3.8 \(\int \frac{a+b \tanh ^{-1}(c x)}{x^2} \, dx\)

Optimal. Leaf size=36 \[ -\frac{a+b \tanh ^{-1}(c x)}{x}-\frac{1}{2} b c \log \left (1-c^2 x^2\right )+b c \log (x) \]

[Out]

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

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Rubi [A]  time = 0.0262703, antiderivative size = 36, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.417, Rules used = {5916, 266, 36, 29, 31} \[ -\frac{a+b \tanh ^{-1}(c x)}{x}-\frac{1}{2} b c \log \left (1-c^2 x^2\right )+b c \log (x) \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c*x])/x^2,x]

[Out]

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

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 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 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]

Rubi steps

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

Mathematica [A]  time = 0.0079244, size = 39, normalized size = 1.08 \[ -\frac{a}{x}-\frac{1}{2} b c \log \left (1-c^2 x^2\right )+b c \log (x)-\frac{b \tanh ^{-1}(c x)}{x} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTanh[c*x])/x^2,x]

[Out]

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

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Maple [A]  time = 0.01, size = 45, normalized size = 1.3 \begin{align*} -{\frac{a}{x}}-{\frac{b{\it Artanh} \left ( cx \right ) }{x}}-{\frac{cb\ln \left ( cx-1 \right ) }{2}}+cb\ln \left ( cx \right ) -{\frac{cb\ln \left ( cx+1 \right ) }{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctanh(c*x))/x^2,x)

[Out]

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

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Maxima [A]  time = 0.97607, size = 53, normalized size = 1.47 \begin{align*} -\frac{1}{2} \,{\left (c{\left (\log \left (c^{2} x^{2} - 1\right ) - \log \left (x^{2}\right )\right )} + \frac{2 \, \operatorname{artanh}\left (c x\right )}{x}\right )} b - \frac{a}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))/x^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.87, size = 116, normalized size = 3.22 \begin{align*} -\frac{b c x \log \left (c^{2} x^{2} - 1\right ) - 2 \, b c x \log \left (x\right ) + b \log \left (-\frac{c x + 1}{c x - 1}\right ) + 2 \, a}{2 \, x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))/x^2,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.946261, size = 41, normalized size = 1.14 \begin{align*} \begin{cases} - \frac{a}{x} + b c \log{\left (x \right )} - b c \log{\left (x - \frac{1}{c} \right )} - b c \operatorname{atanh}{\left (c x \right )} - \frac{b \operatorname{atanh}{\left (c x \right )}}{x} & \text{for}\: c \neq 0 \\- \frac{a}{x} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atanh(c*x))/x**2,x)

[Out]

Piecewise((-a/x + b*c*log(x) - b*c*log(x - 1/c) - b*c*atanh(c*x) - b*atanh(c*x)/x, Ne(c, 0)), (-a/x, True))

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Giac [A]  time = 1.15233, size = 63, normalized size = 1.75 \begin{align*} -\frac{1}{2} \, b c \log \left (c^{2} x^{2} - 1\right ) + b c \log \left (x\right ) - \frac{b \log \left (-\frac{c x + 1}{c x - 1}\right )}{2 \, x} - \frac{a}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))/x^2,x, algorithm="giac")

[Out]

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