\(\int \frac {\text {arctanh}(\frac {\sqrt {e} x}{\sqrt {d+e x^2}})}{x^{7/2}} \, dx\) [20]

Optimal result

Integrand size = 25, antiderivative size = 145 \[ \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^{7/2}} \, dx=-\frac {4 \sqrt {e} \sqrt {d+e x^2}}{15 d x^{3/2}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{5 x^{5/2}}-\frac {2 e^{5/4} \left (\sqrt {d}+\sqrt {e} x\right ) \sqrt {\frac {d+e x^2}{\left (\sqrt {d}+\sqrt {e} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{e} \sqrt {x}}{\sqrt [4]{d}}\right ),\frac {1}{2}\right )}{15 d^{5/4} \sqrt {d+e x^2}} \] Output:

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.15 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.98 \[ \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^{7/2}} \, dx=-\frac {2 \left (2 \sqrt {e} x \sqrt {d+e x^2}+3 d \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )\right )}{15 d x^{5/2}}-\frac {4 \sqrt {\frac {i \sqrt {d}}{\sqrt {e}}} e^2 \sqrt {1+\frac {d}{e x^2}} x \operatorname {EllipticF}\left (i \text {arcsinh}\left (\frac {\sqrt {\frac {i \sqrt {d}}{\sqrt {e}}}}{\sqrt {x}}\right ),-1\right )}{15 d^{3/2} \sqrt {d+e x^2}} \] Input:

Integrate[ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]]/x^(7/2),x]
 

Output:

(-2*(2*Sqrt[e]*x*Sqrt[d + e*x^2] + 3*d*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2] 
]))/(15*d*x^(5/2)) - (4*Sqrt[(I*Sqrt[d])/Sqrt[e]]*e^2*Sqrt[1 + d/(e*x^2)]* 
x*EllipticF[I*ArcSinh[Sqrt[(I*Sqrt[d])/Sqrt[e]]/Sqrt[x]], -1])/(15*d^(3/2) 
*Sqrt[d + e*x^2])
 

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.03, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {6775, 264, 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.

\(\displaystyle \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^{7/2}} \, dx\)

\(\Big \downarrow \) 6775

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

\(\Big \downarrow \) 264

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

\(\Big \downarrow \) 266

\(\displaystyle \frac {2}{5} \sqrt {e} \left (-\frac {2 e \int \frac {1}{\sqrt {e x^2+d}}d\sqrt {x}}{3 d}-\frac {2 \sqrt {d+e x^2}}{3 d x^{3/2}}\right )-\frac {2 \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{5 x^{5/2}}\)

\(\Big \downarrow \) 761

\(\displaystyle \frac {2}{5} \sqrt {e} \left (-\frac {e^{3/4} \left (\sqrt {d}+\sqrt {e} x\right ) \sqrt {\frac {d+e x^2}{\left (\sqrt {d}+\sqrt {e} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{e} \sqrt {x}}{\sqrt [4]{d}}\right ),\frac {1}{2}\right )}{3 d^{5/4} \sqrt {d+e x^2}}-\frac {2 \sqrt {d+e x^2}}{3 d x^{3/2}}\right )-\frac {2 \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{5 x^{5/2}}\)

Input:

Int[ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]]/x^(7/2),x]
 

Output:

(-2*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/(5*x^(5/2)) + (2*Sqrt[e]*((-2*Sq 
rt[d + e*x^2])/(3*d*x^(3/2)) - (e^(3/4)*(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])/(3*d^(5/4)*Sqrt[d + e*x^2])))/5
 

Defintions of rubi rules used

rule 264

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] - Simp[b*((m + 2*p + 3)/(a*c 
^2*(m + 1)))   Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p 
}, x] && LtQ[m, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
 

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]

\[\int \frac {\operatorname {arctanh}\left (\frac {\sqrt {e}\, x}{\sqrt {e \,x^{2}+d}}\right )}{x^{\frac {7}{2}}}d x\]

Input:

int(arctanh(e^(1/2)*x/(e*x^2+d)^(1/2))/x^(7/2),x)
 

