Integrand size = 13, antiderivative size = 60 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))^3}{x^3} \, dx=3 b^3 x-\frac {3 b \coth ^{-1}(\tanh (a+b x))^2}{2 x}-\frac {\coth ^{-1}(\tanh (a+b x))^3}{2 x^2}-3 b^2 \left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \log (x) \]
3*b^3*x-3/2*b*arccoth(tanh(b*x+a))^2/x-1/2*arccoth(tanh(b*x+a))^3/x^2-3*b^ 2*(b*x-arccoth(tanh(b*x+a)))*ln(x)
Time = 0.03 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.10 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))^3}{x^3} \, dx=b^3 x-\frac {3 b \left (-b x+\coth ^{-1}(\tanh (a+b x))\right )^2}{x}-\frac {\left (-b x+\coth ^{-1}(\tanh (a+b x))\right )^3}{2 x^2}+3 b^2 \left (-b x+\coth ^{-1}(\tanh (a+b x))\right ) \log (x) \]
b^3*x - (3*b*(-(b*x) + ArcCoth[Tanh[a + b*x]])^2)/x - (-(b*x) + ArcCoth[Ta nh[a + b*x]])^3/(2*x^2) + 3*b^2*(-(b*x) + ArcCoth[Tanh[a + b*x]])*Log[x]
Time = 0.25 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.02, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {2599, 2599, 2589, 14}
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.
\(\displaystyle \int \frac {\coth ^{-1}(\tanh (a+b x))^3}{x^3} \, dx\) |
\(\Big \downarrow \) 2599 |
\(\displaystyle \frac {3}{2} b \int \frac {\coth ^{-1}(\tanh (a+b x))^2}{x^2}dx-\frac {\coth ^{-1}(\tanh (a+b x))^3}{2 x^2}\) |
\(\Big \downarrow \) 2599 |
\(\displaystyle \frac {3}{2} b \left (2 b \int \frac {\coth ^{-1}(\tanh (a+b x))}{x}dx-\frac {\coth ^{-1}(\tanh (a+b x))^2}{x}\right )-\frac {\coth ^{-1}(\tanh (a+b x))^3}{2 x^2}\) |
\(\Big \downarrow \) 2589 |
\(\displaystyle \frac {3}{2} b \left (2 b \left (b x-\left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \int \frac {1}{x}dx\right )-\frac {\coth ^{-1}(\tanh (a+b x))^2}{x}\right )-\frac {\coth ^{-1}(\tanh (a+b x))^3}{2 x^2}\) |
\(\Big \downarrow \) 14 |
\(\displaystyle \frac {3}{2} b \left (2 b \left (b x-\log (x) \left (b x-\coth ^{-1}(\tanh (a+b x))\right )\right )-\frac {\coth ^{-1}(\tanh (a+b x))^2}{x}\right )-\frac {\coth ^{-1}(\tanh (a+b x))^3}{2 x^2}\) |
-1/2*ArcCoth[Tanh[a + b*x]]^3/x^2 + (3*b*(-(ArcCoth[Tanh[a + b*x]]^2/x) + 2*b*(b*x - (b*x - ArcCoth[Tanh[a + b*x]])*Log[x])))/2
3.2.54.3.1 Defintions of rubi rules used
Int[(v_)/(u_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[b*(x/a), x] - Simp[(b*u - a*v)/a Int[1/u, x], x] /; NeQ[b*u - a*v, 0]] /; PiecewiseLinearQ[u, v, x]
Int[(u_)^(m_)*(v_)^(n_.), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Sim plify[D[v, x]]}, Simp[u^(m + 1)*(v^n/(a*(m + 1))), x] - Simp[b*(n/(a*(m + 1 ))) Int[u^(m + 1)*v^(n - 1), x], x] /; NeQ[b*u - a*v, 0]] /; FreeQ[{m, n} , x] && PiecewiseLinearQ[u, v, x] && NeQ[m, -1] && ((LtQ[m, -1] && GtQ[n, 0 ] && !(ILtQ[m + n, -2] && (FractionQ[m] || GeQ[2*n + m + 1, 0]))) || (IGtQ [n, 0] && IGtQ[m, 0] && LeQ[n, m]) || (IGtQ[n, 0] && !IntegerQ[m]) || (ILt Q[m, 0] && !IntegerQ[n]))
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.28 (sec) , antiderivative size = 3227, normalized size of antiderivative = 53.78
