\(\int \frac {x^6}{(a x^2+b x^3+c x^4)^{3/2}} \, dx\) [56]

Optimal result

Integrand size = 24, antiderivative size = 201 \[ \int \frac {x^6}{\left (a x^2+b x^3+c x^4\right )^{3/2}} \, dx=\frac {2 x^3 (2 a+b x)}{\left (b^2-4 a c\right ) \sqrt {a x^2+b x^3+c x^4}}-\frac {2 b \sqrt {a x^2+b x^3+c x^4}}{c \left (b^2-4 a c\right )}+\frac {\left (3 b^2-8 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{c^2 \left (b^2-4 a c\right ) x}-\frac {3 b x \sqrt {a+b x+c x^2} \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 c^{5/2} \sqrt {a x^2+b x^3+c x^4}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1937, 1963, 12, 1928, 635, 212} \[ \int \frac {x^6}{\left (a x^2+b x^3+c x^4\right )^{3/2}} \, dx=-\frac {3 b x \sqrt {a+b x+c x^2} \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 c^{5/2} \sqrt {a x^2+b x^3+c x^4}}+\frac {\left (3 b^2-8 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{c^2 x \left (b^2-4 a c\right )}+\frac {2 x^3 (2 a+b x)}{\left (b^2-4 a c\right ) \sqrt {a x^2+b x^3+c x^4}}-\frac {2 b \sqrt {a x^2+b x^3+c x^4}}{c \left (b^2-4 a c\right )} \]

[In]

Int[x^6/(a*x^2 + b*x^3 + c*x^4)^(3/2),x]

[Out]

(2*x^3*(2*a + b*x))/((b^2 - 4*a*c)*Sqrt[a*x^2 + b*x^3 + c*x^4]) - (2*b*Sqrt[a*x^2 + b*x^3 + c*x^4])/(c*(b^2 -
4*a*c)) + ((3*b^2 - 8*a*c)*Sqrt[a*x^2 + b*x^3 + c*x^4])/(c^2*(b^2 - 4*a*c)*x) - (3*b*x*Sqrt[a + b*x + c*x^2]*A
rcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(2*c^(5/2)*Sqrt[a*x^2 + b*x^3 + c*x^4])

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 635

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

Rule 1928

Int[(x_)^(m_.)/Sqrt[(b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.)], x_Symbol] :> Dist[x^(q/2)*(Sqrt[a
+ b*x^(n - q) + c*x^(2*(n - q))]/Sqrt[a*x^q + b*x^n + c*x^(2*n - q)]), Int[x^(m - q/2)/Sqrt[a + b*x^(n - q) +
c*x^(2*(n - q))], x], x] /; FreeQ[{a, b, c, m, n, q}, x] && EqQ[r, 2*n - q] && PosQ[n - q] && ((EqQ[m, 1] &&
EqQ[n, 3] && EqQ[q, 2]) || ((EqQ[m + 1/2] || EqQ[m, 3/2] || EqQ[m, 1/2] || EqQ[m, 5/2]) && EqQ[n, 3] && EqQ[q,
1]))

Rule 1937

Int[(x_)^(m_.)*((b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(p_), x_Symbol] :> Simp[(-x^(m - 2*n +
q + 1))*(2*a + b*x^(n - q))*((a*x^q + b*x^n + c*x^(2*n - q))^(p + 1)/((n - q)*(p + 1)*(b^2 - 4*a*c))), x] + D
ist[1/((n - q)*(p + 1)*(b^2 - 4*a*c)), Int[x^(m - 2*n + q)*(2*a*(m + p*q - 2*(n - q) + 1) + b*(m + p*q + (n -
q)*(2*p + 1) + 1)*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] && LtQ[p, -1] && RationalQ[m, q]
&& GtQ[m + p*q + 1, 2*(n - q)]

