\(\int x^3 (a+b \text {arctanh}(c x^3)) \, dx\) [107]

Optimal result

Integrand size = 14, antiderivative size = 140 \[ \int x^3 \left (a+b \text {arctanh}\left (c x^3\right )\right ) \, dx=\frac {3 b x}{4 c}+\frac {\sqrt {3} b \arctan \left (\frac {1-2 \sqrt [3]{c} x}{\sqrt {3}}\right )}{8 c^{4/3}}-\frac {\sqrt {3} b \arctan \left (\frac {1+2 \sqrt [3]{c} x}{\sqrt {3}}\right )}{8 c^{4/3}}-\frac {b \text {arctanh}\left (\sqrt [3]{c} x\right )}{4 c^{4/3}}+\frac {1}{4} x^4 \left (a+b \text {arctanh}\left (c x^3\right )\right )-\frac {b \text {arctanh}\left (\frac {\sqrt [3]{c} x}{1+c^{2/3} x^2}\right )}{8 c^{4/3}} \] Output:

3/4*b*x/c+1/8*3^(1/2)*b*arctan(1/3*(1-2*c^(1/3)*x)*3^(1/2))/c^(4/3)-1/8*3^ 
(1/2)*b*arctan(1/3*(1+2*c^(1/3)*x)*3^(1/2))/c^(4/3)-1/4*b*arctanh(c^(1/3)* 
x)/c^(4/3)+1/4*x^4*(a+b*arctanh(c*x^3))-1/8*b*arctanh(c^(1/3)*x/(1+c^(2/3) 
*x^2))/c^(4/3)
 

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.40 \[ \int x^3 \left (a+b \text {arctanh}\left (c x^3\right )\right ) \, dx=\frac {3 b x}{4 c}+\frac {a x^4}{4}-\frac {\sqrt {3} b \arctan \left (\frac {-1+2 \sqrt [3]{c} x}{\sqrt {3}}\right )}{8 c^{4/3}}-\frac {\sqrt {3} b \arctan \left (\frac {1+2 \sqrt [3]{c} x}{\sqrt {3}}\right )}{8 c^{4/3}}+\frac {1}{4} b x^4 \text {arctanh}\left (c x^3\right )+\frac {b \log \left (1-\sqrt [3]{c} x\right )}{8 c^{4/3}}-\frac {b \log \left (1+\sqrt [3]{c} x\right )}{8 c^{4/3}}+\frac {b \log \left (1-\sqrt [3]{c} x+c^{2/3} x^2\right )}{16 c^{4/3}}-\frac {b \log \left (1+\sqrt [3]{c} x+c^{2/3} x^2\right )}{16 c^{4/3}} \] Input:

Integrate[x^3*(a + b*ArcTanh[c*x^3]),x]
 

Output:

(3*b*x)/(4*c) + (a*x^4)/4 - (Sqrt[3]*b*ArcTan[(-1 + 2*c^(1/3)*x)/Sqrt[3]]) 
/(8*c^(4/3)) - (Sqrt[3]*b*ArcTan[(1 + 2*c^(1/3)*x)/Sqrt[3]])/(8*c^(4/3)) + 
 (b*x^4*ArcTanh[c*x^3])/4 + (b*Log[1 - c^(1/3)*x])/(8*c^(4/3)) - (b*Log[1 
+ c^(1/3)*x])/(8*c^(4/3)) + (b*Log[1 - c^(1/3)*x + c^(2/3)*x^2])/(16*c^(4/ 
3)) - (b*Log[1 + c^(1/3)*x + c^(2/3)*x^2])/(16*c^(4/3))
 

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.26, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.786, Rules used = {6452, 843, 754, 27, 219, 1142, 25, 27, 1082, 217, 1103}

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.

