Integrand size = 26, antiderivative size = 315 \[ \int \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p x \, dx=-\frac {3 a^6 \left (1+\frac {b \sqrt [3]{x}}{a}\right ) \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p}{b^6 (1+2 p)}+\frac {15 a^6 \left (1+\frac {b \sqrt [3]{x}}{a}\right )^2 \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p}{2 b^6 (1+p)}-\frac {30 a^6 \left (1+\frac {b \sqrt [3]{x}}{a}\right )^3 \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p}{b^6 (3+2 p)}+\frac {15 a^6 \left (1+\frac {b \sqrt [3]{x}}{a}\right )^4 \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p}{b^6 (2+p)}-\frac {15 a^6 \left (1+\frac {b \sqrt [3]{x}}{a}\right )^5 \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p}{b^6 (5+2 p)}+\frac {3 a^6 \left (1+\frac {b \sqrt [3]{x}}{a}\right )^6 \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p}{2 b^6 (3+p)} \]
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Time = 0.09 (sec) , antiderivative size = 315, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {1370, 272, 45} \[ \int \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p x \, dx=\frac {3 a^6 \left (\frac {b \sqrt [3]{x}}{a}+1\right )^6 \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p}{2 b^6 (p+3)}-\frac {15 a^6 \left (\frac {b \sqrt [3]{x}}{a}+1\right )^5 \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p}{b^6 (2 p+5)}+\frac {15 a^6 \left (\frac {b \sqrt [3]{x}}{a}+1\right )^4 \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p}{b^6 (p+2)}-\frac {30 a^6 \left (\frac {b \sqrt [3]{x}}{a}+1\right )^3 \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p}{b^6 (2 p+3)}+\frac {15 a^6 \left (\frac {b \sqrt [3]{x}}{a}+1\right )^2 \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p}{2 b^6 (p+1)}-\frac {3 a^6 \left (\frac {b \sqrt [3]{x}}{a}+1\right ) \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p}{b^6 (2 p+1)} \]
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Rule 45
Rule 272
Rule 1370
Rubi steps \begin{align*} \text {integral}& = \left (\left (1+\frac {b \sqrt [3]{x}}{a}\right )^{-2 p} \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p\right ) \int \left (1+\frac {b \sqrt [3]{x}}{a}\right )^{2 p} x \, dx \\ & = \left (3 \left (1+\frac {b \sqrt [3]{x}}{a}\right )^{-2 p} \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p\right ) \text {Subst}\left (\int x^5 \left (1+\frac {b x}{a}\right )^{2 p} \, dx,x,\sqrt [3]{x}\right ) \\ & = \left (3 \left (1+\frac {b \sqrt [3]{x}}{a}\right )^{-2 p} \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p\right ) \text {Subst}\left (\int \left (-\frac {a^5 \left (1+\frac {b x}{a}\right )^{2 p}}{b^5}+\frac {5 a^5 \left (1+\frac {b x}{a}\right )^{1+2 p}}{b^5}-\frac {10 a^5 \left (1+\frac {b x}{a}\right )^{2+2 p}}{b^5}+\frac {10 a^5 \left (1+\frac {b x}{a}\right )^{3+2 p}}{b^5}-\frac {5 a^5 \left (1+\frac {b x}{a}\right )^{4+2 p}}{b^5}+\frac {a^5 \left (1+\frac {b x}{a}\right )^{5+2 p}}{b^5}\right ) \, dx,x,\sqrt [3]{x}\right ) \\ & = -\frac {3 a^6 \left (1+\frac {b \sqrt [3]{x}}{a}\right ) \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p}{b^6 (1+2 p)}+\frac {15 a^6 \left (1+\frac {b \sqrt [3]{x}}{a}\right )^2 \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p}{2 b^6 (1+p)}-\frac {30 a^6 \left (1+\frac {b \sqrt [3]{x}}{a}\right )^3 \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p}{b^6 (3+2 p)}+\frac {15 a^6 \left (1+\frac {b \sqrt [3]{x}}{a}\right )^4 \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p}{b^6 (2+p)}-\frac {15 a^6 \left (1+\frac {b \sqrt [3]{x}}{a}\right )^5 \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p}{b^6 (5+2 p)}+\frac {3 a^6 \left (1+\frac {b \sqrt [3]{x}}{a}\right )^6 \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p}{2 b^6 (3+p)} \\ \end{align*}
