Integrand size = 30, antiderivative size = 53 \[ \int -\frac {\sqrt {A^2+B^2 \left (1-y^2\right )}}{1-y^2} \, dy=-B \arctan \left (\frac {B y}{\sqrt {A^2+B^2-B^2 y^2}}\right )-A \text {arctanh}\left (\frac {A y}{\sqrt {A^2+B^2-B^2 y^2}}\right ) \]
Time = 0.19 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.38 \[ \int -\frac {\sqrt {A^2+B^2 \left (1-y^2\right )}}{1-y^2} \, dy=2 B \arctan \left (\frac {B y}{\sqrt {A^2+B^2}-\sqrt {A^2+B^2-B^2 y^2}}\right )-A \text {arctanh}\left (\frac {\sqrt {A^2+B^2-B^2 y^2}}{A y}\right ) \]
2*B*ArcTan[(B*y)/(Sqrt[A^2 + B^2] - Sqrt[A^2 + B^2 - B^2*y^2])] - A*ArcTan h[Sqrt[A^2 + B^2 - B^2*y^2]/(A*y)]
Time = 0.22 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.233, Rules used = {25, 2074, 301, 224, 216, 291, 219}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int -\frac {\sqrt {A^2+B^2 \left (1-y^2\right )}}{1-y^2} \, dy\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {\sqrt {A^2+B^2 \left (1-y^2\right )}}{1-y^2}dy\) |
\(\Big \downarrow \) 2074 |
\(\displaystyle -\int \frac {\sqrt {A^2+B^2-B^2 y^2}}{1-y^2}dy\) |
\(\Big \downarrow \) 301 |
\(\displaystyle A^2 \left (-\int \frac {1}{\left (1-y^2\right ) \sqrt {A^2+B^2-B^2 y^2}}dy\right )-B^2 \int \frac {1}{\sqrt {A^2+B^2-B^2 y^2}}dy\) |
\(\Big \downarrow \) 224 |
\(\displaystyle A^2 \left (-\int \frac {1}{\left (1-y^2\right ) \sqrt {A^2+B^2-B^2 y^2}}dy\right )-B^2 \int \frac {1}{\frac {B^2 y^2}{A^2+B^2-B^2 y^2}+1}d\frac {y}{\sqrt {A^2+B^2-B^2 y^2}}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle A^2 \left (-\int \frac {1}{\left (1-y^2\right ) \sqrt {A^2+B^2-B^2 y^2}}dy\right )-B \arctan \left (\frac {B y}{\sqrt {A^2-B^2 y^2+B^2}}\right )\) |
\(\Big \downarrow \) 291 |
\(\displaystyle A^2 \left (-\int \frac {1}{1-\frac {A^2 y^2}{A^2+B^2-B^2 y^2}}d\frac {y}{\sqrt {A^2+B^2-B^2 y^2}}\right )-B \arctan \left (\frac {B y}{\sqrt {A^2-B^2 y^2+B^2}}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -B \arctan \left (\frac {B y}{\sqrt {A^2-B^2 y^2+B^2}}\right )-A \text {arctanh}\left (\frac {A y}{\sqrt {A^2-B^2 y^2+B^2}}\right )\) |
3.1.69.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
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] && (Gt Q[a, 0] || LtQ[b, 0])
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]
Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Subst [Int[1/(c - (b*c - a*d)*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]
Int[((a_) + (b_.)*(x_)^2)^(p_.)/((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[b/ d Int[(a + b*x^2)^(p - 1), x], x] - Simp[(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] || (EqQ[p, 2/3] && E qQ[b*c + 3*a*d, 0]))
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[BinomialDeg ree[u, x] - BinomialDegree[v, x], 0] && !BinomialMatchQ[{u, v}, x]
Time = 0.27 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.68
method | result | size |