\(\displaystyle \int x^3 \left (a+b \text {arctanh}\left (c x^3\right )\right ) \, dx\)

\(\Big \downarrow \) 6452

\(\displaystyle \frac {1}{4} x^4 \left (a+b \text {arctanh}\left (c x^3\right )\right )-\frac {3}{4} b c \int \frac {x^6}{1-c^2 x^6}dx\)

\(\Big \downarrow \) 843

\(\displaystyle \frac {1}{4} x^4 \left (a+b \text {arctanh}\left (c x^3\right )\right )-\frac {3}{4} b c \left (\frac {\int \frac {1}{1-c^2 x^6}dx}{c^2}-\frac {x}{c^2}\right )\)

\(\Big \downarrow \) 754

\(\displaystyle \frac {1}{4} x^4 \left (a+b \text {arctanh}\left (c x^3\right )\right )-\frac {3}{4} b c \left (\frac {\frac {1}{3} \int \frac {1}{1-c^{2/3} x^2}dx+\frac {1}{3} \int \frac {2-\sqrt [3]{c} x}{2 \left (c^{2/3} x^2-\sqrt [3]{c} x+1\right )}dx+\frac {1}{3} \int \frac {\sqrt [3]{c} x+2}{2 \left (c^{2/3} x^2+\sqrt [3]{c} x+1\right )}dx}{c^2}-\frac {x}{c^2}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{4} x^4 \left (a+b \text {arctanh}\left (c x^3\right )\right )-\frac {3}{4} b c \left (\frac {\frac {1}{3} \int \frac {1}{1-c^{2/3} x^2}dx+\frac {1}{6} \int \frac {2-\sqrt [3]{c} x}{c^{2/3} x^2-\sqrt [3]{c} x+1}dx+\frac {1}{6} \int \frac {\sqrt [3]{c} x+2}{c^{2/3} x^2+\sqrt [3]{c} x+1}dx}{c^2}-\frac {x}{c^2}\right )\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{4} x^4 \left (a+b \text {arctanh}\left (c x^3\right )\right )-\frac {3}{4} b c \left (\frac {\frac {1}{6} \int \frac {2-\sqrt [3]{c} x}{c^{2/3} x^2-\sqrt [3]{c} x+1}dx+\frac {1}{6} \int \frac {\sqrt [3]{c} x+2}{c^{2/3} x^2+\sqrt [3]{c} x+1}dx+\frac {\text {arctanh}\left (\sqrt [3]{c} x\right )}{3 \sqrt [3]{c}}}{c^2}-\frac {x}{c^2}\right )\)

\(\Big \downarrow \) 1142

\(\displaystyle \frac {1}{4} x^4 \left (a+b \text {arctanh}\left (c x^3\right )\right )-\frac {3}{4} b c \left (\frac {\frac {1}{6} \left (\frac {3}{2} \int \frac {1}{c^{2/3} x^2-\sqrt [3]{c} x+1}dx-\frac {\int -\frac {\sqrt [3]{c} \left (1-2 \sqrt [3]{c} x\right )}{c^{2/3} x^2-\sqrt [3]{c} x+1}dx}{2 \sqrt [3]{c}}\right )+\frac {1}{6} \left (\frac {3}{2} \int \frac {1}{c^{2/3} x^2+\sqrt [3]{c} x+1}dx+\frac {\int \frac {\sqrt [3]{c} \left (2 \sqrt [3]{c} x+1\right )}{c^{2/3} x^2+\sqrt [3]{c} x+1}dx}{2 \sqrt [3]{c}}\right )+\frac {\text {arctanh}\left (\sqrt [3]{c} x\right )}{3 \sqrt [3]{c}}}{c^2}-\frac {x}{c^2}\right )\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{4} x^4 \left (a+b \text {arctanh}\left (c x^3\right )\right )-\frac {3}{4} b c \left (\frac {\frac {1}{6} \left (\frac {3}{2} \int \frac {1}{c^{2/3} x^2-\sqrt [3]{c} x+1}dx+\frac {\int \frac {\sqrt [3]{c} \left (1-2 \sqrt [3]{c} x\right )}{c^{2/3} x^2-\sqrt [3]{c} x+1}dx}{2 \sqrt [3]{c}}\right )+\frac {1}{6} \left (\frac {3}{2} \int \frac {1}{c^{2/3} x^2+\sqrt [3]{c} x+1}dx+\frac {\int \frac {\sqrt [3]{c} \left (2 \sqrt [3]{c} x+1\right )}{c^{2/3} x^2+\sqrt [3]{c} x+1}dx}{2 \sqrt [3]{c}}\right )+\frac {\text {arctanh}\left (\sqrt [3]{c} x\right )}{3 \sqrt [3]{c}}}{c^2}-\frac {x}{c^2}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{4} x^4 \left (a+b \text {arctanh}\left (c x^3\right )\right )-\frac {3}{4} b c \left (\frac {\frac {1}{6} \left (\frac {3}{2} \int \frac {1}{c^{2/3} x^2-\sqrt [3]{c} x+1}dx+\frac {1}{2} \int \frac {1-2 \sqrt [3]{c} x}{c^{2/3} x^2-\sqrt [3]{c} x+1}dx\right )+\frac {1}{6} \left (\frac {3}{2} \int \frac {1}{c^{2/3} x^2+\sqrt [3]{c} x+1}dx+\frac {1}{2} \int \frac {2 \sqrt [3]{c} x+1}{c^{2/3} x^2+\sqrt [3]{c} x+1}dx\right )+\frac {\text {arctanh}\left (\sqrt [3]{c} x\right )}{3 \sqrt [3]{c}}}{c^2}-\frac {x}{c^2}\right )\)

