3.10 \(\int x^4 \tanh ^{-1}(\frac{\sqrt{e} x}{\sqrt{d+e x^2}}) \, dx\)

Optimal. Leaf size=91 \[ -\frac{d^2 \sqrt{d+e x^2}}{5 e^{5/2}}-\frac{\left (d+e x^2\right )^{5/2}}{25 e^{5/2}}+\frac{2 d \left (d+e x^2\right )^{3/2}}{15 e^{5/2}}+\frac{1}{5} x^5 \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right ) \]

[Out]

-(d^2*Sqrt[d + e*x^2])/(5*e^(5/2)) + (2*d*(d + e*x^2)^(3/2))/(15*e^(5/2)) - (d + e*x^2)^(5/2)/(25*e^(5/2)) + (
x^5*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/5

________________________________________________________________________________________

Rubi [A]  time = 0.0491054, antiderivative size = 91, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {6221, 266, 43} \[ -\frac{d^2 \sqrt{d+e x^2}}{5 e^{5/2}}-\frac{\left (d+e x^2\right )^{5/2}}{25 e^{5/2}}+\frac{2 d \left (d+e x^2\right )^{3/2}}{15 e^{5/2}}+\frac{1}{5} x^5 \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[x^4*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]],x]

[Out]

-(d^2*Sqrt[d + e*x^2])/(5*e^(5/2)) + (2*d*(d + e*x^2)^(3/2))/(15*e^(5/2)) - (d + e*x^2)^(5/2)/(25*e^(5/2)) + (
x^5*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/5

Rule 6221

Int[ArcTanh[((c_.)*(x_))/Sqrt[(a_.) + (b_.)*(x_)^2]]*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*ArcT
anh[(c*x)/Sqrt[a + b*x^2]])/(d*(m + 1)), x] - Dist[c/(d*(m + 1)), Int[(d*x)^(m + 1)/Sqrt[a + b*x^2], x], x] /;
FreeQ[{a, b, c, d, m}, x] && EqQ[b, c^2] && NeQ[m, -1]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int x^4 \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right ) \, dx &=\frac{1}{5} x^5 \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )-\frac{1}{5} \sqrt{e} \int \frac{x^5}{\sqrt{d+e x^2}} \, dx\\ &=\frac{1}{5} x^5 \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )-\frac{1}{10} \sqrt{e} \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{d+e x}} \, dx,x,x^2\right )\\ &=\frac{1}{5} x^5 \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )-\frac{1}{10} \sqrt{e} \operatorname{Subst}\left (\int \left (\frac{d^2}{e^2 \sqrt{d+e x}}-\frac{2 d \sqrt{d+e x}}{e^2}+\frac{(d+e x)^{3/2}}{e^2}\right ) \, dx,x,x^2\right )\\ &=-\frac{d^2 \sqrt{d+e x^2}}{5 e^{5/2}}+\frac{2 d \left (d+e x^2\right )^{3/2}}{15 e^{5/2}}-\frac{\left (d+e x^2\right )^{5/2}}{25 e^{5/2}}+\frac{1}{5} x^5 \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )\\ \end{align*}

Mathematica [A]  time = 0.0572996, size = 68, normalized size = 0.75 \[ \frac{1}{5} x^5 \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )-\frac{\sqrt{d+e x^2} \left (8 d^2-4 d e x^2+3 e^2 x^4\right )}{75 e^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]],x]

[Out]

-(Sqrt[d + e*x^2]*(8*d^2 - 4*d*e*x^2 + 3*e^2*x^4))/(75*e^(5/2)) + (x^5*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/5

________________________________________________________________________________________

Maple [B]  time = 0.033, size = 176, normalized size = 1.9 \begin{align*}{\frac{{x}^{5}}{5}{\it Artanh} \left ({x\sqrt{e}{\frac{1}{\sqrt{e{x}^{2}+d}}}} \right ) }+{\frac{1}{5\,d}{e}^{{\frac{3}{2}}} \left ({\frac{{x}^{6}}{7\,e}\sqrt{e{x}^{2}+d}}-{\frac{6\,d}{7\,e} \left ({\frac{{x}^{4}}{5\,e}\sqrt{e{x}^{2}+d}}-{\frac{4\,d}{5\,e} \left ({\frac{{x}^{2}}{3\,e}\sqrt{e{x}^{2}+d}}-{\frac{2\,d}{3\,{e}^{2}}\sqrt{e{x}^{2}+d}} \right ) } \right ) } \right ) }-{\frac{1}{5\,d}\sqrt{e} \left ({\frac{{x}^{4}}{7\,e} \left ( e{x}^{2}+d \right ) ^{{\frac{3}{2}}}}-{\frac{4\,d}{7\,e} \left ({\frac{{x}^{2}}{5\,e} \left ( e{x}^{2}+d \right ) ^{{\frac{3}{2}}}}-{\frac{2\,d}{15\,{e}^{2}} \left ( e{x}^{2}+d \right ) ^{{\frac{3}{2}}}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2)),x)

