3.953 \(\int \frac {(a+b x^2+c x^4)^{3/2}}{x^8} \, dx\)

Optimal. Leaf size=447 \[ -\frac {\sqrt [4]{c} \left (\sqrt {a} \sqrt {c} \left (b^2-20 a c\right )+2 b \left (b^2-8 a c\right )\right ) \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{70 a^{7/4} \sqrt {a+b x^2+c x^4}}+\frac {2 b \sqrt [4]{c} \left (b^2-8 a c\right ) \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{35 a^{7/4} \sqrt {a+b x^2+c x^4}}+\frac {2 b \left (b^2-8 a c\right ) \sqrt {a+b x^2+c x^4}}{35 a^2 x}-\frac {2 b \sqrt {c} x \left (b^2-8 a c\right ) \sqrt {a+b x^2+c x^4}}{35 a^2 \left (\sqrt {a}+\sqrt {c} x^2\right )}-\frac {\left (b^2-20 a c\right ) \sqrt {a+b x^2+c x^4}}{35 a x^3}-\frac {\left (a+b x^2+c x^4\right )^{3/2}}{7 x^7}-\frac {3 \left (b+10 c x^2\right ) \sqrt {a+b x^2+c x^4}}{35 x^5} \]

[Out]

-1/7*(c*x^4+b*x^2+a)^(3/2)/x^7-1/35*(-20*a*c+b^2)*(c*x^4+b*x^2+a)^(1/2)/a/x^3+2/35*b*(-8*a*c+b^2)*(c*x^4+b*x^2
+a)^(1/2)/a^2/x-3/35*(10*c*x^2+b)*(c*x^4+b*x^2+a)^(1/2)/x^5-2/35*b*(-8*a*c+b^2)*x*c^(1/2)*(c*x^4+b*x^2+a)^(1/2
)/a^2/(a^(1/2)+x^2*c^(1/2))+2/35*b*c^(1/4)*(-8*a*c+b^2)*(cos(2*arctan(c^(1/4)*x/a^(1/4)))^2)^(1/2)/cos(2*arcta
n(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)/(c*x^4+b*x^2+a)^(1/2)-1/70*c^(1/4)*(cos(2*ar
ctan(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*(-8*a*c+b^2)+(-20*a*c+b^2)*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)/(c*x^4+b*x^2+a)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.39, antiderivative size = 447, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {1117, 1271, 1281, 1197, 1103, 1195} \[ \frac {2 b \left (b^2-8 a c\right ) \sqrt {a+b x^2+c x^4}}{35 a^2 x}-\frac {2 b \sqrt {c} x \left (b^2-8 a c\right ) \sqrt {a+b x^2+c x^4}}{35 a^2 \left (\sqrt {a}+\sqrt {c} x^2\right )}-\frac {\sqrt [4]{c} \left (\sqrt {a} \sqrt {c} \left (b^2-20 a c\right )+2 b \left (b^2-8 a c\right )\right ) \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{70 a^{7/4} \sqrt {a+b x^2+c x^4}}+\frac {2 b \sqrt [4]{c} \left (b^2-8 a c\right ) \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{35 a^{7/4} \sqrt {a+b x^2+c x^4}}-\frac {\left (b^2-20 a c\right ) \sqrt {a+b x^2+c x^4}}{35 a x^3}-\frac {\left (a+b x^2+c x^4\right )^{3/2}}{7 x^7}-\frac {3 \left (b+10 c x^2\right ) \sqrt {a+b x^2+c x^4}}{35 x^5} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2 + c*x^4)^(3/2)/x^8,x]

[Out]

-((b^2 - 20*a*c)*Sqrt[a + b*x^2 + c*x^4])/(35*a*x^3) + (2*b*(b^2 - 8*a*c)*Sqrt[a + b*x^2 + c*x^4])/(35*a^2*x)
- (2*b*Sqrt[c]*(b^2 - 8*a*c)*x*Sqrt[a + b*x^2 + c*x^4])/(35*a^2*(Sqrt[a] + Sqrt[c]*x^2)) - (3*(b + 10*c*x^2)*S
qrt[a + b*x^2 + c*x^4])/(35*x^5) - (a + b*x^2 + c*x^4)^(3/2)/(7*x^7) + (2*b*c^(1/4)*(b^2 - 8*a*c)*(Sqrt[a] + S
qrt[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])/(35*a^(7/4)*Sqrt[a + b*x^2 + c*x^4]) - (c^(1/4)*(Sqrt[a]*Sqrt[c]*(b^2 - 20*a*c) + 2*b
*(b^2 - 8*a*c))*(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])/(70*a^(7/4)*Sqrt[a + b*x^2 + c*x^4])

