Integrand size = 14, antiderivative size = 94 \[ \int x \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2 \, dx=-\frac {1}{2} c \left (a+b \coth ^{-1}\left (\frac {x^2}{c}\right )\right )^2+\frac {1}{2} x^2 \left (a+b \coth ^{-1}\left (\frac {x^2}{c}\right )\right )^2-b c \left (a+b \coth ^{-1}\left (\frac {x^2}{c}\right )\right ) \log \left (2-\frac {2}{1+\frac {c}{x^2}}\right )+\frac {1}{2} b^2 c \operatorname {PolyLog}\left (2,-1+\frac {2}{1+\frac {c}{x^2}}\right ) \]
-1/2*c*(a+b*arccoth(x^2/c))^2+1/2*x^2*(a+b*arccoth(x^2/c))^2-b*c*(a+b*arcc oth(x^2/c))*ln(2-2/(1+c/x^2))+1/2*b^2*c*polylog(2,-1+2/(1+c/x^2))
Time = 0.10 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.14 \[ \int x \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2 \, dx=\frac {1}{2} \left (b^2 \left (-c+x^2\right ) \text {arctanh}\left (\frac {c}{x^2}\right )^2+2 b \text {arctanh}\left (\frac {c}{x^2}\right ) \left (a x^2-b c \log \left (1-e^{-2 \text {arctanh}\left (\frac {c}{x^2}\right )}\right )\right )+a \left (a x^2+b c \log \left (1-\frac {c^2}{x^4}\right )-2 b c \log \left (\frac {c}{x^2}\right )\right )+b^2 c \operatorname {PolyLog}\left (2,e^{-2 \text {arctanh}\left (\frac {c}{x^2}\right )}\right )\right ) \]
(b^2*(-c + x^2)*ArcTanh[c/x^2]^2 + 2*b*ArcTanh[c/x^2]*(a*x^2 - b*c*Log[1 - E^(-2*ArcTanh[c/x^2])]) + a*(a*x^2 + b*c*Log[1 - c^2/x^4] - 2*b*c*Log[c/x ^2]) + b^2*c*PolyLog[2, E^(-2*ArcTanh[c/x^2])])/2
Time = 0.55 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.96, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {6454, 6452, 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.
\(\displaystyle \int x \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2 \, dx\) |
\(\Big \downarrow \) 6454 |
\(\displaystyle -\frac {1}{2} \int x^4 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2d\frac {1}{x^2}\) |
\(\Big \downarrow \) 6452 |
\(\displaystyle \frac {1}{2} \left (x^2 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2-2 b c \int \frac {x^2 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )}{1-\frac {c^2}{x^4}}d\frac {1}{x^2}\right )\) |
\(\Big \downarrow \) 6550 |
\(\displaystyle \frac {1}{2} \left (x^2 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2-2 b c \left (\int \frac {x^2 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )}{\frac {c}{x^2}+1}d\frac {1}{x^2}+\frac {\left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2}{2 b}\right )\right )\) |
\(\Big \downarrow \) 6494 |
\(\displaystyle \frac {1}{2} \left (x^2 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2-2 b c \left (-b c \int \frac {\log \left (2-\frac {2}{\frac {c}{x^2}+1}\right )}{1-\frac {c^2}{x^4}}d\frac {1}{x^2}+\frac {\left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2}{2 b}+\log \left (2-\frac {2}{\frac {c}{x^2}+1}\right ) \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )\right )\right )\) |
\(\Big \downarrow \) 2897 |
\(\displaystyle \frac {1}{2} \left (x^2 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2-2 b c \left (\frac {\left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2}{2 b}+\log \left (2-\frac {2}{\frac {c}{x^2}+1}\right ) \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )-\frac {1}{2} b \operatorname {PolyLog}\left (2,\frac {2}{\frac {c}{x^2}+1}-1\right )\right )\right )\) |
