Optimal. Leaf size=320 \[ -\frac{3 \sqrt [3]{a+b x+c x^2} (d (b+2 c x))^{4/3}}{16 c^2 d^5 \left (b^2-4 a c\right )}-\frac{9 \left (a+b x+c x^2\right )^{7/3}}{4 d^3 \left (b^2-4 a c\right )^2 (b d+2 c d x)^{2/3}}+\frac{9 \left (a+b x+c x^2\right )^{4/3} (d (b+2 c x))^{4/3}}{16 c d^5 \left (b^2-4 a c\right )^2}+\frac{3 \left (a+b x+c x^2\right )^{7/3}}{4 d \left (b^2-4 a c\right ) (b d+2 c d x)^{8/3}}-\frac{3 \log \left ((d (b+2 c x))^{2/3}-2^{2/3} \sqrt [3]{c} d^{2/3} \sqrt [3]{a+b x+c x^2}\right )}{32\ 2^{2/3} c^{7/3} d^{11/3}}-\frac{\sqrt{3} \tan ^{-1}\left (\frac{\frac{\sqrt [3]{2} (d (b+2 c x))^{2/3}}{\sqrt [3]{c} d^{2/3} \sqrt [3]{a+b x+c x^2}}+1}{\sqrt{3}}\right )}{16\ 2^{2/3} c^{7/3} d^{11/3}} \]
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Rubi [A] time = 1.28214, antiderivative size = 468, normalized size of antiderivative = 1.46, number of steps used = 14, number of rules used = 12, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {693, 694, 279, 329, 275, 331, 292, 31, 634, 617, 204, 628} \[ -\frac{3 \sqrt [3]{a+b x+c x^2} (d (b+2 c x))^{4/3}}{16 c^2 d^5 \left (b^2-4 a c\right )}-\frac{9 \left (a+b x+c x^2\right )^{7/3}}{4 d^3 \left (b^2-4 a c\right )^2 (b d+2 c d x)^{2/3}}+\frac{9 \left (a+b x+c x^2\right )^{4/3} (d (b+2 c x))^{4/3}}{16 c d^5 \left (b^2-4 a c\right )^2}+\frac{3 \left (a+b x+c x^2\right )^{7/3}}{4 d \left (b^2-4 a c\right ) (b d+2 c d x)^{8/3}}-\frac{\log \left (-\frac{\sqrt [3]{2} (d (b+2 c x))^{2/3}-2 \sqrt [3]{c} d^{2/3} \sqrt [3]{a+b x+c x^2}}{\sqrt [3]{a+b x+c x^2}}\right )}{16\ 2^{2/3} c^{7/3} d^{11/3}}+\frac{\log \left (\frac{2 \sqrt [3]{2} c^{2/3} d^{4/3} \left (a+b x+c x^2\right )^{2/3}+2^{2/3} \sqrt [3]{c} d^{2/3} \sqrt [3]{a+b x+c x^2} (d (b+2 c x))^{2/3}+(d (b+2 c x))^{4/3}}{\left (a+b x+c x^2\right )^{2/3}}\right )}{32\ 2^{2/3} c^{7/3} d^{11/3}}-\frac{\sqrt{3} \tan ^{-1}\left (\frac{\frac{\sqrt [3]{2} (d (b+2 c x))^{2/3}}{\sqrt [3]{a+b x+c x^2}}+\sqrt [3]{c} d^{2/3}}{\sqrt{3} \sqrt [3]{c} d^{2/3}}\right )}{16\ 2^{2/3} c^{7/3} d^{11/3}} \]
Antiderivative was successfully verified.
