3.107 \(\int \frac {d+e x+f x^2+g x^3}{(a+b x^2+c x^4)^{5/2}} \, dx\)

Optimal. Leaf size=680 \[ -\frac {\sqrt [4]{c} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \left (6 a^{3/2} \sqrt {c} f-3 \sqrt {a} b \sqrt {c} d+a b f-10 a c d+2 b^2 d\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{6 a^{7/4} \left (b-2 \sqrt {a} \sqrt {c}\right ) \left (b^2-4 a c\right ) \sqrt {a+b x^2+c x^4}}-\frac {\sqrt {c} x \sqrt {a+b x^2+c x^4} \left (12 a^2 c f+a b^2 f-16 a b c d+2 b^3 d\right )}{3 a^2 \left (b^2-4 a c\right )^2 \left (\sqrt {a}+\sqrt {c} x^2\right )}+\frac {x \left (c x^2 \left (12 a^2 c f+a b^2 f-16 a b c d+2 b^3 d\right )+4 a^2 b c f+20 a^2 c^2 d+a b^3 f-17 a b^2 c d+2 b^4 d\right )}{3 a^2 \left (b^2-4 a c\right )^2 \sqrt {a+b x^2+c x^4}}+\frac {\sqrt [4]{c} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \left (12 a^2 c f+a b^2 f-16 a b c d+2 b^3 d\right ) E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{3 a^{7/4} \left (b^2-4 a c\right )^2 \sqrt {a+b x^2+c x^4}}+\frac {x \left (c x^2 (b d-2 a f)-a b f-2 a c d+b^2 d\right )}{3 a \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^{3/2}}+\frac {4 \left (b+2 c x^2\right ) (2 c e-b g)}{3 \left (b^2-4 a c\right )^2 \sqrt {a+b x^2+c x^4}}-\frac {-2 a g+x^2 (2 c e-b g)+b e}{3 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^{3/2}} \]

[Out]

1/3*x*(b^2*d-2*a*c*d-a*b*f+c*(-2*a*f+b*d)*x^2)/a/(-4*a*c+b^2)/(c*x^4+b*x^2+a)^(3/2)+1/3*(-b*e+2*a*g-(-b*g+2*c*
e)*x^2)/(-4*a*c+b^2)/(c*x^4+b*x^2+a)^(3/2)+4/3*(-b*g+2*c*e)*(2*c*x^2+b)/(-4*a*c+b^2)^2/(c*x^4+b*x^2+a)^(1/2)+1
/3*x*(2*b^4*d-17*a*b^2*c*d+20*a^2*c^2*d+a*b^3*f+4*a^2*b*c*f+c*(12*a^2*c*f+a*b^2*f-16*a*b*c*d+2*b^3*d)*x^2)/a^2
/(-4*a*c+b^2)^2/(c*x^4+b*x^2+a)^(1/2)-1/3*(12*a^2*c*f+a*b^2*f-16*a*b*c*d+2*b^3*d)*x*c^(1/2)*(c*x^4+b*x^2+a)^(1
/2)/a^2/(-4*a*c+b^2)^2/(a^(1/2)+x^2*c^(1/2))+1/3*c^(1/4)*(12*a^2*c*f+a*b^2*f-16*a*b*c*d+2*b^3*d)*(cos(2*arctan
(c^(1/4)*x/a^(1/4)))^2)^(1/2)/cos(2*arctan(c^(1/4)*x/a^(1/4)))*EllipticE(sin(2*arctan(c^(1/4)*x/a^(1/4))),1/2*
(2-b/a^(1/2)/c^(1/2))^(1/2))*(a^(1/2)+x^2*c^(1/2))*((c*x^4+b*x^2+a)/(a^(1/2)+x^2*c^(1/2))^2)^(1/2)/a^(7/4)/(-4
*a*c+b^2)^2/(c*x^4+b*x^2+a)^(1/2)-1/6*c^(1/4)*(cos(2*arctan(c^(1/4)*x/a^(1/4)))^2)^(1/2)/cos(2*arctan(c^(1/4)*
x/a^(1/4)))*EllipticF(sin(2*arctan(c^(1/4)*x/a^(1/4))),1/2*(2-b/a^(1/2)/c^(1/2))^(1/2))*(a^(1/2)+x^2*c^(1/2))*
(2*b^2*d-10*a*c*d+a*b*f+6*a^(3/2)*f*c^(1/2)-3*b*d*a^(1/2)*c^(1/2))*((c*x^4+b*x^2+a)/(a^(1/2)+x^2*c^(1/2))^2)^(
1/2)/a^(7/4)/(-4*a*c+b^2)/(-2*a^(1/2)*c^(1/2)+b)/(c*x^4+b*x^2+a)^(1/2)

