3.1 \(\int e^{\coth ^{-1}(a x)} x^3 \, dx\)

Optimal. Leaf size=114 \[ \frac {3 x^2 \sqrt {1-\frac {1}{a^2 x^2}}}{8 a^2}+\frac {1}{4} x^4 \sqrt {1-\frac {1}{a^2 x^2}}+\frac {x^3 \sqrt {1-\frac {1}{a^2 x^2}}}{3 a}+\frac {3 \tanh ^{-1}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )}{8 a^4}+\frac {2 x \sqrt {1-\frac {1}{a^2 x^2}}}{3 a^3} \]

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 8, 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} \[ \frac {1}{4} x^4 \sqrt {1-\frac {1}{a^2 x^2}}+\frac {x^3 \sqrt {1-\frac {1}{a^2 x^2}}}{3 a}+\frac {3 x^2 \sqrt {1-\frac {1}{a^2 x^2}}}{8 a^2}+\frac {2 x \sqrt {1-\frac {1}{a^2 x^2}}}{3 a^3}+\frac {3 \tanh ^{-1}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )}{8 a^4} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcCoth[a*x]*x^3,x]

[Out]

(2*Sqrt[1 - 1/(a^2*x^2)]*x)/(3*a^3) + (3*Sqrt[1 - 1/(a^2*x^2)]*x^2)/(8*a^2) + (Sqrt[1 - 1/(a^2*x^2)]*x^3)/(3*a
) + (Sqrt[1 - 1/(a^2*x^2)]*x^4)/4 + (3*ArcTanh[Sqrt[1 - 1/(a^2*x^2)]])/(8*a^4)

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^3 \, dx &=-\operatorname {Subst}\left (\int \frac {1+\frac {x}{a}}{x^5 \sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )\\ &=\frac {1}{4} \sqrt {1-\frac {1}{a^2 x^2}} x^4+\frac {1}{4} \operatorname {Subst}\left (\int \frac {-\frac {4}{a}-\frac {3 x}{a^2}}{x^4 \sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )\\ &=\frac {\sqrt {1-\frac {1}{a^2 x^2}} x^3}{3 a}+\frac {1}{4} \sqrt {1-\frac {1}{a^2 x^2}} x^4-\frac {1}{12} \operatorname {Subst}\left (\int \frac {\frac {9}{a^2}+\frac {8 x}{a^3}}{x^3 \sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )\\ &=\frac {3 \sqrt {1-\frac {1}{a^2 x^2}} x^2}{8 a^2}+\frac {\sqrt {1-\frac {1}{a^2 x^2}} x^3}{3 a}+\frac {1}{4} \sqrt {1-\frac {1}{a^2 x^2}} x^4+\frac {1}{24} \operatorname {Subst}\left (\int \frac {-\frac {16}{a^3}-\frac {9 x}{a^4}}{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^3}+\frac {3 \sqrt {1-\frac {1}{a^2 x^2}} x^2}{8 a^2}+\frac {\sqrt {1-\frac {1}{a^2 x^2}} x^3}{3 a}+\frac {1}{4} \sqrt {1-\frac {1}{a^2 x^2}} x^4-\frac {3 \operatorname {Subst}\left (\int \frac {1}{x \sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )}{8 a^4}\\ &=\frac {2 \sqrt {1-\frac {1}{a^2 x^2}} x}{3 a^3}+\frac {3 \sqrt {1-\frac {1}{a^2 x^2}} x^2}{8 a^2}+\frac {\sqrt {1-\frac {1}{a^2 x^2}} x^3}{3 a}+\frac {1}{4} \sqrt {1-\frac {1}{a^2 x^2}} x^4-\frac {3 \operatorname {Subst}\left (\int \frac {1}{x \sqrt {1-\frac {x}{a^2}}} \, dx,x,\frac {1}{x^2}\right )}{16 a^4}\\ &=\frac {2 \sqrt {1-\frac {1}{a^2 x^2}} x}{3 a^3}+\frac {3 \sqrt {1-\frac {1}{a^2 x^2}} x^2}{8 a^2}+\frac {\sqrt {1-\frac {1}{a^2 x^2}} x^3}{3 a}+\frac {1}{4} \sqrt {1-\frac {1}{a^2 x^2}} x^4+\frac {3 \operatorname {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 {2 \sqrt {1-\frac {1}{a^2 x^2}} x}{3 a^3}+\frac {3 \sqrt {1-\frac {1}{a^2 x^2}} x^2}{8 a^2}+\frac {\sqrt {1-\frac {1}{a^2 x^2}} x^3}{3 a}+\frac {1}{4} \sqrt {1-\frac {1}{a^2 x^2}} x^4+\frac {3 \tanh ^{-1}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )}{8 a^4}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 68, normalized size = 0.60 \[ \frac {9 \log \left (x \left (\sqrt {1-\frac {1}{a^2 x^2}}+1\right )\right )+a x \sqrt {1-\frac {1}{a^2 x^2}} \left (6 a^3 x^3+8 a^2 x^2+9 a x+16\right )}{24 a^4} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^ArcCoth[a*x]*x^3,x]