Output:

int(arctanh(e^(1/2)*x/(e*x^2+d)^(1/2))/x^(7/2),x)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.52 \[ \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^{7/2}} \, dx=-\frac {4 \, e x^{3} {\rm weierstrassPInverse}\left (-\frac {4 \, d}{e}, 0, x\right ) + 4 \, \sqrt {e x^{2} + d} \sqrt {e} x^{\frac {3}{2}} + 3 \, d \sqrt {x} \log \left (\frac {2 \, e x^{2} + 2 \, \sqrt {e x^{2} + d} \sqrt {e} x + d}{d}\right )}{15 \, d x^{3}} \] Input:

integrate(arctanh(e^(1/2)*x/(e*x^2+d)^(1/2))/x^(7/2),x, algorithm="fricas" 
)
 

Output:

-1/15*(4*e*x^3*weierstrassPInverse(-4*d/e, 0, x) + 4*sqrt(e*x^2 + d)*sqrt( 
e)*x^(3/2) + 3*d*sqrt(x)*log((2*e*x^2 + 2*sqrt(e*x^2 + d)*sqrt(e)*x + d)/d 
))/(d*x^3)
 

Sympy [F]

\[ \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^{7/2}} \, dx=\int \frac {\operatorname {atanh}{\left (\frac {\sqrt {e} x}{\sqrt {d + e x^{2}}} \right )}}{x^{\frac {7}{2}}}\, dx \] Input:

integrate(atanh(e**(1/2)*x/(e*x**2+d)**(1/2))/x**(7/2),x)
 

Output:

Integral(atanh(sqrt(e)*x/sqrt(d + e*x**2))/x**(7/2), x)
 

Maxima [F]

\[ \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^{7/2}} \, dx=\int { \frac {\operatorname {artanh}\left (\frac {\sqrt {e} x}{\sqrt {e x^{2} + d}}\right )}{x^{\frac {7}{2}}} \,d x } \] Input:

integrate(arctanh(e^(1/2)*x/(e*x^2+d)^(1/2))/x^(7/2),x, algorithm="maxima" 
)
 

Output:

2*d*sqrt(e)*integrate(-1/5*sqrt(e*x^2 + d)*x/((e^2*x^4 + d*e*x^2)*x^(7/2) 
- (e*x^2 + d)*e^(log(e*x^2 + d) + 7/2*log(x))), x) - 1/5*log(sqrt(e)*x + s 
qrt(e*x^2 + d))/x^(5/2) + 1/5*log(-sqrt(e)*x + sqrt(e*x^2 + d))/x^(5/2)
 

Giac [F]

\[ \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^{7/2}} \, dx=\int { \frac {\operatorname {artanh}\left (\frac {\sqrt {e} x}{\sqrt {e x^{2} + d}}\right )}{x^{\frac {7}{2}}} \,d x } \] Input:

integrate(arctanh(e^(1/2)*x/(e*x^2+d)^(1/2))/x^(7/2),x, algorithm="giac")
 

Output:

integrate(arctanh(sqrt(e)*x/sqrt(e*x^2 + d))/x^(7/2), x)
                                                                                    
                                                                                    
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^{7/2}} \, dx=\int \frac {\mathrm {atanh}\left (\frac {\sqrt {e}\,x}{\sqrt {e\,x^2+d}}\right )}{x^{7/2}} \,d x \] Input:

int(atanh((e^(1/2)*x)/(d + e*x^2)^(1/2))/x^(7/2),x)
 

Output:

int(atanh((e^(1/2)*x)/(d + e*x^2)^(1/2))/x^(7/2), x)
 

Reduce [F]

\[ \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^{7/2}} \, dx=\int \frac {\sqrt {x}\, \mathit {atanh} \left (\frac {\sqrt {e}\, x}{\sqrt {e \,x^{2}+d}}\right )}{x^{4}}d x \] Input:

int(atanh(e^(1/2)*x/(e*x^2+d)^(1/2))/x^(7/2),x)
 

Output:

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