Integrand size = 16, antiderivative size = 140 \[ \int \frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{x} \, dx=\frac {2}{3} \left (a+b \text {arctanh}\left (c x^3\right )\right )^2 \text {arctanh}\left (1-\frac {2}{1-c x^3}\right )-\frac {1}{3} b \left (a+b \text {arctanh}\left (c x^3\right )\right ) \operatorname {PolyLog}\left (2,1-\frac {2}{1-c x^3}\right )+\frac {1}{3} b \left (a+b \text {arctanh}\left (c x^3\right )\right ) \operatorname {PolyLog}\left (2,-1+\frac {2}{1-c x^3}\right )+\frac {1}{6} b^2 \operatorname {PolyLog}\left (3,1-\frac {2}{1-c x^3}\right )-\frac {1}{6} b^2 \operatorname {PolyLog}\left (3,-1+\frac {2}{1-c x^3}\right ) \]
-2/3*(a+b*arctanh(c*x^3))^2*arctanh(-1+2/(-c*x^3+1))-1/3*b*(a+b*arctanh(c* x^3))*polylog(2,1-2/(-c*x^3+1))+1/3*b*(a+b*arctanh(c*x^3))*polylog(2,-1+2/ (-c*x^3+1))+1/6*b^2*polylog(3,1-2/(-c*x^3+1))-1/6*b^2*polylog(3,-1+2/(-c*x ^3+1))
Result contains complex when optimal does not.
Time = 0.32 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.29 \[ \int \frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{x} \, dx=a^2 \log (x)+\frac {1}{3} a b \left (-\operatorname {PolyLog}\left (2,-c x^3\right )+\operatorname {PolyLog}\left (2,c x^3\right )\right )+\frac {1}{3} b^2 \left (\frac {i \pi ^3}{24}-\frac {2}{3} \text {arctanh}\left (c x^3\right )^3-\text {arctanh}\left (c x^3\right )^2 \log \left (1+e^{-2 \text {arctanh}\left (c x^3\right )}\right )+\text {arctanh}\left (c x^3\right )^2 \log \left (1-e^{2 \text {arctanh}\left (c x^3\right )}\right )+\text {arctanh}\left (c x^3\right ) \operatorname {PolyLog}\left (2,-e^{-2 \text {arctanh}\left (c x^3\right )}\right )+\text {arctanh}\left (c x^3\right ) \operatorname {PolyLog}\left (2,e^{2 \text {arctanh}\left (c x^3\right )}\right )+\frac {1}{2} \operatorname {PolyLog}\left (3,-e^{-2 \text {arctanh}\left (c x^3\right )}\right )-\frac {1}{2} \operatorname {PolyLog}\left (3,e^{2 \text {arctanh}\left (c x^3\right )}\right )\right ) \]
a^2*Log[x] + (a*b*(-PolyLog[2, -(c*x^3)] + PolyLog[2, c*x^3]))/3 + (b^2*(( I/24)*Pi^3 - (2*ArcTanh[c*x^3]^3)/3 - ArcTanh[c*x^3]^2*Log[1 + E^(-2*ArcTa nh[c*x^3])] + ArcTanh[c*x^3]^2*Log[1 - E^(2*ArcTanh[c*x^3])] + ArcTanh[c*x ^3]*PolyLog[2, -E^(-2*ArcTanh[c*x^3])] + ArcTanh[c*x^3]*PolyLog[2, E^(2*Ar cTanh[c*x^3])] + PolyLog[3, -E^(-2*ArcTanh[c*x^3])]/2 - PolyLog[3, E^(2*Ar cTanh[c*x^3])]/2))/3
Time = 0.82 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.16, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {6450, 6448, 6614, 6620, 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.
