3.1.28 \(\int e^{\text {csch}^{-1}(a x)} x^3 \, dx\) [28]

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

[Out]

1/3*x^3/a-1/8*arctanh((1+1/a^2/x^2)^(1/2))/a^4+1/8*x^2*(1+1/a^2/x^2)^(1/2)/a^2+1/4*x^4*(1+1/a^2/x^2)^(1/2)

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Rubi [A]
time = 0.03, antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.700, Rules used = {6471, 30, 272, 43, 44, 65, 214} \begin {gather*} \frac {x^2 \sqrt {\frac {1}{a^2 x^2}+1}}{8 a^2}+\frac {1}{4} x^4 \sqrt {\frac {1}{a^2 x^2}+1}-\frac {\tanh ^{-1}\left (\sqrt {\frac {1}{a^2 x^2}+1}\right )}{8 a^4}+\frac {x^3}{3 a} \end {gather*}

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 30

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

Rule 43

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, n
}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && GtQ[n, 0]

Rule 44

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] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

Rule 65

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 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 272

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 6471

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^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} \text {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 {\text {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 {\text {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 {\text {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*}

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Mathematica [A]
time = 0.04, size = 76, normalized size = 1.01 \begin {gather*} \frac {a^2 x^2 \left (3 \sqrt {1+\frac {1}{a^2 x^2}}+8 a x+6 a^2 \sqrt {1+\frac {1}{a^2 x^2}} x^2\right )-3 \log \left (\left (1+\sqrt {1+\frac {1}{a^2 x^2}}\right ) x\right )}{24 a^4} \end {gather*}

Antiderivative was successfully verified.

[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 = 0.04, size = 109, normalized size = 1.45

method	result	size
default	\(-\frac {\sqrt {\frac {a^{2} x^{2}+1}{a^{2} x^{2}}}\, x \left (-2 x \left (\frac {a^{2} x^{2}+1}{a^{2}}\right )^{\frac {3}{2}} a^{4}+x \sqrt {\frac {a^{2} x^{2}+1}{a^{2}}}\, a^{2}+\ln \left (x +\sqrt {\frac {a^{2} x^{2}+1}{a^{2}}}\right )\right )}{8 \sqrt {\frac {a^{2} x^{2}+1}{a^{2}}}\, a^{4}}+\frac {x^{3}}{3 a}\)	\(109\)

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,method=_RETURNVERBOSE)

[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 = 0.25, size = 107, normalized size = 1.43 \begin {gather*} \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 {gather*}

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 = 0.37, size = 79, normalized size = 1.05 \begin {gather*} \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 {gather*}

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 = 3.10, size = 73, normalized size = 0.97 \begin {gather*} \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 {gather*}

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 = 0.43, size = 69, normalized size = 0.92 \begin {gather*} \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 {gather*}

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

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Mupad [B]
time = 2.41, size = 61, normalized size = 0.81 \begin {gather*} \frac {x^4\,\sqrt {\frac {1}{a^2\,x^2}+1}}{4}-\frac {\mathrm {atanh}\left (\sqrt {\frac {1}{a^2\,x^2}+1}\right )}{8\,a^4}+\frac {x^3}{3\,a}+\frac {x^2\,\sqrt {\frac {1}{a^2\,x^2}+1}}{8\,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)),x)

[Out]

(x^4*(1/(a^2*x^2) + 1)^(1/2))/4 - atanh((1/(a^2*x^2) + 1)^(1/2))/(8*a^4) + x^3/(3*a) + (x^2*(1/(a^2*x^2) + 1)^
(1/2))/(8*a^2)

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