3.1.7 \(\int \frac {\text {ArcTan}(\frac {\sqrt {-e} x}{\sqrt {d+e x^2}})}{x^3} \, dx\) [7]

Optimal. Leaf size=57 \[ -\frac {\sqrt {-e} \sqrt {d+e x^2}}{2 d x}-\frac {\text {ArcTan}\left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{2 x^2} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {5259, 270} \begin {gather*} -\frac {\text {ArcTan}\left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{2 x^2}-\frac {\sqrt {-e} \sqrt {d+e x^2}}{2 d x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[ArcTan[(Sqrt[-e]*x)/Sqrt[d + e*x^2]]/x^3,x]

[Out]

-1/2*(Sqrt[-e]*Sqrt[d + e*x^2])/(d*x) - ArcTan[(Sqrt[-e]*x)/Sqrt[d + e*x^2]]/(2*x^2)

Rule 270

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 5259

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^3} \, dx &=-\frac {\tan ^{-1}\left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{2 x^2}+\frac {1}{2} \sqrt {-e} \int \frac {1}{x^2 \sqrt {d+e x^2}} \, dx\\ &=-\frac {\sqrt {-e} \sqrt {d+e x^2}}{2 d x}-\frac {\tan ^{-1}\left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{2 x^2}\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 54, normalized size = 0.95 \begin {gather*} -\frac {\sqrt {-e} x \sqrt {d+e x^2}+d \text {ArcTan}\left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{2 d x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[ArcTan[(Sqrt[-e]*x)/Sqrt[d + e*x^2]]/x^3,x]

[Out]

-1/2*(Sqrt[-e]*x*Sqrt[d + e*x^2] + d*ArcTan[(Sqrt[-e]*x)/Sqrt[d + e*x^2]])/(d*x^2)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(121\) vs. \(2(45)=90\).
time = 0.01, size = 122, normalized size = 2.14

method result size
default \(-\frac {\arctan \left (\frac {x \sqrt {-e}}{\sqrt {e \,x^{2}+d}}\right )}{2 x^{2}}-\frac {\sqrt {-e}\, \sqrt {e}\, \ln \left (x \sqrt {e}+\sqrt {e \,x^{2}+d}\right )}{2 d}+\frac {\sqrt {-e}\, \left (-\frac {\left (e \,x^{2}+d \right )^{\frac {3}{2}}}{d x}+\frac {2 e \left (\frac {x \sqrt {e \,x^{2}+d}}{2}+\frac {d \ln \left (x \sqrt {e}+\sqrt {e \,x^{2}+d}\right )}{2 \sqrt {e}}\right )}{d}\right )}{2 d}\) \(122\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctan(x*(-e)^(1/2)/(e*x^2+d)^(1/2))/x^3,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.28, size = 64, normalized size = 1.12 \begin {gather*} -\frac {\arctan \left (\frac {x \sqrt {-e}}{\sqrt {x^{2} e + d}}\right )}{2 \, x^{2}} - \frac {x^{2} \sqrt {-e} e + d \sqrt {-e}}{2 \, \sqrt {x^{2} e + d} d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(x*(-e)^(1/2)/(e*x^2+d)^(1/2))/x^3,x, algorithm="maxima")

[Out]

-1/2*arctan(x*sqrt(-e)/sqrt(x^2*e + d))/x^2 - 1/2*(x^2*sqrt(-e)*e + d*sqrt(-e))/(sqrt(x^2*e + d)*d*x)

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Fricas [C] Result contains complex when optimal does not.
time = 6.19, size = 56, normalized size = 0.98 \begin {gather*} \frac {-2 i \, \sqrt {x^{2} e + d} x e^{\frac {1}{2}} - i \, d \log \left (\frac {2 \, x^{2} e + 2 \, \sqrt {x^{2} e + d} x e^{\frac {1}{2}} + d}{d}\right )}{4 \, d x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(x*(-e)^(1/2)/(e*x^2+d)^(1/2))/x^3,x, algorithm="fricas")

[Out]

1/4*(-2*I*sqrt(x^2*e + d)*x*e^(1/2) - I*d*log((2*x^2*e + 2*sqrt(x^2*e + d)*x*e^(1/2) + d)/d))/(d*x^2)

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Sympy [A]
time = 2.02, size = 53, normalized size = 0.93 \begin {gather*} - \frac {\operatorname {atan}{\left (\frac {x \sqrt {- e}}{\sqrt {d + e x^{2}}} \right )}}{2 x^{2}} - \frac {\sqrt {e} \sqrt {- e} \sqrt {\frac {d}{e x^{2}} + 1}}{2 d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atan(x*(-e)**(1/2)/(e*x**2+d)**(1/2))/x**3,x)

[Out]

-atan(x*sqrt(-e)/sqrt(d + e*x**2))/(2*x**2) - sqrt(e)*sqrt(-e)*sqrt(d/(e*x**2) + 1)/(2*d)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 106 vs. \(2 (45) = 90\).
time = 0.49, size = 106, normalized size = 1.86 \begin {gather*} -\frac {e^{4} x}{4 \, {\left (\sqrt {-d e} e - \sqrt {-e^{2} x^{2} - d e} {\left | e \right |}\right )} d {\left | e \right |}} - \frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {e x^{2} + d}}\right )}{2 \, x^{2}} + \frac {\sqrt {-d e} e - \sqrt {-e^{2} x^{2} - d e} {\left | e \right |}}{4 \, d x {\left | e \right |}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(x*(-e)^(1/2)/(e*x^2+d)^(1/2))/x^3,x, algorithm="giac")

[Out]

-1/4*e^4*x/((sqrt(-d*e)*e - sqrt(-e^2*x^2 - d*e)*abs(e))*d*abs(e)) - 1/2*arctan(sqrt(-e)*x/sqrt(e*x^2 + d))/x^
2 + 1/4*(sqrt(-d*e)*e - sqrt(-e^2*x^2 - d*e)*abs(e))/(d*x*abs(e))

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {\mathrm {atan}\left (\frac {\sqrt {-e}\,x}{\sqrt {e\,x^2+d}}\right )}{x^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atan(((-e)^(1/2)*x)/(d + e*x^2)^(1/2))/x^3,x)

[Out]

int(atan(((-e)^(1/2)*x)/(d + e*x^2)^(1/2))/x^3, x)

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