Integrand size = 13, antiderivative size = 77 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))^3}{x} \, dx=b x \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2-\frac {1}{2} \left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \coth ^{-1}(\tanh (a+b x))^2+\frac {1}{3} \coth ^{-1}(\tanh (a+b x))^3-\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^3 \log (x) \]
b*x*(b*x-arccoth(tanh(b*x+a)))^2-1/2*(b*x-arccoth(tanh(b*x+a)))*arccoth(ta nh(b*x+a))^2+1/3*arccoth(tanh(b*x+a))^3-(b*x-arccoth(tanh(b*x+a)))^3*ln(x)
Time = 0.10 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.35 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))^3}{x} \, dx=\frac {1}{3} (a+b x)^3+(a+b x) \left (a^2-3 a \left (a+b x-\coth ^{-1}(\tanh (a+b x))\right )+3 \left (a+b x-\coth ^{-1}(\tanh (a+b x))\right )^2\right )-\frac {1}{2} (a+b x)^2 \left (2 a+3 b x-3 \coth ^{-1}(\tanh (a+b x))\right )+\left (-b x+\coth ^{-1}(\tanh (a+b x))\right )^3 \log (b x) \]
(a + b*x)^3/3 + (a + b*x)*(a^2 - 3*a*(a + b*x - ArcCoth[Tanh[a + b*x]]) + 3*(a + b*x - ArcCoth[Tanh[a + b*x]])^2) - ((a + b*x)^2*(2*a + 3*b*x - 3*Ar cCoth[Tanh[a + b*x]]))/2 + (-(b*x) + ArcCoth[Tanh[a + b*x]])^3*Log[b*x]
Time = 0.26 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.03, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {2590, 2590, 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.
\(\displaystyle \int \frac {\coth ^{-1}(\tanh (a+b x))^3}{x} \, dx\) |
\(\Big \downarrow \) 2590 |
\(\displaystyle \frac {1}{3} \coth ^{-1}(\tanh (a+b x))^3-\left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \int \frac {\coth ^{-1}(\tanh (a+b x))^2}{x}dx\) |
\(\Big \downarrow \) 2590 |
\(\displaystyle \frac {1}{3} \coth ^{-1}(\tanh (a+b x))^3-\left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \left (\frac {1}{2} \coth ^{-1}(\tanh (a+b x))^2-\left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \int \frac {\coth ^{-1}(\tanh (a+b x))}{x}dx\right )\) |
\(\Big \downarrow \) 2589 |
\(\displaystyle \frac {1}{3} \coth ^{-1}(\tanh (a+b x))^3-\left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \left (\frac {1}{2} \coth ^{-1}(\tanh (a+b x))^2-\left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \left (b x-\left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \int \frac {1}{x}dx\right )\right )\) |
\(\Big \downarrow \) 14 |
\(\displaystyle \frac {1}{3} \coth ^{-1}(\tanh (a+b x))^3-\left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \left (\frac {1}{2} \coth ^{-1}(\tanh (a+b x))^2-\left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \left (b x-\log (x) \left (b x-\coth ^{-1}(\tanh (a+b x))\right )\right )\right )\) |
ArcCoth[Tanh[a + b*x]]^3/3 - (b*x - ArcCoth[Tanh[a + b*x]])*(ArcCoth[Tanh[ a + b*x]]^2/2 - (b*x - ArcCoth[Tanh[a + b*x]])*(b*x - (b*x - ArcCoth[Tanh[ a + b*x]])*Log[x]))
3.2.52.3.1 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[(v_)^(n_)/(u_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[ D[v, x]]}, Simp[v^n/(a*n), x] - Simp[(b*u - a*v)/a Int[v^(n - 1)/u, x], x ] /; NeQ[b*u - a*v, 0]] /; PiecewiseLinearQ[u, v, x] && GtQ[n, 0] && NeQ[n, 1]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.29 (sec) , antiderivative size = 3294, normalized size of antiderivative = 42.78
ln(x)*ln(exp(b*x+a))^3-9/2*b^2*ln(exp(b*x+a))*x^2-3*b*ln(x)*ln(exp(b*x+a)) ^2*x+11/6*b^3*x^3+3*b^2*ln(x)*ln(exp(b*x+a))*x^2+3*b*ln(exp(b*x+a))^2*x-b^ 3*ln(x)*x^3+(-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*c sgn(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))*csgn(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))*csgn(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*ex p(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*c sgn(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*csg n(I*exp(2*b*x+2*a))^2+3/4*Pi^2*csgn(I*exp(b*x+a))^3*csgn(I*exp(2*b*x+2*...