Rule 1963

Int[(x_)^(m_.)*((c_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.))^(p_.)*((A_) + (B_.)*(x_)^(r_.)), x_Sym
bol] :> Simp[B*x^(m - n + 1)*((a*x^q + b*x^n + c*x^(2*n - q))^(p + 1)/(c*(m + p*q + (n - q)*(2*p + 1) + 1))),
x] - Dist[1/(c*(m + p*q + (n - q)*(2*p + 1) + 1)), Int[x^(m - n + q)*Simp[a*B*(m + p*q - n + q + 1) + (b*B*(m
+ p*q + (n - q)*p + 1) - 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, 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] && GeQ[p, -1] && LtQ[p, 0] && RationalQ[m, q] && GeQ[m + p*q, n - q - 1] && NeQ[m + p*q + (
n - q)*(2*p + 1) + 1, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {2 x^3 (2 a+b x)}{\left (b^2-4 a c\right ) \sqrt {a x^2+b x^3+c x^4}}-\frac {2 \int \frac {x^2 (4 a+2 b x)}{\sqrt {a x^2+b x^3+c x^4}} \, dx}{b^2-4 a c} \\ & = \frac {2 x^3 (2 a+b x)}{\left (b^2-4 a c\right ) \sqrt {a x^2+b x^3+c x^4}}-\frac {2 b \sqrt {a x^2+b x^3+c x^4}}{c \left (b^2-4 a c\right )}+\frac {\int \frac {x \left (2 a b+\left (3 b^2-8 a c\right ) x\right )}{\sqrt {a x^2+b x^3+c x^4}} \, dx}{c \left (b^2-4 a c\right )} \\ & = \frac {2 x^3 (2 a+b x)}{\left (b^2-4 a c\right ) \sqrt {a x^2+b x^3+c x^4}}-\frac {2 b \sqrt {a x^2+b x^3+c x^4}}{c \left (b^2-4 a c\right )}+\frac {\left (3 b^2-8 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{c^2 \left (b^2-4 a c\right ) x}-\frac {\int \frac {3 b \left (b^2-4 a c\right ) x}{2 \sqrt {a x^2+b x^3+c x^4}} \, dx}{c^2 \left (b^2-4 a c\right )} \\ & = \frac {2 x^3 (2 a+b x)}{\left (b^2-4 a c\right ) \sqrt {a x^2+b x^3+c x^4}}-\frac {2 b \sqrt {a x^2+b x^3+c x^4}}{c \left (b^2-4 a c\right )}+\frac {\left (3 b^2-8 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{c^2 \left (b^2-4 a c\right ) x}-\frac {(3 b) \int \frac {x}{\sqrt {a x^2+b x^3+c x^4}} \, dx}{2 c^2} \\ & = \frac {2 x^3 (2 a+b x)}{\left (b^2-4 a c\right ) \sqrt {a x^2+b x^3+c x^4}}-\frac {2 b \sqrt {a x^2+b x^3+c x^4}}{c \left (b^2-4 a c\right )}+\frac {\left (3 b^2-8 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{c^2 \left (b^2-4 a c\right ) x}-\frac {\left (3 b x \sqrt {a+b x+c x^2}\right ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{2 c^2 \sqrt {a x^2+b x^3+c x^4}} \\ & = \frac {2 x^3 (2 a+b x)}{\left (b^2-4 a c\right ) \sqrt {a x^2+b x^3+c x^4}}-\frac {2 b \sqrt {a x^2+b x^3+c x^4}}{c \left (b^2-4 a c\right )}+\frac {\left (3 b^2-8 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{c^2 \left (b^2-4 a c\right ) x}-\frac {\left (3 b x \sqrt {a+b x+c x^2}\right ) \text {Subst}\left (\int \frac {1}{4 c-x^2} \, dx,x,\frac {b+2 c x}{\sqrt {a+b x+c x^2}}\right )}{c^2 \sqrt {a x^2+b x^3+c x^4}} \\ & = \frac {2 x^3 (2 a+b x)}{\left (b^2-4 a c\right ) \sqrt {a x^2+b x^3+c x^4}}-\frac {2 b \sqrt {a x^2+b x^3+c x^4}}{c \left (b^2-4 a c\right )}+\frac {\left (3 b^2-8 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{c^2 \left (b^2-4 a c\right ) x}-\frac {3 b x \sqrt {a+b x+c x^2} \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 c^{5/2} \sqrt {a x^2+b x^3+c x^4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.43 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.71 \[ \int \frac {x^6}{\left (a x^2+b x^3+c x^4\right )^{3/2}} \, dx=\frac {x \left (2 \sqrt {c} \left (8 a^2 c-b^2 x (3 b+c x)+a \left (-3 b^2+10 b c x+4 c^2 x^2\right )\right )-3 b \left (b^2-4 a c\right ) \sqrt {a+x (b+c x)} \log \left (c^2 \left (b+2 c x-2 \sqrt {c} \sqrt {a+x (b+c x)}\right )\right )\right )}{2 c^{5/2} \left (-b^2+4 a c\right ) \sqrt {x^2 (a+x (b+c x))}} \]