Time = 0.24 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.45 \[ \int \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p x \, dx=\frac {3 \left (-\frac {2 a^5}{1+2 p}+\frac {5 a^4 \left (a+b \sqrt [3]{x}\right )}{1+p}-\frac {20 a^3 \left (a+b \sqrt [3]{x}\right )^2}{3+2 p}+\frac {10 a^2 \left (a+b \sqrt [3]{x}\right )^3}{2+p}-\frac {10 a \left (a+b \sqrt [3]{x}\right )^4}{5+2 p}+\frac {\left (a+b \sqrt [3]{x}\right )^5}{3+p}\right ) \left (a+b \sqrt [3]{x}\right ) \left (\left (a+b \sqrt [3]{x}\right )^2\right )^p}{2 b^6} \]
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\[\int \left (a^{2}+2 a b \,x^{\frac {1}{3}}+b^{2} x^{\frac {2}{3}}\right )^{p} x d x\]
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none
Time = 0.33 (sec) , antiderivative size = 297, normalized size of antiderivative = 0.94 \[ \int \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p x \, dx=-\frac {3 \, {\left (30 \, a^{6} - {\left (8 \, b^{6} p^{5} + 60 \, b^{6} p^{4} + 170 \, b^{6} p^{3} + 225 \, b^{6} p^{2} + 137 \, b^{6} p + 30 \, b^{6}\right )} x^{2} - 20 \, {\left (2 \, a^{3} b^{3} p^{3} + 3 \, a^{3} b^{3} p^{2} + a^{3} b^{3} p\right )} x + 2 \, {\left (30 \, a^{4} b^{2} p^{2} + 15 \, a^{4} b^{2} p - {\left (4 \, a b^{5} p^{5} + 20 \, a b^{5} p^{4} + 35 \, a b^{5} p^{3} + 25 \, a b^{5} p^{2} + 6 \, a b^{5} p\right )} x\right )} x^{\frac {2}{3}} - 5 \, {\left (12 \, a^{5} b p - {\left (4 \, a^{2} b^{4} p^{4} + 12 \, a^{2} b^{4} p^{3} + 11 \, a^{2} b^{4} p^{2} + 3 \, a^{2} b^{4} p\right )} x\right )} x^{\frac {1}{3}}\right )} {\left (b^{2} x^{\frac {2}{3}} + 2 \, a b x^{\frac {1}{3}} + a^{2}\right )}^{p}}{2 \, {\left (8 \, b^{6} p^{6} + 84 \, b^{6} p^{5} + 350 \, b^{6} p^{4} + 735 \, b^{6} p^{3} + 812 \, b^{6} p^{2} + 441 \, b^{6} p + 90 \, b^{6}\right )}} \]
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\[ \int \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p x \, dx=\int x \left (\left (a + b \sqrt [3]{x}\right )^{2}\right )^{p}\, dx \]
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Time = 0.20 (sec) , antiderivative size = 198, normalized size of antiderivative = 0.63 \[ \int \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p x \, dx=\frac {3 \, {\left ({\left (8 \, p^{5} + 60 \, p^{4} + 170 \, p^{3} + 225 \, p^{2} + 137 \, p + 30\right )} b^{6} x^{2} + 2 \, {\left (4 \, p^{5} + 20 \, p^{4} + 35 \, p^{3} + 25 \, p^{2} + 6 \, p\right )} a b^{5} x^{\frac {5}{3}} - 5 \, {\left (4 \, p^{4} + 12 \, p^{3} + 11 \, p^{2} + 3 \, p\right )} a^{2} b^{4} x^{\frac {4}{3}} + 20 \, {\left (2 \, p^{3} + 3 \, p^{2} + p\right )} a^{3} b^{3} x - 30 \, {\left (2 \, p^{2} + p\right )} a^{4} b^{2} x^{\frac {2}{3}} + 60 \, a^{5} b p x^{\frac {1}{3}} - 30 \, a^{6}\right )} {\left (b x^{\frac {1}{3}} + a\right )}^{2 \, p}}{2 \, {\left (8 \, p^{6} + 84 \, p^{5} + 350 \, p^{4} + 735 \, p^{3} + 812 \, p^{2} + 441 \, p + 90\right )} b^{6}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 745 vs. \(2 (275) = 550\).
