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

Optimal. Leaf size=127 \[ -\frac {3 \tanh ^{-1}(a x)^2}{32 a^4}+\frac {x^4}{32 \left (1-a^2 x^2\right )^2}+\frac {x^4 \tanh ^{-1}(a x)^2}{4 \left (1-a^2 x^2\right )^2}-\frac {x^3 \tanh ^{-1}(a x)}{8 a \left (1-a^2 x^2\right )^2}-\frac {3}{32 a^4 \left (1-a^2 x^2\right )}+\frac {3 x \tanh ^{-1}(a x)}{16 a^3 \left (1-a^2 x^2\right )} \]

[Out]

1/32*x^4/(-a^2*x^2+1)^2-3/32/a^4/(-a^2*x^2+1)-1/8*x^3*arctanh(a*x)/a/(-a^2*x^2+1)^2+3/16*x*arctanh(a*x)/a^3/(-
a^2*x^2+1)-3/32*arctanh(a*x)^2/a^4+1/4*x^4*arctanh(a*x)^2/(-a^2*x^2+1)^2

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Rubi [A]  time = 0.18, antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {6008, 6002, 5998, 5948} \[ \frac {x^4}{32 \left (1-a^2 x^2\right )^2}-\frac {3}{32 a^4 \left (1-a^2 x^2\right )}+\frac {x^4 \tanh ^{-1}(a x)^2}{4 \left (1-a^2 x^2\right )^2}-\frac {x^3 \tanh ^{-1}(a x)}{8 a \left (1-a^2 x^2\right )^2}+\frac {3 x \tanh ^{-1}(a x)}{16 a^3 \left (1-a^2 x^2\right )}-\frac {3 \tanh ^{-1}(a x)^2}{32 a^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

x^4/(32*(1 - a^2*x^2)^2) - 3/(32*a^4*(1 - a^2*x^2)) - (x^3*ArcTanh[a*x])/(8*a*(1 - a^2*x^2)^2) + (3*x*ArcTanh[
a*x])/(16*a^3*(1 - a^2*x^2)) - (3*ArcTanh[a*x]^2)/(32*a^4) + (x^4*ArcTanh[a*x]^2)/(4*(1 - a^2*x^2)^2)

Rule 5948

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

Rule 5998

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))*(x_)^2*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> -Simp[(b*(d + e*x^2)^(
q + 1))/(4*c^3*d*(q + 1)^2), x] + (Dist[1/(2*c^2*d*(q + 1)), Int[(d + e*x^2)^(q + 1)*(a + b*ArcTanh[c*x]), x],
x] - Simp[(x*(d + e*x^2)^(q + 1)*(a + b*ArcTanh[c*x]))/(2*c^2*d*(q + 1)), x]) /; FreeQ[{a, b, c, d, e}, x] &&
EqQ[c^2*d + e, 0] && LtQ[q, -1] && NeQ[q, -5/2]

Rule 6002

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

Rule 6008

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> Sim
p[((f*x)^(m + 1)*(d + e*x^2)^(q + 1)*(a + b*ArcTanh[c*x])^p)/(d*(m + 1)), x] - Dist[(b*c*p)/(m + 1), Int[(f*x)
^(m + 1)*(d + e*x^2)^q*(a + b*ArcTanh[c*x])^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && EqQ[c^2*d
+ e, 0] && EqQ[m + 2*q + 3, 0] && GtQ[p, 0] && NeQ[m, -1]

