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

Optimal. Leaf size=138 \[ \frac {\left (1-a^2 x^2\right )^2}{60 a^2}+\frac {2 \left (1-a^2 x^2\right )}{45 a^2}+\frac {4 \log \left (1-a^2 x^2\right )}{45 a^2}-\frac {\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2}{6 a^2}+\frac {x \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}{15 a}+\frac {4 x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}{45 a}+\frac {8 x \tanh ^{-1}(a x)}{45 a} \]

[Out]

2/45*(-a^2*x^2+1)/a^2+1/60*(-a^2*x^2+1)^2/a^2+8/45*x*arctanh(a*x)/a+4/45*x*(-a^2*x^2+1)*arctanh(a*x)/a+1/15*x*
(-a^2*x^2+1)^2*arctanh(a*x)/a-1/6*(-a^2*x^2+1)^3*arctanh(a*x)^2/a^2+4/45*ln(-a^2*x^2+1)/a^2

________________________________________________________________________________________

Rubi [A]  time = 0.09, antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {5994, 5942, 5910, 260} \[ \frac {\left (1-a^2 x^2\right )^2}{60 a^2}+\frac {2 \left (1-a^2 x^2\right )}{45 a^2}+\frac {4 \log \left (1-a^2 x^2\right )}{45 a^2}-\frac {\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2}{6 a^2}+\frac {x \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}{15 a}+\frac {4 x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}{45 a}+\frac {8 x \tanh ^{-1}(a x)}{45 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(1 - a^2*x^2))/(45*a^2) + (1 - a^2*x^2)^2/(60*a^2) + (8*x*ArcTanh[a*x])/(45*a) + (4*x*(1 - a^2*x^2)*ArcTanh
[a*x])/(45*a) + (x*(1 - a^2*x^2)^2*ArcTanh[a*x])/(15*a) - ((1 - a^2*x^2)^3*ArcTanh[a*x]^2)/(6*a^2) + (4*Log[1
- a^2*x^2])/(45*a^2)

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]

Rule 5994

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

Rubi steps

\begin {align*} \int x \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2 \, dx &=-\frac {\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2}{6 a^2}+\frac {\int \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x) \, dx}{3 a}\\ &=\frac {\left (1-a^2 x^2\right )^2}{60 a^2}+\frac {x \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}{15 a}-\frac {\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2}{6 a^2}+\frac {4 \int \left (1-a^2 x^2\right ) \tanh ^{-1}(a x) \, dx}{15 a}\\ &=\frac {2 \left (1-a^2 x^2\right )}{45 a^2}+\frac {\left (1-a^2 x^2\right )^2}{60 a^2}+\frac {4 x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}{45 a}+\frac {x \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}{15 a}-\frac {\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2}{6 a^2}+\frac {8 \int \tanh ^{-1}(a x) \, dx}{45 a}\\ &=\frac {2 \left (1-a^2 x^2\right )}{45 a^2}+\frac {\left (1-a^2 x^2\right )^2}{60 a^2}+\frac {8 x \tanh ^{-1}(a x)}{45 a}+\frac {4 x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}{45 a}+\frac {x \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}{15 a}-\frac {\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2}{6 a^2}-\frac {8}{45} \int \frac {x}{1-a^2 x^2} \, dx\\ &=\frac {2 \left (1-a^2 x^2\right )}{45 a^2}+\frac {\left (1-a^2 x^2\right )^2}{60 a^2}+\frac {8 x \tanh ^{-1}(a x)}{45 a}+\frac {4 x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}{45 a}+\frac {x \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}{15 a}-\frac {\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2}{6 a^2}+\frac {4 \log \left (1-a^2 x^2\right )}{45 a^2}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.06, size = 82, normalized size = 0.59 \[ \frac {3 a^4 x^4-14 a^2 x^2+16 \log \left (1-a^2 x^2\right )+30 \left (a^2 x^2-1\right )^3 \tanh ^{-1}(a x)^2+4 a x \left (3 a^4 x^4-10 a^2 x^2+15\right ) \tanh ^{-1}(a x)}{180 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-14*a^2*x^2 + 3*a^4*x^4 + 4*a*x*(15 - 10*a^2*x^2 + 3*a^4*x^4)*ArcTanh[a*x] + 30*(-1 + a^2*x^2)^3*ArcTanh[a*x]
^2 + 16*Log[1 - a^2*x^2])/(180*a^2)

