3.8 \(\int \frac{\tanh ^{-1}(\frac{\sqrt{e} x}{\sqrt{d+e x^2}})}{x^9} \, dx\)

Optimal. Leaf size=131 \[ \frac{2 e^{7/2} \sqrt{d+e x^2}}{35 d^4 x}-\frac{e^{5/2} \sqrt{d+e x^2}}{35 d^3 x^3}+\frac{3 e^{3/2} \sqrt{d+e x^2}}{140 d^2 x^5}-\frac{\sqrt{e} \sqrt{d+e x^2}}{56 d x^7}-\frac{\tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{8 x^8} \]

[Out]

-(Sqrt[e]*Sqrt[d + e*x^2])/(56*d*x^7) + (3*e^(3/2)*Sqrt[d + e*x^2])/(140*d^2*x^5) - (e^(5/2)*Sqrt[d + e*x^2])/
(35*d^3*x^3) + (2*e^(7/2)*Sqrt[d + e*x^2])/(35*d^4*x) - ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]]/(8*x^8)

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Rubi [A]  time = 0.048865, antiderivative size = 131, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {6221, 271, 264} \[ \frac{2 e^{7/2} \sqrt{d+e x^2}}{35 d^4 x}-\frac{e^{5/2} \sqrt{d+e x^2}}{35 d^3 x^3}+\frac{3 e^{3/2} \sqrt{d+e x^2}}{140 d^2 x^5}-\frac{\sqrt{e} \sqrt{d+e x^2}}{56 d x^7}-\frac{\tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{8 x^8} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[e]*Sqrt[d + e*x^2])/(56*d*x^7) + (3*e^(3/2)*Sqrt[d + e*x^2])/(140*d^2*x^5) - (e^(5/2)*Sqrt[d + e*x^2])/
(35*d^3*x^3) + (2*e^(7/2)*Sqrt[d + e*x^2])/(35*d^4*x) - ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]]/(8*x^8)

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 271

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x^(m + 1)*(a + b*x^n)^(p + 1))/(a*(m + 1)), x]
- Dist[(b*(m + n*(p + 1) + 1))/(a*(m + 1)), Int[x^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, m, n, p}, x]
&& ILtQ[Simplify[(m + 1)/n + p + 1], 0] && NeQ[m, -1]

Rule 264

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a
*c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]

Rubi steps

\begin{align*} \int \frac{\tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{x^9} \, dx &=-\frac{\tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{8 x^8}+\frac{1}{8} \sqrt{e} \int \frac{1}{x^8 \sqrt{d+e x^2}} \, dx\\ &=-\frac{\sqrt{e} \sqrt{d+e x^2}}{56 d x^7}-\frac{\tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{8 x^8}-\frac{\left (3 e^{3/2}\right ) \int \frac{1}{x^6 \sqrt{d+e x^2}} \, dx}{28 d}\\ &=-\frac{\sqrt{e} \sqrt{d+e x^2}}{56 d x^7}+\frac{3 e^{3/2} \sqrt{d+e x^2}}{140 d^2 x^5}-\frac{\tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{8 x^8}+\frac{\left (3 e^{5/2}\right ) \int \frac{1}{x^4 \sqrt{d+e x^2}} \, dx}{35 d^2}\\ &=-\frac{\sqrt{e} \sqrt{d+e x^2}}{56 d x^7}+\frac{3 e^{3/2} \sqrt{d+e x^2}}{140 d^2 x^5}-\frac{e^{5/2} \sqrt{d+e x^2}}{35 d^3 x^3}-\frac{\tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{8 x^8}-\frac{\left (2 e^{7/2}\right ) \int \frac{1}{x^2 \sqrt{d+e x^2}} \, dx}{35 d^3}\\ &=-\frac{\sqrt{e} \sqrt{d+e x^2}}{56 d x^7}+\frac{3 e^{3/2} \sqrt{d+e x^2}}{140 d^2 x^5}-\frac{e^{5/2} \sqrt{d+e x^2}}{35 d^3 x^3}+\frac{2 e^{7/2} \sqrt{d+e x^2}}{35 d^4 x}-\frac{\tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{8 x^8}\\ \end{align*}

Mathematica [A]  time = 0.0535337, size = 85, normalized size = 0.65 \[ \frac{\sqrt{e} x \sqrt{d+e x^2} \left (6 d^2 e x^2-5 d^3-8 d e^2 x^4+16 e^3 x^6\right )-35 d^4 \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{280 d^4 x^8} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[e]*x*Sqrt[d + e*x^2]*(-5*d^3 + 6*d^2*e*x^2 - 8*d*e^2*x^4 + 16*e^3*x^6) - 35*d^4*ArcTanh[(Sqrt[e]*x)/Sqrt
[d + e*x^2]])/(280*d^4*x^8)

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Maple [A]  time = 0.034, size = 158, normalized size = 1.2 \begin{align*} -{\frac{1}{8\,{x}^{8}}{\it Artanh} \left ({x\sqrt{e}{\frac{1}{\sqrt{e{x}^{2}+d}}}} \right ) }-{\frac{1}{8\,d}{e}^{{\frac{3}{2}}} \left ( -{\frac{1}{5\,d{x}^{5}}\sqrt{e{x}^{2}+d}}-{\frac{4\,e}{5\,d} \left ( -{\frac{1}{3\,d{x}^{3}}\sqrt{e{x}^{2}+d}}+{\frac{2\,e}{3\,{d}^{2}x}\sqrt{e{x}^{2}+d}} \right ) } \right ) }+{\frac{1}{8\,d}\sqrt{e} \left ( -{\frac{1}{7\,d{x}^{7}} \left ( e{x}^{2}+d \right ) ^{{\frac{3}{2}}}}-{\frac{4\,e}{7\,d} \left ( -{\frac{1}{5\,d{x}^{5}} \left ( e{x}^{2}+d \right ) ^{{\frac{3}{2}}}}+{\frac{2\,e}{15\,{d}^{2}{x}^{3}} \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(arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))/x^9,x)

