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

Optimal. Leaf size=87 \[ \frac{a^3 x^7}{56}+\frac{1}{8} a^4 x^8 \tanh ^{-1}(a x)-\frac{1}{3} a^2 x^6 \tanh ^{-1}(a x)+\frac{x}{24 a^3}-\frac{\tanh ^{-1}(a x)}{24 a^4}-\frac{a x^5}{24}+\frac{x^3}{72 a}+\frac{1}{4} x^4 \tanh ^{-1}(a x) \]

[Out]

x/(24*a^3) + x^3/(72*a) - (a*x^5)/24 + (a^3*x^7)/56 - ArcTanh[a*x]/(24*a^4) + (x^4*ArcTanh[a*x])/4 - (a^2*x^6*
ArcTanh[a*x])/3 + (a^4*x^8*ArcTanh[a*x])/8

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Rubi [A]  time = 0.137494, antiderivative size = 87, 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, 302, 206} \[ \frac{a^3 x^7}{56}+\frac{1}{8} a^4 x^8 \tanh ^{-1}(a x)-\frac{1}{3} a^2 x^6 \tanh ^{-1}(a x)+\frac{x}{24 a^3}-\frac{\tanh ^{-1}(a x)}{24 a^4}-\frac{a x^5}{24}+\frac{x^3}{72 a}+\frac{1}{4} x^4 \tanh ^{-1}(a x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

x/(24*a^3) + x^3/(72*a) - (a*x^5)/24 + (a^3*x^7)/56 - ArcTanh[a*x]/(24*a^4) + (x^4*ArcTanh[a*x])/4 - (a^2*x^6*
ArcTanh[a*x])/3 + (a^4*x^8*ArcTanh[a*x])/8

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 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rule 206

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

Rubi steps

\begin{align*} \int x^3 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x) \, dx &=\int \left (x^3 \tanh ^{-1}(a x)-2 a^2 x^5 \tanh ^{-1}(a x)+a^4 x^7 \tanh ^{-1}(a x)\right ) \, dx\\ &=-\left (\left (2 a^2\right ) \int x^5 \tanh ^{-1}(a x) \, dx\right )+a^4 \int x^7 \tanh ^{-1}(a x) \, dx+\int x^3 \tanh ^{-1}(a x) \, dx\\ &=\frac{1}{4} x^4 \tanh ^{-1}(a x)-\frac{1}{3} a^2 x^6 \tanh ^{-1}(a x)+\frac{1}{8} a^4 x^8 \tanh ^{-1}(a x)-\frac{1}{4} a \int \frac{x^4}{1-a^2 x^2} \, dx+\frac{1}{3} a^3 \int \frac{x^6}{1-a^2 x^2} \, dx-\frac{1}{8} a^5 \int \frac{x^8}{1-a^2 x^2} \, dx\\ &=\frac{1}{4} x^4 \tanh ^{-1}(a x)-\frac{1}{3} a^2 x^6 \tanh ^{-1}(a x)+\frac{1}{8} a^4 x^8 \tanh ^{-1}(a x)-\frac{1}{4} a \int \left (-\frac{1}{a^4}-\frac{x^2}{a^2}+\frac{1}{a^4 \left (1-a^2 x^2\right )}\right ) \, dx+\frac{1}{3} a^3 \int \left (-\frac{1}{a^6}-\frac{x^2}{a^4}-\frac{x^4}{a^2}+\frac{1}{a^6 \left (1-a^2 x^2\right )}\right ) \, dx-\frac{1}{8} a^5 \int \left (-\frac{1}{a^8}-\frac{x^2}{a^6}-\frac{x^4}{a^4}-\frac{x^6}{a^2}+\frac{1}{a^8 \left (1-a^2 x^2\right )}\right ) \, dx\\ &=\frac{x}{24 a^3}+\frac{x^3}{72 a}-\frac{a x^5}{24}+\frac{a^3 x^7}{56}+\frac{1}{4} x^4 \tanh ^{-1}(a x)-\frac{1}{3} a^2 x^6 \tanh ^{-1}(a x)+\frac{1}{8} a^4 x^8 \tanh ^{-1}(a x)-\frac{\int \frac{1}{1-a^2 x^2} \, dx}{8 a^3}-\frac{\int \frac{1}{1-a^2 x^2} \, dx}{4 a^3}+\frac{\int \frac{1}{1-a^2 x^2} \, dx}{3 a^3}\\ &=\frac{x}{24 a^3}+\frac{x^3}{72 a}-\frac{a x^5}{24}+\frac{a^3 x^7}{56}-\frac{\tanh ^{-1}(a x)}{24 a^4}+\frac{1}{4} x^4 \tanh ^{-1}(a x)-\frac{1}{3} a^2 x^6 \tanh ^{-1}(a x)+\frac{1}{8} a^4 x^8 \tanh ^{-1}(a x)\\ \end{align*}

Mathematica [A]  time = 0.0283403, size = 103, normalized size = 1.18 \[ \frac{a^3 x^7}{56}+\frac{1}{8} a^4 x^8 \tanh ^{-1}(a x)-\frac{1}{3} a^2 x^6 \tanh ^{-1}(a x)+\frac{x}{24 a^3}+\frac{\log (1-a x)}{48 a^4}-\frac{\log (a x+1)}{48 a^4}-\frac{a x^5}{24}+\frac{x^3}{72 a}+\frac{1}{4} x^4 \tanh ^{-1}(a x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

x/(24*a^3) + x^3/(72*a) - (a*x^5)/24 + (a^3*x^7)/56 + (x^4*ArcTanh[a*x])/4 - (a^2*x^6*ArcTanh[a*x])/3 + (a^4*x
^8*ArcTanh[a*x])/8 + Log[1 - a*x]/(48*a^4) - Log[1 + a*x]/(48*a^4)

