Integrand size = 41, antiderivative size = 56 \[ \int \frac {\left (-2 q+p x^3\right ) \sqrt {q+p x^3}}{x^2 \left (a q+b x^2+a p x^3\right )} \, dx=\frac {2 \sqrt {q+p x^3}}{a x}+\frac {2 \sqrt {b} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a} \sqrt {q+p x^3}}\right )}{a^{3/2}} \]
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\[ \int \frac {\left (-2 q+p x^3\right ) \sqrt {q+p x^3}}{x^2 \left (a q+b x^2+a p x^3\right )} \, dx=\int \frac {\left (-2 q+p x^3\right ) \sqrt {q+p x^3}}{x^2 \left (a q+b x^2+a p x^3\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {2 \sqrt {q+p x^3}}{a x^2}+\frac {(2 b+3 a p x) \sqrt {q+p x^3}}{a \left (a q+b x^2+a p x^3\right )}\right ) \, dx \\ & = \frac {\int \frac {(2 b+3 a p x) \sqrt {q+p x^3}}{a q+b x^2+a p x^3} \, dx}{a}-\frac {2 \int \frac {\sqrt {q+p x^3}}{x^2} \, dx}{a} \\ & = \frac {2 \sqrt {q+p x^3}}{a x}+\frac {\int \left (\frac {2 b \sqrt {q+p x^3}}{a q+b x^2+a p x^3}+\frac {3 a p x \sqrt {q+p x^3}}{a q+b x^2+a p x^3}\right ) \, dx}{a}-\frac {(3 p) \int \frac {x}{\sqrt {q+p x^3}} \, dx}{a} \\ & = \frac {2 \sqrt {q+p x^3}}{a x}+\frac {(2 b) \int \frac {\sqrt {q+p x^3}}{a q+b x^2+a p x^3} \, dx}{a}-\frac {\left (3 p^{2/3}\right ) \int \frac {\left (1-\sqrt {3}\right ) \sqrt [3]{q}+\sqrt [3]{p} x}{\sqrt {q+p x^3}} \, dx}{a}+(3 p) \int \frac {x \sqrt {q+p x^3}}{a q+b x^2+a p x^3} \, dx+\frac {\left (3 \left (1-\sqrt {3}\right ) p^{2/3} \sqrt [3]{q}\right ) \int \frac {1}{\sqrt {q+p x^3}} \, dx}{a} \\ & = \frac {2 \sqrt {q+p x^3}}{a x}-\frac {6 \sqrt [3]{p} \sqrt {q+p x^3}}{a \left (\left (1+\sqrt {3}\right ) \sqrt [3]{q}+\sqrt [3]{p} x\right )}+\frac {3 \sqrt [4]{3} \sqrt {2-\sqrt {3}} \sqrt [3]{p} \sqrt [3]{q} \left (\sqrt [3]{q}+\sqrt [3]{p} x\right ) \sqrt {\frac {q^{2/3}-\sqrt [3]{p} \sqrt [3]{q} x+p^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{q}+\sqrt [3]{p} x\right )^2}} E\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{q}+\sqrt [3]{p} x}{\left (1+\sqrt {3}\right ) \sqrt [3]{q}+\sqrt [3]{p} x}\right )|-7-4 \sqrt {3}\right )}{a \sqrt {\frac {\sqrt [3]{q} \left (\sqrt [3]{q}+\sqrt [3]{p} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{q}+\sqrt [3]{p} x\right )^2}} \sqrt {q+p x^3}}-\frac {2 \sqrt {2} 3^{3/4} \sqrt [3]{p} \sqrt [3]{q} \left (\sqrt [3]{q}+\sqrt [3]{p} x\right ) \sqrt {\frac {q^{2/3}-\sqrt [3]{p} \sqrt [3]{q} x+p^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{q}+\sqrt [3]{p} x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{q}+\sqrt [3]{p} x}{\left (1+\sqrt {3}\right ) \sqrt [3]{q}+\sqrt [3]{p} x}\right ),-7-4 \sqrt {3}\right )}{a \sqrt {\frac {\sqrt [3]{q} \left (\sqrt [3]{q}+\sqrt [3]{p} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{q}+\sqrt [3]{p} x\right )^2}} \sqrt {q+p x^3}}+\frac {(2 b) \int \frac {\sqrt {q+p x^3}}{a q+b x^2+a p x^3} \, dx}{a}+(3 p) \int \frac {x \sqrt {q+p x^3}}{a q+b x^2+a p x^3} \, dx \\ \end{align*}
