Integrand size = 28, antiderivative size = 133 \[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{29/3}} \, dx=\frac {3 \left (a+b x+c x^2\right )^{7/3}}{13 \left (b^2-4 a c\right ) d (b d+2 c d x)^{26/3}}+\frac {9 \left (a+b x+c x^2\right )^{7/3}}{65 \left (b^2-4 a c\right )^2 d^3 (b d+2 c d x)^{20/3}}+\frac {27 \left (a+b x+c x^2\right )^{7/3}}{455 \left (b^2-4 a c\right )^3 d^5 (b d+2 c d x)^{14/3}} \]
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Time = 0.04 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {707, 696} \[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{29/3}} \, dx=\frac {27 \left (a+b x+c x^2\right )^{7/3}}{455 d^5 \left (b^2-4 a c\right )^3 (b d+2 c d x)^{14/3}}+\frac {9 \left (a+b x+c x^2\right )^{7/3}}{65 d^3 \left (b^2-4 a c\right )^2 (b d+2 c d x)^{20/3}}+\frac {3 \left (a+b x+c x^2\right )^{7/3}}{13 d \left (b^2-4 a c\right ) (b d+2 c d x)^{26/3}} \]
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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}}{13 \left (b^2-4 a c\right ) d (b d+2 c d x)^{26/3}}+\frac {6 \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{23/3}} \, dx}{13 \left (b^2-4 a c\right ) d^2} \\ & = \frac {3 \left (a+b x+c x^2\right )^{7/3}}{13 \left (b^2-4 a c\right ) d (b d+2 c d x)^{26/3}}+\frac {9 \left (a+b x+c x^2\right )^{7/3}}{65 \left (b^2-4 a c\right )^2 d^3 (b d+2 c d x)^{20/3}}+\frac {9 \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{17/3}} \, dx}{65 \left (b^2-4 a c\right )^2 d^4} \\ & = \frac {3 \left (a+b x+c x^2\right )^{7/3}}{13 \left (b^2-4 a c\right ) d (b d+2 c d x)^{26/3}}+\frac {9 \left (a+b x+c x^2\right )^{7/3}}{65 \left (b^2-4 a c\right )^2 d^3 (b d+2 c d x)^{20/3}}+\frac {27 \left (a+b x+c x^2\right )^{7/3}}{455 \left (b^2-4 a c\right )^3 d^5 (b d+2 c d x)^{14/3}} \\ \end{align*}
Time = 4.99 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.92 \[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{29/3}} \, dx=\frac {3 (a+x (b+c x))^{7/3} \left (65 b^4+156 b^3 c x+48 b c^2 x \left (-7 a+6 c x^2\right )+4 b^2 c \left (-91 a+75 c x^2\right )+16 c^2 \left (35 a^2-21 a c x^2+9 c^2 x^4\right )\right )}{455 \left (b^2-4 a c\right )^3 d^9 (b+2 c x)^8 (d (b+2 c x))^{2/3}} \]
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Time = 2.39 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.05
method | result | size |
gosper | \(-\frac {3 \left (2 c x +b \right ) \left (c \,x^{2}+b x +a \right )^{\frac {7}{3}} \left (144 c^{4} x^{4}+288 b \,c^{3} x^{3}-336 x^{2} c^{3} a +300 b^{2} c^{2} x^{2}-336 a b \,c^{2} x +156 b^{3} c x +560 a^{2} c^{2}-364 a \,b^{2} c +65 b^{4}\right )}{455 \left (64 c^{3} a^{3}-48 a^{2} b^{2} c^{2}+12 a \,b^{4} c -b^{6}\right ) \left (2 c d x +b d \right )^{\frac {29}{3}}}\) | \(139\) |
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Leaf count of result is larger than twice the leaf count of optimal. 687 vs. \(2 (115) = 230\).