-1/2/x^2*ln(exp(b*x+a))^3-3/2/x*ln(exp(b*x+a))^2*b-3*ln(x)*x*b^3+3*ln(exp( b*x+a))*ln(x)*b^2+3*b^3*x+(-3/4*Pi^2*csgn(I/(exp(2*b*x+2*a)+1))*csgn(I*exp (b*x+a))*csgn(I*exp(2*b*x+2*a))^2*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1) )^2+3/4*Pi^2*csgn(I/(exp(2*b*x+2*a)+1))^2*csgn(I*exp(2*b*x+2*a))^3-3/2*Pi^ 2*csgn(I*exp(b*x+a))*csgn(I*exp(2*b*x+2*a))^2*csgn(I/(exp(2*b*x+2*a)+1))^2 -3/4*Pi^2-3/4*Pi^2*csgn(I/(exp(2*b*x+2*a)+1))^4*csgn(I*exp(2*b*x+2*a))*csg n(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))-3/4*Pi^2*csgn(I*exp(2*b*x+2*a))*csg n(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^2*csgn(I/(exp(2*b*x+2*a)+1))^2+3/4* Pi^2*csgn(I*exp(b*x+a))^2*csgn(I*exp(2*b*x+2*a))*csgn(I/(exp(2*b*x+2*a)+1) )^2+3/8*Pi^2*csgn(I*exp(2*b*x+2*a))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+ 1))^5-3/4*csgn(I*exp(2*b*x+2*a))*csgn(I*exp(b*x+a))^2*Pi^2+3/2*csgn(I*exp( 2*b*x+2*a))^2*csgn(I*exp(b*x+a))*Pi^2-3/4*Pi^2*csgn(I/(exp(2*b*x+2*a)+1))* csgn(I*exp(2*b*x+2*a))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^4-3/8*Pi^ 2*csgn(I*exp(b*x+a))^2*csgn(I*exp(2*b*x+2*a))*csgn(I*exp(2*b*x+2*a)/(exp(2 *b*x+2*a)+1))^3+3/4*Pi^2*csgn(I*exp(b*x+a))*csgn(I*exp(2*b*x+2*a))^2*csgn( I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^3-3/16*Pi^2*csgn(I/(exp(2*b*x+2*a)+1) )^2*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^4+3/8*Pi^2*csgn(I/(exp(2*b*x +2*a)+1))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^5-3/16*Pi^2*csgn(I*exp (b*x+a))^4*csgn(I*exp(2*b*x+2*a))^2+3/4*Pi^2*csgn(I*exp(b*x+a))^3*csgn(I*e xp(2*b*x+2*a))^3-3/4*Pi^2*csgn(I/(exp(2*b*x+2*a)+1))^3*csgn(I*exp(2*b*x...
Result contains complex when optimal does not.
Time = 0.25 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.28 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))^3}{x^3} \, dx=\frac {16 \, b^{3} x^{3} - 48 \, a^{2} b x + i \, \pi ^{3} + 6 \, \pi ^{2} {\left (2 \, b x + a\right )} - 8 \, a^{3} - 12 i \, \pi {\left (4 \, a b x + a^{2}\right )} - 24 \, {\left (-i \, \pi b^{2} x^{2} - 2 \, a b^{2} x^{2}\right )} \log \left (x\right )}{16 \, x^{2}} \]
1/16*(16*b^3*x^3 - 48*a^2*b*x + I*pi^3 + 6*pi^2*(2*b*x + a) - 8*a^3 - 12*I *pi*(4*a*b*x + a^2) - 24*(-I*pi*b^2*x^2 - 2*a*b^2*x^2)*log(x))/x^2
\[ \int \frac {\coth ^{-1}(\tanh (a+b x))^3}{x^3} \, dx=\int \frac {\operatorname {acoth}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}{x^{3}}\, dx \]
Time = 0.31 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.20 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))^3}{x^3} \, dx=3 \, {\left (b \operatorname {arcoth}\left (\tanh \left (b x + a\right )\right ) \log \left (x\right ) - {\left (b {\left (x + \frac {a}{b}\right )} \log \left (x\right ) - b {\left (x + \frac {a \log \left (x\right )}{b}\right )}\right )} b\right )} b - \frac {3 \, b \operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )^{2}}{2 \, x} - \frac {\operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )^{3}}{2 \, x^{2}} \]
3*(b*arccoth(tanh(b*x + a))*log(x) - (b*(x + a/b)*log(x) - b*(x + a*log(x) /b))*b)*b - 3/2*b*arccoth(tanh(b*x + a))^2/x - 1/2*arccoth(tanh(b*x + a))^ 3/x^2
Result contains complex when optimal does not.
