Integrand size = 49, antiderivative size = 23 \[ \int \frac {\left (a+b x+c x^2+d x^3\right )^p \left (-a+b p x+c (1+2 p) x^2+d (2+3 p) x^3\right )}{x^2} \, dx=\frac {\left (a+b x+c x^2+d x^3\right )^{1+p}}{x} \]
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Time = 0.02 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.020, Rules used = {1604} \[ \int \frac {\left (a+b x+c x^2+d x^3\right )^p \left (-a+b p x+c (1+2 p) x^2+d (2+3 p) x^3\right )}{x^2} \, dx=\frac {\left (a+b x+c x^2+d x^3\right )^{p+1}}{x} \]
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Rule 1604
Rubi steps \begin{align*} \text {integral}& = \frac {\left (a+b x+c x^2+d x^3\right )^{1+p}}{x} \\ \end{align*}
Time = 0.80 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {\left (a+b x+c x^2+d x^3\right )^p \left (-a+b p x+c (1+2 p) x^2+d (2+3 p) x^3\right )}{x^2} \, dx=\frac {(a+x (b+x (c+d x)))^{1+p}}{x} \]
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Time = 0.20 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04
method | result | size |
gosper | \(\frac {\left (x^{3} d +c \,x^{2}+b x +a \right )^{1+p}}{x}\) | \(24\) |
risch | \(\frac {\left (x^{3} d +c \,x^{2}+b x +a \right ) \left (x^{3} d +c \,x^{2}+b x +a \right )^{p}}{x}\) | \(37\) |
norman | \(\frac {a \,{\mathrm e}^{p \ln \left (x^{3} d +c \,x^{2}+b x +a \right )}+b x \,{\mathrm e}^{p \ln \left (x^{3} d +c \,x^{2}+b x +a \right )}+c \,x^{2} {\mathrm e}^{p \ln \left (x^{3} d +c \,x^{2}+b x +a \right )}+x^{3} d \,{\mathrm e}^{p \ln \left (x^{3} d +c \,x^{2}+b x +a \right )}}{x}\) | \(97\) |
parallelrisch | \(\frac {x^{3} \left (x^{3} d +c \,x^{2}+b x +a \right )^{p} d^{2}+x^{2} \left (x^{3} d +c \,x^{2}+b x +a \right )^{p} c d +x \left (x^{3} d +c \,x^{2}+b x +a \right )^{p} b d +\left (x^{3} d +c \,x^{2}+b x +a \right )^{p} a d}{d x}\) | \(97\) |
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none
Time = 0.34 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.57 \[ \int \frac {\left (a+b x+c x^2+d x^3\right )^p \left (-a+b p x+c (1+2 p) x^2+d (2+3 p) x^3\right )}{x^2} \, dx=\frac {{\left (d x^{3} + c x^{2} + b x + a\right )} {\left (d x^{3} + c x^{2} + b x + a\right )}^{p}}{x} \]
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Timed out. \[ \int \frac {\left (a+b x+c x^2+d x^3\right )^p \left (-a+b p x+c (1+2 p) x^2+d (2+3 p) x^3\right )}{x^2} \, dx=\text {Timed out} \]
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none
Time = 0.22 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.57 \[ \int \frac {\left (a+b x+c x^2+d x^3\right )^p \left (-a+b p x+c (1+2 p) x^2+d (2+3 p) x^3\right )}{x^2} \, dx=\frac {{\left (d x^{3} + c x^{2} + b x + a\right )} {\left (d x^{3} + c x^{2} + b x + a\right )}^{p}}{x} \]
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\[ \int \frac {\left (a+b x+c x^2+d x^3\right )^p \left (-a+b p x+c (1+2 p) x^2+d (2+3 p) x^3\right )}{x^2} \, dx=\int { \frac {{\left (d {\left (3 \, p + 2\right )} x^{3} + c {\left (2 \, p + 1\right )} x^{2} + b p x - a\right )} {\left (d x^{3} + c x^{2} + b x + a\right )}^{p}}{x^{2}} \,d x } \]
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Time = 9.65 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+b x+c x^2+d x^3\right )^p \left (-a+b p x+c (1+2 p) x^2+d (2+3 p) x^3\right )}{x^2} \, dx=\frac {{\left (d\,x^3+c\,x^2+b\,x+a\right )}^{p+1}}{x} \]
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