[In]

Integrate[x^6/(a*x^2 + b*x^3 + c*x^4)^(3/2),x]

[Out]

(x*(2*Sqrt[c]*(8*a^2*c - b^2*x*(3*b + c*x) + a*(-3*b^2 + 10*b*c*x + 4*c^2*x^2)) - 3*b*(b^2 - 4*a*c)*Sqrt[a + x
*(b + c*x)]*Log[c^2*(b + 2*c*x - 2*Sqrt[c]*Sqrt[a + x*(b + c*x)])]))/(2*c^(5/2)*(-b^2 + 4*a*c)*Sqrt[x^2*(a + x
*(b + c*x))])

Maple [A] (verified)

Time = 0.19 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.64

method	result	size
pseudoelliptic	\(-\frac {6 \left (\frac {\left (\frac {1}{2} b^{2} x^{2}-5 a b x -4 a^{2}\right ) c^{\frac {3}{2}}}{3}-\frac {2 c^{\frac {5}{2}} a \,x^{2}}{3}+\frac {b \left (2 \left (b^{2} x +a b \right ) \sqrt {c}+\sqrt {c \,x^{2}+b x +a}\, \ln \left (2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {c}+2 c x +b \right ) \left (4 a c -b^{2}\right )\right )}{4}\right )}{c^{\frac {5}{2}} \sqrt {c \,x^{2}+b x +a}\, \left (4 a c -b^{2}\right )}\)	\(129\)
default	\(\frac {x^{3} \left (c \,x^{2}+b x +a \right ) \left (8 a \,c^{\frac {7}{2}} x^{2}-2 c^{\frac {5}{2}} b^{2} x^{2}+20 c^{\frac {5}{2}} a b x -6 c^{\frac {3}{2}} b^{3} x +16 c^{\frac {5}{2}} a^{2}-6 c^{\frac {3}{2}} a \,b^{2}-12 \ln \left (\frac {2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {c}+2 c x +b}{2 \sqrt {c}}\right ) \sqrt {c \,x^{2}+b x +a}\, a b \,c^{2}+3 \ln \left (\frac {2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {c}+2 c x +b}{2 \sqrt {c}}\right ) \sqrt {c \,x^{2}+b x +a}\, b^{3} c \right )}{2 c^{\frac {7}{2}} \left (c \,x^{4}+b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}} \left (4 a c -b^{2}\right )}\)	\(199\)
risch	\(\frac {\left (c \,x^{2}+b x +a \right ) x}{c^{2} \sqrt {x^{2} \left (c \,x^{2}+b x +a \right )}}+\frac {\left (\frac {3 b x}{2 c^{2} \sqrt {c \,x^{2}+b x +a}}-\frac {b^{2}}{4 c^{3} \sqrt {c \,x^{2}+b x +a}}-\frac {b^{3} x}{2 c^{2} \left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}-\frac {b^{4}}{4 c^{3} \left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}-\frac {3 b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {5}{2}}}+\frac {a}{c^{2} \sqrt {c \,x^{2}+b x +a}}\right ) x \sqrt {c \,x^{2}+b x +a}}{\sqrt {x^{2} \left (c \,x^{2}+b x +a \right )}}\)	\(216\)

[In]

int(x^6/(c*x^4+b*x^3+a*x^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

-6/c^(5/2)/(c*x^2+b*x+a)^(1/2)*(1/3*(1/2*b^2*x^2-5*a*b*x-4*a^2)*c^(3/2)-2/3*c^(5/2)*a*x^2+1/4*b*(2*(b^2*x+a*b)
*c^(1/2)+(c*x^2+b*x+a)^(1/2)*ln(2*(c*x^2+b*x+a)^(1/2)*c^(1/2)+2*c*x+b)*(4*a*c-b^2)))/(4*a*c-b^2)

