\(\int x^4 (a+b \text {arctanh}(c x^2))^2 \, dx\) [71]

Optimal result

Integrand size = 16, antiderivative size = 1173 \[ \int x^4 \left (a+b \text {arctanh}\left (c x^2\right )\right )^2 \, dx =\text {Too large to display} \] Output:

8/15*b^2*x/c^2-1/10*b^2*polylog(2,1+2*c^(1/2)*(1-(-c)^(1/2)*x)/((-c)^(1/2) 
-c^(1/2))/(1+c^(1/2)*x))/c^(5/2)+1/20*b^2*x^5*ln(c*x^2+1)^2+1/25*b^2*x^5*l 
n(-c*x^2+1)+1/25*b*x^5*(2*a-b*ln(-c*x^2+1))+1/5*I*b^2*polylog(2,1-2/(1+I*c 
^(1/2)*x))/c^(5/2)-2/25*a*b*x^5+2/15*b^2*x^3*ln(c*x^2+1)/c+1/5*a*b*x^5*ln( 
c*x^2+1)-1/10*b^2*x^5*ln(-c*x^2+1)*ln(c*x^2+1)-1/15*b^2*x^3*ln(-c*x^2+1)/c 
+1/15*b*x^3*(2*a-b*ln(-c*x^2+1))/c+2/15*a*b*x^3/c+1/20*x^5*(2*a-b*ln(-c*x^ 
2+1))^2+1/5*b^2*polylog(2,1-2/(1+c^(1/2)*x))/c^(5/2)+1/5*b^2*polylog(2,1-2 
/(1-c^(1/2)*x))/c^(5/2)-4/15*b^2*arctanh(c^(1/2)*x)/c^(5/2)-1/5*b^2*arctan 
h(c^(1/2)*x)^2/c^(5/2)-4/15*b^2*arctan(c^(1/2)*x)/c^(5/2)-1/10*b^2*polylog 
(2,1-2*c^(1/2)*(1+(-c)^(1/2)*x)/((-c)^(1/2)+c^(1/2))/(1+c^(1/2)*x))/c^(5/2 
)+1/5*I*b^2*polylog(2,1-2/(1-I*c^(1/2)*x))/c^(5/2)+1/5*I*b^2*arctan(c^(1/2 
)*x)^2/c^(5/2)-2/5*b^2*arctanh(c^(1/2)*x)*ln(2/(1+c^(1/2)*x))/c^(5/2)+2/5* 
b^2*arctanh(c^(1/2)*x)*ln(2/(1-c^(1/2)*x))/c^(5/2)+1/5*b^2*arctanh(c^(1/2) 
*x)*ln(2*c^(1/2)*(1+(-c)^(1/2)*x)/((-c)^(1/2)+c^(1/2))/(1+c^(1/2)*x))/c^(5 
/2)+1/5*b^2*arctanh(c^(1/2)*x)*ln(-2*c^(1/2)*(1-(-c)^(1/2)*x)/((-c)^(1/2)- 
c^(1/2))/(1+c^(1/2)*x))/c^(5/2)+1/5*b^2*arctan(c^(1/2)*x)*ln((1+I)*(1-c^(1 
/2)*x)/(1-I*c^(1/2)*x))/c^(5/2)+1/5*b^2*arctan(c^(1/2)*x)*ln((1-I)*(1+c^(1 
/2)*x)/(1-I*c^(1/2)*x))/c^(5/2)+2/5*b^2*arctan(c^(1/2)*x)*ln(2/(1+I*c^(1/2 
)*x))/c^(5/2)-2/5*b^2*arctan(c^(1/2)*x)*ln(2/(1-I*c^(1/2)*x))/c^(5/2)+2/5* 
a*b*arctan(c^(1/2)*x)/c^(5/2)-1/5*b^2*arctanh(c^(1/2)*x)*ln(c*x^2+1)/c^...
 

Mathematica [F]

\[ \int x^4 \left (a+b \text {arctanh}\left (c x^2\right )\right )^2 \, dx=\int x^4 \left (a+b \text {arctanh}\left (c x^2\right )\right )^2 \, dx \] 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 = 2.62 (sec) , antiderivative size = 1173, 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.

