3.1.22 \(\int \frac {\text {arctanh}(\frac {\sqrt {e} x}{\sqrt {d+e x^2}})}{x^{15/2}} \, dx\) [22]

3.1.22.1 Optimal result

Integrand size = 25, antiderivative size = 201 \[ \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^{15/2}} \, dx=-\frac {4 \sqrt {e} \sqrt {d+e x^2}}{143 d x^{11/2}}+\frac {36 e^{3/2} \sqrt {d+e x^2}}{1001 d^2 x^{7/2}}-\frac {60 e^{5/2} \sqrt {d+e x^2}}{1001 d^3 x^{3/2}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{13 x^{13/2}}-\frac {30 e^{13/4} \left (\sqrt {d}+\sqrt {e} x\right ) \sqrt {\frac {d+e x^2}{\left (\sqrt {d}+\sqrt {e} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{e} \sqrt {x}}{\sqrt [4]{d}}\right ),\frac {1}{2}\right )}{1001 d^{13/4} \sqrt {d+e x^2}} \]

output

-2/13*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))/x^(13/2)+36/1001*e^(3/2)*(e*x^2+d 
)^(1/2)/d^2/x^(7/2)-60/1001*e^(5/2)*(e*x^2+d)^(1/2)/d^3/x^(3/2)-4/143*e^(1 
/2)*(e*x^2+d)^(1/2)/d/x^(11/2)-30/1001*e^(13/4)*(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(si 
n(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)/d^(13/4)/(e*x^2+d)^(1/2)
 

3.1.22.2 Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.34 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.81 \[ \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^{15/2}} \, dx=\frac {2 \left (-\frac {2 \sqrt {e} x \sqrt {d+e x^2} \left (7 d^2-9 d e x^2+15 e^2 x^4\right )}{d^3}-77 \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )-\frac {30 \sqrt {\frac {i \sqrt {d}}{\sqrt {e}}} e^4 \sqrt {1+\frac {d}{e x^2}} x^{15/2} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\frac {\sqrt {\frac {i \sqrt {d}}{\sqrt {e}}}}{\sqrt {x}}\right ),-1\right )}{d^{7/2} \sqrt {d+e x^2}}\right )}{1001 x^{13/2}} \]

input

Integrate[ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]]/x^(15/2),x]
 

output

(2*((-2*Sqrt[e]*x*Sqrt[d + e*x^2]*(7*d^2 - 9*d*e*x^2 + 15*e^2*x^4))/d^3 - 
77*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]] - (30*Sqrt[(I*Sqrt[d])/Sqrt[e]]*e^ 
4*Sqrt[1 + d/(e*x^2)]*x^(15/2)*EllipticF[I*ArcSinh[Sqrt[(I*Sqrt[d])/Sqrt[e 
]]/Sqrt[x]], -1])/(d^(7/2)*Sqrt[d + e*x^2])))/(1001*x^(13/2))
 

3.1.22.3 Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.06, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {6775, 264, 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.

\(\displaystyle \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^{15/2}} \, dx\)

\(\Big \downarrow \) 6775

\(\displaystyle \frac {2}{13} \sqrt {e} \int \frac {1}{x^{13/2} \sqrt {e x^2+d}}dx-\frac {2 \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{13 x^{13/2}}\)

\(\Big \downarrow \) 264

\(\displaystyle \frac {2}{13} \sqrt {e} \left (-\frac {9 e \int \frac {1}{x^{9/2} \sqrt {e x^2+d}}dx}{11 d}-\frac {2 \sqrt {d+e x^2}}{11 d x^{11/2}}\right )-\frac {2 \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{13 x^{13/2}}\)

\(\Big \downarrow \) 264

\(\displaystyle \frac {2}{13} \sqrt {e} \left (-\frac {9 e \left (-\frac {5 e \int \frac {1}{x^{5/2} \sqrt {e x^2+d}}dx}{7 d}-\frac {2 \sqrt {d+e x^2}}{7 d x^{7/2}}\right )}{11 d}-\frac {2 \sqrt {d+e x^2}}{11 d x^{11/2}}\right )-\frac {2 \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{13 x^{13/2}}\)

\(\Big \downarrow \) 264

\(\displaystyle \frac {2}{13} \sqrt {e} \left (-\frac {9 e \left (-\frac {5 e \left (-\frac {e \int \frac {1}{\sqrt {x} \sqrt {e x^2+d}}dx}{3 d}-\frac {2 \sqrt {d+e x^2}}{3 d x^{3/2}}\right )}{7 d}-\frac {2 \sqrt {d+e x^2}}{7 d x^{7/2}}\right )}{11 d}-\frac {2 \sqrt {d+e x^2}}{11 d x^{11/2}}\right )-\frac {2 \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{13 x^{13/2}}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {2}{13} \sqrt {e} \left (-\frac {9 e \left (-\frac {5 e \left (-\frac {2 e \int \frac {1}{\sqrt {e x^2+d}}d\sqrt {x}}{3 d}-\frac {2 \sqrt {d+e x^2}}{3 d x^{3/2}}\right )}{7 d}-\frac {2 \sqrt {d+e x^2}}{7 d x^{7/2}}\right )}{11 d}-\frac {2 \sqrt {d+e x^2}}{11 d x^{11/2}}\right )-\frac {2 \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{13 x^{13/2}}\)