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 486, normalized size of antiderivative = 2.42 \[ \int \frac {x^6}{\left (a x^2+b x^3+c x^4\right )^{3/2}} \, dx=\left [\frac {3 \, {\left ({\left (b^{3} c - 4 \, a b c^{2}\right )} x^{3} + {\left (b^{4} - 4 \, a b^{2} c\right )} x^{2} + {\left (a b^{3} - 4 \, a^{2} b c\right )} x\right )} \sqrt {c} \log \left (-\frac {8 \, c^{2} x^{3} + 8 \, b c x^{2} - 4 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (2 \, c x + b\right )} \sqrt {c} + {\left (b^{2} + 4 \, a c\right )} x}{x}\right ) + 4 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (3 \, a b^{2} c - 8 \, a^{2} c^{2} + {\left (b^{2} c^{2} - 4 \, a c^{3}\right )} x^{2} + {\left (3 \, b^{3} c - 10 \, a b c^{2}\right )} x\right )}}{4 \, {\left ({\left (b^{2} c^{4} - 4 \, a c^{5}\right )} x^{3} + {\left (b^{3} c^{3} - 4 \, a b c^{4}\right )} x^{2} + {\left (a b^{2} c^{3} - 4 \, a^{2} c^{4}\right )} x\right )}}, \frac {3 \, {\left ({\left (b^{3} c - 4 \, a b c^{2}\right )} x^{3} + {\left (b^{4} - 4 \, a b^{2} c\right )} x^{2} + {\left (a b^{3} - 4 \, a^{2} b c\right )} x\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{3} + b c x^{2} + a c x\right )}}\right ) + 2 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (3 \, a b^{2} c - 8 \, a^{2} c^{2} + {\left (b^{2} c^{2} - 4 \, a c^{3}\right )} x^{2} + {\left (3 \, b^{3} c - 10 \, a b c^{2}\right )} x\right )}}{2 \, {\left ({\left (b^{2} c^{4} - 4 \, a c^{5}\right )} x^{3} + {\left (b^{3} c^{3} - 4 \, a b c^{4}\right )} x^{2} + {\left (a b^{2} c^{3} - 4 \, a^{2} c^{4}\right )} x\right )}}\right ] \]

[In]

integrate(x^6/(c*x^4+b*x^3+a*x^2)^(3/2),x, algorithm="fricas")

[Out]

[1/4*(3*((b^3*c - 4*a*b*c^2)*x^3 + (b^4 - 4*a*b^2*c)*x^2 + (a*b^3 - 4*a^2*b*c)*x)*sqrt(c)*log(-(8*c^2*x^3 + 8*
b*c*x^2 - 4*sqrt(c*x^4 + b*x^3 + a*x^2)*(2*c*x + b)*sqrt(c) + (b^2 + 4*a*c)*x)/x) + 4*sqrt(c*x^4 + b*x^3 + a*x
^2)*(3*a*b^2*c - 8*a^2*c^2 + (b^2*c^2 - 4*a*c^3)*x^2 + (3*b^3*c - 10*a*b*c^2)*x))/((b^2*c^4 - 4*a*c^5)*x^3 + (
b^3*c^3 - 4*a*b*c^4)*x^2 + (a*b^2*c^3 - 4*a^2*c^4)*x), 1/2*(3*((b^3*c - 4*a*b*c^2)*x^3 + (b^4 - 4*a*b^2*c)*x^2
+ (a*b^3 - 4*a^2*b*c)*x)*sqrt(-c)*arctan(1/2*sqrt(c*x^4 + b*x^3 + a*x^2)*(2*c*x + b)*sqrt(-c)/(c^2*x^3 + b*c*
x^2 + a*c*x)) + 2*sqrt(c*x^4 + b*x^3 + a*x^2)*(3*a*b^2*c - 8*a^2*c^2 + (b^2*c^2 - 4*a*c^3)*x^2 + (3*b^3*c - 10
*a*b*c^2)*x))/((b^2*c^4 - 4*a*c^5)*x^3 + (b^3*c^3 - 4*a*b*c^4)*x^2 + (a*b^2*c^3 - 4*a^2*c^4)*x)]

