3.2.81 \(\int \frac {(a+b \text {arctanh}(\frac {c}{x^2}))^2}{x^6} \, dx\) [181]

3.2.81.1 Optimal result

Integrand size = 16, antiderivative size = 1337 \[ \int \frac {\left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2}{x^6} \, dx =\text {Too large to display} \]

output

-2/5*b^2*arctanh(x/c^(1/2))*ln(2-2*c^(1/2)/(x+c^(1/2)))/c^(5/2)+2/5*a*b*ar 
ctan(x/c^(1/2))/c^(5/2)+1/15*b^2*ln(1-c/x^2)/c/x^3-1/5*b^2*ln(1-c/x^2)/c^2 
/x-1/5*b^2*arctan(x/c^(1/2))*ln(1-c/x^2)/c^(5/2)-1/15*b*(2*a-b*ln(1-c/x^2) 
)/c/x^3-1/5*b*(2*a-b*ln(1-c/x^2))/c^2/x+1/5*b*arctanh(x/c^(1/2))*(2*a-b*ln 
(1-c/x^2))/c^(5/2)-1/5*a*b*ln(1+c/x^2)/x^5-2/15*b^2*ln(1+c/x^2)/c/x^3+1/5* 
b^2*arctan(x/c^(1/2))*ln(1+c/x^2)/c^(5/2)+1/5*b^2*arctanh(x/c^(1/2))*ln(1+ 
c/x^2)/c^(5/2)+1/10*b^2*ln(1-c/x^2)*ln(1+c/x^2)/x^5-2/5*b^2*arctan(x/c^(1/ 
2))*ln(2*c^(1/2)/(-I*x+c^(1/2)))/c^(5/2)+1/5*b^2*arctan(x/c^(1/2))*ln((1+I 
)*(-x+c^(1/2))/(-I*x+c^(1/2)))/c^(5/2)+2/5*b^2*arctanh(x/c^(1/2))*ln(2*c^( 
1/2)/(x+c^(1/2)))/c^(5/2)-1/5*b^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)))/c^(5/2)+1/5*b^2*arctan(x/c^(1/2) 
)*ln((1-I)*(x+c^(1/2))/(-I*x+c^(1/2)))/c^(5/2)-1/5*b^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)))/c^(5/2)+2/5* 
b^2*arctan(x/c^(1/2))*ln(2-2*c^(1/2)/(-I*x+c^(1/2)))/c^(5/2)-2/15*a*b/c/x^ 
3+2/5*a*b/c^2/x-1/5*I*b^2*arctan(x/c^(1/2))^2/c^(5/2)-1/5*I*b^2*polylog(2, 
-I*x/c^(1/2))/c^(5/2)-1/5*I*b^2*polylog(2,-1+2*c^(1/2)/(-I*x+c^(1/2)))/c^( 
5/2)-1/10*I*b^2*polylog(2,1-(1+I)*(-x+c^(1/2))/(-I*x+c^(1/2)))/c^(5/2)-1/1 
0*I*b^2*polylog(2,1+(-1+I)*(x+c^(1/2))/(-I*x+c^(1/2)))/c^(5/2)+1/10*b^2*po 
lylog(2,1-2*(-x+(-c)^(1/2))*c^(1/2)/((-c)^(1/2)-c^(1/2))/(x+c^(1/2)))/c^(5 
/2)+1/10*b^2*polylog(2,1-2*(x+(-c)^(1/2))*c^(1/2)/(x+c^(1/2))/((-c)^(1/...
 

3.2.81.2 Mathematica [F]

\[ \int \frac {\left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2}{x^6} \, dx=\int \frac {\left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2}{x^6} \, dx \]

input

Integrate[(a + b*ArcTanh[c/x^2])^2/x^6,x]
 

output

Integrate[(a + b*ArcTanh[c/x^2])^2/x^6, x]
 

3.2.81.3 Rubi [A] (verified)

Time = 2.64 (sec) , antiderivative size = 1337, 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.