\(\displaystyle \int \frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{x} \, dx\) |
\(\Big \downarrow \) 6450 |
\(\displaystyle \frac {1}{3} \int \frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{x^3}dx^3\) |
\(\Big \downarrow \) 6448 |
\(\displaystyle \frac {1}{3} \left (2 \text {arctanh}\left (1-\frac {2}{1-c x^3}\right ) \left (a+b \text {arctanh}\left (c x^3\right )\right )^2-4 b c \int \frac {\left (a+b \text {arctanh}\left (c x^3\right )\right ) \text {arctanh}\left (1-\frac {2}{1-c x^3}\right )}{1-c^2 x^6}dx^3\right )\) |
\(\Big \downarrow \) 6614 |
\(\displaystyle \frac {1}{3} \left (2 \text {arctanh}\left (1-\frac {2}{1-c x^3}\right ) \left (a+b \text {arctanh}\left (c x^3\right )\right )^2-4 b c \left (\frac {1}{2} \int \frac {\left (a+b \text {arctanh}\left (c x^3\right )\right ) \log \left (2-\frac {2}{1-c x^3}\right )}{1-c^2 x^6}dx^3-\frac {1}{2} \int \frac {\left (a+b \text {arctanh}\left (c x^3\right )\right ) \log \left (\frac {2}{1-c x^3}\right )}{1-c^2 x^6}dx^3\right )\right )\) |
\(\Big \downarrow \) 6620 |
\(\displaystyle \frac {1}{3} \left (2 \text {arctanh}\left (1-\frac {2}{1-c x^3}\right ) \left (a+b \text {arctanh}\left (c x^3\right )\right )^2-4 b c \left (\frac {1}{2} \left (\frac {\operatorname {PolyLog}\left (2,1-\frac {2}{1-c x^3}\right ) \left (a+b \text {arctanh}\left (c x^3\right )\right )}{2 c}-\frac {1}{2} b \int \frac {\operatorname {PolyLog}\left (2,1-\frac {2}{1-c x^3}\right )}{1-c^2 x^6}dx^3\right )+\frac {1}{2} \left (\frac {1}{2} b \int \frac {\operatorname {PolyLog}\left (2,\frac {2}{1-c x^3}-1\right )}{1-c^2 x^6}dx^3-\frac {\operatorname {PolyLog}\left (2,\frac {2}{1-c x^3}-1\right ) \left (a+b \text {arctanh}\left (c x^3\right )\right )}{2 c}\right )\right )\right )\) |
\(\Big \downarrow \) 7164 |
\(\displaystyle \frac {1}{3} \left (2 \text {arctanh}\left (1-\frac {2}{1-c x^3}\right ) \left (a+b \text {arctanh}\left (c x^3\right )\right )^2-4 b c \left (\frac {1}{2} \left (\frac {\operatorname {PolyLog}\left (2,1-\frac {2}{1-c x^3}\right ) \left (a+b \text {arctanh}\left (c x^3\right )\right )}{2 c}-\frac {b \operatorname {PolyLog}\left (3,1-\frac {2}{1-c x^3}\right )}{4 c}\right )+\frac {1}{2} \left (\frac {b \operatorname {PolyLog}\left (3,\frac {2}{1-c x^3}-1\right )}{4 c}-\frac {\operatorname {PolyLog}\left (2,\frac {2}{1-c x^3}-1\right ) \left (a+b \text {arctanh}\left (c x^3\right )\right )}{2 c}\right )\right )\right )\) |
(2*(a + b*ArcTanh[c*x^3])^2*ArcTanh[1 - 2/(1 - c*x^3)] - 4*b*c*((((a + b*A rcTanh[c*x^3])*PolyLog[2, 1 - 2/(1 - c*x^3)])/(2*c) - (b*PolyLog[3, 1 - 2/ (1 - c*x^3)])/(4*c))/2 + (-1/2*((a + b*ArcTanh[c*x^3])*PolyLog[2, -1 + 2/( 1 - c*x^3)])/c + (b*PolyLog[3, -1 + 2/(1 - c*x^3)])/(4*c))/2))/3
3.2.20.3.1 Defintions of rubi rules used
Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_)/(x_), x_Symbol] :> Simp[2*(a + b*ArcTanh[c*x])^p*ArcTanh[1 - 2/(1 - c*x)], x] - Simp[2*b*c*p Int[(a + b *ArcTanh[c*x])^(p - 1)*(ArcTanh[1 - 2/(1 - c*x)]/(1 - c^2*x^2)), x], x] /; FreeQ[{a, b, c}, x] && IGtQ[p, 1]
Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_)]*(b_.))^(p_.)/(x_), x_Symbol] :> Simp[ 1/n Subst[Int[(a + b*ArcTanh[c*x])^p/x, x], x, x^n], x] /; FreeQ[{a, b, c , n}, x] && IGtQ[p, 0]
Int[(ArcTanh[u_]*((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*( x_)^2), x_Symbol] :> Simp[1/2 Int[Log[1 + u]*((a + b*ArcTanh[c*x])^p/(d + e*x^2)), x], x] - Simp[1/2 Int[Log[1 - u]*((a + b*ArcTanh[c*x])^p/(d + e *x^2)), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[c^2*d + e, 0] && EqQ[u^2 - (1 - 2/(1 - c*x))^2, 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]
\[\int \frac {{\left (a +b \,\operatorname {arctanh}\left (c \,x^{3}\right )\right )}^{2}}{x}d x\]
\[ \int \frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{x} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (c x^{3}\right ) + a\right )}^{2}}{x} \,d x } \]
\[ \int \frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{x} \, dx=\int \frac {\left (a + b \operatorname {atanh}{\left (c x^{3} \right )}\right )^{2}}{x}\, dx \]
\[ \int \frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{x} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (c x^{3}\right ) + a\right )}^{2}}{x} \,d x } \]
a^2*log(x) + integrate(1/4*b^2*(log(c*x^3 + 1) - log(-c*x^3 + 1))^2/x + a* b*(log(c*x^3 + 1) - log(-c*x^3 + 1))/x, x)
\[ \int \frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{x} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (c x^{3}\right ) + a\right )}^{2}}{x} \,d x } \]
Timed out. \[ \int \frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{x} \, dx=\int \frac {{\left (a+b\,\mathrm {atanh}\left (c\,x^3\right )\right )}^2}{x} \,d x \]