\(\int \frac {\text {arctanh}(\frac {\sqrt {e} x}{\sqrt {d+e x^2}})}{x^4} \, dx\) [14]

Optimal result

Integrand size = 23, antiderivative size = 85 \[ \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^4} \, dx=-\frac {\sqrt {e} \sqrt {d+e x^2}}{6 d x^2}-\frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{3 x^3}+\frac {e^{3/2} \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{6 d^{3/2}} \] Output:

-1/6*e^(1/2)*(e*x^2+d)^(1/2)/d/x^2-1/3*arctanh(e^(1/2)*x/(e*x^2+d)^(1/2))/ 
x^3+1/6*e^(3/2)*arctanh((e*x^2+d)^(1/2)/d^(1/2))/d^(3/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.08 \[ \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^4} \, dx=-\frac {2 \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )+\frac {\sqrt {e} x \left (\sqrt {d} \sqrt {d+e x^2}+e x^2 \log (x)-e x^2 \log \left (d+\sqrt {d} \sqrt {d+e x^2}\right )\right )}{d^{3/2}}}{6 x^3} \] Input:

Integrate[ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]]/x^4,x]
 

Output:

-1/6*(2*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]] + (Sqrt[e]*x*(Sqrt[d]*Sqrt[d 
+ e*x^2] + e*x^2*Log[x] - e*x^2*Log[d + Sqrt[d]*Sqrt[d + e*x^2]]))/d^(3/2) 
)/x^3
 

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.95, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {6775, 243, 52, 73, 221}

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.

\(\displaystyle \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^4} \, dx\)

\(\Big \downarrow \) 6775

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

\(\Big \downarrow \) 243

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

\(\Big \downarrow \) 52

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

Int[ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]]/x^4,x]
 

Output:

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

Defintions of rubi rules used

rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( 
m + n + 2)/((b*c - a*d)*(m + 1)))   Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], 
x] /; FreeQ[{a, b, c, d, n}, x] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 221

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

rule 243

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2   Subst[In 
t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I 
ntegerQ[(m - 1)/2]
 

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 = 0.03 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.45

method	result	size
default	\(-\frac {\operatorname {arctanh}\left (\frac {\sqrt {e}\, x}{\sqrt {e \,x^{2}+d}}\right )}{3 x^{3}}+\frac {e^{\frac {3}{2}} \ln \left (\frac {2 d +2 \sqrt {d}\, \sqrt {e \,x^{2}+d}}{x}\right )}{3 d^{\frac {3}{2}}}+\frac {\sqrt {e}\, \left (-\frac {\left (e \,x^{2}+d \right )^{\frac {3}{2}}}{2 d \,x^{2}}+\frac {e \left (\sqrt {e \,x^{2}+d}-\sqrt {d}\, \ln \left (\frac {2 d +2 \sqrt {d}\, \sqrt {e \,x^{2}+d}}{x}\right )\right )}{2 d}\right )}{3 d}\)	\(123\)
parts	\(-\frac {\operatorname {arctanh}\left (\frac {\sqrt {e}\, x}{\sqrt {e \,x^{2}+d}}\right )}{3 x^{3}}+\frac {e^{\frac {3}{2}} \ln \left (\frac {2 d +2 \sqrt {d}\, \sqrt {e \,x^{2}+d}}{x}\right )}{3 d^{\frac {3}{2}}}+\frac {\sqrt {e}\, \left (-\frac {\left (e \,x^{2}+d \right )^{\frac {3}{2}}}{2 d \,x^{2}}+\frac {e \left (\sqrt {e \,x^{2}+d}-\sqrt {d}\, \ln \left (\frac {2 d +2 \sqrt {d}\, \sqrt {e \,x^{2}+d}}{x}\right )\right )}{2 d}\right )}{3 d}\)	\(123\)

Input:

int(arctanh(e^(1/2)*x/(e*x^2+d)^(1/2))/x^4,x,method=_RETURNVERBOSE)
 

Output:

-1/3*arctanh(e^(1/2)*x/(e*x^2+d)^(1/2))/x^3+1/3*e^(3/2)/d^(3/2)*ln((2*d+2* 
d^(1/2)*(e*x^2+d)^(1/2))/x)+1/3*e^(1/2)/d*(-1/2/d/x^2*(e*x^2+d)^(3/2)+1/2* 
e/d*((e*x^2+d)^(1/2)-d^(1/2)*ln((2*d+2*d^(1/2)*(e*x^2+d)^(1/2))/x)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 167 vs. \(2 (63) = 126\).