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Rubi [A]  time = 0.51, antiderivative size = 680, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1673, 1178, 1197, 1103, 1195, 1247, 638, 613} \[ \frac {x \left (c x^2 \left (12 a^2 c f+a b^2 f-16 a b c d+2 b^3 d\right )+4 a^2 b c f+20 a^2 c^2 d-17 a b^2 c d+a b^3 f+2 b^4 d\right )}{3 a^2 \left (b^2-4 a c\right )^2 \sqrt {a+b x^2+c x^4}}-\frac {\sqrt {c} x \sqrt {a+b x^2+c x^4} \left (12 a^2 c f+a b^2 f-16 a b c d+2 b^3 d\right )}{3 a^2 \left (b^2-4 a c\right )^2 \left (\sqrt {a}+\sqrt {c} x^2\right )}-\frac {\sqrt [4]{c} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \left (6 a^{3/2} \sqrt {c} f-3 \sqrt {a} b \sqrt {c} d+a b f-10 a c d+2 b^2 d\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{6 a^{7/4} \left (b-2 \sqrt {a} \sqrt {c}\right ) \left (b^2-4 a c\right ) \sqrt {a+b x^2+c x^4}}+\frac {\sqrt [4]{c} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \left (12 a^2 c f+a b^2 f-16 a b c d+2 b^3 d\right ) E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{3 a^{7/4} \left (b^2-4 a c\right )^2 \sqrt {a+b x^2+c x^4}}+\frac {x \left (c x^2 (b d-2 a f)-a b f-2 a c d+b^2 d\right )}{3 a \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^{3/2}}+\frac {4 \left (b+2 c x^2\right ) (2 c e-b g)}{3 \left (b^2-4 a c\right )^2 \sqrt {a+b x^2+c x^4}}-\frac {-2 a g+x^2 (2 c e-b g)+b e}{3 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x + f*x^2 + g*x^3)/(a + b*x^2 + c*x^4)^(5/2),x]

[Out]

(x*(b^2*d - 2*a*c*d - a*b*f + c*(b*d - 2*a*f)*x^2))/(3*a*(b^2 - 4*a*c)*(a + b*x^2 + c*x^4)^(3/2)) - (b*e - 2*a
*g + (2*c*e - b*g)*x^2)/(3*(b^2 - 4*a*c)*(a + b*x^2 + c*x^4)^(3/2)) + (4*(2*c*e - b*g)*(b + 2*c*x^2))/(3*(b^2
- 4*a*c)^2*Sqrt[a + b*x^2 + c*x^4]) + (x*(2*b^4*d - 17*a*b^2*c*d + 20*a^2*c^2*d + a*b^3*f + 4*a^2*b*c*f + c*(2
*b^3*d - 16*a*b*c*d + a*b^2*f + 12*a^2*c*f)*x^2))/(3*a^2*(b^2 - 4*a*c)^2*Sqrt[a + b*x^2 + c*x^4]) - (Sqrt[c]*(
2*b^3*d - 16*a*b*c*d + a*b^2*f + 12*a^2*c*f)*x*Sqrt[a + b*x^2 + c*x^4])/(3*a^2*(b^2 - 4*a*c)^2*(Sqrt[a] + Sqrt
[c]*x^2)) + (c^(1/4)*(2*b^3*d - 16*a*b*c*d + a*b^2*f + 12*a^2*c*f)*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + b*x^2 + c
*x^4)/(Sqrt[a] + Sqrt[c]*x^2)^2]*EllipticE[2*ArcTan[(c^(1/4)*x)/a^(1/4)], (2 - b/(Sqrt[a]*Sqrt[c]))/4])/(3*a^(
7/4)*(b^2 - 4*a*c)^2*Sqrt[a + b*x^2 + c*x^4]) - (c^(1/4)*(2*b^2*d - 3*Sqrt[a]*b*Sqrt[c]*d - 10*a*c*d + a*b*f +
6*a^(3/2)*Sqrt[c]*f)*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + b*x^2 + c*x^4)/(Sqrt[a] + Sqrt[c]*x^2)^2]*EllipticF[2*
ArcTan[(c^(1/4)*x)/a^(1/4)], (2 - b/(Sqrt[a]*Sqrt[c]))/4])/(6*a^(7/4)*(b - 2*Sqrt[a]*Sqrt[c])*(b^2 - 4*a*c)*Sq
rt[a + b*x^2 + c*x^4])

