Integrand size = 14, antiderivative size = 135 \[ \int x \left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right )^3 \, dx=-\frac {3}{2} b c^2 \left (a+b \coth ^{-1}\left (\frac {x}{c}\right )\right )^2+\frac {3}{2} b c x \left (a+b \coth ^{-1}\left (\frac {x}{c}\right )\right )^2-\frac {1}{2} c^2 \left (a+b \coth ^{-1}\left (\frac {x}{c}\right )\right )^3+\frac {1}{2} x^2 \left (a+b \coth ^{-1}\left (\frac {x}{c}\right )\right )^3-3 b^2 c^2 \left (a+b \coth ^{-1}\left (\frac {x}{c}\right )\right ) \log \left (2-\frac {2}{1+\frac {c}{x}}\right )+\frac {3}{2} b^3 c^2 \operatorname {PolyLog}\left (2,-1+\frac {2}{1+\frac {c}{x}}\right ) \]
-3/2*b*c^2*(a+b*arccoth(x/c))^2+3/2*b*c*x*(a+b*arccoth(x/c))^2-1/2*c^2*(a+ b*arccoth(x/c))^3+1/2*x^2*(a+b*arccoth(x/c))^3-3*b^2*c^2*(a+b*arccoth(x/c) )*ln(2-2/(1+c/x))+3/2*b^3*c^2*polylog(2,-1+2/(1+c/x))
Time = 0.23 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.43 \[ \int x \left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right )^3 \, dx=\frac {1}{4} \left (6 b^2 (-c+x) (b c+a (c+x)) \text {arctanh}\left (\frac {c}{x}\right )^2+2 b^3 \left (-c^2+x^2\right ) \text {arctanh}\left (\frac {c}{x}\right )^3+6 b \text {arctanh}\left (\frac {c}{x}\right ) \left (a x (2 b c+a x)-2 b^2 c^2 \log \left (1-e^{-2 \text {arctanh}\left (\frac {c}{x}\right )}\right )\right )+a \left (3 a b c^2 \log \left (1-\frac {c}{x}\right )-12 b^2 c^2 \log \left (\frac {c}{\sqrt {1-\frac {c^2}{x^2}} x}\right )+a \left (6 b c x+2 a x^2-3 b c^2 \log \left (\frac {c+x}{x}\right )\right )\right )+6 b^3 c^2 \operatorname {PolyLog}\left (2,e^{-2 \text {arctanh}\left (\frac {c}{x}\right )}\right )\right ) \]
(6*b^2*(-c + x)*(b*c + a*(c + x))*ArcTanh[c/x]^2 + 2*b^3*(-c^2 + x^2)*ArcT anh[c/x]^3 + 6*b*ArcTanh[c/x]*(a*x*(2*b*c + a*x) - 2*b^2*c^2*Log[1 - E^(-2 *ArcTanh[c/x])]) + a*(3*a*b*c^2*Log[1 - c/x] - 12*b^2*c^2*Log[c/(Sqrt[1 - c^2/x^2]*x)] + a*(6*b*c*x + 2*a*x^2 - 3*b*c^2*Log[(c + x)/x])) + 6*b^3*c^2 *PolyLog[2, E^(-2*ArcTanh[c/x])])/4
Time = 1.03 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.97, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {6454, 6452, 6544, 6452, 6510, 6550, 6494, 2897}
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.
| \(\displaystyle \int x \left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right )^3 \, dx\) |
| \(\Big \downarrow \) 6454 |
| \(\displaystyle -\int x^3 \left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right )^3d\frac {1}{x}\) |
| \(\Big \downarrow \) 6452 |
| \(\displaystyle \frac {1}{2} x^2 \left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right )^3-\frac {3}{2} b c \int \frac {x^2 \left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right )^2}{1-\frac {c^2}{x^2}}d\frac {1}{x}\) |
| \(\Big \downarrow \) 6544 |
| \(\displaystyle \frac {1}{2} x^2 \left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right )^3-\frac {3}{2} b c \left (c^2 \int \frac {\left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right )^2}{1-\frac {c^2}{x^2}}d\frac {1}{x}+\int x^2 \left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right )^2d\frac {1}{x}\right )\) |
| \(\Big \downarrow \) 6452 |
| \(\displaystyle \frac {1}{2} x^2 \left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right )^3-\frac {3}{2} b c \left (c^2 \int \frac {\left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right )^2}{1-\frac {c^2}{x^2}}d\frac {1}{x}+2 b c \int \frac {x \left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right )}{1-\frac {c^2}{x^2}}d\frac {1}{x}-x \left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right )^2\right )\) |
| \(\Big \downarrow \) 6510 |
| \(\displaystyle \frac {1}{2} x^2 \left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right )^3-\frac {3}{2} b c \left (2 b c \int \frac {x \left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right )}{1-\frac {c^2}{x^2}}d\frac {1}{x}+\frac {c \left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right )^3}{3 b}-x \left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right )^2\right )\) |
| \(\Big \downarrow \) 6550 |
| \(\displaystyle \frac {1}{2} x^2 \left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right )^3-\frac {3}{2} b c \left (2 b c \left (\int \frac {x \left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right )}{\frac {c}{x}+1}d\frac {1}{x}+\frac {\left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right )^2}{2 b}\right )+\frac {c \left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right )^3}{3 b}-x \left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right )^2\right )\) |
| \(\Big \downarrow \) 6494 |
