3.10 \(\int \frac{a+b \tanh ^{-1}(c x)}{x^4} \, dx\)

Optimal. Leaf size=54 \[ -\frac{a+b \tanh ^{-1}(c x)}{3 x^3}-\frac{1}{6} b c^3 \log \left (1-c^2 x^2\right )+\frac{1}{3} b c^3 \log (x)-\frac{b c}{6 x^2} \]

[Out]

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

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Rubi [A]  time = 0.0360336, antiderivative size = 54, 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} \[ -\frac{a+b \tanh ^{-1}(c x)}{3 x^3}-\frac{1}{6} b c^3 \log \left (1-c^2 x^2\right )+\frac{1}{3} b c^3 \log (x)-\frac{b c}{6 x^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c*x])/x^4,x]

[Out]

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

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

Mathematica [A]  time = 0.008483, size = 59, normalized size = 1.09 \[ -\frac{a}{3 x^3}-\frac{1}{6} b c^3 \log \left (1-c^2 x^2\right )+\frac{1}{3} b c^3 \log (x)-\frac{b c}{6 x^2}-\frac{b \tanh ^{-1}(c x)}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTanh[c*x])/x^4,x]

[Out]

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

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Maple [A]  time = 0.01, size = 59, normalized size = 1.1 \begin{align*} -{\frac{a}{3\,{x}^{3}}}-{\frac{b{\it Artanh} \left ( cx \right ) }{3\,{x}^{3}}}-{\frac{{c}^{3}b\ln \left ( cx-1 \right ) }{6}}-{\frac{bc}{6\,{x}^{2}}}+{\frac{{c}^{3}b\ln \left ( cx \right ) }{3}}-{\frac{{c}^{3}b\ln \left ( cx+1 \right ) }{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctanh(c*x))/x^4,x)

[Out]

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

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Maxima [A]  time = 0.967296, size = 66, normalized size = 1.22 \begin{align*} -\frac{1}{6} \,{\left ({\left (c^{2} \log \left (c^{2} x^{2} - 1\right ) - c^{2} \log \left (x^{2}\right ) + \frac{1}{x^{2}}\right )} c + \frac{2 \, \operatorname{artanh}\left (c x\right )}{x^{3}}\right )} b - \frac{a}{3 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))/x^4,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.92786, size = 140, normalized size = 2.59 \begin{align*} -\frac{b c^{3} x^{3} \log \left (c^{2} x^{2} - 1\right ) - 2 \, b c^{3} x^{3} \log \left (x\right ) + b c x + b \log \left (-\frac{c x + 1}{c x - 1}\right ) + 2 \, a}{6 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))/x^4,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 1.5614, size = 70, normalized size = 1.3 \begin{align*} \begin{cases} - \frac{a}{3 x^{3}} + \frac{b c^{3} \log{\left (x \right )}}{3} - \frac{b c^{3} \log{\left (x - \frac{1}{c} \right )}}{3} - \frac{b c^{3} \operatorname{atanh}{\left (c x \right )}}{3} - \frac{b c}{6 x^{2}} - \frac{b \operatorname{atanh}{\left (c x \right )}}{3 x^{3}} & \text{for}\: c \neq 0 \\- \frac{a}{3 x^{3}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atanh(c*x))/x**4,x)

[Out]

Piecewise((-a/(3*x**3) + b*c**3*log(x)/3 - b*c**3*log(x - 1/c)/3 - b*c**3*atanh(c*x)/3 - b*c/(6*x**2) - b*atan
h(c*x)/(3*x**3), Ne(c, 0)), (-a/(3*x**3), True))

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Giac [A]  time = 1.20603, size = 80, normalized size = 1.48 \begin{align*} -\frac{1}{6} \, b c^{3} \log \left (c^{2} x^{2} - 1\right ) + \frac{1}{3} \, b c^{3} \log \left (x\right ) - \frac{b \log \left (-\frac{c x + 1}{c x - 1}\right )}{6 \, x^{3}} - \frac{b c x + 2 \, a}{6 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))/x^4,x, algorithm="giac")

[Out]

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