3.846 \(\int \frac {\sin ^2(x)}{a+b \sin (2 x)} \, dx\)

Optimal. Leaf size=55 \[ \frac {\tan ^{-1}\left (\frac {a \tan (x)+b}{\sqrt {a^2-b^2}}\right )}{2 \sqrt {a^2-b^2}}-\frac {\log (a+b \sin (2 x))}{4 b} \]

[Out]

-1/4*ln(a+b*sin(2*x))/b+1/2*arctan((b+a*tan(x))/(a^2-b^2)^(1/2))/(a^2-b^2)^(1/2)

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Rubi [A]  time = 0.17, antiderivative size = 70, normalized size of antiderivative = 1.27, number of steps used = 9, number of rules used = 7, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {1075, 12, 634, 618, 204, 628, 260} \[ \frac {\tan ^{-1}\left (\frac {a \tan (x)+b}{\sqrt {a^2-b^2}}\right )}{2 \sqrt {a^2-b^2}}-\frac {\log \left (a \tan ^2(x)+a+2 b \tan (x)\right )}{4 b}-\frac {\log (\cos (x))}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[Sin[x]^2/(a + b*Sin[2*x]),x]

[Out]

ArcTan[(b + a*Tan[x])/Sqrt[a^2 - b^2]]/(2*Sqrt[a^2 - b^2]) - Log[Cos[x]]/(2*b) - Log[a + 2*b*Tan[x] + a*Tan[x]
^2]/(4*b)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 204

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

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 1075

Int[((A_.) + (C_.)*(x_)^2)/(((a_) + (b_.)*(x_) + (c_.)*(x_)^2)*((d_) + (f_.)*(x_)^2)), x_Symbol] :> With[{q =
c^2*d^2 + b^2*d*f - 2*a*c*d*f + a^2*f^2}, Dist[1/q, Int[(A*c^2*d - a*c*C*d + A*b^2*f - a*A*c*f + a^2*C*f + c*(
-(b*C*d) + A*b*f)*x)/(a + b*x + c*x^2), x], x] + Dist[1/q, Int[(c*C*d^2 - A*c*d*f - a*C*d*f + a*A*f^2 - f*(-(b
*C*d) + A*b*f)*x)/(d + f*x^2), x], x] /; NeQ[q, 0]] /; FreeQ[{a, b, c, d, f, A, C}, x] && NeQ[b^2 - 4*a*c, 0]

Rubi steps

\begin {align*} \int \frac {\sin ^2(x)}{a+b \sin (2 x)} \, dx &=\operatorname {Subst}\left (\int \frac {x^2}{\left (1+x^2\right ) \left (a+2 b x+a x^2\right )} \, dx,x,\tan (x)\right )\\ &=\frac {\operatorname {Subst}\left (\int \frac {2 b x}{1+x^2} \, dx,x,\tan (x)\right )}{4 b^2}+\frac {\operatorname {Subst}\left (\int -\frac {2 a b x}{a+2 b x+a x^2} \, dx,x,\tan (x)\right )}{4 b^2}\\ &=\frac {\operatorname {Subst}\left (\int \frac {x}{1+x^2} \, dx,x,\tan (x)\right )}{2 b}-\frac {a \operatorname {Subst}\left (\int \frac {x}{a+2 b x+a x^2} \, dx,x,\tan (x)\right )}{2 b}\\ &=-\frac {\log (\cos (x))}{2 b}+\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{a+2 b x+a x^2} \, dx,x,\tan (x)\right )-\frac {\operatorname {Subst}\left (\int \frac {2 b+2 a x}{a+2 b x+a x^2} \, dx,x,\tan (x)\right )}{4 b}\\ &=-\frac {\log (\cos (x))}{2 b}-\frac {\log \left (a+2 b \tan (x)+a \tan ^2(x)\right )}{4 b}-\operatorname {Subst}\left (\int \frac {1}{-4 \left (a^2-b^2\right )-x^2} \, dx,x,2 b+2 a \tan (x)\right )\\ &=\frac {\tan ^{-1}\left (\frac {2 b+2 a \tan (x)}{2 \sqrt {a^2-b^2}}\right )}{2 \sqrt {a^2-b^2}}-\frac {\log (\cos (x))}{2 b}-\frac {\log \left (a+2 b \tan (x)+a \tan ^2(x)\right )}{4 b}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 55, normalized size = 1.00 \[ \frac {\tan ^{-1}\left (\frac {a \tan (x)+b}{\sqrt {a^2-b^2}}\right )}{2 \sqrt {a^2-b^2}}-\frac {\log (a+b \sin (2 x))}{4 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[x]^2/(a + b*Sin[2*x]),x]

