Integrand size = 8, antiderivative size = 65 \[ \int x \coth ^{-1}(a+b x) \, dx=\frac {x}{2 b}+\frac {1}{2} x^2 \coth ^{-1}(a+b x)+\frac {(1-a)^2 \log (1-a-b x)}{4 b^2}-\frac {(1+a)^2 \log (1+a+b x)}{4 b^2} \] Output:
1/2*x/b+1/2*x^2*arccoth(b*x+a)+1/4*(1-a)^2*ln(-b*x-a+1)/b^2-1/4*(1+a)^2*ln (b*x+a+1)/b^2
Time = 0.02 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.86 \[ \int x \coth ^{-1}(a+b x) \, dx=\frac {2 b x+2 b^2 x^2 \coth ^{-1}(a+b x)+(-1+a)^2 \log (1-a-b x)-(1+a)^2 \log (1+a+b x)}{4 b^2} \] Input:
Integrate[x*ArcCoth[a + b*x],x]
Output:
(2*b*x + 2*b^2*x^2*ArcCoth[a + b*x] + (-1 + a)^2*Log[1 - a - b*x] - (1 + a )^2*Log[1 + a + b*x])/(4*b^2)
Time = 0.32 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.09, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {6662, 25, 27, 6479, 477, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int x \coth ^{-1}(a+b x) \, dx\) |
\(\Big \downarrow \) 6662 |
\(\displaystyle \frac {\int x \coth ^{-1}(a+b x)d(a+b x)}{b}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int -x \coth ^{-1}(a+b x)d(a+b x)}{b}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\int -b x \coth ^{-1}(a+b x)d(a+b x)}{b^2}\) |
\(\Big \downarrow \) 6479 |
\(\displaystyle -\frac {\frac {1}{2} \int \frac {b^2 x^2}{1-(a+b x)^2}d(a+b x)-\frac {1}{2} b^2 x^2 \coth ^{-1}(a+b x)}{b^2}\) |
\(\Big \downarrow \) 477 |
\(\displaystyle -\frac {\frac {1}{2} \int \left (\frac {(1-a)^2}{2 (-a-b x+1)}+\frac {(a+1)^2}{2 (a+b x+1)}-1\right )d(a+b x)-\frac {1}{2} b^2 x^2 \coth ^{-1}(a+b x)}{b^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\frac {1}{2} \left (-\frac {1}{2} (1-a)^2 \log (-a-b x+1)+\frac {1}{2} (a+1)^2 \log (a+b x+1)-a-b x\right )-\frac {1}{2} b^2 x^2 \coth ^{-1}(a+b x)}{b^2}\) |
Input:
Int[x*ArcCoth[a + b*x],x]
Output:
-((-1/2*(b^2*x^2*ArcCoth[a + b*x]) + (-a - b*x - ((1 - a)^2*Log[1 - a - b* x])/2 + ((1 + a)^2*Log[1 + a + b*x])/2)/2)/b^2)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ a^p Int[ExpandIntegrand[(c + d*x)^n*(1 - Rt[-b/a, 2]*x)^p*(1 + Rt[-b/a, 2 ]*x)^p, x], x], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[p, 0] && IntegerQ[n] & & NiceSqrtQ[-b/a] && !FractionalPowerFactorQ[Rt[-b/a, 2]]
Int[((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_))^(q_.), x_Symbol ] :> Simp[(d + e*x)^(q + 1)*((a + b*ArcCoth[c*x])/(e*(q + 1))), x] - Simp[b *(c/(e*(q + 1))) Int[(d + e*x)^(q + 1)/(1 - c^2*x^2), x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[q, -1]
Int[((a_.) + ArcCoth[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^( m_.), x_Symbol] :> Simp[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] && IG tQ[p, 0]
Time = 0.13 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.97
method | result | size |
parallelrisch | \(-\frac {-\operatorname {arccoth}\left (b x +a \right ) b^{2} x^{2}+\operatorname {arccoth}\left (b x +a \right ) a^{2}+2 \ln \left (b x +a -1\right ) a -b x +2 \,\operatorname {arccoth}\left (b x +a \right ) a +\operatorname {arccoth}\left (b x +a \right )+2 a}{2 b^{2}}\) | \(63\) |
parts | \(\frac {x^{2} \operatorname {arccoth}\left (b x +a \right )}{2}+\frac {b \left (\frac {x}{b^{2}}+\frac {\left (a^{2}-2 a +1\right ) \ln \left (b x +a -1\right )}{2 b^{3}}+\frac {\left (-a^{2}-2 a -1\right ) \ln \left (b x +a +1\right )}{2 b^{3}}\right )}{2}\) | \(64\) |
