3.173 \(\int x^3 (1-a^2 x^2) \tanh ^{-1}(a x)^2 \, dx\)

Optimal. Leaf size=116 \[ -\frac{x^2}{180 a^2}+\frac{7 \log \left (1-a^2 x^2\right )}{90 a^4}-\frac{1}{6} a^2 x^6 \tanh ^{-1}(a x)^2+\frac{x \tanh ^{-1}(a x)}{6 a^3}-\frac{\tanh ^{-1}(a x)^2}{12 a^4}-\frac{1}{15} a x^5 \tanh ^{-1}(a x)+\frac{1}{4} x^4 \tanh ^{-1}(a x)^2+\frac{x^3 \tanh ^{-1}(a x)}{18 a}-\frac{x^4}{60} \]

[Out]

-x^2/(180*a^2) - x^4/60 + (x*ArcTanh[a*x])/(6*a^3) + (x^3*ArcTanh[a*x])/(18*a) - (a*x^5*ArcTanh[a*x])/15 - Arc
Tanh[a*x]^2/(12*a^4) + (x^4*ArcTanh[a*x]^2)/4 - (a^2*x^6*ArcTanh[a*x]^2)/6 + (7*Log[1 - a^2*x^2])/(90*a^4)

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Rubi [A]  time = 0.439878, antiderivative size = 116, normalized size of antiderivative = 1., number of steps used = 26, number of rules used = 8, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {6014, 5916, 5980, 266, 43, 5910, 260, 5948} \[ -\frac{x^2}{180 a^2}+\frac{7 \log \left (1-a^2 x^2\right )}{90 a^4}-\frac{1}{6} a^2 x^6 \tanh ^{-1}(a x)^2+\frac{x \tanh ^{-1}(a x)}{6 a^3}-\frac{\tanh ^{-1}(a x)^2}{12 a^4}-\frac{1}{15} a x^5 \tanh ^{-1}(a x)+\frac{1}{4} x^4 \tanh ^{-1}(a x)^2+\frac{x^3 \tanh ^{-1}(a x)}{18 a}-\frac{x^4}{60} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-x^2/(180*a^2) - x^4/60 + (x*ArcTanh[a*x])/(6*a^3) + (x^3*ArcTanh[a*x])/(18*a) - (a*x^5*ArcTanh[a*x])/15 - Arc
Tanh[a*x]^2/(12*a^4) + (x^4*ArcTanh[a*x]^2)/4 - (a^2*x^6*ArcTanh[a*x]^2)/6 + (7*Log[1 - a^2*x^2])/(90*a^4)

