3.2 \(\int e^{\coth ^{-1}(a x)} x^2 \, dx\)

Optimal. Leaf size=90 \[ \frac {x^2 \sqrt {1-\frac {1}{a^2 x^2}}}{2 a}+\frac {2 x \sqrt {1-\frac {1}{a^2 x^2}}}{3 a^2}+\frac {1}{3} x^3 \sqrt {1-\frac {1}{a^2 x^2}}+\frac {\tanh ^{-1}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )}{2 a^3} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {6169, 835, 807, 266, 63, 208} \[ \frac {1}{3} x^3 \sqrt {1-\frac {1}{a^2 x^2}}+\frac {x^2 \sqrt {1-\frac {1}{a^2 x^2}}}{2 a}+\frac {2 x \sqrt {1-\frac {1}{a^2 x^2}}}{3 a^2}+\frac {\tanh ^{-1}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )}{2 a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[1 - 1/(a^2*x^2)]*x)/(3*a^2) + (Sqrt[1 - 1/(a^2*x^2)]*x^2)/(2*a) + (Sqrt[1 - 1/(a^2*x^2)]*x^3)/3 + ArcT
anh[Sqrt[1 - 1/(a^2*x^2)]]/(2*a^3)

Rule 63

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 208

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

Rule 266

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 807

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Simp[((e*f - d*g
)*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[(c*d*f + a*e*g)/(c*d^2 + a*e^2
), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0]
&& EqQ[Simplify[m + 2*p + 3], 0]

Rule 835

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((e*f - d*g)
*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/((m + 1)*(c*d^2 + a*e^2)), x] + Dist[1/((m + 1)*(c*d^2 + a*e^2)), Int[
(d + e*x)^(m + 1)*(a + c*x^2)^p*Simp[(c*d*f + a*e*g)*(m + 1) - c*(e*f - d*g)*(m + 2*p + 3)*x, x], x], x] /; Fr
eeQ[{a, c, d, e, f, g, p}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[m, -1] && (IntegerQ[m] || IntegerQ[p] || Integer
sQ[2*m, 2*p])

Rule 6169

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*(x_)^(m_.), x_Symbol] :> -Subst[Int[(1 + x/a)^((n + 1)/2)/(x^(m + 2)*(1 - x/
a)^((n - 1)/2)*Sqrt[1 - x^2/a^2]), x], x, 1/x] /; FreeQ[a, x] && IntegerQ[(n - 1)/2] && IntegerQ[m]

Rubi steps

\begin {align*} \int e^{\coth ^{-1}(a x)} x^2 \, dx &=-\operatorname {Subst}\left (\int \frac {1+\frac {x}{a}}{x^4 \sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )\\ &=\frac {1}{3} \sqrt {1-\frac {1}{a^2 x^2}} x^3+\frac {1}{3} \operatorname {Subst}\left (\int \frac {-\frac {3}{a}-\frac {2 x}{a^2}}{x^3 \sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )\\ &=\frac {\sqrt {1-\frac {1}{a^2 x^2}} x^2}{2 a}+\frac {1}{3} \sqrt {1-\frac {1}{a^2 x^2}} x^3-\frac {1}{6} \operatorname {Subst}\left (\int \frac {\frac {4}{a^2}+\frac {3 x}{a^3}}{x^2 \sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )\\ &=\frac {2 \sqrt {1-\frac {1}{a^2 x^2}} x}{3 a^2}+\frac {\sqrt {1-\frac {1}{a^2 x^2}} x^2}{2 a}+\frac {1}{3} \sqrt {1-\frac {1}{a^2 x^2}} x^3-\frac {\operatorname {Subst}\left (\int \frac {1}{x \sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )}{2 a^3}\\ &=\frac {2 \sqrt {1-\frac {1}{a^2 x^2}} x}{3 a^2}+\frac {\sqrt {1-\frac {1}{a^2 x^2}} x^2}{2 a}+\frac {1}{3} \sqrt {1-\frac {1}{a^2 x^2}} x^3-\frac {\operatorname {Subst}\left (\int \frac {1}{x \sqrt {1-\frac {x}{a^2}}} \, dx,x,\frac {1}{x^2}\right )}{4 a^3}\\ &=\frac {2 \sqrt {1-\frac {1}{a^2 x^2}} x}{3 a^2}+\frac {\sqrt {1-\frac {1}{a^2 x^2}} x^2}{2 a}+\frac {1}{3} \sqrt {1-\frac {1}{a^2 x^2}} x^3+\frac {\operatorname {Subst}\left (\int \frac {1}{a^2-a^2 x^2} \, dx,x,\sqrt {1-\frac {1}{a^2 x^2}}\right )}{2 a}\\ &=\frac {2 \sqrt {1-\frac {1}{a^2 x^2}} x}{3 a^2}+\frac {\sqrt {1-\frac {1}{a^2 x^2}} x^2}{2 a}+\frac {1}{3} \sqrt {1-\frac {1}{a^2 x^2}} x^3+\frac {\tanh ^{-1}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )}{2 a^3}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 60, normalized size = 0.67 \[ \frac {a x \sqrt {1-\frac {1}{a^2 x^2}} \left (2 a^2 x^2+3 a x+4\right )+3 \log \left (x \left (\sqrt {1-\frac {1}{a^2 x^2}}+1\right )\right )}{6 a^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(a*Sqrt[1 - 1/(a^2*x^2)]*x*(4 + 3*a*x + 2*a^2*x^2) + 3*Log[(1 + Sqrt[1 - 1/(a^2*x^2)])*x])/(6*a^3)