Rubi steps

\begin {align*} \int \frac {x^3 \tanh ^{-1}(a x)^2}{\left (1-a^2 x^2\right )^3} \, dx &=\frac {x^4 \tanh ^{-1}(a x)^2}{4 \left (1-a^2 x^2\right )^2}-\frac {1}{2} a \int \frac {x^4 \tanh ^{-1}(a x)}{\left (1-a^2 x^2\right )^3} \, dx\\ &=\frac {x^4}{32 \left (1-a^2 x^2\right )^2}-\frac {x^3 \tanh ^{-1}(a x)}{8 a \left (1-a^2 x^2\right )^2}+\frac {x^4 \tanh ^{-1}(a x)^2}{4 \left (1-a^2 x^2\right )^2}+\frac {3 \int \frac {x^2 \tanh ^{-1}(a x)}{\left (1-a^2 x^2\right )^2} \, dx}{8 a}\\ &=\frac {x^4}{32 \left (1-a^2 x^2\right )^2}-\frac {3}{32 a^4 \left (1-a^2 x^2\right )}-\frac {x^3 \tanh ^{-1}(a x)}{8 a \left (1-a^2 x^2\right )^2}+\frac {3 x \tanh ^{-1}(a x)}{16 a^3 \left (1-a^2 x^2\right )}+\frac {x^4 \tanh ^{-1}(a x)^2}{4 \left (1-a^2 x^2\right )^2}-\frac {3 \int \frac {\tanh ^{-1}(a x)}{1-a^2 x^2} \, dx}{16 a^3}\\ &=\frac {x^4}{32 \left (1-a^2 x^2\right )^2}-\frac {3}{32 a^4 \left (1-a^2 x^2\right )}-\frac {x^3 \tanh ^{-1}(a x)}{8 a \left (1-a^2 x^2\right )^2}+\frac {3 x \tanh ^{-1}(a x)}{16 a^3 \left (1-a^2 x^2\right )}-\frac {3 \tanh ^{-1}(a x)^2}{32 a^4}+\frac {x^4 \tanh ^{-1}(a x)^2}{4 \left (1-a^2 x^2\right )^2}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 71, normalized size = 0.56 \[ \frac {\left (6 a x-10 a^3 x^3\right ) \tanh ^{-1}(a x)+5 a^2 x^2+\left (5 a^4 x^4+6 a^2 x^2-3\right ) \tanh ^{-1}(a x)^2-4}{32 a^4 \left (a^2 x^2-1\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4 + 5*a^2*x^2 + (6*a*x - 10*a^3*x^3)*ArcTanh[a*x] + (-3 + 6*a^2*x^2 + 5*a^4*x^4)*ArcTanh[a*x]^2)/(32*a^4*(-1
+ a^2*x^2)^2)

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fricas [A]  time = 0.70, size = 99, normalized size = 0.78 \[ \frac {20 \, a^{2} x^{2} + {\left (5 \, a^{4} x^{4} + 6 \, a^{2} x^{2} - 3\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{2} - 4 \, {\left (5 \, a^{3} x^{3} - 3 \, a x\right )} \log \left (-\frac {a x + 1}{a x - 1}\right ) - 16}{128 \, {\left (a^{8} x^{4} - 2 \, a^{6} x^{2} + a^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/128*(20*a^2*x^2 + (5*a^4*x^4 + 6*a^2*x^2 - 3)*log(-(a*x + 1)/(a*x - 1))^2 - 4*(5*a^3*x^3 - 3*a*x)*log(-(a*x
+ 1)/(a*x - 1)) - 16)/(a^8*x^4 - 2*a^6*x^2 + a^4)

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giac [B]  time = 0.17, size = 250, normalized size = 1.97 \[ \frac {1}{512} \, {\left (2 \, {\left (\frac {{\left (a x - 1\right )}^{2} {\left (\frac {4 \, {\left (a x + 1\right )}}{a x - 1} + 1\right )}}{{\left (a x + 1\right )}^{2} a^{5}} + \frac {{\left (a x + 1\right )}^{2}}{{\left (a x - 1\right )}^{2} a^{5}} + \frac {4 \, {\left (a x + 1\right )}}{{\left (a x - 1\right )} a^{5}}\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{2} + 2 \, {\left (\frac {{\left (a x - 1\right )}^{2} {\left (\frac {8 \, {\left (a x + 1\right )}}{a x - 1} + 1\right )}}{{\left (a x + 1\right )}^{2} a^{5}} - \frac {{\left (a x + 1\right )}^{2}}{{\left (a x - 1\right )}^{2} a^{5}} - \frac {8 \, {\left (a x + 1\right )}}{{\left (a x - 1\right )} a^{5}}\right )} \log \left (-\frac {a x + 1}{a x - 1}\right ) + \frac {{\left (a x - 1\right )}^{2} {\left (\frac {16 \, {\left (a x + 1\right )}}{a x - 1} + 1\right )}}{{\left (a x + 1\right )}^{2} a^{5}} + \frac {{\left (a x + 1\right )}^{2}}{{\left (a x - 1\right )}^{2} a^{5}} + \frac {16 \, {\left (a x + 1\right )}}{{\left (a x - 1\right )} a^{5}}\right )} a \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/512*(2*((a*x - 1)^2*(4*(a*x + 1)/(a*x - 1) + 1)/((a*x + 1)^2*a^5) + (a*x + 1)^2/((a*x - 1)^2*a^5) + 4*(a*x +
1)/((a*x - 1)*a^5))*log(-(a*x + 1)/(a*x - 1))^2 + 2*((a*x - 1)^2*(8*(a*x + 1)/(a*x - 1) + 1)/((a*x + 1)^2*a^5
) - (a*x + 1)^2/((a*x - 1)^2*a^5) - 8*(a*x + 1)/((a*x - 1)*a^5))*log(-(a*x + 1)/(a*x - 1)) + (a*x - 1)^2*(16*(
a*x + 1)/(a*x - 1) + 1)/((a*x + 1)^2*a^5) + (a*x + 1)^2/((a*x - 1)^2*a^5) + 16*(a*x + 1)/((a*x - 1)*a^5))*a

