Integrand size = 14, antiderivative size = 102 \[ \int \frac {(a+b \text {arctanh}(c x))^3}{x^2} \, dx=c (a+b \text {arctanh}(c x))^3-\frac {(a+b \text {arctanh}(c x))^3}{x}+3 b c (a+b \text {arctanh}(c x))^2 \log \left (2-\frac {2}{1+c x}\right )-3 b^2 c (a+b \text {arctanh}(c x)) \operatorname {PolyLog}\left (2,-1+\frac {2}{1+c x}\right )-\frac {3}{2} b^3 c \operatorname {PolyLog}\left (3,-1+\frac {2}{1+c x}\right ) \]
c*(a+b*arctanh(c*x))^3-(a+b*arctanh(c*x))^3/x+3*b*c*(a+b*arctanh(c*x))^2*l n(2-2/(c*x+1))-3*b^2*c*(a+b*arctanh(c*x))*polylog(2,-1+2/(c*x+1))-3/2*b^3* c*polylog(3,-1+2/(c*x+1))
Result contains complex when optimal does not.
Time = 0.27 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.92 \[ \int \frac {(a+b \text {arctanh}(c x))^3}{x^2} \, dx=-\frac {a^3}{x}-\frac {3 a^2 b \text {arctanh}(c x)}{x}+3 a^2 b c \log (x)-\frac {3}{2} a^2 b c \log \left (1-c^2 x^2\right )+3 a b^2 c \left (\text {arctanh}(c x) \left (\text {arctanh}(c x)-\frac {\text {arctanh}(c x)}{c x}+2 \log \left (1-e^{-2 \text {arctanh}(c x)}\right )\right )-\operatorname {PolyLog}\left (2,e^{-2 \text {arctanh}(c x)}\right )\right )+b^3 c \left (\frac {i \pi ^3}{8}-\text {arctanh}(c x)^3-\frac {\text {arctanh}(c x)^3}{c x}+3 \text {arctanh}(c x)^2 \log \left (1-e^{2 \text {arctanh}(c x)}\right )+3 \text {arctanh}(c x) \operatorname {PolyLog}\left (2,e^{2 \text {arctanh}(c x)}\right )-\frac {3}{2} \operatorname {PolyLog}\left (3,e^{2 \text {arctanh}(c x)}\right )\right ) \]
-(a^3/x) - (3*a^2*b*ArcTanh[c*x])/x + 3*a^2*b*c*Log[x] - (3*a^2*b*c*Log[1 - c^2*x^2])/2 + 3*a*b^2*c*(ArcTanh[c*x]*(ArcTanh[c*x] - ArcTanh[c*x]/(c*x) + 2*Log[1 - E^(-2*ArcTanh[c*x])]) - PolyLog[2, E^(-2*ArcTanh[c*x])]) + b^ 3*c*((I/8)*Pi^3 - ArcTanh[c*x]^3 - ArcTanh[c*x]^3/(c*x) + 3*ArcTanh[c*x]^2 *Log[1 - E^(2*ArcTanh[c*x])] + 3*ArcTanh[c*x]*PolyLog[2, E^(2*ArcTanh[c*x] )] - (3*PolyLog[3, E^(2*ArcTanh[c*x])])/2)
Time = 0.79 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.13, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {6452, 6550, 6494, 6618, 7164}
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.
