Integrand size = 16, antiderivative size = 94 \[ \int x^3 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2 \, dx=\frac {1}{2} b c x^2 \left (a+b \coth ^{-1}\left (\frac {x^2}{c}\right )\right )-\frac {1}{4} c^2 \left (a+b \coth ^{-1}\left (\frac {x^2}{c}\right )\right )^2+\frac {1}{4} x^4 \left (a+b \coth ^{-1}\left (\frac {x^2}{c}\right )\right )^2+\frac {1}{4} b^2 c^2 \log \left (1-\frac {c^2}{x^4}\right )+b^2 c^2 \log (x) \]
1/2*b*c*x^2*(a+b*arccoth(x^2/c))-1/4*c^2*(a+b*arccoth(x^2/c))^2+1/4*x^4*(a +b*arccoth(x^2/c))^2+1/4*b^2*c^2*ln(1-c^2/x^4)+b^2*c^2*ln(x)
Time = 0.04 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.11 \[ \int x^3 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2 \, dx=\frac {1}{4} \left (2 a b c x^2+a^2 x^4+2 b x^2 \left (b c+a x^2\right ) \text {arctanh}\left (\frac {c}{x^2}\right )+b^2 \left (-c^2+x^4\right ) \text {arctanh}\left (\frac {c}{x^2}\right )^2+b (a+b) c^2 \log \left (-c+x^2\right )-a b c^2 \log \left (c+x^2\right )+b^2 c^2 \log \left (c+x^2\right )\right ) \]
(2*a*b*c*x^2 + a^2*x^4 + 2*b*x^2*(b*c + a*x^2)*ArcTanh[c/x^2] + b^2*(-c^2 + x^4)*ArcTanh[c/x^2]^2 + b*(a + b)*c^2*Log[-c + x^2] - a*b*c^2*Log[c + x^ 2] + b^2*c^2*Log[c + x^2])/4
Time = 0.60 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.94, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.562, Rules used = {6454, 6452, 6544, 6452, 243, 47, 14, 16, 6510}
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.
| \(\displaystyle \int x^3 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2 \, dx\) |
| \(\Big \downarrow \) 6454 |
| \(\displaystyle -\frac {1}{2} \int x^6 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2d\frac {1}{x^2}\) |
| \(\Big \downarrow \) 6452 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{2} x^4 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2-b c \int \frac {x^4 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )}{1-\frac {c^2}{x^4}}d\frac {1}{x^2}\right )\) |
| \(\Big \downarrow \) 6544 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{2} x^4 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2-b c \left (c^2 \int \frac {a+b \text {arctanh}\left (\frac {c}{x^2}\right )}{1-\frac {c^2}{x^4}}d\frac {1}{x^2}+\int x^4 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )d\frac {1}{x^2}\right )\right )\) |
| \(\Big \downarrow \) 6452 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{2} x^4 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2-b c \left (c^2 \int \frac {a+b \text {arctanh}\left (\frac {c}{x^2}\right )}{1-\frac {c^2}{x^4}}d\frac {1}{x^2}+b c \int \frac {x^2}{1-\frac {c^2}{x^4}}d\frac {1}{x^2}-x^2 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )\right )\right )\) |
| \(\Big \downarrow \) 243 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{2} x^4 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2-b c \left (c^2 \int \frac {a+b \text {arctanh}\left (\frac {c}{x^2}\right )}{1-\frac {c^2}{x^4}}d\frac {1}{x^2}+\frac {1}{2} b c \int \frac {x^2}{1-\frac {c^2}{x^4}}d\frac {1}{x^4}-x^2 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )\right )\right )\) |
| \(\Big \downarrow \) 47 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{2} x^4 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2-b c \left (c^2 \int \frac {a+b \text {arctanh}\left (\frac {c}{x^2}\right )}{1-\frac {c^2}{x^4}}d\frac {1}{x^2}+\frac {1}{2} b c \left (c^2 \int \frac {1}{1-\frac {c^2}{x^4}}d\frac {1}{x^4}+\int x^2d\frac {1}{x^4}\right )-x^2 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )\right )\right )\) |
| \(\Big \downarrow \) 14 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{2} x^4 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2-b c \left (c^2 \int \frac {a+b \text {arctanh}\left (\frac {c}{x^2}\right )}{1-\frac {c^2}{x^4}}d\frac {1}{x^2}+\frac {1}{2} b c \left (c^2 \int \frac {1}{1-\frac {c^2}{x^4}}d\frac {1}{x^4}+\log \left (\frac {1}{x^4}\right )\right )-x^2 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )\right )\right )\) |
| \(\Big \downarrow \) 16 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{2} x^4 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2-b c \left (c^2 \int \frac {a+b \text {arctanh}\left (\frac {c}{x^2}\right )}{1-\frac {c^2}{x^4}}d\frac {1}{x^2}-x^2 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )+\frac {1}{2} b c \left (\log \left (\frac {1}{x^4}\right )-\log \left (1-\frac {c^2}{x^4}\right )\right )\right )\right )\) |
| \(\Big \downarrow \) 6510 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{2} x^4 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2-b c \left (-\left (x^2 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )\right )+\frac {c \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2}{2 b}+\frac {1}{2} b c \left (\log \left (\frac {1}{x^4}\right )-\log \left (1-\frac {c^2}{x^4}\right )\right )\right )\right )\) |
((x^4*(a + b*ArcTanh[c/x^2])^2)/2 - b*c*(-(x^2*(a + b*ArcTanh[c/x^2])) + ( c*(a + b*ArcTanh[c/x^2])^2)/(2*b) + (b*c*(-Log[1 - c^2/x^4] + Log[x^(-4)]) )/2))/2
3.2.71.3.1 Defintions of rubi rules used
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Simp[b/(b*c - a*d) Int[1/(a + b*x), x], x] - Simp[d/(b*c - a*d) Int[1/(c + d*x), x ], x] /; FreeQ[{a, b, c, d}, x]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
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_.)/((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]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.50 (sec) , antiderivative size = 805, normalized size of antiderivative = 8.56
\[\text {Expression too large to display}\]
1/4*a^2*x^4-b^2*(-1/4*x^4*arctanh(c/x^2)^2+c*(-1/2*x^2*arctanh(c/x^2)+1/4* arctanh(c/x^2)*c*ln(1+c/x^2)-1/4*arctanh(c/x^2)*c*ln(c/x^2-1)-1/2*c*(c*(Su m(-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,ind ex=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,i ndex=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(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=Ro otOf(_Z^2*c-1)))+1/2*ln(1+c/x^2)+1/2*ln(c/x^2-1)-2*ln(1/x)-c*(Sum(-1/4*(ln (1/x-_alpha)*ln(1+c/x^2)-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(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+_al pha)/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...
