3.1.43 \(\int \frac {1}{(a+b \cos ^2(x))^3} \, dx\) [43]

Optimal. Leaf size=107 \[ -\frac {\left (8 a^2+8 a b+3 b^2\right ) \text {ArcTan}\left (\frac {\sqrt {a+b} \cot (x)}{\sqrt {a}}\right )}{8 a^{5/2} (a+b)^{5/2}}-\frac {b \cos (x) \sin (x)}{4 a (a+b) \left (a+b \cos ^2(x)\right )^2}-\frac {3 b (2 a+b) \cos (x) \sin (x)}{8 a^2 (a+b)^2 \left (a+b \cos ^2(x)\right )} \]

[Out]

-1/8*(8*a^2+8*a*b+3*b^2)*arctan(cot(x)*(a+b)^(1/2)/a^(1/2))/a^(5/2)/(a+b)^(5/2)-1/4*b*cos(x)*sin(x)/a/(a+b)/(a
+b*cos(x)^2)^2-3/8*b*(2*a+b)*cos(x)*sin(x)/a^2/(a+b)^2/(a+b*cos(x)^2)

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Rubi [A]
time = 0.08, antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3263, 3252, 12, 3260, 211} \begin {gather*} -\frac {3 b (2 a+b) \sin (x) \cos (x)}{8 a^2 (a+b)^2 \left (a+b \cos ^2(x)\right )}-\frac {\left (8 a^2+8 a b+3 b^2\right ) \text {ArcTan}\left (\frac {\sqrt {a+b} \cot (x)}{\sqrt {a}}\right )}{8 a^{5/2} (a+b)^{5/2}}-\frac {b \sin (x) \cos (x)}{4 a (a+b) \left (a+b \cos ^2(x)\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[x]^2)^(-3),x]

[Out]

-1/8*((8*a^2 + 8*a*b + 3*b^2)*ArcTan[(Sqrt[a + b]*Cot[x])/Sqrt[a]])/(a^(5/2)*(a + b)^(5/2)) - (b*Cos[x]*Sin[x]
)/(4*a*(a + b)*(a + b*Cos[x]^2)^2) - (3*b*(2*a + b)*Cos[x]*Sin[x])/(8*a^2*(a + b)^2*(a + b*Cos[x]^2))

Rule 12

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

Rule 211

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

Rule 3252

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp
[(-(A*b - a*B))*Cos[e + f*x]*Sin[e + f*x]*((a + b*Sin[e + f*x]^2)^(p + 1)/(2*a*f*(a + b)*(p + 1))), x] - Dist[
1/(2*a*(a + b)*(p + 1)), Int[(a + b*Sin[e + f*x]^2)^(p + 1)*Simp[a*B - A*(2*a*(p + 1) + b*(2*p + 3)) + 2*(A*b
- a*B)*(p + 2)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, B}, x] && LtQ[p, -1] && NeQ[a + b, 0]

Rule 3260

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(-1), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist
[ff/f, Subst[Int[1/(a + (a + b)*ff^2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x]

Rule 3263

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Simp[(-b)*Cos[e + f*x]*Sin[e + f*x]*((a + b*Si
n[e + f*x]^2)^(p + 1)/(2*a*f*(p + 1)*(a + b))), x] + Dist[1/(2*a*(p + 1)*(a + b)), Int[(a + b*Sin[e + f*x]^2)^
(p + 1)*Simp[2*a*(p + 1) + b*(2*p + 3) - 2*b*(p + 2)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f}, x] && N
eQ[a + b, 0] && LtQ[p, -1]

