3.2.10 \(\int \frac {a+b \text {arctanh}(c x^3)}{x^6} \, dx\) [110]

3.2.10.1 Optimal result

Integrand size = 14, antiderivative size = 115 \[ \int \frac {a+b \text {arctanh}\left (c x^3\right )}{x^6} \, dx=-\frac {3 b c}{10 x^2}-\frac {1}{10} \sqrt {3} b c^{5/3} \arctan \left (\frac {1+2 c^{2/3} x^2}{\sqrt {3}}\right )-\frac {a+b \text {arctanh}\left (c x^3\right )}{5 x^5}-\frac {1}{10} b c^{5/3} \log \left (1-c^{2/3} x^2\right )+\frac {1}{20} b c^{5/3} \log \left (1+c^{2/3} x^2+c^{4/3} x^4\right ) \]

output

-3/10*b*c/x^2+1/5*(-a-b*arctanh(c*x^3))/x^5-1/10*b*c^(5/3)*ln(1-c^(2/3)*x^ 
2)+1/20*b*c^(5/3)*ln(1+c^(2/3)*x^2+c^(4/3)*x^4)-1/10*b*c^(5/3)*arctan(1/3* 
(1+2*c^(2/3)*x^2)*3^(1/2))*3^(1/2)
 

3.2.10.2 Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.70 \[ \int \frac {a+b \text {arctanh}\left (c x^3\right )}{x^6} \, dx=-\frac {a}{5 x^5}-\frac {3 b c}{10 x^2}-\frac {1}{10} \sqrt {3} b c^{5/3} \arctan \left (\frac {-1+2 \sqrt [3]{c} x}{\sqrt {3}}\right )+\frac {1}{10} \sqrt {3} b c^{5/3} \arctan \left (\frac {1+2 \sqrt [3]{c} x}{\sqrt {3}}\right )-\frac {b \text {arctanh}\left (c x^3\right )}{5 x^5}-\frac {1}{10} b c^{5/3} \log \left (1-\sqrt [3]{c} x\right )-\frac {1}{10} b c^{5/3} \log \left (1+\sqrt [3]{c} x\right )+\frac {1}{20} b c^{5/3} \log \left (1-\sqrt [3]{c} x+c^{2/3} x^2\right )+\frac {1}{20} b c^{5/3} \log \left (1+\sqrt [3]{c} x+c^{2/3} x^2\right ) \]

input

Integrate[(a + b*ArcTanh[c*x^3])/x^6,x]
 

output

-1/5*a/x^5 - (3*b*c)/(10*x^2) - (Sqrt[3]*b*c^(5/3)*ArcTan[(-1 + 2*c^(1/3)* 
x)/Sqrt[3]])/10 + (Sqrt[3]*b*c^(5/3)*ArcTan[(1 + 2*c^(1/3)*x)/Sqrt[3]])/10 
 - (b*ArcTanh[c*x^3])/(5*x^5) - (b*c^(5/3)*Log[1 - c^(1/3)*x])/10 - (b*c^( 
5/3)*Log[1 + c^(1/3)*x])/10 + (b*c^(5/3)*Log[1 - c^(1/3)*x + c^(2/3)*x^2]) 
/20 + (b*c^(5/3)*Log[1 + c^(1/3)*x + c^(2/3)*x^2])/20
 

3.2.10.3 Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.10, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.714, Rules used = {6452, 807, 847, 821, 16, 1142, 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.

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

\(\Big \downarrow \) 6452

\(\displaystyle \frac {3}{5} b c \int \frac {1}{x^3 \left (1-c^2 x^6\right )}dx-\frac {a+b \text {arctanh}\left (c x^3\right )}{5 x^5}\)

\(\Big \downarrow \) 807

\(\displaystyle \frac {3}{10} b c \int \frac {1}{x^4 \left (1-c^2 x^6\right )}dx^2-\frac {a+b \text {arctanh}\left (c x^3\right )}{5 x^5}\)

\(\Big \downarrow \) 847

\(\displaystyle \frac {3}{10} b c \left (c^2 \int \frac {x^2}{1-c^2 x^6}dx^2-\frac {1}{x^2}\right )-\frac {a+b \text {arctanh}\left (c x^3\right )}{5 x^5}\)

