3.2.38 \(\int \frac {a+b x^2+c x^4}{\sqrt {d-e x} \sqrt {d+e x}} \, dx\)

Optimal. Leaf size=128 \[ -\frac {\tan ^{-1}\left (\frac {\sqrt {d-e x}}{\sqrt {d+e x}}\right ) \left (8 a e^4+4 b d^2 e^2+3 c d^4\right )}{4 e^5}-\frac {x \sqrt {d-e x} \sqrt {d+e x} \left (4 b e^2+3 c d^2\right )}{8 e^4}+\frac {c x^3 (e x-d) \sqrt {d+e x}}{4 e^2 \sqrt {d-e x}} \]

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Rubi [A]  time = 0.09, antiderivative size = 179, normalized size of antiderivative = 1.40, number of steps used = 5, number of rules used = 5, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.156, Rules used = {520, 1159, 388, 217, 203} \begin {gather*} \frac {\sqrt {d^2-e^2 x^2} \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right ) \left (8 a e^4+4 b d^2 e^2+3 c d^4\right )}{8 e^5 \sqrt {d-e x} \sqrt {d+e x}}-\frac {x \left (d^2-e^2 x^2\right ) \left (4 b e^2+3 c d^2\right )}{8 e^4 \sqrt {d-e x} \sqrt {d+e x}}-\frac {c x^3 \left (d^2-e^2 x^2\right )}{4 e^2 \sqrt {d-e x} \sqrt {d+e x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2 + c*x^4)/(Sqrt[d - e*x]*Sqrt[d + e*x]),x]

[Out]

-((3*c*d^2 + 4*b*e^2)*x*(d^2 - e^2*x^2))/(8*e^4*Sqrt[d - e*x]*Sqrt[d + e*x]) - (c*x^3*(d^2 - e^2*x^2))/(4*e^2*
Sqrt[d - e*x]*Sqrt[d + e*x]) + ((3*c*d^4 + 4*b*d^2*e^2 + 8*a*e^4)*Sqrt[d^2 - e^2*x^2]*ArcTan[(e*x)/Sqrt[d^2 -
e^2*x^2]])/(8*e^5*Sqrt[d - e*x]*Sqrt[d + e*x])

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 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 388

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

Rule 520

Int[(u_.)*((c_) + (d_.)*(x_)^(n_.) + (e_.)*(x_)^(n2_.))^(q_.)*((a1_) + (b1_.)*(x_)^(non2_.))^(p_.)*((a2_) + (b
2_.)*(x_)^(non2_.))^(p_.), x_Symbol] :> Dist[((a1 + b1*x^(n/2))^FracPart[p]*(a2 + b2*x^(n/2))^FracPart[p])/(a1
*a2 + b1*b2*x^n)^FracPart[p], Int[u*(a1*a2 + b1*b2*x^n)^p*(c + d*x^n + e*x^(2*n))^q, x], x] /; FreeQ[{a1, b1,
a2, b2, c, d, e, n, p, q}, x] && EqQ[non2, n/2] && EqQ[n2, 2*n] && EqQ[a2*b1 + a1*b2, 0]