Result contains complex when optimal does not.
Time = 0.25 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.97 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))^3}{x} \, dx=\frac {1}{3} \, b^{3} x^{3} + \frac {3}{2} \, a b^{2} x^{2} - \frac {3}{4} \, \pi ^{2} b x + 3 \, a^{2} b x + \frac {3}{4} i \, \pi {\left (b^{2} x^{2} + 4 \, a b x\right )} - \frac {1}{8} \, {\left (i \, \pi ^{3} + 6 \, \pi ^{2} a - 12 i \, \pi a^{2} - 8 \, a^{3}\right )} \log \left (x\right ) \]
1/3*b^3*x^3 + 3/2*a*b^2*x^2 - 3/4*pi^2*b*x + 3*a^2*b*x + 3/4*I*pi*(b^2*x^2 + 4*a*b*x) - 1/8*(I*pi^3 + 6*pi^2*a - 12*I*pi*a^2 - 8*a^3)*log(x)
\[ \int \frac {\coth ^{-1}(\tanh (a+b x))^3}{x} \, dx=\int \frac {\operatorname {acoth}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}{x}\, dx \]
Result contains complex when optimal does not.
Time = 0.54 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.96 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))^3}{x} \, dx=\frac {1}{3} \, b^{3} x^{3} - \frac {3}{4} \, {\left (i \, \pi b^{2} - 2 \, a b^{2}\right )} x^{2} - \frac {3}{4} \, {\left (\pi ^{2} b + 4 i \, \pi a b - 4 \, a^{2} b\right )} x + \frac {1}{8} \, {\left (i \, \pi ^{3} - 6 \, \pi ^{2} a - 12 i \, \pi a^{2} + 8 \, a^{3}\right )} \log \left (x\right ) \]
1/3*b^3*x^3 - 3/4*(I*pi*b^2 - 2*a*b^2)*x^2 - 3/4*(pi^2*b + 4*I*pi*a*b - 4* a^2*b)*x + 1/8*(I*pi^3 - 6*pi^2*a - 12*I*pi*a^2 + 8*a^3)*log(x)
Result contains complex when optimal does not.
Time = 0.30 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.96 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))^3}{x} \, dx=\frac {1}{3} \, b^{3} x^{3} - \frac {3}{4} \, {\left (-i \, \pi b^{2} - 2 \, a b^{2}\right )} x^{2} - \frac {3}{4} \, {\left (\pi ^{2} b - 4 i \, \pi a b - 4 \, a^{2} b\right )} x + \frac {1}{8} \, {\left (-i \, \pi ^{3} - 6 \, \pi ^{2} a + 12 i \, \pi a^{2} + 8 \, a^{3}\right )} \log \left (x\right ) \]
1/3*b^3*x^3 - 3/4*(-I*pi*b^2 - 2*a*b^2)*x^2 - 3/4*(pi^2*b - 4*I*pi*a*b - 4 *a^2*b)*x + 1/8*(-I*pi^3 - 6*pi^2*a + 12*I*pi*a^2 + 8*a^3)*log(x)
Time = 0.16 (sec) , antiderivative size = 306, normalized size of antiderivative = 3.97 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))^3}{x} \, dx=\frac {b^3\,x^3}{3}-\ln \left (x\right )\,\left (\frac {{\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x\right )}^3}{8}-a^3-\frac {3\,a\,{\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x\right )}^2}{4}+\frac {3\,a^2\,\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x\right )}{2}\right )-\frac {3\,b^2\,x^2\,\left (\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x\right )}{4}+\frac {3\,b\,x\,{\left (\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x\right )}^2}{4} \]
(b^3*x^3)/3 - log(x)*((2*a - 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*b*x)^3/8 - a^3 - (3*a*(2 *a - 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*b*x)^2)/4 + (3*a^2*(2*a - 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*b *x))/2) - (3*b^2*x^2*(log(-2/(exp(2*a)*exp(2*b*x) - 1)) - log((2*exp(2*a)* exp(2*b*x))/(exp(2*a)*exp(2*b*x) - 1)) + 2*b*x))/4 + (3*b*x*(log(-2/(exp(2 *a)*exp(2*b*x) - 1)) - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) - 1)) + 2*b*x)^2)/4