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

Optimal. Leaf size=86 \[ \frac{a^3 x^6}{42}+\frac{4 \log \left (1-a^2 x^2\right )}{105 a^3}+\frac{1}{7} a^4 x^7 \tanh ^{-1}(a x)-\frac{2}{5} a^2 x^5 \tanh ^{-1}(a x)-\frac{9 a x^4}{140}+\frac{4 x^2}{105 a}+\frac{1}{3} x^3 \tanh ^{-1}(a x) \]

[Out]

(4*x^2)/(105*a) - (9*a*x^4)/140 + (a^3*x^6)/42 + (x^3*ArcTanh[a*x])/3 - (2*a^2*x^5*ArcTanh[a*x])/5 + (a^4*x^7*
ArcTanh[a*x])/7 + (4*Log[1 - a^2*x^2])/(105*a^3)

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Rubi [A]  time = 0.161448, antiderivative size = 86, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 4, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {6012, 5916, 266, 43} \[ \frac{a^3 x^6}{42}+\frac{4 \log \left (1-a^2 x^2\right )}{105 a^3}+\frac{1}{7} a^4 x^7 \tanh ^{-1}(a x)-\frac{2}{5} a^2 x^5 \tanh ^{-1}(a x)-\frac{9 a x^4}{140}+\frac{4 x^2}{105 a}+\frac{1}{3} x^3 \tanh ^{-1}(a x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*x^2)/(105*a) - (9*a*x^4)/140 + (a^3*x^6)/42 + (x^3*ArcTanh[a*x])/3 - (2*a^2*x^5*ArcTanh[a*x])/5 + (a^4*x^7*
ArcTanh[a*x])/7 + (4*Log[1 - a^2*x^2])/(105*a^3)

Rule 6012

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

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 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])

Rubi steps

\begin{align*} \int x^2 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x) \, dx &=\int \left (x^2 \tanh ^{-1}(a x)-2 a^2 x^4 \tanh ^{-1}(a x)+a^4 x^6 \tanh ^{-1}(a x)\right ) \, dx\\ &=-\left (\left (2 a^2\right ) \int x^4 \tanh ^{-1}(a x) \, dx\right )+a^4 \int x^6 \tanh ^{-1}(a x) \, dx+\int x^2 \tanh ^{-1}(a x) \, dx\\ &=\frac{1}{3} x^3 \tanh ^{-1}(a x)-\frac{2}{5} a^2 x^5 \tanh ^{-1}(a x)+\frac{1}{7} a^4 x^7 \tanh ^{-1}(a x)-\frac{1}{3} a \int \frac{x^3}{1-a^2 x^2} \, dx+\frac{1}{5} \left (2 a^3\right ) \int \frac{x^5}{1-a^2 x^2} \, dx-\frac{1}{7} a^5 \int \frac{x^7}{1-a^2 x^2} \, dx\\ &=\frac{1}{3} x^3 \tanh ^{-1}(a x)-\frac{2}{5} a^2 x^5 \tanh ^{-1}(a x)+\frac{1}{7} a^4 x^7 \tanh ^{-1}(a x)-\frac{1}{6} a \operatorname{Subst}\left (\int \frac{x}{1-a^2 x} \, dx,x,x^2\right )+\frac{1}{5} a^3 \operatorname{Subst}\left (\int \frac{x^2}{1-a^2 x} \, dx,x,x^2\right )-\frac{1}{14} a^5 \operatorname{Subst}\left (\int \frac{x^3}{1-a^2 x} \, dx,x,x^2\right )\\ &=\frac{1}{3} x^3 \tanh ^{-1}(a x)-\frac{2}{5} a^2 x^5 \tanh ^{-1}(a x)+\frac{1}{7} a^4 x^7 \tanh ^{-1}(a x)-\frac{1}{6} a \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{1}{5} a^3 \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{1}{14} a^5 \operatorname{Subst}\left (\int \left (-\frac{1}{a^6}-\frac{x}{a^4}-\frac{x^2}{a^2}-\frac{1}{a^6 \left (-1+a^2 x\right )}\right ) \, dx,x,x^2\right )\\ &=\frac{4 x^2}{105 a}-\frac{9 a x^4}{140}+\frac{a^3 x^6}{42}+\frac{1}{3} x^3 \tanh ^{-1}(a x)-\frac{2}{5} a^2 x^5 \tanh ^{-1}(a x)+\frac{1}{7} a^4 x^7 \tanh ^{-1}(a x)+\frac{4 \log \left (1-a^2 x^2\right )}{105 a^3}\\ \end{align*}

Mathematica [A]  time = 0.0213349, size = 86, normalized size = 1. \[ \frac{a^3 x^6}{42}+\frac{4 \log \left (1-a^2 x^2\right )}{105 a^3}+\frac{1}{7} a^4 x^7 \tanh ^{-1}(a x)-\frac{2}{5} a^2 x^5 \tanh ^{-1}(a x)-\frac{9 a x^4}{140}+\frac{4 x^2}{105 a}+\frac{1}{3} x^3 \tanh ^{-1}(a x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*x^2)/(105*a) - (9*a*x^4)/140 + (a^3*x^6)/42 + (x^3*ArcTanh[a*x])/3 - (2*a^2*x^5*ArcTanh[a*x])/5 + (a^4*x^7*
ArcTanh[a*x])/7 + (4*Log[1 - a^2*x^2])/(105*a^3)

