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

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

[Out]

-1/8*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))/x^8+3/140*e^(3/2)*(e*x^2+d)^(1/2)/d^2/x^5-1/35*e^(5/2)*(e*x^2+d)^(1/2)
/d^3/x^3+2/35*e^(7/2)*(e*x^2+d)^(1/2)/d^4/x-1/56*e^(1/2)*(e*x^2+d)^(1/2)/d/x^7

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Rubi [A]  time = 0.05, antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, 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 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]

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 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]

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*}

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Mathematica [A]  time = 0.06, size = 85, normalized size = 0.65 \[ \frac {\sqrt {e} x \sqrt {d+e x^2} \left (-5 d^3+6 d^2 e x^2-8 d e^2 x^4+16 e^3 x^6\right )-35 d^4 \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{280 d^4 x^8} \]

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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fricas [A]  time = 1.13, size = 89, normalized size = 0.68 \[ -\frac {35 \, d^{4} \log \left (\frac {2 \, e x^{2} + 2 \, \sqrt {e x^{2} + d} \sqrt {e} x + d}{d}\right ) - 2 \, {\left (16 \, e^{3} x^{7} - 8 \, d e^{2} x^{5} + 6 \, d^{2} e x^{3} - 5 \, d^{3} x\right )} \sqrt {e x^{2} + d} \sqrt {e}}{560 \, d^{4} x^{8}} \]

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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giac [A]  time = 1.39, size = 161, normalized size = 1.23 \[ -\frac {\log \left (-\frac {\frac {x e^{\frac {1}{2}}}{\sqrt {x^{2} e + d}} + 1}{\frac {x e^{\frac {1}{2}}}{\sqrt {x^{2} e + d}} - 1}\right )}{16 \, x^{8}} + \frac {4 \, {\left (35 \, {\left (x e^{\frac {1}{2}} - \sqrt {x^{2} e + d}\right )}^{6} d^{3} e^{3} - 21 \, {\left (x e^{\frac {1}{2}} - \sqrt {x^{2} e + d}\right )}^{4} d^{4} e^{3} + 7 \, {\left (x e^{\frac {1}{2}} - \sqrt {x^{2} e + d}\right )}^{2} d^{5} e^{3} - d^{6} e^{3}\right )} e}{35 \, {\left ({\left (x e^{\frac {1}{2}} - \sqrt {x^{2} e + d}\right )}^{2} - d\right )}^{7} d^{3}} \]

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)

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maple [A]  time = 0.04, size = 158, normalized size = 1.21 \[ -\frac {\arctanh \left (\frac {x \sqrt {e}}{\sqrt {e \,x^{2}+d}}\right )}{8 x^{8}}-\frac {e^{\frac {3}{2}} \left (-\frac {\sqrt {e \,x^{2}+d}}{5 d \,x^{5}}-\frac {4 e \left (-\frac {\sqrt {e \,x^{2}+d}}{3 d \,x^{3}}+\frac {2 e \sqrt {e \,x^{2}+d}}{3 d^{2} x}\right )}{5 d}\right )}{8 d}+\frac {\sqrt {e}\, \left (-\frac {\left (e \,x^{2}+d \right )^{\frac {3}{2}}}{7 d \,x^{7}}-\frac {4 e \left (-\frac {\left (e \,x^{2}+d \right )^{\frac {3}{2}}}{5 d \,x^{5}}+\frac {2 e \left (e \,x^{2}+d \right )^{\frac {3}{2}}}{15 d^{2} x^{3}}\right )}{7 d}\right )}{8 d} \]

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 = 0.34, size = 125, normalized size = 0.95 \[ \frac {{\left (8 \, e^{3} x^{6} + 4 \, d e^{2} x^{4} - d^{2} e x^{2} + 3 \, d^{3}\right )} e^{\frac {3}{2}}}{120 \, \sqrt {e x^{2} + d} d^{4} x^{5}} - \frac {\operatorname {artanh}\left (\frac {\sqrt {e} x}{\sqrt {e x^{2} + d}}\right )}{8 \, x^{8}} - \frac {{\left (8 \, e^{3} x^{6} - 4 \, d e^{2} x^{4} + 3 \, d^{2} e x^{2} + 15 \, d^{3}\right )} \sqrt {e x^{2} + d} \sqrt {e}}{840 \, d^{4} x^{7}} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {atanh}\left (\frac {\sqrt {e}\,x}{\sqrt {e\,x^2+d}}\right )}{x^9} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh((e^(1/2)*x)/(d + e*x^2)^(1/2))/x^9,x)

[Out]

int(atanh((e^(1/2)*x)/(d + e*x^2)^(1/2))/x^9, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {atanh}{\left (\frac {\sqrt {e} x}{\sqrt {d + e x^{2}}} \right )}}{x^{9}}\, dx \]

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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