[Out]

-1/8*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))/x^8-1/8*e^(3/2)/d*(-1/5/d/x^5*(e*x^2+d)^(1/2)-4/5*e/d*(-1/3/d/x^3*(e*x
^2+d)^(1/2)+2/3*e/d^2/x*(e*x^2+d)^(1/2)))+1/8*e^(1/2)/d*(-1/7/d/x^7*(e*x^2+d)^(3/2)-4/7*e/d*(-1/5/d/x^5*(e*x^2
+d)^(3/2)+2/15*e/d^2/x^3*(e*x^2+d)^(3/2)))

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Maxima [A]  time = 0.979119, size = 169, normalized size = 1.29 \begin{align*} \frac{{\left (8 \, e^{3} x^{6} + 4 \, d e^{2} x^{4} - d^{2} e x^{2} + 3 \, d^{3}\right )} e^{\frac{3}{2}}}{120 \, \sqrt{e x^{2} + d} d^{4} x^{5}} - \frac{\operatorname{artanh}\left (\frac{\sqrt{e} x}{\sqrt{e x^{2} + d}}\right )}{8 \, x^{8}} - \frac{{\left (8 \, e^{3} x^{6} - 4 \, d e^{2} x^{4} + 3 \, d^{2} e x^{2} + 15 \, d^{3}\right )} \sqrt{e x^{2} + d} \sqrt{e}}{840 \, d^{4} x^{7}} \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^9,x, algorithm="maxima")

[Out]

1/120*(8*e^3*x^6 + 4*d*e^2*x^4 - d^2*e*x^2 + 3*d^3)*e^(3/2)/(sqrt(e*x^2 + d)*d^4*x^5) - 1/8*arctanh(sqrt(e)*x/
sqrt(e*x^2 + d))/x^8 - 1/840*(8*e^3*x^6 - 4*d*e^2*x^4 + 3*d^2*e*x^2 + 15*d^3)*sqrt(e*x^2 + d)*sqrt(e)/(d^4*x^7
)

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Fricas [A]  time = 2.43524, size = 212, normalized size = 1.62 \begin{align*} -\frac{35 \, d^{4} \log \left (\frac{2 \, e x^{2} + 2 \, \sqrt{e x^{2} + d} \sqrt{e} x + d}{d}\right ) - 2 \,{\left (16 \, e^{3} x^{7} - 8 \, d e^{2} x^{5} + 6 \, d^{2} e x^{3} - 5 \, d^{3} x\right )} \sqrt{e x^{2} + d} \sqrt{e}}{560 \, d^{4} x^{8}} \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^9,x, algorithm="fricas")

[Out]

-1/560*(35*d^4*log((2*e*x^2 + 2*sqrt(e*x^2 + d)*sqrt(e)*x + d)/d) - 2*(16*e^3*x^7 - 8*d*e^2*x^5 + 6*d^2*e*x^3
- 5*d^3*x)*sqrt(e*x^2 + d)*sqrt(e))/(d^4*x^8)

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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^{9}}\, 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**9,x)

[Out]

Integral(atanh(sqrt(e)*x/sqrt(d + e*x**2))/x**9, x)

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Giac [A]  time = 1.54826, size = 217, normalized size = 1.66 \begin{align*} -\frac{\log \left (-\frac{\frac{x e^{\frac{1}{2}}}{\sqrt{x^{2} e + d}} + 1}{\frac{x e^{\frac{1}{2}}}{\sqrt{x^{2} e + d}} - 1}\right )}{16 \, x^{8}} + \frac{4 \,{\left (35 \,{\left (x e^{\frac{1}{2}} - \sqrt{x^{2} e + d}\right )}^{6} d^{3} e^{3} - 21 \,{\left (x e^{\frac{1}{2}} - \sqrt{x^{2} e + d}\right )}^{4} d^{4} e^{3} + 7 \,{\left (x e^{\frac{1}{2}} - \sqrt{x^{2} e + d}\right )}^{2} d^{5} e^{3} - d^{6} e^{3}\right )} e}{35 \,{\left ({\left (x e^{\frac{1}{2}} - \sqrt{x^{2} e + d}\right )}^{2} - d\right )}^{7} d^{3}} \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^9,x, algorithm="giac")

[Out]

-1/16*log(-(x*e^(1/2)/sqrt(x^2*e + d) + 1)/(x*e^(1/2)/sqrt(x^2*e + d) - 1))/x^8 + 4/35*(35*(x*e^(1/2) - sqrt(x
^2*e + d))^6*d^3*e^3 - 21*(x*e^(1/2) - sqrt(x^2*e + d))^4*d^4*e^3 + 7*(x*e^(1/2) - sqrt(x^2*e + d))^2*d^5*e^3
- d^6*e^3)*e/(((x*e^(1/2) - sqrt(x^2*e + d))^2 - d)^7*d^3)