3.15 \(\int \frac {\tanh ^{-1}(\frac {\sqrt {e} x}{\sqrt {d+e x^2}})}{x^6} \, dx\)

Optimal. Leaf size=111 \[ -\frac {3 e^{5/2} \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{40 d^{5/2}}+\frac {3 e^{3/2} \sqrt {d+e x^2}}{40 d^2 x^2}-\frac {\tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{5 x^5}-\frac {\sqrt {e} \sqrt {d+e x^2}}{20 d x^4} \]

[Out]

-1/5*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))/x^5-3/40*e^(5/2)*arctanh((e*x^2+d)^(1/2)/d^(1/2))/d^(5/2)+3/40*e^(3/2)
*(e*x^2+d)^(1/2)/d^2/x^2-1/20*e^(1/2)*(e*x^2+d)^(1/2)/d/x^4

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Rubi [A]  time = 0.06, antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 6, 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} \[ \frac {3 e^{3/2} \sqrt {d+e x^2}}{40 d^2 x^2}-\frac {3 e^{5/2} \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{40 d^{5/2}}-\frac {\sqrt {e} \sqrt {d+e x^2}}{20 d x^4}-\frac {\tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{5 x^5} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]]/x^6,x]

[Out]

-(Sqrt[e]*Sqrt[d + e*x^2])/(20*d*x^4) + (3*e^(3/2)*Sqrt[d + e*x^2])/(40*d^2*x^2) - ArcTanh[(Sqrt[e]*x)/Sqrt[d
+ e*x^2]]/(5*x^5) - (3*e^(5/2)*ArcTanh[Sqrt[d + e*x^2]/Sqrt[d]])/(40*d^(5/2))

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^6} \, dx &=-\frac {\tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{5 x^5}+\frac {1}{5} \sqrt {e} \int \frac {1}{x^5 \sqrt {d+e x^2}} \, dx\\ &=-\frac {\tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{5 x^5}+\frac {1}{10} \sqrt {e} \operatorname {Subst}\left (\int \frac {1}{x^3 \sqrt {d+e x}} \, dx,x,x^2\right )\\ &=-\frac {\sqrt {e} \sqrt {d+e x^2}}{20 d x^4}-\frac {\tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{5 x^5}-\frac {\left (3 e^{3/2}\right ) \operatorname {Subst}\left (\int \frac {1}{x^2 \sqrt {d+e x}} \, dx,x,x^2\right )}{40 d}\\ &=-\frac {\sqrt {e} \sqrt {d+e x^2}}{20 d x^4}+\frac {3 e^{3/2} \sqrt {d+e x^2}}{40 d^2 x^2}-\frac {\tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{5 x^5}+\frac {\left (3 e^{5/2}\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {d+e x}} \, dx,x,x^2\right )}{80 d^2}\\ &=-\frac {\sqrt {e} \sqrt {d+e x^2}}{20 d x^4}+\frac {3 e^{3/2} \sqrt {d+e x^2}}{40 d^2 x^2}-\frac {\tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{5 x^5}+\frac {\left (3 e^{3/2}\right ) \operatorname {Subst}\left (\int \frac {1}{-\frac {d}{e}+\frac {x^2}{e}} \, dx,x,\sqrt {d+e x^2}\right )}{40 d^2}\\ &=-\frac {\sqrt {e} \sqrt {d+e x^2}}{20 d x^4}+\frac {3 e^{3/2} \sqrt {d+e x^2}}{40 d^2 x^2}-\frac {\tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{5 x^5}-\frac {3 e^{5/2} \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{40 d^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.16, size = 107, normalized size = 0.96 \[ \frac {\frac {\sqrt {e} x \left (-3 e^2 x^4 \log \left (\sqrt {d} \sqrt {d+e x^2}+d\right )+\sqrt {d} \sqrt {d+e x^2} \left (3 e x^2-2 d\right )+3 e^2 x^4 \log (x)\right )}{d^{5/2}}-8 \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{40 x^5} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]]/x^6,x]

