Integrand size = 19, antiderivative size = 101 \[ \int \frac {\sqrt [3]{c+d x}}{(a+b x)^{13/3}} \, dx=-\frac {3 (c+d x)^{4/3}}{10 (b c-a d) (a+b x)^{10/3}}+\frac {9 d (c+d x)^{4/3}}{35 (b c-a d)^2 (a+b x)^{7/3}}-\frac {27 d^2 (c+d x)^{4/3}}{140 (b c-a d)^3 (a+b x)^{4/3}} \]
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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} \[ \int \frac {\sqrt [3]{c+d x}}{(a+b x)^{13/3}} \, dx=-\frac {27 d^2 (c+d x)^{4/3}}{140 (a+b x)^{4/3} (b c-a d)^3}+\frac {9 d (c+d x)^{4/3}}{35 (a+b x)^{7/3} (b c-a d)^2}-\frac {3 (c+d x)^{4/3}}{10 (a+b x)^{10/3} (b c-a d)} \]
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Rule 37
Rule 47
Rubi steps \begin{align*} \text {integral}& = -\frac {3 (c+d x)^{4/3}}{10 (b c-a d) (a+b x)^{10/3}}-\frac {(3 d) \int \frac {\sqrt [3]{c+d x}}{(a+b x)^{10/3}} \, dx}{5 (b c-a d)} \\ & = -\frac {3 (c+d x)^{4/3}}{10 (b c-a d) (a+b x)^{10/3}}+\frac {9 d (c+d x)^{4/3}}{35 (b c-a d)^2 (a+b x)^{7/3}}+\frac {\left (9 d^2\right ) \int \frac {\sqrt [3]{c+d x}}{(a+b x)^{7/3}} \, dx}{35 (b c-a d)^2} \\ & = -\frac {3 (c+d x)^{4/3}}{10 (b c-a d) (a+b x)^{10/3}}+\frac {9 d (c+d x)^{4/3}}{35 (b c-a d)^2 (a+b x)^{7/3}}-\frac {27 d^2 (c+d x)^{4/3}}{140 (b c-a d)^3 (a+b x)^{4/3}} \\ \end{align*}
Time = 0.28 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.76 \[ \int \frac {\sqrt [3]{c+d x}}{(a+b x)^{13/3}} \, dx=-\frac {3 (c+d x)^{4/3} \left (35 a^2 d^2+10 a b d (-4 c+3 d x)+b^2 \left (14 c^2-12 c d x+9 d^2 x^2\right )\right )}{140 (b c-a d)^3 (a+b x)^{10/3}} \]
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Time = 0.29 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.04
method | result | size |
gosper | \(\frac {3 \left (d x +c \right )^{\frac {4}{3}} \left (9 d^{2} x^{2} b^{2}+30 x a b \,d^{2}-12 x \,b^{2} c d +35 a^{2} d^{2}-40 a b c d +14 b^{2} c^{2}\right )}{140 \left (b x +a \right )^{\frac {10}{3}} \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right )}\) | \(105\) |
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Leaf count of result is larger than twice the leaf count of optimal. 337 vs. \(2 (83) = 166\).
Time = 0.23 (sec) , antiderivative size = 337, normalized size of antiderivative = 3.34 \[ \int \frac {\sqrt [3]{c+d x}}{(a+b x)^{13/3}} \, dx=-\frac {3 \, {\left (9 \, b^{2} d^{3} x^{3} + 14 \, b^{2} c^{3} - 40 \, a b c^{2} d + 35 \, a^{2} c d^{2} - 3 \, {\left (b^{2} c d^{2} - 10 \, a b d^{3}\right )} x^{2} + {\left (2 \, b^{2} c^{2} d - 10 \, a b c d^{2} + 35 \, a^{2} d^{3}\right )} x\right )} {\left (b x + a\right )}^{\frac {2}{3}} {\left (d x + c\right )}^{\frac {1}{3}}}{140 \, {\left (a^{4} b^{3} c^{3} - 3 \, a^{5} b^{2} c^{2} d + 3 \, a^{6} b c d^{2} - a^{7} d^{3} + {\left (b^{7} c^{3} - 3 \, a b^{6} c^{2} d + 3 \, a^{2} b^{5} c d^{2} - a^{3} b^{4} d^{3}\right )} x^{4} + 4 \, {\left (a b^{6} c^{3} - 3 \, a^{2} b^{5} c^{2} d + 3 \, a^{3} b^{4} c d^{2} - a^{4} b^{3} d^{3}\right )} x^{3} + 6 \, {\left (a^{2} b^{5} c^{3} - 3 \, a^{3} b^{4} c^{2} d + 3 \, a^{4} b^{3} c d^{2} - a^{5} b^{2} d^{3}\right )} x^{2} + 4 \, {\left (a^{3} b^{4} c^{3} - 3 \, a^{4} b^{3} c^{2} d + 3 \, a^{5} b^{2} c d^{2} - a^{6} b d^{3}\right )} x\right )}} \]
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\[ \int \frac {\sqrt [3]{c+d x}}{(a+b x)^{13/3}} \, dx=\int \frac {\sqrt [3]{c + d x}}{\left (a + b x\right )^{\frac {13}{3}}}\, dx \]
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\[ \int \frac {\sqrt [3]{c+d x}}{(a+b x)^{13/3}} \, dx=\int { \frac {{\left (d x + c\right )}^{\frac {1}{3}}}{{\left (b x + a\right )}^{\frac {13}{3}}} \,d x } \]
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\[ \int \frac {\sqrt [3]{c+d x}}{(a+b x)^{13/3}} \, dx=\int { \frac {{\left (d x + c\right )}^{\frac {1}{3}}}{{\left (b x + a\right )}^{\frac {13}{3}}} \,d x } \]
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Time = 0.98 (sec) , antiderivative size = 203, normalized size of antiderivative = 2.01 \[ \int \frac {\sqrt [3]{c+d x}}{(a+b x)^{13/3}} \, dx=\frac {{\left (c+d\,x\right )}^{1/3}\,\left (\frac {105\,a^2\,c\,d^2-120\,a\,b\,c^2\,d+42\,b^2\,c^3}{140\,b^3\,{\left (a\,d-b\,c\right )}^3}+\frac {x\,\left (105\,a^2\,d^3-30\,a\,b\,c\,d^2+6\,b^2\,c^2\,d\right )}{140\,b^3\,{\left (a\,d-b\,c\right )}^3}+\frac {27\,d^3\,x^3}{140\,b\,{\left (a\,d-b\,c\right )}^3}+\frac {9\,d^2\,x^2\,\left (10\,a\,d-b\,c\right )}{140\,b^2\,{\left (a\,d-b\,c\right )}^3}\right )}{x^3\,{\left (a+b\,x\right )}^{1/3}+\frac {a^3\,{\left (a+b\,x\right )}^{1/3}}{b^3}+\frac {3\,a\,x^2\,{\left (a+b\,x\right )}^{1/3}}{b}+\frac {3\,a^2\,x\,{\left (a+b\,x\right )}^{1/3}}{b^2}} \]
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