3.1.77 \(\int \frac {\coth ^{-1}(a+b x)}{c+d x} \, dx\) [77]

Optimal. Leaf size=120 \[ -\frac {\coth ^{-1}(a+b x) \log \left (\frac {2}{1+a+b x}\right )}{d}+\frac {\coth ^{-1}(a+b x) \log \left (\frac {2 b (c+d x)}{(b c+d-a d) (1+a+b x)}\right )}{d}+\frac {\text {PolyLog}\left (2,1-\frac {2}{1+a+b x}\right )}{2 d}-\frac {\text {PolyLog}\left (2,1-\frac {2 b (c+d x)}{(b c+d-a d) (1+a+b x)}\right )}{2 d} \]

[Out]

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

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Rubi [A]
time = 0.09, antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {6247, 6058, 2449, 2352, 2497} \begin {gather*} -\frac {\text {Li}_2\left (1-\frac {2 b (c+d x)}{(b c-a d+d) (a+b x+1)}\right )}{2 d}+\frac {\coth ^{-1}(a+b x) \log \left (\frac {2 b (c+d x)}{(a+b x+1) (-a d+b c+d)}\right )}{d}+\frac {\text {Li}_2\left (1-\frac {2}{a+b x+1}\right )}{2 d}-\frac {\log \left (\frac {2}{a+b x+1}\right ) \coth ^{-1}(a+b x)}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[ArcCoth[a + b*x]/(c + d*x),x]

[Out]

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

Rule 2352

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

Rule 2449

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 2497

Int[Log[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[Pq^m*((1 - u)/D[u, x])]}, Simp[C*PolyLog[2, 1 - u
], x] /; FreeQ[C, x]] /; IntegerQ[m] && PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponen
ts[u, x][[2]], Expon[Pq, x]]

Rule 6058

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

Rule 6247

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

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Mathematica [A]
time = 0.05, size = 185, normalized size = 1.54 \begin {gather*} \frac {\log \left (\frac {d (1-a-b x)}{b c+d-a d}\right ) \log (c+d x)}{2 d}-\frac {\log \left (\frac {-1+a+b x}{a+b x}\right ) \log (c+d x)}{2 d}-\frac {\log \left (-\frac {d (1+a+b x)}{b c-d-a d}\right ) \log (c+d x)}{2 d}+\frac {\log \left (\frac {1+a+b x}{a+b x}\right ) \log (c+d x)}{2 d}-\frac {\text {PolyLog}\left (2,\frac {b (c+d x)}{b c-d-a d}\right )}{2 d}+\frac {\text {PolyLog}\left (2,\frac {b (c+d x)}{b c+d-a d}\right )}{2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.90, size = 185, normalized size = 1.54

method	result	size
risch	\(-\frac {\dilog \left (\frac {\left (b x +a -1\right ) d -a d +c b +d}{-a d +c b +d}\right )}{2 d}-\frac {\ln \left (b x +a -1\right ) \ln \left (\frac {\left (b x +a -1\right ) d -a d +c b +d}{-a d +c b +d}\right )}{2 d}+\frac {\dilog \left (\frac {\left (b x +a +1\right ) d -a d +c b -d}{-a d +c b -d}\right )}{2 d}+\frac {\ln \left (b x +a +1\right ) \ln \left (\frac {\left (b x +a +1\right ) d -a d +c b -d}{-a d +c b -d}\right )}{2 d}\)	\(164\)
derivativedivides	\(\frac {\frac {b \ln \left (a d -c b -d \left (b x +a \right )\right ) \mathrm {arccoth}\left (b x +a \right )}{d}-\frac {b \left (\frac {\left (\dilog \left (\frac {-d \left (b x +a \right )-d}{-a d +c b -d}\right )+\ln \left (a d -c b -d \left (b x +a \right )\right ) \ln \left (\frac {-d \left (b x +a \right )-d}{-a d +c b -d}\right )\right ) d}{2}-\frac {\left (\dilog \left (\frac {-d \left (b x +a \right )+d}{-a d +c b +d}\right )+\ln \left (a d -c b -d \left (b x +a \right )\right ) \ln \left (\frac {-d \left (b x +a \right )+d}{-a d +c b +d}\right )\right ) d}{2}\right )}{d^{2}}}{b}\)	\(185\)
default	\(\frac {\frac {b \ln \left (a d -c b -d \left (b x +a \right )\right ) \mathrm {arccoth}\left (b x +a \right )}{d}-\frac {b \left (\frac {\left (\dilog \left (\frac {-d \left (b x +a \right )-d}{-a d +c b -d}\right )+\ln \left (a d -c b -d \left (b x +a \right )\right ) \ln \left (\frac {-d \left (b x +a \right )-d}{-a d +c b -d}\right )\right ) d}{2}-\frac {\left (\dilog \left (\frac {-d \left (b x +a \right )+d}{-a d +c b +d}\right )+\ln \left (a d -c b -d \left (b x +a \right )\right ) \ln \left (\frac {-d \left (b x +a \right )+d}{-a d +c b +d}\right )\right ) d}{2}\right )}{d^{2}}}{b}\)	\(185\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccoth(b*x+a)/(d*x+c),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.27, size = 192, normalized size = 1.60 \begin {gather*} -\frac {1}{2} \, b {\left (\frac {\log \left (b x + a - 1\right ) \log \left (\frac {b d x + a d - d}{b c - a d + d} + 1\right ) + {\rm Li}_2\left (-\frac {b d x + a d - d}{b c - a d + d}\right )}{b d} - \frac {\log \left (b x + a + 1\right ) \log \left (\frac {b d x + a d + d}{b c - a d - d} + 1\right ) + {\rm Li}_2\left (-\frac {b d x + a d + d}{b c - a d - d}\right )}{b d}\right )} - \frac {b {\left (\frac {\log \left (b x + a + 1\right )}{b} - \frac {\log \left (b x + a - 1\right )}{b}\right )} \log \left (d x + c\right )}{2 \, d} + \frac {\operatorname {arcoth}\left (b x + a\right ) \log \left (d x + c\right )}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*b*((log(b*x + a - 1)*log((b*d*x + a*d - d)/(b*c - a*d + d) + 1) + dilog(-(b*d*x + a*d - d)/(b*c - a*d + d
)))/(b*d) - (log(b*x + a + 1)*log((b*d*x + a*d + d)/(b*c - a*d - d) + 1) + dilog(-(b*d*x + a*d + d)/(b*c - a*d
- d)))/(b*d)) - 1/2*b*(log(b*x + a + 1)/b - log(b*x + a - 1)/b)*log(d*x + c)/d + arccoth(b*x + a)*log(d*x + c
)/d

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\operatorname {acoth}{\left (a + b x \right )}}{c + d x}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(acoth(a + b*x)/(c + d*x), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\mathrm {acoth}\left (a+b\,x\right )}{c+d\,x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acoth(a + b*x)/(c + d*x),x)

[Out]

int(acoth(a + b*x)/(c + d*x), x)

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