3.15 \(\int \frac{\tanh ^{-1}(\frac{\sqrt{e} x}{\sqrt{d+e x^2}})}{x^6} \, dx\)

Optimal. Leaf size=111 \[ \frac{3 e^{3/2} \sqrt{d+e x^2}}{40 d^2 x^2}-\frac{3 e^{5/2} \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{40 d^{5/2}}-\frac{\sqrt{e} \sqrt{d+e x^2}}{20 d x^4}-\frac{\tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{5 x^5} \]

[Out]

-(Sqrt[e]*Sqrt[d + e*x^2])/(20*d*x^4) + (3*e^(3/2)*Sqrt[d + e*x^2])/(40*d^2*x^2) - ArcTanh[(Sqrt[e]*x)/Sqrt[d
+ e*x^2]]/(5*x^5) - (3*e^(5/2)*ArcTanh[Sqrt[d + e*x^2]/Sqrt[d]])/(40*d^(5/2))

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Rubi [A]  time = 0.0570099, antiderivative size = 111, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {6221, 266, 51, 63, 208} \[ \frac{3 e^{3/2} \sqrt{d+e x^2}}{40 d^2 x^2}-\frac{3 e^{5/2} \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{40 d^{5/2}}-\frac{\sqrt{e} \sqrt{d+e x^2}}{20 d x^4}-\frac{\tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{5 x^5} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]]/x^6,x]

[Out]

-(Sqrt[e]*Sqrt[d + e*x^2])/(20*d*x^4) + (3*e^(3/2)*Sqrt[d + e*x^2])/(40*d^2*x^2) - ArcTanh[(Sqrt[e]*x)/Sqrt[d
+ e*x^2]]/(5*x^5) - (3*e^(5/2)*ArcTanh[Sqrt[d + e*x^2]/Sqrt[d]])/(40*d^(5/2))

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 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

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

Rubi steps

\begin{align*} \int \frac{\tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{x^6} \, dx &=-\frac{\tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{5 x^5}+\frac{1}{5} \sqrt{e} \int \frac{1}{x^5 \sqrt{d+e x^2}} \, dx\\ &=-\frac{\tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{5 x^5}+\frac{1}{10} \sqrt{e} \operatorname{Subst}\left (\int \frac{1}{x^3 \sqrt{d+e x}} \, dx,x,x^2\right )\\ &=-\frac{\sqrt{e} \sqrt{d+e x^2}}{20 d x^4}-\frac{\tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{5 x^5}-\frac{\left (3 e^{3/2}\right ) \operatorname{Subst}\left (\int \frac{1}{x^2 \sqrt{d+e x}} \, dx,x,x^2\right )}{40 d}\\ &=-\frac{\sqrt{e} \sqrt{d+e x^2}}{20 d x^4}+\frac{3 e^{3/2} \sqrt{d+e x^2}}{40 d^2 x^2}-\frac{\tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{5 x^5}+\frac{\left (3 e^{5/2}\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{d+e x}} \, dx,x,x^2\right )}{80 d^2}\\ &=-\frac{\sqrt{e} \sqrt{d+e x^2}}{20 d x^4}+\frac{3 e^{3/2} \sqrt{d+e x^2}}{40 d^2 x^2}-\frac{\tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{5 x^5}+\frac{\left (3 e^{3/2}\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{d}{e}+\frac{x^2}{e}} \, dx,x,\sqrt{d+e x^2}\right )}{40 d^2}\\ &=-\frac{\sqrt{e} \sqrt{d+e x^2}}{20 d x^4}+\frac{3 e^{3/2} \sqrt{d+e x^2}}{40 d^2 x^2}-\frac{\tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{5 x^5}-\frac{3 e^{5/2} \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{40 d^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.109866, size = 107, normalized size = 0.96 \[ \frac{\frac{\sqrt{e} x \left (-3 e^2 x^4 \log \left (\sqrt{d} \sqrt{d+e x^2}+d\right )+\sqrt{d} \sqrt{d+e x^2} \left (3 e x^2-2 d\right )+3 e^2 x^4 \log (x)\right )}{d^{5/2}}-8 \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{40 x^5} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]]/x^6,x]

[Out]

(-8*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]] + (Sqrt[e]*x*(Sqrt[d]*Sqrt[d + e*x^2]*(-2*d + 3*e*x^2) + 3*e^2*x^4*Lo
g[x] - 3*e^2*x^4*Log[d + Sqrt[d]*Sqrt[d + e*x^2]]))/d^(5/2))/(40*x^5)

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Maple [A]  time = 0.033, size = 130, normalized size = 1.2 \begin{align*} -{\frac{1}{5\,{x}^{5}}{\it Artanh} \left ({x\sqrt{e}{\frac{1}{\sqrt{e{x}^{2}+d}}}} \right ) }+{\frac{1}{10\,{d}^{2}{x}^{2}}{e}^{{\frac{3}{2}}}\sqrt{e{x}^{2}+d}}-{\frac{3}{40}{e}^{{\frac{5}{2}}}\ln \left ({\frac{1}{x} \left ( 2\,d+2\,\sqrt{d}\sqrt{e{x}^{2}+d} \right ) } \right ){d}^{-{\frac{5}{2}}}}-{\frac{1}{20\,{d}^{2}{x}^{4}}\sqrt{e} \left ( e{x}^{2}+d \right ) ^{{\frac{3}{2}}}}+{\frac{1}{40\,{d}^{3}{x}^{2}}{e}^{{\frac{3}{2}}} \left ( e{x}^{2}+d \right ) ^{{\frac{3}{2}}}}-{\frac{1}{40\,{d}^{3}}{e}^{{\frac{5}{2}}}\sqrt{e{x}^{2}+d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))/x^6,x)