pseudoelliptic | \(-\frac {A \ln \left (\frac {A y +\sqrt {-B^{2} y^{2}+A^{2}+B^{2}}}{y}\right )}{2}+\frac {A \ln \left (\frac {A y -\sqrt {-B^{2} y^{2}+A^{2}+B^{2}}}{y}\right )}{2}+B \arctan \left (\frac {\sqrt {-B^{2} y^{2}+A^{2}+B^{2}}}{B y}\right )\) | \(89\) |
default | \(\frac {\sqrt {-B^{2} \left (y -1\right )^{2}-2 B^{2} \left (y -1\right )+A^{2}}}{2}-\frac {B^{2} \arctan \left (\frac {\sqrt {B^{2}}\, y}{\sqrt {-B^{2} \left (y -1\right )^{2}-2 B^{2} \left (y -1\right )+A^{2}}}\right )}{2 \sqrt {B^{2}}}-\frac {A^{2} \ln \left (\frac {2 A^{2}-2 B^{2} \left (y -1\right )+2 \sqrt {A^{2}}\, \sqrt {-B^{2} \left (y -1\right )^{2}-2 B^{2} \left (y -1\right )+A^{2}}}{y -1}\right )}{2 \sqrt {A^{2}}}-\frac {\sqrt {-B^{2} \left (1+y \right )^{2}+2 B^{2} \left (1+y \right )+A^{2}}}{2}-\frac {B^{2} \arctan \left (\frac {\sqrt {B^{2}}\, y}{\sqrt {-B^{2} \left (1+y \right )^{2}+2 B^{2} \left (1+y \right )+A^{2}}}\right )}{2 \sqrt {B^{2}}}+\frac {A^{2} \ln \left (\frac {2 A^{2}+2 B^{2} \left (1+y \right )+2 \sqrt {A^{2}}\, \sqrt {-B^{2} \left (1+y \right )^{2}+2 B^{2} \left (1+y \right )+A^{2}}}{1+y}\right )}{2 \sqrt {A^{2}}}\) | \(262\) |
-1/2*A*ln((A*y+(-B^2*y^2+A^2+B^2)^(1/2))/y)+1/2*A*ln((A*y-(-B^2*y^2+A^2+B^ 2)^(1/2))/y)+B*arctan(1/B/y*(-B^2*y^2+A^2+B^2)^(1/2))
Leaf count of result is larger than twice the leaf count of optimal. 128 vs. \(2 (49) = 98\).
Time = 0.26 (sec) , antiderivative size = 128, normalized size of antiderivative = 2.42 \[ \int -\frac {\sqrt {A^2+B^2 \left (1-y^2\right )}}{1-y^2} \, dy=B \arctan \left (\frac {\sqrt {-B^{2} y^{2} + A^{2} + B^{2}}}{B y}\right ) - \frac {1}{4} \, A \log \left (-\frac {{\left (A^{2} - B^{2}\right )} y^{2} + 2 \, \sqrt {-B^{2} y^{2} + A^{2} + B^{2}} A y + A^{2} + B^{2}}{y^{2}}\right ) + \frac {1}{4} \, A \log \left (-\frac {{\left (A^{2} - B^{2}\right )} y^{2} - 2 \, \sqrt {-B^{2} y^{2} + A^{2} + B^{2}} A y + A^{2} + B^{2}}{y^{2}}\right ) \]
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)
\[ \int -\frac {\sqrt {A^2+B^2 \left (1-y^2\right )}}{1-y^2} \, dy=\int \frac {\sqrt {A^{2} - B^{2} y^{2} + B^{2}}}{\left (y - 1\right ) \left (y + 1\right )}\, dy \]
Leaf count of result is larger than twice the leaf count of optimal. 115 vs. \(2 (49) = 98\).
Time = 0.29 (sec) , antiderivative size = 115, normalized size of antiderivative = 2.17 \[ \int -\frac {\sqrt {A^2+B^2 \left (1-y^2\right )}}{1-y^2} \, dy=-B \arcsin \left (\frac {B^{2} y}{\sqrt {A^{2} B^{2} + B^{4}}}\right ) + \frac {1}{2} \, A \log \left (B^{2} + \frac {A^{2}}{y + 1} + \frac {\sqrt {-B^{2} y^{2} + A^{2} + B^{2}} A}{y + 1}\right ) - \frac {1}{2} \, A \log \left (-B^{2} + \frac {2 \, A^{2}}{{\left | 2 \, y - 2 \right |}} + \frac {2 \, \sqrt {-B^{2} y^{2} + A^{2} + B^{2}} A}{{\left | 2 \, y - 2 \right |}}\right ) \]
-B*arcsin(B^2*y/sqrt(A^2*B^2 + B^4)) + 1/2*A*log(B^2 + A^2/(y + 1) + sqrt( -B^2*y^2 + A^2 + B^2)*A/(y + 1)) - 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))
Leaf count of result is larger than twice the leaf count of optimal. 295 vs. \(2 (49) = 98\).
