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

Optimal. Leaf size=136 \[ -\frac {\left (a^2+1\right ) \coth ^{-1}(a+b x)^2}{2 b^2}+\frac {a \text {Li}_2\left (-\frac {a+b x+1}{-a-b x+1}\right )}{b^2}+\frac {\log \left (1-(a+b x)^2\right )}{2 b^2}-\frac {a \coth ^{-1}(a+b x)^2}{b^2}+\frac {(a+b x) \coth ^{-1}(a+b x)}{b^2}+\frac {2 a \log \left (\frac {2}{-a-b x+1}\right ) \coth ^{-1}(a+b x)}{b^2}+\frac {1}{2} x^2 \coth ^{-1}(a+b x)^2 \]

[Out]

(b*x+a)*arccoth(b*x+a)/b^2-a*arccoth(b*x+a)^2/b^2-1/2*(a^2+1)*arccoth(b*x+a)^2/b^2+1/2*x^2*arccoth(b*x+a)^2+2*
a*arccoth(b*x+a)*ln(2/(-b*x-a+1))/b^2+1/2*ln(1-(b*x+a)^2)/b^2+a*polylog(2,(-b*x-a-1)/(-b*x-a+1))/b^2

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Rubi [A]  time = 0.21, antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 10, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {6112, 5929, 5911, 260, 6049, 5949, 5985, 5919, 2402, 2315} \[ \frac {a \text {PolyLog}\left (2,-\frac {a+b x+1}{-a-b x+1}\right )}{b^2}-\frac {\left (a^2+1\right ) \coth ^{-1}(a+b x)^2}{2 b^2}+\frac {\log \left (1-(a+b x)^2\right )}{2 b^2}-\frac {a \coth ^{-1}(a+b x)^2}{b^2}+\frac {(a+b x) \coth ^{-1}(a+b x)}{b^2}+\frac {2 a \log \left (\frac {2}{-a-b x+1}\right ) \coth ^{-1}(a+b x)}{b^2}+\frac {1}{2} x^2 \coth ^{-1}(a+b x)^2 \]

Antiderivative was successfully verified.

[In]

Int[x*ArcCoth[a + b*x]^2,x]

[Out]

((a + b*x)*ArcCoth[a + b*x])/b^2 - (a*ArcCoth[a + b*x]^2)/b^2 - ((1 + a^2)*ArcCoth[a + b*x]^2)/(2*b^2) + (x^2*
ArcCoth[a + b*x]^2)/2 + (2*a*ArcCoth[a + b*x]*Log[2/(1 - a - b*x)])/b^2 + Log[1 - (a + b*x)^2]/(2*b^2) + (a*Po
lyLog[2, -((1 + a + b*x)/(1 - a - b*x))])/b^2

Rule 260

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

Rule 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rule 2402

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> -Dist[e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]

Rule 5911

Int[((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcCoth[c*x])^p, x] - Dist[b*c*p, In
t[(x*(a + b*ArcCoth[c*x])^(p - 1))/(1 - c^2*x^2), x], x] /; FreeQ[{a, b, c}, x] && IGtQ[p, 0]

Rule 5919

Int[((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcCoth[c*x])^p*
Log[2/(1 + (e*x)/d)])/e, x] + Dist[(b*c*p)/e, Int[((a + b*ArcCoth[c*x])^(p - 1)*Log[2/(1 + (e*x)/d)])/(1 - c^2
*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[c^2*d^2 - e^2, 0]

Rule 5929

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

Rule 5949

Int[((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(a + b*ArcCoth[c*x])^(p
+ 1)/(b*c*d*(p + 1)), x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[c^2*d + e, 0] && NeQ[p, -1]

Rule 5985

Int[(((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))^(p_.)*(x_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(a + b*ArcCoth[c
*x])^(p + 1)/(b*e*(p + 1)), x] + Dist[1/(c*d), Int[(a + b*ArcCoth[c*x])^p/(1 - c*x), x], x] /; FreeQ[{a, b, c,
d, e}, x] && EqQ[c^2*d + e, 0] && IGtQ[p, 0]

Rule 6049

Int[(((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))^(p_.)*((f_) + (g_.)*(x_))^(m_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :>
Int[ExpandIntegrand[(a + b*ArcCoth[c*x])^p/(d + e*x^2), (f + g*x)^m, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x]
&& IGtQ[p, 0] && EqQ[c^2*d + e, 0] && IGtQ[m, 0]

