3.1.33 \(\int e^{-\coth ^{-1}(a x)} x^3 \, dx\) [33]

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

[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 = 0.09, antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6304, 849, 821, 272, 65, 214} \begin {gather*} \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} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3/E^ArcCoth[a*x],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 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 821

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 849

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 6304

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 &=-\text {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} \text {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} \text {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} \text {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 \text {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 \text {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 \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 {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 = 0.05, size = 68, normalized size = 0.60 \begin {gather*} \frac {a \sqrt {1-\frac {1}{a^2 x^2}} x \left (-16+9 a x-8 a^2 x^2+6 a^3 x^3\right )+9 \log \left (\left (1+\sqrt {1-\frac {1}{a^2 x^2}}\right ) x\right )}{24 a^4} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[x^3/E^ArcCoth[a*x],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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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(192\) vs. \(2(94)=188\).
time = 0.09, size = 193, normalized size = 1.69

method	result	size
risch	\(\frac {\left (6 a^{3} x^{3}-8 a^{2} x^{2}+9 a x -16\right ) \left (a x +1\right ) \sqrt {\frac {a x -1}{a x +1}}}{24 a^{4}}+\frac {3 \ln \left (\frac {a^{2} x}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}-1}\right ) \sqrt {\frac {a x -1}{a x +1}}\, \sqrt {\left (a x +1\right ) \left (a x -1\right )}}{8 a^{3} \sqrt {a^{2}}\, \left (a x -1\right )}\)	\(117\)
default	\(\frac {\sqrt {\frac {a x -1}{a x +1}}\, \left (a x +1\right ) \left (6 \sqrt {a^{2}}\, \left (a^{2} x^{2}-1\right )^{\frac {3}{2}} a x +15 \sqrt {a^{2}}\, \sqrt {a^{2} x^{2}-1}\, a x -8 \left (\left (a x +1\right ) \left (a x -1\right )\right )^{\frac {3}{2}} \sqrt {a^{2}}+24 a \ln \left (\frac {a^{2} x +\sqrt {a^{2}}\, \sqrt {\left (a x +1\right ) \left (a x -1\right )}}{\sqrt {a^{2}}}\right )-15 \ln \left (\frac {a^{2} x +\sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right ) a -24 \sqrt {a^{2}}\, \sqrt {\left (a x +1\right ) \left (a x -1\right )}\right )}{24 \sqrt {\left (a x +1\right ) \left (a x -1\right )}\, a^{4} \sqrt {a^{2}}}\)	\(193\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*((a*x-1)/(a*x+1))^(1/2)*(a*x+1)*(6*(a^2)^(1/2)*(a^2*x^2-1)^(3/2)*a*x+15*(a^2)^(1/2)*(a^2*x^2-1)^(1/2)*a*x
-8*((a*x+1)*(a*x-1))^(3/2)*(a^2)^(1/2)+24*a*ln((a^2*x+(a^2)^(1/2)*((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^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] Leaf count of result is larger than twice the leaf count of optimal. 203 vs. \(2 (94) = 188\).
time = 0.26, size = 203, normalized size = 1.78 \begin {gather*} -\frac {1}{24} \, a {\left (\frac {2 \, {\left (39 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {7}{2}} - 31 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {5}{2}} + 49 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}} - 9 \, \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 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*a*(2*(39*((a*x - 1)/(a*x + 1))^(7/2) - 31*((a*x - 1)/(a*x + 1))^(5/2) + 49*((a*x - 1)/(a*x + 1))^(3/2) -
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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Fricas [A]
time = 0.34, size = 91, normalized size = 0.80 \begin {gather*} \frac {{\left (6 \, a^{4} x^{4} - 2 \, a^{3} x^{3} + a^{2} x^{2} - 7 \, 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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*((6*a^4*x^4 - 2*a^3*x^3 + a^2*x^2 - 7*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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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{3} \sqrt {\frac {a x - 1}{a x + 1}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Ch
eck [abs(sa

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Mupad [B]
time = 1.21, size = 172, normalized size = 1.51 \begin {gather*} \frac {3\,\mathrm {atanh}\left (\sqrt {\frac {a\,x-1}{a\,x+1}}\right )}{4\,a^4}-\frac {\frac {3\,\sqrt {\frac {a\,x-1}{a\,x+1}}}{4}-\frac {49\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/2}}{12}+\frac {31\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{5/2}}{12}-\frac {13\,{\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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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