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

Optimal. Leaf size=104 \[ \frac {\left (1-a^2 x^2\right )^2}{20 a}+\frac {2 \left (1-a^2 x^2\right )}{15 a}+\frac {4 \log \left (1-a^2 x^2\right )}{15 a}+\frac {1}{5} x \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)+\frac {4}{15} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)+\frac {8}{15} x \tanh ^{-1}(a x) \]

[Out]

2/15*(-a^2*x^2+1)/a+1/20*(-a^2*x^2+1)^2/a+8/15*x*arctanh(a*x)+4/15*x*(-a^2*x^2+1)*arctanh(a*x)+1/5*x*(-a^2*x^2
+1)^2*arctanh(a*x)+4/15*ln(-a^2*x^2+1)/a

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Rubi [A]  time = 0.04, antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {5942, 5910, 260} \[ \frac {\left (1-a^2 x^2\right )^2}{20 a}+\frac {2 \left (1-a^2 x^2\right )}{15 a}+\frac {4 \log \left (1-a^2 x^2\right )}{15 a}+\frac {1}{5} x \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)+\frac {4}{15} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)+\frac {8}{15} x \tanh ^{-1}(a x) \]

Antiderivative was successfully verified.

[In]

Int[(1 - a^2*x^2)^2*ArcTanh[a*x],x]

[Out]

(2*(1 - a^2*x^2))/(15*a) + (1 - a^2*x^2)^2/(20*a) + (8*x*ArcTanh[a*x])/15 + (4*x*(1 - a^2*x^2)*ArcTanh[a*x])/1
5 + (x*(1 - a^2*x^2)^2*ArcTanh[a*x])/5 + (4*Log[1 - a^2*x^2])/(15*a)

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 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 5942

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

Rubi steps

\begin {align*} \int \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x) \, dx &=\frac {\left (1-a^2 x^2\right )^2}{20 a}+\frac {1}{5} x \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)+\frac {4}{5} \int \left (1-a^2 x^2\right ) \tanh ^{-1}(a x) \, dx\\ &=\frac {2 \left (1-a^2 x^2\right )}{15 a}+\frac {\left (1-a^2 x^2\right )^2}{20 a}+\frac {4}{15} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)+\frac {1}{5} x \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)+\frac {8}{15} \int \tanh ^{-1}(a x) \, dx\\ &=\frac {2 \left (1-a^2 x^2\right )}{15 a}+\frac {\left (1-a^2 x^2\right )^2}{20 a}+\frac {8}{15} x \tanh ^{-1}(a x)+\frac {4}{15} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)+\frac {1}{5} x \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)-\frac {1}{15} (8 a) \int \frac {x}{1-a^2 x^2} \, dx\\ &=\frac {2 \left (1-a^2 x^2\right )}{15 a}+\frac {\left (1-a^2 x^2\right )^2}{20 a}+\frac {8}{15} x \tanh ^{-1}(a x)+\frac {4}{15} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)+\frac {1}{5} x \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)+\frac {4 \log \left (1-a^2 x^2\right )}{15 a}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 71, normalized size = 0.68 \[ \frac {1}{5} a^4 x^5 \tanh ^{-1}(a x)+\frac {a^3 x^4}{20}-\frac {2}{3} a^2 x^3 \tanh ^{-1}(a x)+\frac {4 \log \left (1-a^2 x^2\right )}{15 a}-\frac {7 a x^2}{30}+x \tanh ^{-1}(a x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-7*a*x^2)/30 + (a^3*x^4)/20 + x*ArcTanh[a*x] - (2*a^2*x^3*ArcTanh[a*x])/3 + (a^4*x^5*ArcTanh[a*x])/5 + (4*Log
[1 - a^2*x^2])/(15*a)

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fricas [A]  time = 0.59, size = 72, normalized size = 0.69 \[ \frac {3 \, a^{4} x^{4} - 14 \, a^{2} x^{2} + 2 \, {\left (3 \, a^{5} x^{5} - 10 \, a^{3} x^{3} + 15 \, a x\right )} \log \left (-\frac {a x + 1}{a x - 1}\right ) + 16 \, \log \left (a^{2} x^{2} - 1\right )}{60 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(3*a^4*x^4 - 14*a^2*x^2 + 2*(3*a^5*x^5 - 10*a^3*x^3 + 15*a*x)*log(-(a*x + 1)/(a*x - 1)) + 16*log(a^2*x^2
- 1))/a

