Integrand size = 16, antiderivative size = 1214 \[ \int x^4 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2 \, dx =\text {Too large to display} \]
8/15*b^2*c^2*x+2/5*a*b*c^(5/2)*arctan(x/c^(1/2))-1/15*b^2*c*x^3*ln(1-c/x^2 )-1/5*b^2*c^(5/2)*arctan(x/c^(1/2))*ln(1-c/x^2)+1/15*b*c*x^3*(2*a-b*ln(1-c /x^2))-1/5*b*c^(5/2)*arctanh(x/c^(1/2))*(2*a-b*ln(1-c/x^2))+2/15*b^2*c*x^3 *ln(1+c/x^2)+1/5*a*b*x^5*ln(1+c/x^2)+1/5*b^2*c^(5/2)*arctan(x/c^(1/2))*ln( 1+c/x^2)-1/5*b^2*c^(5/2)*arctanh(x/c^(1/2))*ln(1+c/x^2)-1/10*b^2*x^5*ln(1- c/x^2)*ln(1+c/x^2)-2/5*b^2*c^(5/2)*arctan(x/c^(1/2))*ln(2*c^(1/2)/(-I*x+c^ (1/2)))+1/5*b^2*c^(5/2)*arctan(x/c^(1/2))*ln((1+I)*(-x+c^(1/2))/(-I*x+c^(1 /2)))-2/5*b^2*c^(5/2)*arctanh(x/c^(1/2))*ln(2*c^(1/2)/(x+c^(1/2)))+1/5*b^2 *c^(5/2)*arctanh(x/c^(1/2))*ln(2*(-x+(-c)^(1/2))*c^(1/2)/((-c)^(1/2)-c^(1/ 2))/(x+c^(1/2)))+1/5*b^2*c^(5/2)*arctan(x/c^(1/2))*ln((1-I)*(x+c^(1/2))/(- I*x+c^(1/2)))+1/5*b^2*c^(5/2)*arctanh(x/c^(1/2))*ln(2*(x+(-c)^(1/2))*c^(1/ 2)/(x+c^(1/2))/((-c)^(1/2)+c^(1/2)))+2/5*b^2*c^(5/2)*arctan(x/c^(1/2))*ln( 2-2*c^(1/2)/(-I*x+c^(1/2)))+2/5*b^2*c^(5/2)*arctanh(x/c^(1/2))*ln(2-2*c^(1 /2)/(x+c^(1/2)))+2/15*a*b*c*x^3-1/5*I*b^2*c^(5/2)*arctan(x/c^(1/2))^2-1/5* I*b^2*c^(5/2)*polylog(2,-I*x/c^(1/2))-1/5*I*b^2*c^(5/2)*polylog(2,-1+2*c^( 1/2)/(-I*x+c^(1/2)))-1/10*I*b^2*c^(5/2)*polylog(2,1-(1+I)*(-x+c^(1/2))/(-I *x+c^(1/2)))-1/10*I*b^2*c^(5/2)*polylog(2,1+(-1+I)*(x+c^(1/2))/(-I*x+c^(1/ 2)))-4/15*b^2*c^(5/2)*arctan(x/c^(1/2))-4/15*b^2*c^(5/2)*arctanh(x/c^(1/2) )+1/5*b^2*c^(5/2)*arctanh(x/c^(1/2))^2+1/20*b^2*x^5*ln(1+c/x^2)^2+1/5*b^2* c^(5/2)*polylog(2,-x/c^(1/2))-1/5*b^2*c^(5/2)*polylog(2,x/c^(1/2))+1/5*...
\[ \int x^4 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2 \, dx=\int x^4 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2 \, dx \]
Time = 2.46 (sec) , antiderivative size = 1214, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {6460, 6457, 2009}
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.
