3.27 $$\int e^{\text{csch}^{-1}(a x)} x^4 \, dx$$

Optimal. Leaf size=54 $\frac{1}{5} x^5 \left (\frac{1}{a^2 x^2}+1\right )^{3/2}-\frac{2 x^3 \left (\frac{1}{a^2 x^2}+1\right )^{3/2}}{15 a^2}+\frac{x^4}{4 a}$

[Out]

(-2*(1 + 1/(a^2*x^2))^(3/2)*x^3)/(15*a^2) + x^4/(4*a) + ((1 + 1/(a^2*x^2))^(3/2)*x^5)/5

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Rubi [A]  time = 0.0294947, antiderivative size = 54, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 10, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.4, Rules used = {6336, 30, 271, 264} $\frac{1}{5} x^5 \left (\frac{1}{a^2 x^2}+1\right )^{3/2}-\frac{2 x^3 \left (\frac{1}{a^2 x^2}+1\right )^{3/2}}{15 a^2}+\frac{x^4}{4 a}$

Antiderivative was successfully veriﬁed.

[In]

Int[E^ArcCsch[a*x]*x^4,x]

[Out]

(-2*(1 + 1/(a^2*x^2))^(3/2)*x^3)/(15*a^2) + x^4/(4*a) + ((1 + 1/(a^2*x^2))^(3/2)*x^5)/5

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 271

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x^(m + 1)*(a + b*x^n)^(p + 1))/(a*(m + 1)), x]
- Dist[(b*(m + n*(p + 1) + 1))/(a*(m + 1)), Int[x^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, m, n, p}, x]
&& ILtQ[Simplify[(m + 1)/n + p + 1], 0] && NeQ[m, -1]

Rule 264

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a
*c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]

Rubi steps

\begin{align*} \int e^{\text{csch}^{-1}(a x)} x^4 \, dx &=\frac{\int x^3 \, dx}{a}+\int \sqrt{1+\frac{1}{a^2 x^2}} x^4 \, dx\\ &=\frac{x^4}{4 a}+\frac{1}{5} \left (1+\frac{1}{a^2 x^2}\right )^{3/2} x^5-\frac{2 \int \sqrt{1+\frac{1}{a^2 x^2}} x^2 \, dx}{5 a^2}\\ &=-\frac{2 \left (1+\frac{1}{a^2 x^2}\right )^{3/2} x^3}{15 a^2}+\frac{x^4}{4 a}+\frac{1}{5} \left (1+\frac{1}{a^2 x^2}\right )^{3/2} x^5\\ \end{align*}

Mathematica [A]  time = 0.0488754, size = 49, normalized size = 0.91 $\frac{x \sqrt{\frac{1}{a^2 x^2}+1} \left (3 a^4 x^4+a^2 x^2-2\right )}{15 a^4}+\frac{x^4}{4 a}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[E^ArcCsch[a*x]*x^4,x]

[Out]

x^4/(4*a) + (Sqrt[1 + 1/(a^2*x^2)]*x*(-2 + a^2*x^2 + 3*a^4*x^4))/(15*a^4)

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Maple [A]  time = 0.21, size = 53, normalized size = 1. \begin{align*}{\frac{x \left ({a}^{2}{x}^{2}+1 \right ) \left ( 3\,{a}^{2}{x}^{2}-2 \right ) }{15\,{a}^{4}}\sqrt{{\frac{{a}^{2}{x}^{2}+1}{{a}^{2}{x}^{2}}}}}+{\frac{{x}^{4}}{4\,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^4,x)

[Out]

1/15*((a^2*x^2+1)/a^2/x^2)^(1/2)*x/a^4*(a^2*x^2+1)*(3*a^2*x^2-2)+1/4*x^4/a

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Maxima [A]  time = 1.00775, size = 68, normalized size = 1.26 \begin{align*} \frac{x^{4}}{4 \, a} + \frac{3 \, a^{2} x^{5}{\left (\frac{1}{a^{2} x^{2}} + 1\right )}^{\frac{5}{2}} - 5 \, x^{3}{\left (\frac{1}{a^{2} x^{2}} + 1\right )}^{\frac{3}{2}}}{15 \, a^{2}} \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^4,x, algorithm="maxima")

[Out]

1/4*x^4/a + 1/15*(3*a^2*x^5*(1/(a^2*x^2) + 1)^(5/2) - 5*x^3*(1/(a^2*x^2) + 1)^(3/2))/a^2

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Fricas [A]  time = 2.56468, size = 115, normalized size = 2.13 \begin{align*} \frac{15 \, a^{3} x^{4} + 4 \,{\left (3 \, a^{4} x^{5} + a^{2} x^{3} - 2 \, x\right )} \sqrt{\frac{a^{2} x^{2} + 1}{a^{2} x^{2}}}}{60 \, 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^4,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 3.26558, size = 63, normalized size = 1.17 \begin{align*} \frac{x^{4} \sqrt{a^{2} x^{2} + 1}}{5 a} + \frac{x^{4}}{4 a} + \frac{x^{2} \sqrt{a^{2} x^{2} + 1}}{15 a^{3}} - \frac{2 \sqrt{a^{2} x^{2} + 1}}{15 a^{5}} \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**4,x)

[Out]

x**4*sqrt(a**2*x**2 + 1)/(5*a) + x**4/(4*a) + x**2*sqrt(a**2*x**2 + 1)/(15*a**3) - 2*sqrt(a**2*x**2 + 1)/(15*a
**5)

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Giac [A]  time = 1.15286, size = 108, normalized size = 2. \begin{align*} \frac{\frac{4 \,{\left (3 \,{\left (a^{2} x^{2} + 1\right )}^{\frac{5}{2}} - 5 \,{\left (a^{2} x^{2} + 1\right )}^{\frac{3}{2}}\right )}{\left | a \right |} \mathrm{sgn}\left (x\right )}{a^{2}} - \frac{15 \,{\left (2 \, a^{2} x^{2} -{\left (a^{2} x^{2} + 1\right )}^{2} + 2\right )}}{a}}{60 \, a^{4}} + \frac{2 \,{\left | a \right |} \mathrm{sgn}\left (x\right )}{15 \, a^{6}} \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^4,x, algorithm="giac")

[Out]

1/60*(4*(3*(a^2*x^2 + 1)^(5/2) - 5*(a^2*x^2 + 1)^(3/2))*abs(a)*sgn(x)/a^2 - 15*(2*a^2*x^2 - (a^2*x^2 + 1)^2 +
2)/a)/a^4 + 2/15*abs(a)*sgn(x)/a^6