\(\int \frac {(a+b \text {arctanh}(\frac {c}{x^2}))^2}{x^4} \, dx\) [180]

Optimal result

Integrand size = 16, antiderivative size = 1263 \[ \int \frac {\left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2}{x^4} \, dx =\text {Too large to display} \] Output:

1/3*b^2*arctanh(x/c^(1/2))*ln(1+c/x^2)/c^(3/2)-1/3*b^2*arctan(x/c^(1/2))*l 
n(1+c/x^2)/c^(3/2)+1/3*b*arctanh(x/c^(1/2))*(2*a-b*ln(1-c/x^2))/c^(3/2)+1/ 
3*b^2*arctan(x/c^(1/2))*ln(1-c/x^2)/c^(3/2)-1/3*I*b^2*polylog(2,1-2*c^(1/2 
)/(c^(1/2)-I*x))/c^(3/2)-1/3*I*b^2*polylog(2,I*x/c^(1/2))/c^(3/2)-1/3*b^2* 
arctanh(x/c^(1/2))*ln(2*c^(1/2)*((-c)^(1/2)+x)/((-c)^(1/2)+c^(1/2))/(c^(1/ 
2)+x))/c^(3/2)-1/3*b^2*arctanh(x/c^(1/2))*ln(2*c^(1/2)*((-c)^(1/2)-x)/((-c 
)^(1/2)-c^(1/2))/(c^(1/2)+x))/c^(3/2)-2/3*b^2*arctanh(x/c^(1/2))*ln(2-2*c^ 
(1/2)/(c^(1/2)+x))/c^(3/2)+2/3*b^2*arctanh(x/c^(1/2))*ln(2*c^(1/2)/(c^(1/2 
)+x))/c^(3/2)+1/6*I*b^2*polylog(2,1+(-1+I)*(c^(1/2)+x)/(c^(1/2)-I*x))/c^(3 
/2)+1/6*I*b^2*polylog(2,1-(1+I)*(c^(1/2)-x)/(c^(1/2)-I*x))/c^(3/2)+1/3*I*b 
^2*polylog(2,-1+2*c^(1/2)/(c^(1/2)-I*x))/c^(3/2)+1/3*I*b^2*polylog(2,-I*x/ 
c^(1/2))/c^(3/2)+1/3*I*b^2*arctan(x/c^(1/2))^2/c^(3/2)-1/3*b^2*arctan(x/c^ 
(1/2))*ln((1+I)*(c^(1/2)-x)/(c^(1/2)-I*x))/c^(3/2)-1/3*b^2*arctan(x/c^(1/2 
))*ln((1-I)*(c^(1/2)+x)/(c^(1/2)-I*x))/c^(3/2)-2/3*b^2*arctan(x/c^(1/2))*l 
n(2-2*c^(1/2)/(c^(1/2)-I*x))/c^(3/2)+2/3*b^2*arctan(x/c^(1/2))*ln(2*c^(1/2 
)/(c^(1/2)-I*x))/c^(3/2)-2/3*a*b*arctan(x/c^(1/2))/c^(3/2)-1/12*(2*a-b*ln( 
1-c/x^2))^2/x^3+4/3*b^2*arctanh(x/c^(1/2))/c^(3/2)-1/3*b^2*arctanh(x/c^(1/ 
2))^2/c^(3/2)+4/3*b^2*arctan(x/c^(1/2))/c^(3/2)+1/3*b^2*polylog(2,x/c^(1/2 
))/c^(3/2)-1/3*b^2*polylog(2,-x/c^(1/2))/c^(3/2)-1/3*b^2*polylog(2,1-2*c^( 
1/2)/(c^(1/2)+x))/c^(3/2)+1/3*b^2*polylog(2,-1+2*c^(1/2)/(c^(1/2)+x))/c...
 

Mathematica [F]

\[ \int \frac {\left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2}{x^4} \, dx=\int \frac {\left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2}{x^4} \, dx \] Input:

Integrate[(a + b*ArcTanh[c/x^2])^2/x^4,x]
 

Output:

Integrate[(a + b*ArcTanh[c/x^2])^2/x^4, x]
 

Rubi [A] (verified)

Time = 2.53 (sec) , antiderivative size = 1263, 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.