Rule 6014

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

Rule 5916

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

Rule 5980

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

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 5910

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

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 5948

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

Rubi steps

\begin{align*} \int x^3 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2 \, dx &=-\left (a^2 \int x^5 \tanh ^{-1}(a x)^2 \, dx\right )+\int x^3 \tanh ^{-1}(a x)^2 \, dx\\ &=\frac{1}{4} x^4 \tanh ^{-1}(a x)^2-\frac{1}{6} a^2 x^6 \tanh ^{-1}(a x)^2-\frac{1}{2} a \int \frac{x^4 \tanh ^{-1}(a x)}{1-a^2 x^2} \, dx+\frac{1}{3} a^3 \int \frac{x^6 \tanh ^{-1}(a x)}{1-a^2 x^2} \, dx\\ &=\frac{1}{4} x^4 \tanh ^{-1}(a x)^2-\frac{1}{6} a^2 x^6 \tanh ^{-1}(a x)^2+\frac{\int x^2 \tanh ^{-1}(a x) \, dx}{2 a}-\frac{\int \frac{x^2 \tanh ^{-1}(a x)}{1-a^2 x^2} \, dx}{2 a}-\frac{1}{3} a \int x^4 \tanh ^{-1}(a x) \, dx+\frac{1}{3} a \int \frac{x^4 \tanh ^{-1}(a x)}{1-a^2 x^2} \, dx\\ &=\frac{x^3 \tanh ^{-1}(a x)}{6 a}-\frac{1}{15} a x^5 \tanh ^{-1}(a x)+\frac{1}{4} x^4 \tanh ^{-1}(a x)^2-\frac{1}{6} a^2 x^6 \tanh ^{-1}(a x)^2-\frac{1}{6} \int \frac{x^3}{1-a^2 x^2} \, dx+\frac{\int \tanh ^{-1}(a x) \, dx}{2 a^3}-\frac{\int \frac{\tanh ^{-1}(a x)}{1-a^2 x^2} \, dx}{2 a^3}-\frac{\int x^2 \tanh ^{-1}(a x) \, dx}{3 a}+\frac{\int \frac{x^2 \tanh ^{-1}(a x)}{1-a^2 x^2} \, dx}{3 a}+\frac{1}{15} a^2 \int \frac{x^5}{1-a^2 x^2} \, dx\\ &=\frac{x \tanh ^{-1}(a x)}{2 a^3}+\frac{x^3 \tanh ^{-1}(a x)}{18 a}-\frac{1}{15} a x^5 \tanh ^{-1}(a x)-\frac{\tanh ^{-1}(a x)^2}{4 a^4}+\frac{1}{4} x^4 \tanh ^{-1}(a x)^2-\frac{1}{6} a^2 x^6 \tanh ^{-1}(a x)^2-\frac{1}{12} \operatorname{Subst}\left (\int \frac{x}{1-a^2 x} \, dx,x,x^2\right )+\frac{1}{9} \int \frac{x^3}{1-a^2 x^2} \, dx-\frac{\int \tanh ^{-1}(a x) \, dx}{3 a^3}+\frac{\int \frac{\tanh ^{-1}(a x)}{1-a^2 x^2} \, dx}{3 a^3}-\frac{\int \frac{x}{1-a^2 x^2} \, dx}{2 a^2}+\frac{1}{30} a^2 \operatorname{Subst}\left (\int \frac{x^2}{1-a^2 x} \, dx,x,x^2\right )\\ &=\frac{x \tanh ^{-1}(a x)}{6 a^3}+\frac{x^3 \tanh ^{-1}(a x)}{18 a}-\frac{1}{15} a x^5 \tanh ^{-1}(a x)-\frac{\tanh ^{-1}(a x)^2}{12 a^4}+\frac{1}{4} x^4 \tanh ^{-1}(a x)^2-\frac{1}{6} a^2 x^6 \tanh ^{-1}(a x)^2+\frac{\log \left (1-a^2 x^2\right )}{4 a^4}+\frac{1}{18} \operatorname{Subst}\left (\int \frac{x}{1-a^2 x} \, dx,x,x^2\right )-\frac{1}{12} \operatorname{Subst}\left (\int \left (-\frac{1}{a^2}-\frac{1}{a^2 \left (-1+a^2 x\right )}\right ) \, dx,x,x^2\right )+\frac{\int \frac{x}{1-a^2 x^2} \, dx}{3 a^2}+\frac{1}{30} a^2 \operatorname{Subst}\left (\int \left (-\frac{1}{a^4}-\frac{x}{a^2}-\frac{1}{a^4 \left (-1+a^2 x\right )}\right ) \, dx,x,x^2\right )\\ &=\frac{x^2}{20 a^2}-\frac{x^4}{60}+\frac{x \tanh ^{-1}(a x)}{6 a^3}+\frac{x^3 \tanh ^{-1}(a x)}{18 a}-\frac{1}{15} a x^5 \tanh ^{-1}(a x)-\frac{\tanh ^{-1}(a x)^2}{12 a^4}+\frac{1}{4} x^4 \tanh ^{-1}(a x)^2-\frac{1}{6} a^2 x^6 \tanh ^{-1}(a x)^2+\frac{2 \log \left (1-a^2 x^2\right )}{15 a^4}+\frac{1}{18} \operatorname{Subst}\left (\int \left (-\frac{1}{a^2}-\frac{1}{a^2 \left (-1+a^2 x\right )}\right ) \, dx,x,x^2\right )\\ &=-\frac{x^2}{180 a^2}-\frac{x^4}{60}+\frac{x \tanh ^{-1}(a x)}{6 a^3}+\frac{x^3 \tanh ^{-1}(a x)}{18 a}-\frac{1}{15} a x^5 \tanh ^{-1}(a x)-\frac{\tanh ^{-1}(a x)^2}{12 a^4}+\frac{1}{4} x^4 \tanh ^{-1}(a x)^2-\frac{1}{6} a^2 x^6 \tanh ^{-1}(a x)^2+\frac{7 \log \left (1-a^2 x^2\right )}{90 a^4}\\ \end{align*}

