3.8.31 \(\int \frac {\sqrt {a+b x^2+c x^4}}{x^9} \, dx\)

Optimal. Leaf size=161 \[ \frac {\left (b^2-4 a c\right ) \left (5 b^2-4 a c\right ) \tanh ^{-1}\left (\frac {2 a+b x^2}{2 \sqrt {a} \sqrt {a+b x^2+c x^4}}\right )}{256 a^{7/2}}-\frac {\left (5 b^2-4 a c\right ) \left (2 a+b x^2\right ) \sqrt {a+b x^2+c x^4}}{128 a^3 x^4}+\frac {5 b \left (a+b x^2+c x^4\right )^{3/2}}{48 a^2 x^6}-\frac {\left (a+b x^2+c x^4\right )^{3/2}}{8 a x^8} \]

________________________________________________________________________________________

Rubi [A]  time = 0.15, antiderivative size = 161, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {1114, 744, 806, 720, 724, 206} \begin {gather*} -\frac {\left (5 b^2-4 a c\right ) \left (2 a+b x^2\right ) \sqrt {a+b x^2+c x^4}}{128 a^3 x^4}+\frac {\left (b^2-4 a c\right ) \left (5 b^2-4 a c\right ) \tanh ^{-1}\left (\frac {2 a+b x^2}{2 \sqrt {a} \sqrt {a+b x^2+c x^4}}\right )}{256 a^{7/2}}+\frac {5 b \left (a+b x^2+c x^4\right )^{3/2}}{48 a^2 x^6}-\frac {\left (a+b x^2+c x^4\right )^{3/2}}{8 a x^8} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*x^2 + c*x^4]/x^9,x]

[Out]

-((5*b^2 - 4*a*c)*(2*a + b*x^2)*Sqrt[a + b*x^2 + c*x^4])/(128*a^3*x^4) - (a + b*x^2 + c*x^4)^(3/2)/(8*a*x^8) +
(5*b*(a + b*x^2 + c*x^4)^(3/2))/(48*a^2*x^6) + ((b^2 - 4*a*c)*(5*b^2 - 4*a*c)*ArcTanh[(2*a + b*x^2)/(2*Sqrt[a
]*Sqrt[a + b*x^2 + c*x^4])])/(256*a^(7/2))

Rule 206

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

Rule 720

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(m + 1)*
(d*b - 2*a*e + (2*c*d - b*e)*x)*(a + b*x + c*x^2)^p)/(2*(m + 1)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[(p*(b^2 -
4*a*c))/(2*(m + 1)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^(m + 2)*(a + b*x + c*x^2)^(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] && NeQ[2*c*d - b*e, 0] && EqQ[m +
2*p + 2, 0] && GtQ[p, 0]

Rule 724

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

Rule 744

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

Rule 806

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Si
mp[((e*f - d*g)*(d + e*x)^(m + 1)*(a + b*x + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 - b*d*e + a*e^2)), x] - Dist[(b
*(e*f + d*g) - 2*(c*d*f + a*e*g))/(2*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x],
x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && EqQ[Sim
plify[m + 2*p + 3], 0]