\(\displaystyle \int \frac {(a+b \text {arctanh}(c x))^3}{x^2} \, dx\) |
\(\Big \downarrow \) 6452 |
\(\displaystyle 3 b c \int \frac {(a+b \text {arctanh}(c x))^2}{x \left (1-c^2 x^2\right )}dx-\frac {(a+b \text {arctanh}(c x))^3}{x}\) |
\(\Big \downarrow \) 6550 |
\(\displaystyle 3 b c \left (\int \frac {(a+b \text {arctanh}(c x))^2}{x (c x+1)}dx+\frac {(a+b \text {arctanh}(c x))^3}{3 b}\right )-\frac {(a+b \text {arctanh}(c x))^3}{x}\) |
\(\Big \downarrow \) 6494 |
\(\displaystyle 3 b c \left (-2 b c \int \frac {(a+b \text {arctanh}(c x)) \log \left (2-\frac {2}{c x+1}\right )}{1-c^2 x^2}dx+\frac {(a+b \text {arctanh}(c x))^3}{3 b}+\log \left (2-\frac {2}{c x+1}\right ) (a+b \text {arctanh}(c x))^2\right )-\frac {(a+b \text {arctanh}(c x))^3}{x}\) |
\(\Big \downarrow \) 6618 |
\(\displaystyle 3 b c \left (-2 b c \left (\frac {\operatorname {PolyLog}\left (2,\frac {2}{c x+1}-1\right ) (a+b \text {arctanh}(c x))}{2 c}-\frac {1}{2} b \int \frac {\operatorname {PolyLog}\left (2,\frac {2}{c x+1}-1\right )}{1-c^2 x^2}dx\right )+\frac {(a+b \text {arctanh}(c x))^3}{3 b}+\log \left (2-\frac {2}{c x+1}\right ) (a+b \text {arctanh}(c x))^2\right )-\frac {(a+b \text {arctanh}(c x))^3}{x}\) |
\(\Big \downarrow \) 7164 |
\(\displaystyle 3 b c \left (-2 b c \left (\frac {\operatorname {PolyLog}\left (2,\frac {2}{c x+1}-1\right ) (a+b \text {arctanh}(c x))}{2 c}+\frac {b \operatorname {PolyLog}\left (3,\frac {2}{c x+1}-1\right )}{4 c}\right )+\frac {(a+b \text {arctanh}(c x))^3}{3 b}+\log \left (2-\frac {2}{c x+1}\right ) (a+b \text {arctanh}(c x))^2\right )-\frac {(a+b \text {arctanh}(c x))^3}{x}\) |
-((a + b*ArcTanh[c*x])^3/x) + 3*b*c*((a + b*ArcTanh[c*x])^3/(3*b) + (a + b *ArcTanh[c*x])^2*Log[2 - 2/(1 + c*x)] - 2*b*c*(((a + b*ArcTanh[c*x])*PolyL og[2, -1 + 2/(1 + c*x)])/(2*c) + (b*PolyLog[3, -1 + 2/(1 + c*x)])/(4*c)))
3.1.31.3.1 Defintions of rubi rules used
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_)]*(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_.)/((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]
Int[(Log[u_]*((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^ 2), x_Symbol] :> Simp[(a + b*ArcTanh[c*x])^p*(PolyLog[2, 1 - u]/(2*c*d)), x ] - Simp[b*(p/2) Int[(a + b*ArcTanh[c*x])^(p - 1)*(PolyLog[2, 1 - u]/(d + e*x^2)), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[c^2*d + e, 0] && EqQ[(1 - u)^2 - (1 - 2/(1 + c*x))^2, 0]
Int[(u_)*PolyLog[n_, v_], x_Symbol] :> With[{w = DerivativeDivides[v, u*v, x]}, Simp[w*PolyLog[n + 1, v], x] /; !FalseQ[w]] /; FreeQ[n, x]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 30.60 (sec) , antiderivative size = 1329, normalized size of antiderivative = 13.03
method | result | size |
parts | \(\text {Expression too large to display}\) | \(1329\) |
