Integrand size = 24, antiderivative size = 39 \[ \int \frac {x^3}{\left (a x^2+b x^3+c x^4\right )^{3/2}} \, dx=-\frac {2 x (b+2 c x)}{\left (b^2-4 a c\right ) \sqrt {a x^2+b x^3+c x^4}} \]
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Time = 0.02 (sec) , antiderivative size = 39, 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 = {1929} \[ \int \frac {x^3}{\left (a x^2+b x^3+c x^4\right )^{3/2}} \, dx=-\frac {2 x (b+2 c x)}{\left (b^2-4 a c\right ) \sqrt {a x^2+b x^3+c x^4}} \]
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Rule 1929
Rubi steps \begin{align*} \text {integral}& = -\frac {2 x (b+2 c x)}{\left (b^2-4 a c\right ) \sqrt {a x^2+b x^3+c x^4}} \\ \end{align*}
Time = 0.25 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.92 \[ \int \frac {x^3}{\left (a x^2+b x^3+c x^4\right )^{3/2}} \, dx=-\frac {2 x (b+2 c x)}{\left (b^2-4 a c\right ) \sqrt {x^2 (a+x (b+c x))}} \]
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Time = 0.10 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.85
method | result | size |
pseudoelliptic | \(\frac {4 c x +2 b}{\sqrt {c \,x^{2}+b x +a}\, \left (4 a c -b^{2}\right )}\) | \(33\) |
gosper | \(\frac {2 \left (c \,x^{2}+b x +a \right ) \left (2 c x +b \right ) x^{3}}{\left (4 a c -b^{2}\right ) \left (c \,x^{4}+b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}}}\) | \(52\) |
default | \(\frac {2 \left (c \,x^{2}+b x +a \right ) \left (2 c x +b \right ) x^{3}}{\left (4 a c -b^{2}\right ) \left (c \,x^{4}+b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}}}\) | \(52\) |
trager | \(\frac {2 \left (2 c x +b \right ) \sqrt {c \,x^{4}+b \,x^{3}+a \,x^{2}}}{\left (c \,x^{2}+b x +a \right ) x \left (4 a c -b^{2}\right )}\) | \(54\) |
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none
Time = 0.28 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.85 \[ \int \frac {x^3}{\left (a x^2+b x^3+c x^4\right )^{3/2}} \, dx=-\frac {2 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (2 \, c x + b\right )}}{{\left (b^{2} c - 4 \, a c^{2}\right )} x^{3} + {\left (b^{3} - 4 \, a b c\right )} x^{2} + {\left (a b^{2} - 4 \, a^{2} c\right )} x} \]
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\[ \int \frac {x^3}{\left (a x^2+b x^3+c x^4\right )^{3/2}} \, dx=\int \frac {x^{3}}{\left (x^{2} \left (a + b x + c x^{2}\right )\right )^{\frac {3}{2}}}\, dx \]
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\[ \int \frac {x^3}{\left (a x^2+b x^3+c x^4\right )^{3/2}} \, dx=\int { \frac {x^{3}}{{\left (c x^{4} + b x^{3} + a x^{2}\right )}^{\frac {3}{2}}} \,d x } \]
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none
Time = 0.30 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.90 \[ \int \frac {x^3}{\left (a x^2+b x^3+c x^4\right )^{3/2}} \, dx=\frac {2 \, \sqrt {a} b \mathrm {sgn}\left (x\right )}{a b^{2} - 4 \, a^{2} c} - \frac {2 \, {\left (\frac {2 \, c x}{b^{2} \mathrm {sgn}\left (x\right ) - 4 \, a c \mathrm {sgn}\left (x\right )} + \frac {b}{b^{2} \mathrm {sgn}\left (x\right ) - 4 \, a c \mathrm {sgn}\left (x\right )}\right )}}{\sqrt {c x^{2} + b x + a}} \]
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Time = 8.53 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.92 \[ \int \frac {x^3}{\left (a x^2+b x^3+c x^4\right )^{3/2}} \, dx=\frac {\left (\frac {4\,c^2\,x}{4\,a\,c^2-b^2\,c}+\frac {2\,b\,c}{4\,a\,c^2-b^2\,c}\right )\,\sqrt {c\,x^4+b\,x^3+a\,x^2}}{x\,\left (c\,x^2+b\,x+a\right )} \]
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