3.2.100 \(\int \frac {(1-a^2 x^2)^2 \tanh ^{-1}(a x)}{x^4} \, dx\) [200]

Optimal. Leaf size=68 \[ -\frac {a}{6 x^2}-\frac {\tanh ^{-1}(a x)}{3 x^3}+\frac {2 a^2 \tanh ^{-1}(a x)}{x}+a^4 x \tanh ^{-1}(a x)-\frac {5}{3} a^3 \log (x)+\frac {4}{3} a^3 \log \left (1-a^2 x^2\right ) \]

[Out]

-1/6*a/x^2-1/3*arctanh(a*x)/x^3+2*a^2*arctanh(a*x)/x+a^4*x*arctanh(a*x)-5/3*a^3*ln(x)+4/3*a^3*ln(-a^2*x^2+1)

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Rubi [A]
time = 0.08, antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.450, Rules used = {6159, 6021, 266, 6037, 272, 46, 36, 29, 31} \begin {gather*} a^4 x \tanh ^{-1}(a x)-\frac {5}{3} a^3 \log (x)+\frac {2 a^2 \tanh ^{-1}(a x)}{x}+\frac {4}{3} a^3 \log \left (1-a^2 x^2\right )-\frac {\tanh ^{-1}(a x)}{3 x^3}-\frac {a}{6 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/6*a/x^2 - ArcTanh[a*x]/(3*x^3) + (2*a^2*ArcTanh[a*x])/x + a^4*x*ArcTanh[a*x] - (5*a^3*Log[x])/3 + (4*a^3*Lo
g[1 - a^2*x^2])/3

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 46

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x
)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[m + n + 2, 0])

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 272

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

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 6037

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

Rule 6159

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

Rubi steps

\begin {align*} \int \frac {\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}{x^4} \, dx &=\int \left (a^4 \tanh ^{-1}(a x)+\frac {\tanh ^{-1}(a x)}{x^4}-\frac {2 a^2 \tanh ^{-1}(a x)}{x^2}\right ) \, dx\\ &=-\left (\left (2 a^2\right ) \int \frac {\tanh ^{-1}(a x)}{x^2} \, dx\right )+a^4 \int \tanh ^{-1}(a x) \, dx+\int \frac {\tanh ^{-1}(a x)}{x^4} \, dx\\ &=-\frac {\tanh ^{-1}(a x)}{3 x^3}+\frac {2 a^2 \tanh ^{-1}(a x)}{x}+a^4 x \tanh ^{-1}(a x)+\frac {1}{3} a \int \frac {1}{x^3 \left (1-a^2 x^2\right )} \, dx-\left (2 a^3\right ) \int \frac {1}{x \left (1-a^2 x^2\right )} \, dx-a^5 \int \frac {x}{1-a^2 x^2} \, dx\\ &=-\frac {\tanh ^{-1}(a x)}{3 x^3}+\frac {2 a^2 \tanh ^{-1}(a x)}{x}+a^4 x \tanh ^{-1}(a x)+\frac {1}{2} a^3 \log \left (1-a^2 x^2\right )+\frac {1}{6} a \text {Subst}\left (\int \frac {1}{x^2 \left (1-a^2 x\right )} \, dx,x,x^2\right )-a^3 \text {Subst}\left (\int \frac {1}{x \left (1-a^2 x\right )} \, dx,x,x^2\right )\\ &=-\frac {\tanh ^{-1}(a x)}{3 x^3}+\frac {2 a^2 \tanh ^{-1}(a x)}{x}+a^4 x \tanh ^{-1}(a x)+\frac {1}{2} a^3 \log \left (1-a^2 x^2\right )+\frac {1}{6} a \text {Subst}\left (\int \left (\frac {1}{x^2}+\frac {a^2}{x}-\frac {a^4}{-1+a^2 x}\right ) \, dx,x,x^2\right )-a^3 \text {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )-a^5 \text {Subst}\left (\int \frac {1}{1-a^2 x} \, dx,x,x^2\right )\\ &=-\frac {a}{6 x^2}-\frac {\tanh ^{-1}(a x)}{3 x^3}+\frac {2 a^2 \tanh ^{-1}(a x)}{x}+a^4 x \tanh ^{-1}(a x)-\frac {5}{3} a^3 \log (x)+\frac {4}{3} a^3 \log \left (1-a^2 x^2\right )\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 68, normalized size = 1.00 \begin {gather*} -\frac {a}{6 x^2}-\frac {\tanh ^{-1}(a x)}{3 x^3}+\frac {2 a^2 \tanh ^{-1}(a x)}{x}+a^4 x \tanh ^{-1}(a x)-\frac {5}{3} a^3 \log (x)+\frac {4}{3} a^3 \log \left (1-a^2 x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/6*a/x^2 - ArcTanh[a*x]/(3*x^3) + (2*a^2*ArcTanh[a*x])/x + a^4*x*ArcTanh[a*x] - (5*a^3*Log[x])/3 + (4*a^3*Lo
g[1 - a^2*x^2])/3