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Maple [A]  time = 0.027, size = 85, normalized size = 1. \begin{align*}{\frac{{a}^{4}{x}^{8}{\it Artanh} \left ( ax \right ) }{8}}-{\frac{{a}^{2}{x}^{6}{\it Artanh} \left ( ax \right ) }{3}}+{\frac{{x}^{4}{\it Artanh} \left ( ax \right ) }{4}}+{\frac{{a}^{3}{x}^{7}}{56}}-{\frac{a{x}^{5}}{24}}+{\frac{{x}^{3}}{72\,a}}+{\frac{x}{24\,{a}^{3}}}+{\frac{\ln \left ( ax-1 \right ) }{48\,{a}^{4}}}-{\frac{\ln \left ( ax+1 \right ) }{48\,{a}^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*a^4*x^8*arctanh(a*x)-1/3*a^2*x^6*arctanh(a*x)+1/4*x^4*arctanh(a*x)+1/56*a^3*x^7-1/24*a*x^5+1/72*x^3/a+1/24
*x/a^3+1/48/a^4*ln(a*x-1)-1/48/a^4*ln(a*x+1)

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Maxima [A]  time = 0.943079, size = 119, normalized size = 1.37 \begin{align*} \frac{1}{1008} \, a{\left (\frac{2 \,{\left (9 \, a^{6} x^{7} - 21 \, a^{4} x^{5} + 7 \, a^{2} x^{3} + 21 \, x\right )}}{a^{4}} - \frac{21 \, \log \left (a x + 1\right )}{a^{5}} + \frac{21 \, \log \left (a x - 1\right )}{a^{5}}\right )} + \frac{1}{24} \,{\left (3 \, a^{4} x^{8} - 8 \, a^{2} x^{6} + 6 \, x^{4}\right )} \operatorname{artanh}\left (a x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1008*a*(2*(9*a^6*x^7 - 21*a^4*x^5 + 7*a^2*x^3 + 21*x)/a^4 - 21*log(a*x + 1)/a^5 + 21*log(a*x - 1)/a^5) + 1/2
4*(3*a^4*x^8 - 8*a^2*x^6 + 6*x^4)*arctanh(a*x)

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Fricas [A]  time = 1.95539, size = 177, normalized size = 2.03 \begin{align*} \frac{18 \, a^{7} x^{7} - 42 \, a^{5} x^{5} + 14 \, a^{3} x^{3} + 42 \, a x + 21 \,{\left (3 \, a^{8} x^{8} - 8 \, a^{6} x^{6} + 6 \, a^{4} x^{4} - 1\right )} \log \left (-\frac{a x + 1}{a x - 1}\right )}{1008 \, a^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1008*(18*a^7*x^7 - 42*a^5*x^5 + 14*a^3*x^3 + 42*a*x + 21*(3*a^8*x^8 - 8*a^6*x^6 + 6*a^4*x^4 - 1)*log(-(a*x +
1)/(a*x - 1)))/a^4

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Sympy [A]  time = 4.00134, size = 76, normalized size = 0.87 \begin{align*} \begin{cases} \frac{a^{4} x^{8} \operatorname{atanh}{\left (a x \right )}}{8} + \frac{a^{3} x^{7}}{56} - \frac{a^{2} x^{6} \operatorname{atanh}{\left (a x \right )}}{3} - \frac{a x^{5}}{24} + \frac{x^{4} \operatorname{atanh}{\left (a x \right )}}{4} + \frac{x^{3}}{72 a} + \frac{x}{24 a^{3}} - \frac{\operatorname{atanh}{\left (a x \right )}}{24 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)**2*atanh(a*x),x)

[Out]

Piecewise((a**4*x**8*atanh(a*x)/8 + a**3*x**7/56 - a**2*x**6*atanh(a*x)/3 - a*x**5/24 + x**4*atanh(a*x)/4 + x*
*3/(72*a) + x/(24*a**3) - atanh(a*x)/(24*a**4), Ne(a, 0)), (0, True))

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Giac [A]  time = 1.18255, size = 135, normalized size = 1.55 \begin{align*} \frac{1}{48} \,{\left (3 \, a^{4} x^{8} - 8 \, a^{2} x^{6} + 6 \, x^{4}\right )} \log \left (-\frac{a x + 1}{a x - 1}\right ) - \frac{\log \left ({\left | a x + 1 \right |}\right )}{48 \, a^{4}} + \frac{\log \left ({\left | a x - 1 \right |}\right )}{48 \, a^{4}} + \frac{9 \, a^{17} x^{7} - 21 \, a^{15} x^{5} + 7 \, a^{13} x^{3} + 21 \, a^{11} x}{504 \, a^{14}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(3*a^4*x^8 - 8*a^2*x^6 + 6*x^4)*log(-(a*x + 1)/(a*x - 1)) - 1/48*log(abs(a*x + 1))/a^4 + 1/48*log(abs(a*x
- 1))/a^4 + 1/504*(9*a^17*x^7 - 21*a^15*x^5 + 7*a^13*x^3 + 21*a^11*x)/a^14