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giac [B]  time = 0.38, size = 255, normalized size = 2.45 \[ \frac {4}{15} \, a {\left (\frac {2 \, \log \left (\frac {{\left | -a x - 1 \right |}}{{\left | a x - 1 \right |}}\right )}{a^{2}} - \frac {2 \, \log \left ({\left | -\frac {a x + 1}{a x - 1} + 1 \right |}\right )}{a^{2}} - \frac {\frac {2 \, {\left (a x + 1\right )}^{3}}{{\left (a x - 1\right )}^{3}} - \frac {7 \, {\left (a x + 1\right )}^{2}}{{\left (a x - 1\right )}^{2}} + \frac {2 \, {\left (a x + 1\right )}}{a x - 1}}{a^{2} {\left (\frac {a x + 1}{a x - 1} - 1\right )}^{4}} + \frac {2 \, {\left (\frac {10 \, {\left (a x + 1\right )}^{2}}{{\left (a x - 1\right )}^{2}} - \frac {5 \, {\left (a x + 1\right )}}{a x - 1} + 1\right )} \log \left (-\frac {\frac {a {\left (\frac {a x + 1}{a x - 1} + 1\right )}}{\frac {{\left (a x + 1\right )} a}{a x - 1} - a} + 1}{\frac {a {\left (\frac {a x + 1}{a x - 1} + 1\right )}}{\frac {{\left (a x + 1\right )} a}{a x - 1} - a} - 1}\right )}{a^{2} {\left (\frac {a x + 1}{a x - 1} - 1\right )}^{5}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/15*a*(2*log(abs(-a*x - 1)/abs(a*x - 1))/a^2 - 2*log(abs(-(a*x + 1)/(a*x - 1) + 1))/a^2 - (2*(a*x + 1)^3/(a*x
- 1)^3 - 7*(a*x + 1)^2/(a*x - 1)^2 + 2*(a*x + 1)/(a*x - 1))/(a^2*((a*x + 1)/(a*x - 1) - 1)^4) + 2*(10*(a*x +
1)^2/(a*x - 1)^2 - 5*(a*x + 1)/(a*x - 1) + 1)*log(-(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^2*((a*x + 1)/(a*x - 1) - 1)^5))

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maple [A]  time = 0.03, size = 68, normalized size = 0.65 \[ \frac {a^{4} \arctanh \left (a x \right ) x^{5}}{5}-\frac {2 a^{2} \arctanh \left (a x \right ) x^{3}}{3}+x \arctanh \left (a x \right )+\frac {x^{4} a^{3}}{20}-\frac {7 a \,x^{2}}{30}+\frac {4 \ln \left (a x -1\right )}{15 a}+\frac {4 \ln \left (a x +1\right )}{15 a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*a^4*arctanh(a*x)*x^5-2/3*a^2*arctanh(a*x)*x^3+x*arctanh(a*x)+1/20*x^4*a^3-7/30*a*x^2+4/15/a*ln(a*x-1)+4/15
/a*ln(a*x+1)

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maxima [A]  time = 0.30, size = 66, normalized size = 0.63 \[ \frac {1}{60} \, {\left (3 \, a^{2} x^{4} - 14 \, x^{2} + \frac {16 \, \log \left (a x + 1\right )}{a^{2}} + \frac {16 \, \log \left (a x - 1\right )}{a^{2}}\right )} a + \frac {1}{15} \, {\left (3 \, a^{4} x^{5} - 10 \, a^{2} x^{3} + 15 \, x\right )} \operatorname {artanh}\left (a x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(3*a^2*x^4 - 14*x^2 + 16*log(a*x + 1)/a^2 + 16*log(a*x - 1)/a^2)*a + 1/15*(3*a^4*x^5 - 10*a^2*x^3 + 15*x)
*arctanh(a*x)

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mupad [B]  time = 0.91, size = 60, normalized size = 0.58 \[ x\,\mathrm {atanh}\left (a\,x\right )-\frac {7\,a\,x^2}{30}+\frac {4\,\ln \left (a^2\,x^2-1\right )}{15\,a}+\frac {a^3\,x^4}{20}-\frac {2\,a^2\,x^3\,\mathrm {atanh}\left (a\,x\right )}{3}+\frac {a^4\,x^5\,\mathrm {atanh}\left (a\,x\right )}{5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(a*x)*(a^2*x^2 - 1)^2,x)

[Out]

x*atanh(a*x) - (7*a*x^2)/30 + (4*log(a^2*x^2 - 1))/(15*a) + (a^3*x^4)/20 - (2*a^2*x^3*atanh(a*x))/3 + (a^4*x^5
*atanh(a*x))/5

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sympy [A]  time = 1.24, size = 75, normalized size = 0.72 \[ \begin {cases} \frac {a^{4} x^{5} \operatorname {atanh}{\left (a x \right )}}{5} + \frac {a^{3} x^{4}}{20} - \frac {2 a^{2} x^{3} \operatorname {atanh}{\left (a x \right )}}{3} - \frac {7 a x^{2}}{30} + x \operatorname {atanh}{\left (a x \right )} + \frac {8 \log {\left (x - \frac {1}{a} \right )}}{15 a} + \frac {8 \operatorname {atanh}{\left (a x \right )}}{15 a} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a**4*x**5*atanh(a*x)/5 + a**3*x**4/20 - 2*a**2*x**3*atanh(a*x)/3 - 7*a*x**2/30 + x*atanh(a*x) + 8*l
og(x - 1/a)/(15*a) + 8*atanh(a*x)/(15*a), Ne(a, 0)), (0, True))

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