Integrand size = 10, antiderivative size = 61 \[ \int \frac {\coth ^{-1}(a x)^2}{x^3} \, dx=-\frac {a \coth ^{-1}(a x)}{x}+\frac {1}{2} a^2 \coth ^{-1}(a x)^2-\frac {\coth ^{-1}(a x)^2}{2 x^2}+a^2 \log (x)-\frac {1}{2} a^2 \log \left (1-a^2 x^2\right ) \] Output:
-a*arccoth(a*x)/x+1/2*a^2*arccoth(a*x)^2-1/2*arccoth(a*x)^2/x^2+a^2*ln(x)- 1/2*a^2*ln(-a^2*x^2+1)
Time = 0.01 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.93 \[ \int \frac {\coth ^{-1}(a x)^2}{x^3} \, dx=-\frac {a \coth ^{-1}(a x)}{x}+\frac {\left (-1+a^2 x^2\right ) \coth ^{-1}(a x)^2}{2 x^2}+a^2 \log (x)-\frac {1}{2} a^2 \log \left (1-a^2 x^2\right ) \] Input:
Integrate[ArcCoth[a*x]^2/x^3,x]
Output:
-((a*ArcCoth[a*x])/x) + ((-1 + a^2*x^2)*ArcCoth[a*x]^2)/(2*x^2) + a^2*Log[ x] - (a^2*Log[1 - a^2*x^2])/2
Time = 0.47 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.98, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.800, Rules used = {6453, 6545, 6453, 243, 47, 14, 16, 6511}
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.
\(\displaystyle \int \frac {\coth ^{-1}(a x)^2}{x^3} \, dx\) |
\(\Big \downarrow \) 6453 |
\(\displaystyle a \int \frac {\coth ^{-1}(a x)}{x^2 \left (1-a^2 x^2\right )}dx-\frac {\coth ^{-1}(a x)^2}{2 x^2}\) |
\(\Big \downarrow \) 6545 |
\(\displaystyle a \left (a^2 \int \frac {\coth ^{-1}(a x)}{1-a^2 x^2}dx+\int \frac {\coth ^{-1}(a x)}{x^2}dx\right )-\frac {\coth ^{-1}(a x)^2}{2 x^2}\) |
\(\Big \downarrow \) 6453 |
\(\displaystyle a \left (a \int \frac {1}{x \left (1-a^2 x^2\right )}dx+a^2 \int \frac {\coth ^{-1}(a x)}{1-a^2 x^2}dx-\frac {\coth ^{-1}(a x)}{x}\right )-\frac {\coth ^{-1}(a x)^2}{2 x^2}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle a \left (\frac {1}{2} a \int \frac {1}{x^2 \left (1-a^2 x^2\right )}dx^2+a^2 \int \frac {\coth ^{-1}(a x)}{1-a^2 x^2}dx-\frac {\coth ^{-1}(a x)}{x}\right )-\frac {\coth ^{-1}(a x)^2}{2 x^2}\) |
\(\Big \downarrow \) 47 |
\(\displaystyle a \left (\frac {1}{2} a \left (a^2 \int \frac {1}{1-a^2 x^2}dx^2+\int \frac {1}{x^2}dx^2\right )+a^2 \int \frac {\coth ^{-1}(a x)}{1-a^2 x^2}dx-\frac {\coth ^{-1}(a x)}{x}\right )-\frac {\coth ^{-1}(a x)^2}{2 x^2}\) |
\(\Big \downarrow \) 14 |
\(\displaystyle a \left (\frac {1}{2} a \left (a^2 \int \frac {1}{1-a^2 x^2}dx^2+\log \left (x^2\right )\right )+a^2 \int \frac {\coth ^{-1}(a x)}{1-a^2 x^2}dx-\frac {\coth ^{-1}(a x)}{x}\right )-\frac {\coth ^{-1}(a x)^2}{2 x^2}\) |
\(\Big \downarrow \) 16 |
\(\displaystyle a \left (a^2 \int \frac {\coth ^{-1}(a x)}{1-a^2 x^2}dx+\frac {1}{2} a \left (\log \left (x^2\right )-\log \left (1-a^2 x^2\right )\right )-\frac {\coth ^{-1}(a x)}{x}\right )-\frac {\coth ^{-1}(a x)^2}{2 x^2}\) |
\(\Big \downarrow \) 6511 |
\(\displaystyle a \left (\frac {1}{2} a \left (\log \left (x^2\right )-\log \left (1-a^2 x^2\right )\right )+\frac {1}{2} a \coth ^{-1}(a x)^2-\frac {\coth ^{-1}(a x)}{x}\right )-\frac {\coth ^{-1}(a x)^2}{2 x^2}\) |
Input:
Int[ArcCoth[a*x]^2/x^3,x]
Output:
-1/2*ArcCoth[a*x]^2/x^2 + a*(-(ArcCoth[a*x]/x) + (a*ArcCoth[a*x]^2)/2 + (a *(Log[x^2] - Log[1 - a^2*x^2]))/2)
