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

Optimal. Leaf size=37 \[ -\frac{a+b \tanh ^{-1}(c x)}{2 x^2}+\frac{1}{2} b c^2 \tanh ^{-1}(c x)-\frac{b c}{2 x} \]

[Out]

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

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Rubi [A]  time = 0.0216618, antiderivative size = 37, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {5916, 325, 206} \[ -\frac{a+b \tanh ^{-1}(c x)}{2 x^2}+\frac{1}{2} b c^2 \tanh ^{-1}(c x)-\frac{b c}{2 x} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c*x])/x^3,x]

[Out]

-(b*c)/(2*x) + (b*c^2*ArcTanh[c*x])/2 - (a + b*ArcTanh[c*x])/(2*x^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 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.0083632, size = 59, normalized size = 1.59 \[ -\frac{a}{2 x^2}-\frac{1}{4} b c^2 \log (1-c x)+\frac{1}{4} b c^2 \log (c x+1)-\frac{b \tanh ^{-1}(c x)}{2 x^2}-\frac{b c}{2 x} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTanh[c*x])/x^3,x]

[Out]

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

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Maple [A]  time = 0.01, size = 49, normalized size = 1.3 \begin{align*} -{\frac{a}{2\,{x}^{2}}}-{\frac{b{\it Artanh} \left ( cx \right ) }{2\,{x}^{2}}}-{\frac{bc}{2\,x}}-{\frac{{c}^{2}b\ln \left ( cx-1 \right ) }{4}}+{\frac{{c}^{2}b\ln \left ( cx+1 \right ) }{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.978649, size = 61, normalized size = 1.65 \begin{align*} \frac{1}{4} \,{\left ({\left (c \log \left (c x + 1\right ) - c \log \left (c x - 1\right ) - \frac{2}{x}\right )} c - \frac{2 \, \operatorname{artanh}\left (c x\right )}{x^{2}}\right )} b - \frac{a}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.93332, size = 95, normalized size = 2.57 \begin{align*} -\frac{2 \, b c x -{\left (b c^{2} x^{2} - b\right )} \log \left (-\frac{c x + 1}{c x - 1}\right ) + 2 \, a}{4 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.739748, size = 36, normalized size = 0.97 \begin{align*} - \frac{a}{2 x^{2}} + \frac{b c^{2} \operatorname{atanh}{\left (c x \right )}}{2} - \frac{b c}{2 x} - \frac{b \operatorname{atanh}{\left (c x \right )}}{2 x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a/(2*x**2) + b*c**2*atanh(c*x)/2 - b*c/(2*x) - b*atanh(c*x)/(2*x**2)

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Giac [A]  time = 1.2016, size = 77, normalized size = 2.08 \begin{align*} \frac{1}{4} \, b c^{2} \log \left (c x + 1\right ) - \frac{1}{4} \, b c^{2} \log \left (c x - 1\right ) - \frac{b \log \left (-\frac{c x + 1}{c x - 1}\right )}{4 \, x^{2}} - \frac{b c x + a}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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