Integrand size = 30, antiderivative size = 103 \[ \int \frac {x^2 \left (c+d x^3+e x^6+f x^9\right )}{\left (a+b x^3\right )^2} \, dx=\frac {(b e-2 a f) x^3}{3 b^3}+\frac {f x^6}{6 b^2}-\frac {b^3 c-a b^2 d+a^2 b e-a^3 f}{3 b^4 \left (a+b x^3\right )}+\frac {\left (b^2 d-2 a b e+3 a^2 f\right ) \log \left (a+b x^3\right )}{3 b^4} \]
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Time = 0.10 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {1833, 1864} \[ \int \frac {x^2 \left (c+d x^3+e x^6+f x^9\right )}{\left (a+b x^3\right )^2} \, dx=\frac {\log \left (a+b x^3\right ) \left (3 a^2 f-2 a b e+b^2 d\right )}{3 b^4}-\frac {a^3 (-f)+a^2 b e-a b^2 d+b^3 c}{3 b^4 \left (a+b x^3\right )}+\frac {x^3 (b e-2 a f)}{3 b^3}+\frac {f x^6}{6 b^2} \]
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Rule 1833
Rule 1864
Rubi steps \begin{align*} \text {integral}& = \frac {1}{3} \text {Subst}\left (\int \frac {c+d x+e x^2+f x^3}{(a+b x)^2} \, dx,x,x^3\right ) \\ & = \frac {1}{3} \text {Subst}\left (\int \left (\frac {b e-2 a f}{b^3}+\frac {f x}{b^2}+\frac {b^3 c-a b^2 d+a^2 b e-a^3 f}{b^3 (a+b x)^2}+\frac {b^2 d-2 a b e+3 a^2 f}{b^3 (a+b x)}\right ) \, dx,x,x^3\right ) \\ & = \frac {(b e-2 a f) x^3}{3 b^3}+\frac {f x^6}{6 b^2}-\frac {b^3 c-a b^2 d+a^2 b e-a^3 f}{3 b^4 \left (a+b x^3\right )}+\frac {\left (b^2 d-2 a b e+3 a^2 f\right ) \log \left (a+b x^3\right )}{3 b^4} \\ \end{align*}
Time = 0.06 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.90 \[ \int \frac {x^2 \left (c+d x^3+e x^6+f x^9\right )}{\left (a+b x^3\right )^2} \, dx=\frac {2 b (b e-2 a f) x^3+b^2 f x^6+\frac {2 \left (-b^3 c+a b^2 d-a^2 b e+a^3 f\right )}{a+b x^3}+2 \left (b^2 d-2 a b e+3 a^2 f\right ) \log \left (a+b x^3\right )}{6 b^4} \]
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Time = 1.54 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.97
method | result | size |
norman | \(\frac {\frac {3 f \,a^{3}-2 a^{2} b e +a \,b^{2} d -b^{3} c}{3 b^{4}}+\frac {f \,x^{9}}{6 b}-\frac {\left (3 a f -2 b e \right ) x^{6}}{6 b^{2}}}{b \,x^{3}+a}+\frac {\left (3 a^{2} f -2 a e b +b^{2} d \right ) \ln \left (b \,x^{3}+a \right )}{3 b^{4}}\) | \(100\) |
default | \(\frac {\left (-f \,x^{3} b +2 a f -b e \right )^{2}}{6 b^{4} f}+\frac {\frac {\left (3 a^{2} f -2 a e b +b^{2} d \right ) \ln \left (b \,x^{3}+a \right )}{b}-\frac {-f \,a^{3}+a^{2} b e -a \,b^{2} d +b^{3} c}{b \left (b \,x^{3}+a \right )}}{3 b^{3}}\) | \(102\) |
parallelrisch | \(\frac {b^{3} f \,x^{9}-3 x^{6} a \,b^{2} f +2 x^{6} b^{3} e +6 \ln \left (b \,x^{3}+a \right ) x^{3} a^{2} b f -4 \ln \left (b \,x^{3}+a \right ) x^{3} a \,b^{2} e +2 \ln \left (b \,x^{3}+a \right ) x^{3} b^{3} d +6 \ln \left (b \,x^{3}+a \right ) a^{3} f -4 \ln \left (b \,x^{3}+a \right ) a^{2} b e +2 \ln \left (b \,x^{3}+a \right ) a \,b^{2} d +6 f \,a^{3}-4 a^{2} b e +2 a \,b^{2} d -2 b^{3} c}{6 b^{4} \left (b \,x^{3}+a \right )}\) | \(166\) |
risch | \(\frac {f \,x^{6}}{6 b^{2}}-\frac {2 f a \,x^{3}}{3 b^{3}}+\frac {e \,x^{3}}{3 b^{2}}+\frac {2 f \,a^{2}}{3 b^{4}}-\frac {2 a e}{3 b^{3}}+\frac {e^{2}}{6 b^{2} f}+\frac {f \,a^{3}}{3 b^{4} \left (b \,x^{3}+a \right )}-\frac {a^{2} e}{3 b^{3} \left (b \,x^{3}+a \right )}+\frac {a d}{3 b^{2} \left (b \,x^{3}+a \right )}-\frac {c}{3 b \left (b \,x^{3}+a \right )}+\frac {\ln \left (b \,x^{3}+a \right ) a^{2} f}{b^{4}}-\frac {2 \ln \left (b \,x^{3}+a \right ) a e}{3 b^{3}}+\frac {\ln \left (b \,x^{3}+a \right ) d}{3 b^{2}}\) | \(169\) |