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 \coth ^{-1}(a+b x)^2 \, dx &=\frac {\operatorname {Subst}\left (\int \left (-\frac {a}{b}+\frac {x}{b}\right ) \coth ^{-1}(x)^2 \, dx,x,a+b x\right )}{b}\\ &=\frac {1}{2} x^2 \coth ^{-1}(a+b x)^2-\operatorname {Subst}\left (\int \left (-\frac {\coth ^{-1}(x)}{b^2}+\frac {\left (1+a^2-2 a x\right ) \coth ^{-1}(x)}{b^2 \left (1-x^2\right )}\right ) \, dx,x,a+b x\right )\\ &=\frac {1}{2} x^2 \coth ^{-1}(a+b x)^2+\frac {\operatorname {Subst}\left (\int \coth ^{-1}(x) \, dx,x,a+b x\right )}{b^2}-\frac {\operatorname {Subst}\left (\int \frac {\left (1+a^2-2 a x\right ) \coth ^{-1}(x)}{1-x^2} \, dx,x,a+b x\right )}{b^2}\\ &=\frac {(a+b x) \coth ^{-1}(a+b x)}{b^2}+\frac {1}{2} x^2 \coth ^{-1}(a+b x)^2-\frac {\operatorname {Subst}\left (\int \frac {x}{1-x^2} \, dx,x,a+b x\right )}{b^2}-\frac {\operatorname {Subst}\left (\int \left (\frac {\left (1+a^2\right ) \coth ^{-1}(x)}{1-x^2}-\frac {2 a x \coth ^{-1}(x)}{1-x^2}\right ) \, dx,x,a+b x\right )}{b^2}\\ &=\frac {(a+b x) \coth ^{-1}(a+b x)}{b^2}+\frac {1}{2} x^2 \coth ^{-1}(a+b x)^2+\frac {\log \left (1-(a+b x)^2\right )}{2 b^2}+\frac {(2 a) \operatorname {Subst}\left (\int \frac {x \coth ^{-1}(x)}{1-x^2} \, dx,x,a+b x\right )}{b^2}-\frac {\left (1+a^2\right ) \operatorname {Subst}\left (\int \frac {\coth ^{-1}(x)}{1-x^2} \, dx,x,a+b x\right )}{b^2}\\ &=\frac {(a+b x) \coth ^{-1}(a+b x)}{b^2}-\frac {a \coth ^{-1}(a+b x)^2}{b^2}-\frac {\left (1+a^2\right ) \coth ^{-1}(a+b x)^2}{2 b^2}+\frac {1}{2} x^2 \coth ^{-1}(a+b x)^2+\frac {\log \left (1-(a+b x)^2\right )}{2 b^2}+\frac {(2 a) \operatorname {Subst}\left (\int \frac {\coth ^{-1}(x)}{1-x} \, dx,x,a+b x\right )}{b^2}\\ &=\frac {(a+b x) \coth ^{-1}(a+b x)}{b^2}-\frac {a \coth ^{-1}(a+b x)^2}{b^2}-\frac {\left (1+a^2\right ) \coth ^{-1}(a+b x)^2}{2 b^2}+\frac {1}{2} x^2 \coth ^{-1}(a+b x)^2+\frac {2 a \coth ^{-1}(a+b x) \log \left (\frac {2}{1-a-b x}\right )}{b^2}+\frac {\log \left (1-(a+b x)^2\right )}{2 b^2}-\frac {(2 a) \operatorname {Subst}\left (\int \frac {\log \left (\frac {2}{1-x}\right )}{1-x^2} \, dx,x,a+b x\right )}{b^2}\\ &=\frac {(a+b x) \coth ^{-1}(a+b x)}{b^2}-\frac {a \coth ^{-1}(a+b x)^2}{b^2}-\frac {\left (1+a^2\right ) \coth ^{-1}(a+b x)^2}{2 b^2}+\frac {1}{2} x^2 \coth ^{-1}(a+b x)^2+\frac {2 a \coth ^{-1}(a+b x) \log \left (\frac {2}{1-a-b x}\right )}{b^2}+\frac {\log \left (1-(a+b x)^2\right )}{2 b^2}+\frac {(2 a) \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1-a-b x}\right )}{b^2}\\ &=\frac {(a+b x) \coth ^{-1}(a+b x)}{b^2}-\frac {a \coth ^{-1}(a+b x)^2}{b^2}-\frac {\left (1+a^2\right ) \coth ^{-1}(a+b x)^2}{2 b^2}+\frac {1}{2} x^2 \coth ^{-1}(a+b x)^2+\frac {2 a \coth ^{-1}(a+b x) \log \left (\frac {2}{1-a-b x}\right )}{b^2}+\frac {\log \left (1-(a+b x)^2\right )}{2 b^2}+\frac {a \text {Li}_2\left (1-\frac {2}{1-a-b x}\right )}{b^2}\\ \end {align*}