[Out]

1/5*x^5*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))+1/5*e^(3/2)/d*(1/7*x^6/e*(e*x^2+d)^(1/2)-6/7*d/e*(1/5*x^4/e*(e*x^2+
d)^(1/2)-4/5*d/e*(1/3*x^2/e*(e*x^2+d)^(1/2)-2/3*d/e^2*(e*x^2+d)^(1/2))))-1/5*e^(1/2)/d*(1/7*x^4*(e*x^2+d)^(3/2
)/e-4/7*d/e*(1/5*x^2*(e*x^2+d)^(3/2)/e-2/15*d/e^2*(e*x^2+d)^(3/2)))

________________________________________________________________________________________

Maxima [A]  time = 0.983253, size = 171, normalized size = 1.88 \begin{align*} \frac{1}{5} \, x^{5} \operatorname{artanh}\left (\frac{\sqrt{e} x}{\sqrt{e x^{2} + d}}\right ) - \frac{15 \,{\left (e x^{2} + d\right )}^{\frac{7}{2}} - 42 \,{\left (e x^{2} + d\right )}^{\frac{5}{2}} d + 35 \,{\left (e x^{2} + d\right )}^{\frac{3}{2}} d^{2}}{525 \, d e^{\frac{5}{2}}} + \frac{5 \,{\left (e x^{2} + d\right )}^{\frac{7}{2}} - 21 \,{\left (e x^{2} + d\right )}^{\frac{5}{2}} d + 35 \,{\left (e x^{2} + d\right )}^{\frac{3}{2}} d^{2} - 35 \, \sqrt{e x^{2} + d} d^{3}}{175 \, d e^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2)),x, algorithm="maxima")

[Out]

1/5*x^5*arctanh(sqrt(e)*x/sqrt(e*x^2 + d)) - 1/525*(15*(e*x^2 + d)^(7/2) - 42*(e*x^2 + d)^(5/2)*d + 35*(e*x^2
+ d)^(3/2)*d^2)/(d*e^(5/2)) + 1/175*(5*(e*x^2 + d)^(7/2) - 21*(e*x^2 + d)^(5/2)*d + 35*(e*x^2 + d)^(3/2)*d^2 -
35*sqrt(e*x^2 + d)*d^3)/(d*e^(5/2))

________________________________________________________________________________________

Fricas [A]  time = 2.16161, size = 182, normalized size = 2. \begin{align*} \frac{15 \, e^{3} x^{5} \log \left (\frac{2 \, e x^{2} + 2 \, \sqrt{e x^{2} + d} \sqrt{e} x + d}{d}\right ) - 2 \,{\left (3 \, e^{2} x^{4} - 4 \, d e x^{2} + 8 \, d^{2}\right )} \sqrt{e x^{2} + d} \sqrt{e}}{150 \, e^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2)),x, algorithm="fricas")

[Out]

1/150*(15*e^3*x^5*log((2*e*x^2 + 2*sqrt(e*x^2 + d)*sqrt(e)*x + d)/d) - 2*(3*e^2*x^4 - 4*d*e*x^2 + 8*d^2)*sqrt(
e*x^2 + d)*sqrt(e))/e^3

________________________________________________________________________________________

Sympy [A]  time = 5.2025, size = 90, normalized size = 0.99 \begin{align*} \begin{cases} - \frac{8 d^{2} \sqrt{d + e x^{2}}}{75 e^{\frac{5}{2}}} + \frac{4 d x^{2} \sqrt{d + e x^{2}}}{75 e^{\frac{3}{2}}} + \frac{x^{5} \operatorname{atanh}{\left (\frac{\sqrt{e} x}{\sqrt{d + e x^{2}}} \right )}}{5} - \frac{x^{4} \sqrt{d + e x^{2}}}{25 \sqrt{e}} & \text{for}\: e \neq 0 \\0 & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*atanh(x*e**(1/2)/(e*x**2+d)**(1/2)),x)

[Out]

Piecewise((-8*d**2*sqrt(d + e*x**2)/(75*e**(5/2)) + 4*d*x**2*sqrt(d + e*x**2)/(75*e**(3/2)) + x**5*atanh(sqrt(
e)*x/sqrt(d + e*x**2))/5 - x**4*sqrt(d + e*x**2)/(25*sqrt(e)), Ne(e, 0)), (0, True))

________________________________________________________________________________________

Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: NotImplementedError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2)),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError