Integrand size = 28, antiderivative size = 89 \[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{23/3}} \, dx=\frac {3 \left (a+b x+c x^2\right )^{7/3}}{10 \left (b^2-4 a c\right ) d (b d+2 c d x)^{20/3}}+\frac {9 \left (a+b x+c x^2\right )^{7/3}}{70 \left (b^2-4 a c\right )^2 d^3 (b d+2 c d x)^{14/3}} \]
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Time = 0.03 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {707, 696} \[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{23/3}} \, dx=\frac {9 \left (a+b x+c x^2\right )^{7/3}}{70 d^3 \left (b^2-4 a c\right )^2 (b d+2 c d x)^{14/3}}+\frac {3 \left (a+b x+c x^2\right )^{7/3}}{10 d \left (b^2-4 a c\right ) (b d+2 c d x)^{20/3}} \]
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Rule 696
Rule 707
Rubi steps \begin{align*} \text {integral}& = \frac {3 \left (a+b x+c x^2\right )^{7/3}}{10 \left (b^2-4 a c\right ) d (b d+2 c d x)^{20/3}}+\frac {3 \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{17/3}} \, dx}{10 \left (b^2-4 a c\right ) d^2} \\ & = \frac {3 \left (a+b x+c x^2\right )^{7/3}}{10 \left (b^2-4 a c\right ) d (b d+2 c d x)^{20/3}}+\frac {9 \left (a+b x+c x^2\right )^{7/3}}{70 \left (b^2-4 a c\right )^2 d^3 (b d+2 c d x)^{14/3}} \\ \end{align*}
Time = 4.90 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.83 \[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{23/3}} \, dx=\frac {3 \sqrt [3]{d (b+2 c x)} (a+x (b+c x))^{7/3} \left (5 b^2+6 b c x+2 c \left (-7 a+3 c x^2\right )\right )}{35 \left (b^2-4 a c\right )^2 d^8 (b+2 c x)^7} \]
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Time = 2.59 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.85
method | result | size |
gosper | \(-\frac {3 \left (2 c x +b \right ) \left (c \,x^{2}+b x +a \right )^{\frac {7}{3}} \left (-6 c^{2} x^{2}-6 b c x +14 a c -5 b^{2}\right )}{35 \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right ) \left (2 c d x +b d \right )^{\frac {23}{3}}}\) | \(76\) |
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Leaf count of result is larger than twice the leaf count of optimal. 409 vs. \(2 (77) = 154\).
Time = 0.40 (sec) , antiderivative size = 409, normalized size of antiderivative = 4.60 \[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{23/3}} \, dx=\frac {3 \, {\left (6 \, c^{4} x^{6} + 18 \, b c^{3} x^{5} + {\left (23 \, b^{2} c^{2} - 2 \, a c^{3}\right )} x^{4} + 5 \, a^{2} b^{2} - 14 \, a^{3} c + 4 \, {\left (4 \, b^{3} c - a b c^{2}\right )} x^{3} + {\left (5 \, b^{4} + 8 \, a b^{2} c - 22 \, a^{2} c^{2}\right )} x^{2} + 2 \, {\left (5 \, a b^{3} - 11 \, a^{2} b c\right )} x\right )} {\left (2 \, c d x + b d\right )}^{\frac {1}{3}} {\left (c x^{2} + b x + a\right )}^{\frac {1}{3}}}{35 \, {\left (128 \, {\left (b^{4} c^{7} - 8 \, a b^{2} c^{8} + 16 \, a^{2} c^{9}\right )} d^{8} x^{7} + 448 \, {\left (b^{5} c^{6} - 8 \, a b^{3} c^{7} + 16 \, a^{2} b c^{8}\right )} d^{8} x^{6} + 672 \, {\left (b^{6} c^{5} - 8 \, a b^{4} c^{6} + 16 \, a^{2} b^{2} c^{7}\right )} d^{8} x^{5} + 560 \, {\left (b^{7} c^{4} - 8 \, a b^{5} c^{5} + 16 \, a^{2} b^{3} c^{6}\right )} d^{8} x^{4} + 280 \, {\left (b^{8} c^{3} - 8 \, a b^{6} c^{4} + 16 \, a^{2} b^{4} c^{5}\right )} d^{8} x^{3} + 84 \, {\left (b^{9} c^{2} - 8 \, a b^{7} c^{3} + 16 \, a^{2} b^{5} c^{4}\right )} d^{8} x^{2} + 14 \, {\left (b^{10} c - 8 \, a b^{8} c^{2} + 16 \, a^{2} b^{6} c^{3}\right )} d^{8} x + {\left (b^{11} - 8 \, a b^{9} c + 16 \, a^{2} b^{7} c^{2}\right )} d^{8}\right )}} \]
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Timed out. \[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{23/3}} \, dx=\text {Timed out} \]
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\[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{23/3}} \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )}^{\frac {4}{3}}}{{\left (2 \, c d x + b d\right )}^{\frac {23}{3}}} \,d x } \]
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\[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{23/3}} \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )}^{\frac {4}{3}}}{{\left (2 \, c d x + b d\right )}^{\frac {23}{3}}} \,d x } \]
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Timed out. \[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{23/3}} \, dx=\int \frac {{\left (c\,x^2+b\,x+a\right )}^{4/3}}{{\left (b\,d+2\,c\,d\,x\right )}^{23/3}} \,d x \]
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