3.345 \(\int \frac {\tanh ^{-1}(a x)}{(1-a^2 x^2)^4} \, dx\)

Optimal. Leaf size=134 \[ -\frac {5}{32 a \left (1-a^2 x^2\right )}-\frac {5}{96 a \left (1-a^2 x^2\right )^2}-\frac {1}{36 a \left (1-a^2 x^2\right )^3}+\frac {5 x \tanh ^{-1}(a x)}{16 \left (1-a^2 x^2\right )}+\frac {5 x \tanh ^{-1}(a x)}{24 \left (1-a^2 x^2\right )^2}+\frac {x \tanh ^{-1}(a x)}{6 \left (1-a^2 x^2\right )^3}+\frac {5 \tanh ^{-1}(a x)^2}{32 a} \]

[Out]

-1/36/a/(-a^2*x^2+1)^3-5/96/a/(-a^2*x^2+1)^2-5/32/a/(-a^2*x^2+1)+1/6*x*arctanh(a*x)/(-a^2*x^2+1)^3+5/24*x*arct
anh(a*x)/(-a^2*x^2+1)^2+5/16*x*arctanh(a*x)/(-a^2*x^2+1)+5/32*arctanh(a*x)^2/a

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Rubi [A]  time = 0.07, antiderivative size = 134, 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 = {5960, 5956, 261} \[ -\frac {5}{32 a \left (1-a^2 x^2\right )}-\frac {5}{96 a \left (1-a^2 x^2\right )^2}-\frac {1}{36 a \left (1-a^2 x^2\right )^3}+\frac {5 x \tanh ^{-1}(a x)}{16 \left (1-a^2 x^2\right )}+\frac {5 x \tanh ^{-1}(a x)}{24 \left (1-a^2 x^2\right )^2}+\frac {x \tanh ^{-1}(a x)}{6 \left (1-a^2 x^2\right )^3}+\frac {5 \tanh ^{-1}(a x)^2}{32 a} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[a*x]/(1 - a^2*x^2)^4,x]

[Out]

-1/(36*a*(1 - a^2*x^2)^3) - 5/(96*a*(1 - a^2*x^2)^2) - 5/(32*a*(1 - a^2*x^2)) + (x*ArcTanh[a*x])/(6*(1 - a^2*x
^2)^3) + (5*x*ArcTanh[a*x])/(24*(1 - a^2*x^2)^2) + (5*x*ArcTanh[a*x])/(16*(1 - a^2*x^2)) + (5*ArcTanh[a*x]^2)/
(32*a)

Rule 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rule 5956

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

Rule 5960

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

Rubi steps

\begin {align*} \int \frac {\tanh ^{-1}(a x)}{\left (1-a^2 x^2\right )^4} \, dx &=-\frac {1}{36 a \left (1-a^2 x^2\right )^3}+\frac {x \tanh ^{-1}(a x)}{6 \left (1-a^2 x^2\right )^3}+\frac {5}{6} \int \frac {\tanh ^{-1}(a x)}{\left (1-a^2 x^2\right )^3} \, dx\\ &=-\frac {1}{36 a \left (1-a^2 x^2\right )^3}-\frac {5}{96 a \left (1-a^2 x^2\right )^2}+\frac {x \tanh ^{-1}(a x)}{6 \left (1-a^2 x^2\right )^3}+\frac {5 x \tanh ^{-1}(a x)}{24 \left (1-a^2 x^2\right )^2}+\frac {5}{8} \int \frac {\tanh ^{-1}(a x)}{\left (1-a^2 x^2\right )^2} \, dx\\ &=-\frac {1}{36 a \left (1-a^2 x^2\right )^3}-\frac {5}{96 a \left (1-a^2 x^2\right )^2}+\frac {x \tanh ^{-1}(a x)}{6 \left (1-a^2 x^2\right )^3}+\frac {5 x \tanh ^{-1}(a x)}{24 \left (1-a^2 x^2\right )^2}+\frac {5 x \tanh ^{-1}(a x)}{16 \left (1-a^2 x^2\right )}+\frac {5 \tanh ^{-1}(a x)^2}{32 a}-\frac {1}{16} (5 a) \int \frac {x}{\left (1-a^2 x^2\right )^2} \, dx\\ &=-\frac {1}{36 a \left (1-a^2 x^2\right )^3}-\frac {5}{96 a \left (1-a^2 x^2\right )^2}-\frac {5}{32 a \left (1-a^2 x^2\right )}+\frac {x \tanh ^{-1}(a x)}{6 \left (1-a^2 x^2\right )^3}+\frac {5 x \tanh ^{-1}(a x)}{24 \left (1-a^2 x^2\right )^2}+\frac {5 x \tanh ^{-1}(a x)}{16 \left (1-a^2 x^2\right )}+\frac {5 \tanh ^{-1}(a x)^2}{32 a}\\ \end {align*}

