3.10.31 \(\int \frac {e^{\tanh ^{-1}(a x)} x^2}{(1-a^2 x^2)^{3/2}} \, dx\) [931]

Optimal. Leaf size=43 \[ \frac {1}{2 a^3 (1-a x)}+\frac {3 \log (1-a x)}{4 a^3}+\frac {\log (1+a x)}{4 a^3} \]

[Out]

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

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Rubi [A]
time = 0.08, antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {6285, 90} \begin {gather*} \frac {1}{2 a^3 (1-a x)}+\frac {3 \log (1-a x)}{4 a^3}+\frac {\log (a x+1)}{4 a^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^ArcTanh[a*x]*x^2)/(1 - a^2*x^2)^(3/2),x]

[Out]

1/(2*a^3*(1 - a*x)) + (3*Log[1 - a*x])/(4*a^3) + Log[1 + a*x]/(4*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 6285

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[c^p, Int[x^m*(1 -
a*x)^(p - n/2)*(1 + a*x)^(p + n/2), x], x] /; FreeQ[{a, c, d, m, n, p}, x] && EqQ[a^2*c + d, 0] && (IntegerQ[p
] || GtQ[c, 0])

Rubi steps

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

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Mathematica [A]
time = 0.04, size = 33, normalized size = 0.77 \begin {gather*} \frac {\frac {2}{1-a x}+3 \log (1-a x)+\log (1+a x)}{4 a^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^ArcTanh[a*x]*x^2)/(1 - a^2*x^2)^(3/2),x]

[Out]

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

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Maple [A]
time = 0.05, size = 36, normalized size = 0.84

method	result	size
default	\(\frac {\ln \left (a x +1\right )}{4 a^{3}}-\frac {1}{2 a^{3} \left (a x -1\right )}+\frac {3 \ln \left (a x -1\right )}{4 a^{3}}\)	\(36\)
risch	\(-\frac {1}{2 a^{3} \left (a x -1\right )}+\frac {\ln \left (a x +1\right )}{4 a^{3}}+\frac {3 \ln \left (-a x +1\right )}{4 a^{3}}\)	\(37\)
norman	\(\frac {-\frac {x}{2 a^{2}}-\frac {x^{2}}{2 a}}{a^{2} x^{2}-1}+\frac {3 \ln \left (a x -1\right )}{4 a^{3}}+\frac {\ln \left (a x +1\right )}{4 a^{3}}\)	\(51\)
meijerg	\(\frac {\frac {a^{2} x^{2}}{-a^{2} x^{2}+1}+\ln \left (-a^{2} x^{2}+1\right )}{2 a^{3}}-\frac {\frac {x \left (-a^{2}\right )^{\frac {3}{2}}}{a^{2} \left (-a^{2} x^{2}+1\right )}-\frac {\left (-a^{2}\right )^{\frac {3}{2}} \arctanh \left (a x \right )}{a^{3}}}{2 a^{2} \sqrt {-a^{2}}}\)	\(91\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.25, size = 38, normalized size = 0.88 \begin {gather*} -\frac {1}{2 \, {\left (a^{4} x - a^{3}\right )}} + \frac {\log \left (a x + 1\right )}{4 \, a^{3}} + \frac {3 \, \log \left (a x - 1\right )}{4 \, a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.35, size = 42, normalized size = 0.98 \begin {gather*} \frac {{\left (a x - 1\right )} \log \left (a x + 1\right ) + 3 \, {\left (a x - 1\right )} \log \left (a x - 1\right ) - 2}{4 \, {\left (a^{4} x - a^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.11, size = 34, normalized size = 0.79 \begin {gather*} - \frac {1}{2 a^{4} x - 2 a^{3}} + \frac {\frac {3 \log {\left (x - \frac {1}{a} \right )}}{4} + \frac {\log {\left (x + \frac {1}{a} \right )}}{4}}{a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.43, size = 37, normalized size = 0.86 \begin {gather*} \frac {\log \left ({\left | a x + 1 \right |}\right )}{4 \, a^{3}} + \frac {3 \, \log \left ({\left | a x - 1 \right |}\right )}{4 \, a^{3}} - \frac {1}{2 \, {\left (a x - 1\right )} a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.95, size = 39, normalized size = 0.91 \begin {gather*} \frac {3\,\ln \left (a\,x-1\right )}{4\,a^3}+\frac {\ln \left (a\,x+1\right )}{4\,a^3}-\frac {1}{2\,\left (a^4\,x-a^3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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