\(\displaystyle \int \frac {\left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2}{x^6} \, dx\)

\(\Big \downarrow \) 6460

\(\displaystyle \int \frac {\left (a+b \coth ^{-1}\left (\frac {x^2}{c}\right )\right )^2}{x^6}dx\)

\(\Big \downarrow \) 6457

\(\displaystyle \int \left (\frac {\left (2 a-b \log \left (1-\frac {c}{x^2}\right )\right )^2}{4 x^6}-\frac {b \log \left (\frac {c}{x^2}+1\right ) \left (b \log \left (1-\frac {c}{x^2}\right )-2 a\right )}{2 x^6}+\frac {b^2 \log ^2\left (\frac {c}{x^2}+1\right )}{4 x^6}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {i \arctan \left (\frac {x}{\sqrt {c}}\right )^2 b^2}{5 c^{5/2}}-\frac {\text {arctanh}\left (\frac {x}{\sqrt {c}}\right )^2 b^2}{5 c^{5/2}}-\frac {\log ^2\left (\frac {c}{x^2}+1\right ) b^2}{20 x^5}-\frac {4 \arctan \left (\frac {x}{\sqrt {c}}\right ) b^2}{15 c^{5/2}}+\frac {4 \text {arctanh}\left (\frac {x}{\sqrt {c}}\right ) b^2}{15 c^{5/2}}+\frac {2 \arctan \left (\frac {x}{\sqrt {c}}\right ) \log \left (2-\frac {2 \sqrt {c}}{\sqrt {c}-i x}\right ) b^2}{5 c^{5/2}}-\frac {\arctan \left (\frac {x}{\sqrt {c}}\right ) \log \left (1-\frac {c}{x^2}\right ) b^2}{5 c^{5/2}}-\frac {\log \left (1-\frac {c}{x^2}\right ) b^2}{5 c^2 x}+\frac {\log \left (1-\frac {c}{x^2}\right ) b^2}{15 c x^3}-\frac {\log \left (1-\frac {c}{x^2}\right ) b^2}{25 x^5}+\frac {\arctan \left (\frac {x}{\sqrt {c}}\right ) \log \left (\frac {c}{x^2}+1\right ) b^2}{5 c^{5/2}}+\frac {\text {arctanh}\left (\frac {x}{\sqrt {c}}\right ) \log \left (\frac {c}{x^2}+1\right ) b^2}{5 c^{5/2}}+\frac {\log \left (1-\frac {c}{x^2}\right ) \log \left (\frac {c}{x^2}+1\right ) b^2}{10 x^5}-\frac {2 \log \left (\frac {c}{x^2}+1\right ) b^2}{15 c x^3}-\frac {2 \arctan \left (\frac {x}{\sqrt {c}}\right ) \log \left (\frac {2 \sqrt {c}}{\sqrt {c}-i x}\right ) b^2}{5 c^{5/2}}+\frac {\arctan \left (\frac {x}{\sqrt {c}}\right ) \log \left (\frac {(1+i) \left (\sqrt {c}-x\right )}{\sqrt {c}-i x}\right ) b^2}{5 c^{5/2}}+\frac {2 \text {arctanh}\left (\frac {x}{\sqrt {c}}\right ) \log \left (\frac {2 \sqrt {c}}{x+\sqrt {c}}\right ) b^2}{5 c^{5/2}}-\frac {\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 ) b^2}{5 c^{5/2}}-\frac {\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 ) b^2}{5 c^{5/2}}+\frac {\arctan \left (\frac {x}{\sqrt {c}}\right ) \log \left (\frac {(1-i) \left (x+\sqrt {c}\right )}{\sqrt {c}-i x}\right ) b^2}{5 c^{5/2}}-\frac {2 \text {arctanh}\left (\frac {x}{\sqrt {c}}\right ) \log \left (2-\frac {2 \sqrt {c}}{x+\sqrt {c}}\right ) b^2}{5 c^{5/2}}+\frac {i \operatorname {PolyLog}\left (2,1-\frac {2 \sqrt {c}}{\sqrt {c}-i x}\right ) b^2}{5 c^{5/2}}-\frac {i \operatorname {PolyLog}\left (2,\frac {2 \sqrt {c}}{\sqrt {c}-i x}-1\right ) b^2}{5 c^{5/2}}-\frac {i \operatorname {PolyLog}\left (2,1-\frac {(1+i) \left (\sqrt {c}-x\right )}{\sqrt {c}-i x}\right ) b^2}{10 c^{5/2}}-\frac {\operatorname {PolyLog}\left (2,-\frac {x}{\sqrt {c}}\right ) b^2}{5 c^{5/2}}-\frac {i \operatorname {PolyLog}\left (2,-\frac {i x}{\sqrt {c}}\right ) b^2}{5 c^{5/2}}+\frac {i \operatorname {PolyLog}\left (2,\frac {i x}{\sqrt {c}}\right ) b^2}{5 c^{5/2}}+\frac {\operatorname {PolyLog}\left (2,\frac {x}{\sqrt {c}}\right ) b^2}{5 c^{5/2}}-\frac {\operatorname {PolyLog}\left (2,1-\frac {2 \sqrt {c}}{x+\sqrt {c}}\right ) b^2}{5 c^{5/2}}+\frac {\operatorname {PolyLog}\left (2,\frac {2 \sqrt {c}}{x+\sqrt {c}}-1\right ) b^2}{5 c^{5/2}}+\frac {\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 ) b^2}{10 c^{5/2}}+\frac {\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 ) b^2}{10 c^{5/2}}-\frac {i \operatorname {PolyLog}\left (2,1-\frac {(1-i) \left (x+\sqrt {c}\right )}{\sqrt {c}-i x}\right ) b^2}{10 c^{5/2}}-\frac {8 b^2}{15 c^2 x}+\frac {2 a \arctan \left (\frac {x}{\sqrt {c}}\right ) b}{5 c^{5/2}}+\frac {\text {arctanh}\left (\frac {x}{\sqrt {c}}\right ) \left (2 a-b \log \left (1-\frac {c}{x^2}\right )\right ) b}{5 c^{5/2}}-\frac {\left (2 a-b \log \left (1-\frac {c}{x^2}\right )\right ) b}{5 c^2 x}-\frac {\left (2 a-b \log \left (1-\frac {c}{x^2}\right )\right ) b}{15 c x^3}-\frac {\left (2 a-b \log \left (1-\frac {c}{x^2}\right )\right ) b}{25 x^5}-\frac {a \log \left (\frac {c}{x^2}+1\right ) b}{5 x^5}+\frac {2 a b}{5 c^2 x}-\frac {2 a b}{15 c x^3}+\frac {2 a b}{25 x^5}-\frac {\left (2 a-b \log \left (1-\frac {c}{x^2}\right )\right )^2}{20 x^5}\)

