3.14 \(\int \frac{\tanh ^{-1}(\frac{\sqrt{e} x}{\sqrt{d+e x^2}})}{x^4} \, dx\)

Optimal. Leaf size=85 \[ \frac{e^{3/2} \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{6 d^{3/2}}-\frac{\sqrt{e} \sqrt{d+e x^2}}{6 d x^2}-\frac{\tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{3 x^3} \]

[Out]

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

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Rubi [A]  time = 0.0447361, antiderivative size = 85, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {6221, 266, 51, 63, 208} \[ \frac{e^{3/2} \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{6 d^{3/2}}-\frac{\sqrt{e} \sqrt{d+e x^2}}{6 d x^2}-\frac{\tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]]/x^4,x]

[Out]

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

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 266

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

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

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

Mathematica [A]  time = 0.0881199, size = 92, normalized size = 1.08 \[ -\frac{\frac{\sqrt{e} x \left (\sqrt{d} \sqrt{d+e x^2}-e x^2 \log \left (\sqrt{d} \sqrt{d+e x^2}+d\right )+e x^2 \log (x)\right )}{d^{3/2}}+2 \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{6 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]]/x^4,x]

[Out]

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

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Maple [A]  time = 0.032, size = 90, normalized size = 1.1 \begin{align*} -{\frac{1}{3\,{x}^{3}}{\it Artanh} \left ({x\sqrt{e}{\frac{1}{\sqrt{e{x}^{2}+d}}}} \right ) }+{\frac{1}{6}{e}^{{\frac{3}{2}}}\ln \left ({\frac{1}{x} \left ( 2\,d+2\,\sqrt{d}\sqrt{e{x}^{2}+d} \right ) } \right ){d}^{-{\frac{3}{2}}}}-{\frac{1}{6\,{d}^{2}{x}^{2}}\sqrt{e} \left ( e{x}^{2}+d \right ) ^{{\frac{3}{2}}}}+{\frac{1}{6\,{d}^{2}}{e}^{{\frac{3}{2}}}\sqrt{e{x}^{2}+d}} \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^4,x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} d \sqrt{e} \int -\frac{\sqrt{e x^{2} + d}}{3 \,{\left (e^{2} x^{7} + d e x^{5} -{\left (e x^{5} + d x^{3}\right )}{\left (e x^{2} + d\right )}\right )}}\,{d x} - \frac{\log \left (\sqrt{e} x + \sqrt{e x^{2} + d}\right ) - \log \left (-\sqrt{e} x + \sqrt{e x^{2} + d}\right )}{6 \, x^{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^4,x, algorithm="maxima")

[Out]

d*sqrt(e)*integrate(-1/3*sqrt(e*x^2 + d)/(e^2*x^7 + d*e*x^5 - (e*x^5 + d*x^3)*(e*x^2 + d)), x) - 1/6*(log(sqrt
(e)*x + sqrt(e*x^2 + d)) - log(-sqrt(e)*x + sqrt(e*x^2 + d)))/x^3

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Fricas [B]  time = 2.54526, size = 783, normalized size = 9.21 \begin{align*} \left [\frac{e x^{3} \sqrt{\frac{e}{d}} \log \left (-\frac{e^{2} x^{2} + 2 \, \sqrt{e x^{2} + d} d \sqrt{e} \sqrt{\frac{e}{d}} + 2 \, d e}{x^{2}}\right ) - 2 \, d x^{3} \log \left (\frac{e x + \sqrt{e x^{2} + d} \sqrt{e}}{x}\right ) + 2 \, d x^{3} \log \left (\frac{e x - \sqrt{e x^{2} + d} \sqrt{e}}{x}\right ) - 2 \, \sqrt{e x^{2} + d} \sqrt{e} x + 2 \,{\left (d x^{3} - d\right )} \log \left (\frac{2 \, e x^{2} + 2 \, \sqrt{e x^{2} + d} \sqrt{e} x + d}{d}\right )}{12 \, d x^{3}}, -\frac{e x^{3} \sqrt{-\frac{e}{d}} \arctan \left (\frac{\sqrt{e x^{2} + d} d \sqrt{e} \sqrt{-\frac{e}{d}}}{e^{2} x^{2} + d e}\right ) + d x^{3} \log \left (\frac{e x + \sqrt{e x^{2} + d} \sqrt{e}}{x}\right ) - d x^{3} \log \left (\frac{e x - \sqrt{e x^{2} + d} \sqrt{e}}{x}\right ) + \sqrt{e x^{2} + d} \sqrt{e} x -{\left (d x^{3} - d\right )} \log \left (\frac{2 \, e x^{2} + 2 \, \sqrt{e x^{2} + d} \sqrt{e} x + d}{d}\right )}{6 \, d x^{3}}\right ] \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^4,x, algorithm="fricas")

[Out]

[1/12*(e*x^3*sqrt(e/d)*log(-(e^2*x^2 + 2*sqrt(e*x^2 + d)*d*sqrt(e)*sqrt(e/d) + 2*d*e)/x^2) - 2*d*x^3*log((e*x
+ sqrt(e*x^2 + d)*sqrt(e))/x) + 2*d*x^3*log((e*x - sqrt(e*x^2 + d)*sqrt(e))/x) - 2*sqrt(e*x^2 + d)*sqrt(e)*x +
2*(d*x^3 - d)*log((2*e*x^2 + 2*sqrt(e*x^2 + d)*sqrt(e)*x + d)/d))/(d*x^3), -1/6*(e*x^3*sqrt(-e/d)*arctan(sqrt
(e*x^2 + d)*d*sqrt(e)*sqrt(-e/d)/(e^2*x^2 + d*e)) + d*x^3*log((e*x + sqrt(e*x^2 + d)*sqrt(e))/x) - d*x^3*log((
e*x - sqrt(e*x^2 + d)*sqrt(e))/x) + sqrt(e*x^2 + d)*sqrt(e)*x - (d*x^3 - d)*log((2*e*x^2 + 2*sqrt(e*x^2 + d)*s
qrt(e)*x + d)/d))/(d*x^3)]

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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^{4}}\, 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**4,x)

[Out]

Integral(atanh(sqrt(e)*x/sqrt(d + e*x**2))/x**4, x)

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: NotImplementedError} \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^4,x, algorithm="giac")

[Out]

Exception raised: NotImplementedError