derivativedivides | \(\text {Expression too large to display}\) | \(1331\) |
default | \(\text {Expression too large to display}\) | \(1331\) |
-1/x*a^3+b^3*c*(-1/c/x*arctanh(c*x)^3-3/2*arctanh(c*x)^2*ln(c*x-1)+3*ln(c* x)*arctanh(c*x)^2-3/2*arctanh(c*x)^2*ln(c*x+1)+3*arctanh(c*x)^2*ln((c*x+1) /(-c^2*x^2+1)^(1/2))-arctanh(c*x)^3+3/4*(2*I*Pi*csgn(I*(-(c*x+1)^2/(c^2*x^ 2-1)-1))*csgn(I/(1-(c*x+1)^2/(c^2*x^2-1)))*csgn(I*(-(c*x+1)^2/(c^2*x^2-1)- 1)/(1-(c*x+1)^2/(c^2*x^2-1)))+2*I*Pi*csgn(I/(1-(c*x+1)^2/(c^2*x^2-1)))^3-I *Pi*csgn(I*(c*x+1)^2/(c^2*x^2-1))*csgn(I*(c*x+1)^2/(c^2*x^2-1)/(1-(c*x+1)^ 2/(c^2*x^2-1)))^2+I*Pi*csgn(I/(1-(c*x+1)^2/(c^2*x^2-1)))*csgn(I*(c*x+1)^2/ (c^2*x^2-1)/(1-(c*x+1)^2/(c^2*x^2-1)))^2-I*Pi*csgn(I/(1-(c*x+1)^2/(c^2*x^2 -1)))*csgn(I*(c*x+1)^2/(c^2*x^2-1))*csgn(I*(c*x+1)^2/(c^2*x^2-1)/(1-(c*x+1 )^2/(c^2*x^2-1)))-2*I*Pi*csgn(I/(1-(c*x+1)^2/(c^2*x^2-1)))*csgn(I*(-(c*x+1 )^2/(c^2*x^2-1)-1)/(1-(c*x+1)^2/(c^2*x^2-1)))^2+I*Pi*csgn(I*(c*x+1)/(-c^2* x^2+1)^(1/2))^2*csgn(I*(c*x+1)^2/(c^2*x^2-1))-2*I*Pi*csgn(I/(1-(c*x+1)^2/( c^2*x^2-1)))^2+2*I*Pi*csgn(I*(c*x+1)/(-c^2*x^2+1)^(1/2))*csgn(I*(c*x+1)^2/ (c^2*x^2-1))^2+2*I*Pi-2*I*Pi*csgn(I*(-(c*x+1)^2/(c^2*x^2-1)-1))*csgn(I*(-( c*x+1)^2/(c^2*x^2-1)-1)/(1-(c*x+1)^2/(c^2*x^2-1)))^2+2*I*Pi*csgn(I*(-(c*x+ 1)^2/(c^2*x^2-1)-1)/(1-(c*x+1)^2/(c^2*x^2-1)))^3+I*Pi*csgn(I*(c*x+1)^2/(c^ 2*x^2-1)/(1-(c*x+1)^2/(c^2*x^2-1)))^3+I*Pi*csgn(I*(c*x+1)^2/(c^2*x^2-1))^3 +4*ln(2))*arctanh(c*x)^2-3*arctanh(c*x)^2*ln((c*x+1)^2/(-c^2*x^2+1)-1)+3*a rctanh(c*x)^2*ln(1-(c*x+1)/(-c^2*x^2+1)^(1/2))+6*arctanh(c*x)*polylog(2,(c *x+1)/(-c^2*x^2+1)^(1/2))-6*polylog(3,(c*x+1)/(-c^2*x^2+1)^(1/2))+3*arc...
\[ \int \frac {(a+b \text {arctanh}(c x))^3}{x^2} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (c x\right ) + a\right )}^{3}}{x^{2}} \,d x } \]
\[ \int \frac {(a+b \text {arctanh}(c x))^3}{x^2} \, dx=\int \frac {\left (a + b \operatorname {atanh}{\left (c x \right )}\right )^{3}}{x^{2}}\, dx \]
\[ \int \frac {(a+b \text {arctanh}(c x))^3}{x^2} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (c x\right ) + a\right )}^{3}}{x^{2}} \,d x } \]
-3/2*(c*(log(c^2*x^2 - 1) - log(x^2)) + 2*arctanh(c*x)/x)*a^2*b - a^3/x - 1/8*((b^3*c*x - b^3)*log(-c*x + 1)^3 + 3*(2*a*b^2 + (b^3*c*x + b^3)*log(c* x + 1))*log(-c*x + 1)^2)/x - integrate(-1/8*((b^3*c*x - b^3)*log(c*x + 1)^ 3 + 6*(a*b^2*c*x - a*b^2)*log(c*x + 1)^2 + 3*(4*a*b^2*c*x - (b^3*c*x - b^3 )*log(c*x + 1)^2 + 2*(b^3*c^2*x^2 + 2*a*b^2 - (2*a*b^2*c - b^3*c)*x)*log(c *x + 1))*log(-c*x + 1))/(c*x^3 - x^2), x)
\[ \int \frac {(a+b \text {arctanh}(c x))^3}{x^2} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (c x\right ) + a\right )}^{3}}{x^{2}} \,d x } \]
Timed out. \[ \int \frac {(a+b \text {arctanh}(c x))^3}{x^2} \, dx=\int \frac {{\left (a+b\,\mathrm {atanh}\left (c\,x\right )\right )}^3}{x^2} \,d x \]