3.1.50 \(\int e^{2 \text {csch}^{-1}(a x)} x^3 \, dx\) [50]

Optimal. Leaf size=38 \[ \frac {x^2}{a^2}+\frac {2 \left (1+\frac {1}{a^2 x^2}\right )^{3/2} x^3}{3 a}+\frac {x^4}{4} \]

[Out]

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

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Rubi [A]
time = 0.15, antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6473, 6874, 270} \begin {gather*} \frac {x^2}{a^2}+\frac {2 x^3 \left (\frac {1}{a^2 x^2}+1\right )^{3/2}}{3 a}+\frac {x^4}{4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 270

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 6473

Int[E^(ArcCsch[u_]*(n_.))*(x_)^(m_.), x_Symbol] :> Int[x^m*(1/u + Sqrt[1 + 1/u^2])^n, x] /; FreeQ[m, x] && Int
egerQ[n]

Rule 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

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

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Mathematica [A]
time = 0.03, size = 44, normalized size = 1.16 \begin {gather*} \frac {x^2}{a^2}+\frac {x^4}{4}+\frac {2 \sqrt {1+\frac {1}{a^2 x^2}} \left (x+a^2 x^3\right )}{3 a^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.12, size = 59, normalized size = 1.55

method	result	size
default	\(\frac {\left (a^{2} x^{2}+1\right )^{2}}{4 a^{4}}+\frac {2 \sqrt {\frac {a^{2} x^{2}+1}{a^{2} x^{2}}}\, x \left (a^{2} x^{2}+1\right )}{3 a^{3}}+\frac {x^{2}}{2 a^{2}}\)	\(59\)
trager	\(\frac {\frac {\left (a^{2} x^{3}+a^{2} x^{2}+a^{2} x +a^{2}+4 x +4\right ) \left (-1+x \right )}{4}+\frac {2 \left (a^{2} x^{2}+1\right ) x \sqrt {-\frac {-a^{2} x^{2}-1}{a^{2} x^{2}}}}{3 a}}{a^{2}}\)	\(73\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.26, size = 32, normalized size = 0.84 \begin {gather*} \frac {1}{4} \, x^{4} + \frac {2 \, x^{3} {\left (\frac {1}{a^{2} x^{2}} + 1\right )}^{\frac {3}{2}}}{3 \, a} + \frac {x^{2}}{a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.33, size = 49, normalized size = 1.29 \begin {gather*} \frac {3 \, a^{3} x^{4} + 12 \, a x^{2} + 8 \, {\left (a^{2} x^{3} + x\right )} \sqrt {\frac {a^{2} x^{2} + 1}{a^{2} x^{2}}}}{12 \, a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*a^3*x^4 + 12*a*x^2 + 8*(a^2*x^3 + x)*sqrt((a^2*x^2 + 1)/(a^2*x^2)))/a^3

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Sympy [A]
time = 1.73, size = 51, normalized size = 1.34 \begin {gather*} \frac {x^{4}}{4} + \frac {2 x^{2} \sqrt {a^{2} x^{2} + 1}}{3 a^{2}} + \frac {x^{2}}{a^{2}} + \frac {2 \sqrt {a^{2} x^{2} + 1}}{3 a^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 66 vs. \(2 (32) = 64\).
time = 0.43, size = 66, normalized size = 1.74 \begin {gather*} \frac {a^{2} x^{2} + 1}{2 \, a^{4}} - \frac {2 \, {\left | a \right |} \mathrm {sgn}\left (x\right )}{3 \, a^{5}} + \frac {8 \, {\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}} a^{2} {\left | a \right |} \mathrm {sgn}\left (x\right ) + 3 \, {\left (a^{2} x^{2} + 1\right )}^{2} a^{3}}{12 \, a^{7}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(a^2*x^2 + 1)/a^4 - 2/3*abs(a)*sgn(x)/a^5 + 1/12*(8*(a^2*x^2 + 1)^(3/2)*a^2*abs(a)*sgn(x) + 3*(a^2*x^2 + 1
)^2*a^3)/a^7

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Mupad [B]
time = 2.14, size = 40, normalized size = 1.05 \begin {gather*} \sqrt {\frac {1}{a^2\,x^2}+1}\,\left (\frac {2\,x}{3\,a^3}+\frac {2\,x^3}{3\,a}\right )+\frac {x^4}{4}+\frac {x^2}{a^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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