\(\int x^{9/2} \text {arctanh}(\frac {\sqrt {e} x}{\sqrt {d+e x^2}}) \, dx\) [16]

Optimal result

Integrand size = 25, antiderivative size = 196 \[ \int x^{9/2} \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right ) \, dx=-\frac {60 d^2 \sqrt {x} \sqrt {d+e x^2}}{847 e^{5/2}}+\frac {36 d x^{5/2} \sqrt {d+e x^2}}{847 e^{3/2}}-\frac {4 x^{9/2} \sqrt {d+e x^2}}{121 \sqrt {e}}+\frac {2}{11} x^{11/2} \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )+\frac {30 d^{11/4} \left (\sqrt {d}+\sqrt {e} x\right ) \sqrt {\frac {d+e x^2}{\left (\sqrt {d}+\sqrt {e} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{e} \sqrt {x}}{\sqrt [4]{d}}\right ),\frac {1}{2}\right )}{847 e^{11/4} \sqrt {d+e x^2}} \] Output:

-60/847*d^2*x^(1/2)*(e*x^2+d)^(1/2)/e^(5/2)+36/847*d*x^(5/2)*(e*x^2+d)^(1/ 
2)/e^(3/2)-4/121*x^(9/2)*(e*x^2+d)^(1/2)/e^(1/2)+2/11*x^(11/2)*arctanh(e^( 
1/2)*x/(e*x^2+d)^(1/2))+30/847*d^(11/4)*(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^(11/4)/(e*x^2+d)^(1/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.35 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.82 \[ \int x^{9/2} \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right ) \, dx=\frac {2}{847} \sqrt {x} \left (-\frac {2 \sqrt {d+e x^2} \left (15 d^2-9 d e x^2+7 e^2 x^4\right )}{e^{5/2}}+77 x^5 \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )\right )+\frac {60 d^{5/2} \sqrt {\frac {i \sqrt {d}}{\sqrt {e}}} \sqrt {1+\frac {d}{e x^2}} x \operatorname {EllipticF}\left (i \text {arcsinh}\left (\frac {\sqrt {\frac {i \sqrt {d}}{\sqrt {e}}}}{\sqrt {x}}\right ),-1\right )}{847 e^2 \sqrt {d+e x^2}} \] Input:

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

Output:

(2*Sqrt[x]*((-2*Sqrt[d + e*x^2]*(15*d^2 - 9*d*e*x^2 + 7*e^2*x^4))/e^(5/2) 
+ 77*x^5*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]]))/847 + (60*d^(5/2)*Sqrt[(I* 
Sqrt[d])/Sqrt[e]]*Sqrt[1 + d/(e*x^2)]*x*EllipticF[I*ArcSinh[Sqrt[(I*Sqrt[d 
])/Sqrt[e]]/Sqrt[x]], -1])/(847*e^2*Sqrt[d + e*x^2])
 

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.09, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {6775, 262, 262, 262, 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.

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

\(\Big \downarrow \) 6775

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

\(\Big \downarrow \) 262

\(\displaystyle \frac {2}{11} x^{11/2} \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )-\frac {2}{11} \sqrt {e} \left (\frac {2 x^{9/2} \sqrt {d+e x^2}}{11 e}-\frac {9 d \int \frac {x^{7/2}}{\sqrt {e x^2+d}}dx}{11 e}\right )\)

\(\Big \downarrow \) 262

\(\displaystyle \frac {2}{11} x^{11/2} \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )-\frac {2}{11} \sqrt {e} \left (\frac {2 x^{9/2} \sqrt {d+e x^2}}{11 e}-\frac {9 d \left (\frac {2 x^{5/2} \sqrt {d+e x^2}}{7 e}-\frac {5 d \int \frac {x^{3/2}}{\sqrt {e x^2+d}}dx}{7 e}\right )}{11 e}\right )\)

