\(\int x^{3/2} \text {arctanh}(\frac {\sqrt {e} x}{\sqrt {d+e x^2}}) \, dx\) [24]

Optimal result

Integrand size = 25, antiderivative size = 269 \[ \int x^{3/2} \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right ) \, dx=-\frac {4 x^{3/2} \sqrt {d+e x^2}}{25 \sqrt {e}}+\frac {12 d \sqrt {x} \sqrt {d+e x^2}}{25 e \left (\sqrt {d}+\sqrt {e} x\right )}+\frac {2}{5} x^{5/2} \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )-\frac {12 d^{5/4} \left (\sqrt {d}+\sqrt {e} x\right ) \sqrt {\frac {d+e x^2}{\left (\sqrt {d}+\sqrt {e} x\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{e} \sqrt {x}}{\sqrt [4]{d}}\right )|\frac {1}{2}\right )}{25 e^{5/4} \sqrt {d+e x^2}}+\frac {6 d^{5/4} \left (\sqrt {d}+\sqrt {e} x\right ) \sqrt {\frac {d+e x^2}{\left (\sqrt {d}+\sqrt {e} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{e} \sqrt {x}}{\sqrt [4]{d}}\right ),\frac {1}{2}\right )}{25 e^{5/4} \sqrt {d+e x^2}} \] Output:

-4/25*x^(3/2)*(e*x^2+d)^(1/2)/e^(1/2)+12/25*d*x^(1/2)*(e*x^2+d)^(1/2)/e/(d 
^(1/2)+e^(1/2)*x)+2/5*x^(5/2)*arctanh(e^(1/2)*x/(e*x^2+d)^(1/2))-12/25*d^( 
5/4)*(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))/e^(5/4)/(e*x^2+d)^(1/ 
2)+6/25*d^(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))/e^(5/4)/( 
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 = 109, normalized size of antiderivative = 0.41 \[ \int x^{3/2} \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right ) \, dx=\frac {2 x^{3/2} \left (-2 \left (d+e x^2\right )+5 \sqrt {e} x \sqrt {d+e x^2} \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )+2 d \sqrt {1+\frac {e x^2}{d}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-\frac {e x^2}{d}\right )\right )}{25 \sqrt {e} \sqrt {d+e x^2}} \] Input:

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

Output:

(2*x^(3/2)*(-2*(d + e*x^2) + 5*Sqrt[e]*x*Sqrt[d + e*x^2]*ArcTanh[(Sqrt[e]* 
x)/Sqrt[d + e*x^2]] + 2*d*Sqrt[1 + (e*x^2)/d]*Hypergeometric2F1[1/2, 3/4, 
7/4, -((e*x^2)/d)]))/(25*Sqrt[e]*Sqrt[d + e*x^2])
 

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 285, normalized size of antiderivative = 1.06, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {6775, 262, 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.

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

\(\Big \downarrow \) 6775

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

\(\Big \downarrow \) 262

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

\(\Big \downarrow \) 266

\(\displaystyle \frac {2}{5} x^{5/2} \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )-\frac {2}{5} \sqrt {e} \left (\frac {2 x^{3/2} \sqrt {d+e x^2}}{5 e}-\frac {6 d \int \frac {x}{\sqrt {e x^2+d}}d\sqrt {x}}{5 e}\right )\)

\(\Big \downarrow \) 834

\(\displaystyle \frac {2}{5} x^{5/2} \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )-\frac {2}{5} \sqrt {e} \left (\frac {2 x^{3/2} \sqrt {d+e x^2}}{5 e}-\frac {6 d \left (\frac {\sqrt {d} \int \frac {1}{\sqrt {e x^2+d}}d\sqrt {x}}{\sqrt {e}}-\frac {\sqrt {d} \int \frac {\sqrt {d}-\sqrt {e} x}{\sqrt {d} \sqrt {e x^2+d}}d\sqrt {x}}{\sqrt {e}}\right )}{5 e}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2}{5} x^{5/2} \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )-\frac {2}{5} \sqrt {e} \left (\frac {2 x^{3/2} \sqrt {d+e x^2}}{5 e}-\frac {6 d \left (\frac {\sqrt {d} \int \frac {1}{\sqrt {e x^2+d}}d\sqrt {x}}{\sqrt {e}}-\frac {\int \frac {\sqrt {d}-\sqrt {e} x}{\sqrt {e x^2+d}}d\sqrt {x}}{\sqrt {e}}\right )}{5 e}\right )\)

