### 3.28 $$\int e^{\text{csch}^{-1}(a x)} x^3 \, dx$$

Optimal. Leaf size=75 $\frac{1}{4} x^4 \sqrt{\frac{1}{a^2 x^2}+1}+\frac{x^2 \sqrt{\frac{1}{a^2 x^2}+1}}{8 a^2}-\frac{\tanh ^{-1}\left (\sqrt{\frac{1}{a^2 x^2}+1}\right )}{8 a^4}+\frac{x^3}{3 a}$

[Out]

(Sqrt[1 + 1/(a^2*x^2)]*x^2)/(8*a^2) + x^3/(3*a) + (Sqrt[1 + 1/(a^2*x^2)]*x^4)/4 - ArcTanh[Sqrt[1 + 1/(a^2*x^2)
]]/(8*a^4)

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Rubi [A]  time = 0.0440765, antiderivative size = 75, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 10, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.7, Rules used = {6336, 30, 266, 47, 51, 63, 208} $\frac{1}{4} x^4 \sqrt{\frac{1}{a^2 x^2}+1}+\frac{x^2 \sqrt{\frac{1}{a^2 x^2}+1}}{8 a^2}-\frac{\tanh ^{-1}\left (\sqrt{\frac{1}{a^2 x^2}+1}\right )}{8 a^4}+\frac{x^3}{3 a}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(Sqrt[1 + 1/(a^2*x^2)]*x^2)/(8*a^2) + x^3/(3*a) + (Sqrt[1 + 1/(a^2*x^2)]*x^4)/4 - ArcTanh[Sqrt[1 + 1/(a^2*x^2)
]]/(8*a^4)

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 51

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] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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}(a x)} x^3 \, dx &=\frac{\int x^2 \, dx}{a}+\int \sqrt{1+\frac{1}{a^2 x^2}} x^3 \, dx\\ &=\frac{x^3}{3 a}-\frac{1}{2} \operatorname{Subst}\left (\int \frac{\sqrt{1+\frac{x}{a^2}}}{x^3} \, dx,x,\frac{1}{x^2}\right )\\ &=\frac{x^3}{3 a}+\frac{1}{4} \sqrt{1+\frac{1}{a^2 x^2}} x^4-\frac{\operatorname{Subst}\left (\int \frac{1}{x^2 \sqrt{1+\frac{x}{a^2}}} \, dx,x,\frac{1}{x^2}\right )}{8 a^2}\\ &=\frac{\sqrt{1+\frac{1}{a^2 x^2}} x^2}{8 a^2}+\frac{x^3}{3 a}+\frac{1}{4} \sqrt{1+\frac{1}{a^2 x^2}} x^4+\frac{\operatorname{Subst}\left (\int \frac{1}{x \sqrt{1+\frac{x}{a^2}}} \, dx,x,\frac{1}{x^2}\right )}{16 a^4}\\ &=\frac{\sqrt{1+\frac{1}{a^2 x^2}} x^2}{8 a^2}+\frac{x^3}{3 a}+\frac{1}{4} \sqrt{1+\frac{1}{a^2 x^2}} x^4+\frac{\operatorname{Subst}\left (\int \frac{1}{-a^2+a^2 x^2} \, dx,x,\sqrt{1+\frac{1}{a^2 x^2}}\right )}{8 a^2}\\ &=\frac{\sqrt{1+\frac{1}{a^2 x^2}} x^2}{8 a^2}+\frac{x^3}{3 a}+\frac{1}{4} \sqrt{1+\frac{1}{a^2 x^2}} x^4-\frac{\tanh ^{-1}\left (\sqrt{1+\frac{1}{a^2 x^2}}\right )}{8 a^4}\\ \end{align*}

Mathematica [A]  time = 0.0552299, size = 76, normalized size = 1.01 $\frac{a^2 x^2 \left (6 a^2 x^2 \sqrt{\frac{1}{a^2 x^2}+1}+3 \sqrt{\frac{1}{a^2 x^2}+1}+8 a x\right )-3 \log \left (x \left (\sqrt{\frac{1}{a^2 x^2}+1}+1\right )\right )}{24 a^4}$

