3.1.69 \(\int -\frac {\sqrt {A^2+B^2 (1-y^2)}}{1-y^2} \, dy\) [69]

Optimal. Leaf size=53 \[ -B \tan ^{-1}\left (\frac {B y}{\sqrt {A^2+B^2-B^2 y^2}}\right )-A \tanh ^{-1}\left (\frac {A y}{\sqrt {A^2+B^2-B^2 y^2}}\right ) \]

[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 = 0.04, 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 = {1999, 399, 223, 209, 385, 212} \begin {gather*} -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 ) \end {gather*}

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 209

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

Rule 212

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

Rule 223

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 385

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 399

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

Rule 1999

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 \text {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 \text {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 [A]
time = 0.08, size = 98, normalized size = 1.85 \begin {gather*} \frac {B \left (-A \tan ^{-1}\left (\frac {B^2 \left (-1+y^2\right )+\sqrt {-B^2} y \sqrt {A^2+B^2-B^2 y^2}}{A B}\right )+B \log \left (-\sqrt {-B^2} y+\sqrt {A^2+B^2-B^2 y^2}\right )\right )}{\sqrt {-B^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathics [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {cought exception: maximum recursion depth exceeded while calling a Python object} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

cought exception: maximum recursion depth exceeded while calling a Python object

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(261\) vs. \(2(49)=98\).
time = 0.05, size = 262, normalized size = 4.94

method	result	size
default	\(\frac {\sqrt {-B^{2} \left (y -1\right )^{2}-2 B^{2} \left (y -1\right )+A^{2}}}{2}-\frac {B^{2} \arctan \left (\frac {\sqrt {B^{2}}\, y}{\sqrt {-B^{2} \left (y -1\right )^{2}-2 B^{2} \left (y -1\right )+A^{2}}}\right )}{2 \sqrt {B^{2}}}-\frac {A^{2} \ln \left (\frac {2 A^{2}-2 B^{2} \left (y -1\right )+2 \sqrt {A^{2}}\, \sqrt {-B^{2} \left (y -1\right )^{2}-2 B^{2} \left (y -1\right )+A^{2}}}{y -1}\right )}{2 \sqrt {A^{2}}}-\frac {\sqrt {-B^{2} \left (1+y \right )^{2}+2 B^{2} \left (1+y \right )+A^{2}}}{2}-\frac {B^{2} \arctan \left (\frac {\sqrt {B^{2}}\, y}{\sqrt {-B^{2} \left (1+y \right )^{2}+2 B^{2} \left (1+y \right )+A^{2}}}\right )}{2 \sqrt {B^{2}}}+\frac {A^{2} \ln \left (\frac {2 A^{2}+2 B^{2} \left (1+y \right )+2 \sqrt {A^{2}}\, \sqrt {-B^{2} \left (1+y \right )^{2}+2 B^{2} \left (1+y \right )+A^{2}}}{1+y}\right )}{2 \sqrt {A^{2}}}\)	\(262\)

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,method=_RETURNVERBOSE)

[Out]

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))
-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)
)

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 123 vs. \(2 (49) = 98\).
time = 0.35, size = 123, normalized size = 2.32 \begin {gather*} -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 ) \end {gather*}

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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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 128 vs. \(2 (49) = 98\).
time = 0.34, size = 128, normalized size = 2.42 \begin {gather*} 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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {A^{2} - B^{2} y^{2} + B^{2}}}{\left (y - 1\right ) \left (y + 1\right )}\, dy \end {gather*}

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 [B] Leaf count of result is larger than twice the leaf count of optimal. 295 vs. \(2 (49) = 98\).
time = 0.02, size = 355, normalized size = 6.70 \begin {gather*} \frac {B^{2} \left (\frac {1}{2} \pi \mathrm {sign}\left (y\right )-\arctan \left (\frac {B^{2} y \left (\left (-\frac {-2 B \sqrt {A^{2}+B^{2}}-2 \sqrt {-B^{2} y^{2}+A^{2}+B^{2}} \left |B\right |}{2 \left (B^{2} y\right )}\right )^{2}-1\right )}{-2 B \sqrt {A^{2}+B^{2}}-2 \sqrt {-B^{2} y^{2}+A^{2}+B^{2}} \left |B\right |}\right )\right )}{\left |B\right |}-\frac {A B^{2} \ln \left |B \left (-\frac {-2 B \sqrt {A^{2}+B^{2}}-2 \sqrt {-B^{2} y^{2}+A^{2}+B^{2}} \left |B\right |}{2 \left (B^{2} y\right )}+\frac {2 B^{2} y}{-2 B \sqrt {A^{2}+B^{2}}-2 \sqrt {-B^{2} y^{2}+A^{2}+B^{2}} \left |B\right |}\right )+2 A\right |}{2 \left (B \left |B\right |\right )}+\frac {A B^{2} \ln \left |B \left (-\frac {-2 B \sqrt {A^{2}+B^{2}}-2 \sqrt {-B^{2} y^{2}+A^{2}+B^{2}} \left |B\right |}{2 \left (B^{2} y\right )}+\frac {2 B^{2} y}{-2 B \sqrt {A^{2}+B^{2}}-2 \sqrt {-B^{2} y^{2}+A^{2}+B^{2}} \left |B\right |}\right )-2 A\right |}{2 \left (B \left |B\right |\right )} \end {gather*}

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]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \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 . \end {gather*}

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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