\(\int \frac {(a x^2+b x^3+c x^4)^{3/2}}{x^9} \, dx\) [48]

Optimal result

Integrand size = 24, antiderivative size = 249 \[ \int \frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{x^9} \, dx=-\frac {\left (b^2-8 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{80 a x^4}+\frac {b \left (5 b^2-28 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{320 a^2 x^3}-\frac {\left (15 b^4-100 a b^2 c+128 a^2 c^2\right ) \sqrt {a x^2+b x^3+c x^4}}{640 a^3 x^2}-\frac {3 (b+4 c x) \sqrt {a x^2+b x^3+c x^4}}{40 x^5}-\frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{5 x^8}+\frac {3 b \left (b^2-4 a c\right )^2 \text {arctanh}\left (\frac {x (2 a+b x)}{2 \sqrt {a} \sqrt {a x^2+b x^3+c x^4}}\right )}{256 a^{7/2}} \]

[Out]

-1/5*(c*x^4+b*x^3+a*x^2)^(3/2)/x^8+3/256*b*(-4*a*c+b^2)^2*arctanh(1/2*x*(b*x+2*a)/a^(1/2)/(c*x^4+b*x^3+a*x^2)^
(1/2))/a^(7/2)-1/80*(-8*a*c+b^2)*(c*x^4+b*x^3+a*x^2)^(1/2)/a/x^4+1/320*b*(-28*a*c+5*b^2)*(c*x^4+b*x^3+a*x^2)^(
1/2)/a^2/x^3-1/640*(128*a^2*c^2-100*a*b^2*c+15*b^4)*(c*x^4+b*x^3+a*x^2)^(1/2)/a^3/x^2-3/40*(4*c*x+b)*(c*x^4+b*
x^3+a*x^2)^(1/2)/x^5

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1934, 1955, 1965, 12, 1918, 212} \[ \int \frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{x^9} \, dx=\frac {3 b \left (b^2-4 a c\right )^2 \text {arctanh}\left (\frac {x (2 a+b x)}{2 \sqrt {a} \sqrt {a x^2+b x^3+c x^4}}\right )}{256 a^{7/2}}+\frac {b \left (5 b^2-28 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{320 a^2 x^3}-\frac {\left (128 a^2 c^2-100 a b^2 c+15 b^4\right ) \sqrt {a x^2+b x^3+c x^4}}{640 a^3 x^2}-\frac {\left (b^2-8 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{80 a x^4}-\frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{5 x^8}-\frac {3 (b+4 c x) \sqrt {a x^2+b x^3+c x^4}}{40 x^5} \]

[In]

Int[(a*x^2 + b*x^3 + c*x^4)^(3/2)/x^9,x]

[Out]

-1/80*((b^2 - 8*a*c)*Sqrt[a*x^2 + b*x^3 + c*x^4])/(a*x^4) + (b*(5*b^2 - 28*a*c)*Sqrt[a*x^2 + b*x^3 + c*x^4])/(
320*a^2*x^3) - ((15*b^4 - 100*a*b^2*c + 128*a^2*c^2)*Sqrt[a*x^2 + b*x^3 + c*x^4])/(640*a^3*x^2) - (3*(b + 4*c*
x)*Sqrt[a*x^2 + b*x^3 + c*x^4])/(40*x^5) - (a*x^2 + b*x^3 + c*x^4)^(3/2)/(5*x^8) + (3*b*(b^2 - 4*a*c)^2*ArcTan
h[(x*(2*a + b*x))/(2*Sqrt[a]*Sqrt[a*x^2 + b*x^3 + c*x^4])])/(256*a^(7/2))

Rule 12

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

Rule 212

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

Rule 1918

Int[1/Sqrt[(a_.)*(x_)^2 + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(r_.)], x_Symbol] :> Dist[-2/(n - 2), Subst[Int[1/(4*a
- x^2), x], x, x*((2*a + b*x^(n - 2))/Sqrt[a*x^2 + b*x^n + c*x^r])], x] /; FreeQ[{a, b, c, n, r}, x] && EqQ[r
, 2*n - 2] && PosQ[n - 2] && NeQ[b^2 - 4*a*c, 0]

Rule 1934

Int[(x_)^(m_.)*((b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(p_), x_Symbol] :> Simp[x^(m + 1)*((a*
x^q + b*x^n + c*x^(2*n - q))^p/(m + p*q + 1)), x] - Dist[(n - q)*(p/(m + p*q + 1)), Int[x^(m + n)*(b + 2*c*x^(
n - q))*(a*x^q + b*x^n + c*x^(2*n - q))^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && EqQ[r, 2*n - q] && PosQ[n -
q] &&  !IntegerQ[p] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n, 0] && GtQ[p, 0] && RationalQ[m, q] && LeQ[m + p*q + 1, -
(n - q) + 1] && NeQ[m + p*q + 1, 0]

