3.69 \(\int -\frac{\sqrt{A^2+B^2 (1-y^2)}}{1-y^2} \, dy\)

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]

-(B*ArcTan[(B*y)/Sqrt[A^2 + B^2 - B^2*y^2]]) - A*ArcTanh[(A*y)/Sqrt[A^2 + B^2 - B^2*y^2]]

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Rubi [A]  time = 0.0648788, 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, Rules used = {1974, 402, 217, 203, 377, 206} \[ -B \tan ^{-1}\left (\frac{B y}{\sqrt{A^2-B^2 y^2+B^2}}\right )-A \tanh ^{-1}\left (\frac{A y}{\sqrt{A^2-B^2 y^2+B^2}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[-(Sqrt[A^2 + B^2*(1 - y^2)]/(1 - y^2)),y]

[Out]

-(B*ArcTan[(B*y)/Sqrt[A^2 + B^2 - B^2*y^2]]) - A*ArcTanh[(A*y)/Sqrt[A^2 + B^2 - B^2*y^2]]

Rule 1974

Int[(u_)^(p_.)*(v_)^(q_.), x_Symbol] :> Int[ExpandToSum[u, x]^p*ExpandToSum[v, x]^q, x] /; FreeQ[{p, q}, x] &&
BinomialQ[{u, v}, x] && EqQ[BinomialDegree[u, x] - BinomialDegree[v, x], 0] &&  !BinomialMatchQ[{u, v}, x]

Rule 402

Int[((a_) + (b_.)*(x_)^2)^(p_.)/((c_) + (d_.)*(x_)^2), x_Symbol] :> Dist[b/d, Int[(a + b*x^2)^(p - 1), x], x]
- Dist[(b*c - a*d)/d, Int[(a + b*x^2)^(p - 1)/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d,
0] && GtQ[p, 0] && (EqQ[p, 1/2] || EqQ[Denominator[p], 4])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int -\frac{\sqrt{A^2+B^2 \left (1-y^2\right )}}{1-y^2} \, dy &=-\int \frac{\sqrt{A^2+B^2-B^2 y^2}}{1-y^2} \, dy\\ &=-\left (A^2 \int \frac{1}{\left (1-y^2\right ) \sqrt{A^2+B^2-B^2 y^2}} \, dy\right )-B^2 \int \frac{1}{\sqrt{A^2+B^2-B^2 y^2}} \, dy\\ &=-\left (A^2 \operatorname{Subst}\left (\int \frac{1}{1-A^2 y^2} \, dy,y,\frac{y}{\sqrt{A^2+B^2-B^2 y^2}}\right )\right )-B^2 \operatorname{Subst}\left (\int \frac{1}{1+B^2 y^2} \, dy,y,\frac{y}{\sqrt{A^2+B^2-B^2 y^2}}\right )\\ &=-B \tan ^{-1}\left (\frac{B y}{\sqrt{A^2+B^2-B^2 y^2}}\right )-A \tanh ^{-1}\left (\frac{A y}{\sqrt{A^2+B^2-B^2 y^2}}\right )\\ \end{align*}

Mathematica [C]  time = 0.0549715, 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]

(A*Log[1 - y] - A*Log[1 + y] - (2*I)*B*Log[2*((-I)*B*y + Sqrt[A^2 + B^2 - B^2*y^2])] - A*Log[A^2 + B^2 - B^2*y
+ A*Sqrt[A^2 + B^2 - B^2*y^2]] + A*Log[A^2 + B^2*(1 + y) + A*Sqrt[A^2 + B^2 - B^2*y^2]])/2

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Maple [B]  time = 0.006, size = 262, normalized size = 4.9 \begin{align*} -{\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}}}}}+{\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}}}}} \end{align*}

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]

-1/2*(-B^2*(1+y)^2+2*B^2*(1+y)+A^2)^(1/2)-1/2*B^2/(B^2)^(1/2)*arctan((B^2)^(1/2)*y/(-B^2*(1+y)^2+2*B^2*(1+y)+A
^2)^(1/2))+1/2*A^2/(A^2)^(1/2)*ln((2*A^2+2*B^2*(1+y)+2*(A^2)^(1/2)*(-B^2*(1+y)^2+2*B^2*(1+y)+A^2)^(1/2))/(1+y)
)+1/2*(-B^2*(y-1)^2-2*B^2*(y-1)+A^2)^(1/2)-1/2*B^2/(B^2)^(1/2)*arctan((B^2)^(1/2)*y/(-B^2*(y-1)^2-2*B^2*(y-1)+
A^2)^(1/2))-1/2*A^2/(A^2)^(1/2)*ln((2*A^2-2*B^2*(y-1)+2*(A^2)^(1/2)*(-B^2*(y-1)^2-2*B^2*(y-1)+A^2)^(1/2))/(y-1
))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.6256, size = 284, normalized size = 5.36 \begin{align*} 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 ) \end{align*}

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, algorithm="fricas")

[Out]

B*arctan(sqrt(-B^2*y^2 + A^2 + B^2)/(B*y)) - 1/4*A*log(-((A^2 - B^2)*y^2 + 2*sqrt(-B^2*y^2 + A^2 + B^2)*A*y +
A^2 + B^2)/y^2) + 1/4*A*log(-((A^2 - B^2)*y^2 - 2*sqrt(-B^2*y^2 + A^2 + B^2)*A*y + A^2 + B^2)/y^2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{A^{2} - B^{2} y^{2} + B^{2}}}{\left (y - 1\right ) \left (y + 1\right )}\, dy \end{align*}

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]

Integral(sqrt(A**2 - B**2*y**2 + B**2)/((y - 1)*(y + 1)), y)

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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, algorithm="giac")

[Out]

Timed out