\(\int \frac {\text {arctanh}(\frac {\sqrt {e} x}{\sqrt {d+e x^2}})}{x^9} \, dx\) [8]

Optimal result

Integrand size = 23, antiderivative size = 131 \[ \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^9} \, dx=-\frac {\sqrt {e} \sqrt {d+e x^2}}{56 d x^7}+\frac {3 e^{3/2} \sqrt {d+e x^2}}{140 d^2 x^5}-\frac {e^{5/2} \sqrt {d+e x^2}}{35 d^3 x^3}+\frac {2 e^{7/2} \sqrt {d+e x^2}}{35 d^4 x}-\frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{8 x^8} \] Output:

-1/56*e^(1/2)*(e*x^2+d)^(1/2)/d/x^7+3/140*e^(3/2)*(e*x^2+d)^(1/2)/d^2/x^5- 
1/35*e^(5/2)*(e*x^2+d)^(1/2)/d^3/x^3+2/35*e^(7/2)*(e*x^2+d)^(1/2)/d^4/x-1/ 
8*arctanh(e^(1/2)*x/(e*x^2+d)^(1/2))/x^8
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.65 \[ \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^9} \, dx=\frac {\sqrt {e} x \sqrt {d+e x^2} \left (-5 d^3+6 d^2 e x^2-8 d e^2 x^4+16 e^3 x^6\right )-35 d^4 \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{280 d^4 x^8} \] Input:

Integrate[ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]]/x^9,x]
 

Output:

(Sqrt[e]*x*Sqrt[d + e*x^2]*(-5*d^3 + 6*d^2*e*x^2 - 8*d*e^2*x^4 + 16*e^3*x^ 
6) - 35*d^4*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/(280*d^4*x^8)
 

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.07, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {6775, 245, 245, 245, 242}

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.

\(\displaystyle \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^9} \, dx\)

\(\Big \downarrow \) 6775

\(\displaystyle \frac {1}{8} \sqrt {e} \int \frac {1}{x^8 \sqrt {e x^2+d}}dx-\frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{8 x^8}\)

\(\Big \downarrow \) 245

\(\displaystyle \frac {1}{8} \sqrt {e} \left (-\frac {6 e \int \frac {1}{x^6 \sqrt {e x^2+d}}dx}{7 d}-\frac {\sqrt {d+e x^2}}{7 d x^7}\right )-\frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{8 x^8}\)

\(\Big \downarrow \) 245

\(\displaystyle \frac {1}{8} \sqrt {e} \left (-\frac {6 e \left (-\frac {4 e \int \frac {1}{x^4 \sqrt {e x^2+d}}dx}{5 d}-\frac {\sqrt {d+e x^2}}{5 d x^5}\right )}{7 d}-\frac {\sqrt {d+e x^2}}{7 d x^7}\right )-\frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{8 x^8}\)

\(\Big \downarrow \) 245

\(\displaystyle \frac {1}{8} \sqrt {e} \left (-\frac {6 e \left (-\frac {4 e \left (-\frac {2 e \int \frac {1}{x^2 \sqrt {e x^2+d}}dx}{3 d}-\frac {\sqrt {d+e x^2}}{3 d x^3}\right )}{5 d}-\frac {\sqrt {d+e x^2}}{5 d x^5}\right )}{7 d}-\frac {\sqrt {d+e x^2}}{7 d x^7}\right )-\frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{8 x^8}\)

\(\Big \downarrow \) 242

\(\displaystyle \frac {1}{8} \sqrt {e} \left (-\frac {6 e \left (-\frac {4 e \left (\frac {2 e \sqrt {d+e x^2}}{3 d^2 x}-\frac {\sqrt {d+e x^2}}{3 d x^3}\right )}{5 d}-\frac {\sqrt {d+e x^2}}{5 d x^5}\right )}{7 d}-\frac {\sqrt {d+e x^2}}{7 d x^7}\right )-\frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{8 x^8}\)

Input:

Int[ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]]/x^9,x]
 

Output:

(Sqrt[e]*(-1/7*Sqrt[d + e*x^2]/(d*x^7) - (6*e*(-1/5*Sqrt[d + e*x^2]/(d*x^5 
) - (4*e*(-1/3*Sqrt[d + e*x^2]/(d*x^3) + (2*e*Sqrt[d + e*x^2])/(3*d^2*x))) 
/(5*d)))/(7*d)))/8 - ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]]/(8*x^8)
 

