\(\int e^{4 \coth ^{-1}(a x)} x^2 \, dx\) [26]

Optimal result

Integrand size = 12, antiderivative size = 47 \[ \int e^{4 \coth ^{-1}(a x)} x^2 \, dx=\frac {8 x}{a^2}+\frac {2 x^2}{a}+\frac {x^3}{3}+\frac {4}{a^3 (1-a x)}+\frac {12 \log (1-a x)}{a^3} \]

[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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6302, 6261, 90} \[ \int e^{4 \coth ^{-1}(a x)} x^2 \, dx=\frac {4}{a^3 (1-a x)}+\frac {12 \log (1-a x)}{a^3}+\frac {8 x}{a^2}+\frac {2 x^2}{a}+\frac {x^3}{3} \]

[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 90

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 6261

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 6302

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*} \text {integral}& = \int e^{4 \text {arctanh}(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*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00 \[ \int e^{4 \coth ^{-1}(a x)} x^2 \, dx=\frac {8 x}{a^2}+\frac {2 x^2}{a}+\frac {x^3}{3}+\frac {4}{a^3 (1-a x)}+\frac {12 \log (1-a x)}{a^3} \]

[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

Maple [A] (verified)

Time = 0.44 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.94

method	result	size
risch	\(\frac {x^{3}}{3}+\frac {2 x^{2}}{a}+\frac {8 x}{a^{2}}-\frac {4}{a^{3} \left (a x -1\right )}+\frac {12 \ln \left (a x -1\right )}{a^{3}}\)	\(44\)
norman	\(\frac {\frac {5 x^{3}}{3}+\frac {a \,x^{4}}{3}+\frac {6 x^{2}}{a}-\frac {12}{a^{3}}}{a x -1}+\frac {12 \ln \left (a x -1\right )}{a^{3}}\)	\(46\)
default	\(\frac {\frac {1}{3} a^{2} x^{3}+2 a \,x^{2}+8 x}{a^{2}}-\frac {4}{a^{3} \left (a x -1\right )}+\frac {12 \ln \left (a x -1\right )}{a^{3}}\)	\(47\)
parallelrisch	\(\frac {a^{4} x^{4}+5 a^{3} x^{3}-36+18 a^{2} x^{2}+36 a \ln \left (a x -1\right ) x -36 \ln \left (a x -1\right )}{3 a^{3} \left (a x -1\right )}\)	\(56\)
meijerg	\(-\frac {-\frac {a x \left (-5 a^{3} x^{3}-10 a^{2} x^{2}-30 a x +60\right )}{15 \left (-a x +1\right )}-4 \ln \left (-a x +1\right )}{a^{3}}+\frac {\frac {2 a x \left (-2 a^{2} x^{2}-6 a x +12\right )}{-4 a x +4}+6 \ln \left (-a x +1\right )}{a^{3}}-\frac {-\frac {a x \left (-3 a x +6\right )}{3 \left (-a x +1\right )}-2 \ln \left (-a x +1\right )}{a^{3}}\)	\(125\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.21 \[ \int e^{4 \coth ^{-1}(a x)} x^2 \, dx=\frac {a^{4} x^{4} + 5 \, a^{3} x^{3} + 18 \, a^{2} x^{2} - 24 \, a x + 36 \, {\left (a x - 1\right )} \log \left (a x - 1\right ) - 12}{3 \, {\left (a^{4} x - a^{3}\right )}} \]

[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)

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.83 \[ \int e^{4 \coth ^{-1}(a x)} x^2 \, dx=\frac {x^{3}}{3} - \frac {4}{a^{4} x - a^{3}} + \frac {2 x^{2}}{a} + \frac {8 x}{a^{2}} + \frac {12 \log {\left (a x - 1 \right )}}{a^{3}} \]

[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

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.04 \[ \int e^{4 \coth ^{-1}(a x)} x^2 \, dx=-\frac {4}{a^{4} x - a^{3}} + \frac {a^{2} x^{3} + 6 \, a x^{2} + 24 \, x}{3 \, a^{2}} + \frac {12 \, \log \left (a x - 1\right )}{a^{3}} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.47 \[ \int e^{4 \coth ^{-1}(a x)} x^2 \, dx=\frac {{\left (a x - 1\right )}^{3} {\left (\frac {9}{a x - 1} + \frac {39}{{\left (a x - 1\right )}^{2}} + 1\right )}}{3 \, a^{3}} - \frac {12 \, \log \left (\frac {{\left | a x - 1 \right |}}{{\left (a x - 1\right )}^{2} {\left | a \right |}}\right )}{a^{3}} - \frac {4}{{\left (a x - 1\right )} a^{3}} \]

[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)

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.04 \[ \int e^{4 \coth ^{-1}(a x)} x^2 \, dx=\frac {12\,\ln \left (a\,x-1\right )}{a^3}-\frac {4}{a\,\left (a^3\,x-a^2\right )}+\frac {8\,x}{a^2}+\frac {x^3}{3}+\frac {2\,x^2}{a} \]

[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