\(\Big \downarrow \) 1082

\(\displaystyle \frac {1}{4} x^4 \left (a+b \text {arctanh}\left (c x^3\right )\right )-\frac {3}{4} b c \left (\frac {\frac {1}{6} \left (\frac {1}{2} \int \frac {1-2 \sqrt [3]{c} x}{c^{2/3} x^2-\sqrt [3]{c} x+1}dx+\frac {3 \int \frac {1}{-\left (1-2 \sqrt [3]{c} x\right )^2-3}d\left (1-2 \sqrt [3]{c} x\right )}{\sqrt [3]{c}}\right )+\frac {1}{6} \left (\frac {1}{2} \int \frac {2 \sqrt [3]{c} x+1}{c^{2/3} x^2+\sqrt [3]{c} x+1}dx-\frac {3 \int \frac {1}{-\left (2 \sqrt [3]{c} x+1\right )^2-3}d\left (2 \sqrt [3]{c} x+1\right )}{\sqrt [3]{c}}\right )+\frac {\text {arctanh}\left (\sqrt [3]{c} x\right )}{3 \sqrt [3]{c}}}{c^2}-\frac {x}{c^2}\right )\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {1}{4} x^4 \left (a+b \text {arctanh}\left (c x^3\right )\right )-\frac {3}{4} b c \left (\frac {\frac {1}{6} \left (\frac {1}{2} \int \frac {1-2 \sqrt [3]{c} x}{c^{2/3} x^2-\sqrt [3]{c} x+1}dx-\frac {\sqrt {3} \arctan \left (\frac {1-2 \sqrt [3]{c} x}{\sqrt {3}}\right )}{\sqrt [3]{c}}\right )+\frac {1}{6} \left (\frac {1}{2} \int \frac {2 \sqrt [3]{c} x+1}{c^{2/3} x^2+\sqrt [3]{c} x+1}dx+\frac {\sqrt {3} \arctan \left (\frac {2 \sqrt [3]{c} x+1}{\sqrt {3}}\right )}{\sqrt [3]{c}}\right )+\frac {\text {arctanh}\left (\sqrt [3]{c} x\right )}{3 \sqrt [3]{c}}}{c^2}-\frac {x}{c^2}\right )\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {1}{4} x^4 \left (a+b \text {arctanh}\left (c x^3\right )\right )-\frac {3}{4} b c \left (\frac {\frac {1}{6} \left (-\frac {\sqrt {3} \arctan \left (\frac {1-2 \sqrt [3]{c} x}{\sqrt {3}}\right )}{\sqrt [3]{c}}-\frac {\log \left (c^{2/3} x^2-\sqrt [3]{c} x+1\right )}{2 \sqrt [3]{c}}\right )+\frac {1}{6} \left (\frac {\sqrt {3} \arctan \left (\frac {2 \sqrt [3]{c} x+1}{\sqrt {3}}\right )}{\sqrt [3]{c}}+\frac {\log \left (c^{2/3} x^2+\sqrt [3]{c} x+1\right )}{2 \sqrt [3]{c}}\right )+\frac {\text {arctanh}\left (\sqrt [3]{c} x\right )}{3 \sqrt [3]{c}}}{c^2}-\frac {x}{c^2}\right )\)

Input:

Int[x^3*(a + b*ArcTanh[c*x^3]),x]
 

Output:

(x^4*(a + b*ArcTanh[c*x^3]))/4 - (3*b*c*(-(x/c^2) + (ArcTanh[c^(1/3)*x]/(3 
*c^(1/3)) + (-((Sqrt[3]*ArcTan[(1 - 2*c^(1/3)*x)/Sqrt[3]])/c^(1/3)) - Log[ 
1 - c^(1/3)*x + c^(2/3)*x^2]/(2*c^(1/3)))/6 + ((Sqrt[3]*ArcTan[(1 + 2*c^(1 
/3)*x)/Sqrt[3]])/c^(1/3) + Log[1 + c^(1/3)*x + c^(2/3)*x^2]/(2*c^(1/3)))/6 
)/c^2))/4
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 217

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( 
-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & 
& (LtQ[a, 0] || LtQ[b, 0])
 

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 754

Int[((a_) + (b_.)*(x_)^(n_))^(-1), x_Symbol] :> Module[{r = Numerator[Rt[-a 
/b, n]], s = Denominator[Rt[-a/b, n]], k, u}, Simp[u = Int[(r - s*Cos[(2*k* 
Pi)/n]*x)/(r^2 - 2*r*s*Cos[(2*k*Pi)/n]*x + s^2*x^2), x] + Int[(r + s*Cos[(2 
*k*Pi)/n]*x)/(r^2 + 2*r*s*Cos[(2*k*Pi)/n]*x + s^2*x^2), x]; 2*(r^2/(a*n)) 
 Int[1/(r^2 - s^2*x^2), x] + 2*(r/(a*n))   Sum[u, {k, 1, (n - 2)/4}], x]] / 
; FreeQ[{a, b}, x] && IGtQ[(n - 2)/4, 0] && NegQ[a/b]
 

rule 843

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n 
 - 1)*(c*x)^(m - n + 1)*((a + b*x^n)^(p + 1)/(b*(m + n*p + 1))), x] - Simp[ 
a*c^n*((m - n + 1)/(b*(m + n*p + 1)))   Int[(c*x)^(m - n)*(a + b*x^n)^p, x] 
, x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n* 
p + 1, 0] && IntBinomialQ[a, b, c, n, m, p, x]
 

rule 1082

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S 
implify[a*(c/b^2)]}, Simp[-2/b   Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b 
)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /; Fre 
eQ[{a, b, c}, x]
 

rule 1103

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, 
e}, x] && EqQ[2*c*d - b*e, 0]
 

rule 1142

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[(2*c*d - b*e)/(2*c)   Int[1/(a + b*x + c*x^2), x], x] + Simp[e/(2*c) 
Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x]
 

rule 6452

Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] : 
> Simp[x^(m + 1)*((a + b*ArcTanh[c*x^n])^p/(m + 1)), x] - Simp[b*c*n*(p/(m 
+ 1))   Int[x^(m + n)*((a + b*ArcTanh[c*x^n])^(p - 1)/(1 - c^2*x^(2*n))), x 
], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0] && (EqQ[p, 1] || (EqQ[n, 1 
] && IntegerQ[m])) && NeQ[m, -1]
 

Maple [A] (verified)

