3.106 \(\int \frac{a+b \tanh ^{-1}(c x^3)}{x^{10}} \, dx\)

Optimal. Leaf size=56 \[ -\frac{a+b \tanh ^{-1}\left (c x^3\right )}{9 x^9}-\frac{1}{18} b c^3 \log \left (1-c^2 x^6\right )+\frac{1}{3} b c^3 \log (x)-\frac{b c}{18 x^6} \]

[Out]

-(b*c)/(18*x^6) - (a + b*ArcTanh[c*x^3])/(9*x^9) + (b*c^3*Log[x])/3 - (b*c^3*Log[1 - c^2*x^6])/18

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Rubi [A]  time = 0.0363813, antiderivative size = 56, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {6097, 266, 44} \[ -\frac{a+b \tanh ^{-1}\left (c x^3\right )}{9 x^9}-\frac{1}{18} b c^3 \log \left (1-c^2 x^6\right )+\frac{1}{3} b c^3 \log (x)-\frac{b c}{18 x^6} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c*x^3])/x^10,x]

[Out]

-(b*c)/(18*x^6) - (a + b*ArcTanh[c*x^3])/(9*x^9) + (b*c^3*Log[x])/3 - (b*c^3*Log[1 - c^2*x^6])/18

Rule 6097

Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcTa
nh[c*x^n]))/(d*(m + 1)), x] - Dist[(b*c*n)/(d*(m + 1)), Int[(x^(n - 1)*(d*x)^(m + 1))/(1 - c^2*x^(2*n)), x], x
] /; FreeQ[{a, b, c, d, m, n}, x] && 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}\left (c x^3\right )}{x^{10}} \, dx &=-\frac{a+b \tanh ^{-1}\left (c x^3\right )}{9 x^9}+\frac{1}{3} (b c) \int \frac{1}{x^7 \left (1-c^2 x^6\right )} \, dx\\ &=-\frac{a+b \tanh ^{-1}\left (c x^3\right )}{9 x^9}+\frac{1}{18} (b c) \operatorname{Subst}\left (\int \frac{1}{x^2 \left (1-c^2 x\right )} \, dx,x,x^6\right )\\ &=-\frac{a+b \tanh ^{-1}\left (c x^3\right )}{9 x^9}+\frac{1}{18} (b c) \operatorname{Subst}\left (\int \left (\frac{1}{x^2}+\frac{c^2}{x}-\frac{c^4}{-1+c^2 x}\right ) \, dx,x,x^6\right )\\ &=-\frac{b c}{18 x^6}-\frac{a+b \tanh ^{-1}\left (c x^3\right )}{9 x^9}+\frac{1}{3} b c^3 \log (x)-\frac{1}{18} b c^3 \log \left (1-c^2 x^6\right )\\ \end{align*}

Mathematica [A]  time = 0.0117102, size = 61, normalized size = 1.09 \[ -\frac{a}{9 x^9}-\frac{1}{18} b c^3 \log \left (1-c^2 x^6\right )+\frac{1}{3} b c^3 \log (x)-\frac{b c}{18 x^6}-\frac{b \tanh ^{-1}\left (c x^3\right )}{9 x^9} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTanh[c*x^3])/x^10,x]

[Out]

-a/(9*x^9) - (b*c)/(18*x^6) - (b*ArcTanh[c*x^3])/(9*x^9) + (b*c^3*Log[x])/3 - (b*c^3*Log[1 - c^2*x^6])/18

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Maple [A]  time = 0.014, size = 63, normalized size = 1.1 \begin{align*} -{\frac{a}{9\,{x}^{9}}}-{\frac{b{\it Artanh} \left ( c{x}^{3} \right ) }{9\,{x}^{9}}}-{\frac{bc}{18\,{x}^{6}}}+{\frac{b{c}^{3}\ln \left ( x \right ) }{3}}-{\frac{b{c}^{3}\ln \left ( c{x}^{3}-1 \right ) }{18}}-{\frac{b{c}^{3}\ln \left ( c{x}^{3}+1 \right ) }{18}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctanh(c*x^3))/x^10,x)

[Out]

-1/9*a/x^9-1/9*b/x^9*arctanh(c*x^3)-1/18*b*c/x^6+1/3*b*c^3*ln(x)-1/18*b*c^3*ln(c*x^3-1)-1/18*b*c^3*ln(c*x^3+1)

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Maxima [A]  time = 0.964576, size = 69, normalized size = 1.23 \begin{align*} -\frac{1}{18} \,{\left ({\left (c^{2} \log \left (c^{2} x^{6} - 1\right ) - c^{2} \log \left (x^{6}\right ) + \frac{1}{x^{6}}\right )} c + \frac{2 \, \operatorname{artanh}\left (c x^{3}\right )}{x^{9}}\right )} b - \frac{a}{9 \, x^{9}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x^3))/x^10,x, algorithm="maxima")

[Out]

-1/18*((c^2*log(c^2*x^6 - 1) - c^2*log(x^6) + 1/x^6)*c + 2*arctanh(c*x^3)/x^9)*b - 1/9*a/x^9

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Fricas [A]  time = 2.15736, size = 150, normalized size = 2.68 \begin{align*} -\frac{b c^{3} x^{9} \log \left (c^{2} x^{6} - 1\right ) - 6 \, b c^{3} x^{9} \log \left (x\right ) + b c x^{3} + b \log \left (-\frac{c x^{3} + 1}{c x^{3} - 1}\right ) + 2 \, a}{18 \, x^{9}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x^3))/x^10,x, algorithm="fricas")

[Out]

-1/18*(b*c^3*x^9*log(c^2*x^6 - 1) - 6*b*c^3*x^9*log(x) + b*c*x^3 + b*log(-(c*x^3 + 1)/(c*x^3 - 1)) + 2*a)/x^9

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atanh(c*x**3))/x**10,x)

[Out]

Timed out

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Giac [A]  time = 1.17147, size = 88, normalized size = 1.57 \begin{align*} -\frac{1}{18} \, b c^{3} \log \left (c^{2} x^{6} - 1\right ) + \frac{1}{3} \, b c^{3} \log \left (x\right ) - \frac{b \log \left (-\frac{c x^{3} + 1}{c x^{3} - 1}\right )}{18 \, x^{9}} - \frac{b c x^{3} + 2 \, a}{18 \, x^{9}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x^3))/x^10,x, algorithm="giac")

[Out]

-1/18*b*c^3*log(c^2*x^6 - 1) + 1/3*b*c^3*log(x) - 1/18*b*log(-(c*x^3 + 1)/(c*x^3 - 1))/x^9 - 1/18*(b*c*x^3 + 2
*a)/x^9