3.62 \(\int x^3 \coth ^{-1}(a+b x) \, dx\)

Optimal. Leaf size=101 \[ \frac{\left (6 a^2+1\right ) x}{4 b^3}+\frac{(a+b x)^3}{12 b^4}-\frac{a (a+b x)^2}{2 b^4}+\frac{(1-a)^4 \log (-a-b x+1)}{8 b^4}-\frac{(a+1)^4 \log (a+b x+1)}{8 b^4}+\frac{1}{4} x^4 \coth ^{-1}(a+b x) \]

[Out]

((1 + 6*a^2)*x)/(4*b^3) - (a*(a + b*x)^2)/(2*b^4) + (a + b*x)^3/(12*b^4) + (x^4*ArcCoth[a + b*x])/4 + ((1 - a)
^4*Log[1 - a - b*x])/(8*b^4) - ((1 + a)^4*Log[1 + a + b*x])/(8*b^4)

________________________________________________________________________________________

Rubi [A]  time = 0.126333, antiderivative size = 101, 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} \[ \frac{\left (6 a^2+1\right ) x}{4 b^3}+\frac{(a+b x)^3}{12 b^4}-\frac{a (a+b x)^2}{2 b^4}+\frac{(1-a)^4 \log (-a-b x+1)}{8 b^4}-\frac{(a+1)^4 \log (a+b x+1)}{8 b^4}+\frac{1}{4} x^4 \coth ^{-1}(a+b x) \]

Antiderivative was successfully verified.

[In]

Int[x^3*ArcCoth[a + b*x],x]

[Out]

((1 + 6*a^2)*x)/(4*b^3) - (a*(a + b*x)^2)/(2*b^4) + (a + b*x)^3/(12*b^4) + (x^4*ArcCoth[a + b*x])/4 + ((1 - a)
^4*Log[1 - a - b*x])/(8*b^4) - ((1 + a)^4*Log[1 + a + b*x])/(8*b^4)

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^3 \coth ^{-1}(a+b x) \, dx &=\frac{\operatorname{Subst}\left (\int \left (-\frac{a}{b}+\frac{x}{b}\right )^3 \coth ^{-1}(x) \, dx,x,a+b x\right )}{b}\\ &=\frac{1}{4} x^4 \coth ^{-1}(a+b x)-\frac{1}{4} \operatorname{Subst}\left (\int \frac{\left (-\frac{a}{b}+\frac{x}{b}\right )^4}{1-x^2} \, dx,x,a+b x\right )\\ &=\frac{1}{4} x^4 \coth ^{-1}(a+b x)-\frac{1}{4} \operatorname{Subst}\left (\int \left (-\frac{1+6 a^2}{b^4}+\frac{4 a x}{b^4}-\frac{x^2}{b^4}+\frac{1+6 a^2+a^4-4 a \left (1+a^2\right ) x}{b^4 \left (1-x^2\right )}\right ) \, dx,x,a+b x\right )\\ &=\frac{\left (1+6 a^2\right ) x}{4 b^3}-\frac{a (a+b x)^2}{2 b^4}+\frac{(a+b x)^3}{12 b^4}+\frac{1}{4} x^4 \coth ^{-1}(a+b x)-\frac{\operatorname{Subst}\left (\int \frac{1+6 a^2+a^4-4 a \left (1+a^2\right ) x}{1-x^2} \, dx,x,a+b x\right )}{4 b^4}\\ &=\frac{\left (1+6 a^2\right ) x}{4 b^3}-\frac{a (a+b x)^2}{2 b^4}+\frac{(a+b x)^3}{12 b^4}+\frac{1}{4} x^4 \coth ^{-1}(a+b x)-\frac{(1-a)^4 \operatorname{Subst}\left (\int \frac{1}{1-x} \, dx,x,a+b x\right )}{8 b^4}+\frac{(1+a)^4 \operatorname{Subst}\left (\int \frac{1}{-1-x} \, dx,x,a+b x\right )}{8 b^4}\\ &=\frac{\left (1+6 a^2\right ) x}{4 b^3}-\frac{a (a+b x)^2}{2 b^4}+\frac{(a+b x)^3}{12 b^4}+\frac{1}{4} x^4 \coth ^{-1}(a+b x)+\frac{(1-a)^4 \log (1-a-b x)}{8 b^4}-\frac{(1+a)^4 \log (1+a+b x)}{8 b^4}\\ \end{align*}

Mathematica [A]  time = 0.038819, size = 81, normalized size = 0.8 \[ \frac{6 \left (3 a^2+1\right ) b x-6 a b^2 x^2+6 b^4 x^4 \coth ^{-1}(a+b x)+3 (a-1)^4 \log (-a-b x+1)-3 (a+1)^4 \log (a+b x+1)+2 b^3 x^3}{24 b^4} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*ArcCoth[a + b*x],x]

[Out]

(6*(1 + 3*a^2)*b*x - 6*a*b^2*x^2 + 2*b^3*x^3 + 6*b^4*x^4*ArcCoth[a + b*x] + 3*(-1 + a)^4*Log[1 - a - b*x] - 3*
(1 + a)^4*Log[1 + a + b*x])/(24*b^4)