input

Int[(a + b*ArcTanh[c/x^2])^2/x^6,x]
 

output

(2*a*b)/(25*x^5) - (2*a*b)/(15*c*x^3) + (2*a*b)/(5*c^2*x) - (8*b^2)/(15*c^ 
2*x) + (2*a*b*ArcTan[x/Sqrt[c]])/(5*c^(5/2)) - (4*b^2*ArcTan[x/Sqrt[c]])/( 
15*c^(5/2)) - ((I/5)*b^2*ArcTan[x/Sqrt[c]]^2)/c^(5/2) + (4*b^2*ArcTanh[x/S 
qrt[c]])/(15*c^(5/2)) - (b^2*ArcTanh[x/Sqrt[c]]^2)/(5*c^(5/2)) + (2*b^2*Ar 
cTan[x/Sqrt[c]]*Log[2 - (2*Sqrt[c])/(Sqrt[c] - I*x)])/(5*c^(5/2)) - (b^2*L 
og[1 - c/x^2])/(25*x^5) + (b^2*Log[1 - c/x^2])/(15*c*x^3) - (b^2*Log[1 - c 
/x^2])/(5*c^2*x) - (b^2*ArcTan[x/Sqrt[c]]*Log[1 - c/x^2])/(5*c^(5/2)) - (b 
*(2*a - b*Log[1 - c/x^2]))/(25*x^5) - (b*(2*a - b*Log[1 - c/x^2]))/(15*c*x 
^3) - (b*(2*a - b*Log[1 - c/x^2]))/(5*c^2*x) + (b*ArcTanh[x/Sqrt[c]]*(2*a 
- b*Log[1 - c/x^2]))/(5*c^(5/2)) - (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*Log[1 + c/x^2])/(15*c*x^3) + (b^2*Arc 
Tan[x/Sqrt[c]]*Log[1 + c/x^2])/(5*c^(5/2)) + (b^2*ArcTanh[x/Sqrt[c]]*Log[1 
 + c/x^2])/(5*c^(5/2)) + (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) - (2*b^2*ArcTan[x/Sqrt[c]]*Log[(2*Sqrt[c])/( 
Sqrt[c] - I*x)])/(5*c^(5/2)) + (b^2*ArcTan[x/Sqrt[c]]*Log[((1 + I)*(Sqrt[c 
] - x))/(Sqrt[c] - I*x)])/(5*c^(5/2)) + (2*b^2*ArcTanh[x/Sqrt[c]]*Log[(2*S 
qrt[c])/(Sqrt[c] + x)])/(5*c^(5/2)) - (b^2*ArcTanh[x/Sqrt[c]]*Log[(2*Sqrt[ 
c]*(Sqrt[-c] - x))/((Sqrt[-c] - Sqrt[c])*(Sqrt[c] + x))])/(5*c^(5/2)) - (b 
^2*ArcTanh[x/Sqrt[c]]*Log[(2*Sqrt[c]*(Sqrt[-c] + x))/((Sqrt[-c] + Sqrt[c]) 
*(Sqrt[c] + x))])/(5*c^(5/2)) + (b^2*ArcTan[x/Sqrt[c]]*Log[((1 - I)*(Sq...
 