________________________________________________________________________________________

fricas [A]  time = 0.68, size = 116, normalized size = 0.84 \[ \frac {6 \, a^{4} x^{4} - 28 \, a^{2} x^{2} + 15 \, {\left (a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, a^{2} x^{2} - 1\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{2} + 4 \, {\left (3 \, a^{5} x^{5} - 10 \, a^{3} x^{3} + 15 \, a x\right )} \log \left (-\frac {a x + 1}{a x - 1}\right ) + 32 \, \log \left (a^{2} x^{2} - 1\right )}{360 \, a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/360*(6*a^4*x^4 - 28*a^2*x^2 + 15*(a^6*x^6 - 3*a^4*x^4 + 3*a^2*x^2 - 1)*log(-(a*x + 1)/(a*x - 1))^2 + 4*(3*a^
5*x^5 - 10*a^3*x^3 + 15*a*x)*log(-(a*x + 1)/(a*x - 1)) + 32*log(a^2*x^2 - 1))/a^2

________________________________________________________________________________________

giac [B]  time = 0.19, size = 473, normalized size = 3.43 \[ \frac {4}{45} \, a {\left (\frac {2 \, {\left (\frac {10 \, {\left (a x + 1\right )}^{2}}{{\left (a x - 1\right )}^{2}} - \frac {5 \, {\left (a x + 1\right )}}{a x - 1} + 1\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )}{\frac {{\left (a x + 1\right )}^{5} a^{3}}{{\left (a x - 1\right )}^{5}} - \frac {5 \, {\left (a x + 1\right )}^{4} a^{3}}{{\left (a x - 1\right )}^{4}} + \frac {10 \, {\left (a x + 1\right )}^{3} a^{3}}{{\left (a x - 1\right )}^{3}} - \frac {10 \, {\left (a x + 1\right )}^{2} a^{3}}{{\left (a x - 1\right )}^{2}} + \frac {5 \, {\left (a x + 1\right )} a^{3}}{a x - 1} - a^{3}} + \frac {30 \, {\left (a x + 1\right )}^{3} \log \left (-\frac {a x + 1}{a x - 1}\right )^{2}}{{\left (\frac {{\left (a x + 1\right )}^{6} a^{3}}{{\left (a x - 1\right )}^{6}} - \frac {6 \, {\left (a x + 1\right )}^{5} a^{3}}{{\left (a x - 1\right )}^{5}} + \frac {15 \, {\left (a x + 1\right )}^{4} a^{3}}{{\left (a x - 1\right )}^{4}} - \frac {20 \, {\left (a x + 1\right )}^{3} a^{3}}{{\left (a x - 1\right )}^{3}} + \frac {15 \, {\left (a x + 1\right )}^{2} a^{3}}{{\left (a x - 1\right )}^{2}} - \frac {6 \, {\left (a x + 1\right )} a^{3}}{a x - 1} + a^{3}\right )} {\left (a x - 1\right )}^{3}} - \frac {\frac {2 \, {\left (a x + 1\right )}^{3}}{{\left (a x - 1\right )}^{3}} - \frac {7 \, {\left (a x + 1\right )}^{2}}{{\left (a x - 1\right )}^{2}} + \frac {2 \, {\left (a x + 1\right )}}{a x - 1}}{\frac {{\left (a x + 1\right )}^{4} a^{3}}{{\left (a x - 1\right )}^{4}} - \frac {4 \, {\left (a x + 1\right )}^{3} a^{3}}{{\left (a x - 1\right )}^{3}} + \frac {6 \, {\left (a x + 1\right )}^{2} a^{3}}{{\left (a x - 1\right )}^{2}} - \frac {4 \, {\left (a x + 1\right )} a^{3}}{a x - 1} + a^{3}} - \frac {2 \, \log \left (-\frac {a x + 1}{a x - 1} + 1\right )}{a^{3}} + \frac {2 \, \log \left (-\frac {a x + 1}{a x - 1}\right )}{a^{3}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/45*a*(2*(10*(a*x + 1)^2/(a*x - 1)^2 - 5*(a*x + 1)/(a*x - 1) + 1)*log(-(a*x + 1)/(a*x - 1))/((a*x + 1)^5*a^3/
(a*x - 1)^5 - 5*(a*x + 1)^4*a^3/(a*x - 1)^4 + 10*(a*x + 1)^3*a^3/(a*x - 1)^3 - 10*(a*x + 1)^2*a^3/(a*x - 1)^2
+ 5*(a*x + 1)*a^3/(a*x - 1) - a^3) + 30*(a*x + 1)^3*log(-(a*x + 1)/(a*x - 1))^2/(((a*x + 1)^6*a^3/(a*x - 1)^6
- 6*(a*x + 1)^5*a^3/(a*x - 1)^5 + 15*(a*x + 1)^4*a^3/(a*x - 1)^4 - 20*(a*x + 1)^3*a^3/(a*x - 1)^3 + 15*(a*x +
1)^2*a^3/(a*x - 1)^2 - 6*(a*x + 1)*a^3/(a*x - 1) + a^3)*(a*x - 1)^3) - (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*x + 1)^4*a^3/(a*x - 1)^4 - 4*(a*x + 1)^3*a^3/(a*x - 1)^3 + 6*(a*x
+ 1)^2*a^3/(a*x - 1)^2 - 4*(a*x + 1)*a^3/(a*x - 1) + a^3) - 2*log(-(a*x + 1)/(a*x - 1) + 1)/a^3 + 2*log(-(a*x
+ 1)/(a*x - 1))/a^3)