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Time = 0.26 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.39 \[ \int \frac {x^2 \left (c+d x^3+e x^6+f x^9\right )}{\left (a+b x^3\right )^2} \, dx=\frac {b^{3} f x^{9} + {\left (2 \, b^{3} e - 3 \, a b^{2} f\right )} x^{6} - 2 \, b^{3} c + 2 \, a b^{2} d - 2 \, a^{2} b e + 2 \, a^{3} f + 2 \, {\left (a b^{2} e - 2 \, a^{2} b f\right )} x^{3} + 2 \, {\left (a b^{2} d - 2 \, a^{2} b e + 3 \, a^{3} f + {\left (b^{3} d - 2 \, a b^{2} e + 3 \, a^{2} b f\right )} x^{3}\right )} \log \left (b x^{3} + a\right )}{6 \, {\left (b^{5} x^{3} + a b^{4}\right )}} \]
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Time = 5.43 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.97 \[ \int \frac {x^2 \left (c+d x^3+e x^6+f x^9\right )}{\left (a+b x^3\right )^2} \, dx=x^{3} \left (- \frac {2 a f}{3 b^{3}} + \frac {e}{3 b^{2}}\right ) + \frac {a^{3} f - a^{2} b e + a b^{2} d - b^{3} c}{3 a b^{4} + 3 b^{5} x^{3}} + \frac {f x^{6}}{6 b^{2}} + \frac {\left (3 a^{2} f - 2 a b e + b^{2} d\right ) \log {\left (a + b x^{3} \right )}}{3 b^{4}} \]
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Time = 0.20 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.95 \[ \int \frac {x^2 \left (c+d x^3+e x^6+f x^9\right )}{\left (a+b x^3\right )^2} \, dx=-\frac {b^{3} c - a b^{2} d + a^{2} b e - a^{3} f}{3 \, {\left (b^{5} x^{3} + a b^{4}\right )}} + \frac {b f x^{6} + 2 \, {\left (b e - 2 \, a f\right )} x^{3}}{6 \, b^{3}} + \frac {{\left (b^{2} d - 2 \, a b e + 3 \, a^{2} f\right )} \log \left (b x^{3} + a\right )}{3 \, b^{4}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 205 vs. \(2 (95) = 190\).
Time = 0.27 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.99 \[ \int \frac {x^2 \left (c+d x^3+e x^6+f x^9\right )}{\left (a+b x^3\right )^2} \, dx=-\frac {1}{6} \, f {\left (\frac {{\left (b x^{3} + a\right )}^{2} {\left (\frac {6 \, a}{b x^{3} + a} - 1\right )}}{b^{4}} + \frac {6 \, a^{2} \log \left (\frac {{\left | b x^{3} + a \right |}}{{\left (b x^{3} + a\right )}^{2} {\left | b \right |}}\right )}{b^{4}} - \frac {2 \, a^{3}}{{\left (b x^{3} + a\right )} b^{4}}\right )} + \frac {1}{3} \, e {\left (\frac {2 \, a \log \left (\frac {{\left | b x^{3} + a \right |}}{{\left (b x^{3} + a\right )}^{2} {\left | b \right |}}\right )}{b^{3}} + \frac {b x^{3} + a}{b^{3}} - \frac {a^{2}}{{\left (b x^{3} + a\right )} b^{3}}\right )} - \frac {d {\left (\frac {\log \left (\frac {{\left | b x^{3} + a \right |}}{{\left (b x^{3} + a\right )}^{2} {\left | b \right |}}\right )}{b} - \frac {a}{{\left (b x^{3} + a\right )} b}\right )}}{3 \, b} - \frac {c}{3 \, {\left (b x^{3} + a\right )} b} \]
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Time = 0.09 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.00 \[ \int \frac {x^2 \left (c+d x^3+e x^6+f x^9\right )}{\left (a+b x^3\right )^2} \, dx=x^3\,\left (\frac {e}{3\,b^2}-\frac {2\,a\,f}{3\,b^3}\right )+\frac {f\,x^6}{6\,b^2}-\frac {-f\,a^3+e\,a^2\,b-d\,a\,b^2+c\,b^3}{3\,b\,\left (b^4\,x^3+a\,b^3\right )}+\frac {\ln \left (b\,x^3+a\right )\,\left (3\,f\,a^2-2\,e\,a\,b+d\,b^2\right )}{3\,b^4} \]
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