3.4.4 \(\int \frac {x \tanh ^{-1}(a x)}{(1-a^2 x^2)^3} \, dx\) [304]

Optimal. Leaf size=75 \[ -\frac {x}{16 a \left (1-a^2 x^2\right )^2}-\frac {3 x}{32 a \left (1-a^2 x^2\right )}-\frac {3 \tanh ^{-1}(a x)}{32 a^2}+\frac {\tanh ^{-1}(a x)}{4 a^2 \left (1-a^2 x^2\right )^2} \]

[Out]

-1/16*x/a/(-a^2*x^2+1)^2-3/32*x/a/(-a^2*x^2+1)-3/32*arctanh(a*x)/a^2+1/4*arctanh(a*x)/a^2/(-a^2*x^2+1)^2

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Rubi [A]
time = 0.03, antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6141, 205, 212} \begin {gather*} -\frac {3 x}{32 a \left (1-a^2 x^2\right )}-\frac {x}{16 a \left (1-a^2 x^2\right )^2}+\frac {\tanh ^{-1}(a x)}{4 a^2 \left (1-a^2 x^2\right )^2}-\frac {3 \tanh ^{-1}(a x)}{32 a^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/16*x/(a*(1 - a^2*x^2)^2) - (3*x)/(32*a*(1 - a^2*x^2)) - (3*ArcTanh[a*x])/(32*a^2) + ArcTanh[a*x]/(4*a^2*(1
- a^2*x^2)^2)

Rule 205

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 1)/(a*n*(p + 1))), x] + Dist[(n*(p
+ 1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (
IntegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[
p])

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 6141

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*(x_)*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> Simp[(d + e*x^2)^
(q + 1)*((a + b*ArcTanh[c*x])^p/(2*e*(q + 1))), x] + Dist[b*(p/(2*c*(q + 1))), Int[(d + e*x^2)^q*(a + b*ArcTan
h[c*x])^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, q}, x] && EqQ[c^2*d + e, 0] && GtQ[p, 0] && NeQ[q, -1]

Rubi steps

\begin {align*} \int \frac {x \tanh ^{-1}(a x)}{\left (1-a^2 x^2\right )^3} \, dx &=\frac {\tanh ^{-1}(a x)}{4 a^2 \left (1-a^2 x^2\right )^2}-\frac {\int \frac {1}{\left (1-a^2 x^2\right )^3} \, dx}{4 a}\\ &=-\frac {x}{16 a \left (1-a^2 x^2\right )^2}+\frac {\tanh ^{-1}(a x)}{4 a^2 \left (1-a^2 x^2\right )^2}-\frac {3 \int \frac {1}{\left (1-a^2 x^2\right )^2} \, dx}{16 a}\\ &=-\frac {x}{16 a \left (1-a^2 x^2\right )^2}-\frac {3 x}{32 a \left (1-a^2 x^2\right )}+\frac {\tanh ^{-1}(a x)}{4 a^2 \left (1-a^2 x^2\right )^2}-\frac {3 \int \frac {1}{1-a^2 x^2} \, dx}{32 a}\\ &=-\frac {x}{16 a \left (1-a^2 x^2\right )^2}-\frac {3 x}{32 a \left (1-a^2 x^2\right )}-\frac {3 \tanh ^{-1}(a x)}{32 a^2}+\frac {\tanh ^{-1}(a x)}{4 a^2 \left (1-a^2 x^2\right )^2}\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 88, normalized size = 1.17 \begin {gather*} -\frac {x}{16 a \left (-1+a^2 x^2\right )^2}+\frac {3 x}{32 a \left (-1+a^2 x^2\right )}+\frac {\tanh ^{-1}(a x)}{4 a^2 \left (-1+a^2 x^2\right )^2}+\frac {3 \log (1-a x)}{64 a^2}-\frac {3 \log (1+a x)}{64 a^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/16*x/(a*(-1 + a^2*x^2)^2) + (3*x)/(32*a*(-1 + a^2*x^2)) + ArcTanh[a*x]/(4*a^2*(-1 + a^2*x^2)^2) + (3*Log[1
- a*x])/(64*a^2) - (3*Log[1 + a*x])/(64*a^2)

