\(\int \frac {x^4}{(a x^2+b x^3+c x^4)^{3/2}} \, dx\) [58]

Optimal result

Integrand size = 24, antiderivative size = 40 \[ \int \frac {x^4}{\left (a x^2+b x^3+c x^4\right )^{3/2}} \, dx=\frac {2 x (2 a+b x)}{\left (b^2-4 a c\right ) \sqrt {a x^2+b x^3+c x^4}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {1930} \[ \int \frac {x^4}{\left (a x^2+b x^3+c x^4\right )^{3/2}} \, dx=\frac {2 x (2 a+b x)}{\left (b^2-4 a c\right ) \sqrt {a x^2+b x^3+c x^4}} \]

[In]

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

[Out]

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

Rule 1930

Int[(x_)^(m_.)/((b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(3/2), x_Symbol] :> Simp[x^((n - 1)/2)
*((4*a + 2*b*x)/((b^2 - 4*a*c)*Sqrt[a*x^(n - 1) + b*x^n + c*x^(n + 1)])), x] /; FreeQ[{a, b, c, n}, x] && EqQ[
m, (3*n - 1)/2] && EqQ[q, n - 1] && EqQ[r, n + 1] && NeQ[b^2 - 4*a*c, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {2 x (2 a+b x)}{\left (b^2-4 a c\right ) \sqrt {a x^2+b x^3+c x^4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.92 \[ \int \frac {x^4}{\left (a x^2+b x^3+c x^4\right )^{3/2}} \, dx=\frac {2 x (2 a+b x)}{\left (b^2-4 a c\right ) \sqrt {x^2 (a+x (b+c x))}} \]

[In]

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

[Out]

(2*x*(2*a + b*x))/((b^2 - 4*a*c)*Sqrt[x^2*(a + x*(b + c*x))])

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.85

method	result	size
pseudoelliptic	\(\frac {-2 b x -4 a}{\sqrt {c \,x^{2}+b x +a}\, \left (4 a c -b^{2}\right )}\)	\(34\)
gosper	\(-\frac {2 \left (c \,x^{2}+b x +a \right ) \left (b x +2 a \right ) x^{3}}{\left (4 a c -b^{2}\right ) \left (c \,x^{4}+b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}}}\)	\(53\)
default	\(-\frac {2 \left (c \,x^{2}+b x +a \right ) \left (b x +2 a \right ) x^{3}}{\left (4 a c -b^{2}\right ) \left (c \,x^{4}+b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}}}\)	\(53\)
trager	\(-\frac {2 \left (b x +2 a \right ) \sqrt {c \,x^{4}+b \,x^{3}+a \,x^{2}}}{\left (c \,x^{2}+b x +a \right ) x \left (4 a c -b^{2}\right )}\)	\(55\)

[In]

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

[Out]

(-2*b*x-4*a)/(c*x^2+b*x+a)^(1/2)/(4*a*c-b^2)

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.82 \[ \int \frac {x^4}{\left (a x^2+b x^3+c x^4\right )^{3/2}} \, dx=\frac {2 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (b x + 2 \, a\right )}}{{\left (b^{2} c - 4 \, a c^{2}\right )} x^{3} + {\left (b^{3} - 4 \, a b c\right )} x^{2} + {\left (a b^{2} - 4 \, a^{2} c\right )} x} \]

[In]

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

[Out]

2*sqrt(c*x^4 + b*x^3 + a*x^2)*(b*x + 2*a)/((b^2*c - 4*a*c^2)*x^3 + (b^3 - 4*a*b*c)*x^2 + (a*b^2 - 4*a^2*c)*x)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.72 \[ \int \frac {x^4}{\left (a x^2+b x^3+c x^4\right )^{3/2}} \, dx=\frac {2 \, {\left (\frac {b x}{b^{2} \mathrm {sgn}\left (x\right ) - 4 \, a c \mathrm {sgn}\left (x\right )} + \frac {2 \, a}{b^{2} \mathrm {sgn}\left (x\right ) - 4 \, a c \mathrm {sgn}\left (x\right )}\right )}}{\sqrt {c x^{2} + b x + a}} - \frac {4 \, \sqrt {a} \mathrm {sgn}\left (x\right )}{b^{2} - 4 \, a c} \]

[In]

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

[Out]

2*(b*x/(b^2*sgn(x) - 4*a*c*sgn(x)) + 2*a/(b^2*sgn(x) - 4*a*c*sgn(x)))/sqrt(c*x^2 + b*x + a) - 4*sqrt(a)*sgn(x)
/(b^2 - 4*a*c)

Mupad [B] (verification not implemented)

Time = 8.68 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.88 \[ \int \frac {x^4}{\left (a x^2+b x^3+c x^4\right )^{3/2}} \, dx=-\frac {\left (\frac {4\,a\,c}{4\,a\,c^2-b^2\,c}+\frac {2\,b\,c\,x}{4\,a\,c^2-b^2\,c}\right )\,\sqrt {c\,x^4+b\,x^3+a\,x^2}}{x\,\left (c\,x^2+b\,x+a\right )} \]

[In]

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

[Out]

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