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

\(\Big \downarrow \) 6456

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(8*b^2*x)/(15*c^2) + (2*a*b*x^3)/(15*c) - (2*a*b*x^5)/25 + (2*a*b*ArcTan[S 
qrt[c]*x])/(5*c^(5/2)) - (4*b^2*ArcTan[Sqrt[c]*x])/(15*c^(5/2)) + ((I/5)*b 
^2*ArcTan[Sqrt[c]*x]^2)/c^(5/2) - (4*b^2*ArcTanh[Sqrt[c]*x])/(15*c^(5/2)) 
- (b^2*ArcTanh[Sqrt[c]*x]^2)/(5*c^(5/2)) + (2*b^2*ArcTanh[Sqrt[c]*x]*Log[2 
/(1 - Sqrt[c]*x)])/(5*c^(5/2)) - (2*b^2*ArcTan[Sqrt[c]*x]*Log[2/(1 - I*Sqr 
t[c]*x)])/(5*c^(5/2)) + (b^2*ArcTan[Sqrt[c]*x]*Log[((1 + I)*(1 - Sqrt[c]*x 
))/(1 - I*Sqrt[c]*x)])/(5*c^(5/2)) + (2*b^2*ArcTan[Sqrt[c]*x]*Log[2/(1 + I 
*Sqrt[c]*x)])/(5*c^(5/2)) - (2*b^2*ArcTanh[Sqrt[c]*x]*Log[2/(1 + Sqrt[c]*x 
)])/(5*c^(5/2)) + (b^2*ArcTanh[Sqrt[c]*x]*Log[(-2*Sqrt[c]*(1 - Sqrt[-c]*x) 
)/((Sqrt[-c] - Sqrt[c])*(1 + Sqrt[c]*x))])/(5*c^(5/2)) + (b^2*ArcTanh[Sqrt 
[c]*x]*Log[(2*Sqrt[c]*(1 + Sqrt[-c]*x))/((Sqrt[-c] + Sqrt[c])*(1 + Sqrt[c] 
*x))])/(5*c^(5/2)) + (b^2*ArcTan[Sqrt[c]*x]*Log[((1 - I)*(1 + Sqrt[c]*x))/ 
(1 - I*Sqrt[c]*x)])/(5*c^(5/2)) - (b^2*x^3*Log[1 - c*x^2])/(15*c) + (b^2*x 
^5*Log[1 - c*x^2])/25 - (b^2*ArcTan[Sqrt[c]*x]*Log[1 - c*x^2])/(5*c^(5/2)) 
 + (b*x^3*(2*a - b*Log[1 - c*x^2]))/(15*c) + (b*x^5*(2*a - b*Log[1 - c*x^2 
]))/25 - (b*ArcTanh[Sqrt[c]*x]*(2*a - b*Log[1 - c*x^2]))/(5*c^(5/2)) + (x^ 
5*(2*a - b*Log[1 - c*x^2])^2)/20 + (2*b^2*x^3*Log[1 + c*x^2])/(15*c) + (a* 
b*x^5*Log[1 + c*x^2])/5 + (b^2*ArcTan[Sqrt[c]*x]*Log[1 + c*x^2])/(5*c^(5/2 
)) - (b^2*ArcTanh[Sqrt[c]*x]*Log[1 + c*x^2])/(5*c^(5/2)) - (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 + (b^2*Poly...
 

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]

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

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]

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

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

\[ \int x^4 \left (a+b \text {arctanh}\left (c x^2\right )\right )^2 \, dx=\int { {\left (b \operatorname {artanh}\left (c x^{2}\right ) + a\right )}^{2} x^{4} \,d x } \] 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) + c*(4*x^3/c^2 + 6*arctan(sqrt(c) 
*x)/c^(7/2) + 3*log((c*x - sqrt(c))/(c*x + sqrt(c)))/c^(7/2)))*a*b + 1/20* 
(x^5*log(-c*x^2 + 1)^2 - 5*integrate(-1/5*(5*(c*x^6 - x^4)*log(c*x^2 + 1)^ 
2 - 2*(2*c*x^6 + 5*(c*x^6 - x^4)*log(c*x^2 + 1))*log(-c*x^2 + 1))/(c*x^2 - 
 1), x))*b^2
 

Giac [F]

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

\[ \int x^4 \left (a+b \text {arctanh}\left (c x^2\right )\right )^2 \, dx=\frac {6 \sqrt {c}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}}\right ) a b -4 \sqrt {c}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}}\right ) b^{2}+3 \mathit {atanh} \left (c \,x^{2}\right )^{2} b^{2} c^{3} x^{5}-3 \mathit {atanh} \left (c \,x^{2}\right )^{2} b^{2} c x +6 \sqrt {c}\, \mathit {atanh} \left (c \,x^{2}\right ) a b +4 \sqrt {c}\, \mathit {atanh} \left (c \,x^{2}\right ) b^{2}+6 \mathit {atanh} \left (c \,x^{2}\right ) a b \,c^{3} x^{5}+4 \mathit {atanh} \left (c \,x^{2}\right ) b^{2} c^{2} x^{3}+6 \sqrt {c}\, \mathrm {log}\left (\sqrt {c}\, x -1\right ) a b +4 \sqrt {c}\, \mathrm {log}\left (\sqrt {c}\, x -1\right ) b^{2}-3 \sqrt {c}\, \mathrm {log}\left (c \,x^{2}+1\right ) a b -2 \sqrt {c}\, \mathrm {log}\left (c \,x^{2}+1\right ) b^{2}+3 \left (\int \mathit {atanh} \left (c \,x^{2}\right )^{2}d x \right ) b^{2} c +3 a^{2} c^{3} x^{5}+4 a b \,c^{2} x^{3}+8 b^{2} c x}{15 c^{3}} \] Input:

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

Output:

(6*sqrt(c)*atan((c*x)/sqrt(c))*a*b - 4*sqrt(c)*atan((c*x)/sqrt(c))*b**2 + 
3*atanh(c*x**2)**2*b**2*c**3*x**5 - 3*atanh(c*x**2)**2*b**2*c*x + 6*sqrt(c 
)*atanh(c*x**2)*a*b + 4*sqrt(c)*atanh(c*x**2)*b**2 + 6*atanh(c*x**2)*a*b*c 
**3*x**5 + 4*atanh(c*x**2)*b**2*c**2*x**3 + 6*sqrt(c)*log(sqrt(c)*x - 1)*a 
*b + 4*sqrt(c)*log(sqrt(c)*x - 1)*b**2 - 3*sqrt(c)*log(c*x**2 + 1)*a*b - 2 
*sqrt(c)*log(c*x**2 + 1)*b**2 + 3*int(atanh(c*x**2)**2,x)*b**2*c + 3*a**2* 
c**3*x**5 + 4*a*b*c**2*x**3 + 8*b**2*c*x)/(15*c**3)