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

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

[Out]

-1/4*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))/x^4+1/6*e^(3/2)*(e*x^2+d)^(1/2)/d^2/x-1/12*e^(1/2)*(e*x^2+d)^(1/2)/d/x
^3

________________________________________________________________________________________

Rubi [A]  time = 0.03, antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, 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 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^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 = 0.05, size = 63, normalized size = 0.80 \[ \frac {\sqrt {e} x \sqrt {d+e x^2} \left (2 e x^2-d\right )-3 d^2 \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{12 d^2 x^4} \]

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)

________________________________________________________________________________________

fricas [A]  time = 0.70, size = 67, normalized size = 0.85 \[ -\frac {3 \, d^{2} \log \left (\frac {2 \, e x^{2} + 2 \, \sqrt {e x^{2} + d} \sqrt {e} x + d}{d}\right ) - 2 \, {\left (2 \, e x^{3} - d x\right )} \sqrt {e x^{2} + d} \sqrt {e}}{24 \, d^{2} x^{4}} \]

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)

________________________________________________________________________________________

giac [A]  time = 0.39, size = 107, normalized size = 1.35 \[ \frac {{\left (3 \, {\left (x e^{\frac {1}{2}} - \sqrt {x^{2} e + d}\right )}^{2} d e - d^{2} e\right )} e}{3 \, {\left ({\left (x e^{\frac {1}{2}} - \sqrt {x^{2} e + d}\right )}^{2} - d\right )}^{3} d} - \frac {\log \left (-\frac {\frac {x e^{\frac {1}{2}}}{\sqrt {x^{2} e + d}} + 1}{\frac {x e^{\frac {1}{2}}}{\sqrt {x^{2} e + d}} - 1}\right )}{8 \, x^{4}} \]

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

________________________________________________________________________________________

maple [A]  time = 0.03, size = 62, normalized size = 0.78 \[ -\frac {\arctanh \left (\frac {x \sqrt {e}}{\sqrt {e \,x^{2}+d}}\right )}{4 x^{4}}+\frac {e^{\frac {3}{2}} \sqrt {e \,x^{2}+d}}{4 d^{2} x}-\frac {\sqrt {e}\, \left (e \,x^{2}+d \right )^{\frac {3}{2}}}{12 d^{2} x^{3}} \]

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)

________________________________________________________________________________________

maxima [A]  time = 0.35, size = 61, normalized size = 0.77 \[ \frac {\sqrt {e x^{2} + d} e^{\frac {3}{2}}}{4 \, d^{2} x} - \frac {{\left (e x^{2} + d\right )}^{\frac {3}{2}} \sqrt {e}}{12 \, d^{2} x^{3}} - \frac {\operatorname {artanh}\left (\frac {\sqrt {e} x}{\sqrt {e x^{2} + d}}\right )}{4 \, x^{4}} \]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {atanh}\left (\frac {\sqrt {e}\,x}{\sqrt {e\,x^2+d}}\right )}{x^5} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh((e^(1/2)*x)/(d + e*x^2)^(1/2))/x^5,x)

[Out]

int(atanh((e^(1/2)*x)/(d + e*x^2)^(1/2))/x^5, x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {atanh}{\left (\frac {\sqrt {e} x}{\sqrt {d + e x^{2}}} \right )}}{x^{5}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(x*e**(1/2)/(e*x**2+d)**(1/2))/x**5,x)

[Out]

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

________________________________________________________________________________________