Rule 1159

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

Rubi steps

\begin {align*} \int \frac {a+b x^2+c x^4}{\sqrt {d-e x} \sqrt {d+e x}} \, dx &=\frac {\sqrt {d^2-e^2 x^2} \int \frac {a+b x^2+c x^4}{\sqrt {d^2-e^2 x^2}} \, dx}{\sqrt {d-e x} \sqrt {d+e x}}\\ &=-\frac {c x^3 \left (d^2-e^2 x^2\right )}{4 e^2 \sqrt {d-e x} \sqrt {d+e x}}-\frac {\sqrt {d^2-e^2 x^2} \int \frac {-4 a e^2-\left (3 c d^2+4 b e^2\right ) x^2}{\sqrt {d^2-e^2 x^2}} \, dx}{4 e^2 \sqrt {d-e x} \sqrt {d+e x}}\\ &=-\frac {\left (3 c d^2+4 b e^2\right ) x \left (d^2-e^2 x^2\right )}{8 e^4 \sqrt {d-e x} \sqrt {d+e x}}-\frac {c x^3 \left (d^2-e^2 x^2\right )}{4 e^2 \sqrt {d-e x} \sqrt {d+e x}}-\frac {\left (\left (-8 a e^4+d^2 \left (-3 c d^2-4 b e^2\right )\right ) \sqrt {d^2-e^2 x^2}\right ) \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx}{8 e^4 \sqrt {d-e x} \sqrt {d+e x}}\\ &=-\frac {\left (3 c d^2+4 b e^2\right ) x \left (d^2-e^2 x^2\right )}{8 e^4 \sqrt {d-e x} \sqrt {d+e x}}-\frac {c x^3 \left (d^2-e^2 x^2\right )}{4 e^2 \sqrt {d-e x} \sqrt {d+e x}}-\frac {\left (\left (-8 a e^4+d^2 \left (-3 c d^2-4 b e^2\right )\right ) \sqrt {d^2-e^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{1+e^2 x^2} \, dx,x,\frac {x}{\sqrt {d^2-e^2 x^2}}\right )}{8 e^4 \sqrt {d-e x} \sqrt {d+e x}}\\ &=-\frac {\left (3 c d^2+4 b e^2\right ) x \left (d^2-e^2 x^2\right )}{8 e^4 \sqrt {d-e x} \sqrt {d+e x}}-\frac {c x^3 \left (d^2-e^2 x^2\right )}{4 e^2 \sqrt {d-e x} \sqrt {d+e x}}+\frac {\left (3 c d^4+4 b d^2 e^2+8 a e^4\right ) \sqrt {d^2-e^2 x^2} \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{8 e^5 \sqrt {d-e x} \sqrt {d+e x}}\\ \end {align*}

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Mathematica [A]  time = 0.56, size = 157, normalized size = 1.23 \begin {gather*} -\frac {16 \tan ^{-1}\left (\frac {\sqrt {d-e x}}{\sqrt {d+e x}}\right ) \left (a e^4+b d^2 e^2+c d^4\right )+e x \sqrt {d-e x} \sqrt {d+e x} \left (4 b e^2+3 c d^2+2 c e^2 x^2\right )-\frac {2 d^{5/2} \sqrt {\frac {e x}{d}+1} \left (4 b e^2+5 c d^2\right ) \sin ^{-1}\left (\frac {\sqrt {d-e x}}{\sqrt {2} \sqrt {d}}\right )}{\sqrt {d+e x}}}{8 e^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2 + c*x^4)/(Sqrt[d - e*x]*Sqrt[d + e*x]),x]

[Out]

-1/8*(e*x*Sqrt[d - e*x]*Sqrt[d + e*x]*(3*c*d^2 + 4*b*e^2 + 2*c*e^2*x^2) - (2*d^(5/2)*(5*c*d^2 + 4*b*e^2)*Sqrt[
1 + (e*x)/d]*ArcSin[Sqrt[d - e*x]/(Sqrt[2]*Sqrt[d])])/Sqrt[d + e*x] + 16*(c*d^4 + b*d^2*e^2 + a*e^4)*ArcTan[Sq
rt[d - e*x]/Sqrt[d + e*x]])/e^5

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IntegrateAlgebraic [A]  time = 0.19, size = 207, normalized size = 1.62 \begin {gather*} \frac {\tan ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d-e x}}\right ) \left (8 a e^4+4 b d^2 e^2+3 c d^4\right )}{4 e^5}-\frac {d^2 \sqrt {d+e x} \left (\frac {d+e x}{d-e x}-1\right ) \left (\frac {4 b e^2 (d+e x)^2}{(d-e x)^2}+\frac {8 b e^2 (d+e x)}{d-e x}+4 b e^2+\frac {5 c d^2 (d+e x)^2}{(d-e x)^2}+\frac {2 c d^2 (d+e x)}{d-e x}+5 c d^2\right )}{4 e^5 \sqrt {d-e x} \left (\frac {d+e x}{d-e x}+1\right )^4} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(a + b*x^2 + c*x^4)/(Sqrt[d - e*x]*Sqrt[d + e*x]),x]

[Out]

