Optimal. Leaf size=320 \[ -\frac {3 \sqrt [3]{a+b x+c x^2} (d (b+2 c x))^{4/3}}{16 c^2 d^5 \left (b^2-4 a c\right )}+\frac {9 \left (a+b x+c x^2\right )^{4/3} (d (b+2 c x))^{4/3}}{16 c d^5 \left (b^2-4 a c\right )^2}-\frac {9 \left (a+b x+c x^2\right )^{7/3}}{4 d^3 \left (b^2-4 a c\right )^2 (b d+2 c d x)^{2/3}}+\frac {3 \left (a+b x+c x^2\right )^{7/3}}{4 d \left (b^2-4 a c\right ) (b d+2 c d x)^{8/3}}-\frac {3 \log \left ((d (b+2 c x))^{2/3}-2^{2/3} \sqrt [3]{c} d^{2/3} \sqrt [3]{a+b x+c x^2}\right )}{32\ 2^{2/3} c^{7/3} d^{11/3}}-\frac {\sqrt {3} \tan ^{-1}\left (\frac {\frac {\sqrt [3]{2} (d (b+2 c x))^{2/3}}{\sqrt [3]{c} d^{2/3} \sqrt [3]{a+b x+c x^2}}+1}{\sqrt {3}}\right )}{16\ 2^{2/3} c^{7/3} d^{11/3}} \]
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Rubi [A] time = 1.28, 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} \begin {gather*} -\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}} \end {gather*}
Antiderivative was successfully verified.
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Rule 31
Rule 204
Rule 275
Rule 279
Rule 292
Rule 329
Rule 331
Rule 617
Rule 628
Rule 634
Rule 693
Rule 694
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*}
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Mathematica [C] time = 0.09, size = 104, normalized size = 0.32 \begin {gather*} \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}} \end {gather*}
Antiderivative was successfully verified.
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IntegrateAlgebraic [A] time = 68.71, size = 633, normalized size = 1.98 \begin {gather*} \frac {2^{2/3} c^{4/3} d^{2/3} (b+2 c x)^{2/3} \sqrt [3]{c (a+x (b+c x))} \left (\frac {4 a c-b^2+(b+2 c x)^2}{c}\right )^{4/3} \left ((b+2 c x)^{2/3}-2^{2/3} \sqrt [3]{c (a+x (b+c x))}\right ) \left (2 \sqrt [3]{2} (c (a+x (b+c x)))^{2/3}+2^{2/3} (b+2 c x)^{2/3} \sqrt [3]{c (a+x (b+c x))}+b \sqrt [3]{b+2 c x}+2 c x \sqrt [3]{b+2 c x}\right )^2 \left (\frac {3 \sqrt [3]{4 a c-b^2+(b+2 c x)^2} \left (-4 a c+b^2-5 (b+2 c x)^2\right )}{64\ 2^{2/3} c^{7/3} d^{11/3} (b+2 c x)^{8/3}}-\frac {\log \left (\sqrt [3]{4 a c-b^2+(b+2 c x)^2}-(b+2 c x)^{2/3}\right )}{16\ 2^{2/3} c^{7/3} d^{11/3}}+\frac {\log \left ((b+2 c x)^{2/3} \sqrt [3]{4 a c-b^2+(b+2 c x)^2}+\left (4 a c-b^2+(b+2 c x)^2\right )^{2/3}+(b+2 c x)^{4/3}\right )}{32\ 2^{2/3} c^{7/3} d^{11/3}}-\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} (b+2 c x)^{2/3}}{2 \sqrt [3]{4 a c-b^2+(b+2 c x)^2}+(b+2 c x)^{2/3}}\right )}{16\ 2^{2/3} c^{7/3} d^{11/3}}\right )}{\left (b^2-4 a c\right ) \left (4 a c-b^2+(b+2 c x)^2\right )^{4/3} \left (2\ 2^{2/3} c x \sqrt [3]{b+2 c x} \sqrt [3]{c (a+x (b+c x))}+2 \sqrt [3]{2} (b+2 c x)^{2/3} (c (a+x (b+c x)))^{2/3}+2^{2/3} b \sqrt [3]{b+2 c x} \sqrt [3]{c (a+x (b+c x))}+4 a c+4 b c x+4 c^2 x^2\right ) (d (b+2 c x))^{2/3}} \end {gather*}
Antiderivative was successfully verified.
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fricas [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 1.34, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (c \,x^{2}+b x +a \right )^{\frac {4}{3}}}{\left (2 c d x +b d \right )^{\frac {11}{3}}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (c\,x^2+b\,x+a\right )}^{4/3}}{{\left (b\,d+2\,c\,d\,x\right )}^{11/3}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b x + c x^{2}\right )^{\frac {4}{3}}}{\left (d \left (b + 2 c x\right )\right )^{\frac {11}{3}}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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