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

Optimal. Leaf size=119 \[ -\frac {3 (-e)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{40 d^{5/2}}-\frac {3 (-e)^{3/2} \sqrt {d+e x^2}}{40 d^2 x^2}-\frac {\tan ^{-1}\left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{5 x^5}-\frac {\sqrt {-e} \sqrt {d+e x^2}}{20 d x^4} \]

[Out]

-1/5*arctan(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 = 119, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {5151, 266, 51, 63, 208} \[ -\frac {3 (-e)^{3/2} \sqrt {d+e x^2}}{40 d^2 x^2}-\frac {3 (-e)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{40 d^{5/2}}-\frac {\sqrt {-e} \sqrt {d+e x^2}}{20 d x^4}-\frac {\tan ^{-1}\left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{5 x^5} \]

Antiderivative was successfully verified.

[In]

Int[ArcTan[(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) - ArcTan[(Sqrt[-e]*x)/Sqr
t[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 5151

Int[ArcTan[((c_.)*(x_))/Sqrt[(a_.) + (b_.)*(x_)^2]]*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*ArcTa
n[(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] /; F
reeQ[{a, b, c, d, m}, x] && EqQ[b + c^2, 0] && NeQ[m, -1]

Rubi steps

\begin {align*} \int \frac {\tan ^{-1}\left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{x^6} \, dx &=-\frac {\tan ^{-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 {\tan ^{-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 {\tan ^{-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 {\tan ^{-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 {\tan ^{-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 {\tan ^{-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 = 0.13, size = 114, normalized size = 0.96 \[ -\frac {3 e^{5/2} \tan ^{-1}\left (\frac {\sqrt {d} \sqrt {-e}}{\sqrt {e} \sqrt {d+e x^2}}\right )}{40 d^{5/2}}+\sqrt {-e} \left (\frac {3 e}{40 d^2 x^2}-\frac {1}{20 d x^4}\right ) \sqrt {d+e x^2}-\frac {\tan ^{-1}\left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{5 x^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Sqrt[-e]*(-1/20*1/(d*x^4) + (3*e)/(40*d^2*x^2))*Sqrt[d + e*x^2] - (3*e^(5/2)*ArcTan[(Sqrt[d]*Sqrt[-e])/(Sqrt[e
]*Sqrt[d + e*x^2])])/(40*d^(5/2)) - ArcTan[(Sqrt[-e]*x)/Sqrt[d + e*x^2]]/(5*x^5)

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fricas [A]  time = 0.63, size = 228, normalized size = 1.92 \[ \left [\frac {3 \, e^{2} x^{5} \sqrt {-\frac {e}{d}} \log \left (-\frac {e^{2} x^{2} + 2 \, \sqrt {e x^{2} + d} d \sqrt {-e} \sqrt {-\frac {e}{d}} + 2 \, d e}{x^{2}}\right ) - 16 \, d^{2} \arctan \left (\frac {\sqrt {-e} x}{\sqrt {e x^{2} + d}}\right ) + 2 \, {\left (3 \, e x^{3} - 2 \, d x\right )} \sqrt {e x^{2} + d} \sqrt {-e}}{80 \, d^{2} x^{5}}, -\frac {3 \, e^{2} x^{5} \sqrt {\frac {e}{d}} \arctan \left (\frac {\sqrt {e x^{2} + d} d \sqrt {-e} \sqrt {\frac {e}{d}}}{e^{2} x^{2} + d e}\right ) + 8 \, d^{2} \arctan \left (\frac {\sqrt {-e} x}{\sqrt {e x^{2} + d}}\right ) - {\left (3 \, e x^{3} - 2 \, d x\right )} \sqrt {e x^{2} + d} \sqrt {-e}}{40 \, d^{2} x^{5}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(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) - 16*d^2*arc
tan(sqrt(-e)*x/sqrt(e*x^2 + d)) + 2*(3*e*x^3 - 2*d*x)*sqrt(e*x^2 + d)*sqrt(-e))/(d^2*x^5), -1/40*(3*e^2*x^5*sq
rt(e/d)*arctan(sqrt(e*x^2 + d)*d*sqrt(-e)*sqrt(e/d)/(e^2*x^2 + d*e)) + 8*d^2*arctan(sqrt(-e)*x/sqrt(e*x^2 + d)
) - (3*e*x^3 - 2*d*x)*sqrt(e*x^2 + d)*sqrt(-e))/(d^2*x^5)]

