3.143 \(\int \frac {1}{\sqrt {a+b x^2} \tan ^{-1}(\frac {e x}{\sqrt {-\frac {a e^2}{b}-e^2 x^2}})} \, dx\)

Optimal. Leaf size=64 \[ \frac {\sqrt {e^2 \left (-x^2\right )-\frac {a e^2}{b}} \log \left (\tan ^{-1}\left (\frac {e x}{\sqrt {e^2 \left (-x^2\right )-\frac {a e^2}{b}}}\right )\right )}{e \sqrt {a+b x^2}} \]

[Out]

ln(arctan(e*x/(-a*e^2/b-e^2*x^2)^(1/2)))*(-a*e^2/b-e^2*x^2)^(1/2)/e/(b*x^2+a)^(1/2)

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Rubi [A]  time = 0.11, antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {5157, 5153} \[ \frac {\sqrt {e^2 \left (-x^2\right )-\frac {a e^2}{b}} \log \left (\tan ^{-1}\left (\frac {e x}{\sqrt {e^2 \left (-x^2\right )-\frac {a e^2}{b}}}\right )\right )}{e \sqrt {a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[a + b*x^2]*ArcTan[(e*x)/Sqrt[-((a*e^2)/b) - e^2*x^2]]),x]

[Out]

(Sqrt[-((a*e^2)/b) - e^2*x^2]*Log[ArcTan[(e*x)/Sqrt[-((a*e^2)/b) - e^2*x^2]]])/(e*Sqrt[a + b*x^2])

Rule 5153

Int[1/(ArcTan[((c_.)*(x_))/Sqrt[(a_.) + (b_.)*(x_)^2]]*Sqrt[(a_.) + (b_.)*(x_)^2]), x_Symbol] :> Simp[(1*Log[A
rcTan[(c*x)/Sqrt[a + b*x^2]]])/c, x] /; FreeQ[{a, b, c}, x] && EqQ[b + c^2, 0]

Rule 5157

Int[ArcTan[((c_.)*(x_))/Sqrt[(a_.) + (b_.)*(x_)^2]]^(m_.)/Sqrt[(d_.) + (e_.)*(x_)^2], x_Symbol] :> Dist[Sqrt[a
+ b*x^2]/Sqrt[d + e*x^2], Int[ArcTan[(c*x)/Sqrt[a + b*x^2]]^m/Sqrt[a + b*x^2], x], x] /; FreeQ[{a, b, c, d, e
, m}, x] && EqQ[b + c^2, 0] && EqQ[b*d - a*e, 0]

Rubi steps

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

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Mathematica [A]  time = 0.11, size = 58, normalized size = 0.91 \[ \frac {\sqrt {-\frac {e^2 \left (a+b x^2\right )}{b}} \log \left (\tan ^{-1}\left (\frac {e x}{\sqrt {-\frac {e^2 \left (a+b x^2\right )}{b}}}\right )\right )}{e \sqrt {a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[a + b*x^2]*ArcTan[(e*x)/Sqrt[-((a*e^2)/b) - e^2*x^2]]),x]

[Out]

(Sqrt[-((e^2*(a + b*x^2))/b)]*Log[ArcTan[(e*x)/Sqrt[-((e^2*(a + b*x^2))/b)]]])/(e*Sqrt[a + b*x^2])

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fricas [A]  time = 0.67, size = 83, normalized size = 1.30 \[ \frac {\sqrt {b x^{2} + a} \sqrt {-\frac {b e^{2} x^{2} + a e^{2}}{b}} \log \left (2 \, \arctan \left (\frac {b x \sqrt {-\frac {b e^{2} x^{2} + a e^{2}}{b}}}{b e x^{2} + a e}\right )\right )}{b e x^{2} + a e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(b*x^2 + a)*sqrt(-(b*e^2*x^2 + a*e^2)/b)*log(2*arctan(b*x*sqrt(-(b*e^2*x^2 + a*e^2)/b)/(b*e*x^2 + a*e)))/(
b*e*x^2 + a*e)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {b x^{2} + a} \arctan \left (\frac {e x}{\sqrt {-e^{2} x^{2} - \frac {a e^{2}}{b}}}\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/(sqrt(b*x^2 + a)*arctan(e*x/sqrt(-e^2*x^2 - a*e^2/b))), x)

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maple [F]  time = 0.45, size = 0, normalized size = 0.00 \[ \int \frac {1}{\arctan \left (\frac {e x}{\sqrt {-\frac {a \,e^{2}}{b}-e^{2} x^{2}}}\right ) \sqrt {b \,x^{2}+a}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/arctan(e*x/(-a*e^2/b-e^2*x^2)^(1/2))/(b*x^2+a)^(1/2),x)

[Out]

int(1/arctan(e*x/(-a*e^2/b-e^2*x^2)^(1/2))/(b*x^2+a)^(1/2),x)

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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(1/arctan(e*x/(-a*e^2/b-e^2*x^2)^(1/2))/(b*x^2+a)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: sign: argument cannot be imaginary
; found  sqrt((-_SAGE_VAR_b*_SAGE_VAR_x^2)-_SAGE_VAR_a)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {1}{\mathrm {atan}\left (\frac {e\,x}{\sqrt {-e^2\,x^2-\frac {a\,e^2}{b}}}\right )\,\sqrt {b\,x^2+a}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(atan((e*x)/(- e^2*x^2 - (a*e^2)/b)^(1/2))*(a + b*x^2)^(1/2)),x)

[Out]

int(1/(atan((e*x)/(- e^2*x^2 - (a*e^2)/b)^(1/2))*(a + b*x^2)^(1/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {a + b x^{2}} \operatorname {atan}{\left (\frac {e x}{\sqrt {- \frac {a e^{2}}{b} - e^{2} x^{2}}} \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/atan(e*x/(-a*e**2/b-e**2*x**2)**(1/2))/(b*x**2+a)**(1/2),x)

[Out]

Integral(1/(sqrt(a + b*x**2)*atan(e*x/sqrt(-a*e**2/b - e**2*x**2))), x)

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