Time = 0.52 (sec) , antiderivative size = 687, normalized size of antiderivative = 5.17 \[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{29/3}} \, dx=\frac {3 \, {\left (144 \, c^{6} x^{8} + 576 \, b c^{5} x^{7} + 12 \, {\left (85 \, b^{2} c^{4} - 4 \, a c^{5}\right )} x^{6} + 65 \, a^{2} b^{4} - 364 \, a^{3} b^{2} c + 560 \, a^{4} c^{2} + 36 \, {\left (29 \, b^{3} c^{3} - 4 \, a b c^{4}\right )} x^{5} + {\left (677 \, b^{4} c^{2} - 196 \, a b^{2} c^{3} + 32 \, a^{2} c^{4}\right )} x^{4} + 2 \, {\left (143 \, b^{5} c - 76 \, a b^{3} c^{2} + 32 \, a^{2} b c^{3}\right )} x^{3} + {\left (65 \, b^{6} + 78 \, a b^{4} c - 540 \, a^{2} b^{2} c^{2} + 784 \, a^{3} c^{3}\right )} x^{2} + 2 \, {\left (65 \, a b^{5} - 286 \, a^{2} b^{3} c + 392 \, a^{3} b c^{2}\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}}}{455 \, {\left (512 \, {\left (b^{6} c^{9} - 12 \, a b^{4} c^{10} + 48 \, a^{2} b^{2} c^{11} - 64 \, a^{3} c^{12}\right )} d^{10} x^{9} + 2304 \, {\left (b^{7} c^{8} - 12 \, a b^{5} c^{9} + 48 \, a^{2} b^{3} c^{10} - 64 \, a^{3} b c^{11}\right )} d^{10} x^{8} + 4608 \, {\left (b^{8} c^{7} - 12 \, a b^{6} c^{8} + 48 \, a^{2} b^{4} c^{9} - 64 \, a^{3} b^{2} c^{10}\right )} d^{10} x^{7} + 5376 \, {\left (b^{9} c^{6} - 12 \, a b^{7} c^{7} + 48 \, a^{2} b^{5} c^{8} - 64 \, a^{3} b^{3} c^{9}\right )} d^{10} x^{6} + 4032 \, {\left (b^{10} c^{5} - 12 \, a b^{8} c^{6} + 48 \, a^{2} b^{6} c^{7} - 64 \, a^{3} b^{4} c^{8}\right )} d^{10} x^{5} + 2016 \, {\left (b^{11} c^{4} - 12 \, a b^{9} c^{5} + 48 \, a^{2} b^{7} c^{6} - 64 \, a^{3} b^{5} c^{7}\right )} d^{10} x^{4} + 672 \, {\left (b^{12} c^{3} - 12 \, a b^{10} c^{4} + 48 \, a^{2} b^{8} c^{5} - 64 \, a^{3} b^{6} c^{6}\right )} d^{10} x^{3} + 144 \, {\left (b^{13} c^{2} - 12 \, a b^{11} c^{3} + 48 \, a^{2} b^{9} c^{4} - 64 \, a^{3} b^{7} c^{5}\right )} d^{10} x^{2} + 18 \, {\left (b^{14} c - 12 \, a b^{12} c^{2} + 48 \, a^{2} b^{10} c^{3} - 64 \, a^{3} b^{8} c^{4}\right )} d^{10} x + {\left (b^{15} - 12 \, a b^{13} c + 48 \, a^{2} b^{11} c^{2} - 64 \, a^{3} b^{9} c^{3}\right )} d^{10}\right )}} \]
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Timed out. \[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{29/3}} \, dx=\text {Timed out} \]
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Leaf count of result is larger than twice the leaf count of optimal. 783 vs. \(2 (115) = 230\).
