\(\int \frac {\text {arctanh}(\frac {\sqrt {e} x}{\sqrt {d+e x^2}})}{x^{9/2}} \, dx\) [27]

Optimal result

Integrand size = 25, antiderivative size = 302 \[ \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^{9/2}} \, dx=-\frac {4 \sqrt {e} \sqrt {d+e x^2}}{35 d x^{5/2}}+\frac {12 e^{3/2} \sqrt {d+e x^2}}{35 d^2 \sqrt {x}}-\frac {12 e^2 \sqrt {x} \sqrt {d+e x^2}}{35 d^2 \left (\sqrt {d}+\sqrt {e} x\right )}-\frac {2 \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{7 x^{7/2}}+\frac {12 e^{7/4} \left (\sqrt {d}+\sqrt {e} x\right ) \sqrt {\frac {d+e x^2}{\left (\sqrt {d}+\sqrt {e} x\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{e} \sqrt {x}}{\sqrt [4]{d}}\right )|\frac {1}{2}\right )}{35 d^{7/4} \sqrt {d+e x^2}}-\frac {6 e^{7/4} \left (\sqrt {d}+\sqrt {e} x\right ) \sqrt {\frac {d+e x^2}{\left (\sqrt {d}+\sqrt {e} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{e} \sqrt {x}}{\sqrt [4]{d}}\right ),\frac {1}{2}\right )}{35 d^{7/4} \sqrt {d+e x^2}} \] Output:

-4/35*e^(1/2)*(e*x^2+d)^(1/2)/d/x^(5/2)+12/35*e^(3/2)*(e*x^2+d)^(1/2)/d^2/ 
x^(1/2)-12/35*e^2*x^(1/2)*(e*x^2+d)^(1/2)/d^2/(d^(1/2)+e^(1/2)*x)-2/7*arct 
anh(e^(1/2)*x/(e*x^2+d)^(1/2))/x^(7/2)+12/35*e^(7/4)*(d^(1/2)+e^(1/2)*x)*( 
(e*x^2+d)/(d^(1/2)+e^(1/2)*x)^2)^(1/2)*EllipticE(sin(2*arctan(e^(1/4)*x^(1 
/2)/d^(1/4))),1/2*2^(1/2))/d^(7/4)/(e*x^2+d)^(1/2)-6/35*e^(7/4)*(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^(7/4)/(e*x^2+d)^(1/2)
                                                                                    
                                                                                    
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.07 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.43 \[ \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^{9/2}} \, dx=\frac {4 \sqrt {e} x \left (-d^2+2 d e x^2+3 e^2 x^4\right )-10 d^2 \sqrt {d+e x^2} \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )-4 e^{5/2} x^5 \sqrt {1+\frac {e x^2}{d}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-\frac {e x^2}{d}\right )}{35 d^2 x^{7/2} \sqrt {d+e x^2}} \] Input:

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

Output:

(4*Sqrt[e]*x*(-d^2 + 2*d*e*x^2 + 3*e^2*x^4) - 10*d^2*Sqrt[d + e*x^2]*ArcTa 
nh[(Sqrt[e]*x)/Sqrt[d + e*x^2]] - 4*e^(5/2)*x^5*Sqrt[1 + (e*x^2)/d]*Hyperg 
eometric2F1[1/2, 3/4, 7/4, -((e*x^2)/d)])/(35*d^2*x^(7/2)*Sqrt[d + e*x^2])
 

Rubi [A] (verified)

Time = 0.66 (sec) , antiderivative size = 313, normalized size of antiderivative = 1.04, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {6775, 264, 264, 266, 834, 27, 761, 1510}

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.

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

\(\Big \downarrow \) 6775

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

\(\Big \downarrow \) 264

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

\(\Big \downarrow \) 264

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 834

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 761

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

\(\Big \downarrow \) 1510

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

Input:

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

Output:

(-2*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/(7*x^(7/2)) + (2*Sqrt[e]*((-2*Sq 
rt[d + e*x^2])/(5*d*x^(5/2)) - (3*e*((-2*Sqrt[d + e*x^2])/(d*Sqrt[x]) + (2 
*e*(-((-((Sqrt[x]*Sqrt[d + e*x^2])/(Sqrt[d] + Sqrt[e]*x)) + (d^(1/4)*(Sqrt 
[d] + Sqrt[e]*x)*Sqrt[(d + e*x^2)/(Sqrt[d] + Sqrt[e]*x)^2]*EllipticE[2*Arc 
Tan[(e^(1/4)*Sqrt[x])/d^(1/4)], 1/2])/(e^(1/4)*Sqrt[d + e*x^2]))/Sqrt[e]) 
+ (d^(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])/(2*e^(3/4)*Sqrt[d + 
e*x^2])))/d))/(5*d)))/7
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

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 834

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

rule 1510

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = 
 Rt[c/a, 4]}, Simp[(-d)*x*(Sqrt[a + c*x^4]/(a*(1 + q^2*x^2))), x] + Simp[d* 
(1 + q^2*x^2)*(Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]/(q*Sqrt[a + c*x^4]))*E 
llipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e 
}, x] && PosQ[c/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]

\[\int \frac {\operatorname {arctanh}\left (\frac {\sqrt {e}\, x}{\sqrt {e \,x^{2}+d}}\right )}{x^{\frac {9}{2}}}d x\]

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.32 \[ \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^{9/2}} \, dx=\frac {12 \, e^{2} x^{4} {\rm weierstrassZeta}\left (-\frac {4 \, d}{e}, 0, {\rm weierstrassPInverse}\left (-\frac {4 \, d}{e}, 0, x\right )\right ) - 5 \, d^{2} \sqrt {x} \log \left (\frac {2 \, e x^{2} + 2 \, \sqrt {e x^{2} + d} \sqrt {e} x + d}{d}\right ) + 4 \, {\left (3 \, e x^{3} - d x\right )} \sqrt {e x^{2} + d} \sqrt {e} \sqrt {x}}{35 \, d^{2} x^{4}} \] Input:

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

Output:

1/35*(12*e^2*x^4*weierstrassZeta(-4*d/e, 0, weierstrassPInverse(-4*d/e, 0, 
 x)) - 5*d^2*sqrt(x)*log((2*e*x^2 + 2*sqrt(e*x^2 + d)*sqrt(e)*x + d)/d) + 
4*(3*e*x^3 - d*x)*sqrt(e*x^2 + d)*sqrt(e)*sqrt(x))/(d^2*x^4)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

2*d*sqrt(e)*integrate(-1/7*sqrt(e*x^2 + d)*x/((e^2*x^4 + d*e*x^2)*x^(9/2) 
- (e*x^2 + d)*e^(log(e*x^2 + d) + 9/2*log(x))), x) - 1/7*log(sqrt(e)*x + s 
qrt(e*x^2 + d))/x^(7/2) + 1/7*log(-sqrt(e)*x + sqrt(e*x^2 + d))/x^(7/2)
 

Giac [F]

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

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

Output:

integrate(arctanh(sqrt(e)*x/sqrt(e*x^2 + d))/x^(9/2), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^{9/2}} \, dx=\int \frac {\sqrt {x}\, \mathit {atanh} \left (\frac {\sqrt {e}\, x}{\sqrt {e \,x^{2}+d}}\right )}{x^{5}}d x \] Input:

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

Output:

int((sqrt(x)*atanh((sqrt(e)*x)/sqrt(d + e*x**2)))/x**5,x)