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Rule 693
Rule 694
Rule 279
Rule 329
Rule 275
Rule 331
Rule 292
Rule 31
Rule 634
Rule 617
Rule 204
Rule 628
Rubi steps
\begin{align*} \int \frac{\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{11/3}} \, dx &=\frac{3 \left (a+b x+c x^2\right )^{7/3}}{4 \left (b^2-4 a c\right ) d (b d+2 c d x)^{8/3}}-\frac{3 \int \frac{\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{5/3}} \, dx}{4 \left (b^2-4 a c\right ) d^2}\\ &=\frac{3 \left (a+b x+c x^2\right )^{7/3}}{4 \left (b^2-4 a c\right ) d (b d+2 c d x)^{8/3}}-\frac{9 \left (a+b x+c x^2\right )^{7/3}}{4 \left (b^2-4 a c\right )^2 d^3 (b d+2 c d x)^{2/3}}+\frac{9 \int \sqrt [3]{b d+2 c d x} \left (a+b x+c x^2\right )^{4/3} \, dx}{2 \left (b^2-4 a c\right )^2 d^4}\\ &=\frac{3 \left (a+b x+c x^2\right )^{7/3}}{4 \left (b^2-4 a c\right ) d (b d+2 c d x)^{8/3}}-\frac{9 \left (a+b x+c x^2\right )^{7/3}}{4 \left (b^2-4 a c\right )^2 d^3 (b d+2 c d x)^{2/3}}+\frac{9 \operatorname{Subst}\left (\int \sqrt [3]{x} \left (a-\frac{b^2}{4 c}+\frac{x^2}{4 c d^2}\right )^{4/3} \, dx,x,b d+2 c d x\right )}{4 c \left (b^2-4 a c\right )^2 d^5}\\ &=\frac{9 (d (b+2 c x))^{4/3} \left (a+b x+c x^2\right )^{4/3}}{16 c \left (b^2-4 a c\right )^2 d^5}+\frac{3 \left (a+b x+c x^2\right )^{7/3}}{4 \left (b^2-4 a c\right ) d (b d+2 c d x)^{8/3}}-\frac{9 \left (a+b x+c x^2\right )^{7/3}}{4 \left (b^2-4 a c\right )^2 d^3 (b d+2 c d x)^{2/3}}-\frac{3 \operatorname{Subst}\left (\int \sqrt [3]{x} \sqrt [3]{a-\frac{b^2}{4 c}+\frac{x^2}{4 c d^2}} \, dx,x,b d+2 c d x\right )}{8 c^2 \left (b^2-4 a c\right ) d^5}\\ &=-\frac{3 (d (b+2 c x))^{4/3} \sqrt [3]{a+b x+c x^2}}{16 c^2 \left (b^2-4 a c\right ) d^5}+\frac{9 (d (b+2 c x))^{4/3} \left (a+b x+c x^2\right )^{4/3}}{16 c \left (b^2-4 a c\right )^2 d^5}+\frac{3 \left (a+b x+c x^2\right )^{7/3}}{4 \left (b^2-4 a c\right ) d (b d+2 c d x)^{8/3}}-\frac{9 \left (a+b x+c x^2\right )^{7/3}}{4 \left (b^2-4 a c\right )^2 d^3 (b d+2 c d x)^{2/3}}+\frac{\operatorname{Subst}\left (\int \frac{\sqrt [3]{x}}{\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 )}{32 c^3 d^5}\\ &=-\frac{3 (d (b+2 c x))^{4/3} \sqrt [3]{a+b x+c x^2}}{16 c^2 \left (b^2-4 a c\right ) d^5}+\frac{9 (d (b+2 c x))^{4/3} \left (a+b x+c x^2\right )^{4/3}}{16 c \left (b^2-4 a c\right )^2 d^5}+\frac{3 \left (a+b x+c x^2\right )^{7/3}}{4 \left (b^2-4 a c\right ) d (b d+2 c d x)^{8/3}}-\frac{9 \left (a+b x+c x^2\right )^{7/3}}{4 \left (b^2-4 a c\right )^2 d^3 (b d+2 c d x)^{2/3}}+\frac{3 \operatorname{Subst}\left (\int \frac{x^3}{\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 )}{32 c^3 d^5}\\ &=-\frac{3 (d (b+2 c x))^{4/3} \sqrt [3]{a+b x+c x^2}}{16 c^2 \left (b^2-4 a c\right ) d^5}+\frac{9 (d (b+2 