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

Optimal. Leaf size=30 \[ a x+\frac{b \log \left (1-c^2 x^2\right )}{2 c}+b x \tanh ^{-1}(c x) \]

[Out]

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

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Rubi [A]  time = 0.0125908, antiderivative size = 30, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {5910, 260} \[ a x+\frac{b \log \left (1-c^2 x^2\right )}{2 c}+b x \tanh ^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

Int[a + b*ArcTanh[c*x],x]

[Out]

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

Rule 5910

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcTanh[c*x])^p, x] - Dist[b*c*p, In
t[(x*(a + b*ArcTanh[c*x])^(p - 1))/(1 - c^2*x^2), x], x] /; FreeQ[{a, b, c}, x] && IGtQ[p, 0]

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rubi steps

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

Mathematica [A]  time = 0.0028612, size = 30, normalized size = 1. \[ a x+\frac{b \log \left (1-c^2 x^2\right )}{2 c}+b x \tanh ^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

Integrate[a + b*ArcTanh[c*x],x]

[Out]

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

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Maple [A]  time = 0.002, size = 29, normalized size = 1. \begin{align*} ax+bx{\it Artanh} \left ( cx \right ) +{\frac{b\ln \left ( -{c}^{2}{x}^{2}+1 \right ) }{2\,c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(a+b*arctanh(c*x),x)

[Out]

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

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Maxima [A]  time = 0.975951, size = 41, normalized size = 1.37 \begin{align*} a x + \frac{{\left (2 \, c x \operatorname{artanh}\left (c x\right ) + \log \left (-c^{2} x^{2} + 1\right )\right )} b}{2 \, c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.94706, size = 97, normalized size = 3.23 \begin{align*} \frac{b c x \log \left (-\frac{c x + 1}{c x - 1}\right ) + 2 \, a c x + b \log \left (c^{2} x^{2} - 1\right )}{2 \, c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.386652, size = 27, normalized size = 0.9 \begin{align*} a x + b \left (\begin{cases} x \operatorname{atanh}{\left (c x \right )} + \frac{\log{\left (c x + 1 \right )}}{c} - \frac{\operatorname{atanh}{\left (c x \right )}}{c} & \text{for}\: c \neq 0 \\0 & \text{otherwise} \end{cases}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.17245, size = 54, normalized size = 1.8 \begin{align*} \frac{1}{2} \,{\left (x \log \left (-\frac{c x + 1}{c x - 1}\right ) + \frac{\log \left ({\left | c^{2} x^{2} - 1 \right |}\right )}{c}\right )} b + a x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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