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

Optimal. Leaf size=97 \[ \frac {(c+d x) \left (a+b \coth ^{-1}(c+d x)\right )^2}{d}+\frac {\left (a+b \coth ^{-1}(c+d x)\right )^2}{d}-\frac {2 b \log \left (\frac {2}{-c-d x+1}\right ) \left (a+b \coth ^{-1}(c+d x)\right )}{d}-\frac {b^2 \text {Li}_2\left (-\frac {c+d x+1}{-c-d x+1}\right )}{d} \]

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, 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 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 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 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]

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*}

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Mathematica [A]  time = 0.17, size = 111, normalized size = 1.14 \[ \frac {a \left (a c+a d x-2 b \log \left (\frac {1}{(c+d x) \sqrt {1-\frac {1}{(c+d x)^2}}}\right )\right )+2 b \coth ^{-1}(c+d x) \left (a c+a d x-b \log \left (1-e^{-2 \coth ^{-1}(c+d x)}\right )\right )+b^2 \text {Li}_2\left (e^{-2 \coth ^{-1}(c+d x)}\right )+b^2 (c+d x-1) \coth ^{-1}(c+d x)^2}{d} \]

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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fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (b^{2} \operatorname {arcoth}\left (d x + c\right )^{2} + 2 \, a b \operatorname {arcoth}\left (d x + c\right ) + a^{2}, x\right ) \]

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

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)

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maple [B]  time = 0.24, size = 226, normalized size = 2.33 \[ \mathrm {arccoth}\left (d x +c \right )^{2} x \,b^{2}+\frac {\mathrm {arccoth}\left (d x +c \right )^{2} b^{2} c}{d}+2 \,\mathrm {arccoth}\left (d x +c \right ) x a b +\frac {b^{2} \mathrm {arccoth}\left (d x +c \right )^{2}}{d}-\frac {2 \,\mathrm {arccoth}\left (d x +c \right ) \ln \left (1+\frac {1}{\sqrt {\frac {d x +c -1}{d x +c +1}}}\right ) b^{2}}{d}-\frac {2 \,\mathrm {arccoth}\left (d x +c \right ) \ln \left (1-\frac {1}{\sqrt {\frac {d x +c -1}{d x +c +1}}}\right ) b^{2}}{d}+\frac {2 \,\mathrm {arccoth}\left (d x +c \right ) a b c}{d}+a^{2} x +\frac {a b \ln \left (\left (d x +c \right )^{2}-1\right )}{d}-\frac {2 \polylog \left (2, -\frac {1}{\sqrt {\frac {d x +c -1}{d x +c +1}}}\right ) b^{2}}{d}-\frac {2 \polylog \left (2, \frac {1}{\sqrt {\frac {d x +c -1}{d x +c +1}}}\right ) b^{2}}{d}+\frac {a^{2} c}{d} \]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a^{2} x + \frac {1}{4} \, b^{2} {\left (\frac {d x \log \left (d x + c - 1\right )^{2} + {\left (d x + c + 1\right )} \log \left (d x + c + 1\right )^{2} - 2 \, {\left (d x + c - 1\right )} \log \left (d x + c + 1\right ) \log \left (d x + c - 1\right )}{d} + \int \frac {2 \, {\left (c^{2} + {\left (c d - 3 \, d\right )} x - 2 \, c + 1\right )} \log \left (d x + c - 1\right )}{d^{2} x^{2} + 2 \, c d x + c^{2} - 1}\,{d x}\right )} + \frac {{\left (2 \, {\left (d x + c\right )} \operatorname {arcoth}\left (d x + c\right ) + \log \left (-{\left (d x + c\right )}^{2} + 1\right )\right )} a b}{d} \]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*acoth(c + d*x))^2,x)

[Out]

int((a + b*acoth(c + d*x))^2, x)

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

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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