Rubi steps

\begin {align*} \int \frac {1}{\left (a+b \cos ^2(x)\right )^3} \, dx &=-\frac {b \cos (x) \sin (x)}{4 a (a+b) \left (a+b \cos ^2(x)\right )^2}-\frac {\int \frac {-4 a-3 b+2 b \cos ^2(x)}{\left (a+b \cos ^2(x)\right )^2} \, dx}{4 a (a+b)}\\ &=-\frac {b \cos (x) \sin (x)}{4 a (a+b) \left (a+b \cos ^2(x)\right )^2}-\frac {3 b (2 a+b) \cos (x) \sin (x)}{8 a^2 (a+b)^2 \left (a+b \cos ^2(x)\right )}-\frac {\int \frac {-8 a^2-8 a b-3 b^2}{a+b \cos ^2(x)} \, dx}{8 a^2 (a+b)^2}\\ &=-\frac {b \cos (x) \sin (x)}{4 a (a+b) \left (a+b \cos ^2(x)\right )^2}-\frac {3 b (2 a+b) \cos (x) \sin (x)}{8 a^2 (a+b)^2 \left (a+b \cos ^2(x)\right )}+\frac {\left (8 a^2+8 a b+3 b^2\right ) \int \frac {1}{a+b \cos ^2(x)} \, dx}{8 a^2 (a+b)^2}\\ &=-\frac {b \cos (x) \sin (x)}{4 a (a+b) \left (a+b \cos ^2(x)\right )^2}-\frac {3 b (2 a+b) \cos (x) \sin (x)}{8 a^2 (a+b)^2 \left (a+b \cos ^2(x)\right )}-\frac {\left (8 a^2+8 a b+3 b^2\right ) \text {Subst}\left (\int \frac {1}{a+(a+b) x^2} \, dx,x,\cot (x)\right )}{8 a^2 (a+b)^2}\\ &=-\frac {\left (8 a^2+8 a b+3 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {a+b} \cot (x)}{\sqrt {a}}\right )}{8 a^{5/2} (a+b)^{5/2}}-\frac {b \cos (x) \sin (x)}{4 a (a+b) \left (a+b \cos ^2(x)\right )^2}-\frac {3 b (2 a+b) \cos (x) \sin (x)}{8 a^2 (a+b)^2 \left (a+b \cos ^2(x)\right )}\\ \end {align*}

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Mathematica [A]
time = 0.74, size = 106, normalized size = 0.99 \begin {gather*} \frac {\frac {\left (8 a^2+8 a b+3 b^2\right ) \text {ArcTan}\left (\frac {\sqrt {a} \tan (x)}{\sqrt {a+b}}\right )}{(a+b)^{5/2}}-\frac {\sqrt {a} b \left (16 a^2+16 a b+3 b^2+3 b (2 a+b) \cos (2 x)\right ) \sin (2 x)}{(a+b)^2 (2 a+b+b \cos (2 x))^2}}{8 a^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cos[x]^2)^(-3),x]

[Out]

(((8*a^2 + 8*a*b + 3*b^2)*ArcTan[(Sqrt[a]*Tan[x])/Sqrt[a + b]])/(a + b)^(5/2) - (Sqrt[a]*b*(16*a^2 + 16*a*b +
3*b^2 + 3*b*(2*a + b)*Cos[2*x])*Sin[2*x])/((a + b)^2*(2*a + b + b*Cos[2*x])^2))/(8*a^(5/2))

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Maple [A]
time = 0.22, size = 117, normalized size = 1.09

method	result	size
default	\(\frac {-\frac {b \left (8 a +5 b \right ) \left (\tan ^{3}\left (x \right )\right )}{8 a \left (a^{2}+2 a b +b^{2}\right )}-\frac {\left (8 a +3 b \right ) b \tan \left (x \right )}{8 a^{2} \left (a +b \right )}}{\left (a \left (\tan ^{2}\left (x \right )\right )+a +b \right )^{2}}+\frac {\left (8 a^{2}+8 a b +3 b^{2}\right ) \arctan \left (\frac {a \tan \left (x \right )}{\sqrt {\left (a +b \right ) a}}\right )}{8 \left (a^{2}+2 a b +b^{2}\right ) a^{2} \sqrt {\left (a +b \right ) a}}\)	\(117\)
risch	\(-\frac {i \left (8 a^{2} b \,{\mathrm e}^{6 i x}+8 a \,b^{2} {\mathrm e}^{6 i x}+3 b^{3} {\mathrm e}^{6 i x}+48 a^{3} {\mathrm e}^{4 i x}+72 a^{2} b \,{\mathrm e}^{4 i x}+42 a \,b^{2} {\mathrm e}^{4 i x}+9 b^{3} {\mathrm e}^{4 i x}+40 a^{2} b \,{\mathrm e}^{2 i x}+40 a \,b^{2} {\mathrm e}^{2 i x}+9 b^{3} {\mathrm e}^{2 i x}+6 b^{2} a +3 b^{3}\right )}{4 \left (a +b \right )^{2} a^{2} \left (b \,{\mathrm e}^{4 i x}+4 a \,{\mathrm e}^{2 i x}+2 b \,{\mathrm e}^{2 i x}+b \right )^{2}}-\frac {\ln \left ({\mathrm e}^{2 i x}+\frac {2 i a^{2}+2 i a b +2 a \sqrt {-a^{2}-a b}+b \sqrt {-a^{2}-a b}}{b \sqrt {-a^{2}-a b}}\right )}{2 \sqrt {-a^{2}-a b}\, \left (a +b \right )^{2}}-\frac {\ln \left ({\mathrm e}^{2 i x}+\frac {2 i a^{2}+2 i a b +2 a \sqrt {-a^{2}-a b}+b \sqrt {-a^{2}-a b}}{b \sqrt {-a^{2}-a b}}\right ) b}{2 \sqrt {-a^{2}-a b}\, \left (a +b \right )^{2} a}-\frac {3 \ln \left ({\mathrm e}^{2 i x}+\frac {2 i a^{2}+2 i a b +2 a \sqrt {-a^{2}-a b}+b \sqrt {-a^{2}-a b}}{b \sqrt {-a^{2}-a b}}\right ) b^{2}}{16 \sqrt {-a^{2}-a b}\, \left (a +b \right )^{2} a^{2}}+\frac {\ln \left ({\mathrm e}^{2 i x}+\frac {-2 i a^{2}-2 i a b +2 a \sqrt {-a^{2}-a b}+b \sqrt {-a^{2}-a b}}{b \sqrt {-a^{2}-a b}}\right )}{2 \sqrt {-a^{2}-a b}\, \left (a +b \right )^{2}}+\frac {\ln \left ({\mathrm e}^{2 i x}+\frac {-2 i a^{2}-2 i a b +2 a \sqrt {-a^{2}-a b}+b \sqrt {-a^{2}-a b}}{b \sqrt {-a^{2}-a b}}\right ) b}{2 \sqrt {-a^{2}-a b}\, \left (a +b \right )^{2} a}+\frac {3 \ln \left ({\mathrm e}^{2 i x}+\frac {-2 i a^{2}-2 i a b +2 a \sqrt {-a^{2}-a b}+b \sqrt {-a^{2}-a b}}{b \sqrt {-a^{2}-a b}}\right ) b^{2}}{16 \sqrt {-a^{2}-a b}\, \left (a +b \right )^{2} a^{2}}\)	\(676\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*cos(x)^2)^3,x,method=_RETURNVERBOSE)