Rule 613

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[(-2*(b + 2*c*x))/((b^2 - 4*a*c)*Sqrt[a + b*x
+ c*x^2]), x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 638

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b*d - 2*a*e + (2*c*d -
b*e)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)), x] - Dist[((2*p + 3)*(2*c*d - b*e))/((p + 1)*(b^2
- 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] && NeQ[b^
2 - 4*a*c, 0] && LtQ[p, -1] && NeQ[p, -3/2]

Rule 1103

Int[1/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(
a + b*x^2 + c*x^4)/(a*(1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2 - (b*q^2)/(4*c)])/(2*q*Sqrt[a + b*x^2 + c
*x^4]), x]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]

Rule 1178

Int[((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(x*(a*b*e - d*(b^2 - 2*
a*c) - c*(b*d - 2*a*e)*x^2)*(a + b*x^2 + c*x^4)^(p + 1))/(2*a*(p + 1)*(b^2 - 4*a*c)), x] + Dist[1/(2*a*(p + 1)
*(b^2 - 4*a*c)), Int[Simp[(2*p + 3)*d*b^2 - a*b*e - 2*a*c*d*(4*p + 5) + (4*p + 7)*(d*b - 2*a*e)*c*x^2, x]*(a +
b*x^2 + c*x^4)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e
^2, 0] && LtQ[p, -1] && IntegerQ[2*p]

Rule 1195

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, -Simp[
(d*x*Sqrt[a + b*x^2 + c*x^4])/(a*(1 + q^2*x^2)), x] + Simp[(d*(1 + q^2*x^2)*Sqrt[(a + b*x^2 + c*x^4)/(a*(1 + q
^2*x^2)^2)]*EllipticE[2*ArcTan[q*x], 1/2 - (b*q^2)/(4*c)])/(q*Sqrt[a + b*x^2 + c*x^4]), x] /; EqQ[e + d*q^2, 0
]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]

Rule 1197

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

Rule 1247

Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[
Int[(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x]

Rule 1673

Int[(Pq_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Module[{q = Expon[Pq, x], k}, Int[Sum[Coeff[
Pq, x, 2*k]*x^(2*k), {k, 0, q/2}]*(a + b*x^2 + c*x^4)^p, x] + Int[x*Sum[Coeff[Pq, x, 2*k + 1]*x^(2*k), {k, 0,
(q - 1)/2}]*(a + b*x^2 + c*x^4)^p, x]] /; FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] &&  !PolyQ[Pq, x^2]