derivativedivides | \(\frac {\frac {\left (b x +a \right )^{2} \operatorname {arccoth}\left (b x +a \right )}{2}-\operatorname {arccoth}\left (b x +a \right ) a \left (b x +a \right )+\frac {b x}{2}+\frac {a}{2}-\frac {\left (2 a -1\right ) \ln \left (b x +a -1\right )}{4}+\frac {\left (-2 a -1\right ) \ln \left (b x +a +1\right )}{4}}{b^{2}}\) | \(70\) |
default | \(\frac {\frac {\left (b x +a \right )^{2} \operatorname {arccoth}\left (b x +a \right )}{2}-\operatorname {arccoth}\left (b x +a \right ) a \left (b x +a \right )+\frac {b x}{2}+\frac {a}{2}-\frac {\left (2 a -1\right ) \ln \left (b x +a -1\right )}{4}+\frac {\left (-2 a -1\right ) \ln \left (b x +a +1\right )}{4}}{b^{2}}\) | \(70\) |
risch | \(\frac {x^{2} \ln \left (b x +a +1\right )}{4}-\frac {x^{2} \ln \left (b x +a -1\right )}{4}-\frac {\ln \left (b x +a +1\right ) a^{2}}{4 b^{2}}+\frac {\ln \left (-b x -a +1\right ) a^{2}}{4 b^{2}}-\frac {\ln \left (b x +a +1\right ) a}{2 b^{2}}-\frac {\ln \left (-b x -a +1\right ) a}{2 b^{2}}+\frac {x}{2 b}-\frac {\ln \left (b x +a +1\right )}{4 b^{2}}+\frac {\ln \left (-b x -a +1\right )}{4 b^{2}}\) | \(121\) |
Input:
int(x*arccoth(b*x+a),x,method=_RETURNVERBOSE)
Output:
-1/2*(-arccoth(b*x+a)*b^2*x^2+arccoth(b*x+a)*a^2+2*ln(b*x+a-1)*a-b*x+2*arc coth(b*x+a)*a+arccoth(b*x+a)+2*a)/b^2
Time = 0.10 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.02 \[ \int x \coth ^{-1}(a+b x) \, dx=\frac {b^{2} x^{2} \log \left (\frac {b x + a + 1}{b x + a - 1}\right ) + 2 \, b x - {\left (a^{2} + 2 \, a + 1\right )} \log \left (b x + a + 1\right ) + {\left (a^{2} - 2 \, a + 1\right )} \log \left (b x + a - 1\right )}{4 \, b^{2}} \] Input:
integrate(x*arccoth(b*x+a),x, algorithm="fricas")
Output:
1/4*(b^2*x^2*log((b*x + a + 1)/(b*x + a - 1)) + 2*b*x - (a^2 + 2*a + 1)*lo g(b*x + a + 1) + (a^2 - 2*a + 1)*log(b*x + a - 1))/b^2
Time = 0.51 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.17 \[ \int x \coth ^{-1}(a+b x) \, dx=\begin {cases} - \frac {a^{2} \operatorname {acoth}{\left (a + b x \right )}}{2 b^{2}} - \frac {a \log {\left (a + b x + 1 \right )}}{b^{2}} + \frac {a \operatorname {acoth}{\left (a + b x \right )}}{b^{2}} + \frac {x^{2} \operatorname {acoth}{\left (a + b x \right )}}{2} + \frac {x}{2 b} - \frac {\operatorname {acoth}{\left (a + b x \right )}}{2 b^{2}} & \text {for}\: b \neq 0 \\\frac {x^{2} \operatorname {acoth}{\left (a \right )}}{2} & \text {otherwise} \end {cases} \] Input:
integrate(x*acoth(b*x+a),x)
Output:
Piecewise((-a**2*acoth(a + b*x)/(2*b**2) - a*log(a + b*x + 1)/b**2 + a*aco th(a + b*x)/b**2 + x**2*acoth(a + b*x)/2 + x/(2*b) - acoth(a + b*x)/(2*b** 2), Ne(b, 0)), (x**2*acoth(a)/2, True))
Time = 0.03 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.94 \[ \int x \coth ^{-1}(a+b x) \, dx=\frac {1}{2} \, x^{2} \operatorname {arcoth}\left (b x + a\right ) + \frac {1}{4} \, b {\left (\frac {2 \, x}{b^{2}} - \frac {{\left (a^{2} + 2 \, a + 1\right )} \log \left (b x + a + 1\right )}{b^{3}} + \frac {{\left (a^{2} - 2 \, a + 1\right )} \log \left (b x + a - 1\right )}{b^{3}}\right )} \] Input:
integrate(x*arccoth(b*x+a),x, algorithm="maxima")
Output:
1/2*x^2*arccoth(b*x + a) + 1/4*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)
Leaf count of result is larger than twice the leaf count of optimal. 259 vs. \(2 (55) = 110\).