\(\Big \downarrow \) 821

\(\displaystyle \frac {3}{10} b c \left (c^2 \left (\frac {\int \frac {1}{1-c^{2/3} x^2}dx^2}{3 c^{2/3}}-\frac {\int \frac {1-c^{2/3} x^2}{c^{4/3} x^4+c^{2/3} x^2+1}dx^2}{3 c^{2/3}}\right )-\frac {1}{x^2}\right )-\frac {a+b \text {arctanh}\left (c x^3\right )}{5 x^5}\)

\(\Big \downarrow \) 16

\(\displaystyle \frac {3}{10} b c \left (c^2 \left (-\frac {\int \frac {1-c^{2/3} x^2}{c^{4/3} x^4+c^{2/3} x^2+1}dx^2}{3 c^{2/3}}-\frac {\log \left (1-c^{2/3} x^2\right )}{3 c^{4/3}}\right )-\frac {1}{x^2}\right )-\frac {a+b \text {arctanh}\left (c x^3\right )}{5 x^5}\)

\(\Big \downarrow \) 1142

\(\displaystyle \frac {3}{10} b c \left (c^2 \left (-\frac {\frac {3}{2} \int \frac {1}{c^{4/3} x^4+c^{2/3} x^2+1}dx^2-\frac {\int \frac {c^{2/3} \left (2 c^{2/3} x^2+1\right )}{c^{4/3} x^4+c^{2/3} x^2+1}dx^2}{2 c^{2/3}}}{3 c^{2/3}}-\frac {\log \left (1-c^{2/3} x^2\right )}{3 c^{4/3}}\right )-\frac {1}{x^2}\right )-\frac {a+b \text {arctanh}\left (c x^3\right )}{5 x^5}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {3}{10} b c \left (c^2 \left (-\frac {\frac {3}{2} \int \frac {1}{c^{4/3} x^4+c^{2/3} x^2+1}dx^2-\frac {1}{2} \int \frac {2 c^{2/3} x^2+1}{c^{4/3} x^4+c^{2/3} x^2+1}dx^2}{3 c^{2/3}}-\frac {\log \left (1-c^{2/3} x^2\right )}{3 c^{4/3}}\right )-\frac {1}{x^2}\right )-\frac {a+b \text {arctanh}\left (c x^3\right )}{5 x^5}\)

\(\Big \downarrow \) 1082

\(\displaystyle \frac {3}{10} b c \left (c^2 \left (-\frac {-\frac {3 \int \frac {1}{-x^4-3}d\left (2 c^{2/3} x^2+1\right )}{c^{2/3}}-\frac {1}{2} \int \frac {2 c^{2/3} x^2+1}{c^{4/3} x^4+c^{2/3} x^2+1}dx^2}{3 c^{2/3}}-\frac {\log \left (1-c^{2/3} x^2\right )}{3 c^{4/3}}\right )-\frac {1}{x^2}\right )-\frac {a+b \text {arctanh}\left (c x^3\right )}{5 x^5}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {3}{10} b c \left (c^2 \left (-\frac {\frac {\sqrt {3} \arctan \left (\frac {2 c^{2/3} x^2+1}{\sqrt {3}}\right )}{c^{2/3}}-\frac {1}{2} \int \frac {2 c^{2/3} x^2+1}{c^{4/3} x^4+c^{2/3} x^2+1}dx^2}{3 c^{2/3}}-\frac {\log \left (1-c^{2/3} x^2\right )}{3 c^{4/3}}\right )-\frac {1}{x^2}\right )-\frac {a+b \text {arctanh}\left (c x^3\right )}{5 x^5}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {3}{10} b c \left (c^2 \left (-\frac {\frac {\sqrt {3} \arctan \left (\frac {2 c^{2/3} x^2+1}{\sqrt {3}}\right )}{c^{2/3}}-\frac {\log \left (c^{4/3} x^4+c^{2/3} x^2+1\right )}{2 c^{2/3}}}{3 c^{2/3}}-\frac {\log \left (1-c^{2/3} x^2\right )}{3 c^{4/3}}\right )-\frac {1}{x^2}\right )-\frac {a+b \text {arctanh}\left (c x^3\right )}{5 x^5}\)

input

Int[(a + b*ArcTanh[c*x^3])/x^6,x]
 

output

-1/5*(a + b*ArcTanh[c*x^3])/x^5 + (3*b*c*(-x^(-2) + c^2*(-1/3*Log[1 - c^(2 
/3)*x^2]/c^(4/3) - ((Sqrt[3]*ArcTan[(1 + 2*c^(2/3)*x^2)/Sqrt[3]])/c^(2/3) 
- Log[1 + c^(2/3)*x^2 + c^(4/3)*x^4]/(2*c^(2/3)))/(3*c^(2/3)))))/10
 

