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

Optimal. Leaf size=31 \[ \frac {1}{3} x^3 \left (\frac {1}{a^2 x^2}+1\right )^{3/2}+\frac {x^2}{2 a} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {6336, 30, 264} \[ \frac {1}{3} x^3 \left (\frac {1}{a^2 x^2}+1\right )^{3/2}+\frac {x^2}{2 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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^2 \, dx &=\frac {\int x \, dx}{a}+\int \sqrt {1+\frac {1}{a^2 x^2}} x^2 \, dx\\ &=\frac {x^2}{2 a}+\frac {1}{3} \left (1+\frac {1}{a^2 x^2}\right )^{3/2} x^3\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 38, normalized size = 1.23 \[ \frac {2 \sqrt {\frac {1}{a^2 x^2}+1} \left (a^2 x^3+x\right )+3 a x^2}{6 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*a*x^2 + 2*Sqrt[1 + 1/(a^2*x^2)]*(x + a^2*x^3))/(6*a^2)

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fricas [A]  time = 0.56, size = 41, normalized size = 1.32 \[ \frac {3 \, a x^{2} + 2 \, {\left (a^{2} x^{3} + x\right )} \sqrt {\frac {a^{2} x^{2} + 1}{a^{2} x^{2}}}}{6 \, a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(3*a*x^2 + 2*(a^2*x^3 + x)*sqrt((a^2*x^2 + 1)/(a^2*x^2)))/a^2

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giac [A]  time = 0.14, size = 44, normalized size = 1.42 \[ \frac {{\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}} {\left | a \right |} \mathrm {sgn}\relax (x)}{3 \, a^{4}} + \frac {a^{2} x^{2} + 1}{2 \, a^{3}} - \frac {{\left | a \right |} \mathrm {sgn}\relax (x)}{3 \, a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.04, size = 43, normalized size = 1.39 \[ \frac {\sqrt {\frac {a^{2} x^{2}+1}{a^{2} x^{2}}}\, x \left (a^{2} x^{2}+1\right )}{3 a^{2}}+\frac {x^{2}}{2 a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.31, size = 25, normalized size = 0.81 \[ \frac {1}{3} \, x^{3} {\left (\frac {1}{a^{2} x^{2}} + 1\right )}^{\frac {3}{2}} + \frac {x^{2}}{2 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 2.17, size = 33, normalized size = 1.06 \[ \left (\frac {x}{3\,a^2}+\frac {x^3}{3}\right )\,\sqrt {\frac {1}{a^2\,x^2}+1}+\frac {x^2}{2\,a} \]

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)),x)

[Out]

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

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sympy [A]  time = 2.71, size = 41, normalized size = 1.32 \[ \frac {x^{2} \sqrt {a^{2} x^{2} + 1}}{3 a} + \frac {x^{2}}{2 a} + \frac {\sqrt {a^{2} x^{2} + 1}}{3 a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**2*sqrt(a**2*x**2 + 1)/(3*a) + x**2/(2*a) + sqrt(a**2*x**2 + 1)/(3*a**3)

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