\(\int \coth ^{-1}(1+d+d \coth (a+b x)) \, dx\) [224]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [B] (verified)
   Fricas [B] (verification not implemented)
   Sympy [F]
   Maxima [A] (verification not implemented)
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 12, antiderivative size = 69 \[ \int \coth ^{-1}(1+d+d \coth (a+b x)) \, dx=\frac {b x^2}{2}+x \coth ^{-1}(1+d+d \coth (a+b x))-\frac {1}{2} x \log \left (1-(1+d) e^{2 a+2 b x}\right )-\frac {\operatorname {PolyLog}\left (2,(1+d) e^{2 a+2 b x}\right )}{4 b} \]

[Out]

1/2*b*x^2+x*arccoth(1+d+d*coth(b*x+a))-1/2*x*ln(1-(1+d)*exp(2*b*x+2*a))-1/4*polylog(2,(1+d)*exp(2*b*x+2*a))/b

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {6369, 2215, 2221, 2317, 2438} \[ \int \coth ^{-1}(1+d+d \coth (a+b x)) \, dx=-\frac {\operatorname {PolyLog}\left (2,(d+1) e^{2 a+2 b x}\right )}{4 b}-\frac {1}{2} x \log \left (1-(d+1) e^{2 a+2 b x}\right )+x \coth ^{-1}(d \coth (a+b x)+d+1)+\frac {b x^2}{2} \]

[In]

Int[ArcCoth[1 + d + d*Coth[a + b*x]],x]

[Out]

(b*x^2)/2 + x*ArcCoth[1 + d + d*Coth[a + b*x]] - (x*Log[1 - (1 + d)*E^(2*a + 2*b*x)])/2 - PolyLog[2, (1 + d)*E
^(2*a + 2*b*x)]/(4*b)

Rule 2215

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[(c
+ d*x)^(m + 1)/(a*d*(m + 1)), x] - Dist[b/a, Int[(c + d*x)^m*((F^(g*(e + f*x)))^n/(a + b*(F^(g*(e + f*x)))^n))
, x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2221

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
 (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x]
 - Dist[d*(m/(b*f*g*n*Log[F])), Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2317

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2438

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

Rule 6369

Int[ArcCoth[(c_.) + Coth[(a_.) + (b_.)*(x_)]*(d_.)], x_Symbol] :> Simp[x*ArcCoth[c + d*Coth[a + b*x]], x] + Di
st[b, Int[x/(c - d - c*E^(2*a + 2*b*x)), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[(c - d)^2, 1]

Rubi steps \begin{align*} \text {integral}& = x \coth ^{-1}(1+d+d \coth (a+b x))+b \int \frac {x}{1+(-1-d) e^{2 a+2 b x}} \, dx \\ & = \frac {b x^2}{2}+x \coth ^{-1}(1+d+d \coth (a+b x))+(b (1+d)) \int \frac {e^{2 a+2 b x} x}{1+(-1-d) e^{2 a+2 b x}} \, dx \\ & = \frac {b x^2}{2}+x \coth ^{-1}(1+d+d \coth (a+b x))-\frac {1}{2} x \log \left (1-(1+d) e^{2 a+2 b x}\right )+\frac {1}{2} \int \log \left (1+(-1-d) e^{2 a+2 b x}\right ) \, dx \\ & = \frac {b x^2}{2}+x \coth ^{-1}(1+d+d \coth (a+b x))-\frac {1}{2} x \log \left (1-(1+d) e^{2 a+2 b x}\right )+\frac {\text {Subst}\left (\int \frac {\log (1+(-1-d) x)}{x} \, dx,x,e^{2 a+2 b x}\right )}{4 b} \\ & = \frac {b x^2}{2}+x \coth ^{-1}(1+d+d \coth (a+b x))-\frac {1}{2} x \log \left (1-(1+d) e^{2 a+2 b x}\right )-\frac {\operatorname {PolyLog}\left (2,(1+d) e^{2 a+2 b x}\right )}{4 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.91 \[ \int \coth ^{-1}(1+d+d \coth (a+b x)) \, dx=x \coth ^{-1}(1+d+d \coth (a+b x))+\frac {-2 b x \log \left (1-\frac {e^{-2 (a+b x)}}{1+d}\right )+\operatorname {PolyLog}\left (2,\frac {e^{-2 (a+b x)}}{1+d}\right )}{4 b} \]

[In]

Integrate[ArcCoth[1 + d + d*Coth[a + b*x]],x]

[Out]

x*ArcCoth[1 + d + d*Coth[a + b*x]] + (-2*b*x*Log[1 - 1/((1 + d)*E^(2*(a + b*x)))] + PolyLog[2, 1/((1 + d)*E^(2
*(a + b*x)))])/(4*b)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(254\) vs. \(2(61)=122\).

