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

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

[Out]

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

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Rubi [A]  time = 0.0172118, antiderivative size = 37, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {5916, 321, 206} \[ \frac{1}{2} x^2 \left (a+b \tanh ^{-1}(c x)\right )-\frac{b \tanh ^{-1}(c x)}{2 c^2}+\frac{b x}{2 c} \]

Antiderivative was successfully verified.

[In]

Int[x*(a + b*ArcTanh[c*x]),x]

[Out]

(b*x)/(2*c) - (b*ArcTanh[c*x])/(2*c^2) + (x^2*(a + b*ArcTanh[c*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 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && 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 x \left (a+b \tanh ^{-1}(c x)\right ) \, dx &=\frac{1}{2} x^2 \left (a+b \tanh ^{-1}(c x)\right )-\frac{1}{2} (b c) \int \frac{x^2}{1-c^2 x^2} \, dx\\ &=\frac{b x}{2 c}+\frac{1}{2} x^2 \left (a+b \tanh ^{-1}(c x)\right )-\frac{b \int \frac{1}{1-c^2 x^2} \, dx}{2 c}\\ &=\frac{b x}{2 c}-\frac{b \tanh ^{-1}(c x)}{2 c^2}+\frac{1}{2} x^2 \left (a+b \tanh ^{-1}(c x)\right )\\ \end{align*}

Mathematica [A]  time = 0.0072913, size = 59, normalized size = 1.59 \[ \frac{a x^2}{2}+\frac{b \log (1-c x)}{4 c^2}-\frac{b \log (c x+1)}{4 c^2}+\frac{1}{2} b x^2 \tanh ^{-1}(c x)+\frac{b x}{2 c} \]

Antiderivative was successfully verified.

[In]

Integrate[x*(a + b*ArcTanh[c*x]),x]

[Out]

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

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Maple [A]  time = 0.007, size = 49, normalized size = 1.3 \begin{align*}{\frac{a{x}^{2}}{2}}+{\frac{b{x}^{2}{\it Artanh} \left ( cx \right ) }{2}}+{\frac{bx}{2\,c}}+{\frac{b\ln \left ( cx-1 \right ) }{4\,{c}^{2}}}-{\frac{b\ln \left ( cx+1 \right ) }{4\,{c}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a*x^2+1/2*b*x^2*arctanh(c*x)+1/2*b*x/c+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 = 0.98538, size = 68, normalized size = 1.84 \begin{align*} \frac{1}{2} \, a x^{2} + \frac{1}{4} \,{\left (2 \, x^{2} \operatorname{artanh}\left (c x\right ) + c{\left (\frac{2 \, x}{c^{2}} - \frac{\log \left (c x + 1\right )}{c^{3}} + \frac{\log \left (c x - 1\right )}{c^{3}}\right )}\right )} b \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a*x^2 + 1/4*(2*x^2*arctanh(c*x) + c*(2*x/c^2 - log(c*x + 1)/c^3 + log(c*x - 1)/c^3))*b

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Fricas [A]  time = 1.90822, size = 104, normalized size = 2.81 \begin{align*} \frac{2 \, a c^{2} x^{2} + 2 \, b c x +{\left (b c^{2} x^{2} - b\right )} \log \left (-\frac{c x + 1}{c x - 1}\right )}{4 \, c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.592059, size = 42, normalized size = 1.14 \begin{align*} \begin{cases} \frac{a x^{2}}{2} + \frac{b x^{2} \operatorname{atanh}{\left (c x \right )}}{2} + \frac{b x}{2 c} - \frac{b \operatorname{atanh}{\left (c x \right )}}{2 c^{2}} & \text{for}\: c \neq 0 \\\frac{a x^{2}}{2} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a*x**2/2 + b*x**2*atanh(c*x)/2 + b*x/(2*c) - b*atanh(c*x)/(2*c**2), Ne(c, 0)), (a*x**2/2, True))

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Giac [A]  time = 1.20851, size = 80, normalized size = 2.16 \begin{align*} \frac{1}{4} \, b x^{2} \log \left (-\frac{c x + 1}{c x - 1}\right ) + \frac{1}{2} \, a x^{2} + \frac{b x}{2 \, c} - \frac{b \log \left (c x + 1\right )}{4 \, c^{2}} + \frac{b \log \left (c x - 1\right )}{4 \, c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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