3.12 \(\int \frac{a+b \tanh ^{-1}(c x)}{x^6} \, dx\)

Optimal. Leaf size=65 \[ -\frac{a+b \tanh ^{-1}(c x)}{5 x^5}-\frac{b c^3}{10 x^2}-\frac{1}{10} b c^5 \log \left (1-c^2 x^2\right )+\frac{1}{5} b c^5 \log (x)-\frac{b c}{20 x^4} \]

[Out]

-(b*c)/(20*x^4) - (b*c^3)/(10*x^2) - (a + b*ArcTanh[c*x])/(5*x^5) + (b*c^5*Log[x])/5 - (b*c^5*Log[1 - c^2*x^2]
)/10

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Rubi [A]  time = 0.0413205, antiderivative size = 65, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {5916, 266, 44} \[ -\frac{a+b \tanh ^{-1}(c x)}{5 x^5}-\frac{b c^3}{10 x^2}-\frac{1}{10} b c^5 \log \left (1-c^2 x^2\right )+\frac{1}{5} b c^5 \log (x)-\frac{b c}{20 x^4} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c*x])/x^6,x]

[Out]

-(b*c)/(20*x^4) - (b*c^3)/(10*x^2) - (a + b*ArcTanh[c*x])/(5*x^5) + (b*c^5*Log[x])/5 - (b*c^5*Log[1 - c^2*x^2]
)/10

Rule 5916

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

Rule 266

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 44

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] && L
tQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{a+b \tanh ^{-1}(c x)}{x^6} \, dx &=-\frac{a+b \tanh ^{-1}(c x)}{5 x^5}+\frac{1}{5} (b c) \int \frac{1}{x^5 \left (1-c^2 x^2\right )} \, dx\\ &=-\frac{a+b \tanh ^{-1}(c x)}{5 x^5}+\frac{1}{10} (b c) \operatorname{Subst}\left (\int \frac{1}{x^3 \left (1-c^2 x\right )} \, dx,x,x^2\right )\\ &=-\frac{a+b \tanh ^{-1}(c x)}{5 x^5}+\frac{1}{10} (b c) \operatorname{Subst}\left (\int \left (\frac{1}{x^3}+\frac{c^2}{x^2}+\frac{c^4}{x}-\frac{c^6}{-1+c^2 x}\right ) \, dx,x,x^2\right )\\ &=-\frac{b c}{20 x^4}-\frac{b c^3}{10 x^2}-\frac{a+b \tanh ^{-1}(c x)}{5 x^5}+\frac{1}{5} b c^5 \log (x)-\frac{1}{10} b c^5 \log \left (1-c^2 x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.0089695, size = 70, normalized size = 1.08 \[ -\frac{a}{5 x^5}-\frac{b c^3}{10 x^2}-\frac{1}{10} b c^5 \log \left (1-c^2 x^2\right )+\frac{1}{5} b c^5 \log (x)-\frac{b c}{20 x^4}-\frac{b \tanh ^{-1}(c x)}{5 x^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTanh[c*x])/x^6,x]

[Out]

-a/(5*x^5) - (b*c)/(20*x^4) - (b*c^3)/(10*x^2) - (b*ArcTanh[c*x])/(5*x^5) + (b*c^5*Log[x])/5 - (b*c^5*Log[1 -
c^2*x^2])/10

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Maple [A]  time = 0.01, size = 68, normalized size = 1.1 \begin{align*} -{\frac{a}{5\,{x}^{5}}}-{\frac{b{\it Artanh} \left ( cx \right ) }{5\,{x}^{5}}}-{\frac{{c}^{5}b\ln \left ( cx-1 \right ) }{10}}-{\frac{bc}{20\,{x}^{4}}}-{\frac{b{c}^{3}}{10\,{x}^{2}}}+{\frac{{c}^{5}b\ln \left ( cx \right ) }{5}}-{\frac{{c}^{5}b\ln \left ( cx+1 \right ) }{10}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctanh(c*x))/x^6,x)