[Out]

ArcTan[(b + a*Tan[x])/Sqrt[a^2 - b^2]]/(2*Sqrt[a^2 - b^2]) - Log[a + b*Sin[2*x]]/(4*b)

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fricas [B]  time = 1.91, size = 320, normalized size = 5.82 \[ \left [-\frac {\sqrt {-a^{2} + b^{2}} b \log \left (-\frac {4 \, {\left (2 \, a^{2} - b^{2}\right )} \cos \relax (x)^{4} - 4 \, a b \cos \relax (x) \sin \relax (x) - 4 \, {\left (2 \, a^{2} - b^{2}\right )} \cos \relax (x)^{2} + a^{2} - 2 \, b^{2} + 2 \, {\left (2 \, b \cos \relax (x)^{2} + 2 \, {\left (2 \, a \cos \relax (x)^{3} - a \cos \relax (x)\right )} \sin \relax (x) - b\right )} \sqrt {-a^{2} + b^{2}}}{4 \, b^{2} \cos \relax (x)^{4} - 4 \, b^{2} \cos \relax (x)^{2} - 4 \, a b \cos \relax (x) \sin \relax (x) - a^{2}}\right ) + {\left (a^{2} - b^{2}\right )} \log \left (-4 \, b^{2} \cos \relax (x)^{4} + 4 \, b^{2} \cos \relax (x)^{2} + 4 \, a b \cos \relax (x) \sin \relax (x) + a^{2}\right )}{8 \, {\left (a^{2} b - b^{3}\right )}}, -\frac {2 \, \sqrt {a^{2} - b^{2}} b \arctan \left (-\frac {{\left (2 \, a \cos \relax (x) \sin \relax (x) + b\right )} \sqrt {a^{2} - b^{2}}}{2 \, {\left (a^{2} - b^{2}\right )} \cos \relax (x)^{2} - a^{2} + b^{2}}\right ) + {\left (a^{2} - b^{2}\right )} \log \left (-4 \, b^{2} \cos \relax (x)^{4} + 4 \, b^{2} \cos \relax (x)^{2} + 4 \, a b \cos \relax (x) \sin \relax (x) + a^{2}\right )}{8 \, {\left (a^{2} b - b^{3}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)^2/(a+b*sin(2*x)),x, algorithm="fricas")

[Out]

[-1/8*(sqrt(-a^2 + b^2)*b*log(-(4*(2*a^2 - b^2)*cos(x)^4 - 4*a*b*cos(x)*sin(x) - 4*(2*a^2 - b^2)*cos(x)^2 + a^
2 - 2*b^2 + 2*(2*b*cos(x)^2 + 2*(2*a*cos(x)^3 - a*cos(x))*sin(x) - b)*sqrt(-a^2 + b^2))/(4*b^2*cos(x)^4 - 4*b^
2*cos(x)^2 - 4*a*b*cos(x)*sin(x) - a^2)) + (a^2 - b^2)*log(-4*b^2*cos(x)^4 + 4*b^2*cos(x)^2 + 4*a*b*cos(x)*sin
(x) + a^2))/(a^2*b - b^3), -1/8*(2*sqrt(a^2 - b^2)*b*arctan(-(2*a*cos(x)*sin(x) + b)*sqrt(a^2 - b^2)/(2*(a^2 -
b^2)*cos(x)^2 - a^2 + b^2)) + (a^2 - b^2)*log(-4*b^2*cos(x)^4 + 4*b^2*cos(x)^2 + 4*a*b*cos(x)*sin(x) + a^2))/
(a^2*b - b^3)]

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giac [A]  time = 0.16, size = 77, normalized size = 1.40 \[ \frac {\pi \left \lfloor \frac {x}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (a) + \arctan \left (\frac {a \tan \relax (x) + b}{\sqrt {a^{2} - b^{2}}}\right )}{2 \, \sqrt {a^{2} - b^{2}}} - \frac {\log \left (a \tan \relax (x)^{2} + 2 \, b \tan \relax (x) + a\right )}{4 \, b} + \frac {\log \left (\tan \relax (x)^{2} + 1\right )}{4 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)^2/(a+b*sin(2*x)),x, algorithm="giac")

