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

Optimal. Leaf size=40 \[ \frac {2 x (2 a+b x)}{\left (b^2-4 a c\right ) \sqrt {a x^2+b x^3+c x^4}} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {1916} \[ \frac {2 x (2 a+b x)}{\left (b^2-4 a c\right ) \sqrt {a x^2+b x^3+c x^4}} \]

Antiderivative was successfully verified.

[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 1916

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*} \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}}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 37, normalized size = 0.92 \[ \frac {2 x (2 a+b x)}{\left (b^2-4 a c\right ) \sqrt {x^2 (a+x (b+c x))}} \]

Antiderivative was successfully verified.

[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))])

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fricas [A]  time = 0.62, size = 73, normalized size = 1.82 \[ \frac {2 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (b x + 2 \, a\right )}}{{\left (b^{2} c - 4 \, a c^{2}\right )} x^{3} + {\left (b^{3} - 4 \, a b c\right )} x^{2} + {\left (a b^{2} - 4 \, a^{2} c\right )} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.92, size = 45, normalized size = 1.12 \[ \frac {2 \, {\left (\frac {b}{b^{2} - 4 \, a c} + \frac {2 \, a}{{\left (b^{2} - 4 \, a c\right )} x}\right )}}{\sqrt {c + \frac {b}{x} + \frac {a}{x^{2}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.00, size = 53, normalized size = 1.32 \[ -\frac {2 \left (c \,x^{2}+b x +a \right ) \left (b x +2 a \right ) x^{3}}{\left (4 a c -b^{2}\right ) \left (c \,x^{4}+b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 2.12, size = 75, normalized size = 1.88 \[ -\frac {\left (\frac {4\,a\,c}{4\,a\,c^2-b^2\,c}+\frac {2\,b\,c\,x}{4\,a\,c^2-b^2\,c}\right )\,\sqrt {c\,x^4+b\,x^3+a\,x^2}}{x\,\left (c\,x^2+b\,x+a\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[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
))

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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