Rule 1955

Int[(x_)^(m_.)*((c_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.))^(p_.)*((A_) + (B_.)*(x_)^(r_.)), x_Sym
bol] :> Simp[x^(m + 1)*(A*(m + p*q + (n - q)*(2*p + 1) + 1) + B*(m + p*q + 1)*x^(n - q))*((a*x^q + b*x^n + c*x
^(2*n - q))^p/((m + p*q + 1)*(m + p*q + (n - q)*(2*p + 1) + 1))), x] + Dist[(n - q)*(p/((m + p*q + 1)*(m + p*q
+ (n - q)*(2*p + 1) + 1))), Int[x^(n + m)*Simp[2*a*B*(m + p*q + 1) - A*b*(m + p*q + (n - q)*(2*p + 1) + 1) +
(b*B*(m + p*q + 1) - 2*A*c*(m + p*q + (n - q)*(2*p + 1) + 1))*x^(n - q), x]*(a*x^q + b*x^n + c*x^(2*n - q))^(p
- 1), x], x] /; FreeQ[{a, b, c, A, B}, x] && EqQ[r, n - q] && EqQ[j, 2*n - q] &&  !IntegerQ[p] && NeQ[b^2 - 4
*a*c, 0] && IGtQ[n, 0] && GtQ[p, 0] && RationalQ[m, q] && LeQ[m + p*q, -(n - q)] && NeQ[m + p*q + 1, 0] && NeQ
[m + p*q + (n - q)*(2*p + 1) + 1, 0]