[Out]

1/2*(pi*floor(x/pi + 1/2)*sgn(a) + arctan((a*tan(x) + b)/sqrt(a^2 - b^2)))/sqrt(a^2 - b^2) - 1/4*log(a*tan(x)^
2 + 2*b*tan(x) + a)/b + 1/4*log(tan(x)^2 + 1)/b

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maple [A]  time = 0.19, size = 69, normalized size = 1.25 \[ -\frac {\ln \left (a +2 b \tan \relax (x )+a \left (\tan ^{2}\relax (x )\right )\right )}{4 b}+\frac {\arctan \left (\frac {2 a \tan \relax (x )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{2 \sqrt {a^{2}-b^{2}}}+\frac {\ln \left (1+\tan ^{2}\relax (x )\right )}{4 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(x)^2/(a+b*sin(2*x)),x)

[Out]

-1/4*ln(a+2*b*tan(x)+a*tan(x)^2)/b+1/2/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(x)+2*b)/(a^2-b^2)^(1/2))+1/4/b*ln(1
+tan(x)^2)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)^2/(a+b*sin(2*x)),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?`
for more details)Is 4*b^2-4*a^2 positive or negative?

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mupad [B]  time = 4.18, size = 1108, normalized size = 20.15 \[ \frac {\ln \left ({\mathrm {tan}\relax (x)}^2+1\right )}{4\,b}+\frac {\mathrm {atan}\left (\frac {2\,\mathrm {tan}\relax (x)\,{\left (a^2-b^2\right )}^{3/2}\,\left (\frac {\left (4\,a^2\,b-2\,b^3\right )\,\left (2\,a\,b-\frac {\frac {8\,a\,b^3+\frac {\left (8\,a^2\,b-8\,b^3\right )\,\left (96\,a\,b^4-64\,a^3\,b^2\right )}{2\,\left (16\,b^4-16\,a^2\,b^2\right )}}{4\,\sqrt {a^2-b^2}}+\frac {\left (8\,a^2\,b-8\,b^3\right )\,\left (96\,a\,b^4-64\,a^3\,b^2\right )}{8\,\sqrt {a^2-b^2}\,\left (16\,b^4-16\,a^2\,b^2\right )}}{4\,\sqrt {a^2-b^2}}+\frac {\left (8\,a^2\,b-8\,b^3\right )\,\left (4\,a^3-16\,a\,b^2+\frac {\left (8\,a^2\,b-8\,b^3\right )\,\left (8\,a\,b^3+\frac {\left (8\,a^2\,b-8\,b^3\right )\,\left (96\,a\,b^4-64\,a^3\,b^2\right )}{2\,\left (16\,b^4-16\,a^2\,b^2\right )}\right )}{2\,\left (16\,b^4-16\,a^2\,b^2\right )}\right )}{2\,\left (16\,b^4-16\,a^2\,b^2\right )}-\frac {\left (8\,a^2\,b-8\,b^3\right )\,\left (96\,a\,b^4-64\,a^3\,b^2\right )}{32\,\left (a^2-b^2\right )\,\left (16\,b^4-16\,a^2\,b^2\right )}\right )}{a^3\,{\left (4\,a^2-3\,b^2\right )}^2}-\frac {\left (4\,a^4-5\,a^2\,b^2+2\,b^4\right )\,\left (\frac {4\,a^3-16\,a\,b^2+\frac {\left (8\,a^2\,b-8\,b^3\right )\,\left (8\,a\,b^3+\frac {\left (8\,a^2\,b-8\,b^3\right )\,\left (96\,a\,b^4-64\,a^3\,b^2\right )}{2\,\left (16\,b^4-16\,a^2\,b^2\right )}\right )}{2\,\left (16\,b^4-16\,a^2\,b^2\right )}}{4\,\sqrt {a^2-b^2}}-\frac {96\,a\,b^4-64\,a^3\,b^2}{64\,{\left (a^2-b^2\right )}^{3/2}}+\frac {\left (8\,a^2\,b-8\,b^3\right )\,\left (\frac {8\,a\,b^3+\frac {\left (8\,a^2\,b-8\,b^3\right )\,\left (96\,a\,b^4-64\,a^3\,b^2\right )}{2\,\left (16\,b^4-16\,a^2\,b^2\right )}}{4\,\sqrt {a^2-b^2}}+\frac {\left (8\,a^2\,b-8\,b^3\right )\,\left (96\,a\,b^4-64\,a^3\,b^2\right )}{8\,\sqrt {a^2-b^2}\,\left (16\,b^4-16\,a^2\,b^2\right )}\right )}{2\,\left (16\,b^4-16\,a^2\,b^2\right )}\right )}{a^3\,\sqrt {a^2-b^2}\,{\left (4\,a^2-3\,b^2\right )}^2}\right )}{a}+\frac {2\,\left (a^2-b^2\right )\,\left (\frac {6\,a^2\,b-\frac {8\,a^2\,b^3\,{\left (8\,a^2\,b-8\,b^3\right )}^2}{{\left (16\,b^4-16\,a^2\,b^2\right )}^2}}{4\,\sqrt {a^2-b^2}}+\frac {a^2\,b^3}{2\,{\left (a^2-b^2\right )}^{3/2}}-\frac {4\,a^2\,b^3\,{\left (8\,a^2\,b-8\,b^3\right )}^2}{\sqrt {a^2-b^2}\,{\left (16\,b^4-16\,a^2\,b^2\right )}^2}\right )\,\left (4\,a^4-5\,a^2\,b^2+2\,b^4\right )}{a^4\,{\left (4\,a^2-3\,b^2\right )}^2}-\frac {2\,\left (4\,a^2\,b-2\,b^3\right )\,{\left (a^2-b^2\right )}^{3/2}\,\left (\frac {\left (8\,a^2\,b-8\,b^3\right )\,\left (6\,a^2\,b-\frac {8\,a^2\,b^3\,{\left (8\,a^2\,b-8\,b^3\right )}^2}{{\left (16\,b^4-16\,a^2\,b^2\right )}^2}\right )}{2\,\left (16\,b^4-16\,a^2\,b^2\right )}-a^2+\frac {3\,a^2\,b^3\,\left (8\,a^2\,b-8\,b^3\right )}{\left (a^2-b^2\right )\,\left (16\,b^4-16\,a^2\,b^2\right )}\right )}{a^4\,{\left (4\,a^2-3\,b^2\right )}^2}\right )}{2\,\sqrt {a^2-b^2}}+\frac {\ln \left (a\,{\mathrm {tan}\relax (x)}^2+2\,b\,\mathrm {tan}\relax (x)+a\right )\,\left (8\,a^2\,b-8\,b^3\right )}{2\,\left (16\,b^4-16\,a^2\,b^2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(x)^2/(a + b*sin(2*x)),x)