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maple [B]  time = 0.07, size = 297, normalized size = 2.34 \[ \frac {\arctanh \left (a x \right )^{2}}{16 a^{4} \left (a x -1\right )^{2}}+\frac {3 \arctanh \left (a x \right )^{2}}{16 a^{4} \left (a x -1\right )}+\frac {\arctanh \left (a x \right )^{2}}{16 a^{4} \left (a x +1\right )^{2}}-\frac {3 \arctanh \left (a x \right )^{2}}{16 a^{4} \left (a x +1\right )}-\frac {\arctanh \left (a x \right )}{32 a^{4} \left (a x -1\right )^{2}}-\frac {5 \arctanh \left (a x \right )}{32 a^{4} \left (a x -1\right )}-\frac {5 \arctanh \left (a x \right ) \ln \left (a x -1\right )}{32 a^{4}}+\frac {\arctanh \left (a x \right )}{32 a^{4} \left (a x +1\right )^{2}}-\frac {5 \arctanh \left (a x \right )}{32 a^{4} \left (a x +1\right )}+\frac {5 \arctanh \left (a x \right ) \ln \left (a x +1\right )}{32 a^{4}}-\frac {5 \ln \left (a x -1\right )^{2}}{128 a^{4}}+\frac {5 \ln \left (a x -1\right ) \ln \left (\frac {1}{2}+\frac {a x}{2}\right )}{64 a^{4}}-\frac {5 \ln \left (a x +1\right )^{2}}{128 a^{4}}-\frac {5 \ln \left (-\frac {a x}{2}+\frac {1}{2}\right ) \ln \left (\frac {1}{2}+\frac {a x}{2}\right )}{64 a^{4}}+\frac {5 \ln \left (-\frac {a x}{2}+\frac {1}{2}\right ) \ln \left (a x +1\right )}{64 a^{4}}+\frac {1}{128 a^{4} \left (a x -1\right )^{2}}+\frac {9}{128 a^{4} \left (a x -1\right )}+\frac {1}{128 a^{4} \left (a x +1\right )^{2}}-\frac {9}{128 a^{4} \left (a x +1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16/a^4*arctanh(a*x)^2/(a*x-1)^2+3/16/a^4*arctanh(a*x)^2/(a*x-1)+1/16/a^4*arctanh(a*x)^2/(a*x+1)^2-3/16/a^4*a
rctanh(a*x)^2/(a*x+1)-1/32/a^4*arctanh(a*x)/(a*x-1)^2-5/32/a^4*arctanh(a*x)/(a*x-1)-5/32/a^4*arctanh(a*x)*ln(a
*x-1)+1/32/a^4*arctanh(a*x)/(a*x+1)^2-5/32/a^4*arctanh(a*x)/(a*x+1)+5/32/a^4*arctanh(a*x)*ln(a*x+1)-5/128/a^4*
ln(a*x-1)^2+5/64/a^4*ln(a*x-1)*ln(1/2+1/2*a*x)-5/128/a^4*ln(a*x+1)^2-5/64/a^4*ln(-1/2*a*x+1/2)*ln(1/2+1/2*a*x)
+5/64/a^4*ln(-1/2*a*x+1/2)*ln(a*x+1)+1/128/a^4/(a*x-1)^2+9/128/a^4/(a*x-1)+1/128/a^4/(a*x+1)^2-9/128/a^4/(a*x+
1)

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maxima [B]  time = 0.33, size = 226, normalized size = 1.78 \[ -\frac {1}{32} \, a {\left (\frac {2 \, {\left (5 \, a^{2} x^{3} - 3 \, x\right )}}{a^{8} x^{4} - 2 \, a^{6} x^{2} + a^{4}} - \frac {5 \, \log \left (a x + 1\right )}{a^{5}} + \frac {5 \, \log \left (a x - 1\right )}{a^{5}}\right )} \operatorname {artanh}\left (a x\right ) + \frac {{\left (20 \, a^{2} x^{2} - 5 \, {\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \log \left (a x + 1\right )^{2} + 10 \, {\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \log \left (a x + 1\right ) \log \left (a x - 1\right ) - 5 \, {\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \log \left (a x - 1\right )^{2} - 16\right )} a^{2}}{128 \, {\left (a^{10} x^{4} - 2 \, a^{8} x^{2} + a^{6}\right )}} + \frac {{\left (2 \, a^{2} x^{2} - 1\right )} \operatorname {artanh}\left (a x\right )^{2}}{4 \, {\left (a^{8} x^{4} - 2 \, a^{6} x^{2} + a^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/32*a*(2*(5*a^2*x^3 - 3*x)/(a^8*x^4 - 2*a^6*x^2 + a^4) - 5*log(a*x + 1)/a^5 + 5*log(a*x - 1)/a^5)*arctanh(a*
x) + 1/128*(20*a^2*x^2 - 5*(a^4*x^4 - 2*a^2*x^2 + 1)*log(a*x + 1)^2 + 10*(a^4*x^4 - 2*a^2*x^2 + 1)*log(a*x + 1
)*log(a*x - 1) - 5*(a^4*x^4 - 2*a^2*x^2 + 1)*log(a*x - 1)^2 - 16)*a^2/(a^10*x^4 - 2*a^8*x^2 + a^6) + 1/4*(2*a^
2*x^2 - 1)*arctanh(a*x)^2/(a^8*x^4 - 2*a^6*x^2 + a^4)

