Integrand size = 19, antiderivative size = 101 \[ \int \frac {1}{(a+b x)^{10/3} (c+d x)^{2/3}} \, dx=-\frac {3 \sqrt [3]{c+d x}}{7 (b c-a d) (a+b x)^{7/3}}+\frac {9 d \sqrt [3]{c+d x}}{14 (b c-a d)^2 (a+b x)^{4/3}}-\frac {27 d^2 \sqrt [3]{c+d x}}{14 (b c-a d)^3 \sqrt [3]{a+b x}} \]
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Time = 0.01 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {47, 37} \[ \int \frac {1}{(a+b x)^{10/3} (c+d x)^{2/3}} \, dx=-\frac {27 d^2 \sqrt [3]{c+d x}}{14 \sqrt [3]{a+b x} (b c-a d)^3}+\frac {9 d \sqrt [3]{c+d x}}{14 (a+b x)^{4/3} (b c-a d)^2}-\frac {3 \sqrt [3]{c+d x}}{7 (a+b x)^{7/3} (b c-a d)} \]
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Rule 37
Rule 47
Rubi steps \begin{align*} \text {integral}& = -\frac {3 \sqrt [3]{c+d x}}{7 (b c-a d) (a+b x)^{7/3}}-\frac {(6 d) \int \frac {1}{(a+b x)^{7/3} (c+d x)^{2/3}} \, dx}{7 (b c-a d)} \\ & = -\frac {3 \sqrt [3]{c+d x}}{7 (b c-a d) (a+b x)^{7/3}}+\frac {9 d \sqrt [3]{c+d x}}{14 (b c-a d)^2 (a+b x)^{4/3}}+\frac {\left (9 d^2\right ) \int \frac {1}{(a+b x)^{4/3} (c+d x)^{2/3}} \, dx}{14 (b c-a d)^2} \\ & = -\frac {3 \sqrt [3]{c+d x}}{7 (b c-a d) (a+b x)^{7/3}}+\frac {9 d \sqrt [3]{c+d x}}{14 (b c-a d)^2 (a+b x)^{4/3}}-\frac {27 d^2 \sqrt [3]{c+d x}}{14 (b c-a d)^3 \sqrt [3]{a+b x}} \\ \end{align*}
Time = 0.21 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.74 \[ \int \frac {1}{(a+b x)^{10/3} (c+d x)^{2/3}} \, dx=-\frac {3 \sqrt [3]{c+d x} \left (14 a^2 d^2-7 a b d (c-3 d x)+b^2 \left (2 c^2-3 c d x+9 d^2 x^2\right )\right )}{14 (b c-a d)^3 (a+b x)^{7/3}} \]
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Time = 0.30 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.04
method | result | size |
gosper | \(\frac {3 \left (d x +c \right )^{\frac {1}{3}} \left (9 d^{2} x^{2} b^{2}+21 x a b \,d^{2}-3 x \,b^{2} c d +14 a^{2} d^{2}-7 a b c d +2 b^{2} c^{2}\right )}{14 \left (b x +a \right )^{\frac {7}{3}} \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right )}\) | \(105\) |
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Leaf count of result is larger than twice the leaf count of optimal. 251 vs. \(2 (83) = 166\).
Time = 0.24 (sec) , antiderivative size = 251, normalized size of antiderivative = 2.49 \[ \int \frac {1}{(a+b x)^{10/3} (c+d x)^{2/3}} \, dx=-\frac {3 \, {\left (9 \, b^{2} d^{2} x^{2} + 2 \, b^{2} c^{2} - 7 \, a b c d + 14 \, a^{2} d^{2} - 3 \, {\left (b^{2} c d - 7 \, a b d^{2}\right )} x\right )} {\left (b x + a\right )}^{\frac {2}{3}} {\left (d x + c\right )}^{\frac {1}{3}}}{14 \, {\left (a^{3} b^{3} c^{3} - 3 \, a^{4} b^{2} c^{2} d + 3 \, a^{5} b c d^{2} - a^{6} d^{3} + {\left (b^{6} c^{3} - 3 \, a b^{5} c^{2} d + 3 \, a^{2} b^{4} c d^{2} - a^{3} b^{3} d^{3}\right )} x^{3} + 3 \, {\left (a b^{5} c^{3} - 3 \, a^{2} b^{4} c^{2} d + 3 \, a^{3} b^{3} c d^{2} - a^{4} b^{2} d^{3}\right )} x^{2} + 3 \, {\left (a^{2} b^{4} c^{3} - 3 \, a^{3} b^{3} c^{2} d + 3 \, a^{4} b^{2} c d^{2} - a^{5} b d^{3}\right )} x\right )}} \]
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\[ \int \frac {1}{(a+b x)^{10/3} (c+d x)^{2/3}} \, dx=\int \frac {1}{\left (a + b x\right )^{\frac {10}{3}} \left (c + d x\right )^{\frac {2}{3}}}\, dx \]
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\[ \int \frac {1}{(a+b x)^{10/3} (c+d x)^{2/3}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {10}{3}} {\left (d x + c\right )}^{\frac {2}{3}}} \,d x } \]
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\[ \int \frac {1}{(a+b x)^{10/3} (c+d x)^{2/3}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {10}{3}} {\left (d x + c\right )}^{\frac {2}{3}}} \,d x } \]
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Time = 0.98 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.32 \[ \int \frac {1}{(a+b x)^{10/3} (c+d x)^{2/3}} \, dx=\frac {{\left (c+d\,x\right )}^{1/3}\,\left (\frac {27\,d^2\,x^2}{14\,{\left (a\,d-b\,c\right )}^3}+\frac {42\,a^2\,d^2-21\,a\,b\,c\,d+6\,b^2\,c^2}{14\,b^2\,{\left (a\,d-b\,c\right )}^3}+\frac {9\,d\,x\,\left (7\,a\,d-b\,c\right )}{14\,b\,{\left (a\,d-b\,c\right )}^3}\right )}{x^2\,{\left (a+b\,x\right )}^{1/3}+\frac {a^2\,{\left (a+b\,x\right )}^{1/3}}{b^2}+\frac {2\,a\,x\,{\left (a+b\,x\right )}^{1/3}}{b}} \]
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