Time = 0.13 (sec) , antiderivative size = 259, normalized size of antiderivative = 3.98 \[ \int x \coth ^{-1}(a+b x) \, dx=-\frac {1}{2} \, {\left ({\left (a + 1\right )} b - {\left (a - 1\right )} b\right )} {\left (\frac {a \log \left (\frac {{\left | b x + a + 1 \right |}}{{\left | b x + a - 1 \right |}}\right )}{b^{3}} - \frac {a \log \left ({\left | \frac {b x + a + 1}{b x + a - 1} - 1 \right |}\right )}{b^{3}} + \frac {{\left (\frac {{\left (b x + a + 1\right )} a}{b x + a - 1} - a - \frac {b x + a + 1}{b x + a - 1}\right )} \log \left (-\frac {\frac {1}{a - \frac {{\left (\frac {{\left (b x + a + 1\right )} {\left (a - 1\right )}}{b x + a - 1} - a - 1\right )} b}{\frac {{\left (b x + a + 1\right )} b}{b x + a - 1} - b}} + 1}{\frac {1}{a - \frac {{\left (\frac {{\left (b x + a + 1\right )} {\left (a - 1\right )}}{b x + a - 1} - a - 1\right )} b}{\frac {{\left (b x + a + 1\right )} b}{b x + a - 1} - b}} - 1}\right )}{b^{3} {\left (\frac {b x + a + 1}{b x + a - 1} - 1\right )}^{2}} - \frac {1}{b^{3} {\left (\frac {b x + a + 1}{b x + a - 1} - 1\right )}}\right )} \] Input:
integrate(x*arccoth(b*x+a),x, algorithm="giac")
Output:
-1/2*((a + 1)*b - (a - 1)*b)*(a*log(abs(b*x + a + 1)/abs(b*x + a - 1))/b^3 - a*log(abs((b*x + a + 1)/(b*x + a - 1) - 1))/b^3 + ((b*x + a + 1)*a/(b*x + a - 1) - a - (b*x + a + 1)/(b*x + a - 1))*log(-(1/(a - ((b*x + a + 1)*( a - 1)/(b*x + a - 1) - a - 1)*b/((b*x + a + 1)*b/(b*x + a - 1) - b)) + 1)/ (1/(a - ((b*x + a + 1)*(a - 1)/(b*x + a - 1) - a - 1)*b/((b*x + a + 1)*b/( b*x + a - 1) - b)) - 1))/(b^3*((b*x + a + 1)/(b*x + a - 1) - 1)^2) - 1/(b^ 3*((b*x + a + 1)/(b*x + a - 1) - 1)))
Time = 4.63 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.95 \[ \int x \coth ^{-1}(a+b x) \, dx=\frac {x^2\,\mathrm {acoth}\left (a+b\,x\right )}{2}-\frac {\frac {\mathrm {acoth}\left (a+b\,x\right )}{2}-\frac {b\,x}{2}+\frac {a^2\,\mathrm {acoth}\left (a+b\,x\right )}{2}+\frac {a\,\ln \left (a^2+2\,a\,b\,x+b^2\,x^2-1\right )}{2}}{b^2} \] Input:
int(x*acoth(a + b*x),x)
Output:
(x^2*acoth(a + b*x))/2 - (acoth(a + b*x)/2 - (b*x)/2 + (a^2*acoth(a + b*x) )/2 + (a*log(a^2 + b^2*x^2 + 2*a*b*x - 1))/2)/b^2
Time = 0.16 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.94 \[ \int x \coth ^{-1}(a+b x) \, dx=\frac {-\mathit {acoth} \left (b x +a \right ) a^{2}-2 \mathit {acoth} \left (b x +a \right ) a +\mathit {acoth} \left (b x +a \right ) b^{2} x^{2}-\mathit {acoth} \left (b x +a \right )+2 \,\mathrm {log}\left (b x +a -1\right ) a -b x}{2 b^{2}} \] Input:
int(x*acoth(b*x+a),x)
Output:
( - acoth(a + b*x)*a**2 - 2*acoth(a + b*x)*a + acoth(a + b*x)*b**2*x**2 - acoth(a + b*x) + 2*log(a + b*x - 1)*a - b*x)/(2*b**2)