Time = 0.36 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.31

method	result	size
default	\(\frac {x^{4} a}{4}+\frac {b \,x^{4} \operatorname {arctanh}\left (c \,x^{3}\right )}{4}+\frac {3 b x}{4 c}+\frac {b \ln \left (x -\left (\frac {1}{c}\right )^{\frac {1}{3}}\right )}{8 c^{2} \left (\frac {1}{c}\right )^{\frac {2}{3}}}-\frac {b \ln \left (x^{2}+\left (\frac {1}{c}\right )^{\frac {1}{3}} x +\left (\frac {1}{c}\right )^{\frac {2}{3}}\right )}{16 c^{2} \left (\frac {1}{c}\right )^{\frac {2}{3}}}-\frac {b \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {1}{c}\right )^{\frac {1}{3}}}+1\right )}{3}\right )}{8 c^{2} \left (\frac {1}{c}\right )^{\frac {2}{3}}}-\frac {b \ln \left (x +\left (\frac {1}{c}\right )^{\frac {1}{3}}\right )}{8 c^{2} \left (\frac {1}{c}\right )^{\frac {2}{3}}}+\frac {b \ln \left (x^{2}-\left (\frac {1}{c}\right )^{\frac {1}{3}} x +\left (\frac {1}{c}\right )^{\frac {2}{3}}\right )}{16 c^{2} \left (\frac {1}{c}\right )^{\frac {2}{3}}}-\frac {b \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {1}{c}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{8 c^{2} \left (\frac {1}{c}\right )^{\frac {2}{3}}}\)	\(184\)
parts	\(\frac {x^{4} a}{4}+\frac {b \,x^{4} \operatorname {arctanh}\left (c \,x^{3}\right )}{4}+\frac {3 b x}{4 c}+\frac {b \ln \left (x -\left (\frac {1}{c}\right )^{\frac {1}{3}}\right )}{8 c^{2} \left (\frac {1}{c}\right )^{\frac {2}{3}}}-\frac {b \ln \left (x^{2}+\left (\frac {1}{c}\right )^{\frac {1}{3}} x +\left (\frac {1}{c}\right )^{\frac {2}{3}}\right )}{16 c^{2} \left (\frac {1}{c}\right )^{\frac {2}{3}}}-\frac {b \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {1}{c}\right )^{\frac {1}{3}}}+1\right )}{3}\right )}{8 c^{2} \left (\frac {1}{c}\right )^{\frac {2}{3}}}-\frac {b \ln \left (x +\left (\frac {1}{c}\right )^{\frac {1}{3}}\right )}{8 c^{2} \left (\frac {1}{c}\right )^{\frac {2}{3}}}+\frac {b \ln \left (x^{2}-\left (\frac {1}{c}\right )^{\frac {1}{3}} x +\left (\frac {1}{c}\right )^{\frac {2}{3}}\right )}{16 c^{2} \left (\frac {1}{c}\right )^{\frac {2}{3}}}-\frac {b \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {1}{c}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{8 c^{2} \left (\frac {1}{c}\right )^{\frac {2}{3}}}\)	\(184\)
risch	\(\frac {b \,x^{4} \ln \left (c \,x^{3}+1\right )}{8}+\frac {x^{4} a}{4}-\frac {b \,x^{4} \ln \left (-c \,x^{3}+1\right )}{8}+\frac {3 b x}{4 c}+\frac {b \ln \left (x -\left (\frac {1}{c}\right )^{\frac {1}{3}}\right )}{8 c^{2} \left (\frac {1}{c}\right )^{\frac {2}{3}}}-\frac {b \ln \left (x^{2}+\left (\frac {1}{c}\right )^{\frac {1}{3}} x +\left (\frac {1}{c}\right )^{\frac {2}{3}}\right )}{16 c^{2} \left (\frac {1}{c}\right )^{\frac {2}{3}}}-\frac {b \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {1}{c}\right )^{\frac {1}{3}}}+1\right )}{3}\right )}{8 c^{2} \left (\frac {1}{c}\right )^{\frac {2}{3}}}-\frac {b \ln \left (x +\left (\frac {1}{c}\right )^{\frac {1}{3}}\right )}{8 c^{2} \left (\frac {1}{c}\right )^{\frac {2}{3}}}+\frac {b \ln \left (x^{2}-\left (\frac {1}{c}\right )^{\frac {1}{3}} x +\left (\frac {1}{c}\right )^{\frac {2}{3}}\right )}{16 c^{2} \left (\frac {1}{c}\right )^{\frac {2}{3}}}-\frac {b \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {1}{c}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{8 c^{2} \left (\frac {1}{c}\right )^{\frac {2}{3}}}\)	\(201\)

Input:

int(x^3*(a+b*arctanh(c*x^3)),x,method=_RETURNVERBOSE)
                                                                                    
                                                                                    
 

Output:

1/4*x^4*a+1/4*b*x^4*arctanh(c*x^3)+3/4*b*x/c+1/8*b/c^2/(1/c)^(2/3)*ln(x-(1 
/c)^(1/3))-1/16*b/c^2/(1/c)^(2/3)*ln(x^2+(1/c)^(1/3)*x+(1/c)^(2/3))-1/8*b/ 
c^2/(1/c)^(2/3)*3^(1/2)*arctan(1/3*3^(1/2)*(2/(1/c)^(1/3)*x+1))-1/8*b/c^2/ 
(1/c)^(2/3)*ln(x+(1/c)^(1/3))+1/16*b/c^2/(1/c)^(2/3)*ln(x^2-(1/c)^(1/3)*x+ 
(1/c)^(2/3))-1/8*b/c^2/(1/c)^(2/3)*3^(1/2)*arctan(1/3*3^(1/2)*(2/(1/c)^(1/ 
3)*x-1))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 211 vs. \(2 (104) = 208\).