-1/4*(d^2*Sqrt[d + e*x]*(-1 + (d + e*x)/(d - e*x))*(5*c*d^2 + 4*b*e^2 + (2*c*d^2*(d + e*x))/(d - e*x) + (8*b*e
^2*(d + e*x))/(d - e*x) + (5*c*d^2*(d + e*x)^2)/(d - e*x)^2 + (4*b*e^2*(d + e*x)^2)/(d - e*x)^2))/(e^5*Sqrt[d
- e*x]*(1 + (d + e*x)/(d - e*x))^4) + ((3*c*d^4 + 4*b*d^2*e^2 + 8*a*e^4)*ArcTan[Sqrt[d + e*x]/Sqrt[d - e*x]])/
(4*e^5)

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fricas [A]  time = 1.25, size = 100, normalized size = 0.78 \begin {gather*} -\frac {{\left (2 \, c e^{3} x^{3} + {\left (3 \, c d^{2} e + 4 \, b e^{3}\right )} x\right )} \sqrt {e x + d} \sqrt {-e x + d} + 2 \, {\left (3 \, c d^{4} + 4 \, b d^{2} e^{2} + 8 \, a e^{4}\right )} \arctan \left (\frac {\sqrt {e x + d} \sqrt {-e x + d} - d}{e x}\right )}{8 \, e^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2+a)/(-e*x+d)^(1/2)/(e*x+d)^(1/2),x, algorithm="fricas")

[Out]

-1/8*((2*c*e^3*x^3 + (3*c*d^2*e + 4*b*e^3)*x)*sqrt(e*x + d)*sqrt(-e*x + d) + 2*(3*c*d^4 + 4*b*d^2*e^2 + 8*a*e^
4)*arctan((sqrt(e*x + d)*sqrt(-e*x + d) - d)/(e*x)))/e^5

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giac [A]  time = 0.59, size = 135, normalized size = 1.05 \begin {gather*} \frac {1}{8} \, {\left (2 \, {\left (3 \, c d^{4} + 4 \, b d^{2} e^{2} + 8 \, a e^{4}\right )} \arcsin \left (\frac {\sqrt {2} \sqrt {x e + d}}{2 \, \sqrt {d}}\right ) e^{\left (-4\right )} - {\left ({\left (2 \, {\left ({\left (x e + d\right )} c e^{\left (-4\right )} - 3 \, c d e^{\left (-4\right )}\right )} {\left (x e + d\right )} + {\left (9 \, c d^{2} e^{16} + 4 \, b e^{18}\right )} e^{\left (-20\right )}\right )} {\left (x e + d\right )} - {\left (5 \, c d^{3} e^{16} + 4 \, b d e^{18}\right )} e^{\left (-20\right )}\right )} \sqrt {x e + d} \sqrt {-x e + d}\right )} e^{\left (-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2+a)/(-e*x+d)^(1/2)/(e*x+d)^(1/2),x, algorithm="giac")

[Out]

1/8*(2*(3*c*d^4 + 4*b*d^2*e^2 + 8*a*e^4)*arcsin(1/2*sqrt(2)*sqrt(x*e + d)/sqrt(d))*e^(-4) - ((2*((x*e + d)*c*e
^(-4) - 3*c*d*e^(-4))*(x*e + d) + (9*c*d^2*e^16 + 4*b*e^18)*e^(-20))*(x*e + d) - (5*c*d^3*e^16 + 4*b*d*e^18)*e
^(-20))*sqrt(x*e + d)*sqrt(-x*e + d))*e^(-1)

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maple [C]  time = 0.02, size = 191, normalized size = 1.49 \begin {gather*} -\frac {\sqrt {-e x +d}\, \sqrt {e x +d}\, \left (2 \sqrt {-e^{2} x^{2}+d^{2}}\, c \,e^{3} x^{3} \mathrm {csgn}\relax (e )-8 a \,e^{4} \arctan \left (\frac {e x \,\mathrm {csgn}\relax (e )}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )-4 b \,d^{2} e^{2} \arctan \left (\frac {e x \,\mathrm {csgn}\relax (e )}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )+4 \sqrt {-e^{2} x^{2}+d^{2}}\, b \,e^{3} x \,\mathrm {csgn}\relax (e )-3 c \,d^{4} \arctan \left (\frac {e x \,\mathrm {csgn}\relax (e )}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )+3 \sqrt {-e^{2} x^{2}+d^{2}}\, c \,d^{2} e x \,\mathrm {csgn}\relax (e )\right ) \mathrm {csgn}\relax (e )}{8 \sqrt {-e^{2} x^{2}+d^{2}}\, e^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^4+b*x^2+a)/(-e*x+d)^(1/2)/(e*x+d)^(1/2),x)