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Maple [A]
time = 0.28, size = 67, normalized size = 0.99

method	result	size
derivativedivides	\(a^{3} \left (a x \arctanh \left (a x \right )-\frac {\arctanh \left (a x \right )}{3 a^{3} x^{3}}+\frac {2 \arctanh \left (a x \right )}{a x}+\frac {4 \ln \left (a x -1\right )}{3}+\frac {4 \ln \left (a x +1\right )}{3}-\frac {1}{6 a^{2} x^{2}}-\frac {5 \ln \left (a x \right )}{3}\right )\)	\(67\)
default	\(a^{3} \left (a x \arctanh \left (a x \right )-\frac {\arctanh \left (a x \right )}{3 a^{3} x^{3}}+\frac {2 \arctanh \left (a x \right )}{a x}+\frac {4 \ln \left (a x -1\right )}{3}+\frac {4 \ln \left (a x +1\right )}{3}-\frac {1}{6 a^{2} x^{2}}-\frac {5 \ln \left (a x \right )}{3}\right )\)	\(67\)
risch	\(\frac {\left (3 a^{4} x^{4}+6 a^{2} x^{2}-1\right ) \ln \left (a x +1\right )}{6 x^{3}}-\frac {3 x^{4} \ln \left (-a x +1\right ) a^{4}+10 \ln \left (x \right ) a^{3} x^{3}-8 \ln \left (-a^{2} x^{2}+1\right ) a^{3} x^{3}+6 x^{2} \ln \left (-a x +1\right ) a^{2}+a x -\ln \left (-a x +1\right )}{6 x^{3}}\)	\(108\)
meijerg	\(-\frac {a^{3} \left (-\frac {2 \left (10 a^{2} x^{2}+30\right )}{45 a^{2} x^{2}}-\frac {2 \left (\ln \left (1-\sqrt {a^{2} x^{2}}\right )-\ln \left (1+\sqrt {a^{2} x^{2}}\right )\right )}{3 a^{2} x^{2} \sqrt {a^{2} x^{2}}}+\frac {2 \ln \left (-a^{2} x^{2}+1\right )}{3}+\frac {4}{9}-\frac {4 \ln \left (x \right )}{3}-\frac {4 \ln \left (i a \right )}{3}+\frac {2}{a^{2} x^{2}}\right )}{4}-\frac {a^{3} \left (\frac {2 a^{2} x^{2} \left (\ln \left (1-\sqrt {a^{2} x^{2}}\right )-\ln \left (1+\sqrt {a^{2} x^{2}}\right )\right )}{\sqrt {a^{2} x^{2}}}-2 \ln \left (-a^{2} x^{2}+1\right )\right )}{4}-\frac {a^{3} \left (\frac {2 \ln \left (1-\sqrt {a^{2} x^{2}}\right )-2 \ln \left (1+\sqrt {a^{2} x^{2}}\right )}{\sqrt {a^{2} x^{2}}}-2 \ln \left (-a^{2} x^{2}+1\right )+4 \ln \left (x \right )+4 \ln \left (i a \right )\right )}{2}\)	\(240\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^3*(a*x*arctanh(a*x)-1/3*arctanh(a*x)/a^3/x^3+2*arctanh(a*x)/a/x+4/3*ln(a*x-1)+4/3*ln(a*x+1)-1/6/a^2/x^2-5/3*
ln(a*x))

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Maxima [A]
time = 0.27, size = 66, normalized size = 0.97 \begin {gather*} \frac {1}{6} \, {\left (8 \, a^{2} \log \left (a x + 1\right ) + 8 \, a^{2} \log \left (a x - 1\right ) - 10 \, a^{2} \log \left (x\right ) - \frac {1}{x^{2}}\right )} a + \frac {1}{3} \, {\left (3 \, a^{4} x + \frac {6 \, a^{2} x^{2} - 1}{x^{3}}\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^4,x, algorithm="maxima")

[Out]

1/6*(8*a^2*log(a*x + 1) + 8*a^2*log(a*x - 1) - 10*a^2*log(x) - 1/x^2)*a + 1/3*(3*a^4*x + (6*a^2*x^2 - 1)/x^3)*
arctanh(a*x)