[Out]

(-1/8*b*(8*a+5*b)/a/(a^2+2*a*b+b^2)*tan(x)^3-1/8*(8*a+3*b)/a^2*b/(a+b)*tan(x))/(a*tan(x)^2+a+b)^2+1/8*(8*a^2+8
*a*b+3*b^2)/(a^2+2*a*b+b^2)/a^2/((a+b)*a)^(1/2)*arctan(a*tan(x)/((a+b)*a)^(1/2))

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Maxima [A]
time = 0.47, size = 186, normalized size = 1.74 \begin {gather*} \frac {{\left (8 \, a^{2} + 8 \, a b + 3 \, b^{2}\right )} \arctan \left (\frac {a \tan \left (x\right )}{\sqrt {{\left (a + b\right )} a}}\right )}{8 \, {\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )} \sqrt {{\left (a + b\right )} a}} - \frac {{\left (8 \, a^{2} b + 5 \, a b^{2}\right )} \tan \left (x\right )^{3} + {\left (8 \, a^{2} b + 11 \, a b^{2} + 3 \, b^{3}\right )} \tan \left (x\right )}{8 \, {\left (a^{6} + 4 \, a^{5} b + 6 \, a^{4} b^{2} + 4 \, a^{3} b^{3} + a^{2} b^{4} + {\left (a^{6} + 2 \, a^{5} b + a^{4} b^{2}\right )} \tan \left (x\right )^{4} + 2 \, {\left (a^{6} + 3 \, a^{5} b + 3 \, a^{4} b^{2} + a^{3} b^{3}\right )} \tan \left (x\right )^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cos(x)^2)^3,x, algorithm="maxima")

[Out]