\(\Big \downarrow \) 761

\(\displaystyle \frac {2}{5} x^{5/2} \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )-\frac {2}{5} \sqrt {e} \left (\frac {2 x^{3/2} \sqrt {d+e x^2}}{5 e}-\frac {6 d \left (\frac {\sqrt [4]{d} \left (\sqrt {d}+\sqrt {e} x\right ) \sqrt {\frac {d+e x^2}{\left (\sqrt {d}+\sqrt {e} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{e} \sqrt {x}}{\sqrt [4]{d}}\right ),\frac {1}{2}\right )}{2 e^{3/4} \sqrt {d+e x^2}}-\frac {\int \frac {\sqrt {d}-\sqrt {e} x}{\sqrt {e x^2+d}}d\sqrt {x}}{\sqrt {e}}\right )}{5 e}\right )\)

\(\Big \downarrow \) 1510

\(\displaystyle \frac {2}{5} x^{5/2} \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )-\frac {2}{5} \sqrt {e} \left (\frac {2 x^{3/2} \sqrt {d+e x^2}}{5 e}-\frac {6 d \left (\frac {\sqrt [4]{d} \left (\sqrt {d}+\sqrt {e} x\right ) \sqrt {\frac {d+e x^2}{\left (\sqrt {d}+\sqrt {e} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{e} \sqrt {x}}{\sqrt [4]{d}}\right ),\frac {1}{2}\right )}{2 e^{3/4} \sqrt {d+e x^2}}-\frac {\frac {\sqrt [4]{d} \left (\sqrt {d}+\sqrt {e} x\right ) \sqrt {\frac {d+e x^2}{\left (\sqrt {d}+\sqrt {e} x\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{e} \sqrt {x}}{\sqrt [4]{d}}\right )|\frac {1}{2}\right )}{\sqrt [4]{e} \sqrt {d+e x^2}}-\frac {\sqrt {x} \sqrt {d+e x^2}}{\sqrt {d}+\sqrt {e} x}}{\sqrt {e}}\right )}{5 e}\right )\)

Input:

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

Output:

(2*x^(5/2)*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/5 - (2*Sqrt[e]*((2*x^(3/2 
)*Sqrt[d + e*x^2])/(5*e) - (6*d*(-((-((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*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*ArcTan[(e^(1/4)*Sqrt[x])/d^(1/4) 
], 1/2])/(2*e^(3/4)*Sqrt[d + e*x^2])))/(5*e)))/5
 

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 262

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) 
^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ 
(b*(m + 2*p + 1)))   Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b 
, c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && 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]

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

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.29 \[ \int x^{3/2} \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right ) \, dx=\frac {5 \, e x^{\frac {5}{2}} \log \left (\frac {2 \, e x^{2} + 2 \, \sqrt {e x^{2} + d} \sqrt {e} x + d}{d}\right ) - 4 \, \sqrt {e x^{2} + d} \sqrt {e} x^{\frac {3}{2}} - 12 \, d {\rm weierstrassZeta}\left (-\frac {4 \, d}{e}, 0, {\rm weierstrassPInverse}\left (-\frac {4 \, d}{e}, 0, x\right )\right )}{25 \, e} \] Input:

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

Output:

1/25*(5*e*x^(5/2)*log((2*e*x^2 + 2*sqrt(e*x^2 + d)*sqrt(e)*x + d)/d) - 4*s 
qrt(e*x^2 + d)*sqrt(e)*x^(3/2) - 12*d*weierstrassZeta(-4*d/e, 0, weierstra 
ssPInverse(-4*d/e, 0, x)))/e
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

1/5*x^(5/2)*log(sqrt(e)*x + sqrt(e*x^2 + d)) - 1/5*x^(5/2)*log(-sqrt(e)*x 
+ sqrt(e*x^2 + d)) - 2*d*sqrt(e)*integrate(-1/5*x*e^(1/2*log(e*x^2 + d) + 
3/2*log(x))/(e^2*x^4 + d*e*x^2 - (e*x^2 + d)^2), x)
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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