Time = 0.14 (sec) , antiderivative size = 340, normalized size of antiderivative = 4.00 \[ \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^4} \, dx=\left [\frac {e x^{3} \sqrt {\frac {e}{d}} \log \left (-\frac {e^{2} x^{2} + 2 \, \sqrt {e x^{2} + d} d \sqrt {e} \sqrt {\frac {e}{d}} + 2 \, d e}{x^{2}}\right ) - 2 \, d x^{3} \log \left (\frac {e x + \sqrt {e x^{2} + d} \sqrt {e}}{x}\right ) + 2 \, d x^{3} \log \left (\frac {e x - \sqrt {e x^{2} + d} \sqrt {e}}{x}\right ) - 2 \, \sqrt {e x^{2} + d} \sqrt {e} x + 2 \, {\left (d x^{3} - d\right )} \log \left (\frac {2 \, e x^{2} + 2 \, \sqrt {e x^{2} + d} \sqrt {e} x + d}{d}\right )}{12 \, d x^{3}}, -\frac {e x^{3} \sqrt {-\frac {e}{d}} \arctan \left (\frac {\sqrt {e x^{2} + d} d \sqrt {e} \sqrt {-\frac {e}{d}}}{e^{2} x^{2} + d e}\right ) + d x^{3} \log \left (\frac {e x + \sqrt {e x^{2} + d} \sqrt {e}}{x}\right ) - d x^{3} \log \left (\frac {e x - \sqrt {e x^{2} + d} \sqrt {e}}{x}\right ) + \sqrt {e x^{2} + d} \sqrt {e} x - {\left (d x^{3} - d\right )} \log \left (\frac {2 \, e x^{2} + 2 \, \sqrt {e x^{2} + d} \sqrt {e} x + d}{d}\right )}{6 \, d x^{3}}\right ] \] Input:

integrate(arctanh(e^(1/2)*x/(e*x^2+d)^(1/2))/x^4,x, algorithm="fricas")
 

Output:

[1/12*(e*x^3*sqrt(e/d)*log(-(e^2*x^2 + 2*sqrt(e*x^2 + d)*d*sqrt(e)*sqrt(e/ 
d) + 2*d*e)/x^2) - 2*d*x^3*log((e*x + sqrt(e*x^2 + d)*sqrt(e))/x) + 2*d*x^ 
3*log((e*x - sqrt(e*x^2 + d)*sqrt(e))/x) - 2*sqrt(e*x^2 + d)*sqrt(e)*x + 2 
*(d*x^3 - d)*log((2*e*x^2 + 2*sqrt(e*x^2 + d)*sqrt(e)*x + d)/d))/(d*x^3), 
-1/6*(e*x^3*sqrt(-e/d)*arctan(sqrt(e*x^2 + d)*d*sqrt(e)*sqrt(-e/d)/(e^2*x^ 
2 + d*e)) + d*x^3*log((e*x + sqrt(e*x^2 + d)*sqrt(e))/x) - d*x^3*log((e*x 
- sqrt(e*x^2 + d)*sqrt(e))/x) + sqrt(e*x^2 + d)*sqrt(e)*x - (d*x^3 - d)*lo 
g((2*e*x^2 + 2*sqrt(e*x^2 + d)*sqrt(e)*x + d)/d))/(d*x^3)]
 

Sympy [F]

\[ \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^4} \, dx=\int \frac {\operatorname {atanh}{\left (\frac {\sqrt {e} x}{\sqrt {d + e x^{2}}} \right )}}{x^{4}}\, dx \] Input:

integrate(atanh(e**(1/2)*x/(e*x**2+d)**(1/2))/x**4,x)
 

Output:

Integral(atanh(sqrt(e)*x/sqrt(d + e*x**2))/x**4, x)
 

Maxima [F]

\[ \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^4} \, dx=\int { \frac {\operatorname {artanh}\left (\frac {\sqrt {e} x}{\sqrt {e x^{2} + d}}\right )}{x^{4}} \,d x } \] Input:

integrate(arctanh(e^(1/2)*x/(e*x^2+d)^(1/2))/x^4,x, algorithm="maxima")
 

Output:

d*sqrt(e)*integrate(-1/3*sqrt(e*x^2 + d)/(e^2*x^7 + d*e*x^5 - (e*x^5 + d*x 
^3)*(e*x^2 + d)), x) - 1/6*(log(sqrt(e)*x + sqrt(e*x^2 + d)) - log(-sqrt(e 
)*x + sqrt(e*x^2 + d)))/x^3
 

Giac [F(-1)]

Timed out. \[ \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^4} \, dx=\text {Timed out} \] Input:

integrate(arctanh(e^(1/2)*x/(e*x^2+d)^(1/2))/x^4,x, algorithm="giac")
 

Output:

Timed out
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^4} \, dx=\int \frac {\mathrm {atanh}\left (\frac {\sqrt {e}\,x}{\sqrt {e\,x^2+d}}\right )}{x^4} \,d x \] Input:

int(atanh((e^(1/2)*x)/(d + e*x^2)^(1/2))/x^4,x)
 

Output:

int(atanh((e^(1/2)*x)/(d + e*x^2)^(1/2))/x^4, x)
 

Reduce [F]

\[ \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^4} \, dx=\int \frac {\mathit {atanh} \left (\frac {\sqrt {e}\, x}{\sqrt {e \,x^{2}+d}}\right )}{x^{4}}d x \] Input:

int(atanh(e^(1/2)*x/(e*x^2+d)^(1/2))/x^4,x)
 

Output:

int(atanh((sqrt(e)*x)/sqrt(d + e*x**2))/x**4,x)