3.2.96 \(\int (1-a^2 x^2)^2 \tanh ^{-1}(a x) \, dx\) [196]

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

[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 = 0.03, 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 = {6089, 6021, 266} \begin {gather*} \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) \end {gather*}

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 266

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 6021

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

Rule 6089

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 \begin {gather*} -\frac {7 a x^2}{30}+\frac {a^3 x^4}{20}+x \tanh ^{-1}(a x)-\frac {2}{3} a^2 x^3 \tanh ^{-1}(a x)+\frac {1}{5} a^4 x^5 \tanh ^{-1}(a x)+\frac {4 \log \left (1-a^2 x^2\right )}{15 a} \end {gather*}

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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Maple [A]
time = 0.27, size = 69, normalized size = 0.66

method	result	size
derivativedivides	\(\frac {\frac {\arctanh \left (a x \right ) a^{5} x^{5}}{5}-\frac {2 a^{3} x^{3} \arctanh \left (a x \right )}{3}+a x \arctanh \left (a x \right )+\frac {a^{4} x^{4}}{20}-\frac {7 a^{2} x^{2}}{30}+\frac {4 \ln \left (a x -1\right )}{15}+\frac {4 \ln \left (a x +1\right )}{15}}{a}\)	\(69\)
default	\(\frac {\frac {\arctanh \left (a x \right ) a^{5} x^{5}}{5}-\frac {2 a^{3} x^{3} \arctanh \left (a x \right )}{3}+a x \arctanh \left (a x \right )+\frac {a^{4} x^{4}}{20}-\frac {7 a^{2} x^{2}}{30}+\frac {4 \ln \left (a x -1\right )}{15}+\frac {4 \ln \left (a x +1\right )}{15}}{a}\)	\(69\)
risch	\(\left (\frac {1}{10} a^{4} x^{5}-\frac {1}{3} a^{2} x^{3}+\frac {1}{2} x \right ) \ln \left (a x +1\right )-\frac {a^{4} x^{5} \ln \left (-a x +1\right )}{10}+\frac {a^{3} x^{4}}{20}+\frac {a^{2} x^{3} \ln \left (-a x +1\right )}{3}-\frac {7 a \,x^{2}}{30}-\frac {x \ln \left (-a x +1\right )}{2}+\frac {4 \ln \left (a^{2} x^{2}-1\right )}{15 a}+\frac {49}{180 a}\)	\(103\)
meijerg	\(-\frac {\frac {2 a^{2} x^{2} \left (\ln \left (1-\sqrt {a^{2} x^{2}}\right )-\ln \left (1+\sqrt {a^{2} x^{2}}\right )\right )}{\sqrt {a^{2} x^{2}}}-2 \ln \left (-a^{2} x^{2}+1\right )}{4 a}-\frac {-\frac {a^{2} x^{2} \left (3 a^{2} x^{2}+6\right )}{15}+\frac {2 a^{6} x^{6} \left (\ln \left (1-\sqrt {a^{2} x^{2}}\right )-\ln \left (1+\sqrt {a^{2} x^{2}}\right )\right )}{5 \sqrt {a^{2} x^{2}}}-\frac {2 \ln \left (-a^{2} x^{2}+1\right )}{5}}{4 a}-\frac {\frac {2 a^{2} x^{2}}{3}-\frac {2 a^{4} x^{4} \left (\ln \left (1-\sqrt {a^{2} x^{2}}\right )-\ln \left (1+\sqrt {a^{2} x^{2}}\right )\right )}{3 \sqrt {a^{2} x^{2}}}+\frac {2 \ln \left (-a^{2} x^{2}+1\right )}{3}}{2 a}\)	\(223\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.26, size = 66, normalized size = 0.63 \begin {gather*} \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 ) \end {gather*}

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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Fricas [A]
time = 0.34, size = 72, normalized size = 0.69 \begin {gather*} \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} \end {gather*}

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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Sympy [A]
time = 0.28, size = 75, normalized size = 0.72 \begin {gather*} \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} \end {gather*}

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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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 255 vs. \(2 (88) = 176\).
time = 0.40, size = 255, normalized size = 2.45 \begin {gather*} \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 )} \end {gather*}

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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Mupad [B]
time = 0.91, size = 60, normalized size = 0.58 \begin {gather*} 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} \end {gather*}

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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