Integrand size = 29, antiderivative size = 30 \[ \int \left (a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3\right )^p \, dx=\frac {\left (\frac {a}{b}+x\right ) \left (b^3 \left (\frac {a}{b}+x\right )^3\right )^p}{1+3 p} \]
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Time = 0.01 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {2092, 15, 30} \[ \int \left (a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3\right )^p \, dx=\frac {\left (\frac {a}{b}+x\right ) \left (b^3 \left (\frac {a}{b}+x\right )^3\right )^p}{3 p+1} \]
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Rule 15
Rule 30
Rule 2092
Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int \left (b^3 x^3\right )^p \, dx,x,\frac {a}{b}+x\right ) \\ & = \left (\left (\frac {a}{b}+x\right )^{-3 p} \left (b^3 \left (\frac {a}{b}+x\right )^3\right )^p\right ) \text {Subst}\left (\int x^{3 p} \, dx,x,\frac {a}{b}+x\right ) \\ & = \frac {(a+b x) \left ((a+b x)^3\right )^p}{b (1+3 p)} \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.83 \[ \int \left (a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3\right )^p \, dx=\frac {(a+b x) \left ((a+b x)^3\right )^p}{b (1+3 p)} \]
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Time = 0.05 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87
method | result | size |
risch | \(\frac {\left (b x +a \right ) \left (\left (b x +a \right )^{3}\right )^{p}}{b \left (1+3 p \right )}\) | \(26\) |
gosper | \(\frac {\left (b x +a \right ) \left (b^{3} x^{3}+3 a \,b^{2} x^{2}+3 a^{2} b x +a^{3}\right )^{p}}{b \left (1+3 p \right )}\) | \(46\) |
parallelrisch | \(\frac {x \left (b^{3} x^{3}+3 a \,b^{2} x^{2}+3 a^{2} b x +a^{3}\right )^{p} a b +\left (b^{3} x^{3}+3 a \,b^{2} x^{2}+3 a^{2} b x +a^{3}\right )^{p} a^{2}}{\left (1+3 p \right ) a b}\) | \(82\) |
norman | \(\frac {x \,{\mathrm e}^{p \ln \left (b^{3} x^{3}+3 a \,b^{2} x^{2}+3 a^{2} b x +a^{3}\right )}}{1+3 p}+\frac {a \,{\mathrm e}^{p \ln \left (b^{3} x^{3}+3 a \,b^{2} x^{2}+3 a^{2} b x +a^{3}\right )}}{b \left (1+3 p \right )}\) | \(85\) |
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Time = 0.23 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.43 \[ \int \left (a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3\right )^p \, dx=\frac {{\left (b x + a\right )} {\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3}\right )}^{p}}{3 \, b p + b} \]
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\[ \int \left (a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3\right )^p \, dx=\begin {cases} \frac {x}{\sqrt [3]{a^{3}}} & \text {for}\: b = 0 \wedge p = - \frac {1}{3} \\x \left (a^{3}\right )^{p} & \text {for}\: b = 0 \\\int \frac {1}{\sqrt [3]{a^{3} + 3 a^{2} b x + 3 a b^{2} x^{2} + b^{3} x^{3}}}\, dx & \text {for}\: p = - \frac {1}{3} \\\frac {a \left (a^{3} + 3 a^{2} b x + 3 a b^{2} x^{2} + b^{3} x^{3}\right )^{p}}{3 b p + b} + \frac {b x \left (a^{3} + 3 a^{2} b x + 3 a b^{2} x^{2} + b^{3} x^{3}\right )^{p}}{3 b p + b} & \text {otherwise} \end {cases} \]
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Time = 0.19 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.83 \[ \int \left (a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3\right )^p \, dx=\frac {{\left (b x + a\right )} {\left (b x + a\right )}^{3 \, p}}{b {\left (3 \, p + 1\right )}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 73 vs. \(2 (30) = 60\).
Time = 0.28 (sec) , antiderivative size = 73, normalized size of antiderivative = 2.43 \[ \int \left (a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3\right )^p \, dx=\frac {{\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3}\right )}^{p} b x + {\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3}\right )}^{p} a}{3 \, b p + b} \]
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Time = 9.33 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.73 \[ \int \left (a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3\right )^p \, dx=\left (\frac {x}{3\,p+1}+\frac {a}{b\,\left (3\,p+1\right )}\right )\,{\left (a^3+3\,a^2\,b\,x+3\,a\,b^2\,x^2+b^3\,x^3\right )}^p \]
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