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]

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

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Rubi [A]  time = 0.03, antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, 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 30

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

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 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]

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*}

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Mathematica [A]  time = 0.05, 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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fricas [A]  time = 0.40, size = 53, normalized size = 0.98 \[ \frac {15 \, a^{3} x^{4} + 4 \, {\left (3 \, a^{4} x^{5} + a^{2} x^{3} - 2 \, x\right )} \sqrt {\frac {a^{2} x^{2} + 1}{a^{2} x^{2}}}}{60 \, a^{4}} \]

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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giac [A]  time = 0.14, size = 78, normalized size = 1.44 \[ -\frac {a^{2} x^{2} + 1}{2 \, a^{5}} + \frac {2 \, {\left | a \right |} \mathrm {sgn}\relax (x)}{15 \, a^{6}} + \frac {12 \, {\left (a^{2} x^{2} + 1\right )}^{\frac {5}{2}} {\left | a \right |} \mathrm {sgn}\relax (x) - 20 \, {\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}} {\left | a \right |} \mathrm {sgn}\relax (x) + 15 \, {\left (a^{2} x^{2} + 1\right )}^{2} a}{60 \, a^{6}} \]

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]

-1/2*(a^2*x^2 + 1)/a^5 + 2/15*abs(a)*sgn(x)/a^6 + 1/60*(12*(a^2*x^2 + 1)^(5/2)*abs(a)*sgn(x) - 20*(a^2*x^2 + 1
)^(3/2)*abs(a)*sgn(x) + 15*(a^2*x^2 + 1)^2*a)/a^6

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maple [A]  time = 0.09, size = 53, normalized size = 0.98 \[ \frac {\sqrt {\frac {a^{2} x^{2}+1}{a^{2} x^{2}}}\, x \left (a^{2} x^{2}+1\right ) \left (3 a^{2} x^{2}-2\right )}{15 a^{4}}+\frac {x^{4}}{4 a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1/a/x+(1+1/x^2/a^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 = 0.33, size = 50, normalized size = 0.93 \[ \frac {x^{4}}{4 \, a} + \frac {3 \, a^{2} x^{5} {\left (\frac {1}{a^{2} x^{2}} + 1\right )}^{\frac {5}{2}} - 5 \, x^{3} {\left (\frac {1}{a^{2} x^{2}} + 1\right )}^{\frac {3}{2}}}{15 \, a^{2}} \]

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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mupad [B]  time = 2.18, size = 41, normalized size = 0.76 \[ \sqrt {\frac {1}{a^2\,x^2}+1}\,\left (\frac {x^5}{5}-\frac {2\,x}{15\,a^4}+\frac {x^3}{15\,a^2}\right )+\frac {x^4}{4\,a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 3.01, size = 63, normalized size = 1.17 \[ \frac {x^{4} \sqrt {a^{2} x^{2} + 1}}{5 a} + \frac {x^{4}}{4 a} + \frac {x^{2} \sqrt {a^{2} x^{2} + 1}}{15 a^{3}} - \frac {2 \sqrt {a^{2} x^{2} + 1}}{15 a^{5}} \]

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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