________________________________________________________________________________________

maple [A]  time = 0.05, size = 219, normalized size = 1.59 \[ \frac {a^{4} \arctanh \left (a x \right )^{2} x^{6}}{6}-\frac {a^{2} \arctanh \left (a x \right )^{2} x^{4}}{2}+\frac {\arctanh \left (a x \right )^{2} x^{2}}{2}+\frac {a^{3} \arctanh \left (a x \right ) x^{5}}{15}-\frac {2 a \arctanh \left (a x \right ) x^{3}}{9}+\frac {x \arctanh \left (a x \right )}{3 a}+\frac {\arctanh \left (a x \right ) \ln \left (a x -1\right )}{6 a^{2}}-\frac {\arctanh \left (a x \right ) \ln \left (a x +1\right )}{6 a^{2}}+\frac {\ln \left (a x -1\right )^{2}}{24 a^{2}}-\frac {\ln \left (a x -1\right ) \ln \left (\frac {1}{2}+\frac {a x}{2}\right )}{12 a^{2}}+\frac {\ln \left (a x +1\right )^{2}}{24 a^{2}}+\frac {\ln \left (-\frac {a x}{2}+\frac {1}{2}\right ) \ln \left (\frac {1}{2}+\frac {a x}{2}\right )}{12 a^{2}}-\frac {\ln \left (-\frac {a x}{2}+\frac {1}{2}\right ) \ln \left (a x +1\right )}{12 a^{2}}+\frac {a^{2} x^{4}}{60}-\frac {7 x^{2}}{90}+\frac {4 \ln \left (a x -1\right )}{45 a^{2}}+\frac {4 \ln \left (a x +1\right )}{45 a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*a^4*arctanh(a*x)^2*x^6-1/2*a^2*arctanh(a*x)^2*x^4+1/2*arctanh(a*x)^2*x^2+1/15*a^3*arctanh(a*x)*x^5-2/9*a*a
rctanh(a*x)*x^3+1/3*x*arctanh(a*x)/a+1/6/a^2*arctanh(a*x)*ln(a*x-1)-1/6/a^2*arctanh(a*x)*ln(a*x+1)+1/24/a^2*ln
(a*x-1)^2-1/12/a^2*ln(a*x-1)*ln(1/2+1/2*a*x)+1/24/a^2*ln(a*x+1)^2+1/12/a^2*ln(-1/2*a*x+1/2)*ln(1/2+1/2*a*x)-1/
12/a^2*ln(-1/2*a*x+1/2)*ln(a*x+1)+1/60*a^2*x^4-7/90*x^2+4/45/a^2*ln(a*x-1)+4/45/a^2*ln(a*x+1)