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Mathematica [A]  time = 0.26, size = 106, normalized size = 0.78 \[ \frac {\left (-a^2+2 a+b^2 x^2-1\right ) \coth ^{-1}(a+b x)^2-2 a \text {Li}_2\left (e^{-2 \coth ^{-1}(a+b x)}\right )-2 \log \left (\frac {1}{(a+b x) \sqrt {1-\frac {1}{(a+b x)^2}}}\right )+2 \coth ^{-1}(a+b x) \left (2 a \log \left (1-e^{-2 \coth ^{-1}(a+b x)}\right )+a+b x\right )}{2 b^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x*ArcCoth[a + b*x]^2,x]

[Out]

((-1 + 2*a - a^2 + b^2*x^2)*ArcCoth[a + b*x]^2 + 2*ArcCoth[a + b*x]*(a + b*x + 2*a*Log[1 - E^(-2*ArcCoth[a + b
*x])]) - 2*Log[1/((a + b*x)*Sqrt[1 - (a + b*x)^(-2)])] - 2*a*PolyLog[2, E^(-2*ArcCoth[a + b*x])])/(2*b^2)

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fricas [F]  time = 0.68, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x \operatorname {arcoth}\left (b x + a\right )^{2}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x*arccoth(b*x + a)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x*arccoth(b*x + a)^2, x)

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maple [B]  time = 0.06, size = 365, normalized size = 2.68 \[ \frac {x^{2} \mathrm {arccoth}\left (b x +a \right )^{2}}{2}-\frac {\mathrm {arccoth}\left (b x +a \right )^{2} a^{2}}{2 b^{2}}+\frac {\mathrm {arccoth}\left (b x +a \right ) x}{b}+\frac {\mathrm {arccoth}\left (b x +a \right ) a}{b^{2}}-\frac {\mathrm {arccoth}\left (b x +a \right ) \ln \left (b x +a -1\right ) a}{b^{2}}+\frac {\mathrm {arccoth}\left (b x +a \right ) \ln \left (b x +a -1\right )}{2 b^{2}}-\frac {\mathrm {arccoth}\left (b x +a \right ) \ln \left (b x +a +1\right ) a}{b^{2}}-\frac {\mathrm {arccoth}\left (b x +a \right ) \ln \left (b x +a +1\right )}{2 b^{2}}+\frac {\ln \left (b x +a -1\right )}{2 b^{2}}+\frac {\ln \left (b x +a +1\right )}{2 b^{2}}+\frac {\ln \left (b x +a -1\right )^{2}}{8 b^{2}}-\frac {\ln \left (b x +a -1\right ) \ln \left (\frac {1}{2}+\frac {b x}{2}+\frac {a}{2}\right )}{4 b^{2}}-\frac {\ln \left (b x +a -1\right )^{2} a}{4 b^{2}}+\frac {\dilog \left (\frac {1}{2}+\frac {b x}{2}+\frac {a}{2}\right ) a}{b^{2}}+\frac {\ln \left (b x +a -1\right ) \ln \left (\frac {1}{2}+\frac {b x}{2}+\frac {a}{2}\right ) a}{2 b^{2}}+\frac {\ln \left (b x +a +1\right )^{2} a}{4 b^{2}}-\frac {\ln \left (b x +a +1\right ) \ln \left (-\frac {b x}{2}-\frac {a}{2}+\frac {1}{2}\right ) a}{2 b^{2}}+\frac {\ln \left (-\frac {b x}{2}-\frac {a}{2}+\frac {1}{2}\right ) \ln \left (\frac {1}{2}+\frac {b x}{2}+\frac {a}{2}\right ) a}{2 b^{2}}+\frac {\ln \left (b x +a +1\right )^{2}}{8 b^{2}}-\frac {\ln \left (-\frac {b x}{2}-\frac {a}{2}+\frac {1}{2}\right ) \ln \left (b x +a +1\right )}{4 b^{2}}+\frac {\ln \left (-\frac {b x}{2}-\frac {a}{2}+\frac {1}{2}\right ) \ln \left (\frac {1}{2}+\frac {b x}{2}+\frac {a}{2}\right )}{4 b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2*arccoth(b*x+a)^2-1/2/b^2*arccoth(b*x+a)^2*a^2+1/b*arccoth(b*x+a)*x+1/b^2*arccoth(b*x+a)*a-1/b^2*arccot
h(b*x+a)*ln(b*x+a-1)*a+1/2/b^2*arccoth(b*x+a)*ln(b*x+a-1)-1/b^2*arccoth(b*x+a)*ln(b*x+a+1)*a-1/2/b^2*arccoth(b
*x+a)*ln(b*x+a+1)+1/2/b^2*ln(b*x+a-1)+1/2/b^2*ln(b*x+a+1)+1/8/b^2*ln(b*x+a-1)^2-1/4/b^2*ln(b*x+a-1)*ln(1/2+1/2
*b*x+1/2*a)-1/4/b^2*ln(b*x+a-1)^2*a+1/b^2*dilog(1/2+1/2*b*x+1/2*a)*a+1/2/b^2*ln(b*x+a-1)*ln(1/2+1/2*b*x+1/2*a)
*a+1/4/b^2*ln(b*x+a+1)^2*a-1/2/b^2*ln(b*x+a+1)*ln(-1/2*b*x-1/2*a+1/2)*a+1/2/b^2*ln(-1/2*b*x-1/2*a+1/2)*ln(1/2+
1/2*b*x+1/2*a)*a+1/8/b^2*ln(b*x+a+1)^2-1/4/b^2*ln(-1/2*b*x-1/2*a+1/2)*ln(b*x+a+1)+1/4/b^2*ln(-1/2*b*x-1/2*a+1/
2)*ln(1/2+1/2*b*x+1/2*a)