Mathematica [A]  time = 0.0446643, size = 88, normalized size = 0.76 \[ -\frac{3 a^4 x^4+a^2 x^2-14 \log \left (1-a^2 x^2\right )+2 a x \left (6 a^4 x^4-5 a^2 x^2-15\right ) \tanh ^{-1}(a x)+15 \left (2 a^6 x^6-3 a^4 x^4+1\right ) \tanh ^{-1}(a x)^2}{180 a^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^2*x^2 + 3*a^4*x^4 + 2*a*x*(-15 - 5*a^2*x^2 + 6*a^4*x^4)*ArcTanh[a*x] + 15*(1 - 3*a^4*x^4 + 2*a^6*x^6)*ArcT
anh[a*x]^2 - 14*Log[1 - a^2*x^2])/(180*a^4)

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Maple [B]  time = 0.045, size = 205, normalized size = 1.8 \begin{align*} -{\frac{{a}^{2}{x}^{6} \left ({\it Artanh} \left ( ax \right ) \right ) ^{2}}{6}}+{\frac{{x}^{4} \left ({\it Artanh} \left ( ax \right ) \right ) ^{2}}{4}}-{\frac{a{x}^{5}{\it Artanh} \left ( ax \right ) }{15}}+{\frac{{x}^{3}{\it Artanh} \left ( ax \right ) }{18\,a}}+{\frac{x{\it Artanh} \left ( ax \right ) }{6\,{a}^{3}}}+{\frac{{\it Artanh} \left ( ax \right ) \ln \left ( ax-1 \right ) }{12\,{a}^{4}}}-{\frac{{\it Artanh} \left ( ax \right ) \ln \left ( ax+1 \right ) }{12\,{a}^{4}}}+{\frac{ \left ( \ln \left ( ax-1 \right ) \right ) ^{2}}{48\,{a}^{4}}}-{\frac{\ln \left ( ax-1 \right ) }{24\,{a}^{4}}\ln \left ({\frac{1}{2}}+{\frac{ax}{2}} \right ) }+{\frac{1}{24\,{a}^{4}}\ln \left ( -{\frac{ax}{2}}+{\frac{1}{2}} \right ) \ln \left ({\frac{1}{2}}+{\frac{ax}{2}} \right ) }-{\frac{\ln \left ( ax+1 \right ) }{24\,{a}^{4}}\ln \left ( -{\frac{ax}{2}}+{\frac{1}{2}} \right ) }+{\frac{ \left ( \ln \left ( ax+1 \right ) \right ) ^{2}}{48\,{a}^{4}}}-{\frac{{x}^{4}}{60}}-{\frac{{x}^{2}}{180\,{a}^{2}}}+{\frac{7\,\ln \left ( ax-1 \right ) }{90\,{a}^{4}}}+{\frac{7\,\ln \left ( ax+1 \right ) }{90\,{a}^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(-a^2*x^2+1)*arctanh(a*x)^2,x)

[Out]

-1/6*a^2*x^6*arctanh(a*x)^2+1/4*x^4*arctanh(a*x)^2-1/15*a*x^5*arctanh(a*x)+1/18*x^3*arctanh(a*x)/a+1/6*x*arcta
nh(a*x)/a^3+1/12/a^4*arctanh(a*x)*ln(a*x-1)-1/12/a^4*arctanh(a*x)*ln(a*x+1)+1/48/a^4*ln(a*x-1)^2-1/24/a^4*ln(a
*x-1)*ln(1/2+1/2*a*x)+1/24/a^4*ln(-1/2*a*x+1/2)*ln(1/2+1/2*a*x)-1/24/a^4*ln(-1/2*a*x+1/2)*ln(a*x+1)+1/48/a^4*l
n(a*x+1)^2-1/60*x^4-1/180*x^2/a^2+7/90/a^4*ln(a*x-1)+7/90/a^4*ln(a*x+1)

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Maxima [A]  time = 1.01882, size = 197, normalized size = 1.7 \begin{align*} -\frac{1}{180} \, a{\left (\frac{2 \,{\left (6 \, a^{4} x^{5} - 5 \, a^{2} x^{3} - 15 \, x\right )}}{a^{4}} + \frac{15 \, \log \left (a x + 1\right )}{a^{5}} - \frac{15 \, \log \left (a x - 1\right )}{a^{5}}\right )} \operatorname{artanh}\left (a x\right ) - \frac{1}{12} \,{\left (2 \, a^{2} x^{6} - 3 \, x^{4}\right )} \operatorname{artanh}\left (a x\right )^{2} - \frac{12 \, a^{4} x^{4} + 4 \, a^{2} x^{2} + 2 \,{\left (15 \, \log \left (a x - 1\right ) - 28\right )} \log \left (a x + 1\right ) - 15 \, \log \left (a x + 1\right )^{2} - 15 \, \log \left (a x - 1\right )^{2} - 56 \, \log \left (a x - 1\right )}{720 \, a^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/180*a*(2*(6*a^4*x^5 - 5*a^2*x^3 - 15*x)/a^4 + 15*log(a*x + 1)/a^5 - 15*log(a*x - 1)/a^5)*arctanh(a*x) - 1/1
2*(2*a^2*x^6 - 3*x^4)*arctanh(a*x)^2 - 1/720*(12*a^4*x^4 + 4*a^2*x^2 + 2*(15*log(a*x - 1) - 28)*log(a*x + 1) -
15*log(a*x + 1)^2 - 15*log(a*x - 1)^2 - 56*log(a*x - 1))/a^4

