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/(-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 = 0.06, antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, 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 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])

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]

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*}

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Mathematica [C]  time = 0.06, size = 127, normalized size = 2.40 \[ \frac {1}{2} \left (-2 i B \log \left (2 \left (\sqrt {A^2-B^2 y^2+B^2}-i B y\right )\right )-A \log \left (A \sqrt {A^2-B^2 y^2+B^2}+A^2-B^2 y+B^2\right )+A \log \left (A \sqrt {A^2-B^2 y^2+B^2}+A^2+B^2 (y+1)\right )+A \log (1-y)-A \log (y+1)\right ) \]

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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fricas [B]  time = 0.44, size = 128, normalized size = 2.42 \[ B \arctan \left (\frac {\sqrt {-B^{2} y^{2} + A^{2} + B^{2}}}{B y}\right ) - \frac {1}{4} \, A \log \left (-\frac {{\left (A^{2} - B^{2}\right )} y^{2} + 2 \, \sqrt {-B^{2} y^{2} + A^{2} + B^{2}} A y + A^{2} + B^{2}}{y^{2}}\right ) + \frac {1}{4} \, A \log \left (-\frac {{\left (A^{2} - B^{2}\right )} y^{2} - 2 \, \sqrt {-B^{2} y^{2} + A^{2} + B^{2}} A y + A^{2} + B^{2}}{y^{2}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-(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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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(-(A^2+B^2*(-y^2+1))^(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 = 0.01, size = 262, normalized size = 4.94 \[ \frac {A^{2} \ln \left (\frac {2 A^{2}+2 \left (y +1\right ) B^{2}+2 \sqrt {A^{2}}\, \sqrt {A^{2}-\left (y +1\right )^{2} B^{2}+2 \left (y +1\right ) B^{2}}}{y +1}\right )}{2 \sqrt {A^{2}}}-\frac {A^{2} \ln \left (\frac {2 A^{2}-2 \left (y -1\right ) B^{2}+2 \sqrt {A^{2}}\, \sqrt {A^{2}-\left (y -1\right )^{2} B^{2}-2 \left (y -1\right ) B^{2}}}{y -1}\right )}{2 \sqrt {A^{2}}}-\frac {B^{2} \arctan \left (\frac {\sqrt {B^{2}}\, y}{\sqrt {A^{2}-\left (y +1\right )^{2} B^{2}+2 \left (y +1\right ) B^{2}}}\right )}{2 \sqrt {B^{2}}}-\frac {B^{2} \arctan \left (\frac {\sqrt {B^{2}}\, y}{\sqrt {A^{2}-\left (y -1\right )^{2} B^{2}-2 \left (y -1\right ) B^{2}}}\right )}{2 \sqrt {B^{2}}}-\frac {\sqrt {A^{2}-\left (y +1\right )^{2} B^{2}+2 \left (y +1\right ) B^{2}}}{2}+\frac {\sqrt {A^{2}-\left (y -1\right )^{2} B^{2}-2 \left (y -1\right ) B^{2}}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 1.49, size = 123, normalized size = 2.32 \[ -B \arcsin \left (\frac {B^{2} y}{\sqrt {A^{2} B^{2} + B^{4}}}\right ) + \frac {1}{2} \, A \log \left (B^{2} + \frac {2 \, A^{2}}{{\left | 2 \, y + 2 \right |}} + \frac {2 \, \sqrt {-B^{2} y^{2} + A^{2} + B^{2}} A}{{\left | 2 \, y + 2 \right |}}\right ) - \frac {1}{2} \, A \log \left (-B^{2} + \frac {2 \, A^{2}}{{\left | 2 \, y - 2 \right |}} + \frac {2 \, \sqrt {-B^{2} y^{2} + A^{2} + B^{2}} A}{{\left | 2 \, y - 2 \right |}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-(A^2+B^2*(-y^2+1))^(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*(y^2 - 1))^(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 \[ \int \frac {\sqrt {A^{2} - B^{2} y^{2} + B^{2}}}{\left (y - 1\right ) \left (y + 1\right )}\, dy \]

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]

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

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