\(\int (a+b \text {arctanh}(c x^3)) \, dx\) [108]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [A] (verification not implemented)
Sympy [F(-1)]
Maxima [A] (verification not implemented)
Giac [A] (verification not implemented)
Mupad [B] (verification not implemented)
Reduce [B] (verification not implemented)

Optimal result

Integrand size = 10, antiderivative size = 101 \[ \int \left (a+b \text {arctanh}\left (c x^3\right )\right ) \, dx=a x+\frac {\sqrt {3} b \arctan \left (\frac {1+2 c^{2/3} x^2}{\sqrt {3}}\right )}{2 \sqrt [3]{c}}+b x \text {arctanh}\left (c x^3\right )+\frac {b \log \left (1-c^{2/3} x^2\right )}{2 \sqrt [3]{c}}-\frac {b \log \left (1+c^{2/3} x^2+c^{4/3} x^4\right )}{4 \sqrt [3]{c}} \] Output:

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.35 \[ \int \left (a+b \text {arctanh}\left (c x^3\right )\right ) \, dx=a x+b x \text {arctanh}\left (c x^3\right )-\frac {b \left (-2 \sqrt {3} \arctan \left (\frac {-1+2 \sqrt [3]{c} x}{\sqrt {3}}\right )+2 \sqrt {3} \arctan \left (\frac {1+2 \sqrt [3]{c} x}{\sqrt {3}}\right )-2 \log \left (1-\sqrt [3]{c} x\right )-2 \log \left (1+\sqrt [3]{c} x\right )+\log \left (1-\sqrt [3]{c} x+c^{2/3} x^2\right )+\log \left (1+\sqrt [3]{c} x+c^{2/3} x^2\right )\right )}{4 \sqrt [3]{c}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {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.

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 
Maple [A] (verified)

