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

Optimal. Leaf size=46 \[ \frac{1}{3} x^3 \left (a+b \tanh ^{-1}(c x)\right )+\frac{b \log \left (1-c^2 x^2\right )}{6 c^3}+\frac{b x^2}{6 c} \]

[Out]

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

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Rubi [A]  time = 0.03505, antiderivative size = 46, 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, 43} \[ \frac{1}{3} x^3 \left (a+b \tanh ^{-1}(c x)\right )+\frac{b \log \left (1-c^2 x^2\right )}{6 c^3}+\frac{b x^2}{6 c} \]

Antiderivative was successfully verified.

[In]

Int[x^2*(a + b*ArcTanh[c*x]),x]

[Out]

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

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 43

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.0079837, size = 51, normalized size = 1.11 \[ \frac{a x^3}{3}+\frac{b \log \left (1-c^2 x^2\right )}{6 c^3}+\frac{b x^2}{6 c}+\frac{1}{3} b x^3 \tanh ^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*(a + b*ArcTanh[c*x]),x]

[Out]

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

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Maple [A]  time = 0.007, size = 51, normalized size = 1.1 \begin{align*}{\frac{{x}^{3}a}{3}}+{\frac{b{x}^{3}{\it Artanh} \left ( cx \right ) }{3}}+{\frac{b{x}^{2}}{6\,c}}+{\frac{b\ln \left ( cx-1 \right ) }{6\,{c}^{3}}}+{\frac{b\ln \left ( cx+1 \right ) }{6\,{c}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.968135, size = 59, normalized size = 1.28 \begin{align*} \frac{1}{3} \, a x^{3} + \frac{1}{6} \,{\left (2 \, x^{3} \operatorname{artanh}\left (c x\right ) + c{\left (\frac{x^{2}}{c^{2}} + \frac{\log \left (c^{2} x^{2} - 1\right )}{c^{4}}\right )}\right )} b \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*a*x^3 + 1/6*(2*x^3*arctanh(c*x) + c*(x^2/c^2 + log(c^2*x^2 - 1)/c^4))*b

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Fricas [A]  time = 1.96861, size = 127, normalized size = 2.76 \begin{align*} \frac{b c^{3} x^{3} \log \left (-\frac{c x + 1}{c x - 1}\right ) + 2 \, a c^{3} x^{3} + b c^{2} x^{2} + b \log \left (c^{2} x^{2} - 1\right )}{6 \, c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.944725, size = 58, normalized size = 1.26 \begin{align*} \begin{cases} \frac{a x^{3}}{3} + \frac{b x^{3} \operatorname{atanh}{\left (c x \right )}}{3} + \frac{b x^{2}}{6 c} + \frac{b \log{\left (x - \frac{1}{c} \right )}}{3 c^{3}} + \frac{b \operatorname{atanh}{\left (c x \right )}}{3 c^{3}} & \text{for}\: c \neq 0 \\\frac{a x^{3}}{3} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.23154, size = 72, normalized size = 1.57 \begin{align*} \frac{1}{6} \, b x^{3} \log \left (-\frac{c x + 1}{c x - 1}\right ) + \frac{1}{3} \, a x^{3} + \frac{b x^{2}}{6 \, c} + \frac{b \log \left (c^{2} x^{2} - 1\right )}{6 \, c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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