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

Optimal result

Integrand size = 23, antiderivative size = 101 \[ \int x^3 \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right ) \, dx=\frac {3 d x \sqrt {d+e x^2}}{32 e^{3/2}}-\frac {x^3 \sqrt {d+e x^2}}{16 \sqrt {e}}-\frac {3 d^2 \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{32 e^2}+\frac {1}{4} x^4 \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right ) \] Output:

3/32*d*x*(e*x^2+d)^(1/2)/e^(3/2)-1/16*x^3*(e*x^2+d)^(1/2)/e^(1/2)-3/32*d^2 
*arctanh(e^(1/2)*x/(e*x^2+d)^(1/2))/e^2+1/4*x^4*arctanh(e^(1/2)*x/(e*x^2+d 
)^(1/2))
 

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.87 \[ \int x^3 \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right ) \, dx=\frac {\sqrt {e} x \left (3 d-2 e x^2\right ) \sqrt {d+e x^2}+8 e^2 x^4 \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )-3 d^2 \log \left (\sqrt {e} x+\sqrt {d+e x^2}\right )}{32 e^2} \] Input:

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

Output:

(Sqrt[e]*x*(3*d - 2*e*x^2)*Sqrt[d + e*x^2] + 8*e^2*x^4*ArcTanh[(Sqrt[e]*x) 
/Sqrt[d + e*x^2]] - 3*d^2*Log[Sqrt[e]*x + Sqrt[d + e*x^2]])/(32*e^2)
 

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.14, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {6775, 262, 262, 224, 219}

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.

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

\(\Big \downarrow \) 6775

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

\(\Big \downarrow \) 262

\(\displaystyle \frac {1}{4} x^4 \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )-\frac {1}{4} \sqrt {e} \left (\frac {x^3 \sqrt {d+e x^2}}{4 e}-\frac {3 d \int \frac {x^2}{\sqrt {e x^2+d}}dx}{4 e}\right )\)

\(\Big \downarrow \) 262

\(\displaystyle \frac {1}{4} x^4 \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )-\frac {1}{4} \sqrt {e} \left (\frac {x^3 \sqrt {d+e x^2}}{4 e}-\frac {3 d \left (\frac {x \sqrt {d+e x^2}}{2 e}-\frac {d \int \frac {1}{\sqrt {e x^2+d}}dx}{2 e}\right )}{4 e}\right )\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {1}{4} x^4 \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )-\frac {1}{4} \sqrt {e} \left (\frac {x^3 \sqrt {d+e x^2}}{4 e}-\frac {3 d \left (\frac {x \sqrt {d+e x^2}}{2 e}-\frac {d \int \frac {1}{1-\frac {e x^2}{e x^2+d}}d\frac {x}{\sqrt {e x^2+d}}}{2 e}\right )}{4 e}\right )\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{4} x^4 \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )-\frac {1}{4} \sqrt {e} \left (\frac {x^3 \sqrt {d+e x^2}}{4 e}-\frac {3 d \left (\frac {x \sqrt {d+e x^2}}{2 e}-\frac {d \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{2 e^{3/2}}\right )}{4 e}\right )\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 224

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], 
x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] &&  !GtQ[a, 0]
 

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 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 [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(204\) vs. \(2(77)=154\).

Time = 0.03 (sec) , antiderivative size = 205, normalized size of antiderivative = 2.03

method	result	size
default	\(\frac {x^{4} \operatorname {arctanh}\left (\frac {\sqrt {e}\, x}{\sqrt {e \,x^{2}+d}}\right )}{4}+\frac {e^{\frac {3}{2}} \left (\frac {x^{5} \sqrt {e \,x^{2}+d}}{6 e}-\frac {5 d \left (\frac {x^{3} \sqrt {e \,x^{2}+d}}{4 e}-\frac {3 d \left (\frac {x \sqrt {e \,x^{2}+d}}{2 e}-\frac {d \ln \left (\sqrt {e}\, x +\sqrt {e \,x^{2}+d}\right )}{2 e^{\frac {3}{2}}}\right )}{4 e}\right )}{6 e}\right )}{4 d}-\frac {\sqrt {e}\, \left (\frac {x^{3} \left (e \,x^{2}+d \right )^{\frac {3}{2}}}{6 e}-\frac {d \left (\frac {x \left (e \,x^{2}+d \right )^{\frac {3}{2}}}{4 e}-\frac {d \left (\frac {x \sqrt {e \,x^{2}+d}}{2}+\frac {d \ln \left (\sqrt {e}\, x +\sqrt {e \,x^{2}+d}\right )}{2 \sqrt {e}}\right )}{4 e}\right )}{2 e}\right )}{4 d}\)	\(205\)
parts	\(\frac {x^{4} \operatorname {arctanh}\left (\frac {\sqrt {e}\, x}{\sqrt {e \,x^{2}+d}}\right )}{4}+\frac {e^{\frac {3}{2}} \left (\frac {x^{5} \sqrt {e \,x^{2}+d}}{6 e}-\frac {5 d \left (\frac {x^{3} \sqrt {e \,x^{2}+d}}{4 e}-\frac {3 d \left (\frac {x \sqrt {e \,x^{2}+d}}{2 e}-\frac {d \ln \left (\sqrt {e}\, x +\sqrt {e \,x^{2}+d}\right )}{2 e^{\frac {3}{2}}}\right )}{4 e}\right )}{6 e}\right )}{4 d}-\frac {\sqrt {e}\, \left (\frac {x^{3} \left (e \,x^{2}+d \right )^{\frac {3}{2}}}{6 e}-\frac {d \left (\frac {x \left (e \,x^{2}+d \right )^{\frac {3}{2}}}{4 e}-\frac {d \left (\frac {x \sqrt {e \,x^{2}+d}}{2}+\frac {d \ln \left (\sqrt {e}\, x +\sqrt {e \,x^{2}+d}\right )}{2 \sqrt {e}}\right )}{4 e}\right )}{2 e}\right )}{4 d}\)	\(205\)