Time = 0.30 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.34 \[ \int x^3 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2 \, dx=\frac {1}{4} \, a^{2} x^{4} + \frac {1}{2} \, a b c x^{2} - \frac {1}{4} \, {\left (a b - b^{2}\right )} c^{2} \log \left (x^{2} + c\right ) + \frac {1}{4} \, {\left (a b + b^{2}\right )} c^{2} \log \left (x^{2} - c\right ) + \frac {1}{16} \, {\left (b^{2} x^{4} - b^{2} c^{2}\right )} \log \left (\frac {x^{2} + c}{x^{2} - c}\right )^{2} + \frac {1}{4} \, {\left (a b x^{4} + b^{2} c x^{2}\right )} \log \left (\frac {x^{2} + c}{x^{2} - c}\right ) \]
1/4*a^2*x^4 + 1/2*a*b*c*x^2 - 1/4*(a*b - b^2)*c^2*log(x^2 + c) + 1/4*(a*b + b^2)*c^2*log(x^2 - c) + 1/16*(b^2*x^4 - b^2*c^2)*log((x^2 + c)/(x^2 - c) )^2 + 1/4*(a*b*x^4 + b^2*c*x^2)*log((x^2 + c)/(x^2 - c))
Time = 2.00 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.61 \[ \int x^3 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2 \, dx=\frac {a^{2} x^{4}}{4} - \frac {a b c^{2} \operatorname {atanh}{\left (\frac {c}{x^{2}} \right )}}{2} + \frac {a b c x^{2}}{2} + \frac {a b x^{4} \operatorname {atanh}{\left (\frac {c}{x^{2}} \right )}}{2} + \frac {b^{2} c^{2} \log {\left (x - \sqrt {- c} \right )}}{2} + \frac {b^{2} c^{2} \log {\left (x + \sqrt {- c} \right )}}{2} - \frac {b^{2} c^{2} \operatorname {atanh}^{2}{\left (\frac {c}{x^{2}} \right )}}{4} - \frac {b^{2} c^{2} \operatorname {atanh}{\left (\frac {c}{x^{2}} \right )}}{2} + \frac {b^{2} c x^{2} \operatorname {atanh}{\left (\frac {c}{x^{2}} \right )}}{2} + \frac {b^{2} x^{4} \operatorname {atanh}^{2}{\left (\frac {c}{x^{2}} \right )}}{4} \]
a**2*x**4/4 - a*b*c**2*atanh(c/x**2)/2 + a*b*c*x**2/2 + a*b*x**4*atanh(c/x **2)/2 + b**2*c**2*log(x - sqrt(-c))/2 + b**2*c**2*log(x + sqrt(-c))/2 - b **2*c**2*atanh(c/x**2)**2/4 - b**2*c**2*atanh(c/x**2)/2 + b**2*c*x**2*atan h(c/x**2)/2 + b**2*x**4*atanh(c/x**2)**2/4
Time = 0.20 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.67 \[ \int x^3 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2 \, dx=\frac {1}{4} \, b^{2} x^{4} \operatorname {artanh}\left (\frac {c}{x^{2}}\right )^{2} + \frac {1}{4} \, a^{2} x^{4} + \frac {1}{4} \, {\left (2 \, x^{4} \operatorname {artanh}\left (\frac {c}{x^{2}}\right ) + {\left (2 \, x^{2} - c \log \left (x^{2} + c\right ) + c \log \left (x^{2} - c\right )\right )} c\right )} a b + \frac {1}{16} \, {\left ({\left (\log \left (x^{2} + c\right )^{2} - 2 \, {\left (\log \left (x^{2} + c\right ) - 2\right )} \log \left (x^{2} - c\right ) + \log \left (x^{2} - c\right )^{2} + 4 \, \log \left (x^{2} + c\right )\right )} c^{2} + 4 \, {\left (2 \, x^{2} - c \log \left (x^{2} + c\right ) + c \log \left (x^{2} - c\right )\right )} c \operatorname {artanh}\left (\frac {c}{x^{2}}\right )\right )} b^{2} \]
1/4*b^2*x^4*arctanh(c/x^2)^2 + 1/4*a^2*x^4 + 1/4*(2*x^4*arctanh(c/x^2) + ( 2*x^2 - c*log(x^2 + c) + c*log(x^2 - c))*c)*a*b + 1/16*((log(x^2 + c)^2 - 2*(log(x^2 + c) - 2)*log(x^2 - c) + log(x^2 - c)^2 + 4*log(x^2 + c))*c^2 + 4*(2*x^2 - c*log(x^2 + c) + c*log(x^2 - c))*c*arctanh(c/x^2))*b^2
Leaf count of result is larger than twice the leaf count of optimal. 327 vs. \(2 (86) = 172\).
