Integrand size = 19, antiderivative size = 101 \[ \int \frac {1}{(a+b x)^{11/3} \sqrt [3]{c+d x}} \, dx=-\frac {3 (c+d x)^{2/3}}{8 (b c-a d) (a+b x)^{8/3}}+\frac {9 d (c+d x)^{2/3}}{20 (b c-a d)^2 (a+b x)^{5/3}}-\frac {27 d^2 (c+d x)^{2/3}}{40 (b c-a d)^3 (a+b x)^{2/3}} \]
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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} \[ \int \frac {1}{(a+b x)^{11/3} \sqrt [3]{c+d x}} \, dx=-\frac {27 d^2 (c+d x)^{2/3}}{40 (a+b x)^{2/3} (b c-a d)^3}+\frac {9 d (c+d x)^{2/3}}{20 (a+b x)^{5/3} (b c-a d)^2}-\frac {3 (c+d x)^{2/3}}{8 (a+b x)^{8/3} (b c-a d)} \]
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Rule 37
Rule 47
Rubi steps \begin{align*} \text {integral}& = -\frac {3 (c+d x)^{2/3}}{8 (b c-a d) (a+b x)^{8/3}}-\frac {(3 d) \int \frac {1}{(a+b x)^{8/3} \sqrt [3]{c+d x}} \, dx}{4 (b c-a d)} \\ & = -\frac {3 (c+d x)^{2/3}}{8 (b c-a d) (a+b x)^{8/3}}+\frac {9 d (c+d x)^{2/3}}{20 (b c-a d)^2 (a+b x)^{5/3}}+\frac {\left (9 d^2\right ) \int \frac {1}{(a+b x)^{5/3} \sqrt [3]{c+d x}} \, dx}{20 (b c-a d)^2} \\ & = -\frac {3 (c+d x)^{2/3}}{8 (b c-a d) (a+b x)^{8/3}}+\frac {9 d (c+d x)^{2/3}}{20 (b c-a d)^2 (a+b x)^{5/3}}-\frac {27 d^2 (c+d x)^{2/3}}{40 (b c-a d)^3 (a+b x)^{2/3}} \\ \end{align*}
Time = 0.25 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.76 \[ \int \frac {1}{(a+b x)^{11/3} \sqrt [3]{c+d x}} \, dx=-\frac {3 (c+d x)^{2/3} \left (20 a^2 d^2+8 a b d (-2 c+3 d x)+b^2 \left (5 c^2-6 c d x+9 d^2 x^2\right )\right )}{40 (b c-a d)^3 (a+b x)^{8/3}} \]
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Time = 0.30 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.04
method | result | size |
gosper | \(\frac {3 \left (d x +c \right )^{\frac {2}{3}} \left (9 d^{2} x^{2} b^{2}+24 x a b \,d^{2}-6 x \,b^{2} c d +20 a^{2} d^{2}-16 a b c d +5 b^{2} c^{2}\right )}{40 \left (b x +a \right )^{\frac {8}{3}} \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right )}\) | \(105\) |
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Leaf count of result is larger than twice the leaf count of optimal. 251 vs. \(2 (83) = 166\).
Time = 0.25 (sec) , antiderivative size = 251, normalized size of antiderivative = 2.49 \[ \int \frac {1}{(a+b x)^{11/3} \sqrt [3]{c+d x}} \, dx=-\frac {3 \, {\left (9 \, b^{2} d^{2} x^{2} + 5 \, b^{2} c^{2} - 16 \, a b c d + 20 \, a^{2} d^{2} - 6 \, {\left (b^{2} c d - 4 \, a b d^{2}\right )} x\right )} {\left (b x + a\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {2}{3}}}{40 \, {\left (a^{3} b^{3} c^{3} - 3 \, a^{4} b^{2} c^{2} d + 3 \, a^{5} b c d^{2} - a^{6} d^{3} + {\left (b^{6} c^{3} - 3 \, a b^{5} c^{2} d + 3 \, a^{2} b^{4} c d^{2} - a^{3} b^{3} d^{3}\right )} x^{3} + 3 \, {\left (a b^{5} c^{3} - 3 \, a^{2} b^{4} c^{2} d + 3 \, a^{3} b^{3} c d^{2} - a^{4} b^{2} d^{3}\right )} x^{2} + 3 \, {\left (a^{2} b^{4} c^{3} - 3 \, a^{3} b^{3} c^{2} d + 3 \, a^{4} b^{2} c d^{2} - a^{5} b d^{3}\right )} x\right )}} \]
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\[ \int \frac {1}{(a+b x)^{11/3} \sqrt [3]{c+d x}} \, dx=\int \frac {1}{\left (a + b x\right )^{\frac {11}{3}} \sqrt [3]{c + d x}}\, dx \]
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\[ \int \frac {1}{(a+b x)^{11/3} \sqrt [3]{c+d x}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {11}{3}} {\left (d x + c\right )}^{\frac {1}{3}}} \,d x } \]
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\[ \int \frac {1}{(a+b x)^{11/3} \sqrt [3]{c+d x}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {11}{3}} {\left (d x + c\right )}^{\frac {1}{3}}} \,d x } \]
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Timed out. \[ \int \frac {1}{(a+b x)^{11/3} \sqrt [3]{c+d x}} \, dx=\int \frac {1}{{\left (a+b\,x\right )}^{11/3}\,{\left (c+d\,x\right )}^{1/3}} \,d x \]
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