3.1.12 \(\int e^{2 \tanh ^{-1}(a x)} x^2 \, dx\) [12]

Optimal. Leaf size=34 \[ -\frac {2 x}{a^2}-\frac {x^2}{a}-\frac {x^3}{3}-\frac {2 \log (1-a x)}{a^3} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.02, antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6261, 78} \begin {gather*} -\frac {2 \log (1-a x)}{a^3}-\frac {2 x}{a^2}-\frac {x^2}{a}-\frac {x^3}{3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*x)/a^2 - x^2/a - x^3/3 - (2*Log[1 - a*x])/a^3

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

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]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]
time = 0.01, size = 34, normalized size = 1.00 \begin {gather*} -\frac {2 x}{a^2}-\frac {x^2}{a}-\frac {x^3}{3}-\frac {2 \log (1-a x)}{a^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*x)/a^2 - x^2/a - x^3/3 - (2*Log[1 - a*x])/a^3

________________________________________________________________________________________

Maple [A]
time = 0.76, size = 35, normalized size = 1.03

method	result	size
norman	\(-\frac {x^{3}}{3}-\frac {2 x}{a^{2}}-\frac {x^{2}}{a}-\frac {2 \ln \left (a x -1\right )}{a^{3}}\)	\(32\)
risch	\(-\frac {x^{3}}{3}-\frac {2 x}{a^{2}}-\frac {x^{2}}{a}-\frac {2 \ln \left (a x -1\right )}{a^{3}}\)	\(32\)
default	\(-\frac {\frac {1}{3} a^{2} x^{3}+a \,x^{2}+2 x}{a^{2}}-\frac {2 \ln \left (a x -1\right )}{a^{3}}\)	\(35\)
meijerg	\(\frac {-\frac {2 x \left (-a^{2}\right )^{\frac {5}{2}} \left (5 a^{2} x^{2}+15\right )}{15 a^{4}}+\frac {2 \left (-a^{2}\right )^{\frac {5}{2}} \arctanh \left (a x \right )}{a^{5}}}{2 a^{2} \sqrt {-a^{2}}}+\frac {-a^{2} x^{2}-\ln \left (-a^{2} x^{2}+1\right )}{a^{3}}-\frac {-\frac {2 x \left (-a^{2}\right )^{\frac {3}{2}}}{a^{2}}+\frac {2 \left (-a^{2}\right )^{\frac {3}{2}} \arctanh \left (a x \right )}{a^{3}}}{2 a^{2} \sqrt {-a^{2}}}\)	\(122\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]
time = 0.26, size = 34, normalized size = 1.00 \begin {gather*} -\frac {a^{2} x^{3} + 3 \, a x^{2} + 6 \, x}{3 \, a^{2}} - \frac {2 \, \log \left (a x - 1\right )}{a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]
time = 0.35, size = 33, normalized size = 0.97 \begin {gather*} -\frac {a^{3} x^{3} + 3 \, a^{2} x^{2} + 6 \, a x + 6 \, \log \left (a x - 1\right )}{3 \, a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [A]
time = 0.04, size = 29, normalized size = 0.85 \begin {gather*} - \frac {x^{3}}{3} - \frac {x^{2}}{a} - \frac {2 x}{a^{2}} - \frac {2 \log {\left (a x - 1 \right )}}{a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x**3/3 - x**2/a - 2*x/a**2 - 2*log(a*x - 1)/a**3

________________________________________________________________________________________

Giac [A]
time = 0.41, size = 38, normalized size = 1.12 \begin {gather*} -\frac {a^{3} x^{3} + 3 \, a^{2} x^{2} + 6 \, a x}{3 \, a^{3}} - \frac {2 \, \log \left ({\left | a x - 1 \right |}\right )}{a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Mupad [B]
time = 0.04, size = 31, normalized size = 0.91 \begin {gather*} -\frac {2\,\ln \left (a\,x-1\right )}{a^3}-\frac {2\,x}{a^2}-\frac {x^3}{3}-\frac {x^2}{a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- (2*log(a*x - 1))/a^3 - (2*x)/a^2 - x^3/3 - x^2/a

________________________________________________________________________________________