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giac [A]  time = 0.25, size = 108, normalized size = 0.91 \[ -\frac {1}{40} \, {\left (\frac {3 \, \arctan \left (\frac {\sqrt {-x^{2} e^{2} - d e} e^{\left (-\frac {1}{2}\right )}}{\sqrt {d}}\right ) e^{\frac {7}{2}}}{d^{\frac {5}{2}}} + \frac {{\left (5 \, \sqrt {-x^{2} e^{2} - d e} d e^{5} + 3 \, {\left (-x^{2} e^{2} - d e\right )}^{\frac {3}{2}} e^{4}\right )} e^{\left (-4\right )}}{d^{2} x^{4}}\right )} e^{\left (-1\right )} - \frac {\arctan \left (\frac {x \sqrt {-e}}{\sqrt {x^{2} e + d}}\right )}{5 \, x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(x*(-e)^(1/2)/(e*x^2+d)^(1/2))/x^6,x, algorithm="giac")

[Out]

-1/40*(3*arctan(sqrt(-x^2*e^2 - d*e)*e^(-1/2)/sqrt(d))*e^(7/2)/d^(5/2) + (5*sqrt(-x^2*e^2 - d*e)*d*e^5 + 3*(-x
^2*e^2 - d*e)^(3/2)*e^4)*e^(-4)/(d^2*x^4))*e^(-1) - 1/5*arctan(x*sqrt(-e)/sqrt(x^2*e + d))/x^5

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maple [A]  time = 0.04, size = 150, normalized size = 1.26 \[ -\frac {\arctan \left (\frac {x \sqrt {-e}}{\sqrt {e \,x^{2}+d}}\right )}{5 x^{5}}-\frac {\sqrt {-e}\, \left (e \,x^{2}+d \right )^{\frac {3}{2}}}{20 d^{2} x^{4}}+\frac {\sqrt {-e}\, e \left (e \,x^{2}+d \right )^{\frac {3}{2}}}{40 d^{3} x^{2}}-\frac {3 \sqrt {-e}\, e^{2} \ln \left (\frac {2 d +2 \sqrt {d}\, \sqrt {e \,x^{2}+d}}{x}\right )}{40 d^{\frac {5}{2}}}-\frac {\sqrt {-e}\, e^{2} \sqrt {e \,x^{2}+d}}{40 d^{3}}+\frac {\sqrt {-e}\, e \sqrt {e \,x^{2}+d}}{10 d^{2} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: sign: argument cannot be imaginary
; found sqrt(-_SAGE_VAR_e)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 19.73, size = 148, normalized size = 1.24 \[ - \frac {\operatorname {atan}{\left (\frac {x \sqrt {- e}}{\sqrt {d + e x^{2}}} \right )}}{5 x^{5}} - \frac {\sqrt {- e}}{20 \sqrt {e} x^{5} \sqrt {\frac {d}{e x^{2}} + 1}} + \frac {\sqrt {e} \sqrt {- e}}{40 d x^{3} \sqrt {\frac {d}{e x^{2}} + 1}} + \frac {3 e^{\frac {3}{2}} \sqrt {- e}}{40 d^{2} x \sqrt {\frac {d}{e x^{2}} + 1}} - \frac {3 e^{2} \sqrt {- e} \operatorname {asinh}{\left (\frac {\sqrt {d}}{\sqrt {e} x} \right )}}{40 d^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-atan(x*sqrt(-e)/sqrt(d + e*x**2))/(5*x**5) - sqrt(-e)/(20*sqrt(e)*x**5*sqrt(d/(e*x**2) + 1)) + sqrt(e)*sqrt(-
e)/(40*d*x**3*sqrt(d/(e*x**2) + 1)) + 3*e**(3/2)*sqrt(-e)/(40*d**2*x*sqrt(d/(e*x**2) + 1)) - 3*e**2*sqrt(-e)*a
sinh(sqrt(d)/(sqrt(e)*x))/(40*d**(5/2))

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