Time = 0.30 (sec) , antiderivative size = 327, normalized size of antiderivative = 3.48 \[ \int x^3 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2 \, dx=-\frac {2 \, b^{2} c^{3} \log \left (\frac {x^{2} + c}{x^{2} - c} - 1\right ) - 2 \, b^{2} c^{3} \log \left (\frac {x^{2} + c}{x^{2} - c}\right ) - \frac {{\left (x^{2} + c\right )} b^{2} c^{3} \log \left (\frac {x^{2} + c}{x^{2} - c}\right )^{2}}{{\left (x^{2} - c\right )} {\left (\frac {{\left (x^{2} + c\right )}^{2}}{{\left (x^{2} - c\right )}^{2}} - \frac {2 \, {\left (x^{2} + c\right )}}{x^{2} - c} + 1\right )}} - \frac {2 \, {\left (\frac {2 \, {\left (x^{2} + c\right )} a b c^{3}}{x^{2} - c} + \frac {{\left (x^{2} + c\right )} b^{2} c^{3}}{x^{2} - c} - b^{2} c^{3}\right )} \log \left (\frac {x^{2} + c}{x^{2} - c}\right )}{\frac {{\left (x^{2} + c\right )}^{2}}{{\left (x^{2} - c\right )}^{2}} - \frac {2 \, {\left (x^{2} + c\right )}}{x^{2} - c} + 1} - \frac {4 \, {\left (\frac {{\left (x^{2} + c\right )} a^{2} c^{3}}{x^{2} - c} + \frac {{\left (x^{2} + c\right )} a b c^{3}}{x^{2} - c} - a b c^{3}\right )}}{\frac {{\left (x^{2} + c\right )}^{2}}{{\left (x^{2} - c\right )}^{2}} - \frac {2 \, {\left (x^{2} + c\right )}}{x^{2} - c} + 1}}{4 \, c} \]
-1/4*(2*b^2*c^3*log((x^2 + c)/(x^2 - c) - 1) - 2*b^2*c^3*log((x^2 + c)/(x^ 2 - c)) - (x^2 + c)*b^2*c^3*log((x^2 + c)/(x^2 - c))^2/((x^2 - c)*((x^2 + c)^2/(x^2 - c)^2 - 2*(x^2 + c)/(x^2 - c) + 1)) - 2*(2*(x^2 + c)*a*b*c^3/(x ^2 - c) + (x^2 + c)*b^2*c^3/(x^2 - c) - b^2*c^3)*log((x^2 + c)/(x^2 - c))/ ((x^2 + c)^2/(x^2 - c)^2 - 2*(x^2 + c)/(x^2 - c) + 1) - 4*((x^2 + c)*a^2*c ^3/(x^2 - c) + (x^2 + c)*a*b*c^3/(x^2 - c) - a*b*c^3)/((x^2 + c)^2/(x^2 - c)^2 - 2*(x^2 + c)/(x^2 - c) + 1))/c
Time = 3.81 (sec) , antiderivative size = 247, normalized size of antiderivative = 2.63 \[ \int x^3 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2 \, dx=\frac {a^2\,x^4}{4}-\frac {a\,b\,c^2\,\ln \left (x^2+c\right )}{4}+\frac {a\,b\,c^2\,\ln \left (x^2-c\right )}{4}+\frac {a\,b\,c\,x^2}{2}+\frac {a\,b\,x^4\,\ln \left (x^2+c\right )}{4}-\frac {a\,b\,x^4\,\ln \left (x^2-c\right )}{4}-\frac {b^2\,c^2\,{\ln \left (x^2+c\right )}^2}{16}+\frac {b^2\,c^2\,\ln \left (x^2+c\right )\,\ln \left (x^2-c\right )}{8}+\frac {b^2\,c^2\,\ln \left (x^2+c\right )}{4}-\frac {b^2\,c^2\,{\ln \left (x^2-c\right )}^2}{16}+\frac {b^2\,c^2\,\ln \left (x^2-c\right )}{4}+\frac {b^2\,c\,x^2\,\ln \left (x^2+c\right )}{4}-\frac {b^2\,c\,x^2\,\ln \left (x^2-c\right )}{4}+\frac {b^2\,x^4\,{\ln \left (x^2+c\right )}^2}{16}-\frac {b^2\,x^4\,\ln \left (x^2+c\right )\,\ln \left (x^2-c\right )}{8}+\frac {b^2\,x^4\,{\ln \left (x^2-c\right )}^2}{16} \]
(a^2*x^4)/4 + (b^2*c^2*log(x^2 - c))/4 - (b^2*c^2*log(c + x^2)^2)/16 + (b^ 2*x^4*log(c + x^2)^2)/16 - (b^2*c^2*log(x^2 - c)^2)/16 + (b^2*x^4*log(x^2 - c)^2)/16 + (b^2*c^2*log(c + x^2))/4 + (a*b*x^4*log(c + x^2))/4 + (a*b*c^ 2*log(x^2 - c))/4 + (b^2*c^2*log(c + x^2)*log(x^2 - c))/8 + (a*b*c*x^2)/2 - (a*b*x^4*log(x^2 - c))/4 + (b^2*c*x^2*log(c + x^2))/4 - (b^2*x^4*log(c + x^2)*log(x^2 - c))/8 - (b^2*c*x^2*log(x^2 - c))/4 - (a*b*c^2*log(c + x^2) )/4