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

Optimal. Leaf size=63 \[ -\frac {\tanh ^{-1}(a x)}{12 a^4}+\frac {x}{12 a^3}-\frac {1}{6} a^2 x^6 \tanh ^{-1}(a x)-\frac {a x^5}{30}+\frac {1}{4} x^4 \tanh ^{-1}(a x)+\frac {x^3}{36 a} \]

[Out]

1/12*x/a^3+1/36*x^3/a-1/30*a*x^5-1/12*arctanh(a*x)/a^4+1/4*x^4*arctanh(a*x)-1/6*a^2*x^6*arctanh(a*x)

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Rubi [A]  time = 0.08, antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {6014, 5916, 302, 206} \[ -\frac {1}{6} a^2 x^6 \tanh ^{-1}(a x)+\frac {x}{12 a^3}-\frac {\tanh ^{-1}(a x)}{12 a^4}-\frac {a x^5}{30}+\frac {x^3}{36 a}+\frac {1}{4} x^4 \tanh ^{-1}(a x) \]

Antiderivative was successfully verified.

[In]

Int[x^3*(1 - a^2*x^2)*ArcTanh[a*x],x]

[Out]

x/(12*a^3) + x^3/(36*a) - (a*x^5)/30 - ArcTanh[a*x]/(12*a^4) + (x^4*ArcTanh[a*x])/4 - (a^2*x^6*ArcTanh[a*x])/6

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])

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 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 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]))

Rubi steps

\begin {align*} \int x^3 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x) \, dx &=-\left (a^2 \int x^5 \tanh ^{-1}(a x) \, dx\right )+\int x^3 \tanh ^{-1}(a x) \, dx\\ &=\frac {1}{4} x^4 \tanh ^{-1}(a x)-\frac {1}{6} a^2 x^6 \tanh ^{-1}(a x)-\frac {1}{4} a \int \frac {x^4}{1-a^2 x^2} \, dx+\frac {1}{6} a^3 \int \frac {x^6}{1-a^2 x^2} \, dx\\ &=\frac {1}{4} x^4 \tanh ^{-1}(a x)-\frac {1}{6} a^2 x^6 \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}{6} 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 {x}{12 a^3}+\frac {x^3}{36 a}-\frac {a x^5}{30}+\frac {1}{4} x^4 \tanh ^{-1}(a x)-\frac {1}{6} a^2 x^6 \tanh ^{-1}(a x)+\frac {\int \frac {1}{1-a^2 x^2} \, dx}{6 a^3}-\frac {\int \frac {1}{1-a^2 x^2} \, dx}{4 a^3}\\ &=\frac {x}{12 a^3}+\frac {x^3}{36 a}-\frac {a x^5}{30}-\frac {\tanh ^{-1}(a x)}{12 a^4}+\frac {1}{4} x^4 \tanh ^{-1}(a x)-\frac {1}{6} a^2 x^6 \tanh ^{-1}(a x)\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 79, normalized size = 1.25 \[ \frac {\log (1-a x)}{24 a^4}-\frac {\log (a x+1)}{24 a^4}+\frac {x}{12 a^3}-\frac {1}{6} a^2 x^6 \tanh ^{-1}(a x)-\frac {a x^5}{30}+\frac {1}{4} x^4 \tanh ^{-1}(a x)+\frac {x^3}{36 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

x/(12*a^3) + x^3/(36*a) - (a*x^5)/30 + (x^4*ArcTanh[a*x])/4 - (a^2*x^6*ArcTanh[a*x])/6 + Log[1 - a*x]/(24*a^4)
- Log[1 + a*x]/(24*a^4)

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fricas [A]  time = 0.67, size = 61, normalized size = 0.97 \[ -\frac {12 \, a^{5} x^{5} - 10 \, a^{3} x^{3} - 30 \, a x + 15 \, {\left (2 \, a^{6} x^{6} - 3 \, a^{4} x^{4} + 1\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )}{360 \, a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/360*(12*a^5*x^5 - 10*a^3*x^3 - 30*a*x + 15*(2*a^6*x^6 - 3*a^4*x^4 + 1)*log(-(a*x + 1)/(a*x - 1)))/a^4