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Mathematica [A]  time = 0.18, size = 81, normalized size = 0.60 \[ \frac {45 a^4 x^4-105 a^2 x^2+45 \left (a^2 x^2-1\right )^3 \tanh ^{-1}(a x)^2-6 a x \left (15 a^4 x^4-40 a^2 x^2+33\right ) \tanh ^{-1}(a x)+68}{288 a \left (a^2 x^2-1\right )^3} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTanh[a*x]/(1 - a^2*x^2)^4,x]

[Out]

(68 - 105*a^2*x^2 + 45*a^4*x^4 - 6*a*x*(33 - 40*a^2*x^2 + 15*a^4*x^4)*ArcTanh[a*x] + 45*(-1 + a^2*x^2)^3*ArcTa
nh[a*x]^2)/(288*a*(-1 + a^2*x^2)^3)

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fricas [A]  time = 0.47, size = 131, normalized size = 0.98 \[ \frac {180 \, a^{4} x^{4} - 420 \, a^{2} x^{2} + 45 \, {\left (a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, a^{2} x^{2} - 1\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{2} - 12 \, {\left (15 \, a^{5} x^{5} - 40 \, a^{3} x^{3} + 33 \, a x\right )} \log \left (-\frac {a x + 1}{a x - 1}\right ) + 272}{1152 \, {\left (a^{7} x^{6} - 3 \, a^{5} x^{4} + 3 \, a^{3} x^{2} - a\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(a*x)/(-a^2*x^2+1)^4,x, algorithm="fricas")

[Out]

1/1152*(180*a^4*x^4 - 420*a^2*x^2 + 45*(a^6*x^6 - 3*a^4*x^4 + 3*a^2*x^2 - 1)*log(-(a*x + 1)/(a*x - 1))^2 - 12*
(15*a^5*x^5 - 40*a^3*x^3 + 33*a*x)*log(-(a*x + 1)/(a*x - 1)) + 272)/(a^7*x^6 - 3*a^5*x^4 + 3*a^3*x^2 - a)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {artanh}\left (a x\right )}{{\left (a^{2} x^{2} - 1\right )}^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(a*x)/(-a^2*x^2+1)^4,x, algorithm="giac")

[Out]

integrate(arctanh(a*x)/(a^2*x^2 - 1)^4, x)

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maple [B]  time = 0.07, size = 281, normalized size = 2.10 \[ -\frac {\arctanh \left (a x \right )}{48 a \left (a x -1\right )^{3}}+\frac {\arctanh \left (a x \right )}{16 a \left (a x -1\right )^{2}}-\frac {5 \arctanh \left (a x \right )}{32 a \left (a x -1\right )}-\frac {5 \arctanh \left (a x \right ) \ln \left (a x -1\right )}{32 a}-\frac {\arctanh \left (a x \right )}{48 a \left (a x +1\right )^{3}}-\frac {\arctanh \left (a x \right )}{16 a \left (a x +1\right )^{2}}-\frac {5 \arctanh \left (a x \right )}{32 a \left (a x +1\right )}+\frac {5 \arctanh \left (a x \right ) \ln \left (a x +1\right )}{32 a}-\frac {5 \ln \left (a x -1\right )^{2}}{128 a}+\frac {5 \ln \left (a x -1\right ) \ln \left (\frac {1}{2}+\frac {a x}{2}\right )}{64 a}-\frac {5 \ln \left (a x +1\right )^{2}}{128 a}-\frac {5 \ln \left (-\frac {a x}{2}+\frac {1}{2}\right ) \ln \left (\frac {1}{2}+\frac {a x}{2}\right )}{64 a}+\frac {5 \ln \left (-\frac {a x}{2}+\frac {1}{2}\right ) \ln \left (a x +1\right )}{64 a}+\frac {1}{288 a \left (a x -1\right )^{3}}-\frac {7}{384 a \left (a x -1\right )^{2}}+\frac {37}{384 a \left (a x -1\right )}-\frac {1}{288 a \left (a x +1\right )^{3}}-\frac {7}{384 a \left (a x +1\right )^{2}}-\frac {37}{384 a \left (a x +1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(a*x)/(-a^2*x^2+1)^4,x)

[Out]

-1/48/a*arctanh(a*x)/(a*x-1)^3+1/16/a*arctanh(a*x)/(a*x-1)^2-5/32/a*arctanh(a*x)/(a*x-1)-5/32/a*arctanh(a*x)*l
n(a*x-1)-1/48/a*arctanh(a*x)/(a*x+1)^3-1/16/a*arctanh(a*x)/(a*x+1)^2-5/32/a*arctanh(a*x)/(a*x+1)+5/32/a*arctan
h(a*x)*ln(a*x+1)-5/128/a*ln(a*x-1)^2+5/64/a*ln(a*x-1)*ln(1/2+1/2*a*x)-5/128/a*ln(a*x+1)^2-5/64/a*ln(-1/2*a*x+1
/2)*ln(1/2+1/2*a*x)+5/64/a*ln(-1/2*a*x+1/2)*ln(a*x+1)+1/288/a/(a*x-1)^3-7/384/a/(a*x-1)^2+37/384/a/(a*x-1)-1/2
88/a/(a*x+1)^3-7/384/a/(a*x+1)^2-37/384/a/(a*x+1)

