Integrand size = 19, antiderivative size = 136 \[ \int \frac {1}{(a+b x)^{13/3} (c+d x)^{2/3}} \, dx=-\frac {3 \sqrt [3]{c+d x}}{10 (b c-a d) (a+b x)^{10/3}}+\frac {27 d \sqrt [3]{c+d x}}{70 (b c-a d)^2 (a+b x)^{7/3}}-\frac {81 d^2 \sqrt [3]{c+d x}}{140 (b c-a d)^3 (a+b x)^{4/3}}+\frac {243 d^3 \sqrt [3]{c+d x}}{140 (b c-a d)^4 \sqrt [3]{a+b x}} \]
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Time = 0.02 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {47, 37} \[ \int \frac {1}{(a+b x)^{13/3} (c+d x)^{2/3}} \, dx=\frac {243 d^3 \sqrt [3]{c+d x}}{140 \sqrt [3]{a+b x} (b c-a d)^4}-\frac {81 d^2 \sqrt [3]{c+d x}}{140 (a+b x)^{4/3} (b c-a d)^3}+\frac {27 d \sqrt [3]{c+d x}}{70 (a+b x)^{7/3} (b c-a d)^2}-\frac {3 \sqrt [3]{c+d x}}{10 (a+b x)^{10/3} (b c-a d)} \]
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Rule 37
Rule 47
Rubi steps \begin{align*} \text {integral}& = -\frac {3 \sqrt [3]{c+d x}}{10 (b c-a d) (a+b x)^{10/3}}-\frac {(9 d) \int \frac {1}{(a+b x)^{10/3} (c+d x)^{2/3}} \, dx}{10 (b c-a d)} \\ & = -\frac {3 \sqrt [3]{c+d x}}{10 (b c-a d) (a+b x)^{10/3}}+\frac {27 d \sqrt [3]{c+d x}}{70 (b c-a d)^2 (a+b x)^{7/3}}+\frac {\left (27 d^2\right ) \int \frac {1}{(a+b x)^{7/3} (c+d x)^{2/3}} \, dx}{35 (b c-a d)^2} \\ & = -\frac {3 \sqrt [3]{c+d x}}{10 (b c-a d) (a+b x)^{10/3}}+\frac {27 d \sqrt [3]{c+d x}}{70 (b c-a d)^2 (a+b x)^{7/3}}-\frac {81 d^2 \sqrt [3]{c+d x}}{140 (b c-a d)^3 (a+b x)^{4/3}}-\frac {\left (81 d^3\right ) \int \frac {1}{(a+b x)^{4/3} (c+d x)^{2/3}} \, dx}{140 (b c-a d)^3} \\ & = -\frac {3 \sqrt [3]{c+d x}}{10 (b c-a d) (a+b x)^{10/3}}+\frac {27 d \sqrt [3]{c+d x}}{70 (b c-a d)^2 (a+b x)^{7/3}}-\frac {81 d^2 \sqrt [3]{c+d x}}{140 (b c-a d)^3 (a+b x)^{4/3}}+\frac {243 d^3 \sqrt [3]{c+d x}}{140 (b c-a d)^4 \sqrt [3]{a+b x}} \\ \end{align*}
Time = 0.32 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.85 \[ \int \frac {1}{(a+b x)^{13/3} (c+d x)^{2/3}} \, dx=\frac {3 \sqrt [3]{c+d x} \left (140 a^3 d^3-105 a^2 b d^2 (c-3 d x)+30 a b^2 d \left (2 c^2-3 c d x+9 d^2 x^2\right )+b^3 \left (-14 c^3+18 c^2 d x-27 c d^2 x^2+81 d^3 x^3\right )\right )}{140 (b c-a d)^4 (a+b x)^{10/3}} \]
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Time = 0.71 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.26
method | result | size |
gosper | \(\frac {3 \left (d x +c \right )^{\frac {1}{3}} \left (81 d^{3} x^{3} b^{3}+270 x^{2} a \,b^{2} d^{3}-27 x^{2} b^{3} c \,d^{2}+315 x \,a^{2} b \,d^{3}-90 x a \,b^{2} c \,d^{2}+18 x \,b^{3} c^{2} d +140 a^{3} d^{3}-105 a^{2} b c \,d^{2}+60 a \,b^{2} c^{2} d -14 b^{3} c^{3}\right )}{140 \left (b x +a \right )^{\frac {10}{3}} \left (a^{4} d^{4}-4 a^{3} b c \,d^{3}+6 a^{2} b^{2} c^{2} d^{2}-4 a \,b^{3} c^{3} d +b^{4} c^{4}\right )}\) | \(171\) |
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Leaf count of result is larger than twice the leaf count of optimal. 419 vs. \(2 (112) = 224\).