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Fricas [A]  time = 2.12239, size = 247, normalized size = 2.13 \begin{align*} -\frac{12 \, a^{4} x^{4} + 4 \, a^{2} x^{2} + 15 \,{\left (2 \, a^{6} x^{6} - 3 \, a^{4} x^{4} + 1\right )} \log \left (-\frac{a x + 1}{a x - 1}\right )^{2} + 4 \,{\left (6 \, a^{5} x^{5} - 5 \, a^{3} x^{3} - 15 \, a x\right )} \log \left (-\frac{a x + 1}{a x - 1}\right ) - 56 \, \log \left (a^{2} x^{2} - 1\right )}{720 \, a^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/720*(12*a^4*x^4 + 4*a^2*x^2 + 15*(2*a^6*x^6 - 3*a^4*x^4 + 1)*log(-(a*x + 1)/(a*x - 1))^2 + 4*(6*a^5*x^5 - 5
*a^3*x^3 - 15*a*x)*log(-(a*x + 1)/(a*x - 1)) - 56*log(a^2*x^2 - 1))/a^4

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Sympy [A]  time = 3.20107, size = 114, normalized size = 0.98 \begin{align*} \begin{cases} - \frac{a^{2} x^{6} \operatorname{atanh}^{2}{\left (a x \right )}}{6} - \frac{a x^{5} \operatorname{atanh}{\left (a x \right )}}{15} + \frac{x^{4} \operatorname{atanh}^{2}{\left (a x \right )}}{4} - \frac{x^{4}}{60} + \frac{x^{3} \operatorname{atanh}{\left (a x \right )}}{18 a} - \frac{x^{2}}{180 a^{2}} + \frac{x \operatorname{atanh}{\left (a x \right )}}{6 a^{3}} + \frac{7 \log{\left (x - \frac{1}{a} \right )}}{45 a^{4}} - \frac{\operatorname{atanh}^{2}{\left (a x \right )}}{12 a^{4}} + \frac{7 \operatorname{atanh}{\left (a x \right )}}{45 a^{4}} & \text{for}\: a \neq 0 \\0 & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-a**2*x**6*atanh(a*x)**2/6 - a*x**5*atanh(a*x)/15 + x**4*atanh(a*x)**2/4 - x**4/60 + x**3*atanh(a*x
)/(18*a) - x**2/(180*a**2) + x*atanh(a*x)/(6*a**3) + 7*log(x - 1/a)/(45*a**4) - atanh(a*x)**2/(12*a**4) + 7*at
anh(a*x)/(45*a**4), Ne(a, 0)), (0, True))

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Giac [A]  time = 1.18136, size = 139, normalized size = 1.2 \begin{align*} -\frac{1}{60} \, x^{4} - \frac{1}{48} \,{\left (2 \, a^{2} x^{6} - 3 \, x^{4} + \frac{1}{a^{4}}\right )} \log \left (-\frac{a x + 1}{a x - 1}\right )^{2} - \frac{1}{180} \,{\left (6 \, a x^{5} - \frac{5 \, x^{3}}{a} - \frac{15 \, x}{a^{3}}\right )} \log \left (-\frac{a x + 1}{a x - 1}\right ) - \frac{x^{2}}{180 \, a^{2}} + \frac{7 \, \log \left (a^{2} x^{2} - 1\right )}{90 \, a^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*x^4 - 1/48*(2*a^2*x^6 - 3*x^4 + 1/a^4)*log(-(a*x + 1)/(a*x - 1))^2 - 1/180*(6*a*x^5 - 5*x^3/a - 15*x/a^3
)*log(-(a*x + 1)/(a*x - 1)) - 1/180*x^2/a^2 + 7/90*log(a^2*x^2 - 1)/a^4