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(a^2*x^2*(3*Sqrt[1 + 1/(a^2*x^2)] + 8*a*x + 6*a^2*Sqrt[1 + 1/(a^2*x^2)]*x^2) - 3*Log[(1 + Sqrt[1 + 1/(a^2*x^2)
])*x])/(24*a^4)

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Maple [A]  time = 0.177, size = 109, normalized size = 1.5 \begin{align*} -{\frac{x}{8\,{a}^{4}}\sqrt{{\frac{{a}^{2}{x}^{2}+1}{{a}^{2}{x}^{2}}}} \left ( -2\,x \left ({\frac{{a}^{2}{x}^{2}+1}{{a}^{2}}} \right ) ^{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 ){\frac{1}{\sqrt{{\frac{{a}^{2}{x}^{2}+1}{{a}^{2}}}}}}}+{\frac{{x}^{3}}{3\,a}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*((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)/a^4+1/3*x^3/a

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Maxima [A]  time = 0.973333, size = 144, normalized size = 1.92 \begin{align*} \frac{x^{3}}{3 \, a} + \frac{{\left (\frac{1}{a^{2} x^{2}} + 1\right )}^{\frac{3}{2}} + \sqrt{\frac{1}{a^{2} x^{2}} + 1}}{8 \,{\left (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}\right )}} - \frac{\log \left (\sqrt{\frac{1}{a^{2} x^{2}} + 1} + 1\right )}{16 \, a^{4}} + \frac{\log \left (\sqrt{\frac{1}{a^{2} x^{2}} + 1} - 1\right )}{16 \, a^{4}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x^3/a + 1/8*((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) - 1/16*log(sqrt(1/(a^2*x^2) + 1) + 1)/a^4 + 1/16*log(sqrt(1/(a^2*x^2) + 1) - 1)/a^4

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Fricas [A]  time = 2.64219, size = 171, normalized size = 2.28 \begin{align*} \frac{8 \, a^{3} x^{3} + 3 \,{\left (2 \, a^{4} x^{4} + a^{2} x^{2}\right )} \sqrt{\frac{a^{2} x^{2} + 1}{a^{2} x^{2}}} + 3 \, \log \left (a x \sqrt{\frac{a^{2} x^{2} + 1}{a^{2} x^{2}}} - a x\right )}{24 \, a^{4}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(8*a^3*x^3 + 3*(2*a^4*x^4 + a^2*x^2)*sqrt((a^2*x^2 + 1)/(a^2*x^2)) + 3*log(a*x*sqrt((a^2*x^2 + 1)/(a^2*x^
2)) - a*x))/a^4

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Sympy [A]  time = 5.50681, size = 73, normalized size = 0.97 \begin{align*} \frac{a x^{5}}{4 \sqrt{a^{2} x^{2} + 1}} + \frac{x^{3}}{3 a} + \frac{3 x^{3}}{8 a \sqrt{a^{2} x^{2} + 1}} + \frac{x}{8 a^{3} \sqrt{a^{2} x^{2} + 1}} - \frac{\operatorname{asinh}{\left (a x \right )}}{8 a^{4}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.16368, size = 93, normalized size = 1.24 \begin{align*} \frac{1}{8} \, \sqrt{a^{2} x^{2} + 1}{\left (\frac{2 \, x^{2}{\left | a \right |} \mathrm{sgn}\left (x\right )}{a^{2}} + \frac{{\left | a \right |} \mathrm{sgn}\left (x\right )}{a^{4}}\right )} x + \frac{x^{3}}{3 \, a} + \frac{\log \left (-x{\left | a \right |} + \sqrt{a^{2} x^{2} + 1}\right ) \mathrm{sgn}\left (x\right )}{8 \, a^{4}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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