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

Optimal. Leaf size=68 \[ \frac {\sqrt {e^2 \left (-x^2\right )-\frac {a e^2}{b}} \tan ^{-1}\left (\frac {e x}{\sqrt {e^2 \left (-x^2\right )-\frac {a e^2}{b}}}\right )^2}{2 e \sqrt {a+b x^2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.06, antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {5157, 5155} \[ \frac {\sqrt {e^2 \left (-x^2\right )-\frac {a e^2}{b}} \tan ^{-1}\left (\frac {e x}{\sqrt {e^2 \left (-x^2\right )-\frac {a e^2}{b}}}\right )^2}{2 e \sqrt {a+b x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 5155

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

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

________________________________________________________________________________________

Mathematica [A]  time = 0.06, size = 62, normalized size = 0.91 \[ \frac {\sqrt {-\frac {e^2 \left (a+b x^2\right )}{b}} \tan ^{-1}\left (\frac {e x}{\sqrt {-\frac {e^2 \left (a+b x^2\right )}{b}}}\right )^2}{2 e \sqrt {a+b x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [F]  time = 0.72, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\arctan \left (\frac {b x \sqrt {-\frac {b e^{2} x^{2} + a e^{2}}{b}}}{b e x^{2} + a e}\right )}{\sqrt {b x^{2} + a}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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(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)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________