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Maple [A]  time = 0.029, size = 79, normalized size = 0.9 \begin{align*}{\frac{{a}^{4}{x}^{7}{\it Artanh} \left ( ax \right ) }{7}}-{\frac{2\,{a}^{2}{x}^{5}{\it Artanh} \left ( ax \right ) }{5}}+{\frac{{x}^{3}{\it Artanh} \left ( ax \right ) }{3}}+{\frac{{x}^{6}{a}^{3}}{42}}-{\frac{9\,{x}^{4}a}{140}}+{\frac{4\,{x}^{2}}{105\,a}}+{\frac{4\,\ln \left ( ax-1 \right ) }{105\,{a}^{3}}}+{\frac{4\,\ln \left ( ax+1 \right ) }{105\,{a}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/7*a^4*x^7*arctanh(a*x)-2/5*a^2*x^5*arctanh(a*x)+1/3*x^3*arctanh(a*x)+1/42*x^6*a^3-9/140*x^4*a+4/105*x^2/a+4/
105/a^3*ln(a*x-1)+4/105/a^3*ln(a*x+1)

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Maxima [A]  time = 0.950059, size = 109, normalized size = 1.27 \begin{align*} \frac{1}{420} \, a{\left (\frac{10 \, a^{4} x^{6} - 27 \, a^{2} x^{4} + 16 \, x^{2}}{a^{2}} + \frac{16 \, \log \left (a x + 1\right )}{a^{4}} + \frac{16 \, \log \left (a x - 1\right )}{a^{4}}\right )} + \frac{1}{105} \,{\left (15 \, a^{4} x^{7} - 42 \, a^{2} x^{5} + 35 \, x^{3}\right )} \operatorname{artanh}\left (a x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/420*a*((10*a^4*x^6 - 27*a^2*x^4 + 16*x^2)/a^2 + 16*log(a*x + 1)/a^4 + 16*log(a*x - 1)/a^4) + 1/105*(15*a^4*x
^7 - 42*a^2*x^5 + 35*x^3)*arctanh(a*x)

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Fricas [A]  time = 2.07963, size = 190, normalized size = 2.21 \begin{align*} \frac{10 \, a^{6} x^{6} - 27 \, a^{4} x^{4} + 16 \, a^{2} x^{2} + 2 \,{\left (15 \, a^{7} x^{7} - 42 \, a^{5} x^{5} + 35 \, a^{3} x^{3}\right )} \log \left (-\frac{a x + 1}{a x - 1}\right ) + 16 \, \log \left (a^{2} x^{2} - 1\right )}{420 \, a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/420*(10*a^6*x^6 - 27*a^4*x^4 + 16*a^2*x^2 + 2*(15*a^7*x^7 - 42*a^5*x^5 + 35*a^3*x^3)*log(-(a*x + 1)/(a*x - 1
)) + 16*log(a^2*x^2 - 1))/a^3

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Sympy [A]  time = 3.11666, size = 90, normalized size = 1.05 \begin{align*} \begin{cases} \frac{a^{4} x^{7} \operatorname{atanh}{\left (a x \right )}}{7} + \frac{a^{3} x^{6}}{42} - \frac{2 a^{2} x^{5} \operatorname{atanh}{\left (a x \right )}}{5} - \frac{9 a x^{4}}{140} + \frac{x^{3} \operatorname{atanh}{\left (a x \right )}}{3} + \frac{4 x^{2}}{105 a} + \frac{8 \log{\left (x - \frac{1}{a} \right )}}{105 a^{3}} + \frac{8 \operatorname{atanh}{\left (a x \right )}}{105 a^{3}} & \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**2*(-a**2*x**2+1)**2*atanh(a*x),x)

[Out]

Piecewise((a**4*x**7*atanh(a*x)/7 + a**3*x**6/42 - 2*a**2*x**5*atanh(a*x)/5 - 9*a*x**4/140 + x**3*atanh(a*x)/3
+ 4*x**2/(105*a) + 8*log(x - 1/a)/(105*a**3) + 8*atanh(a*x)/(105*a**3), Ne(a, 0)), (0, True))

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Giac [A]  time = 1.15999, size = 116, normalized size = 1.35 \begin{align*} \frac{1}{210} \,{\left (15 \, a^{4} x^{7} - 42 \, a^{2} x^{5} + 35 \, x^{3}\right )} \log \left (-\frac{a x + 1}{a x - 1}\right ) + \frac{4 \, \log \left ({\left | a^{2} x^{2} - 1 \right |}\right )}{105 \, a^{3}} + \frac{10 \, a^{9} x^{6} - 27 \, a^{7} x^{4} + 16 \, a^{5} x^{2}}{420 \, a^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/210*(15*a^4*x^7 - 42*a^2*x^5 + 35*x^3)*log(-(a*x + 1)/(a*x - 1)) + 4/105*log(abs(a^2*x^2 - 1))/a^3 + 1/420*(
10*a^9*x^6 - 27*a^7*x^4 + 16*a^5*x^2)/a^6