3.63 \(\int x^2 \coth ^{-1}(a+b x) \, dx\)

Optimal. Leaf size=78 \[ \frac{(a+b x)^2}{6 b^3}-\frac{a x}{b^2}+\frac{(1-a)^3 \log (-a-b x+1)}{6 b^3}+\frac{(a+1)^3 \log (a+b x+1)}{6 b^3}+\frac{1}{3} x^3 \coth ^{-1}(a+b x) \]

[Out]

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

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Rubi [A]  time = 0.101837, antiderivative size = 78, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {6112, 5927, 702, 633, 31} \[ \frac{(a+b x)^2}{6 b^3}-\frac{a x}{b^2}+\frac{(1-a)^3 \log (-a-b x+1)}{6 b^3}+\frac{(a+1)^3 \log (a+b x+1)}{6 b^3}+\frac{1}{3} x^3 \coth ^{-1}(a+b x) \]

Antiderivative was successfully verified.

[In]

Int[x^2*ArcCoth[a + b*x],x]

[Out]

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

Rule 6112

Int[((a_.) + ArcCoth[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[
Int[((d*e - c*f)/d + (f*x)/d)^m*(a + b*ArcCoth[x])^p, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] &
& IGtQ[p, 0]

Rule 5927

Int[((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Simp[((d + e*x)^(q + 1)*(a + b
*ArcCoth[c*x]))/(e*(q + 1)), x] - Dist[(b*c)/(e*(q + 1)), Int[(d + e*x)^(q + 1)/(1 - c^2*x^2), x], x] /; FreeQ
[{a, b, c, d, e, q}, x] && NeQ[q, -1]

Rule 702

Int[((d_) + (e_.)*(x_))^(m_)/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[PolynomialDivide[(d + e*x)^m, a + c*x^2,
x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[m, 1] && (NeQ[d, 0] || GtQ[m, 2])

Rule 633

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[-(a*c), 2]}, Dist[e/2 + (c*d)/(2*q),
Int[1/(-q + c*x), x], x] + Dist[e/2 - (c*d)/(2*q), Int[1/(q + c*x), x], x]] /; FreeQ[{a, c, d, e}, x] && NiceS
qrtQ[-(a*c)]

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

Mathematica [A]  time = 0.0233845, size = 92, normalized size = 1.18 \[ \frac{\left (-a^3+3 a^2-3 a+1\right ) \log (-a-b x+1)}{6 b^3}+\frac{\left (a^3+3 a^2+3 a+1\right ) \log (a+b x+1)}{6 b^3}-\frac{2 a x}{3 b^2}+\frac{1}{3} x^3 \coth ^{-1}(a+b x)+\frac{x^2}{6 b} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*ArcCoth[a + b*x],x]

[Out]

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

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Maple [B]  time = 0.036, size = 146, normalized size = 1.9 \begin{align*}{\frac{{x}^{3}{\rm arccoth} \left (bx+a\right )}{3}}+{\frac{{x}^{2}}{6\,b}}-{\frac{2\,ax}{3\,{b}^{2}}}-{\frac{5\,{a}^{2}}{6\,{b}^{3}}}-{\frac{\ln \left ( bx+a-1 \right ){a}^{3}}{6\,{b}^{3}}}+{\frac{\ln \left ( bx+a-1 \right ){a}^{2}}{2\,{b}^{3}}}-{\frac{\ln \left ( bx+a-1 \right ) a}{2\,{b}^{3}}}+{\frac{\ln \left ( bx+a-1 \right ) }{6\,{b}^{3}}}+{\frac{\ln \left ( bx+a+1 \right ){a}^{3}}{6\,{b}^{3}}}+{\frac{\ln \left ( bx+a+1 \right ){a}^{2}}{2\,{b}^{3}}}+{\frac{\ln \left ( bx+a+1 \right ) a}{2\,{b}^{3}}}+{\frac{\ln \left ( bx+a+1 \right ) }{6\,{b}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*arccoth(b*x+a),x)

[Out]

1/3*x^3*arccoth(b*x+a)+1/6*x^2/b-2/3*a*x/b^2-5/6/b^3*a^2-1/6/b^3*ln(b*x+a-1)*a^3+1/2/b^3*ln(b*x+a-1)*a^2-1/2/b
^3*ln(b*x+a-1)*a+1/6/b^3*ln(b*x+a-1)+1/6/b^3*ln(b*x+a+1)*a^3+1/2/b^3*ln(b*x+a+1)*a^2+1/2/b^3*ln(b*x+a+1)*a+1/6
/b^3*ln(b*x+a+1)

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Maxima [A]  time = 0.958009, size = 107, normalized size = 1.37 \begin{align*} \frac{1}{3} \, x^{3} \operatorname{arcoth}\left (b x + a\right ) + \frac{1}{6} \, b{\left (\frac{b x^{2} - 4 \, a x}{b^{3}} + \frac{{\left (a^{3} + 3 \, a^{2} + 3 \, a + 1\right )} \log \left (b x + a + 1\right )}{b^{4}} - \frac{{\left (a^{3} - 3 \, a^{2} + 3 \, a - 1\right )} \log \left (b x + a - 1\right )}{b^{4}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.61077, size = 213, normalized size = 2.73 \begin{align*} \frac{b^{3} x^{3} \log \left (\frac{b x + a + 1}{b x + a - 1}\right ) + b^{2} x^{2} - 4 \, a b x +{\left (a^{3} + 3 \, a^{2} + 3 \, a + 1\right )} \log \left (b x + a + 1\right ) -{\left (a^{3} - 3 \, a^{2} + 3 \, a - 1\right )} \log \left (b x + a - 1\right )}{6 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 3.19662, size = 117, normalized size = 1.5 \begin{align*} \begin{cases} \frac{a^{3} \operatorname{acoth}{\left (a + b x \right )}}{3 b^{3}} + \frac{a^{2} \log{\left (a + b x + 1 \right )}}{b^{3}} - \frac{a^{2} \operatorname{acoth}{\left (a + b x \right )}}{b^{3}} - \frac{2 a x}{3 b^{2}} + \frac{a \operatorname{acoth}{\left (a + b x \right )}}{b^{3}} + \frac{x^{3} \operatorname{acoth}{\left (a + b x \right )}}{3} + \frac{x^{2}}{6 b} + \frac{\log{\left (a + b x + 1 \right )}}{3 b^{3}} - \frac{\operatorname{acoth}{\left (a + b x \right )}}{3 b^{3}} & \text{for}\: b \neq 0 \\\frac{x^{3} \operatorname{acoth}{\left (a \right )}}{3} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*acoth(b*x+a),x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} \operatorname{arcoth}\left (b x + a\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2*arccoth(b*x + a), x)