Sympy [F]

\[ \int \frac {x^6}{\left (a x^2+b x^3+c x^4\right )^{3/2}} \, dx=\int \frac {x^{6}}{\left (x^{2} \left (a + b x + c x^{2}\right )\right )^{\frac {3}{2}}}\, dx \]

[In]

integrate(x**6/(c*x**4+b*x**3+a*x**2)**(3/2),x)

[Out]

Integral(x**6/(x**2*(a + b*x + c*x**2))**(3/2), x)

Maxima [F]

\[ \int \frac {x^6}{\left (a x^2+b x^3+c x^4\right )^{3/2}} \, dx=\int { \frac {x^{6}}{{\left (c x^{4} + b x^{3} + a x^{2}\right )}^{\frac {3}{2}}} \,d x } \]

[In]

integrate(x^6/(c*x^4+b*x^3+a*x^2)^(3/2),x, algorithm="maxima")

[Out]

integrate(x^6/(c*x^4 + b*x^3 + a*x^2)^(3/2), x)

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.15 \[ \int \frac {x^6}{\left (a x^2+b x^3+c x^4\right )^{3/2}} \, dx=-\frac {{\left (3 \, b^{3} \log \left ({\left | b - 2 \, \sqrt {a} \sqrt {c} \right |}\right ) - 12 \, a b c \log \left ({\left | b - 2 \, \sqrt {a} \sqrt {c} \right |}\right ) + 6 \, \sqrt {a} b^{2} \sqrt {c} - 16 \, a^{\frac {3}{2}} c^{\frac {3}{2}}\right )} \mathrm {sgn}\left (x\right )}{2 \, {\left (b^{2} c^{\frac {5}{2}} - 4 \, a c^{\frac {7}{2}}\right )}} + \frac {{\left (\frac {{\left (b^{2} c - 4 \, a c^{2}\right )} x}{b^{2} c^{2} \mathrm {sgn}\left (x\right ) - 4 \, a c^{3} \mathrm {sgn}\left (x\right )} + \frac {3 \, b^{3} - 10 \, a b c}{b^{2} c^{2} \mathrm {sgn}\left (x\right ) - 4 \, a c^{3} \mathrm {sgn}\left (x\right )}\right )} x + \frac {3 \, a b^{2} - 8 \, a^{2} c}{b^{2} c^{2} \mathrm {sgn}\left (x\right ) - 4 \, a c^{3} \mathrm {sgn}\left (x\right )}}{\sqrt {c x^{2} + b x + a}} + \frac {3 \, b \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} + b \right |}\right )}{2 \, c^{\frac {5}{2}} \mathrm {sgn}\left (x\right )} \]

[In]

integrate(x^6/(c*x^4+b*x^3+a*x^2)^(3/2),x, algorithm="giac")

[Out]

-1/2*(3*b^3*log(abs(b - 2*sqrt(a)*sqrt(c))) - 12*a*b*c*log(abs(b - 2*sqrt(a)*sqrt(c))) + 6*sqrt(a)*b^2*sqrt(c)
- 16*a^(3/2)*c^(3/2))*sgn(x)/(b^2*c^(5/2) - 4*a*c^(7/2)) + (((b^2*c - 4*a*c^2)*x/(b^2*c^2*sgn(x) - 4*a*c^3*sg
n(x)) + (3*b^3 - 10*a*b*c)/(b^2*c^2*sgn(x) - 4*a*c^3*sgn(x)))*x + (3*a*b^2 - 8*a^2*c)/(b^2*c^2*sgn(x) - 4*a*c^
3*sgn(x)))/sqrt(c*x^2 + b*x + a) + 3/2*b*log(abs(2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c) + b))/(c^(5/2)*
sgn(x))

Mupad [F(-1)]

Timed out. \[ \int \frac {x^6}{\left (a x^2+b x^3+c x^4\right )^{3/2}} \, dx=\int \frac {x^6}{{\left (c\,x^4+b\,x^3+a\,x^2\right )}^{3/2}} \,d x \]

[In]

int(x^6/(a*x^2 + b*x^3 + c*x^4)^(3/2),x)

[Out]

int(x^6/(a*x^2 + b*x^3 + c*x^4)^(3/2), x)