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 1117

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

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 1271

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

Rule 1281

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

Rubi steps

\begin {align*} \int \frac {\left (a+b x^2+c x^4\right )^{3/2}}{x^8} \, dx &=-\frac {\left (a+b x^2+c x^4\right )^{3/2}}{7 x^7}+\frac {3}{7} \int \frac {\left (b+2 c x^2\right ) \sqrt {a+b x^2+c x^4}}{x^6} \, dx\\ &=-\frac {3 \left (b+10 c x^2\right ) \sqrt {a+b x^2+c x^4}}{35 x^5}-\frac {\left (a+b x^2+c x^4\right )^{3/2}}{7 x^7}+\frac {3}{35} \int \frac {b^2-20 a c-8 b c x^2}{x^4 \sqrt {a+b x^2+c x^4}} \, dx\\ &=-\frac {\left (b^2-20 a c\right ) \sqrt {a+b x^2+c x^4}}{35 a x^3}-\frac {3 \left (b+10 c x^2\right ) \sqrt {a+b x^2+c x^4}}{35 x^5}-\frac {\left (a+b x^2+c x^4\right )^{3/2}}{7 x^7}-\frac {\int \frac {2 b \left (b^2-8 a c\right )+c \left (b^2-20 a c\right ) x^2}{x^2 \sqrt {a+b x^2+c x^4}} \, dx}{35 a}\\ &=-\frac {\left (b^2-20 a c\right ) \sqrt {a+b x^2+c x^4}}{35 a x^3}+\frac {2 b \left (b^2-8 a c\right ) \sqrt {a+b x^2+c x^4}}{35 a^2 x}-\frac {3 \left (b+10 c x^2\right ) \sqrt {a+b x^2+c x^4}}{35 x^5}-\frac {\left (a+b x^2+c x^4\right )^{3/2}}{7 x^7}+\frac {\int \frac {-a c \left (b^2-20 a c\right )-2 b c \left (b^2-8 a c\right ) x^2}{\sqrt {a+b x^2+c x^4}} \, dx}{35 a^2}\\ &=-\frac {\left (b^2-20 a c\right ) \sqrt {a+b x^2+c x^4}}{35 a x^3}+\frac {2 b \left (b^2-8 a c\right ) \sqrt {a+b x^2+c x^4}}{35 a^2 x}-\frac {3 \left (b+10 c x^2\right ) \sqrt {a+b x^2+c x^4}}{35 x^5}-\frac {\left (a+b x^2+c x^4\right )^{3/2}}{7 x^7}+\frac {\left (2 b \sqrt {c} \left (b^2-8 a c\right )\right ) \int \frac {1-\frac {\sqrt {c} x^2}{\sqrt {a}}}{\sqrt {a+b x^2+c x^4}} \, dx}{35 a^{3/2}}-\frac {\left (\sqrt {c} \left (\sqrt {a} \sqrt {c} \left (b^2-20 a c\right )+2 b \left (b^2-8 a c\right )\right )\right ) \int \frac {1}{\sqrt {a+b x^2+c x^4}} \, dx}{35 a^{3/2}}\\ &=-\frac {\left (b^2-20 a c\right ) \sqrt {a+b x^2+c x^4}}{35 a x^3}+\frac {2 b \left (b^2-8 a c\right ) \sqrt {a+b x^2+c x^4}}{35 a^2 x}-\frac {2 b \sqrt {c} \left (b^2-8 a c\right ) x \sqrt {a+b x^2+c x^4}}{35 a^2 \left (\sqrt {a}+\sqrt {c} x^2\right )}-\frac {3 \left (b+10 c x^2\right ) \sqrt {a+b x^2+c x^4}}{35 x^5}-\frac {\left (a+b x^2+c x^4\right )^{3/2}}{7 x^7}+\frac {2 b \sqrt [4]{c} \left (b^2-8 a c\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 )}{35 a^{7/4} \sqrt {a+b x^2+c x^4}}-\frac {\sqrt [4]{c} \left (\sqrt {a} \sqrt {c} \left (b^2-20 a c\right )+2 b \left (b^2-8 a c\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 )}{70 a^{7/4} \sqrt {a+b x^2+c x^4}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 1.56, size = 572, normalized size = 1.28 \[ \frac {i x^7 \left (-20 a^2 c^2+9 a b^2 c-8 a b c \sqrt {b^2-4 a c}+b^3 \sqrt {b^2-4 a c}-b^4\right ) \sqrt {\frac {\sqrt {b^2-4 a c}+b+2 c x^2}{\sqrt {b^2-4 a c}+b}} \sqrt {\frac {-2 \sqrt {b^2-4 a c}+2 b+4 c x^2}{b-\sqrt {b^2-4 a c}}} F\left (i \sinh ^{-1}\left (\sqrt {2} \sqrt {\frac {c}{b+\sqrt {b^2-4 a c}}} x\right )|\frac {b+\sqrt {b^2-4 a c}}{b-\sqrt {b^2-4 a c}}\right )-2 \sqrt {\frac {c}{\sqrt {b^2-4 a c}+b}} \left (5 a^4+a^3 \left (13 b x^2+20 c x^4\right )+3 a^2 \left (3 b^2 x^4+13 b c x^6+5 c^2 x^8\right )+a b x^6 \left (-b^2+17 b c x^2+16 c^2 x^4\right )-2 b^3 x^8 \left (b+c x^2\right )\right )-i b x^7 \left (b^2-8 a c\right ) \left (\sqrt {b^2-4 a c}-b\right ) \sqrt {\frac {\sqrt {b^2-4 a c}+b+2 c x^2}{\sqrt {b^2-4 a c}+b}} \sqrt {\frac {-2 \sqrt {b^2-4 a c}+2 b+4 c x^2}{b-\sqrt {b^2-4 a c}}} E\left (i \sinh ^{-1}\left (\sqrt {2} \sqrt {\frac {c}{b+\sqrt {b^2-4 a c}}} x\right )|\frac {b+\sqrt {b^2-4 a c}}{b-\sqrt {b^2-4 a c}}\right )}{70 a^2 x^7 \sqrt {\frac {c}{\sqrt {b^2-4 a c}+b}} \sqrt {a+b x^2+c x^4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2 + c*x^4)^(3/2)/x^8,x]