\(\displaystyle \int x^4 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2 \, dx\) |
\(\Big \downarrow \) 6460 |
\(\displaystyle \int x^4 \left (a+b \coth ^{-1}\left (\frac {x^2}{c}\right )\right )^2dx\) |
\(\Big \downarrow \) 6457 |
\(\displaystyle \int \left (\frac {1}{4} x^4 \left (2 a-b \log \left (1-\frac {c}{x^2}\right )\right )^2-\frac {1}{2} b x^4 \log \left (\frac {c}{x^2}+1\right ) \left (b \log \left (1-\frac {c}{x^2}\right )-2 a\right )+\frac {1}{4} b^2 x^4 \log ^2\left (\frac {c}{x^2}+1\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{20} \left (2 a-b \log \left (1-\frac {c}{x^2}\right )\right )^2 x^5+\frac {1}{20} b^2 \log ^2\left (\frac {c}{x^2}+1\right ) x^5+\frac {1}{5} a b \log \left (\frac {c}{x^2}+1\right ) x^5-\frac {1}{10} b^2 \log \left (1-\frac {c}{x^2}\right ) \log \left (\frac {c}{x^2}+1\right ) x^5+\frac {2}{15} a b c x^3-\frac {1}{15} b^2 c \log \left (1-\frac {c}{x^2}\right ) x^3+\frac {1}{15} b c \left (2 a-b \log \left (1-\frac {c}{x^2}\right )\right ) x^3+\frac {2}{15} b^2 c \log \left (\frac {c}{x^2}+1\right ) x^3+\frac {8}{15} b^2 c^2 x-\frac {1}{5} i b^2 c^{5/2} \arctan \left (\frac {x}{\sqrt {c}}\right )^2+\frac {1}{5} b^2 c^{5/2} \text {arctanh}\left (\frac {x}{\sqrt {c}}\right )^2-\frac {4}{15} b^2 c^{5/2} \arctan \left (\frac {x}{\sqrt {c}}\right )+\frac {2}{5} a b c^{5/2} \arctan \left (\frac {x}{\sqrt {c}}\right )-\frac {4}{15} b^2 c^{5/2} \text {arctanh}\left (\frac {x}{\sqrt {c}}\right )+\frac {2}{5} b^2 c^{5/2} \arctan \left (\frac {x}{\sqrt {c}}\right ) \log \left (2-\frac {2 \sqrt {c}}{\sqrt {c}-i x}\right )-\frac {1}{5} b^2 c^{5/2} \arctan \left (\frac {x}{\sqrt {c}}\right ) \log \left (1-\frac {c}{x^2}\right )-\frac {1}{5} b c^{5/2} \text {arctanh}\left (\frac {x}{\sqrt {c}}\right ) \left (2 a-b \log \left (1-\frac {c}{x^2}\right )\right )+\frac {1}{5} b^2 c^{5/2} \arctan \left (\frac {x}{\sqrt {c}}\right ) \log \left (\frac {c}{x^2}+1\right )-\frac {1}{5} b^2 c^{5/2} \text {arctanh}\left (\frac {x}{\sqrt {c}}\right ) \log \left (\frac {c}{x^2}+1\right )-\frac {2}{5} b^2 c^{5/2} \arctan \left (\frac {x}{\sqrt {c}}\right ) \log \left (\frac {2 \sqrt {c}}{\sqrt {c}-i x}\right )+\frac {1}{5} b^2 c^{5/2} \arctan \left (\frac {x}{\sqrt {c}}\right ) \log \left (\frac {(1+i) \left (\sqrt {c}-x\right )}{\sqrt {c}-i x}\right )-\frac {2}{5} b^2 c^{5/2} \text {arctanh}\left (\frac {x}{\sqrt {c}}\right ) \log \left (\frac {2 \sqrt {c}}{x+\sqrt {c}}\right )+\frac {1}{5} b^2 c^{5/2} \text {arctanh}\left (\frac {x}{\sqrt {c}}\right ) \log \left (\frac {2 \sqrt {c} \left (\sqrt {-c}-x\right )}{\left (\sqrt {-c}-\sqrt {c}\right ) \left (x+\sqrt {c}\right )}\right )+\frac {1}{5} b^2 c^{5/2} \text {arctanh}\left (\frac {x}{\sqrt {c}}\right ) \log \left (\frac {2 \sqrt {c} \left (x+\sqrt {-c}\right )}{\left (\sqrt {-c}+\sqrt {c}\right ) \left (x+\sqrt {c}\right )}\right )+\frac {1}{5} b^2 c^{5/2} \arctan \left (\frac {x}{\sqrt {c}}\right ) \log \left (\frac {(1-i) \left (x+\sqrt {c}\right )}{\sqrt {c}-i x}\right )+\frac {2}{5} b^2 c^{5/2} \text {arctanh}\left (\frac {x}{\sqrt {c}}\right ) \log \left (2-\frac {2 \sqrt {c}}{x+\sqrt {c}}\right )+\frac {1}{5} i b^2 c^{5/2} \operatorname {PolyLog}\left (2,1-\frac {2 \sqrt {c}}{\sqrt {c}-i x}\right )-\frac {1}{5} i b^2 c^{5/2} \operatorname {PolyLog}\left (2,\frac {2 \sqrt {c}}{\sqrt {c}-i x}-1\right )-\frac {1}{10} i b^2 c^{5/2} \operatorname {PolyLog}\left (2,1-\frac {(1+i) \left (\sqrt {c}-x\right )}{\sqrt {c}-i x}\right )+\frac {1}{5} b^2 c^{5/2} \operatorname {PolyLog}\left (2,-\frac {x}{\sqrt {c}}\right )-\frac {1}{5} i b^2 c^{5/2} \operatorname {PolyLog}\left (2,-\frac {i x}{\sqrt {c}}\right )+\frac {1}{5} i b^2 c^{5/2} \operatorname {PolyLog}\left (2,\frac {i x}{\sqrt {c}}\right )-\frac {1}{5} b^2 c^{5/2} \operatorname {PolyLog}\left (2,\frac {x}{\sqrt {c}}\right )+\frac {1}{5} b^2 c^{5/2} \operatorname {PolyLog}\left (2,1-\frac {2 \sqrt {c}}{x+\sqrt {c}}\right )-\frac {1}{5} b^2 c^{5/2} \operatorname {PolyLog}\left (2,\frac {2 \sqrt {c}}{x+\sqrt {c}}-1\right )-\frac {1}{10} b^2 c^{5/2} \operatorname {PolyLog}\left (2,1-\frac {2 \sqrt {c} \left (\sqrt {-c}-x\right )}{\left (\sqrt {-c}-\sqrt {c}\right ) \left (x+\sqrt {c}\right )}\right )-\frac {1}{10} b^2 c^{5/2} \operatorname {PolyLog}\left (2,1-\frac {2 \sqrt {c} \left (x+\sqrt {-c}\right )}{\left (\sqrt {-c}+\sqrt {c}\right ) \left (x+\sqrt {c}\right )}\right )-\frac {1}{10} i b^2 c^{5/2} \operatorname {PolyLog}\left (2,1-\frac {(1-i) \left (x+\sqrt {c}\right )}{\sqrt {c}-i x}\right )\) |
(8*b^2*c^2*x)/15 + (2*a*b*c*x^3)/15 + (2*a*b*c^(5/2)*ArcTan[x/Sqrt[c]])/5 - (4*b^2*c^(5/2)*ArcTan[x/Sqrt[c]])/15 - (I/5)*b^2*c^(5/2)*ArcTan[x/Sqrt[c ]]^2 - (4*b^2*c^(5/2)*ArcTanh[x/Sqrt[c]])/15 + (b^2*c^(5/2)*ArcTanh[x/Sqrt [c]]^2)/5 + (2*b^2*c^(5/2)*ArcTan[x/Sqrt[c]]*Log[2 - (2*Sqrt[c])/(Sqrt[c] - I*x)])/5 - (b^2*c*x^3*Log[1 - c/x^2])/15 - (b^2*c^(5/2)*ArcTan[x/Sqrt[c] ]*Log[1 - c/x^2])/5 + (b*c*x^3*(2*a - b*Log[1 - c/x^2]))/15 - (b*c^(5/2)*A rcTanh[x/Sqrt[c]]*(2*a - b*Log[1 - c/x^2]))/5 + (x^5*(2*a - b*Log[1 - c/x^ 2])^2)/20 + (2*b^2*c*x^3*Log[1 + c/x^2])/15 + (a*b*x^5*Log[1 + c/x^2])/5 + (b^2*c^(5/2)*ArcTan[x/Sqrt[c]]*Log[1 + c/x^2])/5 - (b^2*c^(5/2)*ArcTanh[x /Sqrt[c]]*Log[1 + c/x^2])/5 - (b^2*x^5*Log[1 - c/x^2]*Log[1 + c/x^2])/10 + (b^2*x^5*Log[1 + c/x^2]^2)/20 - (2*b^2*c^(5/2)*ArcTan[x/Sqrt[c]]*Log[(2*S qrt[c])/(Sqrt[c] - I*x)])/5 + (b^2*c^(5/2)*ArcTan[x/Sqrt[c]]*Log[((1 + I)* (Sqrt[c] - x))/(Sqrt[c] - I*x)])/5 - (2*b^2*c^(5/2)*ArcTanh[x/Sqrt[c]]*Log [(2*Sqrt[c])/(Sqrt[c] + x)])/5 + (b^2*c^(5/2)*ArcTanh[x/Sqrt[c]]*Log[(2*Sq rt[c]*(Sqrt[-c] - x))/((Sqrt[-c] - Sqrt[c])*(Sqrt[c] + x))])/5 + (b^2*c^(5 /2)*ArcTanh[x/Sqrt[c]]*Log[(2*Sqrt[c]*(Sqrt[-c] + x))/((Sqrt[-c] + Sqrt[c] )*(Sqrt[c] + x))])/5 + (b^2*c^(5/2)*ArcTan[x/Sqrt[c]]*Log[((1 - I)*(Sqrt[c ] + x))/(Sqrt[c] - I*x)])/5 + (2*b^2*c^(5/2)*ArcTanh[x/Sqrt[c]]*Log[2 - (2 *Sqrt[c])/(Sqrt[c] + x)])/5 + (I/5)*b^2*c^(5/2)*PolyLog[2, 1 - (2*Sqrt[c]) /(Sqrt[c] - I*x)] - (I/5)*b^2*c^(5/2)*PolyLog[2, -1 + (2*Sqrt[c])/(Sqrt...
3.2.76.3.1 Defintions of rubi rules used
Int[((a_.) + ArcCoth[(c_.)*(x_)^(n_)]*(b_.))^(p_)*(x_)^(m_.), x_Symbol] :> Int[ExpandIntegrand[x^m*(a + b*(Log[1 + 1/(x^n*c)]/2) - b*(Log[1 - 1/(x^n*c )]/2))^p, x], x] /; FreeQ[{a, b, c}, x] && IGtQ[p, 1] && IGtQ[n, 0] && Inte gerQ[m]
Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_)]*(b_.))^(p_)*(x_)^(m_.), x_Symbol] :> Int[x^m*(a + b*ArcCoth[1/(x^n*c)])^p, x] /; FreeQ[{a, b, c, m}, x] && IGtQ[ p, 1] && ILtQ[n, 0]
\[\int x^{4} {\left (a +b \,\operatorname {arctanh}\left (\frac {c}{x^{2}}\right )\right )}^{2}d x\]
\[ \int x^4 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2 \, dx=\int { {\left (b \operatorname {artanh}\left (\frac {c}{x^{2}}\right ) + a\right )}^{2} x^{4} \,d x } \]
\[ \int x^4 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2 \, dx=\int x^{4} \left (a + b \operatorname {atanh}{\left (\frac {c}{x^{2}} \right )}\right )^{2}\, dx \]
\[ \int x^4 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2 \, dx=\int { {\left (b \operatorname {artanh}\left (\frac {c}{x^{2}}\right ) + a\right )}^{2} x^{4} \,d x } \]
1/5*a^2*x^5 + 1/15*(6*x^5*arctanh(c/x^2) + (4*x^3 + 6*c^(3/2)*arctan(x/sqr t(c)) + 3*c^(3/2)*log((x - sqrt(c))/(x + sqrt(c))))*c)*a*b + 1/20*(x^5*log (x^2 - c)^2 - 5*integrate(-1/5*(5*(x^6 - c*x^4)*log(x^2 + c)^2 - 2*(2*x^6 + 5*(x^6 - c*x^4)*log(x^2 + c))*log(x^2 - c))/(x^2 - c), x))*b^2
\[ \int x^4 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2 \, dx=\int { {\left (b \operatorname {artanh}\left (\frac {c}{x^{2}}\right ) + a\right )}^{2} x^{4} \,d x } \]
Timed out. \[ \int x^4 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2 \, dx=\int x^4\,{\left (a+b\,\mathrm {atanh}\left (\frac {c}{x^2}\right )\right )}^2 \,d x \]