(x^2*(a + b*ArcTanh[c/x^2])^2 - 2*b*c*((a + b*ArcTanh[c/x^2])^2/(2*b) + (a + b*ArcTanh[c/x^2])*Log[2 - 2/(1 + c/x^2)] - (b*PolyLog[2, -1 + 2/(1 + c/ x^2)])/2))/2
3.2.72.3.1 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_.)/((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 = 10.81 (sec) , antiderivative size = 889, normalized size of antiderivative = 9.46
method | result | size |
derivativedivides | \(\text {Expression too large to display}\) | \(889\) |
default | \(\text {Expression too large to display}\) | \(889\) |
parts | \(\text {Expression too large to display}\) | \(889\) |
risch | \(\text {Expression too large to display}\) | \(4795\) |
1/2*a^2*x^2-b^2*(-1/2*x^2*arctanh(c/x^2)^2+2*c*(-1/4*arctanh(c/x^2)*ln(c/x ^2-1)-1/4*arctanh(c/x^2)*ln(1+c/x^2)+ln(1/x)*arctanh(c/x^2)-1/2*c*(Sum(1/4 *(ln(1/x-_alpha)*ln(c/x^2-1)-2*c*(1/4/_alpha/c*ln(1/x-_alpha)^2-1/2*_alpha *ln(1/x-_alpha)*ln(1/2*(_alpha+1/x)/_alpha)-1/2*_alpha*dilog(1/2*(_alpha+1 /x)/_alpha)))/c,_alpha=RootOf(_Z^2*c-1))+Sum(-1/4*(ln(1/x-_alpha)*ln(c/x^2 -1)-2*c*(1/2*ln(1/x-_alpha)*(ln((RootOf(_Z^2*c+2*_Z*_alpha*c-2,index=1)-1/ x+_alpha)/RootOf(_Z^2*c+2*_Z*_alpha*c-2,index=1))+ln((RootOf(_Z^2*c+2*_Z*_ alpha*c-2,index=2)-1/x+_alpha)/RootOf(_Z^2*c+2*_Z*_alpha*c-2,index=2)))/c+ 1/2*(dilog((RootOf(_Z^2*c+2*_Z*_alpha*c-2,index=1)-1/x+_alpha)/RootOf(_Z^2 *c+2*_Z*_alpha*c-2,index=1))+dilog((RootOf(_Z^2*c+2*_Z*_alpha*c-2,index=2) -1/x+_alpha)/RootOf(_Z^2*c+2*_Z*_alpha*c-2,index=2)))/c))/c,_alpha=RootOf( _Z^2*c+1))+Sum(1/4*(ln(1/x-_alpha)*ln(1+c/x^2)-2*c*(1/2*ln(1/x-_alpha)*(ln ((RootOf(_Z^2*c+2*_Z*_alpha*c+2,index=1)-1/x+_alpha)/RootOf(_Z^2*c+2*_Z*_a lpha*c+2,index=1))+ln((RootOf(_Z^2*c+2*_Z*_alpha*c+2,index=2)-1/x+_alpha)/ RootOf(_Z^2*c+2*_Z*_alpha*c+2,index=2)))/c+1/2*(dilog((RootOf(_Z^2*c+2*_Z* _alpha*c+2,index=1)-1/x+_alpha)/RootOf(_Z^2*c+2*_Z*_alpha*c+2,index=1))+di log((RootOf(_Z^2*c+2*_Z*_alpha*c+2,index=2)-1/x+_alpha)/RootOf(_Z^2*c+2*_Z *_alpha*c+2,index=2)))/c))/c,_alpha=RootOf(_Z^2*c-1))+Sum(-1/4*(ln(1/x-_al pha)*ln(1+c/x^2)-2*c*(1/4/_alpha/c*ln(1/x-_alpha)^2+1/2*_alpha*ln(1/x-_alp ha)*ln(1/2*(_alpha+1/x)/_alpha)+1/2*_alpha*dilog(1/2*(_alpha+1/x)/_alph...
\[ \int x \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2 \, dx=\int { {\left (b \operatorname {artanh}\left (\frac {c}{x^{2}}\right ) + a\right )}^{2} x \,d x } \]
\[ \int x \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2 \, dx=\int x \left (a + b \operatorname {atanh}{\left (\frac {c}{x^{2}} \right )}\right )^{2}\, dx \]
\[ \int x \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2 \, dx=\int { {\left (b \operatorname {artanh}\left (\frac {c}{x^{2}}\right ) + a\right )}^{2} x \,d x } \]
1/2*a^2*x^2 + 1/2*(2*x^2*arctanh(c/x^2) + c*log(x^4 - c^2))*a*b + 1/8*(x^2 *log(x^2 + c)^2 - 2*(x^2 + c)*log(x^2 + c)*log(x^2 - c) + (x^2 - c)*log(x^ 2 - c)^2 + 2*integrate(2*(3*c*x^3 + c^2*x)*log(x^2 + c)/(x^4 - c^2), x))*b ^2
\[ \int x \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2 \, dx=\int { {\left (b \operatorname {artanh}\left (\frac {c}{x^{2}}\right ) + a\right )}^{2} x \,d x } \]
Timed out. \[ \int x \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2 \, dx=\int x\,{\left (a+b\,\mathrm {atanh}\left (\frac {c}{x^2}\right )\right )}^2 \,d x \]