Time = 1.70 (sec) , antiderivative size = 255, normalized size of antiderivative = 3.70

method result size
derivativedivides \(\frac {-\frac {\operatorname {arccoth}\left (1+d +d \coth \left (b x +a \right )\right ) d \ln \left (-d \coth \left (b x +a \right )+d \right )}{2}+\frac {\operatorname {arccoth}\left (1+d +d \coth \left (b x +a \right )\right ) d \ln \left (d +d \coth \left (b x +a \right )\right )}{2}+\frac {d^{2} \left (\frac {\frac {\ln \left (d +d \coth \left (b x +a \right )\right )^{2}}{4}-\frac {\operatorname {dilog}\left (\frac {d \coth \left (b x +a \right )}{2}+\frac {d}{2}+1\right )}{2}-\frac {\ln \left (d +d \coth \left (b x +a \right )\right ) \ln \left (\frac {d \coth \left (b x +a \right )}{2}+\frac {d}{2}+1\right )}{2}}{d}-\frac {-\frac {\operatorname {dilog}\left (\frac {-d \coth \left (b x +a \right )-d -2}{-2 d -2}\right )}{2}-\frac {\ln \left (-d \coth \left (b x +a \right )+d \right ) \ln \left (\frac {-d \coth \left (b x +a \right )-d -2}{-2 d -2}\right )}{2}+\frac {\operatorname {dilog}\left (-\frac {-d \coth \left (b x +a \right )-d}{2 d}\right )}{2}+\frac {\ln \left (-d \coth \left (b x +a \right )+d \right ) \ln \left (-\frac {-d \coth \left (b x +a \right )-d}{2 d}\right )}{2}}{d}\right )}{2}}{b d}\) \(255\)
default \(\frac {-\frac {\operatorname {arccoth}\left (1+d +d \coth \left (b x +a \right )\right ) d \ln \left (-d \coth \left (b x +a \right )+d \right )}{2}+\frac {\operatorname {arccoth}\left (1+d +d \coth \left (b x +a \right )\right ) d \ln \left (d +d \coth \left (b x +a \right )\right )}{2}+\frac {d^{2} \left (\frac {\frac {\ln \left (d +d \coth \left (b x +a \right )\right )^{2}}{4}-\frac {\operatorname {dilog}\left (\frac {d \coth \left (b x +a \right )}{2}+\frac {d}{2}+1\right )}{2}-\frac {\ln \left (d +d \coth \left (b x +a \right )\right ) \ln \left (\frac {d \coth \left (b x +a \right )}{2}+\frac {d}{2}+1\right )}{2}}{d}-\frac {-\frac {\operatorname {dilog}\left (\frac {-d \coth \left (b x +a \right )-d -2}{-2 d -2}\right )}{2}-\frac {\ln \left (-d \coth \left (b x +a \right )+d \right ) \ln \left (\frac {-d \coth \left (b x +a \right )-d -2}{-2 d -2}\right )}{2}+\frac {\operatorname {dilog}\left (-\frac {-d \coth \left (b x +a \right )-d}{2 d}\right )}{2}+\frac {\ln \left (-d \coth \left (b x +a \right )+d \right ) \ln \left (-\frac {-d \coth \left (b x +a \right )-d}{2 d}\right )}{2}}{d}\right )}{2}}{b d}\) \(255\)
risch \(\text {Expression too large to display}\) \(1028\)

[In]

int(arccoth(1+d+d*coth(b*x+a)),x,method=_RETURNVERBOSE)

[Out]

1/b/d*(-1/2*arccoth(1+d+d*coth(b*x+a))*d*ln(-d*coth(b*x+a)+d)+1/2*arccoth(1+d+d*coth(b*x+a))*d*ln(d+d*coth(b*x
+a))+1/2*d^2*(1/d*(1/4*ln(d+d*coth(b*x+a))^2-1/2*dilog(1/2*d*coth(b*x+a)+1/2*d+1)-1/2*ln(d+d*coth(b*x+a))*ln(1
/2*d*coth(b*x+a)+1/2*d+1))-1/d*(-1/2*dilog((-d*coth(b*x+a)-d-2)/(-2*d-2))-1/2*ln(-d*coth(b*x+a)+d)*ln((-d*coth
(b*x+a)-d-2)/(-2*d-2))+1/2*dilog(-1/2*(-d*coth(b*x+a)-d)/d)+1/2*ln(-d*coth(b*x+a)+d)*ln(-1/2*(-d*coth(b*x+a)-d
)/d))))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 226 vs. \(2 (60) = 120\).