[Out]

-1/8*(-e*x+d)^(1/2)*(e*x+d)^(1/2)*(2*csgn(e)*x^3*c*e^3*(-e^2*x^2+d^2)^(1/2)+4*(-e^2*x^2+d^2)^(1/2)*csgn(e)*e^3
*x*b+3*(-e^2*x^2+d^2)^(1/2)*csgn(e)*e*x*c*d^2-8*arctan(1/(-e^2*x^2+d^2)^(1/2)*e*x*csgn(e))*a*e^4-4*arctan(1/(-
e^2*x^2+d^2)^(1/2)*e*x*csgn(e))*b*d^2*e^2-3*arctan(1/(-e^2*x^2+d^2)^(1/2)*e*x*csgn(e))*c*d^4)*csgn(e)/e^5/(-e^
2*x^2+d^2)^(1/2)

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maxima [A]  time = 1.02, size = 113, normalized size = 0.88 \begin {gather*} -\frac {\sqrt {-e^{2} x^{2} + d^{2}} c x^{3}}{4 \, e^{2}} + \frac {3 \, c d^{4} \arcsin \left (\frac {e x}{d}\right )}{8 \, e^{5}} + \frac {b d^{2} \arcsin \left (\frac {e x}{d}\right )}{2 \, e^{3}} + \frac {a \arcsin \left (\frac {e x}{d}\right )}{e} - \frac {3 \, \sqrt {-e^{2} x^{2} + d^{2}} c d^{2} x}{8 \, e^{4}} - \frac {\sqrt {-e^{2} x^{2} + d^{2}} b x}{2 \, e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2+a)/(-e*x+d)^(1/2)/(e*x+d)^(1/2),x, algorithm="maxima")

[Out]

-1/4*sqrt(-e^2*x^2 + d^2)*c*x^3/e^2 + 3/8*c*d^4*arcsin(e*x/d)/e^5 + 1/2*b*d^2*arcsin(e*x/d)/e^3 + a*arcsin(e*x
/d)/e - 3/8*sqrt(-e^2*x^2 + d^2)*c*d^2*x/e^4 - 1/2*sqrt(-e^2*x^2 + d^2)*b*x/e^2

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mupad [B]  time = 12.86, size = 651, normalized size = 5.09 \begin {gather*} \frac {\frac {14\,b\,d^2\,{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^3}{{\left (\sqrt {d-e\,x}-\sqrt {d}\right )}^3}-\frac {14\,b\,d^2\,{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^5}{{\left (\sqrt {d-e\,x}-\sqrt {d}\right )}^5}+\frac {2\,b\,d^2\,{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^7}{{\left (\sqrt {d-e\,x}-\sqrt {d}\right )}^7}-\frac {2\,b\,d^2\,\left (\sqrt {d+e\,x}-\sqrt {d}\right )}{\sqrt {d-e\,x}-\sqrt {d}}}{e^3\,{\left (\frac {{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^2}{{\left (\sqrt {d-e\,x}-\sqrt {d}\right )}^2}+1\right )}^4}-\frac {4\,a\,\mathrm {atan}\left (\frac {e\,\left (\sqrt {d-e\,x}-\sqrt {d}\right )}{\sqrt {e^2}\,\left (\sqrt {d+e\,x}-\sqrt {d}\right )}\right )}{\sqrt {e^2}}-\frac {\frac {23\,c\,d^4\,{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^3}{2\,{\left (\sqrt {d-e\,x}-\sqrt {d}\right )}^3}-\frac {333\,c\,d^4\,{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^5}{2\,{\left (\sqrt {d-e\,x}-\sqrt {d}\right )}^5}+\frac {671\,c\,d^4\,{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^7}{2\,{\left (\sqrt {d-e\,x}-\sqrt {d}\right )}^7}-\frac {671\,c\,d^4\,{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^9}{2\,{\left (\sqrt {d-e\,x}-\sqrt {d}\right )}^9}+\frac {333\,c\,d^4\,{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^{11}}{2\,{\left (\sqrt {d-e\,x}-\sqrt {d}\right )}^{11}}-\frac {23\,c\,d^4\,{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^{13}}{2\,{\left (\sqrt {d-e\,x}-\sqrt {d}\right )}^{13}}-\frac {3\,c\,d^4\,{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^{15}}{2\,{\left (\sqrt {d-e\,x}-\sqrt {d}\right )}^{15}}+\frac {3\,c\,d^4\,\left (\sqrt {d+e\,x}-\sqrt {d}\right )}{2\,\left (\sqrt {d-e\,x}-\sqrt {d}\right )}}{e^5\,{\left (\frac {{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^2}{{\left (\sqrt {d-e\,x}-\sqrt {d}\right )}^2}+1\right )}^8}+\frac {2\,b\,d^2\,\mathrm {atan}\left (\frac {\sqrt {d+e\,x}-\sqrt {d}}{\sqrt {d-e\,x}-\sqrt {d}}\right )}{e^3}+\frac {3\,c\,d^4\,\mathrm {atan}\left (\frac {\sqrt {d+e\,x}-\sqrt {d}}{\sqrt {d-e\,x}-\sqrt {d}}\right )}{2\,e^5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2 + c*x^4)/((d + e*x)^(1/2)*(d - e*x)^(1/2)),x)