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giac [B]  time = 0.21, size = 227, normalized size = 3.60 \[ -\frac {1}{45} \, a {\left (\frac {\frac {45 \, {\left (a x + 1\right )}^{3}}{{\left (a x - 1\right )}^{3}} - \frac {25 \, {\left (a x + 1\right )}^{2}}{{\left (a x - 1\right )}^{2}} + \frac {35 \, {\left (a x + 1\right )}}{a x - 1} - 7}{a^{5} {\left (\frac {a x + 1}{a x - 1} - 1\right )}^{5}} + \frac {30 \, {\left (\frac {3 \, {\left (a x + 1\right )}^{4}}{{\left (a x - 1\right )}^{4}} + \frac {2 \, {\left (a x + 1\right )}^{3}}{{\left (a x - 1\right )}^{3}} + \frac {3 \, {\left (a x + 1\right )}^{2}}{{\left (a x - 1\right )}^{2}}\right )} \log \left (-\frac {\frac {a {\left (\frac {a x + 1}{a x - 1} + 1\right )}}{\frac {{\left (a x + 1\right )} a}{a x - 1} - a} + 1}{\frac {a {\left (\frac {a x + 1}{a x - 1} + 1\right )}}{\frac {{\left (a x + 1\right )} a}{a x - 1} - a} - 1}\right )}{a^{5} {\left (\frac {a x + 1}{a x - 1} - 1\right )}^{6}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/45*a*((45*(a*x + 1)^3/(a*x - 1)^3 - 25*(a*x + 1)^2/(a*x - 1)^2 + 35*(a*x + 1)/(a*x - 1) - 7)/(a^5*((a*x + 1
)/(a*x - 1) - 1)^5) + 30*(3*(a*x + 1)^4/(a*x - 1)^4 + 2*(a*x + 1)^3/(a*x - 1)^3 + 3*(a*x + 1)^2/(a*x - 1)^2)*l
og(-(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^5*((a*x + 1)/(a*x - 1) - 1)^6))

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maple [A]  time = 0.03, size = 65, normalized size = 1.03 \[ -\frac {a^{2} x^{6} \arctanh \left (a x \right )}{6}+\frac {x^{4} \arctanh \left (a x \right )}{4}-\frac {a \,x^{5}}{30}+\frac {x^{3}}{36 a}+\frac {x}{12 a^{3}}+\frac {\ln \left (a x -1\right )}{24 a^{4}}-\frac {\ln \left (a x +1\right )}{24 a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*a^2*x^6*arctanh(a*x)+1/4*x^4*arctanh(a*x)-1/30*a*x^5+1/36*x^3/a+1/12*x/a^3+1/24/a^4*ln(a*x-1)-1/24/a^4*ln
(a*x+1)

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maxima [A]  time = 0.31, size = 72, normalized size = 1.14 \[ -\frac {1}{360} \, a {\left (\frac {2 \, {\left (6 \, a^{4} x^{5} - 5 \, a^{2} x^{3} - 15 \, x\right )}}{a^{4}} + \frac {15 \, \log \left (a x + 1\right )}{a^{5}} - \frac {15 \, \log \left (a x - 1\right )}{a^{5}}\right )} - \frac {1}{12} \, {\left (2 \, a^{2} x^{6} - 3 \, x^{4}\right )} \operatorname {artanh}\left (a x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/360*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) - 1/12*(2*a^2*x^6
- 3*x^4)*arctanh(a*x)

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mupad [B]  time = 0.95, size = 51, normalized size = 0.81 \[ \frac {\frac {a\,x}{12}-\frac {\mathrm {atanh}\left (a\,x\right )}{12}+\frac {a^3\,x^3}{36}}{a^4}-\frac {a\,x^5}{30}+\frac {x^4\,\mathrm {atanh}\left (a\,x\right )}{4}-\frac {a^2\,x^6\,\mathrm {atanh}\left (a\,x\right )}{6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-x^3*atanh(a*x)*(a^2*x^2 - 1),x)

[Out]

((a*x)/12 - atanh(a*x)/12 + (a^3*x^3)/36)/a^4 - (a*x^5)/30 + (x^4*atanh(a*x))/4 - (a^2*x^6*atanh(a*x))/6

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sympy [A]  time = 1.55, size = 54, normalized size = 0.86 \[ \begin {cases} - \frac {a^{2} x^{6} \operatorname {atanh}{\left (a x \right )}}{6} - \frac {a x^{5}}{30} + \frac {x^{4} \operatorname {atanh}{\left (a x \right )}}{4} + \frac {x^{3}}{36 a} + \frac {x}{12 a^{3}} - \frac {\operatorname {atanh}{\left (a x \right )}}{12 a^{4}} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-a**2*x**6*atanh(a*x)/6 - a*x**5/30 + x**4*atanh(a*x)/4 + x**3/(36*a) + x/(12*a**3) - atanh(a*x)/(1
2*a**4), Ne(a, 0)), (0, True))

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