Time = 0.35 (sec) , antiderivative size = 295, normalized size of antiderivative = 5.57 \[ \int -\frac {\sqrt {A^2+B^2 \left (1-y^2\right )}}{1-y^2} \, dy=\frac {{\left (\pi \mathrm {sgn}\left (y\right ) - 2 \, \arctan \left (-\frac {B^{2} y {\left (\frac {{\left (\sqrt {A^{2} + B^{2}} B + \sqrt {-B^{2} y^{2} + A^{2} + B^{2}} {\left | B \right |}\right )}^{2}}{B^{4} y^{2}} - 1\right )}}{2 \, {\left (\sqrt {A^{2} + B^{2}} B + \sqrt {-B^{2} y^{2} + A^{2} + B^{2}} {\left | B \right |}\right )}}\right )\right )} B^{2}}{2 \, {\left | B \right |}} - \frac {A B \log \left ({\left | -{\left (\frac {B^{2} y}{\sqrt {A^{2} + B^{2}} B + \sqrt {-B^{2} y^{2} + A^{2} + B^{2}} {\left | B \right |}} - \frac {\sqrt {A^{2} + B^{2}} B + \sqrt {-B^{2} y^{2} + A^{2} + B^{2}} {\left | B \right |}}{B^{2} y}\right )} B + 2 \, A \right |}\right )}{2 \, {\left | B \right |}} + \frac {A B \log \left ({\left | -{\left (\frac {B^{2} y}{\sqrt {A^{2} + B^{2}} B + \sqrt {-B^{2} y^{2} + A^{2} + B^{2}} {\left | B \right |}} - \frac {\sqrt {A^{2} + B^{2}} B + \sqrt {-B^{2} y^{2} + A^{2} + B^{2}} {\left | B \right |}}{B^{2} y}\right )} B - 2 \, A \right |}\right )}{2 \, {\left | B \right |}} \]
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)
Timed out. \[ \int -\frac {\sqrt {A^2+B^2 \left (1-y^2\right )}}{1-y^2} \, dy=\left \{\begin {array}{cl} \int \frac {\sqrt {-B^2\,y^2}}{y^2-1} \,d y & \text {\ if\ \ }A^2+B^2=0\\ \ln \left (2\,y\,\sqrt {-B^2}+2\,\sqrt {A^2-B^2\,y^2+B^2}\right )\,\sqrt {-B^2}+\mathrm {atan}\left (\frac {y\,\sqrt {A^2}\,1{}\mathrm {i}}{\sqrt {A^2-B^2\,y^2+B^2}}\right )\,\sqrt {A^2}\,1{}\mathrm {i} & \text {\ if\ \ }A^2+B^2\neq 0 \end {array}\right . \]
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))
Time = 0.01 (sec) , antiderivative size = 126, normalized size of antiderivative = 2.38 \[ \int -\frac {\sqrt {A^2+B^2 \left (1-y^2\right )}}{1-y^2} \, dy=-\mathit {asin} \left (\frac {b y}{\sqrt {a^{2}+b^{2}}}\right ) b +\mathit {atan} \left (\frac {\tan \left (\frac {\mathit {asin} \left (\frac {b y}{\sqrt {a^{2}+b^{2}}}\right )}{2}\right ) b i}{\sqrt {a^{2}+b^{2}}-a}\right ) a i -\frac {\mathrm {log}\left (-\sqrt {a^{2}+b^{2}}+\tan \left (\frac {\mathit {asin} \left (\frac {b y}{\sqrt {a^{2}+b^{2}}}\right )}{2}\right ) b -a \right ) a}{2}+\frac {\mathrm {log}\left (\sqrt {a^{2}+b^{2}}+\tan \left (\frac {\mathit {asin} \left (\frac {b y}{\sqrt {a^{2}+b^{2}}}\right )}{2}\right ) b +a \right ) a}{2} \]