3.2.10.3.1 Defintions of rubi rules used

rule 16

Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + 
b*x, x]]/b), x] /; FreeQ[{a, b, c}, 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 807

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m 
+ 1, n]}, Simp[1/k   Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p, x], x, 
x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]
 

rule 821

Int[(x_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> Simp[-(3*Rt[a, 3]*Rt[b, 3])^(- 
1)   Int[1/(Rt[a, 3] + Rt[b, 3]*x), x], x] + Simp[1/(3*Rt[a, 3]*Rt[b, 3]) 
 Int[(Rt[a, 3] + Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3]^2 
*x^2), x], x] /; FreeQ[{a, b}, x]
 

rule 847

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

3.2.10.4 Maple [A] (verified)

Time = 0.46 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.50

method	result	size
default	\(-\frac {a}{5 x^{5}}-\frac {b \,\operatorname {arctanh}\left (c \,x^{3}\right )}{5 x^{5}}-\frac {3 b c}{10 x^{2}}-\frac {b c \ln \left (x -\left (\frac {1}{c}\right )^{\frac {1}{3}}\right )}{10 \left (\frac {1}{c}\right )^{\frac {2}{3}}}+\frac {b c \ln \left (x^{2}+\left (\frac {1}{c}\right )^{\frac {1}{3}} x +\left (\frac {1}{c}\right )^{\frac {2}{3}}\right )}{20 \left (\frac {1}{c}\right )^{\frac {2}{3}}}+\frac {b c \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {1}{c}\right )^{\frac {1}{3}}}+1\right )}{3}\right )}{10 \left (\frac {1}{c}\right )^{\frac {2}{3}}}-\frac {b c \ln \left (x +\left (\frac {1}{c}\right )^{\frac {1}{3}}\right )}{10 \left (\frac {1}{c}\right )^{\frac {2}{3}}}+\frac {b c \ln \left (x^{2}-\left (\frac {1}{c}\right )^{\frac {1}{3}} x +\left (\frac {1}{c}\right )^{\frac {2}{3}}\right )}{20 \left (\frac {1}{c}\right )^{\frac {2}{3}}}-\frac {b c \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {1}{c}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{10 \left (\frac {1}{c}\right )^{\frac {2}{3}}}\)	\(172\)
parts	\(-\frac {a}{5 x^{5}}-\frac {b \,\operatorname {arctanh}\left (c \,x^{3}\right )}{5 x^{5}}-\frac {3 b c}{10 x^{2}}-\frac {b c \ln \left (x -\left (\frac {1}{c}\right )^{\frac {1}{3}}\right )}{10 \left (\frac {1}{c}\right )^{\frac {2}{3}}}+\frac {b c \ln \left (x^{2}+\left (\frac {1}{c}\right )^{\frac {1}{3}} x +\left (\frac {1}{c}\right )^{\frac {2}{3}}\right )}{20 \left (\frac {1}{c}\right )^{\frac {2}{3}}}+\frac {b c \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {1}{c}\right )^{\frac {1}{3}}}+1\right )}{3}\right )}{10 \left (\frac {1}{c}\right )^{\frac {2}{3}}}-\frac {b c \ln \left (x +\left (\frac {1}{c}\right )^{\frac {1}{3}}\right )}{10 \left (\frac {1}{c}\right )^{\frac {2}{3}}}+\frac {b c \ln \left (x^{2}-\left (\frac {1}{c}\right )^{\frac {1}{3}} x +\left (\frac {1}{c}\right )^{\frac {2}{3}}\right )}{20 \left (\frac {1}{c}\right )^{\frac {2}{3}}}-\frac {b c \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {1}{c}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{10 \left (\frac {1}{c}\right )^{\frac {2}{3}}}\)	\(172\)
risch	\(-\frac {b \ln \left (c \,x^{3}+1\right )}{10 x^{5}}-\frac {a}{5 x^{5}}+\frac {b \ln \left (-c \,x^{3}+1\right )}{10 x^{5}}-\frac {3 b c}{10 x^{2}}-\frac {b c \ln \left (x -\left (\frac {1}{c}\right )^{\frac {1}{3}}\right )}{10 \left (\frac {1}{c}\right )^{\frac {2}{3}}}+\frac {b c \ln \left (x^{2}+\left (\frac {1}{c}\right )^{\frac {1}{3}} x +\left (\frac {1}{c}\right )^{\frac {2}{3}}\right )}{20 \left (\frac {1}{c}\right )^{\frac {2}{3}}}+\frac {b c \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {1}{c}\right )^{\frac {1}{3}}}+1\right )}{3}\right )}{10 \left (\frac {1}{c}\right )^{\frac {2}{3}}}-\frac {b c \ln \left (x +\left (\frac {1}{c}\right )^{\frac {1}{3}}\right )}{10 \left (\frac {1}{c}\right )^{\frac {2}{3}}}+\frac {b c \ln \left (x^{2}-\left (\frac {1}{c}\right )^{\frac {1}{3}} x +\left (\frac {1}{c}\right )^{\frac {2}{3}}\right )}{20 \left (\frac {1}{c}\right )^{\frac {2}{3}}}-\frac {b c \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {1}{c}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{10 \left (\frac {1}{c}\right )^{\frac {2}{3}}}\)	\(189\)

input

int((a+b*arctanh(c*x^3))/x^6,x,method=_RETURNVERBOSE)
 

output

-1/5*a/x^5-1/5*b/x^5*arctanh(c*x^3)-3/10*b*c/x^2-1/10*b*c/(1/c)^(2/3)*ln(x 
-(1/c)^(1/3))+1/20*b*c/(1/c)^(2/3)*ln(x^2+(1/c)^(1/3)*x+(1/c)^(2/3))+1/10* 
b*c/(1/c)^(2/3)*3^(1/2)*arctan(1/3*3^(1/2)*(2/(1/c)^(1/3)*x+1))-1/10*b*c/( 
1/c)^(2/3)*ln(x+(1/c)^(1/3))+1/20*b*c/(1/c)^(2/3)*ln(x^2-(1/c)^(1/3)*x+(1/ 
c)^(2/3))-1/10*b*c/(1/c)^(2/3)*3^(1/2)*arctan(1/3*3^(1/2)*(2/(1/c)^(1/3)*x 
-1))
 

3.2.10.5 Fricas [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.31 \[ \int \frac {a+b \text {arctanh}\left (c x^3\right )}{x^6} \, dx=-\frac {2 \, \sqrt {3} \left (-c^{2}\right )^{\frac {1}{3}} b c x^{5} \arctan \left (\frac {2}{3} \, \sqrt {3} \left (-c^{2}\right )^{\frac {1}{3}} x^{2} - \frac {1}{3} \, \sqrt {3}\right ) + \left (-c^{2}\right )^{\frac {1}{3}} b c x^{5} \log \left (c^{2} x^{4} + \left (-c^{2}\right )^{\frac {2}{3}} x^{2} - \left (-c^{2}\right )^{\frac {1}{3}}\right ) - 2 \, \left (-c^{2}\right )^{\frac {1}{3}} b c x^{5} \log \left (c^{2} x^{2} - \left (-c^{2}\right )^{\frac {2}{3}}\right ) + 6 \, b c x^{3} + 2 \, b \log \left (-\frac {c x^{3} + 1}{c x^{3} - 1}\right ) + 4 \, a}{20 \, x^{5}} \]

input

integrate((a+b*arctanh(c*x^3))/x^6,x, algorithm="fricas")
 

output

-1/20*(2*sqrt(3)*(-c^2)^(1/3)*b*c*x^5*arctan(2/3*sqrt(3)*(-c^2)^(1/3)*x^2 
- 1/3*sqrt(3)) + (-c^2)^(1/3)*b*c*x^5*log(c^2*x^4 + (-c^2)^(2/3)*x^2 - (-c 
^2)^(1/3)) - 2*(-c^2)^(1/3)*b*c*x^5*log(c^2*x^2 - (-c^2)^(2/3)) + 6*b*c*x^ 
3 + 2*b*log(-(c*x^3 + 1)/(c*x^3 - 1)) + 4*a)/x^5
 

3.2.10.6 Sympy [F(-1)]

Timed out. \[ \int \frac {a+b \text {arctanh}\left (c x^3\right )}{x^6} \, dx=\text {Timed out} \]

input

integrate((a+b*atanh(c*x**3))/x**6,x)
 

output

Timed out
 

3.2.10.7 Maxima [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.87 \[ \int \frac {a+b \text {arctanh}\left (c x^3\right )}{x^6} \, dx=-\frac {1}{20} \, {\left ({\left (2 \, \sqrt {3} c^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, c^{\frac {4}{3}} x^{2} + c^{\frac {2}{3}}\right )}}{3 \, c^{\frac {2}{3}}}\right ) - c^{\frac {2}{3}} \log \left (c^{\frac {4}{3}} x^{4} + c^{\frac {2}{3}} x^{2} + 1\right ) + 2 \, c^{\frac {2}{3}} \log \left (\frac {c^{\frac {2}{3}} x^{2} - 1}{c^{\frac {2}{3}}}\right ) + \frac {6}{x^{2}}\right )} c + \frac {4 \, \operatorname {artanh}\left (c x^{3}\right )}{x^{5}}\right )} b - \frac {a}{5 \, x^{5}} \]

input

integrate((a+b*arctanh(c*x^3))/x^6,x, algorithm="maxima")
 

output

-1/20*((2*sqrt(3)*c^(2/3)*arctan(1/3*sqrt(3)*(2*c^(4/3)*x^2 + c^(2/3))/c^( 
2/3)) - c^(2/3)*log(c^(4/3)*x^4 + c^(2/3)*x^2 + 1) + 2*c^(2/3)*log((c^(2/3 
)*x^2 - 1)/c^(2/3)) + 6/x^2)*c + 4*arctanh(c*x^3)/x^5)*b - 1/5*a/x^5
 

3.2.10.8 Giac [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.09 \[ \int \frac {a+b \text {arctanh}\left (c x^3\right )}{x^6} \, dx=-\frac {1}{20} \, b c^{3} {\left (\frac {2 \, \sqrt {3} {\left | c \right |}^{\frac {2}{3}} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{2} + \frac {1}{{\left | c \right |}^{\frac {2}{3}}}\right )} {\left | c \right |}^{\frac {2}{3}}\right )}{c^{2}} - \frac {{\left | c \right |}^{\frac {2}{3}} \log \left (x^{4} + \frac {x^{2}}{{\left | c \right |}^{\frac {2}{3}}} + \frac {1}{{\left | c \right |}^{\frac {4}{3}}}\right )}{c^{2}} + \frac {2 \, \log \left ({\left | x^{2} - \frac {1}{{\left | c \right |}^{\frac {2}{3}}} \right |}\right )}{{\left | c \right |}^{\frac {4}{3}}}\right )} - \frac {b \log \left (-\frac {c x^{3} + 1}{c x^{3} - 1}\right )}{10 \, x^{5}} - \frac {3 \, b c x^{3} + 2 \, a}{10 \, x^{5}} \]

input

integrate((a+b*arctanh(c*x^3))/x^6,x, algorithm="giac")
 

output

-1/20*b*c^3*(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)) - 1/10*b*lo 
g(-(c*x^3 + 1)/(c*x^3 - 1))/x^5 - 1/10*(3*b*c*x^3 + 2*a)/x^5
 

3.2.10.9 Mupad [B] (verification not implemented)

Time = 6.23 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.17 \[ \int \frac {a+b \text {arctanh}\left (c x^3\right )}{x^6} \, dx=\frac {b\,\ln \left (1-c\,x^3\right )}{10\,x^5}-\frac {b\,c^{5/3}\,\ln \left (c^{2/3}\,x^2-1\right )}{10}-\frac {b\,\ln \left (c\,x^3+1\right )}{10\,x^5}-\frac {\frac {3\,b\,c\,x^3}{2}+a}{5\,x^5}+\frac {b\,c^{5/3}\,\ln \left (\sqrt {3}\,c^{2/3}\,x^2-c^{2/3}\,x^2\,1{}\mathrm {i}-2{}\mathrm {i}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{20}-\frac {b\,c^{5/3}\,\ln \left (-c^{2/3}\,x^2\,1{}\mathrm {i}-\sqrt {3}\,c^{2/3}\,x^2-2{}\mathrm {i}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{20} \]

input

int((a + b*atanh(c*x^3))/x^6,x)
 

output

(b*log(1 - c*x^3))/(10*x^5) - (b*c^(5/3)*log(c^(2/3)*x^2 - 1))/10 - (b*log 
(c*x^3 + 1))/(10*x^5) - (a + (3*b*c*x^3)/2)/(5*x^5) + (b*c^(5/3)*log(3^(1/ 
2)*c^(2/3)*x^2 - c^(2/3)*x^2*1i - 2i)*(3^(1/2)*1i + 1))/20 - (b*c^(5/3)*lo 
g(- c^(2/3)*x^2*1i - 3^(1/2)*c^(2/3)*x^2 - 2i)*(3^(1/2)*1i - 1))/20