Integrand size = 12, antiderivative size = 39 \[ \int (a+b x) \coth ^{-1}(a+b x) \, dx=\frac {x}{2}+\frac {(a+b x)^2 \coth ^{-1}(a+b x)}{2 b}-\frac {\text {arctanh}(a+b x)}{2 b} \] Output:
1/2*x+1/2*(b*x+a)^2*arccoth(b*x+a)/b-1/2*arctanh(b*x+a)/b
Time = 0.03 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.69 \[ \int (a+b x) \coth ^{-1}(a+b x) \, dx=\frac {2 b x+2 b x (2 a+b x) \coth ^{-1}(a+b x)-\left (-1+a^2\right ) \log (1-a-b x)-\log (1+a+b x)+a^2 \log (1+a+b x)}{4 b} \] Input:
Integrate[(a + b*x)*ArcCoth[a + b*x],x]
Output:
(2*b*x + 2*b*x*(2*a + b*x)*ArcCoth[a + b*x] - (-1 + a^2)*Log[1 - a - b*x] - Log[1 + a + b*x] + a^2*Log[1 + a + b*x])/(4*b)
Time = 0.24 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6658, 6453, 262, 219}
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.
\(\displaystyle \int (a+b x) \coth ^{-1}(a+b x) \, dx\) |
\(\Big \downarrow \) 6658 |
\(\displaystyle \frac {\int (a+b x) \coth ^{-1}(a+b x)d(a+b x)}{b}\) |
\(\Big \downarrow \) 6453 |
\(\displaystyle \frac {\frac {1}{2} (a+b x)^2 \coth ^{-1}(a+b x)-\frac {1}{2} \int \frac {(a+b x)^2}{1-(a+b x)^2}d(a+b x)}{b}\) |
\(\Big \downarrow \) 262 |
\(\displaystyle \frac {\frac {1}{2} \left (-\int \frac {1}{1-(a+b x)^2}d(a+b x)+a+b x\right )+\frac {1}{2} (a+b x)^2 \coth ^{-1}(a+b x)}{b}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\frac {1}{2} (-\text {arctanh}(a+b x)+a+b x)+\frac {1}{2} (a+b x)^2 \coth ^{-1}(a+b x)}{b}\) |
Input:
Int[(a + b*x)*ArcCoth[a + b*x],x]
Output:
(((a + b*x)^2*ArcCoth[a + b*x])/2 + (a + b*x - ArcTanh[a + b*x])/2)/b
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) ^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ (b*(m + 2*p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b , c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c , 2, m, p, x]
Int[((a_.) + ArcCoth[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] : > Simp[x^(m + 1)*((a + b*ArcCoth[c*x^n])^p/(m + 1)), x] - Simp[b*c*n*(p/(m + 1)) Int[x^(m + n)*((a + b*ArcCoth[c*x^n])^(p - 1)/(1 - c^2*x^(2*n))), x ], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0] && (EqQ[p, 1] || (EqQ[n, 1 ] && IntegerQ[m])) && NeQ[m, -1]
Int[((a_.) + ArcCoth[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^( m_.), x_Symbol] :> Simp[1/d Subst[Int[(f*(x/d))^m*(a + b*ArcCoth[x])^p, x ], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[d*e - c*f, 0] && IGtQ[p, 0]
Time = 0.17 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.18
method | result | size |
derivativedivides | \(\frac {\frac {\left (b x +a \right )^{2} \operatorname {arccoth}\left (b x +a \right )}{2}+\frac {b x}{2}+\frac {a}{2}+\frac {\ln \left (b x +a -1\right )}{4}-\frac {\ln \left (b x +a +1\right )}{4}}{b}\) | \(46\) |
default | \(\frac {\frac {\left (b x +a \right )^{2} \operatorname {arccoth}\left (b x +a \right )}{2}+\frac {b x}{2}+\frac {a}{2}+\frac {\ln \left (b x +a -1\right )}{4}-\frac {\ln \left (b x +a +1\right )}{4}}{b}\) | \(46\) |
parallelrisch | \(-\frac {-b^{3} \operatorname {arccoth}\left (b x +a \right ) x^{2}-2 x \,\operatorname {arccoth}\left (b x +a \right ) a \,b^{2}-\operatorname {arccoth}\left (b x +a \right ) a^{2} b -b^{2} x +\operatorname {arccoth}\left (b x +a \right ) b +2 a b}{2 b^{2}}\) | \(64\) |