Defintions of rubi rules used

rule 242

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^ 
(m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] /; FreeQ[{a, b, c, m, p}, x 
] && EqQ[m + 2*p + 3, 0] && NeQ[m, -1]
 

rule 245

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x^(m + 1)*((a + 
 b*x^2)^(p + 1)/(a*(m + 1))), x] - Simp[b*((m + 2*(p + 1) + 1)/(a*(m + 1))) 
   Int[x^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, m, p}, x] && ILtQ[Si 
mplify[(m + 1)/2 + p + 1], 0] && NeQ[m, -1]
 

rule 6775

Int[ArcTanh[((c_.)*(x_))/Sqrt[(a_.) + (b_.)*(x_)^2]]*((d_.)*(x_))^(m_.), x_ 
Symbol] :> Simp[(d*x)^(m + 1)*(ArcTanh[(c*x)/Sqrt[a + b*x^2]]/(d*(m + 1))), 
 x] - Simp[c/(d*(m + 1))   Int[(d*x)^(m + 1)/Sqrt[a + b*x^2], x], x] /; Fre 
eQ[{a, b, c, d, m}, x] && EqQ[b, c^2] && NeQ[m, -1]
 

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.21

method	result	size
default	\(-\frac {\operatorname {arctanh}\left (\frac {\sqrt {e}\, x}{\sqrt {e \,x^{2}+d}}\right )}{8 x^{8}}-\frac {e^{\frac {3}{2}} \left (-\frac {\sqrt {e \,x^{2}+d}}{5 d \,x^{5}}-\frac {4 e \left (-\frac {\sqrt {e \,x^{2}+d}}{3 d \,x^{3}}+\frac {2 e \sqrt {e \,x^{2}+d}}{3 d^{2} x}\right )}{5 d}\right )}{8 d}+\frac {\sqrt {e}\, \left (-\frac {\left (e \,x^{2}+d \right )^{\frac {3}{2}}}{7 d \,x^{7}}-\frac {4 e \left (-\frac {\left (e \,x^{2}+d \right )^{\frac {3}{2}}}{5 d \,x^{5}}+\frac {2 e \left (e \,x^{2}+d \right )^{\frac {3}{2}}}{15 d^{2} x^{3}}\right )}{7 d}\right )}{8 d}\)	\(158\)
parts	\(-\frac {\operatorname {arctanh}\left (\frac {\sqrt {e}\, x}{\sqrt {e \,x^{2}+d}}\right )}{8 x^{8}}-\frac {e^{\frac {3}{2}} \left (-\frac {\sqrt {e \,x^{2}+d}}{5 d \,x^{5}}-\frac {4 e \left (-\frac {\sqrt {e \,x^{2}+d}}{3 d \,x^{3}}+\frac {2 e \sqrt {e \,x^{2}+d}}{3 d^{2} x}\right )}{5 d}\right )}{8 d}+\frac {\sqrt {e}\, \left (-\frac {\left (e \,x^{2}+d \right )^{\frac {3}{2}}}{7 d \,x^{7}}-\frac {4 e \left (-\frac {\left (e \,x^{2}+d \right )^{\frac {3}{2}}}{5 d \,x^{5}}+\frac {2 e \left (e \,x^{2}+d \right )^{\frac {3}{2}}}{15 d^{2} x^{3}}\right )}{7 d}\right )}{8 d}\)	\(158\)

Input:

int(arctanh(e^(1/2)*x/(e*x^2+d)^(1/2))/x^9,x,method=_RETURNVERBOSE)
 

Output:

-1/8*arctanh(e^(1/2)*x/(e*x^2+d)^(1/2))/x^8-1/8*e^(3/2)/d*(-1/5/d/x^5*(e*x 
^2+d)^(1/2)-4/5*e/d*(-1/3/d/x^3*(e*x^2+d)^(1/2)+2/3*e/d^2/x*(e*x^2+d)^(1/2 
)))+1/8*e^(1/2)/d*(-1/7/d/x^7*(e*x^2+d)^(3/2)-4/7*e/d*(-1/5/d/x^5*(e*x^2+d 
)^(3/2)+2/15*e/d^2/x^3*(e*x^2+d)^(3/2)))
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.68 \[ \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^9} \, dx=-\frac {35 \, d^{4} \log \left (\frac {2 \, e x^{2} + 2 \, \sqrt {e x^{2} + d} \sqrt {e} x + d}{d}\right ) - 2 \, {\left (16 \, e^{3} x^{7} - 8 \, d e^{2} x^{5} + 6 \, d^{2} e x^{3} - 5 \, d^{3} x\right )} \sqrt {e x^{2} + d} \sqrt {e}}{560 \, d^{4} x^{8}} \] Input:

integrate(arctanh(e^(1/2)*x/(e*x^2+d)^(1/2))/x^9,x, algorithm="fricas")
 

Output:

-1/560*(35*d^4*log((2*e*x^2 + 2*sqrt(e*x^2 + d)*sqrt(e)*x + d)/d) - 2*(16* 
e^3*x^7 - 8*d*e^2*x^5 + 6*d^2*e*x^3 - 5*d^3*x)*sqrt(e*x^2 + d)*sqrt(e))/(d 
^4*x^8)
 

Sympy [F]

\[ \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^9} \, dx=\int \frac {\operatorname {atanh}{\left (\frac {\sqrt {e} x}{\sqrt {d + e x^{2}}} \right )}}{x^{9}}\, dx \] Input:

integrate(atanh(e**(1/2)*x/(e*x**2+d)**(1/2))/x**9,x)
 

Output:

Integral(atanh(sqrt(e)*x/sqrt(d + e*x**2))/x**9, x)
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.95 \[ \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^9} \, dx=\frac {{\left (8 \, e^{3} x^{6} + 4 \, d e^{2} x^{4} - d^{2} e x^{2} + 3 \, d^{3}\right )} e^{\frac {3}{2}}}{120 \, \sqrt {e x^{2} + d} d^{4} x^{5}} - \frac {\operatorname {artanh}\left (\frac {\sqrt {e} x}{\sqrt {e x^{2} + d}}\right )}{8 \, x^{8}} - \frac {{\left (8 \, e^{3} x^{6} - 4 \, d e^{2} x^{4} + 3 \, d^{2} e x^{2} + 15 \, d^{3}\right )} \sqrt {e x^{2} + d} \sqrt {e}}{840 \, d^{4} x^{7}} \] Input:

integrate(arctanh(e^(1/2)*x/(e*x^2+d)^(1/2))/x^9,x, algorithm="maxima")
 

Output:

1/120*(8*e^3*x^6 + 4*d*e^2*x^4 - d^2*e*x^2 + 3*d^3)*e^(3/2)/(sqrt(e*x^2 + 
d)*d^4*x^5) - 1/8*arctanh(sqrt(e)*x/sqrt(e*x^2 + d))/x^8 - 1/840*(8*e^3*x^ 
6 - 4*d*e^2*x^4 + 3*d^2*e*x^2 + 15*d^3)*sqrt(e*x^2 + d)*sqrt(e)/(d^4*x^7)
 

Giac [F(-1)]

Timed out. \[ \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^9} \, dx=\text {Timed out} \] Input:

integrate(arctanh(e^(1/2)*x/(e*x^2+d)^(1/2))/x^9,x, algorithm="giac")
 

Output:

Timed out
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^9} \, dx=\int \frac {\mathrm {atanh}\left (\frac {\sqrt {e}\,x}{\sqrt {e\,x^2+d}}\right )}{x^9} \,d x \] Input:

int(atanh((e^(1/2)*x)/(d + e*x^2)^(1/2))/x^9,x)
 

Output:

int(atanh((e^(1/2)*x)/(d + e*x^2)^(1/2))/x^9, x)
 

Reduce [F]

\[ \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^9} \, dx=\int \frac {\mathit {atanh} \left (\frac {\sqrt {e}\, x}{\sqrt {e \,x^{2}+d}}\right )}{x^{9}}d x \] Input:

int(atanh(e^(1/2)*x/(e*x^2+d)^(1/2))/x^9,x)
                                                                                    
                                                                                    
 

Output:

int(atanh((sqrt(e)*x)/sqrt(d + e*x**2))/x**9,x)