Rule 1114

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

Rubi steps

\begin {align*} \int \frac {\sqrt {a+b x^2+c x^4}}{x^9} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {\sqrt {a+b x+c x^2}}{x^5} \, dx,x,x^2\right )\\ &=-\frac {\left (a+b x^2+c x^4\right )^{3/2}}{8 a x^8}-\frac {\operatorname {Subst}\left (\int \frac {\left (\frac {5 b}{2}+c x\right ) \sqrt {a+b x+c x^2}}{x^4} \, dx,x,x^2\right )}{8 a}\\ &=-\frac {\left (a+b x^2+c x^4\right )^{3/2}}{8 a x^8}+\frac {5 b \left (a+b x^2+c x^4\right )^{3/2}}{48 a^2 x^6}+\frac {\left (5 b^2-4 a c\right ) \operatorname {Subst}\left (\int \frac {\sqrt {a+b x+c x^2}}{x^3} \, dx,x,x^2\right )}{32 a^2}\\ &=-\frac {\left (5 b^2-4 a c\right ) \left (2 a+b x^2\right ) \sqrt {a+b x^2+c x^4}}{128 a^3 x^4}-\frac {\left (a+b x^2+c x^4\right )^{3/2}}{8 a x^8}+\frac {5 b \left (a+b x^2+c x^4\right )^{3/2}}{48 a^2 x^6}-\frac {\left (\left (b^2-4 a c\right ) \left (5 b^2-4 a c\right )\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x+c x^2}} \, dx,x,x^2\right )}{256 a^3}\\ &=-\frac {\left (5 b^2-4 a c\right ) \left (2 a+b x^2\right ) \sqrt {a+b x^2+c x^4}}{128 a^3 x^4}-\frac {\left (a+b x^2+c x^4\right )^{3/2}}{8 a x^8}+\frac {5 b \left (a+b x^2+c x^4\right )^{3/2}}{48 a^2 x^6}+\frac {\left (\left (b^2-4 a c\right ) \left (5 b^2-4 a c\right )\right ) \operatorname {Subst}\left (\int \frac {1}{4 a-x^2} \, dx,x,\frac {2 a+b x^2}{\sqrt {a+b x^2+c x^4}}\right )}{128 a^3}\\ &=-\frac {\left (5 b^2-4 a c\right ) \left (2 a+b x^2\right ) \sqrt {a+b x^2+c x^4}}{128 a^3 x^4}-\frac {\left (a+b x^2+c x^4\right )^{3/2}}{8 a x^8}+\frac {5 b \left (a+b x^2+c x^4\right )^{3/2}}{48 a^2 x^6}+\frac {\left (b^2-4 a c\right ) \left (5 b^2-4 a c\right ) \tanh ^{-1}\left (\frac {2 a+b x^2}{2 \sqrt {a} \sqrt {a+b x^2+c x^4}}\right )}{256 a^{7/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.10, size = 141, normalized size = 0.88 \begin {gather*} \frac {3 \left (16 a^2 c^2-24 a b^2 c+5 b^4\right ) \tanh ^{-1}\left (\frac {2 a+b x^2}{2 \sqrt {a} \sqrt {a+b x^2+c x^4}}\right )-\frac {2 \sqrt {a} \sqrt {a+b x^2+c x^4} \left (48 a^3+8 a^2 x^2 \left (b+3 c x^2\right )-2 a b x^4 \left (5 b+26 c x^2\right )+15 b^3 x^6\right )}{x^8}}{768 a^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + b*x^2 + c*x^4]/x^9,x]

[Out]

((-2*Sqrt[a]*Sqrt[a + b*x^2 + c*x^4]*(48*a^3 + 15*b^3*x^6 + 8*a^2*x^2*(b + 3*c*x^2) - 2*a*b*x^4*(5*b + 26*c*x^
2)))/x^8 + 3*(5*b^4 - 24*a*b^2*c + 16*a^2*c^2)*ArcTanh[(2*a + b*x^2)/(2*Sqrt[a]*Sqrt[a + b*x^2 + c*x^4])])/(76
8*a^(7/2))

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.74, size = 141, normalized size = 0.88 \begin {gather*} \frac {\left (-16 a^2 c^2+24 a b^2 c-5 b^4\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x^2-\sqrt {a+b x^2+c x^4}}{\sqrt {a}}\right )}{128 a^{7/2}}+\frac {\sqrt {a+b x^2+c x^4} \left (-48 a^3-8 a^2 b x^2-24 a^2 c x^4+10 a b^2 x^4+52 a b c x^6-15 b^3 x^6\right )}{384 a^3 x^8} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[Sqrt[a + b*x^2 + c*x^4]/x^9,x]

[Out]

(Sqrt[a + b*x^2 + c*x^4]*(-48*a^3 - 8*a^2*b*x^2 + 10*a*b^2*x^4 - 24*a^2*c*x^4 - 15*b^3*x^6 + 52*a*b*c*x^6))/(3
84*a^3*x^8) + ((-5*b^4 + 24*a*b^2*c - 16*a^2*c^2)*ArcTanh[(Sqrt[c]*x^2 - Sqrt[a + b*x^2 + c*x^4])/Sqrt[a]])/(1
28*a^(7/2))