Input:

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

Output:

1/4*x^4*arctanh(e^(1/2)*x/(e*x^2+d)^(1/2))+1/4*e^(3/2)/d*(1/6*x^5/e*(e*x^2 
+d)^(1/2)-5/6*d/e*(1/4*x^3/e*(e*x^2+d)^(1/2)-3/4*d/e*(1/2*x/e*(e*x^2+d)^(1 
/2)-1/2*d/e^(3/2)*ln(e^(1/2)*x+(e*x^2+d)^(1/2)))))-1/4*e^(1/2)/d*(1/6*x^3* 
(e*x^2+d)^(3/2)/e-1/2*d/e*(1/4*x*(e*x^2+d)^(3/2)/e-1/4*d/e*(1/2*x*(e*x^2+d 
)^(1/2)+1/2*d/e^(1/2)*ln(e^(1/2)*x+(e*x^2+d)^(1/2)))))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.74 \[ \int x^3 \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right ) \, dx=-\frac {2 \, {\left (2 \, e x^{3} - 3 \, d x\right )} \sqrt {e x^{2} + d} \sqrt {e} - {\left (8 \, e^{2} x^{4} - 3 \, d^{2}\right )} \log \left (\frac {2 \, e x^{2} + 2 \, \sqrt {e x^{2} + d} \sqrt {e} x + d}{d}\right )}{64 \, e^{2}} \] Input:

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

Output:

-1/64*(2*(2*e*x^3 - 3*d*x)*sqrt(e*x^2 + d)*sqrt(e) - (8*e^2*x^4 - 3*d^2)*l 
og((2*e*x^2 + 2*sqrt(e*x^2 + d)*sqrt(e)*x + d)/d))/e^2
 

Sympy [A] (verification not implemented)

Time = 0.43 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.94 \[ \int x^3 \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right ) \, dx=\begin {cases} - \frac {3 d^{2} \operatorname {atanh}{\left (\frac {\sqrt {e} x}{\sqrt {d + e x^{2}}} \right )}}{32 e^{2}} + \frac {3 d x \sqrt {d + e x^{2}}}{32 e^{\frac {3}{2}}} + \frac {x^{4} \operatorname {atanh}{\left (\frac {\sqrt {e} x}{\sqrt {d + e x^{2}}} \right )}}{4} - \frac {x^{3} \sqrt {d + e x^{2}}}{16 \sqrt {e}} & \text {for}\: e \neq 0 \\0 & \text {otherwise} \end {cases} \] Input:

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

Output:

Piecewise((-3*d**2*atanh(sqrt(e)*x/sqrt(d + e*x**2))/(32*e**2) + 3*d*x*sqr 
t(d + e*x**2)/(32*e**(3/2)) + x**4*atanh(sqrt(e)*x/sqrt(d + e*x**2))/4 - x 
**3*sqrt(d + e*x**2)/(16*sqrt(e)), Ne(e, 0)), (0, True))
 

Maxima [F]

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

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

Output:

1/8*x^4*log(sqrt(e)*x + sqrt(e*x^2 + d)) - 1/8*x^4*log(-sqrt(e)*x + sqrt(e 
*x^2 + d)) - 1/2*d*sqrt(e)*integrate(-1/2*sqrt(e*x^2 + d)*x^4/(e^2*x^4 + d 
*e*x^2 - (e*x^2 + d)^2), x)
 

Giac [F(-1)]

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

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

Output:

Timed out
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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