Rubi steps

\begin {align*} \int \frac {d+e x+f x^2+g x^3}{\left (a+b x^2+c x^4\right )^{5/2}} \, dx &=\int \frac {d+f x^2}{\left (a+b x^2+c x^4\right )^{5/2}} \, dx+\int \frac {x \left (e+g x^2\right )}{\left (a+b x^2+c x^4\right )^{5/2}} \, dx\\ &=\frac {x \left (b^2 d-2 a c d-a b f+c (b d-2 a f) x^2\right )}{3 a \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^{3/2}}+\frac {1}{2} \operatorname {Subst}\left (\int \frac {e+g x}{\left (a+b x+c x^2\right )^{5/2}} \, dx,x,x^2\right )-\frac {\int \frac {-2 b^2 d+10 a c d-a b f-3 c (b d-2 a f) x^2}{\left (a+b x^2+c x^4\right )^{3/2}} \, dx}{3 a \left (b^2-4 a c\right )}\\ &=\frac {x \left (b^2 d-2 a c d-a b f+c (b d-2 a f) x^2\right )}{3 a \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^{3/2}}-\frac {b e-2 a g+(2 c e-b g) x^2}{3 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^{3/2}}+\frac {x \left (2 b^4 d-17 a b^2 c d+20 a^2 c^2 d+a b^3 f+4 a^2 b c f+c \left (2 b^3 d-16 a b c d+a b^2 f+12 a^2 c f\right ) x^2\right )}{3 a^2 \left (b^2-4 a c\right )^2 \sqrt {a+b x^2+c x^4}}+\frac {\int \frac {-a c \left (b^2 d-20 a c d+8 a b f\right )-c \left (2 b^3 d-16 a b c d+a b^2 f+12 a^2 c f\right ) x^2}{\sqrt {a+b x^2+c x^4}} \, dx}{3 a^2 \left (b^2-4 a c\right )^2}-\frac {(2 (2 c e-b g)) \operatorname {Subst}\left (\int \frac {1}{\left (a+b x+c x^2\right )^{3/2}} \, dx,x,x^2\right )}{3 \left (b^2-4 a c\right )}\\ &=\frac {x \left (b^2 d-2 a c d-a b f+c (b d-2 a f) x^2\right )}{3 a \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^{3/2}}-\frac {b e-2 a g+(2 c e-b g) x^2}{3 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^{3/2}}+\frac {4 (2 c e-b g) \left (b+2 c x^2\right )}{3 \left (b^2-4 a c\right )^2 \sqrt {a+b x^2+c x^4}}+\frac {x \left (2 b^4 d-17 a b^2 c d+20 a^2 c^2 d+a b^3 f+4 a^2 b c f+c \left (2 b^3 d-16 a b c d+a b^2 f+12 a^2 c f\right ) x^2\right )}{3 a^2 \left (b^2-4 a c\right )^2 \sqrt {a+b x^2+c x^4}}+\frac {\left (\sqrt {c} \left (2 b^3 d-16 a b c d+a b^2 f+12 a^2 c f\right )\right ) \int \frac {1-\frac {\sqrt {c} x^2}{\sqrt {a}}}{\sqrt {a+b x^2+c x^4}} \, dx}{3 a^{3/2} \left (b^2-4 a c\right )^2}-\frac {\left (\sqrt {c} \left (2 b^3 d-16 a b c d+a b^2 f+12 a^2 c f+\sqrt {a} \sqrt {c} \left (b^2 d-20 a c d+8 a b f\right )\right )\right ) \int \frac {1}{\sqrt {a+b x^2+c x^4}} \, dx}{3 a^{3/2} \left (b^2-4 a c\right )^2}\\ &=\frac {x \left (b^2 d-2 a c d-a b f+c (b d-2 a f) x^2\right )}{3 a \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^{3/2}}-\frac {b e-2 a g+(2 c e-b g) x^2}{3 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^{3/2}}+\frac {4 (2 c e-b g) \left (b+2 c x^2\right )}{3 \left (b^2-4 a c\right )^2 \sqrt {a+b x^2+c x^4}}+\frac {x \left (2 b^4 d-17 a b^2 c d+20 a^2 c^2 d+a b^3 f+4 a^2 b c f+c \left (2 b^3 d-16 a b c d+a b^2 f+12 a^2 c f\right ) x^2\right )}{3 a^2 \left (b^2-4 a c\right )^2 \sqrt {a+b x^2+c x^4}}-\frac {\sqrt {c} \left (2 b^3 d-16 a b c d+a b^2 f+12 a^2 c f\right ) x \sqrt {a+b x^2+c x^4}}{3 a^2 \left (b^2-4 a c\right )^2 \left (\sqrt {a}+\sqrt {c} x^2\right )}+\frac {\sqrt [4]{c} \left (2 b^3 d-16 a b c d+a b^2 f+12 a^2 c f\right ) \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{3 a^{7/4} \left (b^2-4 a c\right )^2 \sqrt {a+b x^2+c x^4}}-\frac {\sqrt [4]{c} \left (2 b^3 d-16 a b c d+a b^2 f+12 a^2 c f+\sqrt {a} \sqrt {c} \left (b^2 d-20 a c d+8 a b f\right )\right ) \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{6 a^{7/4} \left (b^2-4 a c\right )^2 \sqrt {a+b x^2+c x^4}}\\ \end {align*}

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Mathematica [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {\$Aborted} \]

Verification is Not applicable to the result.