\(\displaystyle \int \frac {\left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2}{x^4} \, dx\)

\(\Big \downarrow \) 6460

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

\(\Big \downarrow \) 6457

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {i \arctan \left (\frac {x}{\sqrt {c}}\right )^2 b^2}{3 c^{3/2}}-\frac {\text {arctanh}\left (\frac {x}{\sqrt {c}}\right )^2 b^2}{3 c^{3/2}}-\frac {\log ^2\left (\frac {c}{x^2}+1\right ) b^2}{12 x^3}+\frac {4 \arctan \left (\frac {x}{\sqrt {c}}\right ) b^2}{3 c^{3/2}}+\frac {4 \text {arctanh}\left (\frac {x}{\sqrt {c}}\right ) b^2}{3 c^{3/2}}-\frac {2 \arctan \left (\frac {x}{\sqrt {c}}\right ) \log \left (2-\frac {2 \sqrt {c}}{\sqrt {c}-i x}\right ) b^2}{3 c^{3/2}}+\frac {\arctan \left (\frac {x}{\sqrt {c}}\right ) \log \left (1-\frac {c}{x^2}\right ) b^2}{3 c^{3/2}}+\frac {\log \left (1-\frac {c}{x^2}\right ) b^2}{3 c x}-\frac {\log \left (1-\frac {c}{x^2}\right ) b^2}{9 x^3}-\frac {\arctan \left (\frac {x}{\sqrt {c}}\right ) \log \left (\frac {c}{x^2}+1\right ) b^2}{3 c^{3/2}}+\frac {\text {arctanh}\left (\frac {x}{\sqrt {c}}\right ) \log \left (\frac {c}{x^2}+1\right ) b^2}{3 c^{3/2}}+\frac {\log \left (1-\frac {c}{x^2}\right ) \log \left (\frac {c}{x^2}+1\right ) b^2}{6 x^3}-\frac {2 \log \left (\frac {c}{x^2}+1\right ) b^2}{3 c x}+\frac {2 \arctan \left (\frac {x}{\sqrt {c}}\right ) \log \left (\frac {2 \sqrt {c}}{\sqrt {c}-i x}\right ) b^2}{3 c^{3/2}}-\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}{3 c^{3/2}}+\frac {2 \text {arctanh}\left (\frac {x}{\sqrt {c}}\right ) \log \left (\frac {2 \sqrt {c}}{x+\sqrt {c}}\right ) b^2}{3 c^{3/2}}-\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}{3 c^{3/2}}-\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}{3 c^{3/2}}-\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}{3 c^{3/2}}-\frac {2 \text {arctanh}\left (\frac {x}{\sqrt {c}}\right ) \log \left (2-\frac {2 \sqrt {c}}{x+\sqrt {c}}\right ) b^2}{3 c^{3/2}}-\frac {i \operatorname {PolyLog}\left (2,1-\frac {2 \sqrt {c}}{\sqrt {c}-i x}\right ) b^2}{3 c^{3/2}}+\frac {i \operatorname {PolyLog}\left (2,\frac {2 \sqrt {c}}{\sqrt {c}-i x}-1\right ) b^2}{3 c^{3/2}}+\frac {i \operatorname {PolyLog}\left (2,1-\frac {(1+i) \left (\sqrt {c}-x\right )}{\sqrt {c}-i x}\right ) b^2}{6 c^{3/2}}-\frac {\operatorname {PolyLog}\left (2,-\frac {x}{\sqrt {c}}\right ) b^2}{3 c^{3/2}}+\frac {i \operatorname {PolyLog}\left (2,-\frac {i x}{\sqrt {c}}\right ) b^2}{3 c^{3/2}}-\frac {i \operatorname {PolyLog}\left (2,\frac {i x}{\sqrt {c}}\right ) b^2}{3 c^{3/2}}+\frac {\operatorname {PolyLog}\left (2,\frac {x}{\sqrt {c}}\right ) b^2}{3 c^{3/2}}-\frac {\operatorname {PolyLog}\left (2,1-\frac {2 \sqrt {c}}{x+\sqrt {c}}\right ) b^2}{3 c^{3/2}}+\frac {\operatorname {PolyLog}\left (2,\frac {2 \sqrt {c}}{x+\sqrt {c}}-1\right ) b^2}{3 c^{3/2}}+\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}{6 c^{3/2}}+\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}{6 c^{3/2}}+\frac {i \operatorname {PolyLog}\left (2,1-\frac {(1-i) \left (x+\sqrt {c}\right )}{\sqrt {c}-i x}\right ) b^2}{6 c^{3/2}}-\frac {2 a \arctan \left (\frac {x}{\sqrt {c}}\right ) b}{3 c^{3/2}}+\frac {\text {arctanh}\left (\frac {x}{\sqrt {c}}\right ) \left (2 a-b \log \left (1-\frac {c}{x^2}\right )\right ) b}{3 c^{3/2}}-\frac {\left (2 a-b \log \left (1-\frac {c}{x^2}\right )\right ) b}{3 c x}-\frac {\left (2 a-b \log \left (1-\frac {c}{x^2}\right )\right ) b}{9 x^3}-\frac {a \log \left (\frac {c}{x^2}+1\right ) b}{3 x^3}-\frac {2 a b}{3 c x}+\frac {2 a b}{9 x^3}-\frac {\left (2 a-b \log \left (1-\frac {c}{x^2}\right )\right )^2}{12 x^3}\)

