Optimal. Leaf size=581 \[ -\frac{\left (-4 a c+b^2-(b+2 c x)^2\right ) \sqrt [3]{d (b+2 c x)} \left (2 \sqrt [3]{c} d^{2/3}-\frac{\sqrt [3]{2} (d (b+2 c x))^{2/3}}{\sqrt [3]{a+b x+c x^2}}\right ) \sqrt{\frac{\frac{2^{2/3} \sqrt [3]{c} d^{2/3} (d (b+2 c x))^{2/3}}{\sqrt [3]{a+b x+c x^2}}+\frac{(d (b+2 c x))^{4/3}}{\left (a+b x+c x^2\right )^{2/3}}+2 \sqrt [3]{2} c^{2/3} d^{4/3}}{\left (2^{2/3} \sqrt [3]{c} d^{2/3}-\frac{\left (1+\sqrt{3}\right ) (d (b+2 c x))^{2/3}}{\sqrt [3]{a+b x+c x^2}}\right )^2}} \text{EllipticF}\left (\cos ^{-1}\left (\frac{2^{2/3} \sqrt [3]{c} d^{2/3}-\frac{\left (1-\sqrt{3}\right ) (d (b+2 c x))^{2/3}}{\sqrt [3]{a+b x+c x^2}}}{2^{2/3} \sqrt [3]{c} d^{2/3}-\frac{\left (1+\sqrt{3}\right ) (d (b+2 c x))^{2/3}}{\sqrt [3]{a+b x+c x^2}}}\right ),\frac{1}{4} \left (2+\sqrt{3}\right )\right )}{40 \sqrt [4]{3} c^{10/3} d^{11/3} \left (a+b x+c x^2\right )^{2/3} \sqrt{-\frac{(d (b+2 c x))^{2/3} \left (2^{2/3} \sqrt [3]{c} d^{2/3}-\frac{(d (b+2 c x))^{2/3}}{\sqrt [3]{a+b x+c x^2}}\right )}{\sqrt [3]{a+b x+c x^2} \left (2^{2/3} \sqrt [3]{c} d^{2/3}-\frac{\left (1+\sqrt{3}\right ) (d (b+2 c x))^{2/3}}{\sqrt [3]{a+b x+c x^2}}\right )^2}}}+\frac{\sqrt [3]{a+b x+c x^2} \sqrt [3]{d (b+2 c x)}}{5 c^2 d^3}-\frac{3 \left (a+b x+c x^2\right )^{4/3}}{10 c d (d (b+2 c x))^{5/3}} \]
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Rubi [A] time = 1.94999, antiderivative size = 581, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {694, 277, 279, 329, 241, 225} \[ -\frac{\left (-4 a c+b^2-(b+2 c x)^2\right ) \sqrt [3]{d (b+2 c x)} \left (2 \sqrt [3]{c} d^{2/3}-\frac{\sqrt [3]{2} (d (b+2 c x))^{2/3}}{\sqrt [3]{a+b x+c x^2}}\right ) \sqrt{\frac{\frac{2^{2/3} \sqrt [3]{c} d^{2/3} (d (b+2 c x))^{2/3}}{\sqrt [3]{a+b x+c x^2}}+\frac{(d (b+2 c x))^{4/3}}{\left (a+b x+c x^2\right )^{2/3}}+2 \sqrt [3]{2} c^{2/3} d^{4/3}}{\left (2^{2/3} \sqrt [3]{c} d^{2/3}-\frac{\left (1+\sqrt{3}\right ) (d (b+2 c x))^{2/3}}{\sqrt [3]{a+b x+c x^2}}\right )^2}} F\left (\cos ^{-1}\left (\frac{2^{2/3} \sqrt [3]{c} d^{2/3}-\frac{\left (1-\sqrt{3}\right ) (d (b+2 c x))^{2/3}}{\sqrt [3]{c x^2+b x+a}}}{2^{2/3} \sqrt [3]{c} d^{2/3}-\frac{\left (1+\sqrt{3}\right ) (d (b+2 c x))^{2/3}}{\sqrt [3]{c x^2+b x+a}}}\right )|\frac{1}{4} \left (2+\sqrt{3}\right )\right )}{40 \sqrt [4]{3} c^{10/3} d^{11/3} \left (a+b x+c x^2\right )^{2/3} \sqrt{-\frac{(d (b+2 c x))^{2/3} \left (2^{2/3} \sqrt [3]{c} d^{2/3}-\frac{(d (b+2 c x))^{2/3}}{\sqrt [3]{a+b x+c x^2}}\right )}{\sqrt [3]{a+b x+c x^2} \left (2^{2/3} \sqrt [3]{c} d^{2/3}-\frac{\left (1+\sqrt{3}\right ) (d (b+2 c x))^{2/3}}{\sqrt [3]{a+b x+c x^2}}\right )^2}}}+\frac{\sqrt [3]{a+b x+c x^2} \sqrt [3]{d (b+2 c x)}}{5 c^2 d^3}-\frac{3 \left (a+b x+c x^2\right )^{4/3}}{10 c d (d (b+2 c x))^{5/3}} \]
Antiderivative was successfully verified.