parts | \(\frac {\operatorname {arccoth}\left (b x +a \right ) b \,x^{2}}{2}+\operatorname {arccoth}\left (b x +a \right ) x a +\frac {b \left (\frac {x}{b}+\frac {\left (-a^{2}+1\right ) \ln \left (b x +a -1\right )}{2 b^{2}}+\frac {\left (a^{2}-1\right ) \ln \left (b x +a +1\right )}{2 b^{2}}\right )}{2}\) | \(68\) |
risch | \(\left (\frac {1}{4} b \,x^{2}+\frac {1}{2} x a \right ) \ln \left (b x +a +1\right )-\frac {b \,x^{2} \ln \left (b x +a -1\right )}{4}-\frac {a x \ln \left (b x +a -1\right )}{2}-\frac {\ln \left (b x +a -1\right ) a^{2}}{4 b}+\frac {\ln \left (-b x -a -1\right ) a^{2}}{4 b}+\frac {x}{2}+\frac {\ln \left (b x +a -1\right )}{4 b}-\frac {\ln \left (-b x -a -1\right )}{4 b}\) | \(108\) |
orering | \(\frac {\left (2 b^{3} x^{3}+5 a \,b^{2} x^{2}+4 a^{2} b x +a^{3}-2 b x -a \right ) \operatorname {arccoth}\left (b x +a \right )}{2 \left (b x +a \right ) b}-\frac {x \left (b x +a -1\right ) \left (b x +a +1\right ) \left (\operatorname {arccoth}\left (b x +a \right ) b -\frac {\left (b x +a \right ) b}{\left (b x +a \right )^{2}-1}\right )}{2 b \left (b x +a \right )}\) | \(108\) |
Input:
int((b*x+a)*arccoth(b*x+a),x,method=_RETURNVERBOSE)
Output:
1/b*(1/2*(b*x+a)^2*arccoth(b*x+a)+1/2*b*x+1/2*a+1/4*ln(b*x+a-1)-1/4*ln(b*x +a+1))
Time = 0.08 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.13 \[ \int (a+b x) \coth ^{-1}(a+b x) \, dx=\frac {2 \, b x + {\left (b^{2} x^{2} + 2 \, a b x + a^{2} - 1\right )} \log \left (\frac {b x + a + 1}{b x + a - 1}\right )}{4 \, b} \] Input:
integrate((b*x+a)*arccoth(b*x+a),x, algorithm="fricas")
Output:
1/4*(2*b*x + (b^2*x^2 + 2*a*b*x + a^2 - 1)*log((b*x + a + 1)/(b*x + a - 1) ))/b
Time = 0.41 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.44 \[ \int (a+b x) \coth ^{-1}(a+b x) \, dx=\begin {cases} \frac {a^{2} \operatorname {acoth}{\left (a + b x \right )}}{2 b} + a x \operatorname {acoth}{\left (a + b x \right )} + \frac {b x^{2} \operatorname {acoth}{\left (a + b x \right )}}{2} + \frac {x}{2} - \frac {\operatorname {acoth}{\left (a + b x \right )}}{2 b} & \text {for}\: b \neq 0 \\a x \operatorname {acoth}{\left (a \right )} & \text {otherwise} \end {cases} \] Input:
integrate((b*x+a)*acoth(b*x+a),x)
Output:
Piecewise((a**2*acoth(a + b*x)/(2*b) + a*x*acoth(a + b*x) + b*x**2*acoth(a + b*x)/2 + x/2 - acoth(a + b*x)/(2*b), Ne(b, 0)), (a*x*acoth(a), True))
Time = 0.03 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.59 \[ \int (a+b x) \coth ^{-1}(a+b x) \, dx=\frac {1}{4} \, b {\left (\frac {2 \, x}{b} + \frac {{\left (a^{2} - 1\right )} \log \left (b x + a + 1\right )}{b^{2}} - \frac {{\left (a^{2} - 1\right )} \log \left (b x + a - 1\right )}{b^{2}}\right )} + \frac {1}{2} \, {\left (b x^{2} + 2 \, a x\right )} \operatorname {arcoth}\left (b x + a\right ) \] Input:
integrate((b*x+a)*arccoth(b*x+a),x, algorithm="maxima")
Output:
1/4*b*(2*x/b + (a^2 - 1)*log(b*x + a + 1)/b^2 - (a^2 - 1)*log(b*x + a - 1) /b^2) + 1/2*(b*x^2 + 2*a*x)*arccoth(b*x + a)
Leaf count of result is larger than twice the leaf count of optimal. 188 vs. \(2 (33) = 66\).
Time = 0.13 (sec) , antiderivative size = 188, normalized size of antiderivative = 4.82 \[ \int (a+b x) \coth ^{-1}(a+b x) \, dx=\frac {1}{2} \, {\left ({\left (a + 1\right )} b - {\left (a - 1\right )} b\right )} {\left (\frac {1}{b^{2} {\left (\frac {b x + a + 1}{b x + a - 1} - 1\right )}} + \frac {{\left (b x + a + 1\right )} \log \left (-\frac {\frac {1}{a - \frac {{\left (\frac {{\left (b x + a + 1\right )} {\left (a - 1\right )}}{b x + a - 1} - a - 1\right )} b}{\frac {{\left (b x + a + 1\right )} b}{b x + a - 1} - b}} + 1}{\frac {1}{a - \frac {{\left (\frac {{\left (b x + a + 1\right )} {\left (a - 1\right )}}{b x + a - 1} - a - 1\right )} b}{\frac {{\left (b x + a + 1\right )} b}{b x + a - 1} - b}} - 1}\right )}{{\left (b x + a - 1\right )} b^{2} {\left (\frac {b x + a + 1}{b x + a - 1} - 1\right )}^{2}}\right )} \] Input:
integrate((b*x+a)*arccoth(b*x+a),x, algorithm="giac")
Output:
1/2*((a + 1)*b - (a - 1)*b)*(1/(b^2*((b*x + a + 1)/(b*x + a - 1) - 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*x + a - 1)*b^2*((b*x + a + 1)/(b*x + a - 1) - 1)^2))
Time = 4.65 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.28 \[ \int (a+b x) \coth ^{-1}(a+b x) \, dx=\frac {x}{2}-\frac {\frac {\mathrm {acoth}\left (a+b\,x\right )}{2}-\frac {a^2\,\mathrm {acoth}\left (a+b\,x\right )}{2}}{b}+a\,x\,\mathrm {acoth}\left (a+b\,x\right )+\frac {b\,x^2\,\mathrm {acoth}\left (a+b\,x\right )}{2} \] Input:
int(acoth(a + b*x)*(a + b*x),x)
Output:
x/2 - (acoth(a + b*x)/2 - (a^2*acoth(a + b*x))/2)/b + a*x*acoth(a + b*x) + (b*x^2*acoth(a + b*x))/2
Time = 0.15 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.33 \[ \int (a+b x) \coth ^{-1}(a+b x) \, dx=\frac {\mathit {acoth} \left (b x +a \right ) a^{2}+2 \mathit {acoth} \left (b x +a \right ) a b x +\mathit {acoth} \left (b x +a \right ) b^{2} x^{2}-\mathit {acoth} \left (b x +a \right )-b x}{2 b} \] Input:
int((b*x+a)*acoth(b*x+a),x)
Output:
(acoth(a + b*x)*a**2 + 2*acoth(a + b*x)*a*b*x + acoth(a + b*x)*b**2*x**2 - acoth(a + b*x) - b*x)/(2*b)