\(\int \frac {\text {arctanh}(\frac {\sqrt {e} x}{\sqrt {d+e x^2}})}{x^{11/2}} \, dx\) [21]

Optimal result

Integrand size = 25, antiderivative size = 173 \[ \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^{11/2}} \, dx=-\frac {4 \sqrt {e} \sqrt {d+e x^2}}{63 d x^{7/2}}+\frac {20 e^{3/2} \sqrt {d+e x^2}}{189 d^2 x^{3/2}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{9 x^{9/2}}+\frac {10 e^{9/4} \left (\sqrt {d}+\sqrt {e} x\right ) \sqrt {\frac {d+e x^2}{\left (\sqrt {d}+\sqrt {e} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{e} \sqrt {x}}{\sqrt [4]{d}}\right ),\frac {1}{2}\right )}{189 d^{9/4} \sqrt {d+e x^2}} \] Output:

-4/63*e^(1/2)*(e*x^2+d)^(1/2)/d/x^(7/2)+20/189*e^(3/2)*(e*x^2+d)^(1/2)/d^2 
/x^(3/2)-2/9*arctanh(e^(1/2)*x/(e*x^2+d)^(1/2))/x^(9/2)+10/189*e^(9/4)*(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^(9/4)/(e*x^2+d)^(1/2)
                                                                                    
                                                                                    
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.23 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.89 \[ \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^{11/2}} \, dx=\frac {4 \sqrt {e} x \sqrt {d+e x^2} \left (-3 d+5 e x^2\right )-42 d^2 \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{189 d^2 x^{9/2}}+\frac {20 \sqrt {\frac {i \sqrt {d}}{\sqrt {e}}} e^3 \sqrt {1+\frac {d}{e x^2}} x \operatorname {EllipticF}\left (i \text {arcsinh}\left (\frac {\sqrt {\frac {i \sqrt {d}}{\sqrt {e}}}}{\sqrt {x}}\right ),-1\right )}{189 d^{5/2} \sqrt {d+e x^2}} \] Input:

Integrate[ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]]/x^(11/2),x]
 

Output:

(4*Sqrt[e]*x*Sqrt[d + e*x^2]*(-3*d + 5*e*x^2) - 42*d^2*ArcTanh[(Sqrt[e]*x) 
/Sqrt[d + e*x^2]])/(189*d^2*x^(9/2)) + (20*Sqrt[(I*Sqrt[d])/Sqrt[e]]*e^3*S 
qrt[1 + d/(e*x^2)]*x*EllipticF[I*ArcSinh[Sqrt[(I*Sqrt[d])/Sqrt[e]]/Sqrt[x] 
], -1])/(189*d^(5/2)*Sqrt[d + e*x^2])
 

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.05, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {6775, 264, 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.

\(\displaystyle \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^{11/2}} \, dx\)

\(\Big \downarrow \) 6775

\(\displaystyle \frac {2}{9} \sqrt {e} \int \frac {1}{x^{9/2} \sqrt {e x^2+d}}dx-\frac {2 \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{9 x^{9/2}}\)

\(\Big \downarrow \) 264

\(\displaystyle \frac {2}{9} \sqrt {e} \left (-\frac {5 e \int \frac {1}{x^{5/2} \sqrt {e x^2+d}}dx}{7 d}-\frac {2 \sqrt {d+e x^2}}{7 d x^{7/2}}\right )-\frac {2 \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{9 x^{9/2}}\)

\(\Big \downarrow \) 264

\(\displaystyle \frac {2}{9} \sqrt {e} \left (-\frac {5 e \left (-\frac {e \int \frac {1}{\sqrt {x} \sqrt {e x^2+d}}dx}{3 d}-\frac {2 \sqrt {d+e x^2}}{3 d x^{3/2}}\right )}{7 d}-\frac {2 \sqrt {d+e x^2}}{7 d x^{7/2}}\right )-\frac {2 \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{9 x^{9/2}}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {2}{9} \sqrt {e} \left (-\frac {5 e \left (-\frac {2 e \int \frac {1}{\sqrt {e x^2+d}}d\sqrt {x}}{3 d}-\frac {2 \sqrt {d+e x^2}}{3 d x^{3/2}}\right )}{7 d}-\frac {2 \sqrt {d+e x^2}}{7 d x^{7/2}}\right )-\frac {2 \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{9 x^{9/2}}\)

\(\Big \downarrow \) 761

\(\displaystyle \frac {2}{9} \sqrt {e} \left (-\frac {5 e \left (-\frac {e^{3/4} \left (\sqrt {d}+\sqrt {e} x\right ) \sqrt {\frac {d+e x^2}{\left (\sqrt {d}+\sqrt {e} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{e} \sqrt {x}}{\sqrt [4]{d}}\right ),\frac {1}{2}\right )}{3 d^{5/4} \sqrt {d+e x^2}}-\frac {2 \sqrt {d+e x^2}}{3 d x^{3/2}}\right )}{7 d}-\frac {2 \sqrt {d+e x^2}}{7 d x^{7/2}}\right )-\frac {2 \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{9 x^{9/2}}\)