[Out]

((14*b*d^2*((d + e*x)^(1/2) - d^(1/2))^3)/((d - e*x)^(1/2) - d^(1/2))^3 - (14*b*d^2*((d + e*x)^(1/2) - d^(1/2)
)^5)/((d - e*x)^(1/2) - d^(1/2))^5 + (2*b*d^2*((d + e*x)^(1/2) - d^(1/2))^7)/((d - e*x)^(1/2) - d^(1/2))^7 - (
2*b*d^2*((d + e*x)^(1/2) - d^(1/2)))/((d - e*x)^(1/2) - d^(1/2)))/(e^3*(((d + e*x)^(1/2) - d^(1/2))^2/((d - e*
x)^(1/2) - d^(1/2))^2 + 1)^4) - (4*a*atan((e*((d - e*x)^(1/2) - d^(1/2)))/((e^2)^(1/2)*((d + e*x)^(1/2) - d^(1
/2)))))/(e^2)^(1/2) - ((23*c*d^4*((d + e*x)^(1/2) - d^(1/2))^3)/(2*((d - e*x)^(1/2) - d^(1/2))^3) - (333*c*d^4
*((d + e*x)^(1/2) - d^(1/2))^5)/(2*((d - e*x)^(1/2) - d^(1/2))^5) + (671*c*d^4*((d + e*x)^(1/2) - d^(1/2))^7)/
(2*((d - e*x)^(1/2) - d^(1/2))^7) - (671*c*d^4*((d + e*x)^(1/2) - d^(1/2))^9)/(2*((d - e*x)^(1/2) - d^(1/2))^9
) + (333*c*d^4*((d + e*x)^(1/2) - d^(1/2))^11)/(2*((d - e*x)^(1/2) - d^(1/2))^11) - (23*c*d^4*((d + e*x)^(1/2)
- d^(1/2))^13)/(2*((d - e*x)^(1/2) - d^(1/2))^13) - (3*c*d^4*((d + e*x)^(1/2) - d^(1/2))^15)/(2*((d - e*x)^(1
/2) - d^(1/2))^15) + (3*c*d^4*((d + e*x)^(1/2) - d^(1/2)))/(2*((d - e*x)^(1/2) - d^(1/2))))/(e^5*(((d + e*x)^(
1/2) - d^(1/2))^2/((d - e*x)^(1/2) - d^(1/2))^2 + 1)^8) + (2*b*d^2*atan(((d + e*x)^(1/2) - d^(1/2))/((d - e*x)
^(1/2) - d^(1/2))))/e^3 + (3*c*d^4*atan(((d + e*x)^(1/2) - d^(1/2))/((d - e*x)^(1/2) - d^(1/2))))/(2*e^5)

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sympy [F(-1)]  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((c*x**4+b*x**2+a)/(-e*x+d)**(1/2)/(e*x+d)**(1/2),x)

[Out]

Timed out

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