________________________________________________________________________________________

maxima [A]  time = 0.31, size = 93, normalized size = 0.67 \[ \frac {{\left (a^{2} x^{2} - 1\right )}^{3} \operatorname {artanh}\left (a x\right )^{2}}{6 \, a^{2}} + \frac {{\left (3 \, a^{2} x^{4} - 14 \, x^{2} + \frac {16 \, \log \left (a x + 1\right )}{a^{2}} + \frac {16 \, \log \left (a x - 1\right )}{a^{2}}\right )} a + 4 \, {\left (3 \, a^{4} x^{5} - 10 \, a^{2} x^{3} + 15 \, x\right )} \operatorname {artanh}\left (a x\right )}{180 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(a^2*x^2 - 1)^3*arctanh(a*x)^2/a^2 + 1/180*((3*a^2*x^4 - 14*x^2 + 16*log(a*x + 1)/a^2 + 16*log(a*x - 1)/a^
2)*a + 4*(3*a^4*x^5 - 10*a^2*x^3 + 15*x)*arctanh(a*x))/a

________________________________________________________________________________________

mupad [B]  time = 1.00, size = 111, normalized size = 0.80 \[ \frac {x^2\,{\mathrm {atanh}\left (a\,x\right )}^2}{2}-\frac {{\mathrm {atanh}\left (a\,x\right )}^2}{6\,a^2}-\frac {7\,x^2}{90}+\frac {4\,\ln \left (a^2\,x^2-1\right )}{45\,a^2}+\frac {a^2\,x^4}{60}+\frac {x\,\mathrm {atanh}\left (a\,x\right )}{3\,a}-\frac {2\,a\,x^3\,\mathrm {atanh}\left (a\,x\right )}{9}+\frac {a^3\,x^5\,\mathrm {atanh}\left (a\,x\right )}{15}-\frac {a^2\,x^4\,{\mathrm {atanh}\left (a\,x\right )}^2}{2}+\frac {a^4\,x^6\,{\mathrm {atanh}\left (a\,x\right )}^2}{6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*atanh(a*x)^2*(a^2*x^2 - 1)^2,x)

[Out]

(x^2*atanh(a*x)^2)/2 - atanh(a*x)^2/(6*a^2) - (7*x^2)/90 + (4*log(a^2*x^2 - 1))/(45*a^2) + (a^2*x^4)/60 + (x*a
tanh(a*x))/(3*a) - (2*a*x^3*atanh(a*x))/9 + (a^3*x^5*atanh(a*x))/15 - (a^2*x^4*atanh(a*x)^2)/2 + (a^4*x^6*atan
h(a*x)^2)/6

________________________________________________________________________________________

sympy [A]  time = 2.11, size = 133, normalized size = 0.96 \[ \begin {cases} \frac {a^{4} x^{6} \operatorname {atanh}^{2}{\left (a x \right )}}{6} + \frac {a^{3} x^{5} \operatorname {atanh}{\left (a x \right )}}{15} - \frac {a^{2} x^{4} \operatorname {atanh}^{2}{\left (a x \right )}}{2} + \frac {a^{2} x^{4}}{60} - \frac {2 a x^{3} \operatorname {atanh}{\left (a x \right )}}{9} + \frac {x^{2} \operatorname {atanh}^{2}{\left (a x \right )}}{2} - \frac {7 x^{2}}{90} + \frac {x \operatorname {atanh}{\left (a x \right )}}{3 a} + \frac {8 \log {\left (x - \frac {1}{a} \right )}}{45 a^{2}} - \frac {\operatorname {atanh}^{2}{\left (a x \right )}}{6 a^{2}} + \frac {8 \operatorname {atanh}{\left (a x \right )}}{45 a^{2}} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a**4*x**6*atanh(a*x)**2/6 + a**3*x**5*atanh(a*x)/15 - a**2*x**4*atanh(a*x)**2/2 + a**2*x**4/60 - 2*
a*x**3*atanh(a*x)/9 + x**2*atanh(a*x)**2/2 - 7*x**2/90 + x*atanh(a*x)/(3*a) + 8*log(x - 1/a)/(45*a**2) - atanh
(a*x)**2/(6*a**2) + 8*atanh(a*x)/(45*a**2), Ne(a, 0)), (0, True))

________________________________________________________________________________________