3.1.25 \(\int \frac {\text {arctanh}(\frac {\sqrt {e} x}{\sqrt {d+e x^2}})}{\sqrt {x}} \, dx\) [25]

3.1.25.1 Optimal result

Integrand size = 25, antiderivative size = 232 \[ \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{\sqrt {x}} \, dx=-\frac {4 \sqrt {x} \sqrt {d+e x^2}}{\sqrt {d}+\sqrt {e} x}+2 \sqrt {x} \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )+\frac {4 \sqrt [4]{d} \left (\sqrt {d}+\sqrt {e} x\right ) \sqrt {\frac {d+e x^2}{\left (\sqrt {d}+\sqrt {e} x\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{e} \sqrt {x}}{\sqrt [4]{d}}\right )|\frac {1}{2}\right )}{\sqrt [4]{e} \sqrt {d+e x^2}}-\frac {2 \sqrt [4]{d} \left (\sqrt {d}+\sqrt {e} x\right ) \sqrt {\frac {d+e x^2}{\left (\sqrt {d}+\sqrt {e} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{e} \sqrt {x}}{\sqrt [4]{d}}\right ),\frac {1}{2}\right )}{\sqrt [4]{e} \sqrt {d+e x^2}} \]

output

2*x^(1/2)*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))-4*x^(1/2)*(e*x^2+d)^(1/2)/(d^ 
(1/2)+x*e^(1/2))+4*d^(1/4)*(cos(2*arctan(e^(1/4)*x^(1/2)/d^(1/4)))^2)^(1/2 
)/cos(2*arctan(e^(1/4)*x^(1/2)/d^(1/4)))*EllipticE(sin(2*arctan(e^(1/4)*x^ 
(1/2)/d^(1/4))),1/2*2^(1/2))*(d^(1/2)+x*e^(1/2))*((e*x^2+d)/(d^(1/2)+x*e^( 
1/2))^2)^(1/2)/e^(1/4)/(e*x^2+d)^(1/2)-2*d^(1/4)*(cos(2*arctan(e^(1/4)*x^( 
1/2)/d^(1/4)))^2)^(1/2)/cos(2*arctan(e^(1/4)*x^(1/2)/d^(1/4)))*EllipticF(s 
in(2*arctan(e^(1/4)*x^(1/2)/d^(1/4))),1/2*2^(1/2))*(d^(1/2)+x*e^(1/2))*((e 
*x^2+d)/(d^(1/2)+x*e^(1/2))^2)^(1/2)/e^(1/4)/(e*x^2+d)^(1/2)
 

3.1.25.2 Mathematica [C] (verified)

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

Time = 0.08 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.37 \[ \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{\sqrt {x}} \, dx=2 \sqrt {x} \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )-\frac {4 \sqrt {e} x^{3/2} \sqrt {1+\frac {e x^2}{d}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-\frac {e x^2}{d}\right )}{3 \sqrt {d+e x^2}} \]

input

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

output

2*Sqrt[x]*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]] - (4*Sqrt[e]*x^(3/2)*Sqrt[1 
 + (e*x^2)/d]*Hypergeometric2F1[1/2, 3/4, 7/4, -((e*x^2)/d)])/(3*Sqrt[d + 
e*x^2])
 

3.1.25.3 Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.07, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {6775, 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.

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

\(\Big \downarrow \) 6775

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 834

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 761

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

\(\Big \downarrow \) 1510

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

input

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

output

2*Sqrt[x]*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]] - 4*Sqrt[e]*(-((-((Sqrt[x]* 
Sqrt[d + e*x^2])/(Sqrt[d] + Sqrt[e]*x)) + (d^(1/4)*(Sqrt[d] + Sqrt[e]*x)*S 
qrt[(d + e*x^2)/(Sqrt[d] + Sqrt[e]*x)^2]*EllipticE[2*ArcTan[(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*ArcTa 
n[(e^(1/4)*Sqrt[x])/d^(1/4)], 1/2])/(2*e^(3/4)*Sqrt[d + e*x^2]))
 

3.1.25.3.1 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 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]
 

3.1.25.4 Maple [F]

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

input

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

output

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

3.1.25.5 Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.09 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.22 \[ \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{\sqrt {x}} \, dx=\sqrt {x} \log \left (\frac {2 \, e x^{2} + 2 \, \sqrt {e x^{2} + d} \sqrt {e} x + d}{d}\right ) + 4 \, {\rm weierstrassZeta}\left (-\frac {4 \, d}{e}, 0, {\rm weierstrassPInverse}\left (-\frac {4 \, d}{e}, 0, x\right )\right ) \]

input

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

output

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

3.1.25.6 Sympy [F]

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

input

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

output

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

3.1.25.7 Maxima [F]

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

input

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

output

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

3.1.25.8 Giac [F]

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

input

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

output

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

3.1.25.9 Mupad [F(-1)]

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

input

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

output

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