Integrand size = 10, antiderivative size = 78 \[ \int x^2 \coth ^{-1}(a+b x) \, dx=-\frac {a x}{b^2}+\frac {(a+b x)^2}{6 b^3}+\frac {1}{3} x^3 \coth ^{-1}(a+b x)+\frac {(1-a)^3 \log (1-a-b x)}{6 b^3}+\frac {(1+a)^3 \log (1+a+b x)}{6 b^3} \]
-a*x/b^2+1/6*(b*x+a)^2/b^3+1/3*x^3*arccoth(b*x+a)+1/6*(1-a)^3*ln(-b*x-a+1) /b^3+1/6*(1+a)^3*ln(b*x+a+1)/b^3
Time = 0.02 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.18 \[ \int x^2 \coth ^{-1}(a+b x) \, dx=-\frac {2 a x}{3 b^2}+\frac {x^2}{6 b}+\frac {1}{3} x^3 \coth ^{-1}(a+b x)+\frac {\left (1-3 a+3 a^2-a^3\right ) \log (1-a-b x)}{6 b^3}+\frac {\left (1+3 a+3 a^2+a^3\right ) \log (1+a+b x)}{6 b^3} \]
(-2*a*x)/(3*b^2) + x^2/(6*b) + (x^3*ArcCoth[a + b*x])/3 + ((1 - 3*a + 3*a^ 2 - a^3)*Log[1 - a - b*x])/(6*b^3) + ((1 + 3*a + 3*a^2 + a^3)*Log[1 + a + b*x])/(6*b^3)
Time = 0.32 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.05, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6662, 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.
\(\displaystyle \int x^2 \coth ^{-1}(a+b x) \, dx\) |
\(\Big \downarrow \) 6662 |
\(\displaystyle \frac {\int x^2 \coth ^{-1}(a+b x)d(a+b x)}{b}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int b^2 x^2 \coth ^{-1}(a+b x)d(a+b x)}{b^3}\) |
\(\Big \downarrow \) 6479 |
\(\displaystyle \frac {\frac {1}{3} \int -\frac {b^3 x^3}{1-(a+b x)^2}d(a+b x)+\frac {1}{3} b^3 x^3 \coth ^{-1}(a+b x)}{b^3}\) |
\(\Big \downarrow \) 477 |
\(\displaystyle \frac {\frac {1}{3} \int \left (-\frac {(1-a)^3}{2 (-a-b x+1)}-2 a+b x+\frac {(a+1)^3}{2 (a+b x+1)}\right )d(a+b x)+\frac {1}{3} b^3 x^3 \coth ^{-1}(a+b x)}{b^3}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {1}{3} b^3 x^3 \coth ^{-1}(a+b x)+\frac {1}{3} \left (\frac {1}{2} (a+b x)^2-3 a (a+b x)+\frac {1}{2} (1-a)^3 \log (-a-b x+1)+\frac {1}{2} (a+1)^3 \log (a+b x+1)\right )}{b^3}\) |
((b^3*x^3*ArcCoth[a + b*x])/3 + (-3*a*(a + b*x) + (a + b*x)^2/2 + ((1 - a) ^3*Log[1 - a - b*x])/2 + ((1 + a)^3*Log[1 + a + b*x])/2)/3)/b^3
3.1.63.3.1 Defintions of rubi rules used
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 = 0.10 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.09
method | result | size |
parts | \(\frac {x^{3} \operatorname {arccoth}\left (b x +a \right )}{3}+\frac {b \left (-\frac {-\frac {1}{2} b \,x^{2}+2 a x}{b^{3}}+\frac {\left (-a^{3}+3 a^{2}-3 a +1\right ) \ln \left (b x +a -1\right )}{2 b^{4}}+\frac {\left (a^{3}+3 a^{2}+3 a +1\right ) \ln \left (b x +a +1\right )}{2 b^{4}}\right )}{3}\) | \(85\) |
parallelrisch | \(\frac {2 \,\operatorname {arccoth}\left (b x +a \right ) x^{3} b^{3}+1+b^{2} x^{2}+2 \,\operatorname {arccoth}\left (b x +a \right ) a^{3}+6 \ln \left (b x +a -1\right ) a^{2}-4 a b x +6 \,\operatorname {arccoth}\left (b x +a \right ) a^{2}+6 \,\operatorname {arccoth}\left (b x +a \right ) a +7 a^{2}+2 \ln \left (b x +a -1\right )+2 \,\operatorname {arccoth}\left (b x +a \right )}{6 b^{3}}\) | \(99\) |
derivativedivides | \(\frac {-\frac {\operatorname {arccoth}\left (b x +a \right ) a^{3}}{3}+\operatorname {arccoth}\left (b x +a \right ) a^{2} \left (b x +a \right )-\operatorname {arccoth}\left (b x +a \right ) a \left (b x +a \right )^{2}+\frac {\operatorname {arccoth}\left (b x +a \right ) \left (b x +a \right )^{3}}{3}-\left (b x +a \right ) a +\frac {\left (b x +a \right )^{2}}{6}-\frac {\left (a^{3}-3 a^{2}+3 a -1\right ) \ln \left (b x +a -1\right )}{6}+\frac {\left (a^{3}+3 a^{2}+3 a +1\right ) \ln \left (b x +a +1\right )}{6}}{b^{3}}\) | \(124\) |