________________________________________________________________________________________

fricas [A]  time = 0.77, size = 325, normalized size = 2.02 \begin {gather*} \left [\frac {3 \, {\left (5 \, b^{4} - 24 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {a} x^{8} \log \left (-\frac {{\left (b^{2} + 4 \, a c\right )} x^{4} + 8 \, a b x^{2} + 4 \, \sqrt {c x^{4} + b x^{2} + a} {\left (b x^{2} + 2 \, a\right )} \sqrt {a} + 8 \, a^{2}}{x^{4}}\right ) - 4 \, {\left ({\left (15 \, a b^{3} - 52 \, a^{2} b c\right )} x^{6} + 8 \, a^{3} b x^{2} - 2 \, {\left (5 \, a^{2} b^{2} - 12 \, a^{3} c\right )} x^{4} + 48 \, a^{4}\right )} \sqrt {c x^{4} + b x^{2} + a}}{1536 \, a^{4} x^{8}}, -\frac {3 \, {\left (5 \, b^{4} - 24 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {-a} x^{8} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2} + a} {\left (b x^{2} + 2 \, a\right )} \sqrt {-a}}{2 \, {\left (a c x^{4} + a b x^{2} + a^{2}\right )}}\right ) + 2 \, {\left ({\left (15 \, a b^{3} - 52 \, a^{2} b c\right )} x^{6} + 8 \, a^{3} b x^{2} - 2 \, {\left (5 \, a^{2} b^{2} - 12 \, a^{3} c\right )} x^{4} + 48 \, a^{4}\right )} \sqrt {c x^{4} + b x^{2} + a}}{768 \, a^{4} x^{8}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2+a)^(1/2)/x^9,x, algorithm="fricas")

[Out]

[1/1536*(3*(5*b^4 - 24*a*b^2*c + 16*a^2*c^2)*sqrt(a)*x^8*log(-((b^2 + 4*a*c)*x^4 + 8*a*b*x^2 + 4*sqrt(c*x^4 +
b*x^2 + a)*(b*x^2 + 2*a)*sqrt(a) + 8*a^2)/x^4) - 4*((15*a*b^3 - 52*a^2*b*c)*x^6 + 8*a^3*b*x^2 - 2*(5*a^2*b^2 -
12*a^3*c)*x^4 + 48*a^4)*sqrt(c*x^4 + b*x^2 + a))/(a^4*x^8), -1/768*(3*(5*b^4 - 24*a*b^2*c + 16*a^2*c^2)*sqrt(
-a)*x^8*arctan(1/2*sqrt(c*x^4 + b*x^2 + a)*(b*x^2 + 2*a)*sqrt(-a)/(a*c*x^4 + a*b*x^2 + a^2)) + 2*((15*a*b^3 -
52*a^2*b*c)*x^6 + 8*a^3*b*x^2 - 2*(5*a^2*b^2 - 12*a^3*c)*x^4 + 48*a^4)*sqrt(c*x^4 + b*x^2 + a))/(a^4*x^8)]

________________________________________________________________________________________

giac [B]  time = 0.28, size = 617, normalized size = 3.83 \begin {gather*} -\frac {{\left (5 \, b^{4} - 24 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \arctan \left (-\frac {\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}}{\sqrt {-a}}\right )}{128 \, \sqrt {-a} a^{3}} + \frac {15 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{7} b^{4} - 72 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{7} a b^{2} c + 48 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{7} a^{2} c^{2} - 55 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{5} a b^{4} + 264 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{5} a^{2} b^{2} c + 336 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{5} a^{3} c^{2} + 1152 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{4} a^{3} b c^{\frac {3}{2}} + 73 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{3} a^{2} b^{4} + 648 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{3} a^{3} b^{2} c + 336 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{3} a^{4} c^{2} + 384 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{2} a^{3} b^{3} \sqrt {c} + 256 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{2} a^{4} b c^{\frac {3}{2}} + 15 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} a^{3} b^{4} + 312 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} a^{4} b^{2} c + 48 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} a^{5} c^{2} + 128 \, a^{5} b c^{\frac {3}{2}}}{384 \, {\left ({\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{2} - a\right )}^{4} a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2+a)^(1/2)/x^9,x, algorithm="giac")

[Out]

-1/128*(5*b^4 - 24*a*b^2*c + 16*a^2*c^2)*arctan(-(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))/sqrt(-a))/(sqrt(-a)*a
^3) + 1/384*(15*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^7*b^4 - 72*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^7*a
*b^2*c + 48*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^7*a^2*c^2 - 55*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^5*a
*b^4 + 264*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^5*a^2*b^2*c + 336*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^5
*a^3*c^2 + 1152*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^4*a^3*b*c^(3/2) + 73*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2
+ a))^3*a^2*b^4 + 648*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^3*a^3*b^2*c + 336*(sqrt(c)*x^2 - sqrt(c*x^4 + b
*x^2 + a))^3*a^4*c^2 + 384*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^2*a^3*b^3*sqrt(c) + 256*(sqrt(c)*x^2 - sqrt
(c*x^4 + b*x^2 + a))^2*a^4*b*c^(3/2) + 15*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))*a^3*b^4 + 312*(sqrt(c)*x^2 -
sqrt(c*x^4 + b*x^2 + a))*a^4*b^2*c + 48*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))*a^5*c^2 + 128*a^5*b*c^(3/2))/
(((sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^2 - a)^4*a^3)