Time = 0.19 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.98

method result size
default \(a x +b x \,\operatorname {arctanh}\left (c \,x^{3}\right )+\frac {b \ln \left (x^{2}-\left (\frac {1}{c^{2}}\right )^{\frac {1}{3}}\right )}{2 c \left (\frac {1}{c^{2}}\right )^{\frac {1}{3}}}-\frac {b \ln \left (x^{4}+\left (\frac {1}{c^{2}}\right )^{\frac {1}{3}} x^{2}+\left (\frac {1}{c^{2}}\right )^{\frac {2}{3}}\right )}{4 c \left (\frac {1}{c^{2}}\right )^{\frac {1}{3}}}+\frac {b \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x^{2}}{\left (\frac {1}{c^{2}}\right )^{\frac {1}{3}}}+1\right )}{3}\right )}{2 c \left (\frac {1}{c^{2}}\right )^{\frac {1}{3}}}\) \(99\)
parts \(a x +b x \,\operatorname {arctanh}\left (c \,x^{3}\right )+\frac {b \ln \left (x^{2}-\left (\frac {1}{c^{2}}\right )^{\frac {1}{3}}\right )}{2 c \left (\frac {1}{c^{2}}\right )^{\frac {1}{3}}}-\frac {b \ln \left (x^{4}+\left (\frac {1}{c^{2}}\right )^{\frac {1}{3}} x^{2}+\left (\frac {1}{c^{2}}\right )^{\frac {2}{3}}\right )}{4 c \left (\frac {1}{c^{2}}\right )^{\frac {1}{3}}}+\frac {b \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x^{2}}{\left (\frac {1}{c^{2}}\right )^{\frac {1}{3}}}+1\right )}{3}\right )}{2 c \left (\frac {1}{c^{2}}\right )^{\frac {1}{3}}}\) \(99\)
risch \(a x +\frac {b x \ln \left (c \,x^{3}+1\right )}{2}+\frac {b \ln \left (x +\left (\frac {1}{c}\right )^{\frac {1}{3}}\right )}{2 c \left (\frac {1}{c}\right )^{\frac {2}{3}}}-\frac {b \ln \left (x^{2}-\left (\frac {1}{c}\right )^{\frac {1}{3}} x +\left (\frac {1}{c}\right )^{\frac {2}{3}}\right )}{4 c \left (\frac {1}{c}\right )^{\frac {2}{3}}}-\frac {b \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (-2 x +\left (\frac {1}{c}\right )^{\frac {1}{3}}\right )}{3 \left (\frac {1}{c}\right )^{\frac {1}{3}}}\right )}{2 c \left (\frac {1}{c}\right )^{\frac {2}{3}}}-\frac {b x \ln \left (-c \,x^{3}+1\right )}{2}+\frac {b \ln \left (x -\left (\frac {1}{c}\right )^{\frac {1}{3}}\right )}{2 c \left (\frac {1}{c}\right )^{\frac {2}{3}}}-\frac {b \ln \left (x^{2}+\left (\frac {1}{c}\right )^{\frac {1}{3}} x +\left (\frac {1}{c}\right )^{\frac {2}{3}}\right )}{4 c \left (\frac {1}{c}\right )^{\frac {2}{3}}}-\frac {b \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (2 x +\left (\frac {1}{c}\right )^{\frac {1}{3}}\right )}{3 \left (\frac {1}{c}\right )^{\frac {1}{3}}}\right )}{2 c \left (\frac {1}{c}\right )^{\frac {2}{3}}}\) \(195\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 260, normalized size of antiderivative = 2.57 \[ \int \left (a+b \text {arctanh}\left (c x^3\right )\right ) \, dx=\left [\frac {\sqrt {3} b c \sqrt {-\frac {1}{c^{\frac {2}{3}}}} \log \left (\frac {2 \, c^{2} x^{6} - 3 \, c^{\frac {2}{3}} x^{2} + \sqrt {3} {\left (2 \, c^{\frac {5}{3}} x^{4} - c x^{2} - c^{\frac {1}{3}}\right )} \sqrt {-\frac {1}{c^{\frac {2}{3}}}} + 1}{c^{2} x^{6} - 1}\right ) + 2 \, b c x \log \left (-\frac {c x^{3} + 1}{c x^{3} - 1}\right ) + 4 \, a c x - b c^{\frac {2}{3}} \log \left (c^{2} x^{4} + c^{\frac {4}{3}} x^{2} + c^{\frac {2}{3}}\right ) + 2 \, b c^{\frac {2}{3}} \log \left (c x^{2} - c^{\frac {1}{3}}\right )}{4 \, c}, \frac {2 \, b c x \log \left (-\frac {c x^{3} + 1}{c x^{3} - 1}\right ) + 2 \, \sqrt {3} b c^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, c x^{2} + c^{\frac {1}{3}}\right )}}{3 \, c^{\frac {1}{3}}}\right ) + 4 \, a c x - b c^{\frac {2}{3}} \log \left (c^{2} x^{4} + c^{\frac {4}{3}} x^{2} + c^{\frac {2}{3}}\right ) + 2 \, b c^{\frac {2}{3}} \log \left (c x^{2} - c^{\frac {1}{3}}\right )}{4 \, c}\right ] \] Input:

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.89 \[ \int \left (a+b \text {arctanh}\left (c x^3\right )\right ) \, dx=\frac {1}{4} \, {\left (c {\left (\frac {2 \, \sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (2 \, c^{\frac {4}{3}} x^{2} + c^{\frac {2}{3}}\right )}}{3 \, c^{\frac {2}{3}}}\right )}{c^{\frac {4}{3}}} - \frac {\log \left (c^{\frac {4}{3}} x^{4} + c^{\frac {2}{3}} x^{2} + 1\right )}{c^{\frac {4}{3}}} + \frac {2 \, \log \left (\frac {c^{\frac {2}{3}} x^{2} - 1}{c^{\frac {2}{3}}}\right )}{c^{\frac {4}{3}}}\right )} + 4 \, x \operatorname {artanh}\left (c x^{3}\right )\right )} b + a x \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.08 \[ \int \left (a+b \text {arctanh}\left (c x^3\right )\right ) \, dx=\frac {1}{4} \, {\left (c {\left (\frac {2 \, \sqrt {3} {\left | c \right |}^{\frac {2}{3}} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{2} + \frac {1}{{\left | c \right |}^{\frac {2}{3}}}\right )} {\left | c \right |}^{\frac {2}{3}}\right )}{c^{2}} - \frac {{\left | c \right |}^{\frac {2}{3}} \log \left (x^{4} + \frac {x^{2}}{{\left | c \right |}^{\frac {2}{3}}} + \frac {1}{{\left | c \right |}^{\frac {4}{3}}}\right )}{c^{2}} + \frac {2 \, \log \left ({\left | x^{2} - \frac {1}{{\left | c \right |}^{\frac {2}{3}}} \right |}\right )}{{\left | c \right |}^{\frac {4}{3}}}\right )} + 2 \, x \log \left (-\frac {c x^{3} + 1}{c x^{3} - 1}\right )\right )} b + a x \] Input:

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

Output:

1/4*(c*(2*sqrt(3)*abs(c)^(2/3)*arctan(1/3*sqrt(3)*(2*x^2 + 1/abs(c)^(2/3)) 
*abs(c)^(2/3))/c^2 - abs(c)^(2/3)*log(x^4 + x^2/abs(c)^(2/3) + 1/abs(c)^(4 
/3))/c^2 + 2*log(abs(x^2 - 1/abs(c)^(2/3)))/abs(c)^(4/3)) + 2*x*log(-(c*x^ 
3 + 1)/(c*x^3 - 1)))*b + a*x
 

Mupad [B] (verification not implemented)

Time = 5.54 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.06 \[ \int \left (a+b \text {arctanh}\left (c x^3\right )\right ) \, dx=a\,x+\frac {b\,\ln \left (c^{2/3}\,x^2-1\right )}{2\,c^{1/3}}-\frac {\ln \left (4\,c^{2/3}\,x^2+2-\sqrt {3}\,2{}\mathrm {i}\right )\,\left (b+\sqrt {3}\,b\,1{}\mathrm {i}\right )}{4\,c^{1/3}}-\frac {\ln \left (4\,c^{2/3}\,x^2+2+\sqrt {3}\,2{}\mathrm {i}\right )\,\left (b-\sqrt {3}\,b\,1{}\mathrm {i}\right )}{4\,c^{1/3}}+\frac {b\,x\,\ln \left (c\,x^3+1\right )}{2}-\frac {b\,x\,\ln \left (1-c\,x^3\right )}{2} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.18 \[ \int \left (a+b \text {arctanh}\left (c x^3\right )\right ) \, dx=\frac {2 \sqrt {3}\, \mathit {atan} \left (\frac {2 c^{\frac {1}{3}} x -1}{\sqrt {3}}\right ) b -2 \sqrt {3}\, \mathit {atan} \left (\frac {2 c^{\frac {1}{3}} x +1}{\sqrt {3}}\right ) b +4 c^{\frac {1}{3}} \mathit {atanh} \left (c \,x^{3}\right ) b x +2 \mathit {atanh} \left (c \,x^{3}\right ) b +4 c^{\frac {1}{3}} a x -2 \,\mathrm {log}\left (c^{\frac {2}{3}} x^{2}-c^{\frac {1}{3}} x +1\right ) b +\mathrm {log}\left (c^{\frac {2}{3}} x +c^{\frac {1}{3}}\right ) b +3 \,\mathrm {log}\left (c^{\frac {2}{3}} x -c^{\frac {1}{3}}\right ) b}{4 c^{\frac {1}{3}}} \] Input:

int(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*x + 2*atanh(c*x**3)*b + 4* 
c**(1/3)*a*x - 2*log(c**(2/3)*x**2 - c**(1/3)*x + 1)*b + log(c**(2/3)*x + 
c**(1/3))*b + 3*log(c**(2/3)*x - c**(1/3))*b)/(4*c**(1/3))