Integrand size = 14, antiderivative size = 54 \[ \int (a+b x)^2 \coth ^{-1}(a+b x) \, dx=\frac {(a+b x)^2}{6 b}+\frac {(a+b x)^3 \coth ^{-1}(a+b x)}{3 b}+\frac {\log \left (1-(a+b x)^2\right )}{6 b} \] Output:
1/6*(b*x+a)^2/b+1/3*(b*x+a)^3*arccoth(b*x+a)/b+1/6*ln(1-(b*x+a)^2)/b
Time = 0.03 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.78 \[ \int (a+b x)^2 \coth ^{-1}(a+b x) \, dx=\frac {(a+b x)^2+2 (a+b x)^3 \coth ^{-1}(a+b x)+\log \left (1-(a+b x)^2\right )}{6 b} \] Input:
Integrate[(a + b*x)^2*ArcCoth[a + b*x],x]
Output:
((a + b*x)^2 + 2*(a + b*x)^3*ArcCoth[a + b*x] + Log[1 - (a + b*x)^2])/(6*b )
Time = 0.28 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.76, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {6658, 6453, 243, 49, 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.
\(\displaystyle \int (a+b x)^2 \coth ^{-1}(a+b x) \, dx\) |
\(\Big \downarrow \) 6658 |
\(\displaystyle \frac {\int (a+b x)^2 \coth ^{-1}(a+b x)d(a+b x)}{b}\) |
\(\Big \downarrow \) 6453 |
\(\displaystyle \frac {\frac {1}{3} (a+b x)^3 \coth ^{-1}(a+b x)-\frac {1}{3} \int \frac {(a+b x)^3}{1-(a+b x)^2}d(a+b x)}{b}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {\frac {1}{3} (a+b x)^3 \coth ^{-1}(a+b x)-\frac {1}{6} \int \frac {(a+b x)^2}{-a-b x+1}d(a+b x)^2}{b}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {\frac {1}{3} (a+b x)^3 \coth ^{-1}(a+b x)-\frac {1}{6} \int \left (\frac {1}{-a-b x+1}-1\right )d(a+b x)^2}{b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {1}{6} (\log (-a-b x+1)+a+b x)+\frac {1}{3} (a+b x)^3 \coth ^{-1}(a+b x)}{b}\) |
Input:
Int[(a + b*x)^2*ArcCoth[a + b*x],x]
Output:
(((a + b*x)^3*ArcCoth[a + b*x])/3 + (a + b*x + Log[1 - a - b*x])/6)/b
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
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 = 0.23 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.89
method | result | size |
derivativedivides | \(\frac {\frac {\operatorname {arccoth}\left (b x +a \right ) \left (b x +a \right )^{3}}{3}+\frac {\left (b x +a \right )^{2}}{6}+\frac {\ln \left (b x +a -1\right )}{6}+\frac {\ln \left (b x +a +1\right )}{6}}{b}\) | \(48\) |
default | \(\frac {\frac {\operatorname {arccoth}\left (b x +a \right ) \left (b x +a \right )^{3}}{3}+\frac {\left (b x +a \right )^{2}}{6}+\frac {\ln \left (b x +a -1\right )}{6}+\frac {\ln \left (b x +a +1\right )}{6}}{b}\) | \(48\) |
parts | \(\frac {\operatorname {arccoth}\left (b x +a \right ) b^{2} x^{3}}{3}+\operatorname {arccoth}\left (b x +a \right ) b a \,x^{2}+\operatorname {arccoth}\left (b x +a \right ) a^{2} x +\frac {\operatorname {arccoth}\left (b x +a \right ) a^{3}}{3 b}+\frac {b \,x^{2}}{6}+\frac {x a}{3}+\frac {\ln \left (b x +a -1\right )}{6 b}+\frac {\ln \left (b x +a +1\right )}{6 b}\) | \(87\) |
parallelrisch | \(-\frac {-2 b^{4} \operatorname {arccoth}\left (b x +a \right ) x^{3}-6 x^{2} \operatorname {arccoth}\left (b x +a \right ) a \,b^{3}-6 x \,\operatorname {arccoth}\left (b x +a \right ) a^{2} b^{2}-b^{3} x^{2}-2 \,\operatorname {arccoth}\left (b x +a \right ) a^{3} b -2 x a \,b^{2}+5 a^{2} b -2 \ln \left (b x +a -1\right ) b -2 \,\operatorname {arccoth}\left (b x +a \right ) b -b}{6 b^{2}}\) | \(106\) |