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Maple [A]
time = 0.58, size = 75, normalized size = 1.00

method	result	size
derivativedivides	\(\frac {\frac {\arctanh \left (a x \right )}{4 \left (a^{2} x^{2}-1\right )^{2}}-\frac {1}{64 \left (a x -1\right )^{2}}+\frac {3}{64 \left (a x -1\right )}+\frac {3 \ln \left (a x -1\right )}{64}+\frac {1}{64 \left (a x +1\right )^{2}}+\frac {3}{64 \left (a x +1\right )}-\frac {3 \ln \left (a x +1\right )}{64}}{a^{2}}\)	\(75\)
default	\(\frac {\frac {\arctanh \left (a x \right )}{4 \left (a^{2} x^{2}-1\right )^{2}}-\frac {1}{64 \left (a x -1\right )^{2}}+\frac {3}{64 \left (a x -1\right )}+\frac {3 \ln \left (a x -1\right )}{64}+\frac {1}{64 \left (a x +1\right )^{2}}+\frac {3}{64 \left (a x +1\right )}-\frac {3 \ln \left (a x +1\right )}{64}}{a^{2}}\)	\(75\)
risch	\(\frac {\ln \left (a x +1\right )}{8 a^{2} \left (a^{2} x^{2}-1\right )^{2}}+\frac {3 x^{4} \ln \left (-a x +1\right ) a^{4}-3 \ln \left (a x +1\right ) a^{4} x^{4}+6 a^{3} x^{3}-6 x^{2} \ln \left (-a x +1\right ) a^{2}+6 a^{2} x^{2} \ln \left (a x +1\right )-10 a x -5 \ln \left (-a x +1\right )-3 \ln \left (a x +1\right )}{64 a^{2} \left (a x +1\right ) \left (a x -1\right ) \left (a^{2} x^{2}-1\right )}\)	\(142\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/a^2*(1/4/(a^2*x^2-1)^2*arctanh(a*x)-1/64/(a*x-1)^2+3/64/(a*x-1)+3/64*ln(a*x-1)+1/64/(a*x+1)^2+3/64/(a*x+1)-3
/64*ln(a*x+1))

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Maxima [A]
time = 0.27, size = 82, normalized size = 1.09 \begin {gather*} \frac {\frac {2 \, {\left (3 \, a^{2} x^{3} - 5 \, x\right )}}{a^{4} x^{4} - 2 \, a^{2} x^{2} + 1} - \frac {3 \, \log \left (a x + 1\right )}{a} + \frac {3 \, \log \left (a x - 1\right )}{a}}{64 \, a} + \frac {\operatorname {artanh}\left (a x\right )}{4 \, {\left (a^{2} x^{2} - 1\right )}^{2} a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*(2*(3*a^2*x^3 - 5*x)/(a^4*x^4 - 2*a^2*x^2 + 1) - 3*log(a*x + 1)/a + 3*log(a*x - 1)/a)/a + 1/4*arctanh(a*x
)/((a^2*x^2 - 1)^2*a^2)

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Fricas [A]
time = 0.34, size = 71, normalized size = 0.95 \begin {gather*} \frac {6 \, a^{3} x^{3} - 10 \, a x - {\left (3 \, a^{4} x^{4} - 6 \, a^{2} x^{2} - 5\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )}{64 \, {\left (a^{6} x^{4} - 2 \, a^{4} x^{2} + a^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*(6*a^3*x^3 - 10*a*x - (3*a^4*x^4 - 6*a^2*x^2 - 5)*log(-(a*x + 1)/(a*x - 1)))/(a^6*x^4 - 2*a^4*x^2 + a^2)

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 158 vs. \(2 (61) = 122\).
time = 0.85, size = 158, normalized size = 2.11 \begin {gather*} \begin {cases} - \frac {3 a^{4} x^{4} \operatorname {atanh}{\left (a x \right )}}{32 a^{6} x^{4} - 64 a^{4} x^{2} + 32 a^{2}} + \frac {3 a^{3} x^{3}}{32 a^{6} x^{4} - 64 a^{4} x^{2} + 32 a^{2}} + \frac {6 a^{2} x^{2} \operatorname {atanh}{\left (a x \right )}}{32 a^{6} x^{4} - 64 a^{4} x^{2} + 32 a^{2}} - \frac {5 a x}{32 a^{6} x^{4} - 64 a^{4} x^{2} + 32 a^{2}} + \frac {5 \operatorname {atanh}{\left (a x \right )}}{32 a^{6} x^{4} - 64 a^{4} x^{2} + 32 a^{2}} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-3*a**4*x**4*atanh(a*x)/(32*a**6*x**4 - 64*a**4*x**2 + 32*a**2) + 3*a**3*x**3/(32*a**6*x**4 - 64*a*
*4*x**2 + 32*a**2) + 6*a**2*x**2*atanh(a*x)/(32*a**6*x**4 - 64*a**4*x**2 + 32*a**2) - 5*a*x/(32*a**6*x**4 - 64
*a**4*x**2 + 32*a**2) + 5*atanh(a*x)/(32*a**6*x**4 - 64*a**4*x**2 + 32*a**2), Ne(a, 0)), (0, True))

