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

Optimal. Leaf size=97 \[ -\frac{b^2 \text{PolyLog}\left (2,-\frac{c+d x+1}{-c-d x+1}\right )}{d}+\frac{(c+d x) \left (a+b \coth ^{-1}(c+d x)\right )^2}{d}+\frac{\left (a+b \coth ^{-1}(c+d x)\right )^2}{d}-\frac{2 b \log \left (\frac{2}{-c-d x+1}\right ) \left (a+b \coth ^{-1}(c+d x)\right )}{d} \]

[Out]

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

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Rubi [A]  time = 0.11621, antiderivative size = 97, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {6104, 5911, 5985, 5919, 2402, 2315} \[ -\frac{b^2 \text{PolyLog}\left (2,-\frac{c+d x+1}{-c-d x+1}\right )}{d}+\frac{(c+d x) \left (a+b \coth ^{-1}(c+d x)\right )^2}{d}+\frac{\left (a+b \coth ^{-1}(c+d x)\right )^2}{d}-\frac{2 b \log \left (\frac{2}{-c-d x+1}\right ) \left (a+b \coth ^{-1}(c+d x)\right )}{d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcCoth[c + d*x])^2,x]

[Out]

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

Rule 6104

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

Rubi steps

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

Mathematica [A]  time = 0.148008, size = 111, normalized size = 1.14 \[ \frac{b^2 \text{PolyLog}\left (2,e^{-2 \coth ^{-1}(c+d x)}\right )+a \left (a c+a d x-2 b \log \left (\frac{1}{(c+d x) \sqrt{1-\frac{1}{(c+d x)^2}}}\right )\right )+2 b \coth ^{-1}(c+d x) \left (a c+a d x-b \log \left (1-e^{-2 \coth ^{-1}(c+d x)}\right )\right )+b^2 (c+d x-1) \coth ^{-1}(c+d x)^2}{d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [B]  time = 0.111, size = 226, normalized size = 2.3 \begin{align*} \left ({\rm arccoth} \left (dx+c\right ) \right ) ^{2}x{b}^{2}+{\frac{ \left ({\rm arccoth} \left (dx+c\right ) \right ) ^{2}{b}^{2}c}{d}}+2\,{\rm arccoth} \left (dx+c\right )xab-2\,{\frac{{\rm arccoth} \left (dx+c\right ){b}^{2}}{d}\ln \left ( 1-{\frac{1}{\sqrt{{\frac{dx+c-1}{dx+c+1}}}}} \right ) }-2\,{\frac{{\rm arccoth} \left (dx+c\right ){b}^{2}}{d}\ln \left ( 1+{\frac{1}{\sqrt{{\frac{dx+c-1}{dx+c+1}}}}} \right ) }+{\frac{ \left ({\rm arccoth} \left (dx+c\right ) \right ) ^{2}{b}^{2}}{d}}+2\,{\frac{{\rm arccoth} \left (dx+c\right )abc}{d}}+{a}^{2}x-2\,{\frac{{b}^{2}}{d}{\it polylog} \left ( 2,{\frac{1}{\sqrt{{\frac{dx+c-1}{dx+c+1}}}}} \right ) }-2\,{\frac{{b}^{2}}{d}{\it polylog} \left ( 2,-{\frac{1}{\sqrt{{\frac{dx+c-1}{dx+c+1}}}}} \right ) }+{\frac{ab\ln \left ( \left ( dx+c \right ) ^{2}-1 \right ) }{d}}+{\frac{{a}^{2}c}{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a^{2} x + \frac{1}{4} \, b^{2}{\left (\frac{d x \log \left (d x + c - 1\right )^{2} +{\left (d x + c + 1\right )} \log \left (d x + c + 1\right )^{2} - 2 \,{\left (d x + c - 1\right )} \log \left (d x + c + 1\right ) \log \left (d x + c - 1\right )}{d} + \int \frac{2 \,{\left (c^{2} +{\left (c d - 3 \, d\right )} x - 2 \, c + 1\right )} \log \left (d x + c - 1\right )}{d^{2} x^{2} + 2 \, c d x + c^{2} - 1}\,{d x}\right )} + \frac{{\left (2 \,{\left (d x + c\right )} \operatorname{arcoth}\left (d x + c\right ) + \log \left (-{\left (d x + c\right )}^{2} + 1\right )\right )} a b}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^2*x + 1/4*b^2*((d*x*log(d*x + c - 1)^2 + (d*x + c + 1)*log(d*x + c + 1)^2 - 2*(d*x + c - 1)*log(d*x + c + 1)
*log(d*x + c - 1))/d + integrate(2*(c^2 + (c*d - 3*d)*x - 2*c + 1)*log(d*x + c - 1)/(d^2*x^2 + 2*c*d*x + c^2 -
1), x)) + (2*(d*x + c)*arccoth(d*x + c) + log(-(d*x + c)^2 + 1))*a*b/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(b^2*arccoth(d*x + c)^2 + 2*a*b*arccoth(d*x + c) + a^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*acoth(c + d*x))**2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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