[Out]

(-2*Sqrt[c/(b + Sqrt[b^2 - 4*a*c])]*(5*a^4 - 2*b^3*x^8*(b + c*x^2) + a^3*(13*b*x^2 + 20*c*x^4) + a*b*x^6*(-b^2
+ 17*b*c*x^2 + 16*c^2*x^4) + 3*a^2*(3*b^2*x^4 + 13*b*c*x^6 + 5*c^2*x^8)) - I*b*(b^2 - 8*a*c)*(-b + Sqrt[b^2 -
4*a*c])*x^7*Sqrt[(b + Sqrt[b^2 - 4*a*c] + 2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])]*Sqrt[(2*b - 2*Sqrt[b^2 - 4*a*c] +
4*c*x^2)/(b - Sqrt[b^2 - 4*a*c])]*EllipticE[I*ArcSinh[Sqrt[2]*Sqrt[c/(b + Sqrt[b^2 - 4*a*c])]*x], (b + Sqrt[b
^2 - 4*a*c])/(b - Sqrt[b^2 - 4*a*c])] + I*(-b^4 + 9*a*b^2*c - 20*a^2*c^2 + b^3*Sqrt[b^2 - 4*a*c] - 8*a*b*c*Sqr
t[b^2 - 4*a*c])*x^7*Sqrt[(b + Sqrt[b^2 - 4*a*c] + 2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])]*Sqrt[(2*b - 2*Sqrt[b^2 - 4
*a*c] + 4*c*x^2)/(b - Sqrt[b^2 - 4*a*c])]*EllipticF[I*ArcSinh[Sqrt[2]*Sqrt[c/(b + Sqrt[b^2 - 4*a*c])]*x], (b +
Sqrt[b^2 - 4*a*c])/(b - Sqrt[b^2 - 4*a*c])])/(70*a^2*Sqrt[c/(b + Sqrt[b^2 - 4*a*c])]*x^7*Sqrt[a + b*x^2 + c*x
^4])