default | \(\frac {-\frac {\operatorname {arccoth}\left (b x +a \right ) a^{3}}{3}+\operatorname {arccoth}\left (b x +a \right ) a^{2} \left (b x +a \right )-\operatorname {arccoth}\left (b x +a \right ) a \left (b x +a \right )^{2}+\frac {\operatorname {arccoth}\left (b x +a \right ) \left (b x +a \right )^{3}}{3}-\left (b x +a \right ) a +\frac {\left (b x +a \right )^{2}}{6}-\frac {\left (a^{3}-3 a^{2}+3 a -1\right ) \ln \left (b x +a -1\right )}{6}+\frac {\left (a^{3}+3 a^{2}+3 a +1\right ) \ln \left (b x +a +1\right )}{6}}{b^{3}}\) | \(124\) |
risch | \(\frac {x^{3} \ln \left (b x +a +1\right )}{6}-\frac {\ln \left (b x +a -1\right ) x^{3}}{6}+\frac {\ln \left (-b x -a -1\right ) a^{3}}{6 b^{3}}-\frac {\ln \left (b x +a -1\right ) a^{3}}{6 b^{3}}+\frac {x^{2}}{6 b}+\frac {\ln \left (-b x -a -1\right ) a^{2}}{2 b^{3}}+\frac {\ln \left (b x +a -1\right ) a^{2}}{2 b^{3}}-\frac {2 x a}{3 b^{2}}+\frac {\ln \left (-b x -a -1\right ) a}{2 b^{3}}-\frac {\ln \left (b x +a -1\right ) a}{2 b^{3}}+\frac {\ln \left (-b x -a -1\right )}{6 b^{3}}+\frac {\ln \left (b x +a -1\right )}{6 b^{3}}\) | \(163\) |
1/3*x^3*arccoth(b*x+a)+1/3*b*(-1/b^3*(-1/2*b*x^2+2*a*x)+1/2*(-a^3+3*a^2-3* a+1)/b^4*ln(b*x+a-1)+1/2*(a^3+3*a^2+3*a+1)/b^4*ln(b*x+a+1))
Time = 0.27 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.08 \[ \int x^2 \coth ^{-1}(a+b x) \, dx=\frac {b^{3} x^{3} \log \left (\frac {b x + a + 1}{b x + a - 1}\right ) + b^{2} x^{2} - 4 \, a b x + {\left (a^{3} + 3 \, a^{2} + 3 \, a + 1\right )} \log \left (b x + a + 1\right ) - {\left (a^{3} - 3 \, a^{2} + 3 \, a - 1\right )} \log \left (b x + a - 1\right )}{6 \, b^{3}} \]
1/6*(b^3*x^3*log((b*x + a + 1)/(b*x + a - 1)) + b^2*x^2 - 4*a*b*x + (a^3 + 3*a^2 + 3*a + 1)*log(b*x + a + 1) - (a^3 - 3*a^2 + 3*a - 1)*log(b*x + a - 1))/b^3
Time = 0.38 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.50 \[ \int x^2 \coth ^{-1}(a+b x) \, dx=\begin {cases} \frac {a^{3} \operatorname {acoth}{\left (a + b x \right )}}{3 b^{3}} + \frac {a^{2} \log {\left (a + b x + 1 \right )}}{b^{3}} - \frac {a^{2} \operatorname {acoth}{\left (a + b x \right )}}{b^{3}} - \frac {2 a x}{3 b^{2}} + \frac {a \operatorname {acoth}{\left (a + b x \right )}}{b^{3}} + \frac {x^{3} \operatorname {acoth}{\left (a + b x \right )}}{3} + \frac {x^{2}}{6 b} + \frac {\log {\left (a + b x + 1 \right )}}{3 b^{3}} - \frac {\operatorname {acoth}{\left (a + b x \right )}}{3 b^{3}} & \text {for}\: b \neq 0 \\\frac {x^{3} \operatorname {acoth}{\left (a \right )}}{3} & \text {otherwise} \end {cases} \]
Piecewise((a**3*acoth(a + b*x)/(3*b**3) + a**2*log(a + b*x + 1)/b**3 - a** 2*acoth(a + b*x)/b**3 - 2*a*x/(3*b**2) + a*acoth(a + b*x)/b**3 + x**3*acot h(a + b*x)/3 + x**2/(6*b) + log(a + b*x + 1)/(3*b**3) - acoth(a + b*x)/(3* b**3), Ne(b, 0)), (x**3*acoth(a)/3, True))
Time = 0.21 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.01 \[ \int x^2 \coth ^{-1}(a+b x) \, dx=\frac {1}{3} \, x^{3} \operatorname {arcoth}\left (b x + a\right ) + \frac {1}{6} \, b {\left (\frac {b x^{2} - 4 \, a x}{b^{3}} + \frac {{\left (a^{3} + 3 \, a^{2} + 3 \, a + 1\right )} \log \left (b x + a + 1\right )}{b^{4}} - \frac {{\left (a^{3} - 3 \, a^{2} + 3 \, a - 1\right )} \log \left (b x + a - 1\right )}{b^{4}}\right )} \]
1/3*x^3*arccoth(b*x + a) + 1/6*b*((b*x^2 - 4*a*x)/b^3 + (a^3 + 3*a^2 + 3*a + 1)*log(b*x + a + 1)/b^4 - (a^3 - 3*a^2 + 3*a - 1)*log(b*x + a - 1)/b^4)
Leaf count of result is larger than twice the leaf count of optimal. 360 vs. \(2 (68) = 136\).
