3.2.76 \(\int (1-a^2 x^2) \tanh ^{-1}(a x)^2 \, dx\) [176]

Optimal. Leaf size=115 \[ -\frac {x}{3}+\frac {\left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}{3 a}+\frac {2 \tanh ^{-1}(a x)^2}{3 a}+\frac {2}{3} x \tanh ^{-1}(a x)^2+\frac {1}{3} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2-\frac {4 \tanh ^{-1}(a x) \log \left (\frac {2}{1-a x}\right )}{3 a}-\frac {2 \text {PolyLog}\left (2,1-\frac {2}{1-a x}\right )}{3 a} \]

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.412, Rules used = {6091, 6021, 6131, 6055, 2449, 2352, 8} \begin {gather*} \frac {1}{3} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2+\frac {\left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}{3 a}-\frac {2 \text {Li}_2\left (1-\frac {2}{1-a x}\right )}{3 a}+\frac {2}{3} x \tanh ^{-1}(a x)^2+\frac {2 \tanh ^{-1}(a x)^2}{3 a}-\frac {4 \log \left (\frac {2}{1-a x}\right ) \tanh ^{-1}(a x)}{3 a}-\frac {x}{3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 - a^2*x^2)*ArcTanh[a*x]^2,x]

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2352

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-e^(-1))*PolyLog[2, 1 - c*x], x] /; FreeQ[{c, d, e
}, x] && EqQ[e + c*d, 0]

Rule 2449

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> Dist[-e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]

Rule 6021

Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcTanh[c*x^n])^p, x] - Dist[b
*c*n*p, Int[x^n*((a + b*ArcTanh[c*x^n])^(p - 1)/(1 - c^2*x^(2*n))), x], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[p
, 0] && (EqQ[n, 1] || EqQ[p, 1])

Rule 6055

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

Rule 6091

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

Rule 6131

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

Rubi steps

\begin {align*} \int \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2 \, dx &=\frac {\left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}{3 a}+\frac {1}{3} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2-\frac {\int 1 \, dx}{3}+\frac {2}{3} \int \tanh ^{-1}(a x)^2 \, dx\\ &=-\frac {x}{3}+\frac {\left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}{3 a}+\frac {2}{3} x \tanh ^{-1}(a x)^2+\frac {1}{3} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2-\frac {1}{3} (4 a) \int \frac {x \tanh ^{-1}(a x)}{1-a^2 x^2} \, dx\\ &=-\frac {x}{3}+\frac {\left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}{3 a}+\frac {2 \tanh ^{-1}(a x)^2}{3 a}+\frac {2}{3} x \tanh ^{-1}(a x)^2+\frac {1}{3} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2-\frac {4}{3} \int \frac {\tanh ^{-1}(a x)}{1-a x} \, dx\\ &=-\frac {x}{3}+\frac {\left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}{3 a}+\frac {2 \tanh ^{-1}(a x)^2}{3 a}+\frac {2}{3} x \tanh ^{-1}(a x)^2+\frac {1}{3} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2-\frac {4 \tanh ^{-1}(a x) \log \left (\frac {2}{1-a x}\right )}{3 a}+\frac {4}{3} \int \frac {\log \left (\frac {2}{1-a x}\right )}{1-a^2 x^2} \, dx\\ &=-\frac {x}{3}+\frac {\left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}{3 a}+\frac {2 \tanh ^{-1}(a x)^2}{3 a}+\frac {2}{3} x \tanh ^{-1}(a x)^2+\frac {1}{3} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2-\frac {4 \tanh ^{-1}(a x) \log \left (\frac {2}{1-a x}\right )}{3 a}-\frac {4 \text {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1-a x}\right )}{3 a}\\ &=-\frac {x}{3}+\frac {\left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}{3 a}+\frac {2 \tanh ^{-1}(a x)^2}{3 a}+\frac {2}{3} x \tanh ^{-1}(a x)^2+\frac {1}{3} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2-\frac {4 \tanh ^{-1}(a x) \log \left (\frac {2}{1-a x}\right )}{3 a}-\frac {2 \text {Li}_2\left (1-\frac {2}{1-a x}\right )}{3 a}\\ \end {align*}