Time = 0.27 (sec) , antiderivative size = 226, normalized size of antiderivative = 3.28 \[ \int \coth ^{-1}(1+d+d \coth (a+b x)) \, dx=\frac {b^{2} x^{2} + b x \log \left (\frac {d \cosh \left (b x + a\right ) + {\left (d + 2\right )} \sinh \left (b x + a\right )}{d \cosh \left (b x + a\right ) + d \sinh \left (b x + a\right )}\right ) + a \log \left (2 \, {\left (d + 1\right )} \cosh \left (b x + a\right ) + 2 \, {\left (d + 1\right )} \sinh \left (b x + a\right ) + 2 \, \sqrt {d + 1}\right ) + a \log \left (2 \, {\left (d + 1\right )} \cosh \left (b x + a\right ) + 2 \, {\left (d + 1\right )} \sinh \left (b x + a\right ) - 2 \, \sqrt {d + 1}\right ) - {\left (b x + a\right )} \log \left (\sqrt {d + 1} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )} + 1\right ) - {\left (b x + a\right )} \log \left (-\sqrt {d + 1} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )} + 1\right ) - {\rm Li}_2\left (\sqrt {d + 1} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) - {\rm Li}_2\left (-\sqrt {d + 1} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right )}{2 \, b} \]

[In]

integrate(arccoth(1+d+d*coth(b*x+a)),x, algorithm="fricas")

[Out]

1/2*(b^2*x^2 + b*x*log((d*cosh(b*x + a) + (d + 2)*sinh(b*x + a))/(d*cosh(b*x + a) + d*sinh(b*x + a))) + a*log(
2*(d + 1)*cosh(b*x + a) + 2*(d + 1)*sinh(b*x + a) + 2*sqrt(d + 1)) + a*log(2*(d + 1)*cosh(b*x + a) + 2*(d + 1)
*sinh(b*x + a) - 2*sqrt(d + 1)) - (b*x + a)*log(sqrt(d + 1)*(cosh(b*x + a) + sinh(b*x + a)) + 1) - (b*x + a)*l
og(-sqrt(d + 1)*(cosh(b*x + a) + sinh(b*x + a)) + 1) - dilog(sqrt(d + 1)*(cosh(b*x + a) + sinh(b*x + a))) - di
log(-sqrt(d + 1)*(cosh(b*x + a) + sinh(b*x + a))))/b

Sympy [F]

\[ \int \coth ^{-1}(1+d+d \coth (a+b x)) \, dx=\int \operatorname {acoth}{\left (d \coth {\left (a + b x \right )} + d + 1 \right )}\, dx \]

[In]

integrate(acoth(1+d+d*coth(b*x+a)),x)

[Out]

Integral(acoth(d*coth(a + b*x) + d + 1), x)

Maxima [A] (verification not implemented)

none

Time = 0.75 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.04 \[ \int \coth ^{-1}(1+d+d \coth (a+b x)) \, dx=\frac {1}{4} \, b d {\left (\frac {2 \, x^{2}}{d} - \frac {2 \, b x \log \left (-{\left (d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) + {\rm Li}_2\left ({\left (d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )}\right )}{b^{2} d}\right )} + x \operatorname {arcoth}\left (d \coth \left (b x + a\right ) + d + 1\right ) \]

[In]

integrate(arccoth(1+d+d*coth(b*x+a)),x, algorithm="maxima")

[Out]

1/4*b*d*(2*x^2/d - (2*b*x*log(-(d + 1)*e^(2*b*x + 2*a) + 1) + dilog((d + 1)*e^(2*b*x + 2*a)))/(b^2*d)) + x*arc
coth(d*coth(b*x + a) + d + 1)

Giac [F]

\[ \int \coth ^{-1}(1+d+d \coth (a+b x)) \, dx=\int { \operatorname {arcoth}\left (d \coth \left (b x + a\right ) + d + 1\right ) \,d x } \]

[In]

integrate(arccoth(1+d+d*coth(b*x+a)),x, algorithm="giac")

[Out]

integrate(arccoth(d*coth(b*x + a) + d + 1), x)

Mupad [F(-1)]

Timed out. \[ \int \coth ^{-1}(1+d+d \coth (a+b x)) \, dx=\int \mathrm {acoth}\left (d+d\,\mathrm {coth}\left (a+b\,x\right )+1\right ) \,d x \]

[In]

int(acoth(d + d*coth(a + b*x) + 1),x)

[Out]

int(acoth(d + d*coth(a + b*x) + 1), x)