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mupad [B]  time = 1.53, size = 372, normalized size = 2.93 \[ {\ln \left (a\,x+1\right )}^2\,\left (\frac {5}{128\,a^4}-\frac {\frac {1}{16\,a^5}-\frac {x^2}{8\,a^3}}{\frac {1}{a}-2\,a\,x^2+a^3\,x^4}\right )-\ln \left (1-a\,x\right )\,\left (\frac {\frac {3\,x}{8}+a\,x^2-\frac {3}{4\,a}-\frac {5\,a^2\,x^3}{8}}{8\,a^7\,x^4-16\,a^5\,x^2+8\,a^3}+\frac {\frac {3\,x}{8}-a\,x^2+\frac {3}{4\,a}-\frac {5\,a^2\,x^3}{8}}{8\,a^7\,x^4-16\,a^5\,x^2+8\,a^3}-\ln \left (a\,x+1\right )\,\left (\frac {\frac {1}{4\,a^4}-\frac {x^2}{2\,a^2}}{2\,a^4\,x^4-4\,a^2\,x^2+2}-\frac {5\,\left (a^4\,x^4-2\,a^2\,x^2+1\right )}{32\,a^4\,\left (2\,a^4\,x^4-4\,a^2\,x^2+2\right )}\right )\right )-{\ln \left (1-a\,x\right )}^2\,\left (\frac {\frac {1}{4\,a^4}-\frac {x^2}{2\,a^2}}{4\,a^4\,x^4-8\,a^2\,x^2+4}-\frac {5}{128\,a^4}\right )-\frac {\frac {2}{a^2}-\frac {5\,x^2}{2}}{16\,a^6\,x^4-32\,a^4\,x^2+16\,a^2}+\frac {\ln \left (a\,x+1\right )\,\left (\frac {3\,x}{32\,a^4}-\frac {5\,x^3}{32\,a^2}\right )}{\frac {1}{a}-2\,a\,x^2+a^3\,x^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x^3*atanh(a*x)^2)/(a^2*x^2 - 1)^3,x)

[Out]

log(a*x + 1)^2*(5/(128*a^4) - (1/(16*a^5) - x^2/(8*a^3))/(1/a - 2*a*x^2 + a^3*x^4)) - log(1 - a*x)*(((3*x)/8 +
a*x^2 - 3/(4*a) - (5*a^2*x^3)/8)/(8*a^3 - 16*a^5*x^2 + 8*a^7*x^4) + ((3*x)/8 - a*x^2 + 3/(4*a) - (5*a^2*x^3)/
8)/(8*a^3 - 16*a^5*x^2 + 8*a^7*x^4) - log(a*x + 1)*((1/(4*a^4) - x^2/(2*a^2))/(2*a^4*x^4 - 4*a^2*x^2 + 2) - (5
*(a^4*x^4 - 2*a^2*x^2 + 1))/(32*a^4*(2*a^4*x^4 - 4*a^2*x^2 + 2)))) - log(1 - a*x)^2*((1/(4*a^4) - x^2/(2*a^2))
/(4*a^4*x^4 - 8*a^2*x^2 + 4) - 5/(128*a^4)) - (2/a^2 - (5*x^2)/2)/(16*a^2 - 32*a^4*x^2 + 16*a^6*x^4) + (log(a*
x + 1)*((3*x)/(32*a^4) - (5*x^3)/(32*a^2)))/(1/a - 2*a*x^2 + a^3*x^4)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \int \frac {x^{3} \operatorname {atanh}^{2}{\left (a x \right )}}{a^{6} x^{6} - 3 a^{4} x^{4} + 3 a^{2} x^{2} - 1}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(x**3*atanh(a*x)**2/(a**6*x**6 - 3*a**4*x**4 + 3*a**2*x**2 - 1), x)

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