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

Optimal. Leaf size=54 \[ \frac{(a+b x)^2}{6 b}+\frac{\log \left (1-(a+b x)^2\right )}{6 b}+\frac{(a+b x)^3 \coth ^{-1}(a+b x)}{3 b} \]

[Out]

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

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Rubi [A]  time = 0.0479171, antiderivative size = 54, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {6108, 5917, 266, 43} \[ \frac{(a+b x)^2}{6 b}+\frac{\log \left (1-(a+b x)^2\right )}{6 b}+\frac{(a+b x)^3 \coth ^{-1}(a+b x)}{3 b} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^2*ArcCoth[a + b*x],x]

[Out]

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

Rule 6108

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

Rule 5917

Int[((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcC
oth[c*x])^p)/(d*(m + 1)), x] - Dist[(b*c*p)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcCoth[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 (a+b x)^2 \coth ^{-1}(a+b x) \, dx &=\frac{\operatorname{Subst}\left (\int x^2 \coth ^{-1}(x) \, dx,x,a+b x\right )}{b}\\ &=\frac{(a+b x)^3 \coth ^{-1}(a+b x)}{3 b}-\frac{\operatorname{Subst}\left (\int \frac{x^3}{1-x^2} \, dx,x,a+b x\right )}{3 b}\\ &=\frac{(a+b x)^3 \coth ^{-1}(a+b x)}{3 b}-\frac{\operatorname{Subst}\left (\int \frac{x}{1-x} \, dx,x,(a+b x)^2\right )}{6 b}\\ &=\frac{(a+b x)^3 \coth ^{-1}(a+b x)}{3 b}-\frac{\operatorname{Subst}\left (\int \left (-1+\frac{1}{1-x}\right ) \, dx,x,(a+b x)^2\right )}{6 b}\\ &=\frac{(a+b x)^2}{6 b}+\frac{(a+b x)^3 \coth ^{-1}(a+b x)}{3 b}+\frac{\log \left (1-(a+b x)^2\right )}{6 b}\\ \end{align*}

Mathematica [A]  time = 0.0345896, size = 42, normalized size = 0.78 \[ \frac{(a+b x)^2+\log \left (1-(a+b x)^2\right )+2 (a+b x)^3 \coth ^{-1}(a+b x)}{6 b} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^2*ArcCoth[a + b*x],x]

[Out]

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

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Maple [A]  time = 0.034, size = 95, normalized size = 1.8 \begin{align*}{\frac{{b}^{2}{\rm arccoth} \left (bx+a\right ){x}^{3}}{3}}+b{\rm arccoth} \left (bx+a\right ){x}^{2}a+{\rm arccoth} \left (bx+a\right )x{a}^{2}+{\frac{{\rm arccoth} \left (bx+a\right ){a}^{3}}{3\,b}}+{\frac{b{x}^{2}}{6}}+{\frac{ax}{3}}+{\frac{{a}^{2}}{6\,b}}+{\frac{\ln \left ( bx+a-1 \right ) }{6\,b}}+{\frac{\ln \left ( bx+a+1 \right ) }{6\,b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.95623, size = 109, normalized size = 2.02 \begin{align*} \frac{1}{6} \, b{\left (\frac{b x^{2} + 2 \, a x}{b} + \frac{{\left (a^{3} + 1\right )} \log \left (b x + a + 1\right )}{b^{2}} - \frac{{\left (a^{3} - 1\right )} \log \left (b x + a - 1\right )}{b^{2}}\right )} + \frac{1}{3} \,{\left (b^{2} x^{3} + 3 \, a b x^{2} + 3 \, a^{2} x\right )} \operatorname{arcoth}\left (b x + a\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.87158, size = 211, normalized size = 3.91 \begin{align*} \frac{b^{2} x^{2} + 2 \, a b x +{\left (a^{3} + 1\right )} \log \left (b x + a + 1\right ) -{\left (a^{3} - 1\right )} \log \left (b x + a - 1\right ) +{\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x\right )} \log \left (\frac{b x + a + 1}{b x + a - 1}\right )}{6 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 3.13341, size = 97, normalized size = 1.8 \begin{align*} \begin{cases} \frac{a^{3} \operatorname{acoth}{\left (a + b x \right )}}{3 b} + a^{2} x \operatorname{acoth}{\left (a + b x \right )} + a b x^{2} \operatorname{acoth}{\left (a + b x \right )} + \frac{a x}{3} + \frac{b^{2} x^{3} \operatorname{acoth}{\left (a + b x \right )}}{3} + \frac{b x^{2}}{6} + \frac{\log{\left (\frac{a}{b} + x + \frac{1}{b} \right )}}{3 b} - \frac{\operatorname{acoth}{\left (a + b x \right )}}{3 b} & \text{for}\: b \neq 0 \\a^{2} x \operatorname{acoth}{\left (a \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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