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Simp[b/(b*c - a*d) Int[1/(a + b*x), x], x] - Simp[d/(b*c - a*d) Int[1/(c + d*x), x ], x] /; FreeQ[{a, b, c, d}, x]
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_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2), x_Symb ol] :> Simp[(a + b*ArcCoth[c*x])^(p + 1)/(b*c*d*(p + 1)), x] /; FreeQ[{a, b , c, d, e, p}, x] && EqQ[c^2*d + e, 0] && NeQ[p, -1]
Int[(((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + ( e_.)*(x_)^2), x_Symbol] :> Simp[1/d Int[(f*x)^m*(a + b*ArcCoth[c*x])^p, x ], x] - Simp[e/(d*f^2) Int[(f*x)^(m + 2)*((a + b*ArcCoth[c*x])^p/(d + e*x ^2)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && GtQ[p, 0] && LtQ[m, -1]
Time = 0.15 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.18
method | result | size |
parallelrisch | \(\frac {x^{2} a^{2} \operatorname {arccoth}\left (x a \right )^{2}+2 a^{2} \ln \left (x \right ) x^{2}-2 x^{2} \ln \left (x a -1\right ) a^{2}-2 \,\operatorname {arccoth}\left (x a \right ) a^{2} x^{2}-2 x a \,\operatorname {arccoth}\left (x a \right )-\operatorname {arccoth}\left (x a \right )^{2}}{2 x^{2}}\) | \(72\) |
risch | \(\frac {\left (a^{2} x^{2}-1\right ) \ln \left (x a +1\right )^{2}}{8 x^{2}}-\frac {\left (x^{2} \ln \left (x a -1\right ) a^{2}+2 x a -\ln \left (x a -1\right )\right ) \ln \left (x a +1\right )}{4 x^{2}}+\frac {a^{2} x^{2} \ln \left (x a -1\right )^{2}+8 a^{2} \ln \left (x \right ) x^{2}-4 a^{2} \ln \left (a^{2} x^{2}-1\right ) x^{2}+4 x \ln \left (x a -1\right ) a -\ln \left (x a -1\right )^{2}}{8 x^{2}}\) | \(130\) |
derivativedivides | \(a^{2} \left (-\frac {\operatorname {arccoth}\left (x a \right )^{2}}{2 x^{2} a^{2}}-\frac {\operatorname {arccoth}\left (x a \right ) \ln \left (x a -1\right )}{2}+\frac {\operatorname {arccoth}\left (x a \right ) \ln \left (x a +1\right )}{2}-\frac {\operatorname {arccoth}\left (x a \right )}{x a}-\frac {\ln \left (x a -1\right )^{2}}{8}+\frac {\ln \left (x a -1\right ) \ln \left (\frac {x a}{2}+\frac {1}{2}\right )}{4}+\ln \left (x a \right )-\frac {\ln \left (x a -1\right )}{2}-\frac {\ln \left (x a +1\right )}{2}+\frac {\left (\ln \left (x a +1\right )-\ln \left (\frac {x a}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {x a}{2}+\frac {1}{2}\right )}{4}-\frac {\ln \left (x a +1\right )^{2}}{8}\right )\) | \(136\) |
default | \(a^{2} \left (-\frac {\operatorname {arccoth}\left (x a \right )^{2}}{2 x^{2} a^{2}}-\frac {\operatorname {arccoth}\left (x a \right ) \ln \left (x a -1\right )}{2}+\frac {\operatorname {arccoth}\left (x a \right ) \ln \left (x a +1\right )}{2}-\frac {\operatorname {arccoth}\left (x a \right )}{x a}-\frac {\ln \left (x a -1\right )^{2}}{8}+\frac {\ln \left (x a -1\right ) \ln \left (\frac {x a}{2}+\frac {1}{2}\right )}{4}+\ln \left (x a \right )-\frac {\ln \left (x a -1\right )}{2}-\frac {\ln \left (x a +1\right )}{2}+\frac {\left (\ln \left (x a +1\right )-\ln \left (\frac {x a}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {x a}{2}+\frac {1}{2}\right )}{4}-\frac {\ln \left (x a +1\right )^{2}}{8}\right )\) | \(136\) |