Time = 0.11 (sec) , antiderivative size = 981, normalized size of antiderivative = 7.01 \[ \int x^3 \left (a+b \text {arctanh}\left (c x^3\right )\right ) \, dx=\text {Too large to display} \] Input:

integrate(x^3*(a+b*arctanh(c*x^3)),x, algorithm="fricas")
 

Output:

[1/16*(2*b*c^2*x^4*log(-(c*x^3 + 1)/(c*x^3 - 1)) + 4*a*c^2*x^4 + sqrt(3)*b 
*c*sqrt((-c)^(1/3)/c)*log((2*c*x^3 - sqrt(3)*(2*c*x^2 + (-c)^(2/3)*x + (-c 
)^(1/3))*sqrt((-c)^(1/3)/c) + 3*(-c)^(1/3)*x - 1)/(c*x^3 + 1)) + sqrt(3)*b 
*c*sqrt(-1/c^(2/3))*log((2*c*x^3 - sqrt(3)*(2*c*x^2 - c^(2/3)*x - c^(1/3)) 
*sqrt(-1/c^(2/3)) - 3*c^(1/3)*x + 1)/(c*x^3 - 1)) + 12*b*c*x + b*(-c)^(2/3 
)*log(c*x^2 - (-c)^(2/3)*x - (-c)^(1/3)) - b*c^(2/3)*log(c*x^2 + c^(2/3)*x 
 + c^(1/3)) - 2*b*(-c)^(2/3)*log(c*x + (-c)^(2/3)) + 2*b*c^(2/3)*log(c*x - 
 c^(2/3)))/c^2, 1/16*(2*b*c^2*x^4*log(-(c*x^3 + 1)/(c*x^3 - 1)) + 4*a*c^2* 
x^4 - 2*sqrt(3)*b*c*sqrt(-(-c)^(1/3)/c)*arctan(1/3*sqrt(3)*(2*(-c)^(2/3)*x 
 + (-c)^(1/3))*sqrt(-(-c)^(1/3)/c)) + sqrt(3)*b*c*sqrt(-1/c^(2/3))*log((2* 
c*x^3 - sqrt(3)*(2*c*x^2 - c^(2/3)*x - c^(1/3))*sqrt(-1/c^(2/3)) - 3*c^(1/ 
3)*x + 1)/(c*x^3 - 1)) + 12*b*c*x + b*(-c)^(2/3)*log(c*x^2 - (-c)^(2/3)*x 
- (-c)^(1/3)) - b*c^(2/3)*log(c*x^2 + c^(2/3)*x + c^(1/3)) - 2*b*(-c)^(2/3 
)*log(c*x + (-c)^(2/3)) + 2*b*c^(2/3)*log(c*x - c^(2/3)))/c^2, 1/16*(2*b*c 
^2*x^4*log(-(c*x^3 + 1)/(c*x^3 - 1)) + 4*a*c^2*x^4 + sqrt(3)*b*c*sqrt((-c) 
^(1/3)/c)*log((2*c*x^3 - sqrt(3)*(2*c*x^2 + (-c)^(2/3)*x + (-c)^(1/3))*sqr 
t((-c)^(1/3)/c) + 3*(-c)^(1/3)*x - 1)/(c*x^3 + 1)) - 2*sqrt(3)*b*c^(2/3)*a 
rctan(1/3*sqrt(3)*(2*c^(2/3)*x + c^(1/3))/c^(1/3)) + 12*b*c*x + b*(-c)^(2/ 
3)*log(c*x^2 - (-c)^(2/3)*x - (-c)^(1/3)) - b*c^(2/3)*log(c*x^2 + c^(2/3)* 
x + c^(1/3)) - 2*b*(-c)^(2/3)*log(c*x + (-c)^(2/3)) + 2*b*c^(2/3)*log(c...
 

Sympy [F(-1)]

Timed out. \[ \int x^3 \left (a+b \text {arctanh}\left (c x^3\right )\right ) \, dx=\text {Timed out} \] Input:

integrate(x**3*(a+b*atanh(c*x**3)),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.16 \[ \int x^3 \left (a+b \text {arctanh}\left (c x^3\right )\right ) \, dx=\frac {1}{4} \, a x^{4} + \frac {1}{16} \, {\left (4 \, x^{4} \operatorname {artanh}\left (c x^{3}\right ) - c {\left (\frac {2 \, \sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (2 \, c^{\frac {2}{3}} x + c^{\frac {1}{3}}\right )}}{3 \, c^{\frac {1}{3}}}\right )}{c^{\frac {7}{3}}} + \frac {2 \, \sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (2 \, c^{\frac {2}{3}} x - c^{\frac {1}{3}}\right )}}{3 \, c^{\frac {1}{3}}}\right )}{c^{\frac {7}{3}}} - \frac {12 \, x}{c^{2}} + \frac {\log \left (c^{\frac {2}{3}} x^{2} + c^{\frac {1}{3}} x + 1\right )}{c^{\frac {7}{3}}} - \frac {\log \left (c^{\frac {2}{3}} x^{2} - c^{\frac {1}{3}} x + 1\right )}{c^{\frac {7}{3}}} + \frac {2 \, \log \left (\frac {c^{\frac {1}{3}} x + 1}{c^{\frac {1}{3}}}\right )}{c^{\frac {7}{3}}} - \frac {2 \, \log \left (\frac {c^{\frac {1}{3}} x - 1}{c^{\frac {1}{3}}}\right )}{c^{\frac {7}{3}}}\right )}\right )} b \] Input:

integrate(x^3*(a+b*arctanh(c*x^3)),x, algorithm="maxima")
 

Output:

1/4*a*x^4 + 1/16*(4*x^4*arctanh(c*x^3) - c*(2*sqrt(3)*arctan(1/3*sqrt(3)*( 
2*c^(2/3)*x + c^(1/3))/c^(1/3))/c^(7/3) + 2*sqrt(3)*arctan(1/3*sqrt(3)*(2* 
c^(2/3)*x - c^(1/3))/c^(1/3))/c^(7/3) - 12*x/c^2 + log(c^(2/3)*x^2 + c^(1/ 
3)*x + 1)/c^(7/3) - log(c^(2/3)*x^2 - c^(1/3)*x + 1)/c^(7/3) + 2*log((c^(1 
/3)*x + 1)/c^(1/3))/c^(7/3) - 2*log((c^(1/3)*x - 1)/c^(1/3))/c^(7/3)))*b
 

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.48 \[ \int x^3 \left (a+b \text {arctanh}\left (c x^3\right )\right ) \, dx=\frac {1}{16} \, b c^{7} {\left (\frac {2 \, \left (-\frac {1}{c}\right )^{\frac {1}{3}} \log \left ({\left | x - \left (-\frac {1}{c}\right )^{\frac {1}{3}} \right |}\right )}{c^{8}} - \frac {2 \, \sqrt {3} {\left | c \right |}^{\frac {2}{3}} \arctan \left (\frac {1}{3} \, \sqrt {3} c^{\frac {1}{3}} {\left (2 \, x + \frac {1}{c^{\frac {1}{3}}}\right )}\right )}{c^{9}} - \frac {2 \, \sqrt {3} \left (-c^{2}\right )^{\frac {1}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {1}{c}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {1}{c}\right )^{\frac {1}{3}}}\right )}{c^{9}} - \frac {{\left | c \right |}^{\frac {2}{3}} \log \left (x^{2} + \frac {x}{c^{\frac {1}{3}}} + \frac {1}{c^{\frac {2}{3}}}\right )}{c^{9}} + \frac {2 \, \log \left ({\left | x - \frac {1}{c^{\frac {1}{3}}} \right |}\right )}{c^{\frac {25}{3}}} - \frac {\left (-c^{2}\right )^{\frac {1}{3}} \log \left (x^{2} + x \left (-\frac {1}{c}\right )^{\frac {1}{3}} + \left (-\frac {1}{c}\right )^{\frac {2}{3}}\right )}{c^{9}}\right )} + \frac {1}{8} \, b x^{4} \log \left (-\frac {c x^{3} + 1}{c x^{3} - 1}\right ) + \frac {1}{4} \, a x^{4} + \frac {3 \, b x}{4 \, c} \] Input:

integrate(x^3*(a+b*arctanh(c*x^3)),x, algorithm="giac")
 

Output:

1/16*b*c^7*(2*(-1/c)^(1/3)*log(abs(x - (-1/c)^(1/3)))/c^8 - 2*sqrt(3)*abs( 
c)^(2/3)*arctan(1/3*sqrt(3)*c^(1/3)*(2*x + 1/c^(1/3)))/c^9 - 2*sqrt(3)*(-c 
^2)^(1/3)*arctan(1/3*sqrt(3)*(2*x + (-1/c)^(1/3))/(-1/c)^(1/3))/c^9 - abs( 
c)^(2/3)*log(x^2 + x/c^(1/3) + 1/c^(2/3))/c^9 + 2*log(abs(x - 1/c^(1/3)))/ 
c^(25/3) - (-c^2)^(1/3)*log(x^2 + x*(-1/c)^(1/3) + (-1/c)^(2/3))/c^9) + 1/ 
8*b*x^4*log(-(c*x^3 + 1)/(c*x^3 - 1)) + 1/4*a*x^4 + 3/4*b*x/c
 

Mupad [B] (verification not implemented)

Time = 4.00 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.89 \[ \int x^3 \left (a+b \text {arctanh}\left (c x^3\right )\right ) \, dx=\frac {a\,x^4}{4}+\frac {b\,\left (-\frac {\mathrm {atan}\left (\frac {c^{1/3}\,x\,\left (\sqrt {3}-\mathrm {i}\right )}{2}\right )}{2}+\frac {\mathrm {atan}\left (\frac {c^{1/3}\,x\,\left (\sqrt {3}+1{}\mathrm {i}\right )}{2}\right )}{2}+\mathrm {atan}\left (c^{1/3}\,x\,1{}\mathrm {i}\right )\right )\,1{}\mathrm {i}}{4\,c^{4/3}}+\frac {3\,b\,x}{4\,c}+\frac {b\,x^4\,\ln \left (c\,x^3+1\right )}{8}-\frac {b\,x^4\,\ln \left (1-c\,x^3\right )}{8}-\frac {\sqrt {3}\,b\,\left (\mathrm {atan}\left (\frac {c^{1/3}\,x\,\left (\sqrt {3}-\mathrm {i}\right )}{2}\right )+\mathrm {atan}\left (\frac {c^{1/3}\,x\,\left (\sqrt {3}+1{}\mathrm {i}\right )}{2}\right )\right )}{8\,c^{4/3}} \] Input:

int(x^3*(a + b*atanh(c*x^3)),x)
 

Output:

(a*x^4)/4 + (b*(atan((c^(1/3)*x*(3^(1/2) + 1i))/2)/2 - atan((c^(1/3)*x*(3^ 
(1/2) - 1i))/2)/2 + atan(c^(1/3)*x*1i))*1i)/(4*c^(4/3)) + (3*b*x)/(4*c) + 
(b*x^4*log(c*x^3 + 1))/8 - (b*x^4*log(1 - c*x^3))/8 - (3^(1/2)*b*(atan((c^ 
(1/3)*x*(3^(1/2) - 1i))/2) + atan((c^(1/3)*x*(3^(1/2) + 1i))/2)))/(8*c^(4/ 
3))
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.80 \[ \int x^3 \left (a+b \text {arctanh}\left (c x^3\right )\right ) \, dx=\frac {-2 \sqrt {3}\, \mathit {atan} \left (\frac {2 c^{\frac {1}{3}} x -1}{\sqrt {3}}\right ) b -2 \sqrt {3}\, \mathit {atan} \left (\frac {2 c^{\frac {1}{3}} x +1}{\sqrt {3}}\right ) b +4 c^{\frac {4}{3}} \mathit {atanh} \left (c \,x^{3}\right ) b \,x^{4}+2 \mathit {atanh} \left (c \,x^{3}\right ) b +4 c^{\frac {4}{3}} a \,x^{4}+12 c^{\frac {1}{3}} b x -3 \,\mathrm {log}\left (c^{\frac {2}{3}} x +c^{\frac {1}{3}}\right ) b +3 \,\mathrm {log}\left (c^{\frac {2}{3}} x -c^{\frac {1}{3}}\right ) b}{16 c^{\frac {4}{3}}} \] Input:

int(x^3*(a+b*atanh(c*x^3)),x)
 

Output:

( - 2*sqrt(3)*atan((2*c**(1/3)*x - 1)/sqrt(3))*b - 2*sqrt(3)*atan((2*c**(1 
/3)*x + 1)/sqrt(3))*b + 4*c**(1/3)*atanh(c*x**3)*b*c*x**4 + 2*atanh(c*x**3 
)*b + 4*c**(1/3)*a*c*x**4 + 12*c**(1/3)*b*x - 3*log(c**(2/3)*x + c**(1/3)) 
*b + 3*log(c**(2/3)*x - c**(1/3))*b)/(16*c**(1/3)*c)