Integrand size = 23, antiderivative size = 125 \[ \int \frac {\left (c+d x+e x^2\right ) \left (a+b x^3\right )^3}{x^2} \, dx=-\frac {a^3 c}{x}+a^3 e x+\frac {3}{2} a^2 b c x^2+a^2 b d x^3+\frac {3}{4} a^2 b e x^4+\frac {3}{5} a b^2 c x^5+\frac {1}{2} a b^2 d x^6+\frac {3}{7} a b^2 e x^7+\frac {1}{8} b^3 c x^8+\frac {1}{9} b^3 d x^9+\frac {1}{10} b^3 e x^{10}+a^3 d \log (x) \]
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Time = 0.06 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {1642} \[ \int \frac {\left (c+d x+e x^2\right ) \left (a+b x^3\right )^3}{x^2} \, dx=-\frac {a^3 c}{x}+a^3 d \log (x)+a^3 e x+\frac {3}{2} a^2 b c x^2+a^2 b d x^3+\frac {3}{4} a^2 b e x^4+\frac {3}{5} a b^2 c x^5+\frac {1}{2} a b^2 d x^6+\frac {3}{7} a b^2 e x^7+\frac {1}{8} b^3 c x^8+\frac {1}{9} b^3 d x^9+\frac {1}{10} b^3 e x^{10} \]
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Rule 1642
Rubi steps \begin{align*} \text {integral}& = \int \left (a^3 e+\frac {a^3 c}{x^2}+\frac {a^3 d}{x}+3 a^2 b c x+3 a^2 b d x^2+3 a^2 b e x^3+3 a b^2 c x^4+3 a b^2 d x^5+3 a b^2 e x^6+b^3 c x^7+b^3 d x^8+b^3 e x^9\right ) \, dx \\ & = -\frac {a^3 c}{x}+a^3 e x+\frac {3}{2} a^2 b c x^2+a^2 b d x^3+\frac {3}{4} a^2 b e x^4+\frac {3}{5} a b^2 c x^5+\frac {1}{2} a b^2 d x^6+\frac {3}{7} a b^2 e x^7+\frac {1}{8} b^3 c x^8+\frac {1}{9} b^3 d x^9+\frac {1}{10} b^3 e x^{10}+a^3 d \log (x) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.00 \[ \int \frac {\left (c+d x+e x^2\right ) \left (a+b x^3\right )^3}{x^2} \, dx=-\frac {a^3 c}{x}+a^3 e x+\frac {3}{2} a^2 b c x^2+a^2 b d x^3+\frac {3}{4} a^2 b e x^4+\frac {3}{5} a b^2 c x^5+\frac {1}{2} a b^2 d x^6+\frac {3}{7} a b^2 e x^7+\frac {1}{8} b^3 c x^8+\frac {1}{9} b^3 d x^9+\frac {1}{10} b^3 e x^{10}+a^3 d \log (x) \]
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Time = 1.51 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.88
method | result | size |
default | \(-\frac {a^{3} c}{x}+a^{3} e x +\frac {3 a^{2} b c \,x^{2}}{2}+a^{2} b d \,x^{3}+\frac {3 a^{2} b e \,x^{4}}{4}+\frac {3 a \,b^{2} c \,x^{5}}{5}+\frac {a \,b^{2} d \,x^{6}}{2}+\frac {3 a \,b^{2} e \,x^{7}}{7}+\frac {b^{3} c \,x^{8}}{8}+\frac {b^{3} d \,x^{9}}{9}+\frac {b^{3} e \,x^{10}}{10}+a^{3} d \ln \left (x \right )\) | \(110\) |
risch | \(-\frac {a^{3} c}{x}+a^{3} e x +\frac {3 a^{2} b c \,x^{2}}{2}+a^{2} b d \,x^{3}+\frac {3 a^{2} b e \,x^{4}}{4}+\frac {3 a \,b^{2} c \,x^{5}}{5}+\frac {a \,b^{2} d \,x^{6}}{2}+\frac {3 a \,b^{2} e \,x^{7}}{7}+\frac {b^{3} c \,x^{8}}{8}+\frac {b^{3} d \,x^{9}}{9}+\frac {b^{3} e \,x^{10}}{10}+a^{3} d \ln \left (x \right )\) | \(110\) |
norman | \(\frac {a^{3} e \,x^{2}+a^{2} b d \,x^{4}-c \,a^{3}+\frac {1}{8} b^{3} c \,x^{9}+\frac {1}{9} b^{3} d \,x^{10}+\frac {1}{10} b^{3} e \,x^{11}+\frac {3}{5} a \,b^{2} c \,x^{6}+\frac {1}{2} a \,b^{2} d \,x^{7}+\frac {3}{7} a \,b^{2} e \,x^{8}+\frac {3}{4} a^{2} b e \,x^{5}+\frac {3}{2} a^{2} x^{3} b c}{x}+a^{3} d \ln \left (x \right )\) | \(114\) |