\(\Big \downarrow \) 761

\(\displaystyle \frac {2}{13} \sqrt {e} \left (-\frac {9 e \left (-\frac {5 e \left (-\frac {e^{3/4} \left (\sqrt {d}+\sqrt {e} x\right ) \sqrt {\frac {d+e x^2}{\left (\sqrt {d}+\sqrt {e} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{e} \sqrt {x}}{\sqrt [4]{d}}\right ),\frac {1}{2}\right )}{3 d^{5/4} \sqrt {d+e x^2}}-\frac {2 \sqrt {d+e x^2}}{3 d x^{3/2}}\right )}{7 d}-\frac {2 \sqrt {d+e x^2}}{7 d x^{7/2}}\right )}{11 d}-\frac {2 \sqrt {d+e x^2}}{11 d x^{11/2}}\right )-\frac {2 \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{13 x^{13/2}}\)

input

Int[ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]]/x^(15/2),x]
 

output

(-2*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/(13*x^(13/2)) + (2*Sqrt[e]*((-2* 
Sqrt[d + e*x^2])/(11*d*x^(11/2)) - (9*e*((-2*Sqrt[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]*EllipticF[2*ArcTan[(e^(1/4)* 
Sqrt[x])/d^(1/4)], 1/2])/(3*d^(5/4)*Sqrt[d + e*x^2])))/(7*d)))/(11*d)))/13
 

3.1.22.3.1 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]
 

3.1.22.4 Maple [F]

\[\int \frac {\operatorname {arctanh}\left (\frac {x \sqrt {e}}{\sqrt {e \,x^{2}+d}}\right )}{x^{\frac {15}{2}}}d x\]

input

int(arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))/x^(15/2),x)
 

output

int(arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))/x^(15/2),x)
 

3.1.22.5 Fricas [C] (verification not implemented)

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

Time = 0.09 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.50 \[ \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^{15/2}} \, dx=-\frac {60 \, e^{3} x^{7} {\rm weierstrassPInverse}\left (-\frac {4 \, d}{e}, 0, x\right ) + 77 \, d^{3} \sqrt {x} \log \left (\frac {2 \, e x^{2} + 2 \, \sqrt {e x^{2} + d} \sqrt {e} x + d}{d}\right ) + 4 \, {\left (15 \, e^{2} x^{5} - 9 \, d e x^{3} + 7 \, d^{2} x\right )} \sqrt {e x^{2} + d} \sqrt {e} \sqrt {x}}{1001 \, d^{3} x^{7}} \]

input

integrate(arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))/x^(15/2),x, algorithm="fricas 
")
 

output

-1/1001*(60*e^3*x^7*weierstrassPInverse(-4*d/e, 0, x) + 77*d^3*sqrt(x)*log 
((2*e*x^2 + 2*sqrt(e*x^2 + d)*sqrt(e)*x + d)/d) + 4*(15*e^2*x^5 - 9*d*e*x^ 
3 + 7*d^2*x)*sqrt(e*x^2 + d)*sqrt(e)*sqrt(x))/(d^3*x^7)
 

3.1.22.6 Sympy [F(-1)]

Timed out. \[ \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^{15/2}} \, dx=\text {Timed out} \]

input

integrate(atanh(x*e**(1/2)/(e*x**2+d)**(1/2))/x**(15/2),x)
 

output

Timed out
 

3.1.22.7 Maxima [F]

\[ \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^{15/2}} \, dx=\int { \frac {\operatorname {artanh}\left (\frac {\sqrt {e} x}{\sqrt {e x^{2} + d}}\right )}{x^{\frac {15}{2}}} \,d x } \]

input

integrate(arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))/x^(15/2),x, algorithm="maxima 
")
 

output

2*d*sqrt(e)*integrate(-1/13*sqrt(e*x^2 + d)*x/((e^2*x^4 + d*e*x^2)*x^(15/2 
) - (e*x^2 + d)*e^(log(e*x^2 + d) + 15/2*log(x))), x) - 1/13*log(sqrt(e)*x 
 + sqrt(e*x^2 + d))/x^(13/2) + 1/13*log(-sqrt(e)*x + sqrt(e*x^2 + d))/x^(1 
3/2)
 

3.1.22.8 Giac [F]

\[ \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^{15/2}} \, dx=\int { \frac {\operatorname {artanh}\left (\frac {\sqrt {e} x}{\sqrt {e x^{2} + d}}\right )}{x^{\frac {15}{2}}} \,d x } \]

input

integrate(arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))/x^(15/2),x, algorithm="giac")
 

output

integrate(arctanh(sqrt(e)*x/sqrt(e*x^2 + d))/x^(15/2), x)
 

3.1.22.9 Mupad [F(-1)]

Timed out. \[ \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^{15/2}} \, dx=\int \frac {\mathrm {atanh}\left (\frac {\sqrt {e}\,x}{\sqrt {e\,x^2+d}}\right )}{x^{15/2}} \,d x \]

input

int(atanh((e^(1/2)*x)/(d + e*x^2)^(1/2))/x^(15/2),x)
 

output

int(atanh((e^(1/2)*x)/(d + e*x^2)^(1/2))/x^(15/2), x)