Time = 0.29 (sec) , antiderivative size = 360, normalized size of antiderivative = 4.62 \[ \int x^2 \coth ^{-1}(a+b x) \, dx=\frac {1}{6} \, {\left ({\left (a + 1\right )} b - {\left (a - 1\right )} b\right )} {\left (\frac {{\left (3 \, a^{2} + 1\right )} \log \left (\frac {{\left | b x + a + 1 \right |}}{{\left | b x + a - 1 \right |}}\right )}{b^{4}} - \frac {{\left (3 \, a^{2} + 1\right )} \log \left ({\left | \frac {b x + a + 1}{b x + a - 1} - 1 \right |}\right )}{b^{4}} - \frac {2 \, {\left (\frac {{\left (b x + a + 1\right )} {\left (3 \, a - 1\right )}}{b x + a - 1} - 3 \, a\right )}}{b^{4} {\left (\frac {b x + a + 1}{b x + a - 1} - 1\right )}^{2}} + \frac {{\left (\frac {3 \, {\left (b x + a + 1\right )}^{2} a^{2}}{{\left (b x + a - 1\right )}^{2}} - \frac {6 \, {\left (b x + a + 1\right )} a^{2}}{b x + a - 1} + 3 \, a^{2} - \frac {6 \, {\left (b x + a + 1\right )}^{2} a}{{\left (b x + a - 1\right )}^{2}} + \frac {6 \, {\left (b x + a + 1\right )} a}{b x + a - 1} + \frac {3 \, {\left (b x + a + 1\right )}^{2}}{{\left (b x + a - 1\right )}^{2}} + 1\right )} \log \left (-\frac {\frac {1}{a - \frac {{\left (\frac {{\left (b x + a + 1\right )} {\left (a - 1\right )}}{b x + a - 1} - a - 1\right )} b}{\frac {{\left (b x + a + 1\right )} b}{b x + a - 1} - b}} + 1}{\frac {1}{a - \frac {{\left (\frac {{\left (b x + a + 1\right )} {\left (a - 1\right )}}{b x + a - 1} - a - 1\right )} b}{\frac {{\left (b x + a + 1\right )} b}{b x + a - 1} - b}} - 1}\right )}{b^{4} {\left (\frac {b x + a + 1}{b x + a - 1} - 1\right )}^{3}}\right )} \]
1/6*((a + 1)*b - (a - 1)*b)*((3*a^2 + 1)*log(abs(b*x + a + 1)/abs(b*x + a - 1))/b^4 - (3*a^2 + 1)*log(abs((b*x + a + 1)/(b*x + a - 1) - 1))/b^4 - 2* ((b*x + a + 1)*(3*a - 1)/(b*x + a - 1) - 3*a)/(b^4*((b*x + a + 1)/(b*x + a - 1) - 1)^2) + (3*(b*x + a + 1)^2*a^2/(b*x + a - 1)^2 - 6*(b*x + a + 1)*a ^2/(b*x + a - 1) + 3*a^2 - 6*(b*x + a + 1)^2*a/(b*x + a - 1)^2 + 6*(b*x + a + 1)*a/(b*x + a - 1) + 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^4*((b*x + a + 1)/(b*x + a - 1) - 1)^3))
Time = 4.65 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.26 \[ \int x^2 \coth ^{-1}(a+b x) \, dx=\frac {x^3\,\ln \left (\frac {1}{a+b\,x}+1\right )}{6}-\frac {x^3\,\ln \left (1-\frac {1}{a+b\,x}\right )}{6}+\frac {x^2}{6\,b}-\frac {\ln \left (a+b\,x-1\right )\,\left (a^3-3\,a^2+3\,a-1\right )}{6\,b^3}+\frac {\ln \left (a+b\,x+1\right )\,\left (a^3+3\,a^2+3\,a+1\right )}{6\,b^3}-\frac {2\,a\,x}{3\,b^2} \]