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Mathematica [A]
time = 0.09, size = 71, normalized size = 0.62 \begin {gather*} -\frac {a x+(-1+a x)^2 (2+a x) \tanh ^{-1}(a x)^2+\tanh ^{-1}(a x) \left (-1+a^2 x^2+4 \log \left (1+e^{-2 \tanh ^{-1}(a x)}\right )\right )-2 \text {PolyLog}\left (2,-e^{-2 \tanh ^{-1}(a x)}\right )}{3 a} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.55, size = 154, normalized size = 1.34

method	result	size
derivativedivides	\(\frac {-\frac {\arctanh \left (a x \right )^{2} a^{3} x^{3}}{3}+\arctanh \left (a x \right )^{2} a x -\frac {a^{2} x^{2} \arctanh \left (a x \right )}{3}+\frac {2 \arctanh \left (a x \right ) \ln \left (a x -1\right )}{3}+\frac {2 \arctanh \left (a x \right ) \ln \left (a x +1\right )}{3}-\frac {a x}{3}-\frac {\ln \left (a x -1\right )}{6}+\frac {\ln \left (a x +1\right )}{6}-\frac {2 \dilog \left (\frac {a x}{2}+\frac {1}{2}\right )}{3}-\frac {\ln \left (a x -1\right ) \ln \left (\frac {a x}{2}+\frac {1}{2}\right )}{3}+\frac {\ln \left (a x -1\right )^{2}}{6}+\frac {\left (\ln \left (a x +1\right )-\ln \left (\frac {a x}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {a x}{2}+\frac {1}{2}\right )}{3}-\frac {\ln \left (a x +1\right )^{2}}{6}}{a}\)	\(154\)
default	\(\frac {-\frac {\arctanh \left (a x \right )^{2} a^{3} x^{3}}{3}+\arctanh \left (a x \right )^{2} a x -\frac {a^{2} x^{2} \arctanh \left (a x \right )}{3}+\frac {2 \arctanh \left (a x \right ) \ln \left (a x -1\right )}{3}+\frac {2 \arctanh \left (a x \right ) \ln \left (a x +1\right )}{3}-\frac {a x}{3}-\frac {\ln \left (a x -1\right )}{6}+\frac {\ln \left (a x +1\right )}{6}-\frac {2 \dilog \left (\frac {a x}{2}+\frac {1}{2}\right )}{3}-\frac {\ln \left (a x -1\right ) \ln \left (\frac {a x}{2}+\frac {1}{2}\right )}{3}+\frac {\ln \left (a x -1\right )^{2}}{6}+\frac {\left (\ln \left (a x +1\right )-\ln \left (\frac {a x}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {a x}{2}+\frac {1}{2}\right )}{3}-\frac {\ln \left (a x +1\right )^{2}}{6}}{a}\)	\(154\)
risch	\(-\frac {x}{3}-\frac {2 \dilog \left (\frac {a x}{2}+\frac {1}{2}\right )}{3 a}-\frac {5 \ln \left (a x -1\right )}{9 a}+\frac {a^{2} \ln \left (-a x +1\right ) \ln \left (a x +1\right ) x^{3}}{6}-\frac {\left (-1+\ln \left (a x +1\right )\right ) \left (a x +1\right ) \ln \left (-a x +1\right )}{2 a}-\frac {x \ln \left (-a x +1\right )}{2}-\frac {a^{2} \ln \left (a x +1\right )^{2} x^{3}}{12}-\frac {a \ln \left (a x +1\right ) x^{2}}{6}-\frac {37}{54 a}+\frac {\ln \left (-a x +1\right )^{2} x}{4}-\frac {\ln \left (-a x +1\right )^{2}}{6 a}-\frac {\ln \left (-a x +1\right )}{9 a}+\frac {\ln \left (a x +1\right )^{2} x}{4}+\frac {\ln \left (a x +1\right )^{2}}{6 a}-\frac {\ln \left (a x +1\right ) x}{2}-\frac {\ln \left (a x +1\right )}{3 a}-\frac {a^{2} \ln \left (-a x +1\right )^{2} x^{3}}{12}+\frac {a \ln \left (-a x +1\right ) x^{2}}{6}+\frac {\ln \left (-a x +1\right ) \ln \left (a x +1\right )}{6 a}-\frac {\ln \left (-\frac {a x}{2}+\frac {1}{2}\right ) \ln \left (a x +1\right )}{3 a}+\frac {\ln \left (-\frac {a x}{2}+\frac {1}{2}\right ) \ln \left (\frac {a x}{2}+\frac {1}{2}\right )}{3 a}+\frac {\left (a x +1\right ) \ln \left (a x +1\right )}{2 a}+\frac {\left (\ln \left (a x +1\right )-\ln \left (\frac {a x}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {a x}{2}+\frac {1}{2}\right )}{a}\)	\(327\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/a*(-1/3*arctanh(a*x)^2*a^3*x^3+arctanh(a*x)^2*a*x-1/3*a^2*x^2*arctanh(a*x)+2/3*arctanh(a*x)*ln(a*x-1)+2/3*ar
ctanh(a*x)*ln(a*x+1)-1/3*a*x-1/6*ln(a*x-1)+1/6*ln(a*x+1)-2/3*dilog(1/2*a*x+1/2)-1/3*ln(a*x-1)*ln(1/2*a*x+1/2)+
1/6*ln(a*x-1)^2+1/3*(ln(a*x+1)-ln(1/2*a*x+1/2))*ln(-1/2*a*x+1/2)-1/6*ln(a*x+1)^2)

