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

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

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, 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 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

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 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 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 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]

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*}

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Mathematica [A]  time = 0.02, 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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fricas [A]  time = 0.56, size = 84, normalized size = 1.08 \[ \frac {b^{3} x^{3} \log \left (\frac {b x + a + 1}{b x + a - 1}\right ) + b^{2} x^{2} - 4 \, a b x + {\left (a^{3} + 3 \, a^{2} + 3 \, a + 1\right )} \log \left (b x + a + 1\right ) - {\left (a^{3} - 3 \, a^{2} + 3 \, a - 1\right )} \log \left (b x + a - 1\right )}{6 \, b^{3}} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} \operatorname {arcoth}\left (b x + a\right )\,{d x} \]

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)

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maple [B]  time = 0.04, size = 146, normalized size = 1.87 \[ \frac {x^{3} \mathrm {arccoth}\left (b x +a \right )}{3}+\frac {x^{2}}{6 b}-\frac {2 a x}{3 b^{2}}-\frac {5 a^{2}}{6 b^{3}}-\frac {\ln \left (b x +a -1\right ) a^{3}}{6 b^{3}}+\frac {\ln \left (b x +a -1\right ) a^{2}}{2 b^{3}}-\frac {\ln \left (b x +a -1\right ) a}{2 b^{3}}+\frac {\ln \left (b x +a -1\right )}{6 b^{3}}+\frac {\ln \left (b x +a +1\right ) a^{3}}{6 b^{3}}+\frac {\ln \left (b x +a +1\right ) a^{2}}{2 b^{3}}+\frac {\ln \left (b x +a +1\right ) a}{2 b^{3}}+\frac {\ln \left (b x +a +1\right )}{6 b^{3}} \]

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 = 0.31, size = 79, normalized size = 1.01 \[ \frac {1}{3} \, x^{3} \operatorname {arcoth}\left (b x + a\right ) + \frac {1}{6} \, b {\left (\frac {b x^{2} - 4 \, a x}{b^{3}} + \frac {{\left (a^{3} + 3 \, a^{2} + 3 \, a + 1\right )} \log \left (b x + a + 1\right )}{b^{4}} - \frac {{\left (a^{3} - 3 \, a^{2} + 3 \, a - 1\right )} \log \left (b x + a - 1\right )}{b^{4}}\right )} \]

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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mupad [B]  time = 1.36, size = 98, normalized size = 1.26 \[ \frac {x^3\,\ln \left (\frac {1}{a+b\,x}+1\right )}{6}-\frac {x^3\,\ln \left (1-\frac {1}{a+b\,x}\right )}{6}+\frac {x^2}{6\,b}-\frac {\ln \left (a+b\,x-1\right )\,\left (a^3-3\,a^2+3\,a-1\right )}{6\,b^3}+\frac {\ln \left (a+b\,x+1\right )\,\left (a^3+3\,a^2+3\,a+1\right )}{6\,b^3}-\frac {2\,a\,x}{3\,b^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*acoth(a + b*x),x)

[Out]

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

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sympy [A]  time = 1.27, size = 117, normalized size = 1.50 \[ \begin {cases} \frac {a^{3} \operatorname {acoth}{\left (a + b x \right )}}{3 b^{3}} + \frac {a^{2} \log {\left (a + b x + 1 \right )}}{b^{3}} - \frac {a^{2} \operatorname {acoth}{\left (a + b x \right )}}{b^{3}} - \frac {2 a x}{3 b^{2}} + \frac {a \operatorname {acoth}{\left (a + b x \right )}}{b^{3}} + \frac {x^{3} \operatorname {acoth}{\left (a + b x \right )}}{3} + \frac {x^{2}}{6 b} + \frac {\log {\left (a + b x + 1 \right )}}{3 b^{3}} - \frac {\operatorname {acoth}{\left (a + b x \right )}}{3 b^{3}} & \text {for}\: b \neq 0 \\\frac {x^{3} \operatorname {acoth}{\relax (a )}}{3} & \text {otherwise} \end {cases} \]

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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