Time = 0.23 (sec) , antiderivative size = 419, normalized size of antiderivative = 3.08 \[ \int \frac {1}{(a+b x)^{13/3} (c+d x)^{2/3}} \, dx=\frac {3 \, {\left (81 \, b^{3} d^{3} x^{3} - 14 \, b^{3} c^{3} + 60 \, a b^{2} c^{2} d - 105 \, a^{2} b c d^{2} + 140 \, a^{3} d^{3} - 27 \, {\left (b^{3} c d^{2} - 10 \, a b^{2} d^{3}\right )} x^{2} + 9 \, {\left (2 \, b^{3} c^{2} d - 10 \, a b^{2} c d^{2} + 35 \, a^{2} b d^{3}\right )} x\right )} {\left (b x + a\right )}^{\frac {2}{3}} {\left (d x + c\right )}^{\frac {1}{3}}}{140 \, {\left (a^{4} b^{4} c^{4} - 4 \, a^{5} b^{3} c^{3} d + 6 \, a^{6} b^{2} c^{2} d^{2} - 4 \, a^{7} b c d^{3} + a^{8} d^{4} + {\left (b^{8} c^{4} - 4 \, a b^{7} c^{3} d + 6 \, a^{2} b^{6} c^{2} d^{2} - 4 \, a^{3} b^{5} c d^{3} + a^{4} b^{4} d^{4}\right )} x^{4} + 4 \, {\left (a b^{7} c^{4} - 4 \, a^{2} b^{6} c^{3} d + 6 \, a^{3} b^{5} c^{2} d^{2} - 4 \, a^{4} b^{4} c d^{3} + a^{5} b^{3} d^{4}\right )} x^{3} + 6 \, {\left (a^{2} b^{6} c^{4} - 4 \, a^{3} b^{5} c^{3} d + 6 \, a^{4} b^{4} c^{2} d^{2} - 4 \, a^{5} b^{3} c d^{3} + a^{6} b^{2} d^{4}\right )} x^{2} + 4 \, {\left (a^{3} b^{5} c^{4} - 4 \, a^{4} b^{4} c^{3} d + 6 \, a^{5} b^{3} c^{2} d^{2} - 4 \, a^{6} b^{2} c d^{3} + a^{7} b d^{4}\right )} x\right )}} \]
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\[ \int \frac {1}{(a+b x)^{13/3} (c+d x)^{2/3}} \, dx=\int \frac {1}{\left (a + b x\right )^{\frac {13}{3}} \left (c + d x\right )^{\frac {2}{3}}}\, dx \]
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\[ \int \frac {1}{(a+b x)^{13/3} (c+d x)^{2/3}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {13}{3}} {\left (d x + c\right )}^{\frac {2}{3}}} \,d x } \]
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\[ \int \frac {1}{(a+b x)^{13/3} (c+d x)^{2/3}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {13}{3}} {\left (d x + c\right )}^{\frac {2}{3}}} \,d x } \]
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Time = 1.50 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.54 \[ \int \frac {1}{(a+b x)^{13/3} (c+d x)^{2/3}} \, dx=\frac {{\left (c+d\,x\right )}^{1/3}\,\left (\frac {243\,d^3\,x^3}{140\,{\left (a\,d-b\,c\right )}^4}+\frac {420\,a^3\,d^3-315\,a^2\,b\,c\,d^2+180\,a\,b^2\,c^2\,d-42\,b^3\,c^3}{140\,b^3\,{\left (a\,d-b\,c\right )}^4}+\frac {27\,d\,x\,\left (35\,a^2\,d^2-10\,a\,b\,c\,d+2\,b^2\,c^2\right )}{140\,b^2\,{\left (a\,d-b\,c\right )}^4}+\frac {81\,d^2\,x^2\,\left (10\,a\,d-b\,c\right )}{140\,b\,{\left (a\,d-b\,c\right )}^4}\right )}{x^3\,{\left (a+b\,x\right )}^{1/3}+\frac {a^3\,{\left (a+b\,x\right )}^{1/3}}{b^3}+\frac {3\,a\,x^2\,{\left (a+b\,x\right )}^{1/3}}{b}+\frac {3\,a^2\,x\,{\left (a+b\,x\right )}^{1/3}}{b^2}} \]
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