Rule 1965

Int[(x_)^(m_.)*((c_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.))^(p_.)*((A_) + (B_.)*(x_)^(r_.)), x_Sym
bol] :> Simp[A*x^(m - q + 1)*((a*x^q + b*x^n + c*x^(2*n - q))^(p + 1)/(a*(m + p*q + 1))), x] + Dist[1/(a*(m +
p*q + 1)), Int[x^(m + n - q)*Simp[a*B*(m + p*q + 1) - A*b*(m + p*q + (n - q)*(p + 1) + 1) - A*c*(m + p*q + 2*(
n - q)*(p + 1) + 1)*x^(n - q), x]*(a*x^q + b*x^n + c*x^(2*n - q))^p, x], x] /; FreeQ[{a, b, c, A, B}, x] && Eq
Q[r, n - q] && EqQ[j, 2*n - q] &&  !IntegerQ[p] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n, 0] && RationalQ[m, p, q] &&
((GeQ[p, -1] && LtQ[p, 0]) || EqQ[m + p*q + (n - q)*(2*p + 1) + 1, 0]) && LeQ[m + p*q, -(n - q)] && NeQ[m + p*
q + 1, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{5 x^8}+\frac {3}{10} \int \frac {(b+2 c x) \sqrt {a x^2+b x^3+c x^4}}{x^6} \, dx \\ & = -\frac {3 (b+4 c x) \sqrt {a x^2+b x^3+c x^4}}{40 x^5}-\frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{5 x^8}+\frac {3}{160} \int \frac {2 \left (b^2-8 a c\right )-4 b c x}{x^3 \sqrt {a x^2+b x^3+c x^4}} \, dx \\ & = -\frac {\left (b^2-8 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{80 a x^4}-\frac {3 (b+4 c x) \sqrt {a x^2+b x^3+c x^4}}{40 x^5}-\frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{5 x^8}-\frac {\int \frac {b \left (5 b^2-28 a c\right )+4 c \left (b^2-8 a c\right ) x}{x^2 \sqrt {a x^2+b x^3+c x^4}} \, dx}{160 a} \\ & = -\frac {\left (b^2-8 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{80 a x^4}+\frac {b \left (5 b^2-28 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{320 a^2 x^3}-\frac {3 (b+4 c x) \sqrt {a x^2+b x^3+c x^4}}{40 x^5}-\frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{5 x^8}+\frac {\int \frac {\frac {1}{2} \left (15 b^4-100 a b^2 c+128 a^2 c^2\right )+b c \left (5 b^2-28 a c\right ) x}{x \sqrt {a x^2+b x^3+c x^4}} \, dx}{320 a^2} \\ & = -\frac {\left (b^2-8 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{80 a x^4}+\frac {b \left (5 b^2-28 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{320 a^2 x^3}-\frac {\left (15 b^4-100 a b^2 c+128 a^2 c^2\right ) \sqrt {a x^2+b x^3+c x^4}}{640 a^3 x^2}-\frac {3 (b+4 c x) \sqrt {a x^2+b x^3+c x^4}}{40 x^5}-\frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{5 x^8}-\frac {\int \frac {15 b \left (b^2-4 a c\right )^2}{4 \sqrt {a x^2+b x^3+c x^4}} \, dx}{320 a^3} \\ & = -\frac {\left (b^2-8 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{80 a x^4}+\frac {b \left (5 b^2-28 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{320 a^2 x^3}-\frac {\left (15 b^4-100 a b^2 c+128 a^2 c^2\right ) \sqrt {a x^2+b x^3+c x^4}}{640 a^3 x^2}-\frac {3 (b+4 c x) \sqrt {a x^2+b x^3+c x^4}}{40 x^5}-\frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{5 x^8}-\frac {\left (3 b \left (b^2-4 a c\right )^2\right ) \int \frac {1}{\sqrt {a x^2+b x^3+c x^4}} \, dx}{256 a^3} \\ & = -\frac {\left (b^2-8 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{80 a x^4}+\frac {b \left (5 b^2-28 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{320 a^2 x^3}-\frac {\left (15 b^4-100 a b^2 c+128 a^2 c^2\right ) \sqrt {a x^2+b x^3+c x^4}}{640 a^3 x^2}-\frac {3 (b+4 c x) \sqrt {a x^2+b x^3+c x^4}}{40 x^5}-\frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{5 x^8}+\frac {\left (3 b \left (b^2-4 a c\right )^2\right ) \text {Subst}\left (\int \frac {1}{4 a-x^2} \, dx,x,\frac {x (2 a+b x)}{\sqrt {a x^2+b x^3+c x^4}}\right )}{128 a^3} \\ & = -\frac {\left (b^2-8 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{80 a x^4}+\frac {b \left (5 b^2-28 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{320 a^2 x^3}-\frac {\left (15 b^4-100 a b^2 c+128 a^2 c^2\right ) \sqrt {a x^2+b x^3+c x^4}}{640 a^3 x^2}-\frac {3 (b+4 c x) \sqrt {a x^2+b x^3+c x^4}}{40 x^5}-\frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{5 x^8}+\frac {3 b \left (b^2-4 a c\right )^2 \tanh ^{-1}\left (\frac {x (2 a+b x)}{2 \sqrt {a} \sqrt {a x^2+b x^3+c x^4}}\right )}{256 a^{7/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.98 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.71 \[ \int \frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{x^9} \, dx=-\frac {\sqrt {x^2 (a+x (b+c x))} \left (\sqrt {a} \sqrt {a+x (b+c x)} \left (128 a^4+15 b^4 x^4-10 a b^2 x^3 (b+10 c x)+16 a^3 x (11 b+16 c x)+8 a^2 x^2 \left (b^2+7 b c x+16 c^2 x^2\right )\right )+15 b \left (b^2-4 a c\right )^2 x^5 \text {arctanh}\left (\frac {\sqrt {c} x-\sqrt {a+x (b+c x)}}{\sqrt {a}}\right )\right )}{640 a^{7/2} x^6 \sqrt {a+x (b+c x)}} \]

[In]

Integrate[(a*x^2 + b*x^3 + c*x^4)^(3/2)/x^9,x]

[Out]

-1/640*(Sqrt[x^2*(a + x*(b + c*x))]*(Sqrt[a]*Sqrt[a + x*(b + c*x)]*(128*a^4 + 15*b^4*x^4 - 10*a*b^2*x^3*(b + 1
0*c*x) + 16*a^3*x*(11*b + 16*c*x) + 8*a^2*x^2*(b^2 + 7*b*c*x + 16*c^2*x^2)) + 15*b*(b^2 - 4*a*c)^2*x^5*ArcTanh
[(Sqrt[c]*x - Sqrt[a + x*(b + c*x)])/Sqrt[a]]))/(a^(7/2)*x^6*Sqrt[a + x*(b + c*x)])