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maxima [B]  time = 0.32, size = 240, normalized size = 1.79 \[ -\frac {1}{96} \, {\left (\frac {2 \, {\left (15 \, a^{4} x^{5} - 40 \, a^{2} x^{3} + 33 \, x\right )}}{a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, a^{2} x^{2} - 1} - \frac {15 \, \log \left (a x + 1\right )}{a} + \frac {15 \, \log \left (a x - 1\right )}{a}\right )} \operatorname {artanh}\left (a x\right ) + \frac {{\left (180 \, a^{4} x^{4} - 420 \, a^{2} x^{2} - 45 \, {\left (a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, a^{2} x^{2} - 1\right )} \log \left (a x + 1\right )^{2} + 90 \, {\left (a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, a^{2} x^{2} - 1\right )} \log \left (a x + 1\right ) \log \left (a x - 1\right ) - 45 \, {\left (a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, a^{2} x^{2} - 1\right )} \log \left (a x - 1\right )^{2} + 272\right )} a}{1152 \, {\left (a^{8} x^{6} - 3 \, a^{6} x^{4} + 3 \, a^{4} x^{2} - a^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(a*x)/(-a^2*x^2+1)^4,x, algorithm="maxima")

[Out]

-1/96*(2*(15*a^4*x^5 - 40*a^2*x^3 + 33*x)/(a^6*x^6 - 3*a^4*x^4 + 3*a^2*x^2 - 1) - 15*log(a*x + 1)/a + 15*log(a
*x - 1)/a)*arctanh(a*x) + 1/1152*(180*a^4*x^4 - 420*a^2*x^2 - 45*(a^6*x^6 - 3*a^4*x^4 + 3*a^2*x^2 - 1)*log(a*x
+ 1)^2 + 90*(a^6*x^6 - 3*a^4*x^4 + 3*a^2*x^2 - 1)*log(a*x + 1)*log(a*x - 1) - 45*(a^6*x^6 - 3*a^4*x^4 + 3*a^2
*x^2 - 1)*log(a*x - 1)^2 + 272)*a/(a^8*x^6 - 3*a^6*x^4 + 3*a^4*x^2 - a^2)

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mupad [B]  time = 1.30, size = 206, normalized size = 1.54 \[ \frac {\frac {34}{3\,a}-\frac {35\,a\,x^2}{2}+\frac {15\,a^3\,x^4}{2}}{48\,a^6\,x^6-144\,a^4\,x^4+144\,a^2\,x^2-48}-\ln \left (1-a\,x\right )\,\left (\frac {5\,\ln \left (a\,x+1\right )}{64\,a}-\frac {\frac {5\,a^4\,x^5}{16}-\frac {5\,a^2\,x^3}{6}+\frac {11\,x}{16}}{2\,a^6\,x^6-6\,a^4\,x^4+6\,a^2\,x^2-2}\right )+\frac {5\,{\ln \left (a\,x+1\right )}^2}{128\,a}+\frac {5\,{\ln \left (1-a\,x\right )}^2}{128\,a}-\frac {\ln \left (a\,x+1\right )\,\left (\frac {11\,x}{32\,a}-\frac {5\,a\,x^3}{12}+\frac {5\,a^3\,x^5}{32}\right )}{3\,a\,x^2-\frac {1}{a}-3\,a^3\,x^4+a^5\,x^6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(a*x)/(a^2*x^2 - 1)^4,x)

[Out]

(34/(3*a) - (35*a*x^2)/2 + (15*a^3*x^4)/2)/(144*a^2*x^2 - 144*a^4*x^4 + 48*a^6*x^6 - 48) - log(1 - a*x)*((5*lo
g(a*x + 1))/(64*a) - ((11*x)/16 - (5*a^2*x^3)/6 + (5*a^4*x^5)/16)/(6*a^2*x^2 - 6*a^4*x^4 + 2*a^6*x^6 - 2)) + (
5*log(a*x + 1)^2)/(128*a) + (5*log(1 - a*x)^2)/(128*a) - (log(a*x + 1)*((11*x)/(32*a) - (5*a*x^3)/12 + (5*a^3*
x^5)/32))/(3*a*x^2 - 1/a - 3*a^3*x^4 + a^5*x^6)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(a*x)/(-a**2*x**2+1)**4,x)

[Out]

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

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