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fricas [A]  time = 0.71, size = 84, normalized size = 0.93 \[ \frac {{\left (2 \, a^{3} x^{3} + 5 \, a^{2} x^{2} + 7 \, a x + 4\right )} \sqrt {\frac {a x - 1}{a x + 1}} + 3 \, \log \left (\sqrt {\frac {a x - 1}{a x + 1}} + 1\right ) - 3 \, \log \left (\sqrt {\frac {a x - 1}{a x + 1}} - 1\right )}{6 \, a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*((2*a^3*x^3 + 5*a^2*x^2 + 7*a*x + 4)*sqrt((a*x - 1)/(a*x + 1)) + 3*log(sqrt((a*x - 1)/(a*x + 1)) + 1) - 3*
log(sqrt((a*x - 1)/(a*x + 1)) - 1))/a^3

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giac [B]  time = 0.16, size = 151, normalized size = 1.68 \[ \frac {1}{6} \, a {\left (\frac {3 \, \log \left (\sqrt {\frac {a x - 1}{a x + 1}} + 1\right )}{a^{4}} - \frac {3 \, \log \left ({\left | \sqrt {\frac {a x - 1}{a x + 1}} - 1 \right |}\right )}{a^{4}} + \frac {2 \, {\left (\frac {4 \, {\left (a x - 1\right )} \sqrt {\frac {a x - 1}{a x + 1}}}{a x + 1} - \frac {3 \, {\left (a x - 1\right )}^{2} \sqrt {\frac {a x - 1}{a x + 1}}}{{\left (a x + 1\right )}^{2}} - 9 \, \sqrt {\frac {a x - 1}{a x + 1}}\right )}}{a^{4} {\left (\frac {a x - 1}{a x + 1} - 1\right )}^{3}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*a*(3*log(sqrt((a*x - 1)/(a*x + 1)) + 1)/a^4 - 3*log(abs(sqrt((a*x - 1)/(a*x + 1)) - 1))/a^4 + 2*(4*(a*x -
1)*sqrt((a*x - 1)/(a*x + 1))/(a*x + 1) - 3*(a*x - 1)^2*sqrt((a*x - 1)/(a*x + 1))/(a*x + 1)^2 - 9*sqrt((a*x - 1
)/(a*x + 1)))/(a^4*((a*x - 1)/(a*x + 1) - 1)^3))

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maple [B]  time = 0.05, size = 173, normalized size = 1.92 \[ \frac {\left (a x -1\right ) \left (3 \sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}\, x a +2 \left (\left (a x -1\right ) \left (a x +1\right )\right )^{\frac {3}{2}} \sqrt {a^{2}}-3 \ln \left (\frac {a^{2} x +\sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right ) a +6 \sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \sqrt {a^{2}}+6 a \ln \left (\frac {a^{2} x +\sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right )\right )}{6 \sqrt {\frac {a x -1}{a x +1}}\, \sqrt {\left (a x -1\right ) \left (a x +1\right )}\, a^{3} \sqrt {a^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.32, size = 166, normalized size = 1.84 \[ -\frac {1}{6} \, a {\left (\frac {2 \, {\left (3 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {5}{2}} - 4 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}} + 9 \, \sqrt {\frac {a x - 1}{a x + 1}}\right )}}{\frac {3 \, {\left (a x - 1\right )} a^{4}}{a x + 1} - \frac {3 \, {\left (a x - 1\right )}^{2} a^{4}}{{\left (a x + 1\right )}^{2}} + \frac {{\left (a x - 1\right )}^{3} a^{4}}{{\left (a x + 1\right )}^{3}} - a^{4}} - \frac {3 \, \log \left (\sqrt {\frac {a x - 1}{a x + 1}} + 1\right )}{a^{4}} + \frac {3 \, \log \left (\sqrt {\frac {a x - 1}{a x + 1}} - 1\right )}{a^{4}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*a*(2*(3*((a*x - 1)/(a*x + 1))^(5/2) - 4*((a*x - 1)/(a*x + 1))^(3/2) + 9*sqrt((a*x - 1)/(a*x + 1)))/(3*(a*
x - 1)*a^4/(a*x + 1) - 3*(a*x - 1)^2*a^4/(a*x + 1)^2 + (a*x - 1)^3*a^4/(a*x + 1)^3 - a^4) - 3*log(sqrt((a*x -
1)/(a*x + 1)) + 1)/a^4 + 3*log(sqrt((a*x - 1)/(a*x + 1)) - 1)/a^4)

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mupad [B]  time = 0.06, size = 133, normalized size = 1.48 \[ \frac {3\,\sqrt {\frac {a\,x-1}{a\,x+1}}-\frac {4\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/2}}{3}+{\left (\frac {a\,x-1}{a\,x+1}\right )}^{5/2}}{a^3+\frac {3\,a^3\,{\left (a\,x-1\right )}^2}{{\left (a\,x+1\right )}^2}-\frac {a^3\,{\left (a\,x-1\right )}^3}{{\left (a\,x+1\right )}^3}-\frac {3\,a^3\,\left (a\,x-1\right )}{a\,x+1}}+\frac {\mathrm {atanh}\left (\sqrt {\frac {a\,x-1}{a\,x+1}}\right )}{a^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{\sqrt {\frac {a x - 1}{a x + 1}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2/sqrt((a*x - 1)/(a*x + 1)), x)

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