\(\int x^4 (a+b \text {arctanh}(\frac {c}{x^2}))^2 \, dx\) [176]

Optimal result

Integrand size = 16, antiderivative size = 1214 \[ \int x^4 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2 \, dx =\text {Too large to display} \] Output:

1/20*b^2*x^5*ln(1+c/x^2)^2+1/5*I*b^2*c^(5/2)*polylog(2,1-2*c^(1/2)/(c^(1/2 
)-I*x))+1/5*I*b^2*c^(5/2)*polylog(2,I*x/c^(1/2))+1/5*b^2*c^(5/2)*arctanh(x 
/c^(1/2))*ln(2*c^(1/2)*((-c)^(1/2)+x)/((-c)^(1/2)+c^(1/2))/(c^(1/2)+x))+1/ 
5*b^2*c^(5/2)*arctanh(x/c^(1/2))*ln(2*c^(1/2)*((-c)^(1/2)-x)/((-c)^(1/2)-c 
^(1/2))/(c^(1/2)+x))+2/5*b^2*c^(5/2)*arctanh(x/c^(1/2))*ln(2-2*c^(1/2)/(c^ 
(1/2)+x))-2/5*b^2*c^(5/2)*arctanh(x/c^(1/2))*ln(2*c^(1/2)/(c^(1/2)+x))+2/5 
*a*b*c^(5/2)*arctan(x/c^(1/2))+1/5*b^2*c^(5/2)*arctan(x/c^(1/2))*ln((1+I)* 
(c^(1/2)-x)/(c^(1/2)-I*x))+1/5*b^2*c^(5/2)*arctan(x/c^(1/2))*ln((1-I)*(c^( 
1/2)+x)/(c^(1/2)-I*x))+2/5*b^2*c^(5/2)*arctan(x/c^(1/2))*ln(2-2*c^(1/2)/(c 
^(1/2)-I*x))-2/5*b^2*c^(5/2)*arctan(x/c^(1/2))*ln(2*c^(1/2)/(c^(1/2)-I*x)) 
-1/5*b^2*c^(5/2)*arctanh(x/c^(1/2))*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*c^(5/2)*arctanh(x/c^(1/2))*(2*a-b*ln(1-c/x^2))-1 
/5*b^2*c^(5/2)*arctan(x/c^(1/2))*ln(1-c/x^2)-1/10*I*b^2*c^(5/2)*polylog(2, 
1+(-1+I)*(c^(1/2)+x)/(c^(1/2)-I*x))-1/10*I*b^2*c^(5/2)*polylog(2,1-(1+I)*( 
c^(1/2)-x)/(c^(1/2)-I*x))-1/5*I*b^2*c^(5/2)*polylog(2,-1+2*c^(1/2)/(c^(1/2 
)-I*x))-1/5*I*b^2*c^(5/2)*polylog(2,-I*x/c^(1/2))-1/5*I*b^2*c^(5/2)*arctan 
(x/c^(1/2))^2+8/15*b^2*c^2*x+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))-4/15*b^2*c^(5/2)*arctanh(x/c^(1/2))+1/5*b^2* 
c^(5/2)*arctanh(x/c^(1/2))^2-4/15*b^2*c^(5/2)*arctan(x/c^(1/2))-1/10*b^2*c 
^(5/2)*polylog(2,1-2*c^(1/2)*((-c)^(1/2)+x)/((-c)^(1/2)+c^(1/2))/(c^(1/...
 

Mathematica [F]

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

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

Output:

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

Rubi [A] (verified)

Time = 3.16 (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 )\)

Input:

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

Output:

(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...
 

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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int x^4 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2 \, dx=\frac {2 \sqrt {c}\, \mathit {atan} \left (\frac {x}{\sqrt {c}}\right ) a b \,c^{2}}{5}-\frac {4 \sqrt {c}\, \mathit {atan} \left (\frac {x}{\sqrt {c}}\right ) b^{2} c^{2}}{15}-\frac {\mathit {atanh} \left (\frac {c}{x^{2}}\right )^{2} b^{2} c^{2} x}{5}+\frac {\mathit {atanh} \left (\frac {c}{x^{2}}\right )^{2} b^{2} x^{5}}{5}+\frac {2 \sqrt {c}\, \mathit {atanh} \left (\frac {c}{x^{2}}\right ) a b \,c^{2}}{5}+\frac {4 \sqrt {c}\, \mathit {atanh} \left (\frac {c}{x^{2}}\right ) b^{2} c^{2}}{15}+\frac {2 \mathit {atanh} \left (\frac {c}{x^{2}}\right ) a b \,x^{5}}{5}+\frac {4 \mathit {atanh} \left (\frac {c}{x^{2}}\right ) b^{2} c \,x^{3}}{15}+\frac {2 \sqrt {c}\, \mathrm {log}\left (\sqrt {c}-x \right ) a b \,c^{2}}{5}+\frac {4 \sqrt {c}\, \mathrm {log}\left (\sqrt {c}-x \right ) b^{2} c^{2}}{15}-\frac {\sqrt {c}\, \mathrm {log}\left (x^{2}+c \right ) a b \,c^{2}}{5}-\frac {2 \sqrt {c}\, \mathrm {log}\left (x^{2}+c \right ) b^{2} c^{2}}{15}+\frac {\left (\int \mathit {atanh} \left (\frac {c}{x^{2}}\right )^{2}d x \right ) b^{2} c^{2}}{5}+\frac {a^{2} x^{5}}{5}+\frac {4 a b c \,x^{3}}{15}+\frac {8 b^{2} c^{2} x}{15} \] Input:

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

Output:

(6*sqrt(c)*atan(x/sqrt(c))*a*b*c**2 - 4*sqrt(c)*atan(x/sqrt(c))*b**2*c**2 
- 3*atanh(c/x**2)**2*b**2*c**2*x + 3*atanh(c/x**2)**2*b**2*x**5 + 6*sqrt(c 
)*atanh(c/x**2)*a*b*c**2 + 4*sqrt(c)*atanh(c/x**2)*b**2*c**2 + 6*atanh(c/x 
**2)*a*b*x**5 + 4*atanh(c/x**2)*b**2*c*x**3 + 6*sqrt(c)*log(sqrt(c) - x)*a 
*b*c**2 + 4*sqrt(c)*log(sqrt(c) - x)*b**2*c**2 - 3*sqrt(c)*log(c + x**2)*a 
*b*c**2 - 2*sqrt(c)*log(c + x**2)*b**2*c**2 + 3*int(atanh(c/x**2)**2,x)*b* 
*2*c**2 + 3*a**2*x**5 + 4*a*b*c*x**3 + 8*b**2*c**2*x)/15