| \(\displaystyle \frac {1}{2} x^2 \left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right )^3-\frac {3}{2} b c \left (2 b c \left (-b c \int \frac {\log \left (2-\frac {2}{\frac {c}{x}+1}\right )}{1-\frac {c^2}{x^2}}d\frac {1}{x}+\frac {\left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right )^2}{2 b}+\log \left (2-\frac {2}{\frac {c}{x}+1}\right ) \left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right )\right )+\frac {c \left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right )^3}{3 b}-x \left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right )^2\right )\) |
| \(\Big \downarrow \) 2897 |
| \(\displaystyle \frac {1}{2} x^2 \left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right )^3-\frac {3}{2} b c \left (2 b c \left (\frac {\left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right )^2}{2 b}+\log \left (2-\frac {2}{\frac {c}{x}+1}\right ) \left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right )-\frac {1}{2} b \operatorname {PolyLog}\left (2,\frac {2}{\frac {c}{x}+1}-1\right )\right )+\frac {c \left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right )^3}{3 b}-x \left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right )^2\right )\) |
(x^2*(a + b*ArcTanh[c/x])^3)/2 - (3*b*c*(-(x*(a + b*ArcTanh[c/x])^2) + (c* (a + b*ArcTanh[c/x])^3)/(3*b) + 2*b*c*((a + b*ArcTanh[c/x])^2/(2*b) + (a + b*ArcTanh[c/x])*Log[2 - 2/(1 + c/x)] - (b*PolyLog[2, -1 + 2/(1 + c/x)])/2 )))/2
3.2.52.3.1 Defintions of rubi rules used
Int[Log[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[Pq^m*((1 - u)/ D[u, x])]}, Simp[C*PolyLog[2, 1 - u], x] /; FreeQ[C, x]] /; IntegerQ[m] && PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponents[u, x][[2]], Expon[Pq, x]]
Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] : > Simp[x^(m + 1)*((a + b*ArcTanh[c*x^n])^p/(m + 1)), x] - Simp[b*c*n*(p/(m + 1)) Int[x^(m + n)*((a + b*ArcTanh[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_.) + ArcTanh[(c_.)*(x_)^(n_)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*ArcTanh[c*x])^p, x ], x, x^n], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 1] && IntegerQ[Simpl ify[(m + 1)/n]]
Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_))), x _Symbol] :> Simp[(a + b*ArcTanh[c*x])^p*(Log[2 - 2/(1 + e*(x/d))]/d), x] - Simp[b*c*(p/d) Int[(a + b*ArcTanh[c*x])^(p - 1)*(Log[2 - 2/(1 + e*(x/d))] /(1 - c^2*x^2)), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[c ^2*d^2 - e^2, 0]
Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2), x_Symb ol] :> Simp[(a + b*ArcTanh[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_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + ( e_.)*(x_)^2), x_Symbol] :> Simp[1/d Int[(f*x)^m*(a + b*ArcTanh[c*x])^p, x ], x] - Simp[e/(d*f^2) Int[(f*x)^(m + 2)*((a + b*ArcTanh[c*x])^p/(d + e*x ^2)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && GtQ[p, 0] && LtQ[m, -1]
Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_)^2)), x_Symbol] :> Simp[(a + b*ArcTanh[c*x])^(p + 1)/(b*d*(p + 1)), x] + Simp[1/ d Int[(a + b*ArcTanh[c*x])^p/(x*(1 + c*x)), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && GtQ[p, 0]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 1.14 (sec) , antiderivative size = 5036, normalized size of antiderivative = 37.30
\[\text {output too large to display}\]
\[ \int x \left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right )^3 \, dx=\int { {\left (b \operatorname {artanh}\left (\frac {c}{x}\right ) + a\right )}^{3} x \,d x } \]
integral(b^3*x*arctanh(c/x)^3 + 3*a*b^2*x*arctanh(c/x)^2 + 3*a^2*b*x*arcta nh(c/x) + a^3*x, x)
\[ \int x \left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right )^3 \, dx=\int x \left (a + b \operatorname {atanh}{\left (\frac {c}{x} \right )}\right )^{3}\, dx \]
\[ \int x \left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right )^3 \, dx=\int { {\left (b \operatorname {artanh}\left (\frac {c}{x}\right ) + a\right )}^{3} x \,d x } \]
3/2*a*b^2*x^2*arctanh(c/x)^2 + 1/2*a^3*x^2 + 3/4*(2*x^2*arctanh(c/x) - (c* log(c + x) - c*log(-c + x) - 2*x)*c)*a^2*b + 3/8*((log(c + x)^2 - 2*(log(c + x) - 2)*log(-c + x) + log(-c + x)^2 + 4*log(c + x))*c^2 - 4*(c*log(c + x) - c*log(-c + x) - 2*x)*c*arctanh(c/x))*a*b^2 + 1/16*(6*c*x*log(c + x)^2 - (c^2 - x^2)*log(c + x)^3 + (c^2 - x^2)*log(-c + x)^3 - 3*(2*c^2 - 2*c*x + (c^2 - x^2)*log(c + x))*log(-c + x)^2 + 3*((c^2 - x^2)*log(c + x)^2 - 4 *(c^2 + c*x)*log(c + x))*log(-c + x) + 2*integrate(-6*(c^3 + 3*c^2*x)*log( c + x)/(c^2 - x^2), x))*b^3
\[ \int x \left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right )^3 \, dx=\int { {\left (b \operatorname {artanh}\left (\frac {c}{x}\right ) + a\right )}^{3} x \,d x } \]
Timed out. \[ \int x \left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right )^3 \, dx=\int x\,{\left (a+b\,\mathrm {atanh}\left (\frac {c}{x}\right )\right )}^3 \,d x \]