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 239 vs. \(2 (64) = 128\).
time = 0.40, size = 239, normalized size = 3.19 \begin {gather*} -\frac {1}{256} \, {\left (2 \, {\left (\frac {{\left (a x - 1\right )}^{2} {\left (\frac {4 \, {\left (a x + 1\right )}}{a x - 1} - 1\right )}}{{\left (a x + 1\right )}^{2} a^{3}} - \frac {\frac {{\left (a x + 1\right )}^{2} a^{3}}{{\left (a x - 1\right )}^{2}} - \frac {4 \, {\left (a x + 1\right )} a^{3}}{a x - 1}}{a^{6}}\right )} \log \left (-\frac {\frac {a {\left (\frac {a x + 1}{a x - 1} + 1\right )}}{\frac {{\left (a x + 1\right )} a}{a x - 1} - a} + 1}{\frac {a {\left (\frac {a x + 1}{a x - 1} + 1\right )}}{\frac {{\left (a x + 1\right )} a}{a x - 1} - a} - 1}\right ) + \frac {{\left (a x - 1\right )}^{2} {\left (\frac {8 \, {\left (a x + 1\right )}}{a x - 1} - 1\right )}}{{\left (a x + 1\right )}^{2} a^{3}} + \frac {\frac {{\left (a x + 1\right )}^{2} a^{3}}{{\left (a x - 1\right )}^{2}} - \frac {8 \, {\left (a x + 1\right )} a^{3}}{a x - 1}}{a^{6}}\right )} a \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/256*(2*((a*x - 1)^2*(4*(a*x + 1)/(a*x - 1) - 1)/((a*x + 1)^2*a^3) - ((a*x + 1)^2*a^3/(a*x - 1)^2 - 4*(a*x +
1)*a^3/(a*x - 1))/a^6)*log(-(a*((a*x + 1)/(a*x - 1) + 1)/((a*x + 1)*a/(a*x - 1) - a) + 1)/(a*((a*x + 1)/(a*x
- 1) + 1)/((a*x + 1)*a/(a*x - 1) - a) - 1)) + (a*x - 1)^2*(8*(a*x + 1)/(a*x - 1) - 1)/((a*x + 1)^2*a^3) + ((a*
x + 1)^2*a^3/(a*x - 1)^2 - 8*(a*x + 1)*a^3/(a*x - 1))/a^6)*a

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Mupad [B]
time = 1.08, size = 105, normalized size = 1.40 \begin {gather*} \frac {\frac {3\,\ln \left (a\,x-1\right )}{64}-\frac {3\,\ln \left (a\,x+1\right )}{64}}{a^2}+\frac {\frac {\mathrm {atanh}\left (a\,x\right )}{4}-x^2\,\left (a^2\,\left (\frac {3\,\ln \left (a\,x-1\right )}{32}-\frac {3\,\ln \left (a\,x+1\right )}{32}\right )-2\,a^2\,\left (\frac {3\,\ln \left (a\,x-1\right )}{64}-\frac {3\,\ln \left (a\,x+1\right )}{64}\right )\right )-\frac {5\,a\,x}{32}+\frac {3\,a^3\,x^3}{32}}{a^2\,{\left (a^2\,x^2-1\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((3*log(a*x - 1))/64 - (3*log(a*x + 1))/64)/a^2 + (atanh(a*x)/4 - x^2*(a^2*((3*log(a*x - 1))/32 - (3*log(a*x +
1))/32) - 2*a^2*((3*log(a*x - 1))/64 - (3*log(a*x + 1))/64)) - (5*a*x)/32 + (3*a^3*x^3)/32)/(a^2*(a^2*x^2 - 1
)^2)

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