3.7 \(\int \frac{\tanh ^{-1}(\frac{\sqrt{e} x}{\sqrt{d+e x^2}})}{x^7} \, dx\)

Optimal. Leaf size=105 \[ -\frac{4 e^{5/2} \sqrt{d+e x^2}}{45 d^3 x}+\frac{2 e^{3/2} \sqrt{d+e x^2}}{45 d^2 x^3}-\frac{\sqrt{e} \sqrt{d+e x^2}}{30 d x^5}-\frac{\tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{6 x^6} \]

[Out]

-(Sqrt[e]*Sqrt[d + e*x^2])/(30*d*x^5) + (2*e^(3/2)*Sqrt[d + e*x^2])/(45*d^2*x^3) - (4*e^(5/2)*Sqrt[d + e*x^2])
/(45*d^3*x) - ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]]/(6*x^6)

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Rubi [A]  time = 0.0369368, antiderivative size = 105, 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, 271, 264} \[ -\frac{4 e^{5/2} \sqrt{d+e x^2}}{45 d^3 x}+\frac{2 e^{3/2} \sqrt{d+e x^2}}{45 d^2 x^3}-\frac{\sqrt{e} \sqrt{d+e x^2}}{30 d x^5}-\frac{\tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{6 x^6} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]]/x^7,x]

[Out]

-(Sqrt[e]*Sqrt[d + e*x^2])/(30*d*x^5) + (2*e^(3/2)*Sqrt[d + e*x^2])/(45*d^2*x^3) - (4*e^(5/2)*Sqrt[d + e*x^2])
/(45*d^3*x) - ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]]/(6*x^6)

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^7} \, dx &=-\frac{\tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{6 x^6}+\frac{1}{6} \sqrt{e} \int \frac{1}{x^6 \sqrt{d+e x^2}} \, dx\\ &=-\frac{\sqrt{e} \sqrt{d+e x^2}}{30 d x^5}-\frac{\tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{6 x^6}-\frac{\left (2 e^{3/2}\right ) \int \frac{1}{x^4 \sqrt{d+e x^2}} \, dx}{15 d}\\ &=-\frac{\sqrt{e} \sqrt{d+e x^2}}{30 d x^5}+\frac{2 e^{3/2} \sqrt{d+e x^2}}{45 d^2 x^3}-\frac{\tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{6 x^6}+\frac{\left (4 e^{5/2}\right ) \int \frac{1}{x^2 \sqrt{d+e x^2}} \, dx}{45 d^2}\\ &=-\frac{\sqrt{e} \sqrt{d+e x^2}}{30 d x^5}+\frac{2 e^{3/2} \sqrt{d+e x^2}}{45 d^2 x^3}-\frac{4 e^{5/2} \sqrt{d+e x^2}}{45 d^3 x}-\frac{\tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{6 x^6}\\ \end{align*}

Mathematica [A]  time = 0.0482387, size = 74, normalized size = 0.7 \[ \frac{\sqrt{e} x \sqrt{d+e x^2} \left (-3 d^2+4 d e x^2-8 e^2 x^4\right )-15 d^3 \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{90 d^3 x^6} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]]/x^7,x]

[Out]

(Sqrt[e]*x*Sqrt[d + e*x^2]*(-3*d^2 + 4*d*e*x^2 - 8*e^2*x^4) - 15*d^3*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/(90
*d^3*x^6)

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Maple [A]  time = 0.032, size = 110, normalized size = 1.1 \begin{align*} -{\frac{1}{6\,{x}^{6}}{\it Artanh} \left ({x\sqrt{e}{\frac{1}{\sqrt{e{x}^{2}+d}}}} \right ) }-{\frac{1}{6\,d}{e}^{{\frac{3}{2}}} \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 ) }+{\frac{1}{6\,d}\sqrt{e} \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 ) } \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^7,x)

[Out]

-1/6*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))/x^6-1/6*e^(3/2)/d*(-1/3/d/x^3*(e*x^2+d)^(1/2)+2/3*e/d^2/x*(e*x^2+d)^(1
/2))+1/6*e^(1/2)/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 = 0.976213, size = 138, normalized size = 1.31 \begin{align*} -\frac{{\left (2 \, e^{2} x^{4} + d e x^{2} - d^{2}\right )} e^{\frac{3}{2}}}{18 \, \sqrt{e x^{2} + d} d^{3} x^{3}} - \frac{\operatorname{artanh}\left (\frac{\sqrt{e} x}{\sqrt{e x^{2} + d}}\right )}{6 \, x^{6}} + \frac{{\left (2 \, e^{2} x^{4} - d e x^{2} - 3 \, d^{2}\right )} \sqrt{e x^{2} + d} \sqrt{e}}{90 \, d^{3} 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^7,x, algorithm="maxima")

[Out]

-1/18*(2*e^2*x^4 + d*e*x^2 - d^2)*e^(3/2)/(sqrt(e*x^2 + d)*d^3*x^3) - 1/6*arctanh(sqrt(e)*x/sqrt(e*x^2 + d))/x
^6 + 1/90*(2*e^2*x^4 - d*e*x^2 - 3*d^2)*sqrt(e*x^2 + d)*sqrt(e)/(d^3*x^5)

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Fricas [A]  time = 2.33867, size = 189, normalized size = 1.8 \begin{align*} -\frac{15 \, d^{3} \log \left (\frac{2 \, e x^{2} + 2 \, \sqrt{e x^{2} + d} \sqrt{e} x + d}{d}\right ) + 2 \,{\left (8 \, e^{2} x^{5} - 4 \, d e x^{3} + 3 \, d^{2} x\right )} \sqrt{e x^{2} + d} \sqrt{e}}{180 \, d^{3} x^{6}} \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^7,x, algorithm="fricas")

[Out]

-1/180*(15*d^3*log((2*e*x^2 + 2*sqrt(e*x^2 + d)*sqrt(e)*x + d)/d) + 2*(8*e^2*x^5 - 4*d*e*x^3 + 3*d^2*x)*sqrt(e
*x^2 + d)*sqrt(e))/(d^3*x^6)

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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^{7}}\, 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**7,x)

[Out]

Integral(atanh(sqrt(e)*x/sqrt(d + e*x**2))/x**7, x)

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Giac [A]  time = 1.42731, size = 181, normalized size = 1.72 \begin{align*} \frac{8 \,{\left (10 \,{\left (x e^{\frac{1}{2}} - \sqrt{x^{2} e + d}\right )}^{4} d^{2} e^{2} - 5 \,{\left (x e^{\frac{1}{2}} - \sqrt{x^{2} e + d}\right )}^{2} d^{3} e^{2} + d^{4} e^{2}\right )} e}{45 \,{\left ({\left (x e^{\frac{1}{2}} - \sqrt{x^{2} e + d}\right )}^{2} - d\right )}^{5} d^{2}} - \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 )}{12 \, x^{6}} \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^7,x, algorithm="giac")

[Out]

8/45*(10*(x*e^(1/2) - sqrt(x^2*e + d))^4*d^2*e^2 - 5*(x*e^(1/2) - sqrt(x^2*e + d))^2*d^3*e^2 + d^4*e^2)*e/(((x
*e^(1/2) - sqrt(x^2*e + d))^2 - d)^5*d^2) - 1/12*log(-(x*e^(1/2)/sqrt(x^2*e + d) + 1)/(x*e^(1/2)/sqrt(x^2*e +
d) - 1))/x^6