Input:

Int[(a + b*ArcTanh[c/x^2])^2/x^4,x]
 

Output:

(2*a*b)/(9*x^3) - (2*a*b)/(3*c*x) - (2*a*b*ArcTan[x/Sqrt[c]])/(3*c^(3/2)) 
+ (4*b^2*ArcTan[x/Sqrt[c]])/(3*c^(3/2)) + ((I/3)*b^2*ArcTan[x/Sqrt[c]]^2)/ 
c^(3/2) + (4*b^2*ArcTanh[x/Sqrt[c]])/(3*c^(3/2)) - (b^2*ArcTanh[x/Sqrt[c]] 
^2)/(3*c^(3/2)) - (2*b^2*ArcTan[x/Sqrt[c]]*Log[2 - (2*Sqrt[c])/(Sqrt[c] - 
I*x)])/(3*c^(3/2)) - (b^2*Log[1 - c/x^2])/(9*x^3) + (b^2*Log[1 - c/x^2])/( 
3*c*x) + (b^2*ArcTan[x/Sqrt[c]]*Log[1 - c/x^2])/(3*c^(3/2)) - (b*(2*a - b* 
Log[1 - c/x^2]))/(9*x^3) - (b*(2*a - b*Log[1 - c/x^2]))/(3*c*x) + (b*ArcTa 
nh[x/Sqrt[c]]*(2*a - b*Log[1 - c/x^2]))/(3*c^(3/2)) - (2*a - b*Log[1 - c/x 
^2])^2/(12*x^3) - (a*b*Log[1 + c/x^2])/(3*x^3) - (2*b^2*Log[1 + c/x^2])/(3 
*c*x) - (b^2*ArcTan[x/Sqrt[c]]*Log[1 + c/x^2])/(3*c^(3/2)) + (b^2*ArcTanh[ 
x/Sqrt[c]]*Log[1 + c/x^2])/(3*c^(3/2)) + (b^2*Log[1 - c/x^2]*Log[1 + c/x^2 
])/(6*x^3) - (b^2*Log[1 + c/x^2]^2)/(12*x^3) + (2*b^2*ArcTan[x/Sqrt[c]]*Lo 
g[(2*Sqrt[c])/(Sqrt[c] - I*x)])/(3*c^(3/2)) - (b^2*ArcTan[x/Sqrt[c]]*Log[( 
(1 + I)*(Sqrt[c] - x))/(Sqrt[c] - I*x)])/(3*c^(3/2)) + (2*b^2*ArcTanh[x/Sq 
rt[c]]*Log[(2*Sqrt[c])/(Sqrt[c] + x)])/(3*c^(3/2)) - (b^2*ArcTanh[x/Sqrt[c 
]]*Log[(2*Sqrt[c]*(Sqrt[-c] - x))/((Sqrt[-c] - Sqrt[c])*(Sqrt[c] + x))])/( 
3*c^(3/2)) - (b^2*ArcTanh[x/Sqrt[c]]*Log[(2*Sqrt[c]*(Sqrt[-c] + x))/((Sqrt 
[-c] + Sqrt[c])*(Sqrt[c] + x))])/(3*c^(3/2)) - (b^2*ArcTan[x/Sqrt[c]]*Log[ 
((1 - I)*(Sqrt[c] + x))/(Sqrt[c] - I*x)])/(3*c^(3/2)) - (2*b^2*ArcTanh[x/S 
qrt[c]]*Log[2 - (2*Sqrt[c])/(Sqrt[c] + x)])/(3*c^(3/2)) - ((I/3)*b^2*Po...
 