[Out]

(a*Sqrt[1 - 1/(a^2*x^2)]*x*(16 + 9*a*x + 8*a^2*x^2 + 6*a^3*x^3) + 9*Log[(1 + Sqrt[1 - 1/(a^2*x^2)])*x])/(24*a^
4)

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fricas [A]  time = 0.53, size = 92, normalized size = 0.81 \[ \frac {{\left (6 \, a^{4} x^{4} + 14 \, a^{3} x^{3} + 17 \, a^{2} x^{2} + 25 \, a x + 16\right )} \sqrt {\frac {a x - 1}{a x + 1}} + 9 \, \log \left (\sqrt {\frac {a x - 1}{a x + 1}} + 1\right ) - 9 \, \log \left (\sqrt {\frac {a x - 1}{a x + 1}} - 1\right )}{24 \, a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*((6*a^4*x^4 + 14*a^3*x^3 + 17*a^2*x^2 + 25*a*x + 16)*sqrt((a*x - 1)/(a*x + 1)) + 9*log(sqrt((a*x - 1)/(a*
x + 1)) + 1) - 9*log(sqrt((a*x - 1)/(a*x + 1)) - 1))/a^4

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giac [A]  time = 0.15, size = 182, normalized size = 1.60 \[ \frac {1}{24} \, a {\left (\frac {9 \, \log \left (\sqrt {\frac {a x - 1}{a x + 1}} + 1\right )}{a^{5}} - \frac {9 \, \log \left ({\left | \sqrt {\frac {a x - 1}{a x + 1}} - 1 \right |}\right )}{a^{5}} - \frac {2 \, {\left (\frac {31 \, {\left (a x - 1\right )} \sqrt {\frac {a x - 1}{a x + 1}}}{a x + 1} - \frac {49 \, {\left (a x - 1\right )}^{2} \sqrt {\frac {a x - 1}{a x + 1}}}{{\left (a x + 1\right )}^{2}} + \frac {9 \, {\left (a x - 1\right )}^{3} \sqrt {\frac {a x - 1}{a x + 1}}}{{\left (a x + 1\right )}^{3}} - 39 \, \sqrt {\frac {a x - 1}{a x + 1}}\right )}}{a^{5} {\left (\frac {a x - 1}{a x + 1} - 1\right )}^{4}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*a*(9*log(sqrt((a*x - 1)/(a*x + 1)) + 1)/a^5 - 9*log(abs(sqrt((a*x - 1)/(a*x + 1)) - 1))/a^5 - 2*(31*(a*x
- 1)*sqrt((a*x - 1)/(a*x + 1))/(a*x + 1) - 49*(a*x - 1)^2*sqrt((a*x - 1)/(a*x + 1))/(a*x + 1)^2 + 9*(a*x - 1)^
3*sqrt((a*x - 1)/(a*x + 1))/(a*x + 1)^3 - 39*sqrt((a*x - 1)/(a*x + 1)))/(a^5*((a*x - 1)/(a*x + 1) - 1)^4))