3.2.81.3.1 Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 6457

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]
 

rule 6460

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]
 

3.2.81.4 Maple [F]

\[\int \frac {{\left (a +b \,\operatorname {arctanh}\left (\frac {c}{x^{2}}\right )\right )}^{2}}{x^{6}}d x\]

input

int((a+b*arctanh(c/x^2))^2/x^6,x)
 

output

int((a+b*arctanh(c/x^2))^2/x^6,x)
 

3.2.81.5 Fricas [F]

\[ \int \frac {\left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2}{x^6} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (\frac {c}{x^{2}}\right ) + a\right )}^{2}}{x^{6}} \,d x } \]

input

integrate((a+b*arctanh(c/x^2))^2/x^6,x, algorithm="fricas")
 

output

integral((b^2*arctanh(c/x^2)^2 + 2*a*b*arctanh(c/x^2) + a^2)/x^6, x)
 

3.2.81.6 Sympy [F]

\[ \int \frac {\left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2}{x^6} \, dx=\int \frac {\left (a + b \operatorname {atanh}{\left (\frac {c}{x^{2}} \right )}\right )^{2}}{x^{6}}\, dx \]

input

integrate((a+b*atanh(c/x**2))**2/x**6,x)
 

output

Integral((a + b*atanh(c/x**2))**2/x**6, x)
 

3.2.81.7 Maxima [F]

\[ \int \frac {\left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2}{x^6} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (\frac {c}{x^{2}}\right ) + a\right )}^{2}}{x^{6}} \,d x } \]

input

integrate((a+b*arctanh(c/x^2))^2/x^6,x, algorithm="maxima")
 

output

1/15*(c*(6*arctan(x/sqrt(c))/c^(7/2) - 3*log((x - sqrt(c))/(x + sqrt(c)))/ 
c^(7/2) - 4/(c^2*x^3)) - 6*arctanh(c/x^2)/x^5)*a*b - 1/20*b^2*(log(x^2 - c 
)^2/x^5 + 5*integrate(-1/5*(5*(x^2 - c)*log(x^2 + c)^2 + 2*(2*x^2 - 5*(x^2 
 - c)*log(x^2 + c))*log(x^2 - c))/(x^8 - c*x^6), x)) - 1/5*a^2/x^5
 

3.2.81.8 Giac [F]

\[ \int \frac {\left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2}{x^6} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (\frac {c}{x^{2}}\right ) + a\right )}^{2}}{x^{6}} \,d x } \]

input

integrate((a+b*arctanh(c/x^2))^2/x^6,x, algorithm="giac")
 

output

integrate((b*arctanh(c/x^2) + a)^2/x^6, x)
 

3.2.81.9 Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2}{x^6} \, dx=\int \frac {{\left (a+b\,\mathrm {atanh}\left (\frac {c}{x^2}\right )\right )}^2}{x^6} \,d x \]

input

int((a + b*atanh(c/x^2))^2/x^6,x)
 

output

int((a + b*atanh(c/x^2))^2/x^6, x)