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.65

method	result	size
pseudoelliptic	\(\frac {\frac {3 x^{5} b \left (a c -\frac {b^{2}}{4}\right )^{2} \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x \sqrt {a}}\right )}{16}+\frac {3 \left (-\frac {x^{2} \left (16 c^{2} x^{2}+7 b c x +b^{2}\right ) a^{\frac {5}{2}}}{15}+\frac {b^{2} x^{3} \left (10 c x +b \right ) a^{\frac {3}{2}}}{12}-\frac {22 x \left (\frac {16 c x}{11}+b \right ) a^{\frac {7}{2}}}{15}-\frac {b^{4} x^{4} \sqrt {a}}{8}-\frac {16 a^{\frac {9}{2}}}{15}\right ) \sqrt {c \,x^{2}+b x +a}}{16}-\frac {3 \ln \left (2\right ) x^{5} b \left (a c -\frac {b^{2}}{4}\right )^{2}}{16}}{a^{\frac {7}{2}} x^{5}}\)	\(161\)
risch	\(-\frac {\left (128 a^{2} c^{2} x^{4}-100 a \,b^{2} c \,x^{4}+15 b^{4} x^{4}+56 a^{2} b c \,x^{3}-10 a \,b^{3} x^{3}+256 a^{3} c \,x^{2}+8 a^{2} b^{2} x^{2}+176 a^{3} b x +128 a^{4}\right ) \sqrt {x^{2} \left (c \,x^{2}+b x +a \right )}}{640 x^{6} a^{3}}+\frac {3 \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right ) b \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right ) \sqrt {x^{2} \left (c \,x^{2}+b x +a \right )}}{256 a^{\frac {7}{2}} x \sqrt {c \,x^{2}+b x +a}}\)	\(192\)
default	\(\frac {\left (c \,x^{4}+b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}} \left (240 c^{2} a^{\frac {7}{2}} \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right ) b \,x^{5}+120 c^{2} \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} a \,b^{2} x^{6}-120 c \,a^{\frac {5}{2}} \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right ) b^{3} x^{5}-80 c^{2} \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} a^{2} b \,x^{5}+120 c^{2} \sqrt {c \,x^{2}+b x +a}\, a^{2} b^{2} x^{6}-10 c \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} b^{4} x^{6}-240 c^{2} \sqrt {c \,x^{2}+b x +a}\, a^{3} b \,x^{5}-120 c \left (c \,x^{2}+b x +a \right )^{\frac {5}{2}} a \,b^{2} x^{4}+100 c \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} a \,b^{3} x^{5}-30 c \sqrt {c \,x^{2}+b x +a}\, a \,b^{4} x^{6}+15 a^{\frac {3}{2}} \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right ) b^{5} x^{5}+80 c \left (c \,x^{2}+b x +a \right )^{\frac {5}{2}} a^{2} b \,x^{3}+180 c \sqrt {c \,x^{2}+b x +a}\, a^{2} b^{3} x^{5}+10 \left (c \,x^{2}+b x +a \right )^{\frac {5}{2}} b^{4} x^{4}-10 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} b^{5} x^{5}+20 \left (c \,x^{2}+b x +a \right )^{\frac {5}{2}} a \,b^{3} x^{3}-30 \sqrt {c \,x^{2}+b x +a}\, a \,b^{5} x^{5}-80 \left (c \,x^{2}+b x +a \right )^{\frac {5}{2}} a^{2} b^{2} x^{2}+160 \left (c \,x^{2}+b x +a \right )^{\frac {5}{2}} a^{3} b x -256 \left (c \,x^{2}+b x +a \right )^{\frac {5}{2}} a^{4}\right )}{1280 x^{8} \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} a^{5}}\)	\(534\)

[In]

int((c*x^4+b*x^3+a*x^2)^(3/2)/x^9,x,method=_RETURNVERBOSE)

[Out]

3/16/a^(7/2)*(x^5*b*(a*c-1/4*b^2)^2*ln((2*a+b*x+2*a^(1/2)*(c*x^2+b*x+a)^(1/2))/x/a^(1/2))+(-1/15*x^2*(16*c^2*x
^2+7*b*c*x+b^2)*a^(5/2)+1/12*b^2*x^3*(10*c*x+b)*a^(3/2)-22/15*x*(16/11*c*x+b)*a^(7/2)-1/8*b^4*x^4*a^(1/2)-16/1
5*a^(9/2))*(c*x^2+b*x+a)^(1/2)-ln(2)*x^5*b*(a*c-1/4*b^2)^2)/x^5

