Integrand size = 18, antiderivative size = 164 \[ \int \frac {\left (a+b x^3+c x^6\right )^p}{x^4} \, dx=-\frac {4^p \left (\frac {b-\sqrt {b^2-4 a c}+2 c x^3}{c x^3}\right )^{-p} \left (\frac {b+\sqrt {b^2-4 a c}+2 c x^3}{c x^3}\right )^{-p} \left (a+b x^3+c x^6\right )^p \operatorname {AppellF1}\left (1-2 p,-p,-p,2 (1-p),-\frac {b-\sqrt {b^2-4 a c}}{2 c x^3},-\frac {b+\sqrt {b^2-4 a c}}{2 c x^3}\right )}{3 (1-2 p) x^3} \]
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Time = 0.09 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1371, 772, 138} \[ \int \frac {\left (a+b x^3+c x^6\right )^p}{x^4} \, dx=-\frac {4^p \left (\frac {-\sqrt {b^2-4 a c}+b+2 c x^3}{c x^3}\right )^{-p} \left (\frac {\sqrt {b^2-4 a c}+b+2 c x^3}{c x^3}\right )^{-p} \left (a+b x^3+c x^6\right )^p \operatorname {AppellF1}\left (1-2 p,-p,-p,2 (1-p),-\frac {b-\sqrt {b^2-4 a c}}{2 c x^3},-\frac {b+\sqrt {b^2-4 a c}}{2 c x^3}\right )}{3 (1-2 p) x^3} \]
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Rule 138
Rule 772
Rule 1371
Rubi steps \begin{align*} \text {integral}& = \frac {1}{3} \text {Subst}\left (\int \frac {\left (a+b x+c x^2\right )^p}{x^2} \, dx,x,x^3\right ) \\ & = -\left (\frac {1}{3} \left (2^{2 p} \left (\frac {1}{x^3}\right )^{2 p} \left (\frac {b-\sqrt {b^2-4 a c}+2 c x^3}{c x^3}\right )^{-p} \left (\frac {b+\sqrt {b^2-4 a c}+2 c x^3}{c x^3}\right )^{-p} \left (a+b x^3+c x^6\right )^p\right ) \text {Subst}\left (\int x^{2-2 (1+p)} \left (1+\frac {\left (b-\sqrt {b^2-4 a c}\right ) x}{2 c}\right )^p \left (1+\frac {\left (b+\sqrt {b^2-4 a c}\right ) x}{2 c}\right )^p \, dx,x,\frac {1}{x^3}\right )\right ) \\ & = -\frac {4^p \left (\frac {b-\sqrt {b^2-4 a c}+2 c x^3}{c x^3}\right )^{-p} \left (\frac {b+\sqrt {b^2-4 a c}+2 c x^3}{c x^3}\right )^{-p} \left (a+b x^3+c x^6\right )^p F_1\left (1-2 p;-p,-p;2 (1-p);-\frac {b-\sqrt {b^2-4 a c}}{2 c x^3},-\frac {b+\sqrt {b^2-4 a c}}{2 c x^3}\right )}{3 (1-2 p) x^3} \\ \end{align*}
Time = 0.28 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.99 \[ \int \frac {\left (a+b x^3+c x^6\right )^p}{x^4} \, dx=\frac {4^p \left (\frac {b-\sqrt {b^2-4 a c}+2 c x^3}{c x^3}\right )^{-p} \left (\frac {b+\sqrt {b^2-4 a c}+2 c x^3}{c x^3}\right )^{-p} \left (a+b x^3+c x^6\right )^p \operatorname {AppellF1}\left (1-2 p,-p,-p,2-2 p,-\frac {b+\sqrt {b^2-4 a c}}{2 c x^3},\frac {-b+\sqrt {b^2-4 a c}}{2 c x^3}\right )}{3 (-1+2 p) x^3} \]
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\[\int \frac {\left (c \,x^{6}+b \,x^{3}+a \right )^{p}}{x^{4}}d x\]
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\[ \int \frac {\left (a+b x^3+c x^6\right )^p}{x^4} \, dx=\int { \frac {{\left (c x^{6} + b x^{3} + a\right )}^{p}}{x^{4}} \,d x } \]
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Timed out. \[ \int \frac {\left (a+b x^3+c x^6\right )^p}{x^4} \, dx=\text {Timed out} \]
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\[ \int \frac {\left (a+b x^3+c x^6\right )^p}{x^4} \, dx=\int { \frac {{\left (c x^{6} + b x^{3} + a\right )}^{p}}{x^{4}} \,d x } \]
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\[ \int \frac {\left (a+b x^3+c x^6\right )^p}{x^4} \, dx=\int { \frac {{\left (c x^{6} + b x^{3} + a\right )}^{p}}{x^{4}} \,d x } \]
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Timed out. \[ \int \frac {\left (a+b x^3+c x^6\right )^p}{x^4} \, dx=\int \frac {{\left (c\,x^6+b\,x^3+a\right )}^p}{x^4} \,d x \]
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