c x))^{4/3} \left (a+b x+c x^2\right )^{4/3}}{16 c \left (b^2-4 a c\right )^2 d^5}+\frac{3 \left (a+b x+c x^2\right )^{7/3}}{4 \left (b^2-4 a c\right ) d (b d+2 c d x)^{8/3}}-\frac{9 \left (a+b x+c x^2\right )^{7/3}}{4 \left (b^2-4 a c\right )^2 d^3 (b d+2 c d x)^{2/3}}+\frac{3 \operatorname{Subst}\left (\int \frac{x}{\left (a-\frac{b^2}{4 c}+\frac{x^3}{4 c d^2}\right )^{2/3}} \, dx,x,(d (b+2 c x))^{2/3}\right )}{64 c^3 d^5}\\ &=-\frac{3 (d (b+2 c x))^{4/3} \sqrt [3]{a+b x+c x^2}}{16 c^2 \left (b^2-4 a c\right ) d^5}+\frac{9 (d (b+2 c x))^{4/3} \left (a+b x+c x^2\right )^{4/3}}{16 c \left (b^2-4 a c\right )^2 d^5}+\frac{3 \left (a+b x+c x^2\right )^{7/3}}{4 \left (b^2-4 a c\right ) d (b d+2 c d x)^{8/3}}-\frac{9 \left (a+b x+c x^2\right )^{7/3}}{4 \left (b^2-4 a c\right )^2 d^3 (b d+2 c d x)^{2/3}}+\frac{3 \operatorname{Subst}\left (\int \frac{x}{1-\frac{x^3}{4 c d^2}} \, dx,x,\frac{(d (b+2 c x))^{2/3}}{\sqrt [3]{a+x (b+c x)}}\right )}{64 c^3 d^5}\\ &=-\frac{3 (d (b+2 c x))^{4/3} \sqrt [3]{a+b x+c x^2}}{16 c^2 \left (b^2-4 a c\right ) d^5}+\frac{9 (d (b+2 c x))^{4/3} \left (a+b x+c x^2\right )^{4/3}}{16 c \left (b^2-4 a c\right )^2 d^5}+\frac{3 \left (a+b x+c x^2\right )^{7/3}}{4 \left (b^2-4 a c\right ) d (b d+2 c d x)^{8/3}}-\frac{9 \left (a+b x+c x^2\right )^{7/3}}{4 \left (b^2-4 a c\right )^2 d^3 (b d+2 c d x)^{2/3}}+\frac{\operatorname{Subst}\left (\int \frac{1}{1-\frac{x}{2^{2/3} \sqrt [3]{c} d^{2/3}}} \, dx,x,\frac{(d (b+2 c x))^{2/3}}{\sqrt [3]{a+x (b+c x)}}\right )}{32 \sqrt [3]{2} c^{8/3} d^{13/3}}-\frac{\operatorname{Subst}\left (\int \frac{1-\frac{x}{2^{2/3} \sqrt [3]{c} d^{2/3}}}{1+\frac{x}{2^{2/3} \sqrt [3]{c} d^{2/3}}+\frac{x^2}{2 \sqrt [3]{2} c^{2/3} d^{4/3}}} \, dx,x,\frac{(d (b+2 c x))^{2/3}}{\sqrt [3]{a+x (b+c x)}}\right )}{32 \sqrt [3]{2} c^{8/3} d^{13/3}}\\ &=-\frac{3 (d (b+2 c x))^{4/3} \sqrt [3]{a+b x+c x^2}}{16 c^2 \left (b^2-4 a c\right ) d^5}+\frac{9 (d (b+2 c x))^{4/3} \left (a+b x+c x^2\right )^{4/3}}{16 c \left (b^2-4 a c\right )^2 d^5}+\frac{3 \left (a+b x+c x^2\right )^{7/3}}{4 \left (b^2-4 a c\right ) d (b d+2 c d x)^{8/3}}-\frac{9 \left (a+b x+c x^2\right )^{7/3}}{4 \left (b^2-4 a c\right )^2 d^3 (b d+2 c d x)^{2/3}}-\frac{\log \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 )}{16\ 2^{2/3} c^{7/3} d^{11/3}}-\frac{3 \operatorname{Subst}\left (\int \frac{1}{1+\frac{x}{2^{2/3} \sqrt [3]{c} d^{2/3}}+\frac{x^2}{2 \sqrt [3]{2} c^{2/3} d^{4/3}}} \, dx,x,\frac{(d (b+2 c x))^{2/3}}{\sqrt [3]{a+x (b+c x)}}\right )}{64 \sqrt [3]{2} c^{8/3} d^{13/3}}+\frac{\operatorname{Subst}\left (\int \frac{\frac{1}{2^{2/3} \sqrt [3]{c} d^{2/3}}+\frac{x}{\sqrt [3]{2} c^{2/3} d^{4/3}}}{1+\frac{x}{2^{2/3} \sqrt [3]{c} d^{2/3}}+\frac{x^2}{2 \sqrt [3]{2} c^{2/3} d^{4/3}}} \, dx,x,\frac{(d (b+2 c x))^{2/3}}{\sqrt [3]{a+x (b+c x)}}\right )}{32\ 2^{2/3} c^{7/3} d^{11/3}}\\ &=-\frac{3 (d (b+2 c x))^{4/3} \sqrt [3]{a+b x+c x^2}}{16 c^2 \left (b^2-4 a c\right ) d^5}+\frac{9 (d (b+2 c x))^{4/3} \left (a+b x+c x^2\right )^{4/3}}{16 c \left (b^2-4 a c\right )^2 d^5}+\frac{3 \left (a+b x+c x^2\right )^{7/3}}{4 \left (b^2-4 a c\right ) d (b d+2 c d x)^{8/3}}-\frac{9 \left (a+b x+c x^2\right )^{7/3}}{4 \left (b^2-4 a c\right )^2 d^3 (b d+2 c d x)^{2/3}}-\frac{\log \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 )}{16\ 2^{2/3} c^{7/3} d^{11/3}}+\frac{\log \left (4 c^{2/3} d^{4/3}+\frac{2^{2/3} (d (b+2 c x))^{4/3}}{(a+x (b+c x))^{2/3}}+\frac{2 \sqrt [3]{2} \sqrt [3]{c} d^{2/3} (d (b+2 c x))^{2/3}}{\sqrt [3]{a+x (b+c x)}}\right )}{32\ 2^{2/3} c^{7/3} d^{11/3}}+\frac{3 \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1+\frac{\sqrt [3]{2} (d (b+2 c x))^{2/3}}{\sqrt [3]{c} d^{2/3} \sqrt [3]{a+x (b+c x)}}\right )}{16\ 2^{2/3} c^{7/3} d^{11/3}}\\ &=-\frac{3 (d (b+2 c x))^{4/3} \sqrt [3]{a+b x+c x^2}}{16 c^2 \left (b^2-4 a c\right ) d^5}+\frac{9 (d (b+2 c x))^{4/3} \left (a+b x+c x^2\right )^{4/3}}{16 c \left (b^2-4 a c\right )^2 d^5}+\frac{3 \left (a+b x+c x^2\right )^{7/3}}{4 \left (b^2-4 a c\right ) d (b d+2 c d x)^{8/3}}-\frac{9 \left (a+b x+c x^2\right )^{7/3}}{4 \left (b^2-4 a c\right )^2 d^3 (b d+2 c d x)^{2/3}}-\frac{\sqrt{3} \tan ^{-1}\left (\frac{1+\frac{\sqrt [3]{2} (d (b+2 c x))^{2/3}}{\sqrt [3]{c} d^{2/3} \sqrt [3]{a+x (b+c x)}}}{\sqrt{3}}\right )}{16\ 2^{2/3} c^{7/3} d^{11/3}}-\frac{\log \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 )}{16\ 2^{2/3} c^{7/3} d^{11/3}}+\frac{\log \left (4 c^{2/3} d^{4/3}+\frac{2^{2/3} (d (b+2 c x))^{4/3}}{(a+x (b+c x))^{2/3}}+\frac{2 \sqrt [3]{2} \sqrt [3]{c} d^{2/3} (d (b+2 c x))^{2/3}}{\sqrt [3]{a+x (b+c x)}}\right )}{32\ 2^{2/3} c^{7/3} d^{11/3}}\\ \end{align*}
Mathematica [C] time = 0.0947293, size = 104, normalized size = 0.32 \[ \frac{3 \left (b^2-4 a c\right ) \sqrt [3]{a+x (b+c x)} \, _2F_1\left (-\frac{4}{3},-\frac{4}{3};-\frac{1}{3};\frac{(b+2 c x)^2}{b^2-4 a c}\right )}{64\ 2^{2/3} c^2 d \sqrt [3]{\frac{c (a+x (b+c x))}{4 a c-b^2}} (d (b+2 c x))^{8/3}} \]
Antiderivative was successfully verified.
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Maple [F] time = 1.25, size = 0, normalized size = 0. \begin{align*} \int{ \left ( c{x}^{2}+bx+a \right ) ^{{\frac{4}{3}}} \left ( 2\,cdx+bd \right ) ^{-{\frac{11}{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{11}{3}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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{11}{3}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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