Fricas [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 394, normalized size of antiderivative = 1.58 \[ \int \frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{x^9} \, dx=\left [\frac {15 \, {\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} \sqrt {a} x^{6} \log \left (-\frac {8 \, a b x^{2} + {\left (b^{2} + 4 \, a c\right )} x^{3} + 8 \, a^{2} x + 4 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (b x + 2 \, a\right )} \sqrt {a}}{x^{3}}\right ) - 4 \, {\left (176 \, a^{4} b x + 128 \, a^{5} + {\left (15 \, a b^{4} - 100 \, a^{2} b^{2} c + 128 \, a^{3} c^{2}\right )} x^{4} - 2 \, {\left (5 \, a^{2} b^{3} - 28 \, a^{3} b c\right )} x^{3} + 8 \, {\left (a^{3} b^{2} + 32 \, a^{4} c\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{3} + a x^{2}}}{2560 \, a^{4} x^{6}}, -\frac {15 \, {\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} \sqrt {-a} x^{6} \arctan \left (\frac {\sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (b x + 2 \, a\right )} \sqrt {-a}}{2 \, {\left (a c x^{3} + a b x^{2} + a^{2} x\right )}}\right ) + 2 \, {\left (176 \, a^{4} b x + 128 \, a^{5} + {\left (15 \, a b^{4} - 100 \, a^{2} b^{2} c + 128 \, a^{3} c^{2}\right )} x^{4} - 2 \, {\left (5 \, a^{2} b^{3} - 28 \, a^{3} b c\right )} x^{3} + 8 \, {\left (a^{3} b^{2} + 32 \, a^{4} c\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{3} + a x^{2}}}{1280 \, a^{4} x^{6}}\right ] \]

[In]

integrate((c*x^4+b*x^3+a*x^2)^(3/2)/x^9,x, algorithm="fricas")

[Out]

[1/2560*(15*(b^5 - 8*a*b^3*c + 16*a^2*b*c^2)*sqrt(a)*x^6*log(-(8*a*b*x^2 + (b^2 + 4*a*c)*x^3 + 8*a^2*x + 4*sqr
t(c*x^4 + b*x^3 + a*x^2)*(b*x + 2*a)*sqrt(a))/x^3) - 4*(176*a^4*b*x + 128*a^5 + (15*a*b^4 - 100*a^2*b^2*c + 12
8*a^3*c^2)*x^4 - 2*(5*a^2*b^3 - 28*a^3*b*c)*x^3 + 8*(a^3*b^2 + 32*a^4*c)*x^2)*sqrt(c*x^4 + b*x^3 + a*x^2))/(a^
4*x^6), -1/1280*(15*(b^5 - 8*a*b^3*c + 16*a^2*b*c^2)*sqrt(-a)*x^6*arctan(1/2*sqrt(c*x^4 + b*x^3 + a*x^2)*(b*x
+ 2*a)*sqrt(-a)/(a*c*x^3 + a*b*x^2 + a^2*x)) + 2*(176*a^4*b*x + 128*a^5 + (15*a*b^4 - 100*a^2*b^2*c + 128*a^3*
c^2)*x^4 - 2*(5*a^2*b^3 - 28*a^3*b*c)*x^3 + 8*(a^3*b^2 + 32*a^4*c)*x^2)*sqrt(c*x^4 + b*x^3 + a*x^2))/(a^4*x^6)
]

Sympy [F]

\[ \int \frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{x^9} \, dx=\int \frac {\left (x^{2} \left (a + b x + c x^{2}\right )\right )^{\frac {3}{2}}}{x^{9}}\, dx \]

[In]

integrate((c*x**4+b*x**3+a*x**2)**(3/2)/x**9,x)

[Out]

Integral((x**2*(a + b*x + c*x**2))**(3/2)/x**9, x)

Maxima [F]

\[ \int \frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{x^9} \, dx=\int { \frac {{\left (c x^{4} + b x^{3} + a x^{2}\right )}^{\frac {3}{2}}}{x^{9}} \,d x } \]

[In]

integrate((c*x^4+b*x^3+a*x^2)^(3/2)/x^9,x, algorithm="maxima")

[Out]

integrate((c*x^4 + b*x^3 + a*x^2)^(3/2)/x^9, x)

Giac [F(-2)]

Exception generated. \[ \int \frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{x^9} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate((c*x^4+b*x^3+a*x^2)^(3/2)/x^9,x, algorithm="giac")

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Not invertible Error: Bad Argument Value

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{x^9} \, dx=\int \frac {{\left (c\,x^4+b\,x^3+a\,x^2\right )}^{3/2}}{x^9} \,d x \]

[In]

int((a*x^2 + b*x^3 + c*x^4)^(3/2)/x^9,x)

[Out]

int((a*x^2 + b*x^3 + c*x^4)^(3/2)/x^9, x)