parts | \(-\frac {\operatorname {arccoth}\left (x a \right )^{2}}{2 x^{2}}-a^{2} \left (\frac {\operatorname {arccoth}\left (x a \right ) \ln \left (x a -1\right )}{2}-\frac {\operatorname {arccoth}\left (x a \right ) \ln \left (x a +1\right )}{2}+\frac {\operatorname {arccoth}\left (x a \right )}{x a}+\frac {\ln \left (x a -1\right )^{2}}{8}-\frac {\ln \left (x a -1\right ) \ln \left (\frac {x a}{2}+\frac {1}{2}\right )}{4}-\ln \left (x a \right )+\frac {\ln \left (x a -1\right )}{2}+\frac {\ln \left (x a +1\right )}{2}-\frac {\left (\ln \left (x a +1\right )-\ln \left (\frac {x a}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {x a}{2}+\frac {1}{2}\right )}{4}+\frac {\ln \left (x a +1\right )^{2}}{8}\right )\) | \(136\) |
Input:
int(arccoth(x*a)^2/x^3,x,method=_RETURNVERBOSE)
Output:
1/2*(x^2*a^2*arccoth(x*a)^2+2*a^2*ln(x)*x^2-2*x^2*ln(a*x-1)*a^2-2*arccoth( x*a)*a^2*x^2-2*x*a*arccoth(x*a)-arccoth(x*a)^2)/x^2
Time = 0.09 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.30 \[ \int \frac {\coth ^{-1}(a x)^2}{x^3} \, dx=-\frac {4 \, a^{2} x^{2} \log \left (a^{2} x^{2} - 1\right ) - 8 \, a^{2} x^{2} \log \left (x\right ) + 4 \, a x \log \left (\frac {a x + 1}{a x - 1}\right ) - {\left (a^{2} x^{2} - 1\right )} \log \left (\frac {a x + 1}{a x - 1}\right )^{2}}{8 \, x^{2}} \] Input:
integrate(arccoth(a*x)^2/x^3,x, algorithm="fricas")
Output:
-1/8*(4*a^2*x^2*log(a^2*x^2 - 1) - 8*a^2*x^2*log(x) + 4*a*x*log((a*x + 1)/ (a*x - 1)) - (a^2*x^2 - 1)*log((a*x + 1)/(a*x - 1))^2)/x^2
Time = 0.28 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.92 \[ \int \frac {\coth ^{-1}(a x)^2}{x^3} \, dx=a^{2} \log {\left (x \right )} - a^{2} \log {\left (a x + 1 \right )} + \frac {a^{2} \operatorname {acoth}^{2}{\left (a x \right )}}{2} + a^{2} \operatorname {acoth}{\left (a x \right )} - \frac {a \operatorname {acoth}{\left (a x \right )}}{x} - \frac {\operatorname {acoth}^{2}{\left (a x \right )}}{2 x^{2}} \] Input:
integrate(acoth(a*x)**2/x**3,x)
Output:
a**2*log(x) - a**2*log(a*x + 1) + a**2*acoth(a*x)**2/2 + a**2*acoth(a*x) - a*acoth(a*x)/x - acoth(a*x)**2/(2*x**2)
Time = 0.03 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.57 \[ \int \frac {\coth ^{-1}(a x)^2}{x^3} \, dx=\frac {1}{8} \, {\left (2 \, {\left (\log \left (a x - 1\right ) - 2\right )} \log \left (a x + 1\right ) - \log \left (a x + 1\right )^{2} - \log \left (a x - 1\right )^{2} - 4 \, \log \left (a x - 1\right ) + 8 \, \log \left (x\right )\right )} a^{2} + \frac {1}{2} \, {\left (a \log \left (a x + 1\right ) - a \log \left (a x - 1\right ) - \frac {2}{x}\right )} a \operatorname {arcoth}\left (a x\right ) - \frac {\operatorname {arcoth}\left (a x\right )^{2}}{2 \, x^{2}} \] Input:
integrate(arccoth(a*x)^2/x^3,x, algorithm="maxima")
Output:
1/8*(2*(log(a*x - 1) - 2)*log(a*x + 1) - log(a*x + 1)^2 - log(a*x - 1)^2 - 4*log(a*x - 1) + 8*log(x))*a^2 + 1/2*(a*log(a*x + 1) - a*log(a*x - 1) - 2 /x)*a*arccoth(a*x) - 1/2*arccoth(a*x)^2/x^2
Leaf count of result is larger than twice the leaf count of optimal. 137 vs. \(2 (55) = 110\).