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maxima [A]  time = 0.34, size = 202, normalized size = 1.49 \[ \frac {1}{2} \, x^{2} \operatorname {arcoth}\left (b x + a\right )^{2} + \frac {1}{8} \, b^{2} {\left (\frac {8 \, {\left (\log \left (b x + a - 1\right ) \log \left (\frac {1}{2} \, b x + \frac {1}{2} \, a + \frac {1}{2}\right ) + {\rm Li}_2\left (-\frac {1}{2} \, b x - \frac {1}{2} \, a + \frac {1}{2}\right )\right )} a}{b^{4}} + \frac {4 \, {\left (a + 1\right )} \log \left (b x + a + 1\right )}{b^{4}} + \frac {{\left (a^{2} + 2 \, a + 1\right )} \log \left (b x + a + 1\right )^{2} - 2 \, {\left (a^{2} + 2 \, a + 1\right )} \log \left (b x + a + 1\right ) \log \left (b x + a - 1\right ) + {\left (a^{2} - 2 \, a + 1\right )} \log \left (b x + a - 1\right )^{2} - 4 \, {\left (a - 1\right )} \log \left (b x + a - 1\right )}{b^{4}}\right )} + \frac {1}{2} \, b {\left (\frac {2 \, x}{b^{2}} - \frac {{\left (a^{2} + 2 \, a + 1\right )} \log \left (b x + a + 1\right )}{b^{3}} + \frac {{\left (a^{2} - 2 \, a + 1\right )} \log \left (b x + a - 1\right )}{b^{3}}\right )} \operatorname {arcoth}\left (b x + a\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2*arccoth(b*x + a)^2 + 1/8*b^2*(8*(log(b*x + a - 1)*log(1/2*b*x + 1/2*a + 1/2) + dilog(-1/2*b*x - 1/2*a
+ 1/2))*a/b^4 + 4*(a + 1)*log(b*x + a + 1)/b^4 + ((a^2 + 2*a + 1)*log(b*x + a + 1)^2 - 2*(a^2 + 2*a + 1)*log(b
*x + a + 1)*log(b*x + a - 1) + (a^2 - 2*a + 1)*log(b*x + a - 1)^2 - 4*(a - 1)*log(b*x + a - 1))/b^4) + 1/2*b*(
2*x/b^2 - (a^2 + 2*a + 1)*log(b*x + a + 1)/b^3 + (a^2 - 2*a + 1)*log(b*x + a - 1)/b^3)*arccoth(b*x + a)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x\,{\mathrm {acoth}\left (a+b\,x\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*acoth(a + b*x)^2,x)

[Out]

int(x*acoth(a + b*x)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*acoth(a + b*x)**2, x)

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