Integrand size = 24, antiderivative size = 239 \[ \int \frac {1}{(d+e x)^{3/2} \sqrt [4]{a+b x+c x^2}} \, dx=\frac {2 \left (b-\sqrt {b^2-4 a c}+2 c x\right ) \sqrt [4]{\frac {\left (2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e\right ) \left (b+\sqrt {b^2-4 a c}+2 c x\right )}{\left (2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e\right ) \left (b-\sqrt {b^2-4 a c}+2 c x\right )}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{4},\frac {1}{2},-\frac {4 c \sqrt {b^2-4 a c} (d+e x)}{\left (2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e\right ) \left (b-\sqrt {b^2-4 a c}+2 c x\right )}\right )}{\left (2 c d-b e+\sqrt {b^2-4 a c} e\right ) \sqrt {d+e x} \sqrt [4]{a+b x+c x^2}} \]
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Time = 0.12 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {740} \[ \int \frac {1}{(d+e x)^{3/2} \sqrt [4]{a+b x+c x^2}} \, dx=\frac {2 \left (-\sqrt {b^2-4 a c}+b+2 c x\right ) \sqrt [4]{\frac {\left (\sqrt {b^2-4 a c}+b+2 c x\right ) \left (2 c d-e \left (b-\sqrt {b^2-4 a c}\right )\right )}{\left (-\sqrt {b^2-4 a c}+b+2 c x\right ) \left (2 c d-e \left (\sqrt {b^2-4 a c}+b\right )\right )}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{4},\frac {1}{2},-\frac {4 c \sqrt {b^2-4 a c} (d+e x)}{\left (2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e\right ) \left (b+2 c x-\sqrt {b^2-4 a c}\right )}\right )}{\sqrt {d+e x} \sqrt [4]{a+b x+c x^2} \left (e \sqrt {b^2-4 a c}-b e+2 c d\right )} \]
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Rule 740
Rubi steps \begin{align*} \text {integral}& = \frac {2 \left (b-\sqrt {b^2-4 a c}+2 c x\right ) \sqrt [4]{\frac {\left (2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e\right ) \left (b+\sqrt {b^2-4 a c}+2 c x\right )}{\left (2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e\right ) \left (b-\sqrt {b^2-4 a c}+2 c x\right )}} \, _2F_1\left (-\frac {1}{2},\frac {1}{4};\frac {1}{2};-\frac {4 c \sqrt {b^2-4 a c} (d+e x)}{\left (2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e\right ) \left (b-\sqrt {b^2-4 a c}+2 c x\right )}\right )}{\left (2 c d-b e+\sqrt {b^2-4 a c} e\right ) \sqrt {d+e x} \sqrt [4]{a+b x+c x^2}} \\ \end{align*}
Time = 10.29 (sec) , antiderivative size = 231, normalized size of antiderivative = 0.97 \[ \int \frac {1}{(d+e x)^{3/2} \sqrt [4]{a+b x+c x^2}} \, dx=-\frac {2 \left (b+\sqrt {b^2-4 a c}+2 c x\right ) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{4},\frac {1}{2},-\frac {4 c \sqrt {b^2-4 a c} (d+e x)}{\left (-2 c d+\left (b+\sqrt {b^2-4 a c}\right ) e\right ) \left (-b+\sqrt {b^2-4 a c}-2 c x\right )}\right )}{\left (-2 c d+\left (b+\sqrt {b^2-4 a c}\right ) e\right ) \left (\frac {\left (2 c d+\left (-b+\sqrt {b^2-4 a c}\right ) e\right ) \left (b+\sqrt {b^2-4 a c}+2 c x\right )}{\left (-2 c d+\left (b+\sqrt {b^2-4 a c}\right ) e\right ) \left (-b+\sqrt {b^2-4 a c}-2 c x\right )}\right )^{3/4} \sqrt {d+e x} \sqrt [4]{a+x (b+c x)}} \]
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\[\int \frac {1}{\left (e x +d \right )^{\frac {3}{2}} \left (c \,x^{2}+b x +a \right )^{\frac {1}{4}}}d x\]
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\[ \int \frac {1}{(d+e x)^{3/2} \sqrt [4]{a+b x+c x^2}} \, dx=\int { \frac {1}{{\left (c x^{2} + b x + a\right )}^{\frac {1}{4}} {\left (e x + d\right )}^{\frac {3}{2}}} \,d x } \]
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\[ \int \frac {1}{(d+e x)^{3/2} \sqrt [4]{a+b x+c x^2}} \, dx=\int \frac {1}{\left (d + e x\right )^{\frac {3}{2}} \sqrt [4]{a + b x + c x^{2}}}\, dx \]
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\[ \int \frac {1}{(d+e x)^{3/2} \sqrt [4]{a+b x+c x^2}} \, dx=\int { \frac {1}{{\left (c x^{2} + b x + a\right )}^{\frac {1}{4}} {\left (e x + d\right )}^{\frac {3}{2}}} \,d x } \]
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\[ \int \frac {1}{(d+e x)^{3/2} \sqrt [4]{a+b x+c x^2}} \, dx=\int { \frac {1}{{\left (c x^{2} + b x + a\right )}^{\frac {1}{4}} {\left (e x + d\right )}^{\frac {3}{2}}} \,d x } \]
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Timed out. \[ \int \frac {1}{(d+e x)^{3/2} \sqrt [4]{a+b x+c x^2}} \, dx=\int \frac {1}{{\left (d+e\,x\right )}^{3/2}\,{\left (c\,x^2+b\,x+a\right )}^{1/4}} \,d x \]
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