Optimal. Leaf size=53 \[ -B \tan ^{-1}\left (\frac{B y}{\sqrt{A^2-B^2 y^2+B^2}}\right )-A \tanh ^{-1}\left (\frac{A y}{\sqrt{A^2-B^2 y^2+B^2}}\right ) \]
[Out]
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Rubi [A] time = 0.128817, antiderivative size = 53, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2 \[ -B \tan ^{-1}\left (\frac{B y}{\sqrt{A^2-B^2 y^2+B^2}}\right )-A \tanh ^{-1}\left (\frac{A y}{\sqrt{A^2-B^2 y^2+B^2}}\right ) \]
Antiderivative was successfully verified.
[In] Int[-(Sqrt[A^2 + B^2*(1 - y^2)]/(1 - y^2)),y]
[Out]
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Rubi in Sympy [A] time = 14.4245, size = 48, normalized size = 0.91 \[ - A \operatorname{atanh}{\left (\frac{A y}{\sqrt{A^{2} - B^{2} y^{2} + B^{2}}} \right )} - B \operatorname{atan}{\left (\frac{B y}{\sqrt{A^{2} - B^{2} y^{2} + B^{2}}} \right )} \]
Verification of antiderivative is not currently implemented for this CAS.
[In] rubi_integrate(-(A**2+B**2*(-y**2+1))**(1/2)/(-y**2+1),y)
[Out]
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Mathematica [C] time = 0.0864869, size = 127, normalized size = 2.4 \[ \frac{1}{2} \left (-2 i B \log \left (2 \left (\sqrt{A^2-B^2 y^2+B^2}-i B y\right )\right )-A \log \left (A \sqrt{A^2-B^2 y^2+B^2}+A^2-B^2 y+B^2\right )+A \log \left (A \sqrt{A^2-B^2 y^2+B^2}+A^2+B^2 (y+1)\right )+A \log (1-y)-A \log (y+1)\right ) \]
Antiderivative was successfully verified.
[In] Integrate[-(Sqrt[A^2 + B^2*(1 - y^2)]/(1 - y^2)),y]
[Out]
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Maple [B] time = 0.016, size = 262, normalized size = 4.9 \[{\frac{1}{2}\sqrt{-{B}^{2} \left ( y-1 \right ) ^{2}-2\,{B}^{2} \left ( y-1 \right ) +{A}^{2}}}-{\frac{{B}^{2}}{2}\arctan \left ({y\sqrt{{B}^{2}}{\frac{1}{\sqrt{-{B}^{2} \left ( y-1 \right ) ^{2}-2\,{B}^{2} \left ( y-1 \right ) +{A}^{2}}}}} \right ){\frac{1}{\sqrt{{B}^{2}}}}}-{\frac{{A}^{2}}{2}\ln \left ({\frac{1}{y-1} \left ( 2\,{A}^{2}-2\,{B}^{2} \left ( y-1 \right ) +2\,\sqrt{{A}^{2}}\sqrt{-{B}^{2} \left ( y-1 \right ) ^{2}-2\,{B}^{2} \left ( y-1 \right ) +{A}^{2}} \right ) } \right ){\frac{1}{\sqrt{{A}^{2}}}}}-{\frac{1}{2}\sqrt{-{B}^{2} \left ( 1+y \right ) ^{2}+2\,{B}^{2} \left ( 1+y \right ) +{A}^{2}}}-{\frac{{B}^{2}}{2}\arctan \left ({y\sqrt{{B}^{2}}{\frac{1}{\sqrt{-{B}^{2} \left ( 1+y \right ) ^{2}+2\,{B}^{2} \left ( 1+y \right ) +{A}^{2}}}}} \right ){\frac{1}{\sqrt{{B}^{2}}}}}+{\frac{{A}^{2}}{2}\ln \left ({\frac{1}{1+y} \left ( 2\,{A}^{2}+2\,{B}^{2} \left ( 1+y \right ) +2\,\sqrt{{A}^{2}}\sqrt{-{B}^{2} \left ( 1+y \right ) ^{2}+2\,{B}^{2} \left ( 1+y \right ) +{A}^{2}} \right ) } \right ){\frac{1}{\sqrt{{A}^{2}}}}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In] int(-(A^2+B^2*(-y^2+1))^(1/2)/(-y^2+1),y)
[Out]
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Maxima [F] time = 0., size = 0, normalized size = 0. \[ \text{Exception raised: ValueError} \]
Verification of antiderivative is not currently implemented for this CAS.
[In] integrate(sqrt(-(y^2 - 1)*B^2 + A^2)/(y^2 - 1),y, algorithm="maxima")
[Out]
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Fricas [A] time = 0.243198, size = 173, normalized size = 3.26 \[ B \arctan \left (\frac{\sqrt{-B^{2} y^{2} + A^{2} + B^{2}}}{B y}\right ) - \frac{1}{4} \, A \log \left (-\frac{{\left (A^{2} - B^{2}\right )} y^{2} + 2 \, \sqrt{-B^{2} y^{2} + A^{2} + B^{2}} A y + A^{2} + B^{2}}{y^{2}}\right ) + \frac{1}{4} \, A \log \left (-\frac{{\left (A^{2} - B^{2}\right )} y^{2} - 2 \, \sqrt{-B^{2} y^{2} + A^{2} + B^{2}} A y + A^{2} + B^{2}}{y^{2}}\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In] integrate(sqrt(-(y^2 - 1)*B^2 + A^2)/(y^2 - 1),y, algorithm="fricas")
[Out]
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Sympy [F] time = 0., size = 0, normalized size = 0. \[ \int \frac{\sqrt{A^{2} - B^{2} y^{2} + B^{2}}}{\left (y - 1\right ) \left (y + 1\right )}\, dy \]
Verification of antiderivative is not currently implemented for this CAS.
[In] integrate(-(A**2+B**2*(-y**2+1))**(1/2)/(-y**2+1),y)
[Out]
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GIAC/XCAS [F(-1)] time = 0., size = 0, normalized size = 0. \[ \text{Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
[In] integrate(sqrt(-(y^2 - 1)*B^2 + A^2)/(y^2 - 1),y, algorithm="giac")
[Out]