3.26 \(\int e^{4 \coth ^{-1}(a x)} x^2 \, dx\)

Optimal. Leaf size=47 \[ \frac {4}{a^3 (1-a x)}+\frac {12 \log (1-a x)}{a^3}+\frac {8 x}{a^2}+\frac {2 x^2}{a}+\frac {x^3}{3} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 47, 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 = {6167, 6126, 88} \[ \frac {8 x}{a^2}+\frac {4}{a^3 (1-a x)}+\frac {12 \log (1-a x)}{a^3}+\frac {2 x^2}{a}+\frac {x^3}{3} \]

Antiderivative was successfully verified.

[In]

Int[E^(4*ArcCoth[a*x])*x^2,x]

[Out]

(8*x)/a^2 + (2*x^2)/a + x^3/3 + 4/(a^3*(1 - a*x)) + (12*Log[1 - a*x])/a^3

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 6126

Int[E^(ArcTanh[(a_.)*(x_)]*(n_))*(x_)^(m_.), x_Symbol] :> Int[(x^m*(1 + a*x)^(n/2))/(1 - a*x)^(n/2), x] /; Fre
eQ[{a, m, n}, x] &&  !IntegerQ[(n - 1)/2]

Rule 6167

Int[E^(ArcCoth[(a_.)*(x_)]*(n_))*(u_.), x_Symbol] :> Dist[(-1)^(n/2), Int[u*E^(n*ArcTanh[a*x]), x], x] /; Free
Q[a, x] && IntegerQ[n/2]

Rubi steps

\begin {align*} \int e^{4 \coth ^{-1}(a x)} x^2 \, dx &=\int e^{4 \tanh ^{-1}(a x)} x^2 \, dx\\ &=\int \frac {x^2 (1+a x)^2}{(1-a x)^2} \, dx\\ &=\int \left (\frac {8}{a^2}+\frac {4 x}{a}+x^2+\frac {4}{a^2 (-1+a x)^2}+\frac {12}{a^2 (-1+a x)}\right ) \, dx\\ &=\frac {8 x}{a^2}+\frac {2 x^2}{a}+\frac {x^3}{3}+\frac {4}{a^3 (1-a x)}+\frac {12 \log (1-a x)}{a^3}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 47, normalized size = 1.00 \[ \frac {4}{a^3 (1-a x)}+\frac {12 \log (1-a x)}{a^3}+\frac {8 x}{a^2}+\frac {2 x^2}{a}+\frac {x^3}{3} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(4*ArcCoth[a*x])*x^2,x]

[Out]

(8*x)/a^2 + (2*x^2)/a + x^3/3 + 4/(a^3*(1 - a*x)) + (12*Log[1 - a*x])/a^3

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fricas [A]  time = 0.55, size = 57, normalized size = 1.21 \[ \frac {a^{4} x^{4} + 5 \, a^{3} x^{3} + 18 \, a^{2} x^{2} - 24 \, a x + 36 \, {\left (a x - 1\right )} \log \left (a x - 1\right ) - 12}{3 \, {\left (a^{4} x - a^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(a^4*x^4 + 5*a^3*x^3 + 18*a^2*x^2 - 24*a*x + 36*(a*x - 1)*log(a*x - 1) - 12)/(a^4*x - a^3)

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giac [A]  time = 0.14, size = 69, normalized size = 1.47 \[ \frac {{\left (a x - 1\right )}^{3} {\left (\frac {9}{a x - 1} + \frac {39}{{\left (a x - 1\right )}^{2}} + 1\right )}}{3 \, a^{3}} - \frac {12 \, \log \left (\frac {{\left | a x - 1 \right |}}{{\left (a x - 1\right )}^{2} {\left | a \right |}}\right )}{a^{3}} - \frac {4}{{\left (a x - 1\right )} a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(a*x - 1)^3*(9/(a*x - 1) + 39/(a*x - 1)^2 + 1)/a^3 - 12*log(abs(a*x - 1)/((a*x - 1)^2*abs(a)))/a^3 - 4/((a
*x - 1)*a^3)

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maple [A]  time = 0.04, size = 44, normalized size = 0.94 \[ \frac {x^{3}}{3}+\frac {2 x^{2}}{a}+\frac {8 x}{a^{2}}+\frac {12 \ln \left (a x -1\right )}{a^{3}}-\frac {4}{a^{3} \left (a x -1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.31, size = 49, normalized size = 1.04 \[ -\frac {4}{a^{4} x - a^{3}} + \frac {a^{2} x^{3} + 6 \, a x^{2} + 24 \, x}{3 \, a^{2}} + \frac {12 \, \log \left (a x - 1\right )}{a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/(a^4*x - a^3) + 1/3*(a^2*x^3 + 6*a*x^2 + 24*x)/a^2 + 12*log(a*x - 1)/a^3

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mupad [B]  time = 1.17, size = 49, normalized size = 1.04 \[ \frac {12\,\ln \left (a\,x-1\right )}{a^3}-\frac {4}{a\,\left (a^3\,x-a^2\right )}+\frac {8\,x}{a^2}+\frac {x^3}{3}+\frac {2\,x^2}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*(a*x + 1)^2)/(a*x - 1)^2,x)

[Out]

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

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sympy [A]  time = 0.15, size = 39, normalized size = 0.83 \[ \frac {x^{3}}{3} - \frac {4}{a^{4} x - a^{3}} + \frac {2 x^{2}}{a} + \frac {8 x}{a^{2}} + \frac {12 \log {\left (a x - 1 \right )}}{a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**3/3 - 4/(a**4*x - a**3) + 2*x**2/a + 8*x/a**2 + 12*log(a*x - 1)/a**3

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