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]
 

Maple [F]

\[\int \frac {{\left (a +b \,\operatorname {arctanh}\left (\frac {c}{x^{2}}\right )\right )}^{2}}{x^{4}}d x\]

Input:

int((a+b*arctanh(c/x^2))^2/x^4,x)
 

Output:

int((a+b*arctanh(c/x^2))^2/x^4,x)
 

Fricas [F]

\[ \int \frac {\left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2}{x^4} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (\frac {c}{x^{2}}\right ) + a\right )}^{2}}{x^{4}} \,d x } \] Input:

integrate((a+b*arctanh(c/x^2))^2/x^4,x, algorithm="fricas")
 

Output:

integral((b^2*arctanh(c/x^2)^2 + 2*a*b*arctanh(c/x^2) + a^2)/x^4, x)
 

Sympy [F]

\[ \int \frac {\left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2}{x^4} \, dx=\int \frac {\left (a + b \operatorname {atanh}{\left (\frac {c}{x^{2}} \right )}\right )^{2}}{x^{4}}\, dx \] Input:

integrate((a+b*atanh(c/x**2))**2/x**4,x)
 

Output:

Integral((a + b*atanh(c/x**2))**2/x**4, x)
 

Maxima [F]

\[ \int \frac {\left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2}{x^4} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (\frac {c}{x^{2}}\right ) + a\right )}^{2}}{x^{4}} \,d x } \] Input:

integrate((a+b*arctanh(c/x^2))^2/x^4,x, algorithm="maxima")
 

Output:

-1/3*(c*(2*arctan(x/sqrt(c))/c^(5/2) + log((x - sqrt(c))/(x + sqrt(c)))/c^ 
(5/2) + 4/(c^2*x)) + 2*arctanh(c/x^2)/x^3)*a*b - 1/12*b^2*(log(x^2 - c)^2/ 
x^3 + 3*integrate(-1/3*(3*(x^2 - c)*log(x^2 + c)^2 + 2*(2*x^2 - 3*(x^2 - c 
)*log(x^2 + c))*log(x^2 - c))/(x^6 - c*x^4), x)) - 1/3*a^2/x^3
 

Giac [F]

\[ \int \frac {\left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2}{x^4} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (\frac {c}{x^{2}}\right ) + a\right )}^{2}}{x^{4}} \,d x } \] Input:

integrate((a+b*arctanh(c/x^2))^2/x^4,x, algorithm="giac")
 

Output:

integrate((b*arctanh(c/x^2) + a)^2/x^4, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2}{x^4} \, dx=\int \frac {{\left (a+b\,\mathrm {atanh}\left (\frac {c}{x^2}\right )\right )}^2}{x^4} \,d x \] Input:

int((a + b*atanh(c/x^2))^2/x^4,x)
 

Output:

int((a + b*atanh(c/x^2))^2/x^4, x)
 

Reduce [F]

\[ \int \frac {\left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2}{x^4} \, dx=\frac {-2 \sqrt {c}\, \mathit {atan} \left (\frac {x}{\sqrt {c}}\right ) a b \,x^{3}-2 \sqrt {c}\, \mathit {atanh} \left (\frac {c}{x^{2}}\right ) a b \,x^{3}-2 \mathit {atanh} \left (\frac {c}{x^{2}}\right ) a b \,c^{2}-2 \sqrt {c}\, \mathrm {log}\left (\sqrt {c}-x \right ) a b \,x^{3}+\sqrt {c}\, \mathrm {log}\left (x^{2}+c \right ) a b \,x^{3}+3 \left (\int \frac {\mathit {atanh} \left (\frac {c}{x^{2}}\right )^{2}}{x^{4}}d x \right ) b^{2} c^{2} x^{3}-a^{2} c^{2}-4 a b c \,x^{2}}{3 c^{2} x^{3}} \] Input:

int((a+b*atanh(c/x^2))^2/x^4,x)
 

Output:

( - 2*sqrt(c)*atan(x/sqrt(c))*a*b*x**3 - 2*sqrt(c)*atanh(c/x**2)*a*b*x**3 
- 2*atanh(c/x**2)*a*b*c**2 - 2*sqrt(c)*log(sqrt(c) - x)*a*b*x**3 + sqrt(c) 
*log(c + x**2)*a*b*x**3 + 3*int(atanh(c/x**2)**2/x**4,x)*b**2*c**2*x**3 - 
a**2*c**2 - 4*a*b*c*x**2)/(3*c**2*x**3)