Integrand size = 28, antiderivative size = 99 \[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{10/3}} \, dx=\frac {3 \left (b^2-4 a c\right ) \sqrt [3]{a+b x+c x^2} \operatorname {Hypergeometric2F1}\left (-\frac {4}{3},-\frac {7}{6},-\frac {1}{6},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{56 c^2 d (d (b+2 c x))^{7/3} \sqrt [3]{1-\frac {(b+2 c x)^2}{b^2-4 a c}}} \]
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Time = 0.09 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {708, 372, 371} \[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{10/3}} \, dx=\frac {3 \left (b^2-4 a c\right ) \sqrt [3]{a+b x+c x^2} \operatorname {Hypergeometric2F1}\left (-\frac {4}{3},-\frac {7}{6},-\frac {1}{6},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{56 c^2 d \sqrt [3]{1-\frac {(b+2 c x)^2}{b^2-4 a c}} (d (b+2 c x))^{7/3}} \]
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Rule 371
Rule 372
Rule 708
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {\left (a-\frac {b^2}{4 c}+\frac {x^2}{4 c d^2}\right )^{4/3}}{x^{10/3}} \, dx,x,b d+2 c d x\right )}{2 c d} \\ & = \frac {\left (\left (a-\frac {b^2}{4 c}\right ) \sqrt [3]{a+b x+c x^2}\right ) \text {Subst}\left (\int \frac {\left (1+\frac {x^2}{4 \left (a-\frac {b^2}{4 c}\right ) c d^2}\right )^{4/3}}{x^{10/3}} \, dx,x,b d+2 c d x\right )}{\sqrt [3]{2} c d \sqrt [3]{4+\frac {(b d+2 c d x)^2}{\left (a-\frac {b^2}{4 c}\right ) c d^2}}} \\ & = \frac {3 \left (b^2-4 a c\right ) \sqrt [3]{a+b x+c x^2} \, _2F_1\left (-\frac {4}{3},-\frac {7}{6};-\frac {1}{6};\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{56 c^2 d (d (b+2 c x))^{7/3} \sqrt [3]{1-\frac {(b+2 c x)^2}{b^2-4 a c}}} \\ \end{align*}
Time = 10.06 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.05 \[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{10/3}} \, dx=\frac {3 \left (b^2-4 a c\right ) \sqrt [3]{a+x (b+c x)} \operatorname {Hypergeometric2F1}\left (-\frac {4}{3},-\frac {7}{6},-\frac {1}{6},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{56\ 2^{2/3} c^2 d (d (b+2 c x))^{7/3} \sqrt [3]{\frac {c (a+x (b+c x))}{-b^2+4 a c}}} \]
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\[\int \frac {\left (c \,x^{2}+b x +a \right )^{\frac {4}{3}}}{\left (2 c d x +b d \right )^{\frac {10}{3}}}d x\]
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\[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{10/3}} \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )}^{\frac {4}{3}}}{{\left (2 \, c d x + b d\right )}^{\frac {10}{3}}} \,d x } \]
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\[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{10/3}} \, dx=\int \frac {\left (a + b x + c x^{2}\right )^{\frac {4}{3}}}{\left (d \left (b + 2 c x\right )\right )^{\frac {10}{3}}}\, dx \]
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\[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{10/3}} \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )}^{\frac {4}{3}}}{{\left (2 \, c d x + b d\right )}^{\frac {10}{3}}} \,d x } \]
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\[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{10/3}} \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )}^{\frac {4}{3}}}{{\left (2 \, c d x + b d\right )}^{\frac {10}{3}}} \,d x } \]
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Timed out. \[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{10/3}} \, dx=\int \frac {{\left (c\,x^2+b\,x+a\right )}^{4/3}}{{\left (b\,d+2\,c\,d\,x\right )}^{10/3}} \,d x \]
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