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Fricas [A]
time = 0.36, size = 72, normalized size = 1.06 \begin {gather*} \frac {8 \, a^{3} x^{3} \log \left (a^{2} x^{2} - 1\right ) - 10 \, a^{3} x^{3} \log \left (x\right ) - a x + {\left (3 \, a^{4} x^{4} + 6 \, a^{2} x^{2} - 1\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )}{6 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(8*a^3*x^3*log(a^2*x^2 - 1) - 10*a^3*x^3*log(x) - a*x + (3*a^4*x^4 + 6*a^2*x^2 - 1)*log(-(a*x + 1)/(a*x -
1)))/x^3

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Sympy [A]
time = 0.47, size = 75, normalized size = 1.10 \begin {gather*} \begin {cases} a^{4} x \operatorname {atanh}{\left (a x \right )} - \frac {5 a^{3} \log {\left (x \right )}}{3} + \frac {8 a^{3} \log {\left (x - \frac {1}{a} \right )}}{3} + \frac {8 a^{3} \operatorname {atanh}{\left (a x \right )}}{3} + \frac {2 a^{2} \operatorname {atanh}{\left (a x \right )}}{x} - \frac {a}{6 x^{2}} - \frac {\operatorname {atanh}{\left (a x \right )}}{3 x^{3}} & \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**4,x)

[Out]

Piecewise((a**4*x*atanh(a*x) - 5*a**3*log(x)/3 + 8*a**3*log(x - 1/a)/3 + 8*a**3*atanh(a*x)/3 + 2*a**2*atanh(a*
x)/x - a/(6*x**2) - atanh(a*x)/(3*x**3), Ne(a, 0)), (0, True))

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 274 vs. \(2 (60) = 120\).
time = 0.39, size = 274, normalized size = 4.03 \begin {gather*} \frac {1}{3} \, {\left (8 \, a^{2} \log \left (\frac {{\left | -a x - 1 \right |}}{{\left | a x - 1 \right |}}\right ) - 3 \, a^{2} \log \left ({\left | -\frac {a x + 1}{a x - 1} + 1 \right |}\right ) - 5 \, a^{2} \log \left ({\left | -\frac {a x + 1}{a x - 1} - 1 \right |}\right ) + {\left (\frac {3 \, a^{2}}{\frac {a x + 1}{a x - 1} - 1} - \frac {\frac {3 \, {\left (a x + 1\right )}^{2} a^{2}}{{\left (a x - 1\right )}^{2}} + \frac {12 \, {\left (a x + 1\right )} a^{2}}{a x - 1} + 5 \, a^{2}}{{\left (\frac {a x + 1}{a x - 1} + 1\right )}^{3}}\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 ) + \frac {2 \, {\left (a x + 1\right )} a^{2}}{{\left (a x - 1\right )} {\left (\frac {a x + 1}{a x - 1} + 1\right )}^{2}}\right )} 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^4,x, algorithm="giac")

[Out]

1/3*(8*a^2*log(abs(-a*x - 1)/abs(a*x - 1)) - 3*a^2*log(abs(-(a*x + 1)/(a*x - 1) + 1)) - 5*a^2*log(abs(-(a*x +
1)/(a*x - 1) - 1)) + (3*a^2/((a*x + 1)/(a*x - 1) - 1) - (3*(a*x + 1)^2*a^2/(a*x - 1)^2 + 12*(a*x + 1)*a^2/(a*x
- 1) + 5*a^2)/((a*x + 1)/(a*x - 1) + 1)^3)*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)) + 2*(a*x + 1)*a^2/((a*x - 1)*((a*x + 1)/(a*x -
1) + 1)^2))*a

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Mupad [B]
time = 0.88, size = 59, normalized size = 0.87 \begin {gather*} \frac {4\,a^3\,\ln \left (a^2\,x^2-1\right )}{3}-\frac {a}{6\,x^2}-\frac {\mathrm {atanh}\left (a\,x\right )}{3\,x^3}-\frac {5\,a^3\,\ln \left (x\right )}{3}+a^4\,x\,\mathrm {atanh}\left (a\,x\right )+\frac {2\,a^2\,\mathrm {atanh}\left (a\,x\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(4*a^3*log(a^2*x^2 - 1))/3 - a/(6*x^2) - atanh(a*x)/(3*x^3) - (5*a^3*log(x))/3 + a^4*x*atanh(a*x) + (2*a^2*ata
nh(a*x))/x

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