Time = 0.30 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.18 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))^3}{x^3} \, dx=b^{3} x + \frac {3}{2} \, {\left (i \, \pi b^{2} + 2 \, a b^{2}\right )} \log \left (x\right ) + \frac {12 \, \pi ^{2} b x - 48 i \, \pi a b x - 48 \, a^{2} b x + i \, \pi ^{3} + 6 \, \pi ^{2} a - 12 i \, \pi a^{2} - 8 \, a^{3}}{16 \, x^{2}} \]
b^3*x + 3/2*(I*pi*b^2 + 2*a*b^2)*log(x) + 1/16*(12*pi^2*b*x - 48*I*pi*a*b* x - 48*a^2*b*x + I*pi^3 + 6*pi^2*a - 12*I*pi*a^2 - 8*a^3)/x^2
Time = 3.99 (sec) , antiderivative size = 383, normalized size of antiderivative = 6.38 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))^3}{x^3} \, dx=\frac {{\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )}^3}{16\,x^2}-\frac {{\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )}^3}{16\,x^2}+\frac {9\,b^2\,\ln \left (\frac {{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )}{4}-\frac {9\,b^2\,\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )}{4}-\frac {3\,b^3\,x}{2}-\frac {3\,b\,{\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )}^2}{8\,x}+\frac {3\,b^2\,\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )\,\ln \left (x\right )}{2}-3\,b^3\,x\,\ln \left (x\right )-\frac {3\,b\,{\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )}^2}{8\,x}-\frac {3\,b^2\,\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )\,\ln \left (x\right )}{2}-\frac {3\,\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )\,{\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )}^2}{16\,x^2}+\frac {3\,{\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )}^2\,\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )}{16\,x^2}+\frac {3\,b\,\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )\,\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )}{4\,x} \]
log(-2/(exp(2*a)*exp(2*b*x) - 1))^3/(16*x^2) - log((2*exp(2*a)*exp(2*b*x)) /(exp(2*a)*exp(2*b*x) - 1))^3/(16*x^2) + (9*b^2*log(exp(2*b*x)/(exp(2*a)*e xp(2*b*x) - 1)))/4 - (9*b^2*log(1/(exp(2*a)*exp(2*b*x) - 1)))/4 - (3*b^3*x )/2 - (3*b*log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) - 1))^2)/(8*x) + (3*b^2*log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) - 1))*log(x))/2 - 3*b^3*x*log(x) - (3*b*log(-2/(exp(2*a)*exp(2*b*x) - 1))^2)/(8*x) - (3*b ^2*log(-2/(exp(2*a)*exp(2*b*x) - 1))*log(x))/2 - (3*log((2*exp(2*a)*exp(2* b*x))/(exp(2*a)*exp(2*b*x) - 1))*log(-2/(exp(2*a)*exp(2*b*x) - 1))^2)/(16* x^2) + (3*log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) - 1))^2*log(-2/ (exp(2*a)*exp(2*b*x) - 1)))/(16*x^2) + (3*b*log((2*exp(2*a)*exp(2*b*x))/(e xp(2*a)*exp(2*b*x) - 1))*log(-2/(exp(2*a)*exp(2*b*x) - 1)))/(4*x)