Time = 0.12 (sec) , antiderivative size = 137, normalized size of antiderivative = 2.25 \[ \int \frac {\coth ^{-1}(a x)^2}{x^3} \, dx=\frac {1}{2} \, {\left (2 \, a \log \left (\frac {a x + 1}{a x - 1} + 1\right ) - 2 \, a \log \left (\frac {a x + 1}{a x - 1}\right ) + \frac {{\left (a x + 1\right )} a \log \left (\frac {a x + 1}{a x - 1}\right )^{2}}{{\left (a x - 1\right )} {\left (\frac {{\left (a x + 1\right )}^{2}}{{\left (a x - 1\right )}^{2}} + \frac {2 \, {\left (a x + 1\right )}}{a x - 1} + 1\right )}} + \frac {2 \, a \log \left (\frac {a x + 1}{a x - 1}\right )}{\frac {a x + 1}{a x - 1} + 1}\right )} a \] Input:
integrate(arccoth(a*x)^2/x^3,x, algorithm="giac")
Output:
1/2*(2*a*log((a*x + 1)/(a*x - 1) + 1) - 2*a*log((a*x + 1)/(a*x - 1)) + (a* x + 1)*a*log((a*x + 1)/(a*x - 1))^2/((a*x - 1)*((a*x + 1)^2/(a*x - 1)^2 + 2*(a*x + 1)/(a*x - 1) + 1)) + 2*a*log((a*x + 1)/(a*x - 1))/((a*x + 1)/(a*x - 1) + 1))*a
Time = 3.78 (sec) , antiderivative size = 145, normalized size of antiderivative = 2.38 \[ \int \frac {\coth ^{-1}(a x)^2}{x^3} \, dx=a^2\,\ln \left (x\right )+{\ln \left (\frac {1}{a\,x}+1\right )}^2\,\left (\frac {a^2}{8}-\frac {1}{8\,x^2}\right )+{\ln \left (1-\frac {1}{a\,x}\right )}^2\,\left (\frac {a^2}{8}-\frac {1}{8\,x^2}\right )-\frac {a^2\,\ln \left (a^2\,x^2-1\right )}{2}+\ln \left (1-\frac {1}{a\,x}\right )\,\left (\frac {4\,a\,x-2}{16\,x^2}+\frac {4\,a\,x+2}{16\,x^2}-\ln \left (\frac {1}{a\,x}+1\right )\,\left (\frac {a^2}{4}-\frac {1}{4\,x^2}\right )\right )-\frac {a\,\ln \left (\frac {1}{a\,x}+1\right )}{2\,x} \] Input:
int(acoth(a*x)^2/x^3,x)
Output:
a^2*log(x) + log(1/(a*x) + 1)^2*(a^2/8 - 1/(8*x^2)) + log(1 - 1/(a*x))^2*( a^2/8 - 1/(8*x^2)) - (a^2*log(a^2*x^2 - 1))/2 + log(1 - 1/(a*x))*((4*a*x - 2)/(16*x^2) + (4*a*x + 2)/(16*x^2) - log(1/(a*x) + 1)*(a^2/4 - 1/(4*x^2)) ) - (a*log(1/(a*x) + 1))/(2*x)
Time = 0.17 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.23 \[ \int \frac {\coth ^{-1}(a x)^2}{x^3} \, dx=\frac {\mathit {acoth} \left (a x \right )^{2} a^{2} x^{2}-\mathit {acoth} \left (a x \right )^{2}+2 \mathit {acoth} \left (a x \right ) a^{2} x^{2}+2 \mathit {acoth} \left (a x \right ) a x -2 \,\mathrm {log}\left (a^{2} x -a \right ) a^{2} x^{2}+2 \,\mathrm {log}\left (x \right ) a^{2} x^{2}}{2 x^{2}} \] Input:
int(acoth(a*x)^2/x^3,x)
Output:
(acoth(a*x)**2*a**2*x**2 - acoth(a*x)**2 + 2*acoth(a*x)*a**2*x**2 + 2*acot h(a*x)*a*x - 2*log(a**2*x - a)*a**2*x**2 + 2*log(x)*a**2*x**2)/(2*x**2)