________________________________________________________________________________________

maple [B]  time = 0.02, size = 387, normalized size = 2.40 \begin {gather*} \frac {\sqrt {c \,x^{4}+b \,x^{2}+a}\, b \,c^{2} x^{2}}{32 a^{3}}-\frac {5 \sqrt {c \,x^{4}+b \,x^{2}+a}\, b^{3} c \,x^{2}}{128 a^{4}}+\frac {c^{2} \ln \left (\frac {b \,x^{2}+2 a +2 \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {a}}{x^{2}}\right )}{16 a^{\frac {3}{2}}}-\frac {3 b^{2} c \ln \left (\frac {b \,x^{2}+2 a +2 \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {a}}{x^{2}}\right )}{32 a^{\frac {5}{2}}}+\frac {5 b^{4} \ln \left (\frac {b \,x^{2}+2 a +2 \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {a}}{x^{2}}\right )}{256 a^{\frac {7}{2}}}-\frac {\sqrt {c \,x^{4}+b \,x^{2}+a}\, c^{2}}{16 a^{2}}+\frac {7 \sqrt {c \,x^{4}+b \,x^{2}+a}\, b^{2} c}{64 a^{3}}-\frac {5 \sqrt {c \,x^{4}+b \,x^{2}+a}\, b^{4}}{128 a^{4}}-\frac {\left (c \,x^{4}+b \,x^{2}+a \right )^{\frac {3}{2}} b c}{32 a^{3} x^{2}}+\frac {5 \left (c \,x^{4}+b \,x^{2}+a \right )^{\frac {3}{2}} b^{3}}{128 a^{4} x^{2}}+\frac {\left (c \,x^{4}+b \,x^{2}+a \right )^{\frac {3}{2}} c}{16 a^{2} x^{4}}-\frac {5 \left (c \,x^{4}+b \,x^{2}+a \right )^{\frac {3}{2}} b^{2}}{64 a^{3} x^{4}}+\frac {5 \left (c \,x^{4}+b \,x^{2}+a \right )^{\frac {3}{2}} b}{48 a^{2} x^{6}}-\frac {\left (c \,x^{4}+b \,x^{2}+a \right )^{\frac {3}{2}}}{8 a \,x^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^4+b*x^2+a)^(1/2)/x^9,x)

[Out]

-1/8*(c*x^4+b*x^2+a)^(3/2)/a/x^8+5/48*b*(c*x^4+b*x^2+a)^(3/2)/a^2/x^6-5/64*b^2/a^3/x^4*(c*x^4+b*x^2+a)^(3/2)+5
/128*b^3/a^4/x^2*(c*x^4+b*x^2+a)^(3/2)-5/128*b^4/a^4*(c*x^4+b*x^2+a)^(1/2)+5/256*b^4/a^(7/2)*ln((b*x^2+2*a+2*(
c*x^4+b*x^2+a)^(1/2)*a^(1/2))/x^2)-5/128*b^3/a^4*c*(c*x^4+b*x^2+a)^(1/2)*x^2+7/64*b^2/a^3*c*(c*x^4+b*x^2+a)^(1
/2)-3/32*b^2/a^(5/2)*c*ln((b*x^2+2*a+2*(c*x^4+b*x^2+a)^(1/2)*a^(1/2))/x^2)+1/16*c/a^2/x^4*(c*x^4+b*x^2+a)^(3/2
)-1/32*c/a^3*b/x^2*(c*x^4+b*x^2+a)^(3/2)+1/32*c^2/a^3*b*(c*x^4+b*x^2+a)^(1/2)*x^2-1/16*c^2/a^2*(c*x^4+b*x^2+a)
^(1/2)+1/16*c^2/a^(3/2)*ln((b*x^2+2*a+2*(c*x^4+b*x^2+a)^(1/2)*a^(1/2))/x^2)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2+a)^(1/2)/x^9,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*a*c-b^2>0)', see `assume?` f
or more details)Is 4*a*c-b^2 positive, negative or zero?

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sqrt {c\,x^4+b\,x^2+a}}{x^9} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2 + c*x^4)^(1/2)/x^9,x)

[Out]

int((a + b*x^2 + c*x^4)^(1/2)/x^9, x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {a + b x^{2} + c x^{4}}}{x^{9}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**4+b*x**2+a)**(1/2)/x**9,x)

[Out]

Integral(sqrt(a + b*x**2 + c*x**4)/x**9, x)

________________________________________________________________________________________