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maple [B]  time = 0.07, size = 193, normalized size = 1.69 \[ \frac {\left (a x -1\right ) \left (6 \left (a^{2} x^{2}-1\right )^{\frac {3}{2}} \sqrt {a^{2}}\, x a +8 \left (\left (a x -1\right ) \left (a x +1\right )\right )^{\frac {3}{2}} \sqrt {a^{2}}+15 \sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}\, x a +24 \sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \sqrt {a^{2}}-15 \ln \left (\frac {a^{2} x +\sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right ) a +24 a \ln \left (\frac {a^{2} x +\sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right )\right )}{24 \sqrt {\frac {a x -1}{a x +1}}\, \sqrt {\left (a x -1\right ) \left (a x +1\right )}\, a^{4} \sqrt {a^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(a*x-1)*(6*(a^2*x^2-1)^(3/2)*(a^2)^(1/2)*x*a+8*((a*x-1)*(a*x+1))^(3/2)*(a^2)^(1/2)+15*(a^2*x^2-1)^(1/2)*(
a^2)^(1/2)*x*a+24*((a*x-1)*(a*x+1))^(1/2)*(a^2)^(1/2)-15*ln((a^2*x+(a^2*x^2-1)^(1/2)*(a^2)^(1/2))/(a^2)^(1/2))
*a+24*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^4/(a^2)^(1/2)

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maxima [B]  time = 0.32, size = 203, normalized size = 1.78 \[ \frac {1}{24} \, a {\left (\frac {2 \, {\left (9 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {7}{2}} - 49 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {5}{2}} + 31 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}} - 39 \, \sqrt {\frac {a x - 1}{a x + 1}}\right )}}{\frac {4 \, {\left (a x - 1\right )} a^{5}}{a x + 1} - \frac {6 \, {\left (a x - 1\right )}^{2} a^{5}}{{\left (a x + 1\right )}^{2}} + \frac {4 \, {\left (a x - 1\right )}^{3} a^{5}}{{\left (a x + 1\right )}^{3}} - \frac {{\left (a x - 1\right )}^{4} a^{5}}{{\left (a x + 1\right )}^{4}} - a^{5}} + \frac {9 \, \log \left (\sqrt {\frac {a x - 1}{a x + 1}} + 1\right )}{a^{5}} - \frac {9 \, \log \left (\sqrt {\frac {a x - 1}{a x + 1}} - 1\right )}{a^{5}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*a*(2*(9*((a*x - 1)/(a*x + 1))^(7/2) - 49*((a*x - 1)/(a*x + 1))^(5/2) + 31*((a*x - 1)/(a*x + 1))^(3/2) - 3
9*sqrt((a*x - 1)/(a*x + 1)))/(4*(a*x - 1)*a^5/(a*x + 1) - 6*(a*x - 1)^2*a^5/(a*x + 1)^2 + 4*(a*x - 1)^3*a^5/(a
*x + 1)^3 - (a*x - 1)^4*a^5/(a*x + 1)^4 - a^5) + 9*log(sqrt((a*x - 1)/(a*x + 1)) + 1)/a^5 - 9*log(sqrt((a*x -
1)/(a*x + 1)) - 1)/a^5)

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mupad [B]  time = 1.26, size = 171, normalized size = 1.50 \[ \frac {\frac {13\,\sqrt {\frac {a\,x-1}{a\,x+1}}}{4}-\frac {31\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/2}}{12}+\frac {49\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{5/2}}{12}-\frac {3\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{7/2}}{4}}{a^4+\frac {6\,a^4\,{\left (a\,x-1\right )}^2}{{\left (a\,x+1\right )}^2}-\frac {4\,a^4\,{\left (a\,x-1\right )}^3}{{\left (a\,x+1\right )}^3}+\frac {a^4\,{\left (a\,x-1\right )}^4}{{\left (a\,x+1\right )}^4}-\frac {4\,a^4\,\left (a\,x-1\right )}{a\,x+1}}+\frac {3\,\mathrm {atanh}\left (\sqrt {\frac {a\,x-1}{a\,x+1}}\right )}{4\,a^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((13*((a*x - 1)/(a*x + 1))^(1/2))/4 - (31*((a*x - 1)/(a*x + 1))^(3/2))/12 + (49*((a*x - 1)/(a*x + 1))^(5/2))/1
2 - (3*((a*x - 1)/(a*x + 1))^(7/2))/4)/(a^4 + (6*a^4*(a*x - 1)^2)/(a*x + 1)^2 - (4*a^4*(a*x - 1)^3)/(a*x + 1)^
3 + (a^4*(a*x - 1)^4)/(a*x + 1)^4 - (4*a^4*(a*x - 1))/(a*x + 1)) + (3*atanh(((a*x - 1)/(a*x + 1))^(1/2)))/(4*a
^4)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{3}}{\sqrt {\frac {a x - 1}{a x + 1}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**3/sqrt((a*x - 1)/(a*x + 1)), x)

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