3.6 \(\int \frac{\tanh ^{-1}(\frac{\sqrt{e} x}{\sqrt{d+e x^2}})}{x^5} \, dx\)

Optimal. Leaf size=79 \[ \frac{e^{3/2} \sqrt{d+e x^2}}{6 d^2 x}-\frac{\sqrt{e} \sqrt{d+e x^2}}{12 d x^3}-\frac{\tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{4 x^4} \]

[Out]

-(Sqrt[e]*Sqrt[d + e*x^2])/(12*d*x^3) + (e^(3/2)*Sqrt[d + e*x^2])/(6*d^2*x) - ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x
^2]]/(4*x^4)

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Rubi [A]  time = 0.0262587, antiderivative size = 79, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {6221, 271, 264} \[ \frac{e^{3/2} \sqrt{d+e x^2}}{6 d^2 x}-\frac{\sqrt{e} \sqrt{d+e x^2}}{12 d x^3}-\frac{\tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{4 x^4} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]]/x^5,x]

[Out]

-(Sqrt[e]*Sqrt[d + e*x^2])/(12*d*x^3) + (e^(3/2)*Sqrt[d + e*x^2])/(6*d^2*x) - ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x
^2]]/(4*x^4)

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^5} \, dx &=-\frac{\tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{4 x^4}+\frac{1}{4} \sqrt{e} \int \frac{1}{x^4 \sqrt{d+e x^2}} \, dx\\ &=-\frac{\sqrt{e} \sqrt{d+e x^2}}{12 d x^3}-\frac{\tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{4 x^4}-\frac{e^{3/2} \int \frac{1}{x^2 \sqrt{d+e x^2}} \, dx}{6 d}\\ &=-\frac{\sqrt{e} \sqrt{d+e x^2}}{12 d x^3}+\frac{e^{3/2} \sqrt{d+e x^2}}{6 d^2 x}-\frac{\tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{4 x^4}\\ \end{align*}

Mathematica [A]  time = 0.0404816, size = 63, normalized size = 0.8 \[ \frac{\sqrt{e} x \sqrt{d+e x^2} \left (2 e x^2-d\right )-3 d^2 \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{12 d^2 x^4} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]]/x^5,x]

[Out]

(Sqrt[e]*x*Sqrt[d + e*x^2]*(-d + 2*e*x^2) - 3*d^2*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/(12*d^2*x^4)

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Maple [A]  time = 0.03, size = 62, normalized size = 0.8 \begin{align*} -{\frac{1}{4\,{x}^{4}}{\it Artanh} \left ({x\sqrt{e}{\frac{1}{\sqrt{e{x}^{2}+d}}}} \right ) }+{\frac{1}{4\,{d}^{2}x}{e}^{{\frac{3}{2}}}\sqrt{e{x}^{2}+d}}-{\frac{1}{12\,{d}^{2}{x}^{3}}\sqrt{e} \left ( e{x}^{2}+d \right ) ^{{\frac{3}{2}}}} \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^5,x)

[Out]

-1/4*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))/x^4+1/4*e^(3/2)*(e*x^2+d)^(1/2)/d^2/x-1/12*e^(1/2)/d^2/x^3*(e*x^2+d)^(
3/2)

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Maxima [A]  time = 0.973572, size = 82, normalized size = 1.04 \begin{align*} \frac{\sqrt{e x^{2} + d} e^{\frac{3}{2}}}{4 \, d^{2} x} - \frac{{\left (e x^{2} + d\right )}^{\frac{3}{2}} \sqrt{e}}{12 \, d^{2} x^{3}} - \frac{\operatorname{artanh}\left (\frac{\sqrt{e} x}{\sqrt{e x^{2} + d}}\right )}{4 \, x^{4}} \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^5,x, algorithm="maxima")

[Out]

1/4*sqrt(e*x^2 + d)*e^(3/2)/(d^2*x) - 1/12*(e*x^2 + d)^(3/2)*sqrt(e)/(d^2*x^3) - 1/4*arctanh(sqrt(e)*x/sqrt(e*
x^2 + d))/x^4

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Fricas [A]  time = 2.24824, size = 162, normalized size = 2.05 \begin{align*} -\frac{3 \, d^{2} \log \left (\frac{2 \, e x^{2} + 2 \, \sqrt{e x^{2} + d} \sqrt{e} x + d}{d}\right ) - 2 \,{\left (2 \, e x^{3} - d x\right )} \sqrt{e x^{2} + d} \sqrt{e}}{24 \, d^{2} x^{4}} \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^5,x, algorithm="fricas")

[Out]

-1/24*(3*d^2*log((2*e*x^2 + 2*sqrt(e*x^2 + d)*sqrt(e)*x + d)/d) - 2*(2*e*x^3 - d*x)*sqrt(e*x^2 + d)*sqrt(e))/(
d^2*x^4)

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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^{5}}\, 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**5,x)

[Out]

Integral(atanh(sqrt(e)*x/sqrt(d + e*x**2))/x**5, x)

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Giac [A]  time = 1.40209, size = 144, normalized size = 1.82 \begin{align*} \frac{{\left (3 \,{\left (x e^{\frac{1}{2}} - \sqrt{x^{2} e + d}\right )}^{2} d e - d^{2} e\right )} e}{3 \,{\left ({\left (x e^{\frac{1}{2}} - \sqrt{x^{2} e + d}\right )}^{2} - d\right )}^{3} d} - \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 )}{8 \, x^{4}} \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^5,x, algorithm="giac")

[Out]

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