\(\int \frac {\text {arctanh}(\frac {\sqrt {e} x}{\sqrt {d+e x^2}})}{x^{3/2}} \, dx\) [19]

Optimal result

Integrand size = 25, antiderivative size = 113 \[ \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^{3/2}} \, dx=-\frac {2 \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{\sqrt {x}}+\frac {2 \sqrt [4]{e} \left (\sqrt {d}+\sqrt {e} x\right ) \sqrt {\frac {d+e x^2}{\left (\sqrt {d}+\sqrt {e} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{e} \sqrt {x}}{\sqrt [4]{d}}\right ),\frac {1}{2}\right )}{\sqrt [4]{d} \sqrt {d+e x^2}} \] Output:

-2*arctanh(e^(1/2)*x/(e*x^2+d)^(1/2))/x^(1/2)+2*e^(1/4)*(d^(1/2)+e^(1/2)*x 
)*((e*x^2+d)/(d^(1/2)+e^(1/2)*x)^2)^(1/2)*InverseJacobiAM(2*arctan(e^(1/4) 
*x^(1/2)/d^(1/4)),1/2*2^(1/2))/d^(1/4)/(e*x^2+d)^(1/2)
                                                                                    
                                                                                    
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.09 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.98 \[ \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^{3/2}} \, dx=-\frac {2 \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{\sqrt {x}}+\frac {4 i \sqrt {e} \sqrt {1+\frac {d}{e x^2}} x \operatorname {EllipticF}\left (i \text {arcsinh}\left (\frac {\sqrt {\frac {i \sqrt {d}}{\sqrt {e}}}}{\sqrt {x}}\right ),-1\right )}{\sqrt {\frac {i \sqrt {d}}{\sqrt {e}}} \sqrt {d+e x^2}} \] Input:

Integrate[ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]]/x^(3/2),x]
 

Output:

(-2*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/Sqrt[x] + ((4*I)*Sqrt[e]*Sqrt[1 
+ d/(e*x^2)]*x*EllipticF[I*ArcSinh[Sqrt[(I*Sqrt[d])/Sqrt[e]]/Sqrt[x]], -1] 
)/(Sqrt[(I*Sqrt[d])/Sqrt[e]]*Sqrt[d + e*x^2])
 

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {6775, 266, 761}

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.

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

\(\Big \downarrow \) 6775

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 761

\(\displaystyle \frac {2 \sqrt [4]{e} \left (\sqrt {d}+\sqrt {e} x\right ) \sqrt {\frac {d+e x^2}{\left (\sqrt {d}+\sqrt {e} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{e} \sqrt {x}}{\sqrt [4]{d}}\right ),\frac {1}{2}\right )}{\sqrt [4]{d} \sqrt {d+e x^2}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{\sqrt {x}}\)

Input:

Int[ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]]/x^(3/2),x]
 

Output:

(-2*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/Sqrt[x] + (2*e^(1/4)*(Sqrt[d] + 
Sqrt[e]*x)*Sqrt[(d + e*x^2)/(Sqrt[d] + Sqrt[e]*x)^2]*EllipticF[2*ArcTan[(e 
^(1/4)*Sqrt[x])/d^(1/4)], 1/2])/(d^(1/4)*Sqrt[d + e*x^2])
 

Defintions of rubi rules used

rule 266

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De 
nominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) 
^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I 
ntBinomialQ[a, b, c, 2, m, p, x]
 

rule 761

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[( 
1 + q^2*x^2)*(Sqrt[(a + b*x^4)/(a*(1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^4]))* 
EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
 

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 [F]

\[\int \frac {\operatorname {arctanh}\left (\frac {\sqrt {e}\, x}{\sqrt {e \,x^{2}+d}}\right )}{x^{\frac {3}{2}}}d x\]

Input:

int(arctanh(e^(1/2)*x/(e*x^2+d)^(1/2))/x^(3/2),x)
 

Output:

int(arctanh(e^(1/2)*x/(e*x^2+d)^(1/2))/x^(3/2),x)
 

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.44 \[ \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^{3/2}} \, dx=\frac {4 \, x {\rm weierstrassPInverse}\left (-\frac {4 \, d}{e}, 0, x\right ) - \sqrt {x} \log \left (\frac {2 \, e x^{2} + 2 \, \sqrt {e x^{2} + d} \sqrt {e} x + d}{d}\right )}{x} \] Input:

integrate(arctanh(e^(1/2)*x/(e*x^2+d)^(1/2))/x^(3/2),x, algorithm="fricas" 
)
 

Output:

(4*x*weierstrassPInverse(-4*d/e, 0, x) - sqrt(x)*log((2*e*x^2 + 2*sqrt(e*x 
^2 + d)*sqrt(e)*x + d)/d))/x
 

Sympy [F]

\[ \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^{3/2}} \, dx=\int \frac {\operatorname {atanh}{\left (\frac {\sqrt {e} x}{\sqrt {d + e x^{2}}} \right )}}{x^{\frac {3}{2}}}\, dx \] Input:

integrate(atanh(e**(1/2)*x/(e*x**2+d)**(1/2))/x**(3/2),x)
 

Output:

Integral(atanh(sqrt(e)*x/sqrt(d + e*x**2))/x**(3/2), x)
 

Maxima [F]

\[ \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^{3/2}} \, dx=\int { \frac {\operatorname {artanh}\left (\frac {\sqrt {e} x}{\sqrt {e x^{2} + d}}\right )}{x^{\frac {3}{2}}} \,d x } \] Input:

integrate(arctanh(e^(1/2)*x/(e*x^2+d)^(1/2))/x^(3/2),x, algorithm="maxima" 
)
 

Output:

2*d*sqrt(e)*integrate(-sqrt(e*x^2 + d)*x/((e^2*x^4 + d*e*x^2)*x^(3/2) - (e 
*x^2 + d)*e^(log(e*x^2 + d) + 3/2*log(x))), x) - log(sqrt(e)*x + sqrt(e*x^ 
2 + d))/sqrt(x) + log(-sqrt(e)*x + sqrt(e*x^2 + d))/sqrt(x)
 

Giac [F]

\[ \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^{3/2}} \, dx=\int { \frac {\operatorname {artanh}\left (\frac {\sqrt {e} x}{\sqrt {e x^{2} + d}}\right )}{x^{\frac {3}{2}}} \,d x } \] Input:

integrate(arctanh(e^(1/2)*x/(e*x^2+d)^(1/2))/x^(3/2),x, algorithm="giac")
 

Output:

integrate(arctanh(sqrt(e)*x/sqrt(e*x^2 + d))/x^(3/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^{3/2}} \, dx=\int \frac {\mathrm {atanh}\left (\frac {\sqrt {e}\,x}{\sqrt {e\,x^2+d}}\right )}{x^{3/2}} \,d x \] Input:

int(atanh((e^(1/2)*x)/(d + e*x^2)^(1/2))/x^(3/2),x)
 

Output:

int(atanh((e^(1/2)*x)/(d + e*x^2)^(1/2))/x^(3/2), x)
 

Reduce [F]

\[ \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^{3/2}} \, dx=\int \frac {\sqrt {x}\, \mathit {atanh} \left (\frac {\sqrt {e}\, x}{\sqrt {e \,x^{2}+d}}\right )}{x^{2}}d x \] Input:

int(atanh(e^(1/2)*x/(e*x^2+d)^(1/2))/x^(3/2),x)
 

Output:

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