________________________________________________________________________________________

Maple [B]  time = 0.035, size = 199, normalized size = 2. \begin{align*}{\frac{a}{4\,{b}^{4}}}+{\frac{x}{4\,{b}^{3}}}+{\frac{{x}^{4}{\rm arccoth} \left (bx+a\right )}{4}}-{\frac{\ln \left ( bx+a+1 \right ) }{8\,{b}^{4}}}+{\frac{\ln \left ( bx+a-1 \right ) }{8\,{b}^{4}}}+{\frac{{x}^{3}}{12\,b}}-{\frac{a{x}^{2}}{4\,{b}^{2}}}+{\frac{3\,{a}^{2}x}{4\,{b}^{3}}}+{\frac{13\,{a}^{3}}{12\,{b}^{4}}}+{\frac{\ln \left ( bx+a-1 \right ){a}^{4}}{8\,{b}^{4}}}-{\frac{\ln \left ( bx+a-1 \right ){a}^{3}}{2\,{b}^{4}}}+{\frac{3\,\ln \left ( bx+a-1 \right ){a}^{2}}{4\,{b}^{4}}}-{\frac{\ln \left ( bx+a-1 \right ) a}{2\,{b}^{4}}}-{\frac{\ln \left ( bx+a+1 \right ){a}^{4}}{8\,{b}^{4}}}-{\frac{\ln \left ( bx+a+1 \right ){a}^{3}}{2\,{b}^{4}}}-{\frac{3\,\ln \left ( bx+a+1 \right ){a}^{2}}{4\,{b}^{4}}}-{\frac{\ln \left ( bx+a+1 \right ) a}{2\,{b}^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*arccoth(b*x+a),x)

[Out]

1/4/b^4*a+1/4*x/b^3+1/4*x^4*arccoth(b*x+a)-1/8/b^4*ln(b*x+a+1)+1/8/b^4*ln(b*x+a-1)+1/12*x^3/b-1/4/b^2*x^2*a+3/
4/b^3*x*a^2+13/12/b^4*a^3+1/8/b^4*ln(b*x+a-1)*a^4-1/2/b^4*ln(b*x+a-1)*a^3+3/4/b^4*ln(b*x+a-1)*a^2-1/2/b^4*ln(b
*x+a-1)*a-1/8/b^4*ln(b*x+a+1)*a^4-1/2/b^4*ln(b*x+a+1)*a^3-3/4/b^4*ln(b*x+a+1)*a^2-1/2/b^4*ln(b*x+a+1)*a

________________________________________________________________________________________

Maxima [A]  time = 0.959818, size = 143, normalized size = 1.42 \begin{align*} \frac{1}{4} \, x^{4} \operatorname{arcoth}\left (b x + a\right ) + \frac{1}{24} \, b{\left (\frac{2 \,{\left (b^{2} x^{3} - 3 \, a b x^{2} + 3 \,{\left (3 \, a^{2} + 1\right )} x\right )}}{b^{4}} - \frac{3 \,{\left (a^{4} + 4 \, a^{3} + 6 \, a^{2} + 4 \, a + 1\right )} \log \left (b x + a + 1\right )}{b^{5}} + \frac{3 \,{\left (a^{4} - 4 \, a^{3} + 6 \, a^{2} - 4 \, a + 1\right )} \log \left (b x + a - 1\right )}{b^{5}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arccoth(b*x+a),x, algorithm="maxima")

[Out]

1/4*x^4*arccoth(b*x + a) + 1/24*b*(2*(b^2*x^3 - 3*a*b*x^2 + 3*(3*a^2 + 1)*x)/b^4 - 3*(a^4 + 4*a^3 + 6*a^2 + 4*
a + 1)*log(b*x + a + 1)/b^5 + 3*(a^4 - 4*a^3 + 6*a^2 - 4*a + 1)*log(b*x + a - 1)/b^5)

________________________________________________________________________________________

Fricas [A]  time = 1.55089, size = 279, normalized size = 2.76 \begin{align*} \frac{3 \, b^{4} x^{4} \log \left (\frac{b x + a + 1}{b x + a - 1}\right ) + 2 \, b^{3} x^{3} - 6 \, a b^{2} x^{2} + 6 \,{\left (3 \, a^{2} + 1\right )} b x - 3 \,{\left (a^{4} + 4 \, a^{3} + 6 \, a^{2} + 4 \, a + 1\right )} \log \left (b x + a + 1\right ) + 3 \,{\left (a^{4} - 4 \, a^{3} + 6 \, a^{2} - 4 \, a + 1\right )} \log \left (b x + a - 1\right )}{24 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arccoth(b*x+a),x, algorithm="fricas")

[Out]

1/24*(3*b^4*x^4*log((b*x + a + 1)/(b*x + a - 1)) + 2*b^3*x^3 - 6*a*b^2*x^2 + 6*(3*a^2 + 1)*b*x - 3*(a^4 + 4*a^
3 + 6*a^2 + 4*a + 1)*log(b*x + a + 1) + 3*(a^4 - 4*a^3 + 6*a^2 - 4*a + 1)*log(b*x + a - 1))/b^4

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*acoth(b*x+a),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3} \operatorname{arcoth}\left (b x + a\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arccoth(b*x+a),x, algorithm="giac")

[Out]

integrate(x^3*arccoth(b*x + a), x)