[Out]

-1/5*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))/x^5+1/10*e^(3/2)*(e*x^2+d)^(1/2)/d^2/x^2-3/40*e^(5/2)/d^(5/2)*ln((2*d+
2*d^(1/2)*(e*x^2+d)^(1/2))/x)-1/20*e^(1/2)/d^2/x^4*(e*x^2+d)^(3/2)+1/40*e^(3/2)/d^3/x^2*(e*x^2+d)^(3/2)-1/40*e
^(5/2)/d^3*(e*x^2+d)^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} d \sqrt{e} \int -\frac{\sqrt{e x^{2} + d}}{5 \,{\left (e^{2} x^{9} + d e x^{7} -{\left (e x^{7} + d x^{5}\right )}{\left (e x^{2} + d\right )}\right )}}\,{d x} - \frac{\log \left (\sqrt{e} x + \sqrt{e x^{2} + d}\right ) - \log \left (-\sqrt{e} x + \sqrt{e x^{2} + d}\right )}{10 \, x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))/x^6,x, algorithm="maxima")

[Out]

d*sqrt(e)*integrate(-1/5*sqrt(e*x^2 + d)/(e^2*x^9 + d*e*x^7 - (e*x^7 + d*x^5)*(e*x^2 + d)), x) - 1/10*(log(sqr
t(e)*x + sqrt(e*x^2 + d)) - log(-sqrt(e)*x + sqrt(e*x^2 + d)))/x^5

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Fricas [B]  time = 2.63659, size = 872, normalized size = 7.86 \begin{align*} \left [\frac{3 \, e^{2} x^{5} \sqrt{\frac{e}{d}} \log \left (-\frac{e^{2} x^{2} - 2 \, \sqrt{e x^{2} + d} d \sqrt{e} \sqrt{\frac{e}{d}} + 2 \, d e}{x^{2}}\right ) - 8 \, d^{2} x^{5} \log \left (\frac{e x + \sqrt{e x^{2} + d} \sqrt{e}}{x}\right ) + 8 \, d^{2} x^{5} \log \left (\frac{e x - \sqrt{e x^{2} + d} \sqrt{e}}{x}\right ) + 2 \,{\left (3 \, e x^{3} - 2 \, d x\right )} \sqrt{e x^{2} + d} \sqrt{e} + 8 \,{\left (d^{2} x^{5} - d^{2}\right )} \log \left (\frac{2 \, e x^{2} + 2 \, \sqrt{e x^{2} + d} \sqrt{e} x + d}{d}\right )}{80 \, d^{2} x^{5}}, \frac{3 \, e^{2} x^{5} \sqrt{-\frac{e}{d}} \arctan \left (\frac{\sqrt{e x^{2} + d} d \sqrt{e} \sqrt{-\frac{e}{d}}}{e^{2} x^{2} + d e}\right ) - 4 \, d^{2} x^{5} \log \left (\frac{e x + \sqrt{e x^{2} + d} \sqrt{e}}{x}\right ) + 4 \, d^{2} x^{5} \log \left (\frac{e x - \sqrt{e x^{2} + d} \sqrt{e}}{x}\right ) +{\left (3 \, e x^{3} - 2 \, d x\right )} \sqrt{e x^{2} + d} \sqrt{e} + 4 \,{\left (d^{2} x^{5} - d^{2}\right )} \log \left (\frac{2 \, e x^{2} + 2 \, \sqrt{e x^{2} + d} \sqrt{e} x + d}{d}\right )}{40 \, d^{2} x^{5}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))/x^6,x, algorithm="fricas")

[Out]

[1/80*(3*e^2*x^5*sqrt(e/d)*log(-(e^2*x^2 - 2*sqrt(e*x^2 + d)*d*sqrt(e)*sqrt(e/d) + 2*d*e)/x^2) - 8*d^2*x^5*log
((e*x + sqrt(e*x^2 + d)*sqrt(e))/x) + 8*d^2*x^5*log((e*x - sqrt(e*x^2 + d)*sqrt(e))/x) + 2*(3*e*x^3 - 2*d*x)*s
qrt(e*x^2 + d)*sqrt(e) + 8*(d^2*x^5 - d^2)*log((2*e*x^2 + 2*sqrt(e*x^2 + d)*sqrt(e)*x + d)/d))/(d^2*x^5), 1/40
*(3*e^2*x^5*sqrt(-e/d)*arctan(sqrt(e*x^2 + d)*d*sqrt(e)*sqrt(-e/d)/(e^2*x^2 + d*e)) - 4*d^2*x^5*log((e*x + sqr
t(e*x^2 + d)*sqrt(e))/x) + 4*d^2*x^5*log((e*x - sqrt(e*x^2 + d)*sqrt(e))/x) + (3*e*x^3 - 2*d*x)*sqrt(e*x^2 + d
)*sqrt(e) + 4*(d^2*x^5 - d^2)*log((2*e*x^2 + 2*sqrt(e*x^2 + d)*sqrt(e)*x + d)/d))/(d^2*x^5)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{atanh}{\left (\frac{\sqrt{e} x}{\sqrt{d + e x^{2}}} \right )}}{x^{6}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(x*e**(1/2)/(e*x**2+d)**(1/2))/x**6,x)

[Out]

Integral(atanh(sqrt(e)*x/sqrt(d + e*x**2))/x**6, x)

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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(arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))/x^6,x, algorithm="giac")

[Out]

Exception raised: NotImplementedError