risch | \(\frac {\left (b x +a \right )^{3} \ln \left (b x +a +1\right )}{6 b}-\frac {b^{2} x^{3} \ln \left (b x +a -1\right )}{6}-\frac {b \ln \left (b x +a -1\right ) x^{2} a}{2}-\frac {a^{2} x \ln \left (b x +a -1\right )}{2}-\frac {\ln \left (b x +a -1\right ) a^{3}}{6 b}+\frac {b \,x^{2}}{6}+\frac {x a}{3}+\frac {\ln \left (b^{2} x^{2}+2 b x a +a^{2}-1\right )}{6 b}\) | \(111\) |
Input:
int((b*x+a)^2*arccoth(b*x+a),x,method=_RETURNVERBOSE)
Output:
1/b*(1/3*arccoth(b*x+a)*(b*x+a)^3+1/6*(b*x+a)^2+1/6*ln(b*x+a-1)+1/6*ln(b*x +a+1))
Time = 0.08 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.59 \[ \int (a+b x)^2 \coth ^{-1}(a+b x) \, dx=\frac {b^{2} x^{2} + 2 \, a b x + {\left (a^{3} + 1\right )} \log \left (b x + a + 1\right ) - {\left (a^{3} - 1\right )} \log \left (b x + a - 1\right ) + {\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x\right )} \log \left (\frac {b x + a + 1}{b x + a - 1}\right )}{6 \, b} \] Input:
integrate((b*x+a)^2*arccoth(b*x+a),x, algorithm="fricas")
Output:
1/6*(b^2*x^2 + 2*a*b*x + (a^3 + 1)*log(b*x + a + 1) - (a^3 - 1)*log(b*x + a - 1) + (b^3*x^3 + 3*a*b^2*x^2 + 3*a^2*b*x)*log((b*x + a + 1)/(b*x + a - 1)))/b
Leaf count of result is larger than twice the leaf count of optimal. 97 vs. \(2 (39) = 78\).
Time = 0.64 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.80 \[ \int (a+b x)^2 \coth ^{-1}(a+b x) \, dx=\begin {cases} \frac {a^{3} \operatorname {acoth}{\left (a + b x \right )}}{3 b} + a^{2} x \operatorname {acoth}{\left (a + b x \right )} + a b x^{2} \operatorname {acoth}{\left (a + b x \right )} + \frac {a x}{3} + \frac {b^{2} x^{3} \operatorname {acoth}{\left (a + b x \right )}}{3} + \frac {b x^{2}}{6} + \frac {\log {\left (\frac {a}{b} + x + \frac {1}{b} \right )}}{3 b} - \frac {\operatorname {acoth}{\left (a + b x \right )}}{3 b} & \text {for}\: b \neq 0 \\a^{2} x \operatorname {acoth}{\left (a \right )} & \text {otherwise} \end {cases} \] Input:
integrate((b*x+a)**2*acoth(b*x+a),x)
Output:
Piecewise((a**3*acoth(a + b*x)/(3*b) + a**2*x*acoth(a + b*x) + a*b*x**2*ac oth(a + b*x) + a*x/3 + b**2*x**3*acoth(a + b*x)/3 + b*x**2/6 + log(a/b + x + 1/b)/(3*b) - acoth(a + b*x)/(3*b), Ne(b, 0)), (a**2*x*acoth(a), True))
Time = 0.02 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.50 \[ \int (a+b x)^2 \coth ^{-1}(a+b x) \, dx=\frac {1}{6} \, b {\left (\frac {b x^{2} + 2 \, a x}{b} + \frac {{\left (a^{3} + 1\right )} \log \left (b x + a + 1\right )}{b^{2}} - \frac {{\left (a^{3} - 1\right )} \log \left (b x + a - 1\right )}{b^{2}}\right )} + \frac {1}{3} \, {\left (b^{2} x^{3} + 3 \, a b x^{2} + 3 \, a^{2} x\right )} \operatorname {arcoth}\left (b x + a\right ) \] Input:
integrate((b*x+a)^2*arccoth(b*x+a),x, algorithm="maxima")
Output:
1/6*b*((b*x^2 + 2*a*x)/b + (a^3 + 1)*log(b*x + a + 1)/b^2 - (a^3 - 1)*log( b*x + a - 1)/b^2) + 1/3*(b^2*x^3 + 3*a*b*x^2 + 3*a^2*x)*arccoth(b*x + a)
Leaf count of result is larger than twice the leaf count of optimal. 255 vs. \(2 (48) = 96\).