[Out]

-1/5*a/x^5-1/5*b/x^5*arctanh(c*x)-1/10*c^5*b*ln(c*x-1)-1/20*b*c/x^4-1/10*b*c^3/x^2+1/5*c^5*b*ln(c*x)-1/10*c^5*
b*ln(c*x+1)

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Maxima [A]  time = 0.972758, size = 82, normalized size = 1.26 \begin{align*} -\frac{1}{20} \,{\left ({\left (2 \, c^{4} \log \left (c^{2} x^{2} - 1\right ) - 2 \, c^{4} \log \left (x^{2}\right ) + \frac{2 \, c^{2} x^{2} + 1}{x^{4}}\right )} c + \frac{4 \, \operatorname{artanh}\left (c x\right )}{x^{5}}\right )} b - \frac{a}{5 \, x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))/x^6,x, algorithm="maxima")

[Out]

-1/20*((2*c^4*log(c^2*x^2 - 1) - 2*c^4*log(x^2) + (2*c^2*x^2 + 1)/x^4)*c + 4*arctanh(c*x)/x^5)*b - 1/5*a/x^5

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Fricas [A]  time = 1.97896, size = 166, normalized size = 2.55 \begin{align*} -\frac{2 \, b c^{5} x^{5} \log \left (c^{2} x^{2} - 1\right ) - 4 \, b c^{5} x^{5} \log \left (x\right ) + 2 \, b c^{3} x^{3} + b c x + 2 \, b \log \left (-\frac{c x + 1}{c x - 1}\right ) + 4 \, a}{20 \, x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))/x^6,x, algorithm="fricas")

[Out]

-1/20*(2*b*c^5*x^5*log(c^2*x^2 - 1) - 4*b*c^5*x^5*log(x) + 2*b*c^3*x^3 + b*c*x + 2*b*log(-(c*x + 1)/(c*x - 1))
+ 4*a)/x^5

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Sympy [A]  time = 2.78227, size = 80, normalized size = 1.23 \begin{align*} \begin{cases} - \frac{a}{5 x^{5}} + \frac{b c^{5} \log{\left (x \right )}}{5} - \frac{b c^{5} \log{\left (x - \frac{1}{c} \right )}}{5} - \frac{b c^{5} \operatorname{atanh}{\left (c x \right )}}{5} - \frac{b c^{3}}{10 x^{2}} - \frac{b c}{20 x^{4}} - \frac{b \operatorname{atanh}{\left (c x \right )}}{5 x^{5}} & \text{for}\: c \neq 0 \\- \frac{a}{5 x^{5}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atanh(c*x))/x**6,x)

[Out]

Piecewise((-a/(5*x**5) + b*c**5*log(x)/5 - b*c**5*log(x - 1/c)/5 - b*c**5*atanh(c*x)/5 - b*c**3/(10*x**2) - b*
c/(20*x**4) - b*atanh(c*x)/(5*x**5), Ne(c, 0)), (-a/(5*x**5), True))

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Giac [A]  time = 1.22231, size = 92, normalized size = 1.42 \begin{align*} -\frac{1}{10} \, b c^{5} \log \left (c^{2} x^{2} - 1\right ) + \frac{1}{5} \, b c^{5} \log \left (x\right ) - \frac{b \log \left (-\frac{c x + 1}{c x - 1}\right )}{10 \, x^{5}} - \frac{2 \, b c^{3} x^{3} + b c x + 4 \, a}{20 \, x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))/x^6,x, algorithm="giac")

[Out]

-1/10*b*c^5*log(c^2*x^2 - 1) + 1/5*b*c^5*log(x) - 1/10*b*log(-(c*x + 1)/(c*x - 1))/x^5 - 1/20*(2*b*c^3*x^3 + b
*c*x + 4*a)/x^5