[In]

Integrate[(d + e*x + f*x^2 + g*x^3)/(a + b*x^2 + c*x^4)^(5/2),x]

[Out]

$Aborted

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fricas [F]  time = 0.96, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {c x^{4} + b x^{2} + a} {\left (g x^{3} + f x^{2} + e x + d\right )}}{c^{3} x^{12} + 3 \, b c^{2} x^{10} + 3 \, {\left (b^{2} c + a c^{2}\right )} x^{8} + {\left (b^{3} + 6 \, a b c\right )} x^{6} + 3 \, a^{2} b x^{2} + 3 \, {\left (a b^{2} + a^{2} c\right )} x^{4} + a^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x^3+f*x^2+e*x+d)/(c*x^4+b*x^2+a)^(5/2),x, algorithm="fricas")

[Out]

integral(sqrt(c*x^4 + b*x^2 + a)*(g*x^3 + f*x^2 + e*x + d)/(c^3*x^12 + 3*b*c^2*x^10 + 3*(b^2*c + a*c^2)*x^8 +
(b^3 + 6*a*b*c)*x^6 + 3*a^2*b*x^2 + 3*(a*b^2 + a^2*c)*x^4 + a^3), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {g x^{3} + f x^{2} + e x + d}{{\left (c x^{4} + b x^{2} + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x^3+f*x^2+e*x+d)/(c*x^4+b*x^2+a)^(5/2),x, algorithm="giac")

[Out]

integrate((g*x^3 + f*x^2 + e*x + d)/(c*x^4 + b*x^2 + a)^(5/2), x)

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maple [B]  time = 0.06, size = 1395, normalized size = 2.05 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((g*x^3+f*x^2+e*x+d)/(c*x^4+b*x^2+a)^(5/2),x)

[Out]

