Integrand size = 22, antiderivative size = 105 \[ \int \frac {\left (a-b x^3\right )^2}{\left (a+b x^3\right )^{13/3}} \, dx=\frac {x \left (a-b x^3\right )^3}{20 a^2 \left (a+b x^3\right )^{10/3}}+\frac {19 x \left (a-b x^3\right )^2}{140 a^2 \left (a+b x^3\right )^{7/3}}+\frac {57 x \left (a-b x^3\right )}{280 a^2 \left (a+b x^3\right )^{4/3}}+\frac {171 x}{280 a^2 \sqrt [3]{a+b x^3}} \]
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Time = 0.02 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {390, 386, 197} \[ \int \frac {\left (a-b x^3\right )^2}{\left (a+b x^3\right )^{13/3}} \, dx=\frac {x \left (a-b x^3\right )^3}{20 a^2 \left (a+b x^3\right )^{10/3}}+\frac {19 x \left (a-b x^3\right )^2}{140 a^2 \left (a+b x^3\right )^{7/3}}+\frac {57 x \left (a-b x^3\right )}{280 a^2 \left (a+b x^3\right )^{4/3}}+\frac {171 x}{280 a^2 \sqrt [3]{a+b x^3}} \]
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Rule 197
Rule 386
Rule 390
Rubi steps \begin{align*} \text {integral}& = \frac {x \left (a-b x^3\right )^3}{20 a^2 \left (a+b x^3\right )^{10/3}}+\frac {19 \int \frac {\left (a-b x^3\right )^2}{\left (a+b x^3\right )^{10/3}} \, dx}{20 a} \\ & = \frac {x \left (a-b x^3\right )^3}{20 a^2 \left (a+b x^3\right )^{10/3}}+\frac {19 x \left (a-b x^3\right )^2}{140 a^2 \left (a+b x^3\right )^{7/3}}+\frac {57 \int \frac {a-b x^3}{\left (a+b x^3\right )^{7/3}} \, dx}{70 a} \\ & = \frac {x \left (a-b x^3\right )^3}{20 a^2 \left (a+b x^3\right )^{10/3}}+\frac {19 x \left (a-b x^3\right )^2}{140 a^2 \left (a+b x^3\right )^{7/3}}+\frac {57 x \left (a-b x^3\right )}{280 a^2 \left (a+b x^3\right )^{4/3}}+\frac {171 \int \frac {1}{\left (a+b x^3\right )^{4/3}} \, dx}{280 a} \\ & = \frac {x \left (a-b x^3\right )^3}{20 a^2 \left (a+b x^3\right )^{10/3}}+\frac {19 x \left (a-b x^3\right )^2}{140 a^2 \left (a+b x^3\right )^{7/3}}+\frac {57 x \left (a-b x^3\right )}{280 a^2 \left (a+b x^3\right )^{4/3}}+\frac {171 x}{280 a^2 \sqrt [3]{a+b x^3}} \\ \end{align*}
Time = 0.66 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.49 \[ \int \frac {\left (a-b x^3\right )^2}{\left (a+b x^3\right )^{13/3}} \, dx=\frac {140 a^3 x+245 a^2 b x^4+230 a b^2 x^7+69 b^3 x^{10}}{140 a^2 \left (a+b x^3\right )^{10/3}} \]
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Time = 4.06 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.46
method | result | size |
gosper | \(\frac {x \left (69 b^{3} x^{9}+230 a \,b^{2} x^{6}+245 a^{2} b \,x^{3}+140 a^{3}\right )}{140 \left (b \,x^{3}+a \right )^{\frac {10}{3}} a^{2}}\) | \(48\) |
trager | \(\frac {x \left (69 b^{3} x^{9}+230 a \,b^{2} x^{6}+245 a^{2} b \,x^{3}+140 a^{3}\right )}{140 \left (b \,x^{3}+a \right )^{\frac {10}{3}} a^{2}}\) | \(48\) |
pseudoelliptic | \(\frac {x \left (69 b^{3} x^{9}+230 a \,b^{2} x^{6}+245 a^{2} b \,x^{3}+140 a^{3}\right )}{140 \left (b \,x^{3}+a \right )^{\frac {10}{3}} a^{2}}\) | \(48\) |
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Time = 0.35 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.87 \[ \int \frac {\left (a-b x^3\right )^2}{\left (a+b x^3\right )^{13/3}} \, dx=\frac {{\left (69 \, b^{3} x^{10} + 230 \, a b^{2} x^{7} + 245 \, a^{2} b x^{4} + 140 \, a^{3} x\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{140 \, {\left (a^{2} b^{4} x^{12} + 4 \, a^{3} b^{3} x^{9} + 6 \, a^{4} b^{2} x^{6} + 4 \, a^{5} b x^{3} + a^{6}\right )}} \]
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Timed out. \[ \int \frac {\left (a-b x^3\right )^2}{\left (a+b x^3\right )^{13/3}} \, dx=\text {Timed out} \]
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none
Time = 0.21 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.48 \[ \int \frac {\left (a-b x^3\right )^2}{\left (a+b x^3\right )^{13/3}} \, dx=-\frac {{\left (7 \, b - \frac {10 \, {\left (b x^{3} + a\right )}}{x^{3}}\right )} b^{2} x^{10}}{70 \, {\left (b x^{3} + a\right )}^{\frac {10}{3}} a^{2}} - \frac {{\left (14 \, b^{2} - \frac {40 \, {\left (b x^{3} + a\right )} b}{x^{3}} + \frac {35 \, {\left (b x^{3} + a\right )}^{2}}{x^{6}}\right )} b x^{10}}{70 \, {\left (b x^{3} + a\right )}^{\frac {10}{3}} a^{2}} - \frac {{\left (14 \, b^{3} - \frac {60 \, {\left (b x^{3} + a\right )} b^{2}}{x^{3}} + \frac {105 \, {\left (b x^{3} + a\right )}^{2} b}{x^{6}} - \frac {140 \, {\left (b x^{3} + a\right )}^{3}}{x^{9}}\right )} x^{10}}{140 \, {\left (b x^{3} + a\right )}^{\frac {10}{3}} a^{2}} \]
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\[ \int \frac {\left (a-b x^3\right )^2}{\left (a+b x^3\right )^{13/3}} \, dx=\int { \frac {{\left (b x^{3} - a\right )}^{2}}{{\left (b x^{3} + a\right )}^{\frac {13}{3}}} \,d x } \]
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Time = 5.51 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.53 \[ \int \frac {\left (a-b x^3\right )^2}{\left (a+b x^3\right )^{13/3}} \, dx=\frac {69\,x}{140\,a^2\,{\left (b\,x^3+a\right )}^{1/3}}-\frac {2\,x}{35\,{\left (b\,x^3+a\right )}^{7/3}}+\frac {23\,x}{140\,a\,{\left (b\,x^3+a\right )}^{4/3}}+\frac {2\,a\,x}{5\,{\left (b\,x^3+a\right )}^{10/3}} \]
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