[Out]

(-8*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]] + (Sqrt[e]*x*(Sqrt[d]*Sqrt[d + e*x^2]*(-2*d + 3*e*x^2) + 3*e^2*x^4*Lo
g[x] - 3*e^2*x^4*Log[d + Sqrt[d]*Sqrt[d + e*x^2]]))/d^(5/2))/(40*x^5)

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fricas [B]  time = 1.50, size = 383, normalized size = 3.45 \[ \left [\frac {3 \, e^{2} x^{5} \sqrt {\frac {e}{d}} \log \left (-\frac {e^{2} x^{2} - 2 \, \sqrt {e x^{2} + d} d \sqrt {e} \sqrt {\frac {e}{d}} + 2 \, d e}{x^{2}}\right ) - 8 \, d^{2} x^{5} \log \left (\frac {e x + \sqrt {e x^{2} + d} \sqrt {e}}{x}\right ) + 8 \, d^{2} x^{5} \log \left (\frac {e x - \sqrt {e x^{2} + d} \sqrt {e}}{x}\right ) + 2 \, {\left (3 \, e x^{3} - 2 \, d x\right )} \sqrt {e x^{2} + d} \sqrt {e} + 8 \, {\left (d^{2} x^{5} - d^{2}\right )} \log \left (\frac {2 \, e x^{2} + 2 \, \sqrt {e x^{2} + d} \sqrt {e} x + d}{d}\right )}{80 \, d^{2} x^{5}}, \frac {3 \, e^{2} x^{5} \sqrt {-\frac {e}{d}} \arctan \left (\frac {\sqrt {e x^{2} + d} d \sqrt {e} \sqrt {-\frac {e}{d}}}{e^{2} x^{2} + d e}\right ) - 4 \, d^{2} x^{5} \log \left (\frac {e x + \sqrt {e x^{2} + d} \sqrt {e}}{x}\right ) + 4 \, d^{2} x^{5} \log \left (\frac {e x - \sqrt {e x^{2} + d} \sqrt {e}}{x}\right ) + {\left (3 \, e x^{3} - 2 \, d x\right )} \sqrt {e x^{2} + d} \sqrt {e} + 4 \, {\left (d^{2} x^{5} - d^{2}\right )} \log \left (\frac {2 \, e x^{2} + 2 \, \sqrt {e x^{2} + d} \sqrt {e} x + d}{d}\right )}{40 \, d^{2} x^{5}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))/x^6,x, algorithm="fricas")

[Out]

[1/80*(3*e^2*x^5*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) - 8*d^2*x^5*log
((e*x + sqrt(e*x^2 + d)*sqrt(e))/x) + 8*d^2*x^5*log((e*x - sqrt(e*x^2 + d)*sqrt(e))/x) + 2*(3*e*x^3 - 2*d*x)*s
qrt(e*x^2 + d)*sqrt(e) + 8*(d^2*x^5 - d^2)*log((2*e*x^2 + 2*sqrt(e*x^2 + d)*sqrt(e)*x + d)/d))/(d^2*x^5), 1/40
*(3*e^2*x^5*sqrt(-e/d)*arctan(sqrt(e*x^2 + d)*d*sqrt(e)*sqrt(-e/d)/(e^2*x^2 + d*e)) - 4*d^2*x^5*log((e*x + sqr
t(e*x^2 + d)*sqrt(e))/x) + 4*d^2*x^5*log((e*x - sqrt(e*x^2 + d)*sqrt(e))/x) + (3*e*x^3 - 2*d*x)*sqrt(e*x^2 + d
)*sqrt(e) + 4*(d^2*x^5 - d^2)*log((2*e*x^2 + 2*sqrt(e*x^2 + d)*sqrt(e)*x + d)/d))/(d^2*x^5)]