1/8*(8*a^2 + 8*a*b + 3*b^2)*arctan(a*tan(x)/sqrt((a + b)*a))/((a^4 + 2*a^3*b + a^2*b^2)*sqrt((a + b)*a)) - 1/8
*((8*a^2*b + 5*a*b^2)*tan(x)^3 + (8*a^2*b + 11*a*b^2 + 3*b^3)*tan(x))/(a^6 + 4*a^5*b + 6*a^4*b^2 + 4*a^3*b^3 +
a^2*b^4 + (a^6 + 2*a^5*b + a^4*b^2)*tan(x)^4 + 2*(a^6 + 3*a^5*b + 3*a^4*b^2 + a^3*b^3)*tan(x)^2)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 276 vs. \(2 (93) = 186\).
time = 0.47, size = 616, normalized size = 5.76 \begin {gather*} \left [-\frac {{\left ({\left (8 \, a^{2} b^{2} + 8 \, a b^{3} + 3 \, b^{4}\right )} \cos \left (x\right )^{4} + 8 \, a^{4} + 8 \, a^{3} b + 3 \, a^{2} b^{2} + 2 \, {\left (8 \, a^{3} b + 8 \, a^{2} b^{2} + 3 \, a b^{3}\right )} \cos \left (x\right )^{2}\right )} \sqrt {-a^{2} - a b} \log \left (\frac {{\left (8 \, a^{2} + 8 \, a b + b^{2}\right )} \cos \left (x\right )^{4} - 2 \, {\left (4 \, a^{2} + 3 \, a b\right )} \cos \left (x\right )^{2} + 4 \, {\left ({\left (2 \, a + b\right )} \cos \left (x\right )^{3} - a \cos \left (x\right )\right )} \sqrt {-a^{2} - a b} \sin \left (x\right ) + a^{2}}{b^{2} \cos \left (x\right )^{4} + 2 \, a b \cos \left (x\right )^{2} + a^{2}}\right ) + 4 \, {\left (3 \, {\left (2 \, a^{3} b^{2} + 3 \, a^{2} b^{3} + a b^{4}\right )} \cos \left (x\right )^{3} + {\left (8 \, a^{4} b + 13 \, a^{3} b^{2} + 5 \, a^{2} b^{3}\right )} \cos \left (x\right )\right )} \sin \left (x\right )}{32 \, {\left (a^{8} + 3 \, a^{7} b + 3 \, a^{6} b^{2} + a^{5} b^{3} + {\left (a^{6} b^{2} + 3 \, a^{5} b^{3} + 3 \, a^{4} b^{4} + a^{3} b^{5}\right )} \cos \left (x\right )^{4} + 2 \, {\left (a^{7} b + 3 \, a^{6} b^{2} + 3 \, a^{5} b^{3} + a^{4} b^{4}\right )} \cos \left (x\right )^{2}\right )}}, -\frac {{\left ({\left (8 \, a^{2} b^{2} + 8 \, a b^{3} + 3 \, b^{4}\right )} \cos \left (x\right )^{4} + 8 \, a^{4} + 8 \, a^{3} b + 3 \, a^{2} b^{2} + 2 \, {\left (8 \, a^{3} b + 8 \, a^{2} b^{2} + 3 \, a b^{3}\right )} \cos \left (x\right )^{2}\right )} \sqrt {a^{2} + a b} \arctan \left (\frac {{\left (2 \, a + b\right )} \cos \left (x\right )^{2} - a}{2 \, \sqrt {a^{2} + a b} \cos \left (x\right ) \sin \left (x\right )}\right ) + 2 \, {\left (3 \, {\left (2 \, a^{3} b^{2} + 3 \, a^{2} b^{3} + a b^{4}\right )} \cos \left (x\right )^{3} + {\left (8 \, a^{4} b + 13 \, a^{3} b^{2} + 5 \, a^{2} b^{3}\right )} \cos \left (x\right )\right )} \sin \left (x\right )}{16 \, {\left (a^{8} + 3 \, a^{7} b + 3 \, a^{6} b^{2} + a^{5} b^{3} + {\left (a^{6} b^{2} + 3 \, a^{5} b^{3} + 3 \, a^{4} b^{4} + a^{3} b^{5}\right )} \cos \left (x\right )^{4} + 2 \, {\left (a^{7} b + 3 \, a^{6} b^{2} + 3 \, a^{5} b^{3} + a^{4} b^{4}\right )} \cos \left (x\right )^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cos(x)^2)^3,x, algorithm="fricas")

[Out]