\(\Big \downarrow \) 262

\(\displaystyle \frac {2}{11} x^{11/2} \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )-\frac {2}{11} \sqrt {e} \left (\frac {2 x^{9/2} \sqrt {d+e x^2}}{11 e}-\frac {9 d \left (\frac {2 x^{5/2} \sqrt {d+e x^2}}{7 e}-\frac {5 d \left (\frac {2 \sqrt {x} \sqrt {d+e x^2}}{3 e}-\frac {d \int \frac {1}{\sqrt {x} \sqrt {e x^2+d}}dx}{3 e}\right )}{7 e}\right )}{11 e}\right )\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {2}{11} x^{11/2} \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )-\frac {2}{11} \sqrt {e} \left (\frac {2 x^{9/2} \sqrt {d+e x^2}}{11 e}-\frac {9 d \left (\frac {2 x^{5/2} \sqrt {d+e x^2}}{7 e}-\frac {5 d \left (\frac {2 \sqrt {x} \sqrt {d+e x^2}}{3 e}-\frac {2 d \int \frac {1}{\sqrt {e x^2+d}}d\sqrt {x}}{3 e}\right )}{7 e}\right )}{11 e}\right )\)

\(\Big \downarrow \) 761

\(\displaystyle \frac {2}{11} x^{11/2} \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )-\frac {2}{11} \sqrt {e} \left (\frac {2 x^{9/2} \sqrt {d+e x^2}}{11 e}-\frac {9 d \left (\frac {2 x^{5/2} \sqrt {d+e x^2}}{7 e}-\frac {5 d \left (\frac {2 \sqrt {x} \sqrt {d+e x^2}}{3 e}-\frac {d^{3/4} \left (\sqrt {d}+\sqrt {e} x\right ) \sqrt {\frac {d+e x^2}{\left (\sqrt {d}+\sqrt {e} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{e} \sqrt {x}}{\sqrt [4]{d}}\right ),\frac {1}{2}\right )}{3 e^{5/4} \sqrt {d+e x^2}}\right )}{7 e}\right )}{11 e}\right )\)

Input:

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

Output:

(2*x^(11/2)*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/11 - (2*Sqrt[e]*((2*x^(9 
/2)*Sqrt[d + e*x^2])/(11*e) - (9*d*((2*x^(5/2)*Sqrt[d + e*x^2])/(7*e) - (5 
*d*((2*Sqrt[x]*Sqrt[d + e*x^2])/(3*e) - (d^(3/4)*(Sqrt[d] + Sqrt[e]*x)*Sqr 
t[(d + e*x^2)/(Sqrt[d] + Sqrt[e]*x)^2]*EllipticF[2*ArcTan[(e^(1/4)*Sqrt[x] 
)/d^(1/4)], 1/2])/(3*e^(5/4)*Sqrt[d + e*x^2])))/(7*e)))/(11*e)))/11
 

Defintions of rubi rules used

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 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]

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

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.48 \[ \int x^{9/2} \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right ) \, dx=\frac {77 \, e^{3} x^{\frac {11}{2}} \log \left (\frac {2 \, e x^{2} + 2 \, \sqrt {e x^{2} + d} \sqrt {e} x + d}{d}\right ) + 60 \, d^{3} {\rm weierstrassPInverse}\left (-\frac {4 \, d}{e}, 0, x\right ) - 4 \, {\left (7 \, e^{2} x^{4} - 9 \, d e x^{2} + 15 \, d^{2}\right )} \sqrt {e x^{2} + d} \sqrt {e} \sqrt {x}}{847 \, e^{3}} \] Input:

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

Output:

1/847*(77*e^3*x^(11/2)*log((2*e*x^2 + 2*sqrt(e*x^2 + d)*sqrt(e)*x + d)/d) 
+ 60*d^3*weierstrassPInverse(-4*d/e, 0, x) - 4*(7*e^2*x^4 - 9*d*e*x^2 + 15 
*d^2)*sqrt(e*x^2 + d)*sqrt(e)*sqrt(x))/e^3
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int x^{9/2} \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right ) \, dx=\int \sqrt {x}\, \mathit {atanh} \left (\frac {\sqrt {e}\, x}{\sqrt {e \,x^{2}+d}}\right ) x^{4}d x \] Input:

int(x^(9/2)*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**4,x)