\(\int \frac {(a+b \text {arctanh}(c x^2))^2}{x^6} \, dx\) [76]

Optimal result

Integrand size = 16, antiderivative size = 1176 \[ \int \frac {\left (a+b \text {arctanh}\left (c x^2\right )\right )^2}{x^6} \, dx =\text {Too large to display} \] Output:

-1/5*a*b*ln(c*x^2+1)/x^5-2/15*b^2*c*ln(c*x^2+1)/x^3+1/10*b^2*ln(-c*x^2+1)* 
ln(c*x^2+1)/x^5+1/15*b^2*c*ln(-c*x^2+1)/x^3-1/5*b^2*c^2*ln(-c*x^2+1)/x-1/1 
5*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-2/15*a*b*c 
/x^3+2/5*a*b*c^2/x+4/15*b^2*c^(5/2)*arctanh(c^(1/2)*x)+1/5*b^2*c^(5/2)*arc 
tanh(c^(1/2)*x)^2-4/15*b^2*c^(5/2)*arctan(c^(1/2)*x)+1/10*b^2*c^(5/2)*poly 
log(2,1-2*c^(1/2)*(1+(-c)^(1/2)*x)/((-c)^(1/2)+c^(1/2))/(1+c^(1/2)*x))+1/1 
0*b^2*c^(5/2)*polylog(2,1+2*c^(1/2)*(1-(-c)^(1/2)*x)/((-c)^(1/2)-c^(1/2))/ 
(1+c^(1/2)*x))-1/5*b^2*c^(5/2)*polylog(2,1-2/(1+c^(1/2)*x))+1/5*I*b^2*c^(5 
/2)*polylog(2,1-2/(1+I*c^(1/2)*x))+1/5*I*b^2*c^(5/2)*polylog(2,1-2/(1-I*c^ 
(1/2)*x))+1/5*I*b^2*c^(5/2)*arctan(c^(1/2)*x)^2-2/5*b^2*c^(5/2)*arctanh(c^ 
(1/2)*x)*ln(2/(1-c^(1/2)*x))+1/5*b^2*c^(5/2)*arctan(c^(1/2)*x)*ln((1+I)*(1 
-c^(1/2)*x)/(1-I*c^(1/2)*x))+1/5*b^2*c^(5/2)*arctan(c^(1/2)*x)*ln((1-I)*(1 
+c^(1/2)*x)/(1-I*c^(1/2)*x))+2/5*b^2*c^(5/2)*arctan(c^(1/2)*x)*ln(2/(1+I*c 
^(1/2)*x))-2/5*b^2*c^(5/2)*arctan(c^(1/2)*x)*ln(2/(1-I*c^(1/2)*x))+2/5*a*b 
*c^(5/2)*arctan(c^(1/2)*x)+1/5*b^2*c^(5/2)*arctanh(c^(1/2)*x)*ln(c*x^2+1)+ 
1/5*b^2*c^(5/2)*arctan(c^(1/2)*x)*ln(c*x^2+1)+1/5*b*c^(5/2)*arctanh(c^(1/2 
)*x)*(2*a-b*ln(-c*x^2+1))-1/5*b^2*c^(5/2)*arctan(c^(1/2)*x)*ln(-c*x^2+1)-1 
/10*I*b^2*c^(5/2)*polylog(2,1+(-1+I)*(1+c^(1/2)*x)/(1-I*c^(1/2)*x))-1/10*I 
*b^2*c^(5/2)*polylog(2,1-(1+I)*(1-c^(1/2)*x)/(1-I*c^(1/2)*x))-1/5*b^2*c^(5 
/2)*arctanh(c^(1/2)*x)*ln(2*c^(1/2)*(1+(-c)^(1/2)*x)/((-c)^(1/2)+c^(1/2...
 

Mathematica [F]

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

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

Output:

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

Rubi [A] (verified)

Time = 2.36 (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}\)

Input:

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

Output:

(-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 -...
 

Defintions of rubi rules used

rule 2009

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

rule 6456

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]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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
 

Giac [F]

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\left (a+b \text {arctanh}\left (c x^2\right )\right )^2}{x^6} \, dx=\frac {6 \sqrt {c}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}}\right ) a b \,c^{2} x^{5}-6 \sqrt {c}\, \mathit {atanh} \left (c \,x^{2}\right ) a b \,c^{2} x^{5}-6 \mathit {atanh} \left (c \,x^{2}\right ) a b -6 \sqrt {c}\, \mathrm {log}\left (\sqrt {c}\, x -1\right ) a b \,c^{2} x^{5}+3 \sqrt {c}\, \mathrm {log}\left (c \,x^{2}+1\right ) a b \,c^{2} x^{5}+15 \left (\int \frac {\mathit {atanh} \left (c \,x^{2}\right )^{2}}{x^{6}}d x \right ) b^{2} x^{5}-3 a^{2}-4 a b c \,x^{2}}{15 x^{5}} \] Input:

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

Output:

(6*sqrt(c)*atan((c*x)/sqrt(c))*a*b*c**2*x**5 - 6*sqrt(c)*atanh(c*x**2)*a*b 
*c**2*x**5 - 6*atanh(c*x**2)*a*b - 6*sqrt(c)*log(sqrt(c)*x - 1)*a*b*c**2*x 
**5 + 3*sqrt(c)*log(c*x**2 + 1)*a*b*c**2*x**5 + 15*int(atanh(c*x**2)**2/x* 
*6,x)*b**2*x**5 - 3*a**2 - 4*a*b*c*x**2)/(15*x**5)