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

Optimal. Leaf size=51 \[ B \tan ^{-1}\left (\frac {B y}{\sqrt {A^2-B^2 y^2+B^2}}\right )+A \tanh ^{-1}\left (\frac {A y}{\sqrt {A^2-B^2 y^2+B^2}}\right ) \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {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 - B^2*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 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 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])

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 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 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])

Rubi steps

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

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Mathematica [C]  time = 0.03, size = 134, normalized size = 2.63 \[ i B \log \left (2 \sqrt {A^2-B^2 y^2+B^2}-2 i B y\right )+\frac {1}{2} A \log \left (A \sqrt {A^2-B^2 y^2+B^2}+A^2-B^2 y+B^2\right )-\frac {1}{2} A \log \left (A \sqrt {A^2-B^2 y^2+B^2}+A^2+B^2 y+B^2\right )-\frac {1}{2} A \log (1-y)+\frac {1}{2} A \log (y+1) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.45, size = 129, normalized size = 2.53 \[ -B \arctan \left (\frac {\sqrt {-B^{2} y^{2} + A^{2} + B^{2}}}{B y}\right ) + \frac {1}{4} \, A \log \left (-\frac {{\left (A^{2} - B^{2}\right )} y^{2} + 2 \, \sqrt {-B^{2} y^{2} + A^{2} + B^{2}} A y + A^{2} + B^{2}}{y^{2}}\right ) - \frac {1}{4} \, A \log \left (-\frac {{\left (A^{2} - B^{2}\right )} y^{2} - 2 \, \sqrt {-B^{2} y^{2} + A^{2} + B^{2}} A y + A^{2} + B^{2}}{y^{2}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-B^2*y^2+A^2+B^2)^(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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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-B^2*y^2+A^2+B^2)^(1/2)/(-y^2+1),y, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Warning, choosing root of [1,0,%%%{-4,[2
,0,0]%%%}+%%%{2,[0,2,2]%%%}+%%%{-4,[0,2,0]%%%},0,%%%{1,[0,4,4]%%%}] at parameters values [88,76,-66]Warning, c
hoosing root of [1,0,%%%{-4,[2,0,0]%%%}+%%%{2,[0,2,2]%%%}+%%%{-4,[0,2,0]%%%},0,%%%{1,[0,4,4]%%%}] at parameter
s values [66,5,-23]-B^2*(1/2*pi*sign(y)-atan(B^2*y*((-1/2*(-2*B*sqrt(A^2+B^2)-2*sqrt(-B^2*y^2+A^2+B^2)*abs(B))
/B^2/y)^2-1)/(-2*B*sqrt(A^2+B^2)-2*sqrt(-B^2*y^2+A^2+B^2)*abs(B))))/abs(B)+1/2*A*B^2*ln(abs(B*(-1/2*(-2*B*sqrt
(A^2+B^2)-2*sqrt(-B^2*y^2+A^2+B^2)*abs(B))/B^2/y+2*B^2*y/(-2*B*sqrt(A^2+B^2)-2*sqrt(-B^2*y^2+A^2+B^2)*abs(B)))
+2*A))/B/abs(B)-1/2*A*B^2*ln(abs(B*(-1/2*(-2*B*sqrt(A^2+B^2)-2*sqrt(-B^2*y^2+A^2+B^2)*abs(B))/B^2/y+2*B^2*y/(-
2*B*sqrt(A^2+B^2)-2*sqrt(-B^2*y^2+A^2+B^2)*abs(B)))-2*A))/B/abs(B)

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maple [B]  time = 0.03, size = 262, normalized size = 5.14 \[ -\frac {A^{2} \ln \left (\frac {2 A^{2}+2 \left (y +1\right ) B^{2}+2 \sqrt {A^{2}}\, \sqrt {A^{2}-\left (y +1\right )^{2} B^{2}+2 \left (y +1\right ) B^{2}}}{y +1}\right )}{2 \sqrt {A^{2}}}+\frac {A^{2} \ln \left (\frac {2 A^{2}-2 \left (y -1\right ) B^{2}+2 \sqrt {A^{2}}\, \sqrt {A^{2}-\left (y -1\right )^{2} B^{2}-2 \left (y -1\right ) B^{2}}}{y -1}\right )}{2 \sqrt {A^{2}}}+\frac {B^{2} \arctan \left (\frac {\sqrt {B^{2}}\, y}{\sqrt {A^{2}-\left (y +1\right )^{2} B^{2}+2 \left (y +1\right ) B^{2}}}\right )}{2 \sqrt {B^{2}}}+\frac {B^{2} \arctan \left (\frac {\sqrt {B^{2}}\, y}{\sqrt {A^{2}-\left (y -1\right )^{2} B^{2}-2 \left (y -1\right ) B^{2}}}\right )}{2 \sqrt {B^{2}}}+\frac {\sqrt {A^{2}-\left (y +1\right )^{2} B^{2}+2 \left (y +1\right ) B^{2}}}{2}-\frac {\sqrt {A^{2}-\left (y -1\right )^{2} B^{2}-2 \left (y -1\right ) B^{2}}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-B^2*y^2+A^2+B^2)^(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 [B]  time = 1.46, size = 122, normalized size = 2.39 \[ B \arcsin \left (\frac {B^{2} y}{\sqrt {A^{2} B^{2} + B^{4}}}\right ) - \frac {1}{2} \, A \log \left (B^{2} + \frac {2 \, A^{2}}{{\left | 2 \, y + 2 \right |}} + \frac {2 \, \sqrt {-B^{2} y^{2} + A^{2} + B^{2}} A}{{\left | 2 \, y + 2 \right |}}\right ) + \frac {1}{2} \, A \log \left (-B^{2} + \frac {2 \, A^{2}}{{\left | 2 \, y - 2 \right |}} + \frac {2 \, \sqrt {-B^{2} y^{2} + A^{2} + B^{2}} A}{{\left | 2 \, y - 2 \right |}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-B^2*y^2+A^2+B^2)^(1/2)/(-y^2+1),y, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \left \{\begin {array}{cl} -\int \frac {\sqrt {-B^2\,y^2}}{y^2-1} \,d y & \text {\ if\ \ }A^2+B^2=0\\ -\ln \left (2\,y\,\sqrt {-B^2}+2\,\sqrt {A^2-B^2\,y^2+B^2}\right )\,\sqrt {-B^2}-\mathrm {atan}\left (\frac {y\,\sqrt {A^2}\,1{}\mathrm {i}}{\sqrt {A^2-B^2\,y^2+B^2}}\right )\,\sqrt {A^2}\,1{}\mathrm {i} & \text {\ if\ \ }A^2+B^2\neq 0 \end {array}\right . \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(A^2 + B^2 - B^2*y^2)^(1/2)/(y^2 - 1),y)

[Out]

piecewise(A^2 + B^2 == 0, -int((-B^2*y^2)^(1/2)/(y^2 - 1), y), A^2 + B^2 ~= 0, - atan((y*(A^2)^(1/2)*1i)/(A^2
+ B^2 - B^2*y^2)^(1/2))*(A^2)^(1/2)*1i - log(2*y*(-B^2)^(1/2) + 2*(A^2 + B^2 - B^2*y^2)^(1/2))*(-B^2)^(1/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \int \frac {\sqrt {A^{2} - B^{2} y^{2} + B^{2}}}{y^{2} - 1}\, dy \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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