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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^6,x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: 1/2*(-1/5/x^5*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)^3-6*exp(1)^2*exp(1/2)^2+6*exp(
1)*exp(1/2)^4-2*exp(1/2)^6)/5/d^3/2/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))+(-15*exp(1)^2+20*exp(1)*exp(1/2)^2-8*exp(1/2
)^4)/20/d^3/2/sqrt(-d)*atan(sqrt(d+x^2*exp(1))/sqrt(-d))+(-7*sqrt(d+x^2*exp(1))*(d+x^2*exp(1))*exp(1)^2+4*sqrt
(d+x^2*exp(1))*(d+x^2*exp(1))*exp(1)*exp(1/2)^2+9*sqrt(d+x^2*exp(1))*d*exp(1)^2-4*sqrt(d+x^2*exp(1))*d*exp(1)*
exp(1/2)^2)/40/d^3/(d+x^2*exp(1)-d)^2))

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maple [A]  time = 0.03, size = 130, normalized size = 1.17 \[ -\frac {\arctanh \left (\frac {x \sqrt {e}}{\sqrt {e \,x^{2}+d}}\right )}{5 x^{5}}+\frac {e^{\frac {3}{2}} \sqrt {e \,x^{2}+d}}{10 d^{2} x^{2}}-\frac {3 e^{\frac {5}{2}} \ln \left (\frac {2 d +2 \sqrt {d}\, \sqrt {e \,x^{2}+d}}{x}\right )}{40 d^{\frac {5}{2}}}-\frac {\sqrt {e}\, \left (e \,x^{2}+d \right )^{\frac {3}{2}}}{20 d^{2} x^{4}}+\frac {e^{\frac {3}{2}} \left (e \,x^{2}+d \right )^{\frac {3}{2}}}{40 d^{3} x^{2}}-\frac {e^{\frac {5}{2}} \sqrt {e \,x^{2}+d}}{40 d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))/x^6,x)

[Out]

-1/5*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))/x^5+1/10*e^(3/2)*(e*x^2+d)^(1/2)/d^2/x^2-3/40*e^(5/2)/d^(5/2)*ln((2*d+
2*d^(1/2)*(e*x^2+d)^(1/2))/x)-1/20*e^(1/2)/d^2/x^4*(e*x^2+d)^(3/2)+1/40*e^(3/2)/d^3/x^2*(e*x^2+d)^(3/2)-1/40*e
^(5/2)/d^3*(e*x^2+d)^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ d \sqrt {e} \int -\frac {\sqrt {e x^{2} + d}}{5 \, {\left (e^{2} x^{9} + d e x^{7} - {\left (e x^{7} + d x^{5}\right )} {\left (e x^{2} + d\right )}\right )}}\,{d x} - \frac {\log \left (\sqrt {e} x + \sqrt {e x^{2} + d}\right ) - \log \left (-\sqrt {e} x + \sqrt {e x^{2} + d}\right )}{10 \, x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))/x^6,x, algorithm="maxima")

[Out]

d*sqrt(e)*integrate(-1/5*sqrt(e*x^2 + d)/(e^2*x^9 + d*e*x^7 - (e*x^7 + d*x^5)*(e*x^2 + d)), x) - 1/10*(log(sqr
t(e)*x + sqrt(e*x^2 + d)) - log(-sqrt(e)*x + sqrt(e*x^2 + d)))/x^5

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {atanh}\left (\frac {\sqrt {e}\,x}{\sqrt {e\,x^2+d}}\right )}{x^6} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh((e^(1/2)*x)/(d + e*x^2)^(1/2))/x^6,x)

[Out]

int(atanh((e^(1/2)*x)/(d + e*x^2)^(1/2))/x^6, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {atanh}{\left (\frac {\sqrt {e} x}{\sqrt {d + e x^{2}}} \right )}}{x^{6}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(x*e**(1/2)/(e*x**2+d)**(1/2))/x**6,x)

[Out]

Integral(atanh(sqrt(e)*x/sqrt(d + e*x**2))/x**6, x)

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