-1/3*g*(8*b*c^2*x^6+12*b^2*c*x^4+12*a*b*c*x^2+3*b^3*x^2+8*a^2*c+2*a*b^2)/(c*x^4+b*x^2+a)^(3/2)/(16*a^2*c^2-8*a
*b^2*c+b^4)+f*((2/3/c/(4*a*c-b^2)*x^3+1/3*b/(4*a*c-b^2)/c^2*x)*(c*x^4+b*x^2+a)^(1/2)/(x^4+b/c*x^2+a/c)^2-2*c*(
-1/6*(12*a*c+b^2)/a/(4*a*c-b^2)^2*x^3-1/6*(4*a*c+b^2)*b/a/(4*a*c-b^2)^2/c*x)/((x^4+b/c*x^2+a/c)*c)^(1/2)+1/4*(
-1/3/a*b/(4*a*c-b^2)-1/3*(4*a*c+b^2)*b/a/(4*a*c-b^2)^2)*2^(1/2)/((-b+(-4*a*c+b^2)^(1/2))/a)^(1/2)*(-2*(-b+(-4*
a*c+b^2)^(1/2))/a*x^2+4)^(1/2)*(2*(b+(-4*a*c+b^2)^(1/2))/a*x^2+4)^(1/2)/(c*x^4+b*x^2+a)^(1/2)*EllipticF(1/2*2^
(1/2)*((-b+(-4*a*c+b^2)^(1/2))/a)^(1/2)*x,1/2*(2*(b+(-4*a*c+b^2)^(1/2))/a*b/c-4)^(1/2))+1/6*c*(12*a*c+b^2)/(4*
a*c-b^2)^2*2^(1/2)/((-b+(-4*a*c+b^2)^(1/2))/a)^(1/2)*(-2*(-b+(-4*a*c+b^2)^(1/2))/a*x^2+4)^(1/2)*(2*(b+(-4*a*c+
b^2)^(1/2))/a*x^2+4)^(1/2)/(c*x^4+b*x^2+a)^(1/2)/(b+(-4*a*c+b^2)^(1/2))*(EllipticF(1/2*2^(1/2)*((-b+(-4*a*c+b^
2)^(1/2))/a)^(1/2)*x,1/2*(2*(b+(-4*a*c+b^2)^(1/2))/a*b/c-4)^(1/2))-EllipticE(1/2*2^(1/2)*((-b+(-4*a*c+b^2)^(1/
2))/a)^(1/2)*x,1/2*(2*(b+(-4*a*c+b^2)^(1/2))/a*b/c-4)^(1/2))))+1/3*e*(16*c^3*x^6+24*b*c^2*x^4+24*a*c^2*x^2+6*b
^2*c*x^2+12*a*b*c-b^3)/(c*x^4+b*x^2+a)^(3/2)/(16*a^2*c^2-8*a*b^2*c+b^4)+d*((-1/3/a*b/(4*a*c-b^2)/c*x^3+1/3*(2*
a*c-b^2)/(4*a*c-b^2)/a/c^2*x)*(c*x^4+b*x^2+a)^(1/2)/(x^4+b/c*x^2+a/c)^2-2*c*(1/3*b*(8*a*c-b^2)/(4*a*c-b^2)^2/a
^2*x^3-1/6*(20*a^2*c^2-17*a*b^2*c+2*b^4)/a^2/(4*a*c-b^2)^2/c*x)/((x^4+b/c*x^2+a/c)*c)^(1/2)+1/4*(2/3*(5*a*c-b^
2)/a^2/(4*a*c-b^2)-1/3*(20*a^2*c^2-17*a*b^2*c+2*b^4)/a^2/(4*a*c-b^2)^2)*2^(1/2)/((-b+(-4*a*c+b^2)^(1/2))/a)^(1
/2)*(-2*(-b+(-4*a*c+b^2)^(1/2))/a*x^2+4)^(1/2)*(2*(b+(-4*a*c+b^2)^(1/2))/a*x^2+4)^(1/2)/(c*x^4+b*x^2+a)^(1/2)*
EllipticF(1/2*2^(1/2)*((-b+(-4*a*c+b^2)^(1/2))/a)^(1/2)*x,1/2*(2*(b+(-4*a*c+b^2)^(1/2))/a*b/c-4)^(1/2))-1/3*b*
c*(8*a*c-b^2)/(4*a*c-b^2)^2/a*2^(1/2)/((-b+(-4*a*c+b^2)^(1/2))/a)^(1/2)*(-2*(-b+(-4*a*c+b^2)^(1/2))/a*x^2+4)^(
1/2)*(2*(b+(-4*a*c+b^2)^(1/2))/a*x^2+4)^(1/2)/(c*x^4+b*x^2+a)^(1/2)/(b+(-4*a*c+b^2)^(1/2))*(EllipticF(1/2*2^(1
/2)*((-b+(-4*a*c+b^2)^(1/2))/a)^(1/2)*x,1/2*(2*(b+(-4*a*c+b^2)^(1/2))/a*b/c-4)^(1/2))-EllipticE(1/2*2^(1/2)*((
-b+(-4*a*c+b^2)^(1/2))/a)^(1/2)*x,1/2*(2*(b+(-4*a*c+b^2)^(1/2))/a*b/c-4)^(1/2))))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {g x^{3} + f x^{2} + e x + d}{{\left (c x^{4} + b x^{2} + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x^3+f*x^2+e*x+d)/(c*x^4+b*x^2+a)^(5/2),x, algorithm="maxima")

[Out]

integrate((g*x^3 + f*x^2 + e*x + d)/(c*x^4 + b*x^2 + a)^(5/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {g\,x^3+f\,x^2+e\,x+d}{{\left (c\,x^4+b\,x^2+a\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x + f*x^2 + g*x^3)/(a + b*x^2 + c*x^4)^(5/2),x)

[Out]

int((d + e*x + f*x^2 + g*x^3)/(a + b*x^2 + c*x^4)^(5/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x**3+f*x**2+e*x+d)/(c*x**4+b*x**2+a)**(5/2),x)

[Out]

Timed out

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