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]

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

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Rubi [A]  time = 0.04, antiderivative size = 85, normalized size of antiderivative = 1.00, 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 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]

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 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^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*}

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Mathematica [A]  time = 0.11, 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]

-1/6*(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))/x^3

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fricas [B]  time = 0.64, size = 340, normalized size = 4.00 \[ \left [\frac {e x^{3} \sqrt {\frac {e}{d}} \log \left (-\frac {e^{2} x^{2} + 2 \, \sqrt {e x^{2} + d} d \sqrt {e} \sqrt {\frac {e}{d}} + 2 \, d e}{x^{2}}\right ) - 2 \, d x^{3} \log \left (\frac {e x + \sqrt {e x^{2} + d} \sqrt {e}}{x}\right ) + 2 \, d x^{3} \log \left (\frac {e x - \sqrt {e x^{2} + d} \sqrt {e}}{x}\right ) - 2 \, \sqrt {e x^{2} + d} \sqrt {e} x + 2 \, {\left (d x^{3} - d\right )} \log \left (\frac {2 \, e x^{2} + 2 \, \sqrt {e x^{2} + d} \sqrt {e} x + d}{d}\right )}{12 \, d x^{3}}, -\frac {e x^{3} \sqrt {-\frac {e}{d}} \arctan \left (\frac {\sqrt {e x^{2} + d} d \sqrt {e} \sqrt {-\frac {e}{d}}}{e^{2} x^{2} + d e}\right ) + d x^{3} \log \left (\frac {e x + \sqrt {e x^{2} + d} \sqrt {e}}{x}\right ) - d x^{3} \log \left (\frac {e x - \sqrt {e x^{2} + d} \sqrt {e}}{x}\right ) + \sqrt {e x^{2} + d} \sqrt {e} x - {\left (d x^{3} - d\right )} \log \left (\frac {2 \, e x^{2} + 2 \, \sqrt {e x^{2} + d} \sqrt {e} x + d}{d}\right )}{6 \, d x^{3}}\right ] \]

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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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

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 >> Unable to parse Giac output: 1/2*(-1/3/x^3*ln((1+(sqrt(exp(1)*x^2+d))
^-1*exp(1/2)*x)/(1-(sqrt(exp(1)*x^2+d))^-1*exp(1/2)*x))-2*d*exp(1/2)*(-(2*exp(1)^2-4*exp(1)*exp(1/2)^2+2*exp(1
/2)^4)/3/d^2/2/sqrt(-d*exp(1/2)^2+d*exp(1))/exp(1/2)*atan((sqrt(d+x^2*exp(1))*exp(1)-sqrt(d+x^2*exp(1))*exp(1/
2)^2)/sqrt(-d*exp(1/2)^2+d*exp(1))/exp(1/2))+(3*exp(1)-2*exp(1/2)^2)/3/d^2/2/sqrt(-d)*atan(sqrt(d+x^2*exp(1))/
sqrt(-d))+sqrt(d+x^2*exp(1))*exp(1)/6/d^2/(d+x^2*exp(1)-d)))

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maple [A]  time = 0.03, size = 90, normalized size = 1.06 \[ -\frac {\arctanh \left (\frac {x \sqrt {e}}{\sqrt {e \,x^{2}+d}}\right )}{3 x^{3}}+\frac {e^{\frac {3}{2}} \ln \left (\frac {2 d +2 \sqrt {d}\, \sqrt {e \,x^{2}+d}}{x}\right )}{6 d^{\frac {3}{2}}}-\frac {\sqrt {e}\, \left (e \,x^{2}+d \right )^{\frac {3}{2}}}{6 d^{2} x^{2}}+\frac {e^{\frac {3}{2}} \sqrt {e \,x^{2}+d}}{6 d^{2}} \]

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 = 0.00, size = 0, normalized size = 0.00 \[ d \sqrt {e} \int -\frac {\sqrt {e x^{2} + d}}{3 \, {\left (e^{2} x^{7} + d e x^{5} - {\left (e x^{5} + d x^{3}\right )} {\left (e x^{2} + d\right )}\right )}}\,{d x} - \frac {\log \left (\sqrt {e} x + \sqrt {e x^{2} + d}\right ) - \log \left (-\sqrt {e} x + \sqrt {e x^{2} + d}\right )}{6 \, x^{3}} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {atanh}\left (\frac {\sqrt {e}\,x}{\sqrt {e\,x^2+d}}\right )}{x^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh((e^(1/2)*x)/(d + e*x^2)^(1/2))/x^4,x)

[Out]

int(atanh((e^(1/2)*x)/(d + e*x^2)^(1/2))/x^4, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {atanh}{\left (\frac {\sqrt {e} x}{\sqrt {d + e x^{2}}} \right )}}{x^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(x*e**(1/2)/(e*x**2+d)**(1/2))/x**4,x)

[Out]

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

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