[-1/32*(((8*a^2*b^2 + 8*a*b^3 + 3*b^4)*cos(x)^4 + 8*a^4 + 8*a^3*b + 3*a^2*b^2 + 2*(8*a^3*b + 8*a^2*b^2 + 3*a*b
^3)*cos(x)^2)*sqrt(-a^2 - a*b)*log(((8*a^2 + 8*a*b + b^2)*cos(x)^4 - 2*(4*a^2 + 3*a*b)*cos(x)^2 + 4*((2*a + b)
*cos(x)^3 - a*cos(x))*sqrt(-a^2 - a*b)*sin(x) + a^2)/(b^2*cos(x)^4 + 2*a*b*cos(x)^2 + a^2)) + 4*(3*(2*a^3*b^2
+ 3*a^2*b^3 + a*b^4)*cos(x)^3 + (8*a^4*b + 13*a^3*b^2 + 5*a^2*b^3)*cos(x))*sin(x))/(a^8 + 3*a^7*b + 3*a^6*b^2
+ a^5*b^3 + (a^6*b^2 + 3*a^5*b^3 + 3*a^4*b^4 + a^3*b^5)*cos(x)^4 + 2*(a^7*b + 3*a^6*b^2 + 3*a^5*b^3 + a^4*b^4)
*cos(x)^2), -1/16*(((8*a^2*b^2 + 8*a*b^3 + 3*b^4)*cos(x)^4 + 8*a^4 + 8*a^3*b + 3*a^2*b^2 + 2*(8*a^3*b + 8*a^2*
b^2 + 3*a*b^3)*cos(x)^2)*sqrt(a^2 + a*b)*arctan(1/2*((2*a + b)*cos(x)^2 - a)/(sqrt(a^2 + a*b)*cos(x)*sin(x)))
+ 2*(3*(2*a^3*b^2 + 3*a^2*b^3 + a*b^4)*cos(x)^3 + (8*a^4*b + 13*a^3*b^2 + 5*a^2*b^3)*cos(x))*sin(x))/(a^8 + 3*
a^7*b + 3*a^6*b^2 + a^5*b^3 + (a^6*b^2 + 3*a^5*b^3 + 3*a^4*b^4 + a^3*b^5)*cos(x)^4 + 2*(a^7*b + 3*a^6*b^2 + 3*
a^5*b^3 + a^4*b^4)*cos(x)^2)]

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cos(x)**2)**3,x)

[Out]

Timed out

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Giac [A]
time = 0.40, size = 149, normalized size = 1.39 \begin {gather*} \frac {{\left (\pi \left \lfloor \frac {x}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (x\right )}{\sqrt {a^{2} + a b}}\right )\right )} {\left (8 \, a^{2} + 8 \, a b + 3 \, b^{2}\right )}}{8 \, {\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )} \sqrt {a^{2} + a b}} - \frac {8 \, a^{2} b \tan \left (x\right )^{3} + 5 \, a b^{2} \tan \left (x\right )^{3} + 8 \, a^{2} b \tan \left (x\right ) + 11 \, a b^{2} \tan \left (x\right ) + 3 \, b^{3} \tan \left (x\right )}{8 \, {\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )} {\left (a \tan \left (x\right )^{2} + a + b\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cos(x)^2)^3,x, algorithm="giac")

[Out]

1/8*(pi*floor(x/pi + 1/2)*sgn(a) + arctan(a*tan(x)/sqrt(a^2 + a*b)))*(8*a^2 + 8*a*b + 3*b^2)/((a^4 + 2*a^3*b +
a^2*b^2)*sqrt(a^2 + a*b)) - 1/8*(8*a^2*b*tan(x)^3 + 5*a*b^2*tan(x)^3 + 8*a^2*b*tan(x) + 11*a*b^2*tan(x) + 3*b
^3*tan(x))/((a^4 + 2*a^3*b + a^2*b^2)*(a*tan(x)^2 + a + b)^2)

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Mupad [B]
time = 2.44, size = 123, normalized size = 1.15 \begin {gather*} \frac {\mathrm {atan}\left (\frac {\sqrt {a}\,\mathrm {tan}\left (x\right )}{\sqrt {a+b}}\right )\,\left (8\,a^2+8\,a\,b+3\,b^2\right )}{8\,a^{5/2}\,{\left (a+b\right )}^{5/2}}-\frac {\frac {\mathrm {tan}\left (x\right )\,\left (3\,b^2+8\,a\,b\right )}{8\,a^2\,\left (a+b\right )}+\frac {{\mathrm {tan}\left (x\right )}^3\,\left (5\,b^2+8\,a\,b\right )}{8\,a\,{\left (a+b\right )}^2}}{2\,a\,b+{\mathrm {tan}\left (x\right )}^2\,\left (2\,a^2+2\,b\,a\right )+a^2\,{\mathrm {tan}\left (x\right )}^4+a^2+b^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*cos(x)^2)^3,x)

[Out]

(atan((a^(1/2)*tan(x))/(a + b)^(1/2))*(8*a*b + 8*a^2 + 3*b^2))/(8*a^(5/2)*(a + b)^(5/2)) - ((tan(x)*(8*a*b + 3
*b^2))/(8*a^2*(a + b)) + (tan(x)^3*(8*a*b + 5*b^2))/(8*a*(a + b)^2))/(2*a*b + tan(x)^2*(2*a*b + 2*a^2) + a^2*t
an(x)^4 + a^2 + b^2)

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