Integrand size = 16, antiderivative size = 1176 \[ \int \frac {\left (a+b \text {arctanh}\left (c x^2\right )\right )^2}{x^6} \, dx =\text {Too large to display} \]
1/5*I*b^2*c^(5/2)*polylog(2,1-2/(1-I*x*c^(1/2)))+1/5*I*b^2*c^(5/2)*polylog (2,1-2/(1+I*x*c^(1/2)))+2/5*a*b*c^(5/2)*arctan(x*c^(1/2))+1/15*b^2*c*ln(-c *x^2+1)/x^3-1/5*b^2*c^2*ln(-c*x^2+1)/x-1/5*b^2*c^(5/2)*arctan(x*c^(1/2))*l n(-c*x^2+1)-1/15*b*c*(2*a-b*ln(-c*x^2+1))/x^3-1/5*b*c^2*(2*a-b*ln(-c*x^2+1 ))/x+1/5*b*c^(5/2)*arctanh(x*c^(1/2))*(2*a-b*ln(-c*x^2+1))-1/5*a*b*ln(c*x^ 2+1)/x^5-2/15*b^2*c*ln(c*x^2+1)/x^3+1/5*b^2*c^(5/2)*arctan(x*c^(1/2))*ln(c *x^2+1)+1/5*b^2*c^(5/2)*arctanh(x*c^(1/2))*ln(c*x^2+1)+1/10*b^2*ln(-c*x^2+ 1)*ln(c*x^2+1)/x^5-2/5*b^2*c^(5/2)*arctanh(x*c^(1/2))*ln(2/(1-x*c^(1/2)))- 2/5*b^2*c^(5/2)*arctan(x*c^(1/2))*ln(2/(1-I*x*c^(1/2)))+1/5*b^2*c^(5/2)*ar ctan(x*c^(1/2))*ln((1+I)*(1-x*c^(1/2))/(1-I*x*c^(1/2)))+2/5*b^2*c^(5/2)*ar ctan(x*c^(1/2))*ln(2/(1+I*x*c^(1/2)))+2/5*b^2*c^(5/2)*arctanh(x*c^(1/2))*l n(2/(1+x*c^(1/2)))-1/5*b^2*c^(5/2)*arctanh(x*c^(1/2))*ln(-2*(1-x*(-c)^(1/2 ))*c^(1/2)/((-c)^(1/2)-c^(1/2))/(1+x*c^(1/2)))-1/5*b^2*c^(5/2)*arctanh(x*c ^(1/2))*ln(2*(1+x*(-c)^(1/2))*c^(1/2)/((-c)^(1/2)+c^(1/2))/(1+x*c^(1/2)))+ 1/5*b^2*c^(5/2)*arctan(x*c^(1/2))*ln((1-I)*(1+x*c^(1/2))/(1-I*x*c^(1/2)))- 2/15*a*b*c/x^3+2/5*a*b*c^2/x-1/10*I*b^2*c^(5/2)*polylog(2,1-(1+I)*(1-x*c^( 1/2))/(1-I*x*c^(1/2)))-1/10*I*b^2*c^(5/2)*polylog(2,1+(-1+I)*(1+x*c^(1/2)) /(1-I*x*c^(1/2)))+1/5*I*b^2*c^(5/2)*arctan(x*c^(1/2))^2-1/20*(2*a-b*ln(-c* x^2+1))^2/x^5-4/15*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*ln(c*x^2+1)^2/...
\[ \int \frac {\left (a+b \text {arctanh}\left (c x^2\right )\right )^2}{x^6} \, dx=\int \frac {\left (a+b \text {arctanh}\left (c x^2\right )\right )^2}{x^6} \, dx \]
Time = 2.22 (sec) , antiderivative size = 1176, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {6456, 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.
\(\displaystyle \int \frac {\left (a+b \text {arctanh}\left (c x^2\right )\right )^2}{x^6} \, dx\) |
\(\Big \downarrow \) 6456 |
\(\displaystyle \int \left (\frac {\left (2 a-b \log \left (1-c x^2\right )\right )^2}{4 x^6}-\frac {b \log \left (c x^2+1\right ) \left (b \log \left (1-c x^2\right )-2 a\right )}{2 x^6}+\frac {b^2 \log ^2\left (c x^2+1\right )}{4 x^6}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{5} i b^2 \arctan \left (\sqrt {c} x\right )^2 c^{5/2}+\frac {1}{5} b^2 \text {arctanh}\left (\sqrt {c} x\right )^2 c^{5/2}-\frac {4}{15} b^2 \arctan \left (\sqrt {c} x\right ) c^{5/2}+\frac {2}{5} a b \arctan \left (\sqrt {c} x\right ) c^{5/2}+\frac {4}{15} b^2 \text {arctanh}\left (\sqrt {c} x\right ) c^{5/2}-\frac {2}{5} b^2 \text {arctanh}\left (\sqrt {c} x\right ) \log \left (\frac {2}{1-\sqrt {c} x}\right ) c^{5/2}-\frac {2}{5} b^2 \arctan \left (\sqrt {c} x\right ) \log \left (\frac {2}{1-i \sqrt {c} x}\right ) c^{5/2}+\frac {1}{5} b^2 \arctan \left (\sqrt {c} x\right ) \log \left (\frac {(1+i) \left (1-\sqrt {c} x\right )}{1-i \sqrt {c} x}\right ) c^{5/2}+\frac {2}{5} b^2 \arctan \left (\sqrt {c} x\right ) \log \left (\frac {2}{i \sqrt {c} x+1}\right ) c^{5/2}+\frac {2}{5} b^2 \text {arctanh}\left (\sqrt {c} x\right ) \log \left (\frac {2}{\sqrt {c} x+1}\right ) c^{5/2}-\frac {1}{5} b^2 \text {arctanh}\left (\sqrt {c} x\right ) \log \left (-\frac {2 \sqrt {c} \left (1-\sqrt {-c} x\right )}{\left (\sqrt {-c}-\sqrt {c}\right ) \left (\sqrt {c} x+1\right )}\right ) c^{5/2}-\frac {1}{5} b^2 \text {arctanh}\left (\sqrt {c} x\right ) \log \left (\frac {2 \sqrt {c} \left (\sqrt {-c} x+1\right )}{\left (\sqrt {-c}+\sqrt {c}\right ) \left (\sqrt {c} x+1\right )}\right ) c^{5/2}+\frac {1}{5} b^2 \arctan \left (\sqrt {c} x\right ) \log \left (\frac {(1-i) \left (\sqrt {c} x+1\right )}{1-i \sqrt {c} x}\right ) c^{5/2}-\frac {1}{5} b^2 \arctan \left (\sqrt {c} x\right ) \log \left (1-c x^2\right ) c^{5/2}+\frac {1}{5} b \text {arctanh}\left (\sqrt {c} x\right ) \left (2 a-b \log \left (1-c x^2\right )\right ) c^{5/2}+\frac {1}{5} b^2 \arctan \left (\sqrt {c} x\right ) \log \left (c x^2+1\right ) c^{5/2}+\frac {1}{5} b^2 \text {arctanh}\left (\sqrt {c} x\right ) \log \left (c x^2+1\right ) c^{5/2}-\frac {1}{5} b^2 \operatorname {PolyLog}\left (2,1-\frac {2}{1-\sqrt {c} x}\right ) c^{5/2}+\frac {1}{5} i b^2 \operatorname {PolyLog}\left (2,1-\frac {2}{1-i \sqrt {c} x}\right ) c^{5/2}-\frac {1}{10} i b^2 \operatorname {PolyLog}\left (2,1-\frac {(1+i) \left (1-\sqrt {c} x\right )}{1-i \sqrt {c} x}\right ) c^{5/2}+\frac {1}{5} i b^2 \operatorname {PolyLog}\left (2,1-\frac {2}{i \sqrt {c} x+1}\right ) c^{5/2}-\frac {1}{5} b^2 \operatorname {PolyLog}\left (2,1-\frac {2}{\sqrt {c} x+1}\right ) c^{5/2}+\frac {1}{10} b^2 \operatorname {PolyLog}\left (2,\frac {2 \sqrt {c} \left (1-\sqrt {-c} x\right )}{\left (\sqrt {-c}-\sqrt {c}\right ) \left (\sqrt {c} x+1\right )}+1\right ) c^{5/2}+\frac {1}{10} b^2 \operatorname {PolyLog}\left (2,1-\frac {2 \sqrt {c} \left (\sqrt {-c} x+1\right )}{\left (\sqrt {-c}+\sqrt {c}\right ) \left (\sqrt {c} x+1\right )}\right ) c^{5/2}-\frac {1}{10} i b^2 \operatorname {PolyLog}\left (2,1-\frac {(1-i) \left (\sqrt {c} x+1\right )}{1-i \sqrt {c} x}\right ) c^{5/2}-\frac {b^2 \log \left (1-c x^2\right ) c^2}{5 x}-\frac {b \left (2 a-b \log \left (1-c x^2\right )\right ) c^2}{5 x}-\frac {8 b^2 c^2}{15 x}+\frac {2 a b c^2}{5 x}+\frac {b^2 \log \left (1-c x^2\right ) c}{15 x^3}-\frac {b \left (2 a-b \log \left (1-c x^2\right )\right ) c}{15 x^3}-\frac {2 b^2 \log \left (c x^2+1\right ) c}{15 x^3}-\frac {2 a b c}{15 x^3}-\frac {\left (2 a-b \log \left (1-c x^2\right )\right )^2}{20 x^5}-\frac {b^2 \log ^2\left (c x^2+1\right )}{20 x^5}+\frac {b^2 \log \left (1-c x^2\right ) \log \left (c x^2+1\right )}{10 x^5}-\frac {a b \log \left (c x^2+1\right )}{5 x^5}\) |
(-2*a*b*c)/(15*x^3) + (2*a*b*c^2)/(5*x) - (8*b^2*c^2)/(15*x) + (2*a*b*c^(5 /2)*ArcTan[Sqrt[c]*x])/5 - (4*b^2*c^(5/2)*ArcTan[Sqrt[c]*x])/15 + (I/5)*b^ 2*c^(5/2)*ArcTan[Sqrt[c]*x]^2 + (4*b^2*c^(5/2)*ArcTanh[Sqrt[c]*x])/15 + (b ^2*c^(5/2)*ArcTanh[Sqrt[c]*x]^2)/5 - (2*b^2*c^(5/2)*ArcTanh[Sqrt[c]*x]*Log [2/(1 - Sqrt[c]*x)])/5 - (2*b^2*c^(5/2)*ArcTan[Sqrt[c]*x]*Log[2/(1 - I*Sqr t[c]*x)])/5 + (b^2*c^(5/2)*ArcTan[Sqrt[c]*x]*Log[((1 + I)*(1 - Sqrt[c]*x)) /(1 - I*Sqrt[c]*x)])/5 + (2*b^2*c^(5/2)*ArcTan[Sqrt[c]*x]*Log[2/(1 + I*Sqr t[c]*x)])/5 + (2*b^2*c^(5/2)*ArcTanh[Sqrt[c]*x]*Log[2/(1 + Sqrt[c]*x)])/5 - (b^2*c^(5/2)*ArcTanh[Sqrt[c]*x]*Log[(-2*Sqrt[c]*(1 - Sqrt[-c]*x))/((Sqrt [-c] - Sqrt[c])*(1 + Sqrt[c]*x))])/5 - (b^2*c^(5/2)*ArcTanh[Sqrt[c]*x]*Log [(2*Sqrt[c]*(1 + Sqrt[-c]*x))/((Sqrt[-c] + Sqrt[c])*(1 + Sqrt[c]*x))])/5 + (b^2*c^(5/2)*ArcTan[Sqrt[c]*x]*Log[((1 - I)*(1 + Sqrt[c]*x))/(1 - I*Sqrt[ c]*x)])/5 + (b^2*c*Log[1 - c*x^2])/(15*x^3) - (b^2*c^2*Log[1 - c*x^2])/(5* x) - (b^2*c^(5/2)*ArcTan[Sqrt[c]*x]*Log[1 - c*x^2])/5 - (b*c*(2*a - b*Log[ 1 - c*x^2]))/(15*x^3) - (b*c^2*(2*a - b*Log[1 - c*x^2]))/(5*x) + (b*c^(5/2 )*ArcTanh[Sqrt[c]*x]*(2*a - b*Log[1 - c*x^2]))/5 - (2*a - b*Log[1 - c*x^2] )^2/(20*x^5) - (a*b*Log[1 + c*x^2])/(5*x^5) - (2*b^2*c*Log[1 + c*x^2])/(15 *x^3) + (b^2*c^(5/2)*ArcTan[Sqrt[c]*x]*Log[1 + c*x^2])/5 + (b^2*c^(5/2)*Ar cTanh[Sqrt[c]*x]*Log[1 + c*x^2])/5 + (b^2*Log[1 - c*x^2]*Log[1 + c*x^2])/( 10*x^5) - (b^2*Log[1 + c*x^2]^2)/(20*x^5) - (b^2*c^(5/2)*PolyLog[2, 1 -...
3.1.76.3.1 Defintions of rubi rules used
Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_)]*(b_.))^(p_)*(x_)^(m_.), x_Symbol] :> Int[ExpandIntegrand[x^m*(a + b*(Log[1 + c*x^n]/2) - b*(Log[1 - c*x^n]/2))^p , x], x] /; FreeQ[{a, b, c}, x] && IGtQ[p, 1] && IGtQ[n, 0] && IntegerQ[m]
\[\int \frac {{\left (a +b \,\operatorname {arctanh}\left (c \,x^{2}\right )\right )}^{2}}{x^{6}}d x\]
\[ \int \frac {\left (a+b \text {arctanh}\left (c x^2\right )\right )^2}{x^6} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (c x^{2}\right ) + a\right )}^{2}}{x^{6}} \,d x } \]
\[ \int \frac {\left (a+b \text {arctanh}\left (c x^2\right )\right )^2}{x^6} \, dx=\int \frac {\left (a + b \operatorname {atanh}{\left (c x^{2} \right )}\right )^{2}}{x^{6}}\, dx \]
\[ \int \frac {\left (a+b \text {arctanh}\left (c x^2\right )\right )^2}{x^6} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (c x^{2}\right ) + a\right )}^{2}}{x^{6}} \,d x } \]
1/15*((6*c^(3/2)*arctan(sqrt(c)*x) - 3*c^(3/2)*log((c*x - sqrt(c))/(c*x + sqrt(c))) - 4/x^3)*c - 6*arctanh(c*x^2)/x^5)*a*b - 1/20*b^2*(log(-c*x^2 + 1)^2/x^5 + 5*integrate(-1/5*(5*(c*x^2 - 1)*log(c*x^2 + 1)^2 + 2*(2*c*x^2 - 5*(c*x^2 - 1)*log(c*x^2 + 1))*log(-c*x^2 + 1))/(c*x^8 - x^6), x)) - 1/5*a ^2/x^5
\[ \int \frac {\left (a+b \text {arctanh}\left (c x^2\right )\right )^2}{x^6} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (c x^{2}\right ) + a\right )}^{2}}{x^{6}} \,d x } \]
Timed out. \[ \int \frac {\left (a+b \text {arctanh}\left (c x^2\right )\right )^2}{x^6} \, dx=\int \frac {{\left (a+b\,\mathrm {atanh}\left (c\,x^2\right )\right )}^2}{x^6} \,d x \]