Time = 0.32 (sec) , antiderivative size = 745, normalized size of antiderivative = 2.37 \[ \int \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p x \, dx=\frac {3 \, {\left (8 \, {\left (b^{2} x^{\frac {2}{3}} + 2 \, a b x^{\frac {1}{3}} + a^{2}\right )}^{p} b^{6} p^{5} x^{2} + 8 \, {\left (b^{2} x^{\frac {2}{3}} + 2 \, a b x^{\frac {1}{3}} + a^{2}\right )}^{p} a b^{5} p^{5} x^{\frac {5}{3}} + 60 \, {\left (b^{2} x^{\frac {2}{3}} + 2 \, a b x^{\frac {1}{3}} + a^{2}\right )}^{p} b^{6} p^{4} x^{2} + 40 \, {\left (b^{2} x^{\frac {2}{3}} + 2 \, a b x^{\frac {1}{3}} + a^{2}\right )}^{p} a b^{5} p^{4} x^{\frac {5}{3}} - 20 \, {\left (b^{2} x^{\frac {2}{3}} + 2 \, a b x^{\frac {1}{3}} + a^{2}\right )}^{p} a^{2} b^{4} p^{4} x^{\frac {4}{3}} + 170 \, {\left (b^{2} x^{\frac {2}{3}} + 2 \, a b x^{\frac {1}{3}} + a^{2}\right )}^{p} b^{6} p^{3} x^{2} + 70 \, {\left (b^{2} x^{\frac {2}{3}} + 2 \, a b x^{\frac {1}{3}} + a^{2}\right )}^{p} a b^{5} p^{3} x^{\frac {5}{3}} - 60 \, {\left (b^{2} x^{\frac {2}{3}} + 2 \, a b x^{\frac {1}{3}} + a^{2}\right )}^{p} a^{2} b^{4} p^{3} x^{\frac {4}{3}} + 40 \, {\left (b^{2} x^{\frac {2}{3}} + 2 \, a b x^{\frac {1}{3}} + a^{2}\right )}^{p} a^{3} b^{3} p^{3} x + 225 \, {\left (b^{2} x^{\frac {2}{3}} + 2 \, a b x^{\frac {1}{3}} + a^{2}\right )}^{p} b^{6} p^{2} x^{2} + 50 \, {\left (b^{2} x^{\frac {2}{3}} + 2 \, a b x^{\frac {1}{3}} + a^{2}\right )}^{p} a b^{5} p^{2} x^{\frac {5}{3}} - 55 \, {\left (b^{2} x^{\frac {2}{3}} + 2 \, a b x^{\frac {1}{3}} + a^{2}\right )}^{p} a^{2} b^{4} p^{2} x^{\frac {4}{3}} + 60 \, {\left (b^{2} x^{\frac {2}{3}} + 2 \, a b x^{\frac {1}{3}} + a^{2}\right )}^{p} a^{3} b^{3} p^{2} x + 137 \, {\left (b^{2} x^{\frac {2}{3}} + 2 \, a b x^{\frac {1}{3}} + a^{2}\right )}^{p} b^{6} p x^{2} - 60 \, {\left (b^{2} x^{\frac {2}{3}} + 2 \, a b x^{\frac {1}{3}} + a^{2}\right )}^{p} a^{4} b^{2} p^{2} x^{\frac {2}{3}} + 12 \, {\left (b^{2} x^{\frac {2}{3}} + 2 \, a b x^{\frac {1}{3}} + a^{2}\right )}^{p} a b^{5} p x^{\frac {5}{3}} - 15 \, {\left (b^{2} x^{\frac {2}{3}} + 2 \, a b x^{\frac {1}{3}} + a^{2}\right )}^{p} a^{2} b^{4} p x^{\frac {4}{3}} + 20 \, {\left (b^{2} x^{\frac {2}{3}} + 2 \, a b x^{\frac {1}{3}} + a^{2}\right )}^{p} a^{3} b^{3} p x + 30 \, {\left (b^{2} x^{\frac {2}{3}} + 2 \, a b x^{\frac {1}{3}} + a^{2}\right )}^{p} b^{6} x^{2} - 30 \, {\left (b^{2} x^{\frac {2}{3}} + 2 \, a b x^{\frac {1}{3}} + a^{2}\right )}^{p} a^{4} b^{2} p x^{\frac {2}{3}} + 60 \, {\left (b^{2} x^{\frac {2}{3}} + 2 \, a b x^{\frac {1}{3}} + a^{2}\right )}^{p} a^{5} b p x^{\frac {1}{3}} - 30 \, {\left (b^{2} x^{\frac {2}{3}} + 2 \, a b x^{\frac {1}{3}} + a^{2}\right )}^{p} a^{6}\right )}}{2 \, {\left (8 \, b^{6} p^{6} + 84 \, b^{6} p^{5} + 350 \, b^{6} p^{4} + 735 \, b^{6} p^{3} + 812 \, b^{6} p^{2} + 441 \, b^{6} p + 90 \, b^{6}\right )}} \]
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Time = 8.76 (sec) , antiderivative size = 390, normalized size of antiderivative = 1.24 \[ \int \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p x \, dx={\left (a^2+b^2\,x^{2/3}+2\,a\,b\,x^{1/3}\right )}^p\,\left (\frac {3\,x^2\,\left (8\,p^5+60\,p^4+170\,p^3+225\,p^2+137\,p+30\right )}{2\,\left (8\,p^6+84\,p^5+350\,p^4+735\,p^3+812\,p^2+441\,p+90\right )}-\frac {45\,a^6}{b^6\,\left (8\,p^6+84\,p^5+350\,p^4+735\,p^3+812\,p^2+441\,p+90\right )}+\frac {90\,a^5\,p\,x^{1/3}}{b^5\,\left (8\,p^6+84\,p^5+350\,p^4+735\,p^3+812\,p^2+441\,p+90\right )}-\frac {15\,a^2\,p\,x^{4/3}\,\left (4\,p^3+12\,p^2+11\,p+3\right )}{2\,b^2\,\left (8\,p^6+84\,p^5+350\,p^4+735\,p^3+812\,p^2+441\,p+90\right )}+\frac {30\,a^3\,p\,x\,\left (2\,p^2+3\,p+1\right )}{b^3\,\left (8\,p^6+84\,p^5+350\,p^4+735\,p^3+812\,p^2+441\,p+90\right )}-\frac {45\,a^4\,p\,x^{2/3}\,\left (2\,p+1\right )}{b^4\,\left (8\,p^6+84\,p^5+350\,p^4+735\,p^3+812\,p^2+441\,p+90\right )}+\frac {3\,a\,p\,x^{5/3}\,\left (4\,p^4+20\,p^3+35\,p^2+25\,p+6\right )}{b\,\left (8\,p^6+84\,p^5+350\,p^4+735\,p^3+812\,p^2+441\,p+90\right )}\right ) \]
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