqrt[1 + 1/
(a^2*x^(2*p))], x] /; FreeQ[{a, m, p}, x]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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 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]

Rubi steps

\begin{align*} \int e^{\text{csch}^{-1}\left (a x^2\right )} x^3 \, dx &=\frac{\int x \, dx}{a}+\int \sqrt{1+\frac{1}{a^2 x^4}} x^3 \, dx\\ &=\frac{x^2}{2 a}-\frac{1}{4} \operatorname{Subst}\left (\int \frac{\sqrt{1+\frac{x}{a^2}}}{x^2} \, dx,x,\frac{1}{x^4}\right )\\ &=\frac{x^2}{2 a}+\frac{1}{4} \sqrt{1+\frac{1}{a^2 x^4}} x^4-\frac{\operatorname{Subst}\left (\int \frac{1}{x \sqrt{1+\frac{x}{a^2}}} \, dx,x,\frac{1}{x^4}\right )}{8 a^2}\\ &=\frac{x^2}{2 a}+\frac{1}{4} \sqrt{1+\frac{1}{a^2 x^4}} x^4-\frac{1}{4} \operatorname{Subst}\left (\int \frac{1}{-a^2+a^2 x^2} \, dx,x,\sqrt{1+\frac{1}{a^2 x^4}}\right )\\ &=\frac{x^2}{2 a}+\frac{1}{4} \sqrt{1+\frac{1}{a^2 x^4}} x^4+\frac{\tanh ^{-1}\left (\sqrt{1+\frac{1}{a^2 x^4}}\right )}{4 a^2}\\ \end{align*}

Mathematica [A]  time = 0.0535126, size = 53, normalized size = 1.02 $\frac{a x^2 \left (a x^2 \sqrt{\frac{1}{a^2 x^4}+1}+2\right )+\log \left (x^2 \left (\sqrt{\frac{1}{a^2 x^4}+1}+1\right )\right )}{4 a^2}$

Warning: Unable to verify antiderivative.

[In]

Integrate[E^ArcCsch[a*x^2]*x^3,x]

[Out]

(a*x^2*(2 + a*Sqrt[1 + 1/(a^2*x^4)]*x^2) + Log[(1 + Sqrt[1 + 1/(a^2*x^4)])*x^2])/(4*a^2)

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Maple [B]  time = 0.29, size = 94, normalized size = 1.8 \begin{align*}{\frac{{x}^{2}}{4\,{a}^{2}}\sqrt{{\frac{{a}^{2}{x}^{4}+1}{{a}^{2}{x}^{4}}}} \left ({x}^{2}\sqrt{{\frac{{a}^{2}{x}^{4}+1}{{a}^{2}}}}{a}^{2}+\ln \left ({x}^{2}+\sqrt{{\frac{{a}^{2}{x}^{4}+1}{{a}^{2}}}} \right ) \right ){\frac{1}{\sqrt{{\frac{{a}^{2}{x}^{4}+1}{{a}^{2}}}}}}}+{\frac{{x}^{2}}{2\,a}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((1/a/x^2+(1+1/a^2/x^4)^(1/2))*x^3,x)

[Out]

1/4*((a^2*x^4+1)/a^2/x^4)^(1/2)*x^2*(x^2*((a^2*x^4+1)/a^2)^(1/2)*a^2+ln(x^2+((a^2*x^4+1)/a^2)^(1/2)))/((a^2*x^
4+1)/a^2)^(1/2)/a^2+1/2*x^2/a

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Maxima [A]  time = 1.03212, size = 109, normalized size = 2.1 \begin{align*} \frac{x^{2}}{2 \, a} + \frac{\sqrt{\frac{1}{a^{2} x^{4}} + 1}}{4 \,{\left (a^{2}{\left (\frac{1}{a^{2} x^{4}} + 1\right )} - a^{2}\right )}} + \frac{\log \left (\sqrt{\frac{1}{a^{2} x^{4}} + 1} + 1\right )}{8 \, a^{2}} - \frac{\log \left (\sqrt{\frac{1}{a^{2} x^{4}} + 1} - 1\right )}{8 \, a^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/a/x^2+(1+1/a^2/x^4)^(1/2))*x^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.59534, size = 149, normalized size = 2.87 \begin{align*} \frac{a^{2} x^{4} \sqrt{\frac{a^{2} x^{4} + 1}{a^{2} x^{4}}} + 2 \, a x^{2} - \log \left (a x^{2} \sqrt{\frac{a^{2} x^{4} + 1}{a^{2} x^{4}}} - a x^{2}\right )}{4 \, a^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/a/x^2+(1+1/a^2/x^4)^(1/2))*x^3,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 4.29354, size = 36, normalized size = 0.69 \begin{align*} \frac{x^{2} \sqrt{a^{2} x^{4} + 1}}{4 a} + \frac{x^{2}}{2 a} + \frac{\operatorname{asinh}{\left (a x^{2} \right )}}{4 a^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/a/x**2+(1+1/a**2/x**4)**(1/2))*x**3,x)

[Out]

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

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Giac [A]  time = 1.13288, size = 74, normalized size = 1.42 \begin{align*} \frac{\sqrt{a^{2} x^{4} + 1} x^{2}{\left | a \right |}}{4 \, a^{2}} + \frac{x^{2}}{2 \, a} - \frac{\log \left (-x^{2}{\left | a \right |} + \sqrt{a^{2} x^{4} + 1}\right )}{4 \, a^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/a/x^2+(1+1/a^2/x^4)^(1/2))*x^3,x, algorithm="giac")

[Out]

1/4*sqrt(a^2*x^4 + 1)*x^2*abs(a)/a^2 + 1/2*x^2/a - 1/4*log(-x^2*abs(a) + sqrt(a^2*x^4 + 1))/a^2