Time = 0.13 (sec) , antiderivative size = 255, normalized size of antiderivative = 4.72 \[ \int (a+b x)^2 \coth ^{-1}(a+b x) \, dx=\frac {1}{6} \, {\left ({\left (a + 1\right )} b - {\left (a - 1\right )} b\right )} {\left (\frac {\log \left (\frac {{\left | b x + a + 1 \right |}}{{\left | b x + a - 1 \right |}}\right )}{b^{2}} - \frac {\log \left ({\left | \frac {b x + a + 1}{b x + a - 1} - 1 \right |}\right )}{b^{2}} + \frac {{\left (\frac {3 \, {\left (b x + a + 1\right )}^{2}}{{\left (b x + a - 1\right )}^{2}} + 1\right )} \log \left (-\frac {\frac {1}{a - \frac {{\left (\frac {{\left (b x + a + 1\right )} {\left (a - 1\right )}}{b x + a - 1} - a - 1\right )} b}{\frac {{\left (b x + a + 1\right )} b}{b x + a - 1} - b}} + 1}{\frac {1}{a - \frac {{\left (\frac {{\left (b x + a + 1\right )} {\left (a - 1\right )}}{b x + a - 1} - a - 1\right )} b}{\frac {{\left (b x + a + 1\right )} b}{b x + a - 1} - b}} - 1}\right )}{b^{2} {\left (\frac {b x + a + 1}{b x + a - 1} - 1\right )}^{3}} + \frac {2 \, {\left (b x + a + 1\right )}}{{\left (b x + a - 1\right )} b^{2} {\left (\frac {b x + a + 1}{b x + a - 1} - 1\right )}^{2}}\right )} \] Input:
integrate((b*x+a)^2*arccoth(b*x+a),x, algorithm="giac")
Output:
1/6*((a + 1)*b - (a - 1)*b)*(log(abs(b*x + a + 1)/abs(b*x + a - 1))/b^2 - log(abs((b*x + a + 1)/(b*x + a - 1) - 1))/b^2 + (3*(b*x + a + 1)^2/(b*x + a - 1)^2 + 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^2*( (b*x + a + 1)/(b*x + a - 1) - 1)^3) + 2*(b*x + a + 1)/((b*x + a - 1)*b^2*( (b*x + a + 1)/(b*x + a - 1) - 1)^2))
Time = 4.12 (sec) , antiderivative size = 114, normalized size of antiderivative = 2.11 \[ \int (a+b x)^2 \coth ^{-1}(a+b x) \, dx=\frac {a\,x}{3}+\ln \left (\frac {1}{a+b\,x}+1\right )\,\left (\frac {a^2\,x}{2}+\frac {a\,b\,x^2}{2}+\frac {b^2\,x^3}{6}\right )+\frac {b\,x^2}{6}-\ln \left (1-\frac {1}{a+b\,x}\right )\,\left (\frac {a^2\,x}{2}+\frac {a\,b\,x^2}{2}+\frac {b^2\,x^3}{6}\right )-\frac {\ln \left (a+b\,x-1\right )\,\left (a^3-1\right )}{6\,b}+\frac {\ln \left (a+b\,x+1\right )\,\left (a^3+1\right )}{6\,b} \] Input:
int(acoth(a + b*x)*(a + b*x)^2,x)
Output:
(a*x)/3 + log(1/(a + b*x) + 1)*((a^2*x)/2 + (b^2*x^3)/6 + (a*b*x^2)/2) + ( b*x^2)/6 - log(1 - 1/(a + b*x))*((a^2*x)/2 + (b^2*x^3)/6 + (a*b*x^2)/2) - (log(a + b*x - 1)*(a^3 - 1))/(6*b) + (log(a + b*x + 1)*(a^3 + 1))/(6*b)
Time = 0.16 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.65 \[ \int (a+b x)^2 \coth ^{-1}(a+b x) \, dx=\frac {2 \mathit {acoth} \left (b x +a \right ) a^{3}+6 \mathit {acoth} \left (b x +a \right ) a^{2} b x +6 \mathit {acoth} \left (b x +a \right ) a \,b^{2} x^{2}+2 \mathit {acoth} \left (b x +a \right ) b^{3} x^{3}+2 \mathit {acoth} \left (b x +a \right )-2 \,\mathrm {log}\left (b x +a -1\right )-2 a b x -b^{2} x^{2}}{6 b} \] Input:
int((b*x+a)^2*acoth(b*x+a),x)
Output:
(2*acoth(a + b*x)*a**3 + 6*acoth(a + b*x)*a**2*b*x + 6*acoth(a + b*x)*a*b* *2*x**2 + 2*acoth(a + b*x)*b**3*x**3 + 2*acoth(a + b*x) - 2*log(a + b*x - 1) - 2*a*b*x - b**2*x**2)/(6*b)