parallelrisch | \(\frac {252 b^{3} e \,x^{11}+280 b^{3} d \,x^{10}+315 b^{3} c \,x^{9}+1080 a \,b^{2} e \,x^{8}+1260 a \,b^{2} d \,x^{7}+1512 a \,b^{2} c \,x^{6}+1890 a^{2} b e \,x^{5}+2520 a^{2} b d \,x^{4}+3780 a^{2} x^{3} b c +2520 a^{3} d \ln \left (x \right ) x +2520 a^{3} e \,x^{2}-2520 c \,a^{3}}{2520 x}\) | \(118\) |
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Time = 0.28 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.94 \[ \int \frac {\left (c+d x+e x^2\right ) \left (a+b x^3\right )^3}{x^2} \, dx=\frac {252 \, b^{3} e x^{11} + 280 \, b^{3} d x^{10} + 315 \, b^{3} c x^{9} + 1080 \, a b^{2} e x^{8} + 1260 \, a b^{2} d x^{7} + 1512 \, a b^{2} c x^{6} + 1890 \, a^{2} b e x^{5} + 2520 \, a^{2} b d x^{4} + 3780 \, a^{2} b c x^{3} + 2520 \, a^{3} e x^{2} + 2520 \, a^{3} d x \log \left (x\right ) - 2520 \, a^{3} c}{2520 \, x} \]
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Time = 0.08 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.02 \[ \int \frac {\left (c+d x+e x^2\right ) \left (a+b x^3\right )^3}{x^2} \, dx=- \frac {a^{3} c}{x} + a^{3} d \log {\left (x \right )} + a^{3} e x + \frac {3 a^{2} b c x^{2}}{2} + a^{2} b d x^{3} + \frac {3 a^{2} b e x^{4}}{4} + \frac {3 a b^{2} c x^{5}}{5} + \frac {a b^{2} d x^{6}}{2} + \frac {3 a b^{2} e x^{7}}{7} + \frac {b^{3} c x^{8}}{8} + \frac {b^{3} d x^{9}}{9} + \frac {b^{3} e x^{10}}{10} \]
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Time = 0.21 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.87 \[ \int \frac {\left (c+d x+e x^2\right ) \left (a+b x^3\right )^3}{x^2} \, dx=\frac {1}{10} \, b^{3} e x^{10} + \frac {1}{9} \, b^{3} d x^{9} + \frac {1}{8} \, b^{3} c x^{8} + \frac {3}{7} \, a b^{2} e x^{7} + \frac {1}{2} \, a b^{2} d x^{6} + \frac {3}{5} \, a b^{2} c x^{5} + \frac {3}{4} \, a^{2} b e x^{4} + a^{2} b d x^{3} + \frac {3}{2} \, a^{2} b c x^{2} + a^{3} e x + a^{3} d \log \left (x\right ) - \frac {a^{3} c}{x} \]
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Time = 0.27 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.88 \[ \int \frac {\left (c+d x+e x^2\right ) \left (a+b x^3\right )^3}{x^2} \, dx=\frac {1}{10} \, b^{3} e x^{10} + \frac {1}{9} \, b^{3} d x^{9} + \frac {1}{8} \, b^{3} c x^{8} + \frac {3}{7} \, a b^{2} e x^{7} + \frac {1}{2} \, a b^{2} d x^{6} + \frac {3}{5} \, a b^{2} c x^{5} + \frac {3}{4} \, a^{2} b e x^{4} + a^{2} b d x^{3} + \frac {3}{2} \, a^{2} b c x^{2} + a^{3} e x + a^{3} d \log \left ({\left | x \right |}\right ) - \frac {a^{3} c}{x} \]
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Time = 0.11 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.87 \[ \int \frac {\left (c+d x+e x^2\right ) \left (a+b x^3\right )^3}{x^2} \, dx=\frac {b^3\,c\,x^8}{8}-\frac {a^3\,c}{x}+\frac {b^3\,d\,x^9}{9}+\frac {b^3\,e\,x^{10}}{10}+a^3\,d\,\ln \left (x\right )+a^3\,e\,x+\frac {3\,a^2\,b\,c\,x^2}{2}+\frac {3\,a\,b^2\,c\,x^5}{5}+a^2\,b\,d\,x^3+\frac {a\,b^2\,d\,x^6}{2}+\frac {3\,a^2\,b\,e\,x^4}{4}+\frac {3\,a\,b^2\,e\,x^7}{7} \]
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