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Maxima [A]
time = 0.26, size = 144, normalized size = 1.25 \begin {gather*} -\frac {1}{6} \, a^{2} {\left (\frac {2 \, a x + \log \left (a x + 1\right )^{2} - 2 \, \log \left (a x + 1\right ) \log \left (a x - 1\right ) - \log \left (a x - 1\right )^{2} + \log \left (a x - 1\right )}{a^{3}} + \frac {4 \, {\left (\log \left (a x - 1\right ) \log \left (\frac {1}{2} \, a x + \frac {1}{2}\right ) + {\rm Li}_2\left (-\frac {1}{2} \, a x + \frac {1}{2}\right )\right )}}{a^{3}} - \frac {\log \left (a x + 1\right )}{a^{3}}\right )} - \frac {1}{3} \, {\left (x^{2} - \frac {2 \, \log \left (a x + 1\right )}{a^{2}} - \frac {2 \, \log \left (a x - 1\right )}{a^{2}}\right )} a \operatorname {artanh}\left (a x\right ) - \frac {1}{3} \, {\left (a^{2} x^{3} - 3 \, x\right )} \operatorname {artanh}\left (a x\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*a^2*((2*a*x + log(a*x + 1)^2 - 2*log(a*x + 1)*log(a*x - 1) - log(a*x - 1)^2 + log(a*x - 1))/a^3 + 4*(log(
a*x - 1)*log(1/2*a*x + 1/2) + dilog(-1/2*a*x + 1/2))/a^3 - log(a*x + 1)/a^3) - 1/3*(x^2 - 2*log(a*x + 1)/a^2 -
2*log(a*x - 1)/a^2)*a*arctanh(a*x) - 1/3*(a^2*x^3 - 3*x)*arctanh(a*x)^2

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-(a^2*x^2 - 1)*arctanh(a*x)^2, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \int a^{2} x^{2} \operatorname {atanh}^{2}{\left (a x \right )}\, dx - \int \left (- \operatorname {atanh}^{2}{\left (a x \right )}\right )\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(a**2*x**2*atanh(a*x)**2, x) - Integral(-atanh(a*x)**2, x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(-(a^2*x^2 - 1)*arctanh(a*x)^2, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} -\int {\mathrm {atanh}\left (a\,x\right )}^2\,\left (a^2\,x^2-1\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-atanh(a*x)^2*(a^2*x^2 - 1),x)

[Out]

-int(atanh(a*x)^2*(a^2*x^2 - 1), x)

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