________________________________________________________________________________________

fricas [F]  time = 0.65, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (c x^{4} + b x^{2} + a\right )}^{\frac {3}{2}}}{x^{8}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2+a)^(3/2)/x^8,x, algorithm="fricas")

[Out]

integral((c*x^4 + b*x^2 + a)^(3/2)/x^8, x)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (c x^{4} + b x^{2} + a\right )}^{\frac {3}{2}}}{x^{8}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2+a)^(3/2)/x^8,x, algorithm="giac")

[Out]

integrate((c*x^4 + b*x^2 + a)^(3/2)/x^8, x)

________________________________________________________________________________________

maple [A]  time = 0.02, size = 495, normalized size = 1.11 \[ -\frac {\left (8 a c -b^{2}\right ) \sqrt {2}\, \sqrt {-\frac {2 \left (-b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}+4}\, \sqrt {\frac {2 \left (b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}+4}\, \left (-\EllipticE \left (\frac {\sqrt {2}\, \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, x}{2}, \frac {\sqrt {\frac {2 \left (b +\sqrt {-4 a c +b^{2}}\right ) b}{a c}-4}}{2}\right )+\EllipticF \left (\frac {\sqrt {2}\, \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, x}{2}, \frac {\sqrt {\frac {2 \left (b +\sqrt {-4 a c +b^{2}}\right ) b}{a c}-4}}{2}\right )\right ) b c}{35 \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, \sqrt {c \,x^{4}+b \,x^{2}+a}\, \left (b +\sqrt {-4 a c +b^{2}}\right ) a}+\frac {\left (c^{2}-\frac {\left (15 a c +b^{2}\right ) c}{35 a}\right ) \sqrt {2}\, \sqrt {-\frac {2 \left (-b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}+4}\, \sqrt {\frac {2 \left (b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}+4}\, \EllipticF \left (\frac {\sqrt {2}\, \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, x}{2}, \frac {\sqrt {\frac {2 \left (b +\sqrt {-4 a c +b^{2}}\right ) b}{a c}-4}}{2}\right )}{4 \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}-\frac {2 \left (8 a c -b^{2}\right ) \sqrt {c \,x^{4}+b \,x^{2}+a}\, b}{35 a^{2} x}-\frac {\left (15 a c +b^{2}\right ) \sqrt {c \,x^{4}+b \,x^{2}+a}}{35 a \,x^{3}}-\frac {8 \sqrt {c \,x^{4}+b \,x^{2}+a}\, b}{35 x^{5}}-\frac {\sqrt {c \,x^{4}+b \,x^{2}+a}\, a}{7 x^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^4+b*x^2+a)^(3/2)/x^8,x)

[Out]

-1/7*a*(c*x^4+b*x^2+a)^(1/2)/x^7-8/35*b*(c*x^4+b*x^2+a)^(1/2)/x^5-1/35*(15*a*c+b^2)/a*(c*x^4+b*x^2+a)^(1/2)/x^
3-2/35*b*(8*a*c-b^2)/a^2*(c*x^4+b*x^2+a)^(1/2)/x+1/4*(c^2-1/35*c*(15*a*c+b^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)*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
))-1/35*b*c*(8*a*c-b^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)))

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (c x^{4} + b x^{2} + a\right )}^{\frac {3}{2}}}{x^{8}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2+a)^(3/2)/x^8,x, algorithm="maxima")

[Out]

integrate((c*x^4 + b*x^2 + a)^(3/2)/x^8, x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (c\,x^4+b\,x^2+a\right )}^{3/2}}{x^8} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2 + c*x^4)^(3/2)/x^8,x)

[Out]

int((a + b*x^2 + c*x^4)^(3/2)/x^8, x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b x^{2} + c x^{4}\right )^{\frac {3}{2}}}{x^{8}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**4+b*x**2+a)**(3/2)/x**8,x)

[Out]

Integral((a + b*x**2 + c*x**4)**(3/2)/x**8, x)

________________________________________________________________________________________