Input:

Int[ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]]/x^(11/2),x]
 

Output:

(-2*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/(9*x^(9/2)) + (2*Sqrt[e]*((-2*Sq 
rt[d + e*x^2])/(7*d*x^(7/2)) - (5*e*((-2*Sqrt[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]*E 
llipticF[2*ArcTan[(e^(1/4)*Sqrt[x])/d^(1/4)], 1/2])/(3*d^(5/4)*Sqrt[d + e* 
x^2])))/(7*d)))/9
 

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]

\[\int \frac {\operatorname {arctanh}\left (\frac {\sqrt {e}\, x}{\sqrt {e \,x^{2}+d}}\right )}{x^{\frac {11}{2}}}d x\]

Input:

int(arctanh(e^(1/2)*x/(e*x^2+d)^(1/2))/x^(11/2),x)
 

Output:

int(arctanh(e^(1/2)*x/(e*x^2+d)^(1/2))/x^(11/2),x)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.52 \[ \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^{11/2}} \, dx=\frac {20 \, e^{2} x^{5} {\rm weierstrassPInverse}\left (-\frac {4 \, d}{e}, 0, x\right ) - 21 \, d^{2} \sqrt {x} \log \left (\frac {2 \, e x^{2} + 2 \, \sqrt {e x^{2} + d} \sqrt {e} x + d}{d}\right ) + 4 \, {\left (5 \, e x^{3} - 3 \, d x\right )} \sqrt {e x^{2} + d} \sqrt {e} \sqrt {x}}{189 \, d^{2} x^{5}} \] Input:

integrate(arctanh(e^(1/2)*x/(e*x^2+d)^(1/2))/x^(11/2),x, algorithm="fricas 
")
 

Output:

1/189*(20*e^2*x^5*weierstrassPInverse(-4*d/e, 0, x) - 21*d^2*sqrt(x)*log(( 
2*e*x^2 + 2*sqrt(e*x^2 + d)*sqrt(e)*x + d)/d) + 4*(5*e*x^3 - 3*d*x)*sqrt(e 
*x^2 + d)*sqrt(e)*sqrt(x))/(d^2*x^5)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^{11/2}} \, dx=\text {Timed out} \] Input:

integrate(atanh(e**(1/2)*x/(e*x**2+d)**(1/2))/x**(11/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^{11/2}} \, dx=\int { \frac {\operatorname {artanh}\left (\frac {\sqrt {e} x}{\sqrt {e x^{2} + d}}\right )}{x^{\frac {11}{2}}} \,d x } \] Input:

integrate(arctanh(e^(1/2)*x/(e*x^2+d)^(1/2))/x^(11/2),x, algorithm="maxima 
")
 

Output:

2*d*sqrt(e)*integrate(-1/9*sqrt(e*x^2 + d)*x/((e^2*x^4 + d*e*x^2)*x^(11/2) 
 - (e*x^2 + d)*e^(log(e*x^2 + d) + 11/2*log(x))), x) - 1/9*log(sqrt(e)*x + 
 sqrt(e*x^2 + d))/x^(9/2) + 1/9*log(-sqrt(e)*x + sqrt(e*x^2 + d))/x^(9/2)
 

Giac [F]

\[ \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^{11/2}} \, dx=\int { \frac {\operatorname {artanh}\left (\frac {\sqrt {e} x}{\sqrt {e x^{2} + d}}\right )}{x^{\frac {11}{2}}} \,d x } \] Input:

integrate(arctanh(e^(1/2)*x/(e*x^2+d)^(1/2))/x^(11/2),x, algorithm="giac")
 

Output:

integrate(arctanh(sqrt(e)*x/sqrt(e*x^2 + d))/x^(11/2), x)
                                                                                    
                                                                                    
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^{11/2}} \, dx=\int \frac {\mathrm {atanh}\left (\frac {\sqrt {e}\,x}{\sqrt {e\,x^2+d}}\right )}{x^{11/2}} \,d x \] Input:

int(atanh((e^(1/2)*x)/(d + e*x^2)^(1/2))/x^(11/2),x)
 

Output:

int(atanh((e^(1/2)*x)/(d + e*x^2)^(1/2))/x^(11/2), x)
 

Reduce [F]

\[ \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^{11/2}} \, dx=\int \frac {\sqrt {x}\, \mathit {atanh} \left (\frac {\sqrt {e}\, x}{\sqrt {e \,x^{2}+d}}\right )}{x^{6}}d x \] Input:

int(atanh(e^(1/2)*x/(e*x^2+d)^(1/2))/x^(11/2),x)
 

Output:

int((sqrt(x)*atanh((sqrt(e)*x)/sqrt(d + e*x**2)))/x**6,x)