Integrand size = 28, antiderivative size = 44 \[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{17/3}} \, dx=\frac {3 \left (a+b x+c x^2\right )^{7/3}}{7 \left (b^2-4 a c\right ) d (b d+2 c d x)^{14/3}} \]
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Time = 0.01 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.036, Rules used = {696} \[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{17/3}} \, dx=\frac {3 \left (a+b x+c x^2\right )^{7/3}}{7 d \left (b^2-4 a c\right ) (b d+2 c d x)^{14/3}} \]
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Rule 696
Rubi steps \begin{align*} \text {integral}& = \frac {3 \left (a+b x+c x^2\right )^{7/3}}{7 \left (b^2-4 a c\right ) d (b d+2 c d x)^{14/3}} \\ \end{align*}
Time = 4.94 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.14 \[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{17/3}} \, dx=\frac {3 \sqrt [3]{d (b+2 c x)} (a+x (b+c x))^{7/3}}{7 \left (b^2-4 a c\right ) d^6 (b+2 c x)^5} \]
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Time = 2.72 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.00
method | result | size |
gosper | \(-\frac {3 \left (2 c x +b \right ) \left (c \,x^{2}+b x +a \right )^{\frac {7}{3}}}{7 \left (4 a c -b^{2}\right ) \left (2 c d x +b d \right )^{\frac {17}{3}}}\) | \(44\) |
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Leaf count of result is larger than twice the leaf count of optimal. 194 vs. \(2 (38) = 76\).
Time = 0.38 (sec) , antiderivative size = 194, normalized size of antiderivative = 4.41 \[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{17/3}} \, dx=\frac {3 \, {\left (c^{2} x^{4} + 2 \, b c x^{3} + 2 \, a b x + {\left (b^{2} + 2 \, a c\right )} x^{2} + a^{2}\right )} {\left (2 \, c d x + b d\right )}^{\frac {1}{3}} {\left (c x^{2} + b x + a\right )}^{\frac {1}{3}}}{7 \, {\left (32 \, {\left (b^{2} c^{5} - 4 \, a c^{6}\right )} d^{6} x^{5} + 80 \, {\left (b^{3} c^{4} - 4 \, a b c^{5}\right )} d^{6} x^{4} + 80 \, {\left (b^{4} c^{3} - 4 \, a b^{2} c^{4}\right )} d^{6} x^{3} + 40 \, {\left (b^{5} c^{2} - 4 \, a b^{3} c^{3}\right )} d^{6} x^{2} + 10 \, {\left (b^{6} c - 4 \, a b^{4} c^{2}\right )} d^{6} x + {\left (b^{7} - 4 \, a b^{5} c\right )} d^{6}\right )}} \]
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Timed out. \[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{17/3}} \, dx=\text {Timed out} \]
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\[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{17/3}} \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )}^{\frac {4}{3}}}{{\left (2 \, c d x + b d\right )}^{\frac {17}{3}}} \,d x } \]
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\[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{17/3}} \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )}^{\frac {4}{3}}}{{\left (2 \, c d x + b d\right )}^{\frac {17}{3}}} \,d x } \]
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Timed out. \[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{17/3}} \, dx=\int \frac {{\left (c\,x^2+b\,x+a\right )}^{4/3}}{{\left (b\,d+2\,c\,d\,x\right )}^{17/3}} \,d x \]
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