[Out]

log(tan(x)^2 + 1)/(4*b) + atan((2*tan(x)*(a^2 - b^2)^(3/2)*(((4*a^2*b - 2*b^3)*(2*a*b - ((8*a*b^3 + ((8*a^2*b
- 8*b^3)*(96*a*b^4 - 64*a^3*b^2))/(2*(16*b^4 - 16*a^2*b^2)))/(4*(a^2 - b^2)^(1/2)) + ((8*a^2*b - 8*b^3)*(96*a*
b^4 - 64*a^3*b^2))/(8*(a^2 - b^2)^(1/2)*(16*b^4 - 16*a^2*b^2)))/(4*(a^2 - b^2)^(1/2)) + ((8*a^2*b - 8*b^3)*(4*
a^3 - 16*a*b^2 + ((8*a^2*b - 8*b^3)*(8*a*b^3 + ((8*a^2*b - 8*b^3)*(96*a*b^4 - 64*a^3*b^2))/(2*(16*b^4 - 16*a^2
*b^2))))/(2*(16*b^4 - 16*a^2*b^2))))/(2*(16*b^4 - 16*a^2*b^2)) - ((8*a^2*b - 8*b^3)*(96*a*b^4 - 64*a^3*b^2))/(
32*(a^2 - b^2)*(16*b^4 - 16*a^2*b^2))))/(a^3*(4*a^2 - 3*b^2)^2) - ((4*a^4 + 2*b^4 - 5*a^2*b^2)*((4*a^3 - 16*a*
b^2 + ((8*a^2*b - 8*b^3)*(8*a*b^3 + ((8*a^2*b - 8*b^3)*(96*a*b^4 - 64*a^3*b^2))/(2*(16*b^4 - 16*a^2*b^2))))/(2
*(16*b^4 - 16*a^2*b^2)))/(4*(a^2 - b^2)^(1/2)) - (96*a*b^4 - 64*a^3*b^2)/(64*(a^2 - b^2)^(3/2)) + ((8*a^2*b -
8*b^3)*((8*a*b^3 + ((8*a^2*b - 8*b^3)*(96*a*b^4 - 64*a^3*b^2))/(2*(16*b^4 - 16*a^2*b^2)))/(4*(a^2 - b^2)^(1/2)
) + ((8*a^2*b - 8*b^3)*(96*a*b^4 - 64*a^3*b^2))/(8*(a^2 - b^2)^(1/2)*(16*b^4 - 16*a^2*b^2))))/(2*(16*b^4 - 16*
a^2*b^2))))/(a^3*(a^2 - b^2)^(1/2)*(4*a^2 - 3*b^2)^2)))/a + (2*(a^2 - b^2)*((6*a^2*b - (8*a^2*b^3*(8*a^2*b - 8
*b^3)^2)/(16*b^4 - 16*a^2*b^2)^2)/(4*(a^2 - b^2)^(1/2)) + (a^2*b^3)/(2*(a^2 - b^2)^(3/2)) - (4*a^2*b^3*(8*a^2*
b - 8*b^3)^2)/((a^2 - b^2)^(1/2)*(16*b^4 - 16*a^2*b^2)^2))*(4*a^4 + 2*b^4 - 5*a^2*b^2))/(a^4*(4*a^2 - 3*b^2)^2
) - (2*(4*a^2*b - 2*b^3)*(a^2 - b^2)^(3/2)*(((8*a^2*b - 8*b^3)*(6*a^2*b - (8*a^2*b^3*(8*a^2*b - 8*b^3)^2)/(16*
b^4 - 16*a^2*b^2)^2))/(2*(16*b^4 - 16*a^2*b^2)) - a^2 + (3*a^2*b^3*(8*a^2*b - 8*b^3))/((a^2 - b^2)*(16*b^4 - 1
6*a^2*b^2))))/(a^4*(4*a^2 - 3*b^2)^2))/(2*(a^2 - b^2)^(1/2)) + (log(a + a*tan(x)^2 + 2*b*tan(x))*(8*a^2*b - 8*
b^3))/(2*(16*b^4 - 16*a^2*b^2))

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sympy [A]  time = 9.31, size = 155, normalized size = 2.82 \[ - \begin {cases} \frac {\log {\left (\frac {a}{b} + \sin {\left (2 x \right )} \right )}}{4 b} & \text {for}\: b \neq 0 \\\frac {\sin {\left (2 x \right )}}{4 a} & \text {otherwise} \end {cases} + \begin {cases} \frac {\sqrt {b^{2}}}{2 b^{2} \tan {\relax (x )} - 2 b \sqrt {b^{2}}} & \text {for}\: a = - \sqrt {b^{2}} \\- \frac {\sqrt {b^{2}}}{2 b^{2} \tan {\relax (x )} + 2 b \sqrt {b^{2}}} & \text {for}\: a = \sqrt {b^{2}} \\\frac {\log {\left (\tan {\relax (x )} \right )}}{4 b} & \text {for}\: a = 0 \\\frac {\log {\left (\tan {\relax (x )} + \frac {b}{a} - \frac {\sqrt {- a^{2} + b^{2}}}{a} \right )}}{4 \sqrt {- a^{2} + b^{2}}} - \frac {\log {\left (\tan {\relax (x )} + \frac {b}{a} + \frac {\sqrt {- a^{2} + b^{2}}}{a} \right )}}{4 \sqrt {- a^{2} + b^{2}}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)**2/(a+b*sin(2*x)),x)

[Out]

-Piecewise((log(a/b + sin(2*x))/(4*b), Ne(b, 0)), (sin(2*x)/(4*a), True)) + Piecewise((sqrt(b**2)/(2*b**2*tan(
x) - 2*b*sqrt(b**2)), Eq(a, -sqrt(b**2))), (-sqrt(b**2)/(2*b**2*tan(x) + 2*b*sqrt(b**2)), Eq(a, sqrt(b**2))),
(log(tan(x))/(4*b), Eq(a, 0)), (log(tan(x) + b/a - sqrt(-a**2 + b**2)/a)/(4*sqrt(-a**2 + b**2)) - log(tan(x) +
b/a + sqrt(-a**2 + b**2)/a)/(4*sqrt(-a**2 + b**2)), True))

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