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Rule 694
Rule 277
Rule 279
Rule 329
Rule 241
Rule 225
Rubi steps
\begin{align*} \int \frac{\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{8/3}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (a-\frac{b^2}{4 c}+\frac{x^2}{4 c d^2}\right )^{4/3}}{x^{8/3}} \, dx,x,b d+2 c d x\right )}{2 c d}\\ &=-\frac{3 \left (a+b x+c x^2\right )^{4/3}}{10 c d (d (b+2 c x))^{5/3}}+\frac{\operatorname{Subst}\left (\int \frac{\sqrt [3]{a-\frac{b^2}{4 c}+\frac{x^2}{4 c d^2}}}{x^{2/3}} \, dx,x,b d+2 c d x\right )}{5 c^2 d^3}\\ &=\frac{\sqrt [3]{d (b+2 c x)} \sqrt [3]{a+b x+c x^2}}{5 c^2 d^3}-\frac{3 \left (a+b x+c x^2\right )^{4/3}}{10 c d (d (b+2 c x))^{5/3}}-\frac{\left (b^2-4 a c\right ) \operatorname{Subst}\left (\int \frac{1}{x^{2/3} \left (a-\frac{b^2}{4 c}+\frac{x^2}{4 c d^2}\right )^{2/3}} \, dx,x,b d+2 c d x\right )}{30 c^3 d^3}\\ &=\frac{\sqrt [3]{d (b+2 c x)} \sqrt [3]{a+b x+c x^2}}{5 c^2 d^3}-\frac{3 \left (a+b x+c x^2\right )^{4/3}}{10 c d (d (b+2 c x))^{5/3}}-\frac{\left (b^2-4 a c\right ) \operatorname{Subst}\left (\int \frac{1}{\left (a-\frac{b^2}{4 c}+\frac{x^6}{4 c d^2}\right )^{2/3}} \, dx,x,\sqrt [3]{d (b+2 c x)}\right )}{10 c^3 d^3}\\ &=\frac{\sqrt [3]{d (b+2 c x)} \sqrt [3]{a+b x+c x^2}}{5 c^2 d^3}-\frac{3 \left (a+b x+c x^2\right )^{4/3}}{10 c d (d (b+2 c x))^{5/3}}-\frac{\left (b^2-4 a c\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-\frac{x^6}{4 c d^2}}} \, dx,x,\frac{\sqrt [3]{d (b+2 c x)}}{\sqrt [6]{a+x (b+c x)}}\right )}{10 c^3 d^3 \sqrt{\frac{a-\frac{b^2}{4 c}}{a+x (b+c x)}} \sqrt{a+x (b+c x)}}\\ &=\frac{\sqrt [3]{d (b+2 c x)} \sqrt [3]{a+b x+c x^2}}{5 c^2 d^3}-\frac{3 \left (a+b x+c x^2\right )^{4/3}}{10 c d (d (b+2 c x))^{5/3}}-\frac{\left (b^2-4 a c\right ) \sqrt [3]{d (b+2 c x)} \left (2^{2/3} \sqrt [3]{c} d^{2/3}-\frac{(d (b+2 c x))^{2/3}}{\sqrt [3]{a+x (b+c x)}}\right ) \sqrt{\frac{2 \sqrt [3]{2} c^{2/3} d^{4/3}+\frac{(d (b+2 c x))^{4/3}}{(a+x (b+c x))^{2/3}}+\frac{2^{2/3} \sqrt [3]{c} d^{2/3} (d (b+2 c x))^{2/3}}{\sqrt [3]{a+x (b+c x)}}}{\left (2^{2/3} \sqrt [3]{c} d^{2/3}-\frac{\left (1+\sqrt{3}\right ) (d (b+2 c x))^{2/3}}{\sqrt [3]{a+x (b+c x)}}\right )^2}} F\left (\cos ^{-1}\left (\frac{2^{2/3} \sqrt [3]{c} d^{2/3}-\frac{\left (1-\sqrt{3}\right ) (d (b+2 c x))^{2/3}}{\sqrt [3]{a+x (b+c x)}}}{2^{2/3} \sqrt [3]{c} d^{2/3}-\frac{\left (1+\sqrt{3}\right ) (d (b+2 c x))^{2/3}}{\sqrt [3]{a+x (b+c x)}}}\right )|\frac{1}{4} \left (2+\sqrt{3}\right )\right )}{5\ 2^{2/3} \sqrt [4]{3} c^{10/3} d^{11/3} \sqrt{\frac{4 a-\frac{b^2}{c}}{a+x (b+c x)}} (a+x (b+c x))^{2/3} \sqrt{4-\frac{(b+2 c x)^2}{c (a+x (b+c x))}} \sqrt{-\frac{(d (b+2 c x))^{2/3} \left (2^{2/3} \sqrt [3]{c} d^{2/3}-\frac{(d (b+2 c x))^{2/3}}{\sqrt [3]{a+x (b+c x)}}\right )}{\sqrt [3]{a+x (b+c x)} \left (2^{2/3} \sqrt [3]{c} d^{2/3}-\frac{\left (1+\sqrt{3}\right ) (d (b+2 c x))^{2/3}}{\sqrt [3]{a+x (b+c x)}}\right )^2}}}\\ \end{align*}
Mathematica [C] time = 0.0645814, size = 104, normalized size = 0.18 \[ \frac{3 \left (b^2-4 a c\right ) \sqrt [3]{a+x (b+c x)} \, _2F_1\left (-\frac{4}{3},-\frac{5}{6};\frac{1}{6};\frac{(b+2 c x)^2}{b^2-4 a c}\right )}{40\ 2^{2/3} c^2 d \sqrt [3]{\frac{c (a+x (b+c x))}{4 a c-b^2}} (d (b+2 c x))^{5/3}} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.776, size = 0, normalized size = 0. \begin{align*} \int{ \left ( c{x}^{2}+bx+a \right ) ^{{\frac{4}{3}}} \left ( 2\,cdx+bd \right ) ^{-{\frac{8}{3}}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (c x^{2} + b x + a\right )}^{\frac{4}{3}}}{{\left (2 \, c d x + b d\right )}^{\frac{8}{3}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (2 \, c d x + b d\right )}^{\frac{1}{3}}{\left (c x^{2} + b x + a\right )}^{\frac{4}{3}}}{8 \, c^{3} d^{3} x^{3} + 12 \, b c^{2} d^{3} x^{2} + 6 \, b^{2} c d^{3} x + b^{3} d^{3}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b x + c x^{2}\right )^{\frac{4}{3}}}{\left (d \left (b + 2 c x\right )\right )^{\frac{8}{3}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (c x^{2} + b x + a\right )}^{\frac{4}{3}}}{{\left (2 \, c d x + b d\right )}^{\frac{8}{3}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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