Time = 0.40 (sec) , antiderivative size = 783, normalized size of antiderivative = 5.89 \[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{29/3}} \, dx=\frac {3 \, {\left (144 \, c^{6} d^{\frac {1}{3}} x^{8} + 576 \, b c^{5} d^{\frac {1}{3}} x^{7} + 12 \, {\left (85 \, b^{2} c^{4} d^{\frac {1}{3}} - 4 \, a c^{5} d^{\frac {1}{3}}\right )} x^{6} + 65 \, a^{2} b^{4} d^{\frac {1}{3}} - 364 \, a^{3} b^{2} c d^{\frac {1}{3}} + 560 \, a^{4} c^{2} d^{\frac {1}{3}} + 36 \, {\left (29 \, b^{3} c^{3} d^{\frac {1}{3}} - 4 \, a b c^{4} d^{\frac {1}{3}}\right )} x^{5} + {\left (677 \, b^{4} c^{2} d^{\frac {1}{3}} - 196 \, a b^{2} c^{3} d^{\frac {1}{3}} + 32 \, a^{2} c^{4} d^{\frac {1}{3}}\right )} x^{4} + 2 \, {\left (143 \, b^{5} c d^{\frac {1}{3}} - 76 \, a b^{3} c^{2} d^{\frac {1}{3}} + 32 \, a^{2} b c^{3} d^{\frac {1}{3}}\right )} x^{3} + {\left (65 \, b^{6} d^{\frac {1}{3}} + 78 \, a b^{4} c d^{\frac {1}{3}} - 540 \, a^{2} b^{2} c^{2} d^{\frac {1}{3}} + 784 \, a^{3} c^{3} d^{\frac {1}{3}}\right )} x^{2} + 2 \, {\left (65 \, a b^{5} d^{\frac {1}{3}} - 286 \, a^{2} b^{3} c d^{\frac {1}{3}} + 392 \, a^{3} b c^{2} d^{\frac {1}{3}}\right )} x\right )} {\left (c x^{2} + b x + a\right )}^{\frac {1}{3}}}{455 \, {\left (b^{14} d^{10} - 12 \, a b^{12} c d^{10} + 48 \, a^{2} b^{10} c^{2} d^{10} - 64 \, a^{3} b^{8} c^{3} d^{10} + 256 \, {\left (b^{6} c^{8} d^{10} - 12 \, a b^{4} c^{9} d^{10} + 48 \, a^{2} b^{2} c^{10} d^{10} - 64 \, a^{3} c^{11} d^{10}\right )} x^{8} + 1024 \, {\left (b^{7} c^{7} d^{10} - 12 \, a b^{5} c^{8} d^{10} + 48 \, a^{2} b^{3} c^{9} d^{10} - 64 \, a^{3} b c^{10} d^{10}\right )} x^{7} + 1792 \, {\left (b^{8} c^{6} d^{10} - 12 \, a b^{6} c^{7} d^{10} + 48 \, a^{2} b^{4} c^{8} d^{10} - 64 \, a^{3} b^{2} c^{9} d^{10}\right )} x^{6} + 1792 \, {\left (b^{9} c^{5} d^{10} - 12 \, a b^{7} c^{6} d^{10} + 48 \, a^{2} b^{5} c^{7} d^{10} - 64 \, a^{3} b^{3} c^{8} d^{10}\right )} x^{5} + 1120 \, {\left (b^{10} c^{4} d^{10} - 12 \, a b^{8} c^{5} d^{10} + 48 \, a^{2} b^{6} c^{6} d^{10} - 64 \, a^{3} b^{4} c^{7} d^{10}\right )} x^{4} + 448 \, {\left (b^{11} c^{3} d^{10} - 12 \, a b^{9} c^{4} d^{10} + 48 \, a^{2} b^{7} c^{5} d^{10} - 64 \, a^{3} b^{5} c^{6} d^{10}\right )} x^{3} + 112 \, {\left (b^{12} c^{2} d^{10} - 12 \, a b^{10} c^{3} d^{10} + 48 \, a^{2} b^{8} c^{4} d^{10} - 64 \, a^{3} b^{6} c^{5} d^{10}\right )} x^{2} + 16 \, {\left (b^{13} c d^{10} - 12 \, a b^{11} c^{2} d^{10} + 48 \, a^{2} b^{9} c^{3} d^{10} - 64 \, a^{3} b^{7} c^{4} d^{10}\right )} x\right )} {\left (2 \, c x + b\right )}^{\frac {2}{3}}} \]
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\[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{29/3}} \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )}^{\frac {4}{3}}}{{\left (2 \, c d x + b d\right )}^{\frac {29}{3}}} \,d x } \]
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Timed out. \[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{29/3}} \, dx=\int \frac {{\left (c\,x^2+b\,x+a\right )}^{4/3}}{{\left (b\,d+2\,c\,d\,x\right )}^{29/3}} \,d x \]
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