Time = 0.79 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.00 \[ \int \frac {\left (-2 q+p x^3\right ) \sqrt {q+p x^3}}{x^2 \left (a q+b x^2+a p x^3\right )} \, dx=\frac {2 \sqrt {q+p x^3}}{a x}+\frac {2 \sqrt {b} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a} \sqrt {q+p x^3}}\right )}{a^{3/2}} \]
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Time = 6.64 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.89
method | result | size |
risch | \(\frac {2 \sqrt {p \,x^{3}+q}}{a x}-\frac {2 b \arctan \left (\frac {a \sqrt {p \,x^{3}+q}}{x \sqrt {a b}}\right )}{a \sqrt {a b}}\) | \(50\) |
default | \(-\frac {2 \left (b \arctan \left (\frac {a \sqrt {p \,x^{3}+q}}{x \sqrt {a b}}\right ) x -\sqrt {p \,x^{3}+q}\, \sqrt {a b}\right )}{a x \sqrt {a b}}\) | \(54\) |
pseudoelliptic | \(-\frac {2 \left (b \arctan \left (\frac {a \sqrt {p \,x^{3}+q}}{x \sqrt {a b}}\right ) x -\sqrt {p \,x^{3}+q}\, \sqrt {a b}\right )}{a x \sqrt {a b}}\) | \(54\) |
elliptic | \(\text {Expression too large to display}\) | \(877\) |
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Timed out. \[ \int \frac {\left (-2 q+p x^3\right ) \sqrt {q+p x^3}}{x^2 \left (a q+b x^2+a p x^3\right )} \, dx=\text {Timed out} \]
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Timed out. \[ \int \frac {\left (-2 q+p x^3\right ) \sqrt {q+p x^3}}{x^2 \left (a q+b x^2+a p x^3\right )} \, dx=\text {Timed out} \]
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\[ \int \frac {\left (-2 q+p x^3\right ) \sqrt {q+p x^3}}{x^2 \left (a q+b x^2+a p x^3\right )} \, dx=\int { \frac {\sqrt {p x^{3} + q} {\left (p x^{3} - 2 \, q\right )}}{{\left (a p x^{3} + b x^{2} + a q\right )} x^{2}} \,d x } \]
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\[ \int \frac {\left (-2 q+p x^3\right ) \sqrt {q+p x^3}}{x^2 \left (a q+b x^2+a p x^3\right )} \, dx=\int { \frac {\sqrt {p x^{3} + q} {\left (p x^{3} - 2 \, q\right )}}{{\left (a p x^{3} + b x^{2} + a q\right )} x^{2}} \,d x } \]
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Time = 12.40 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.82 \[ \int \frac {\left (-2 q+p x^3\right ) \sqrt {q+p x^3}}{x^2 \left (a q+b x^2+a p x^3\right )} \, dx=\frac {2\,\sqrt {p\,x^3+q}}{a\,x}+\frac {\sqrt {b}\,\ln \left (\frac {a^5\,b\,p^4\,x^2-a^6\,p^4\,\left (p\,x^3+q\right )+a^{11/2}\,\sqrt {b}\,p^4\,x\,\sqrt {p\,x^3+q}\,2{}\mathrm {i}}{4\,b^2\,q\,x^2+4\,a\,b\,q\,\left (p\,x^3+q\right )}\right )\,1{}\mathrm {i}}{a^{3/2}} \]
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