Optimal. Leaf size=30 \[ \frac {\left (\frac {a}{b}+x\right ) \left (b^3 \left (\frac {a}{b}+x\right )^3\right )^p}{1+3 p} \]
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Rubi [A]
time = 0.01, antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps
used = 3, number of rules used = 3, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {2092, 15, 30}
\begin {gather*} \frac {\left (\frac {a}{b}+x\right ) \left (b^3 \left (\frac {a}{b}+x\right )^3\right )^p}{3 p+1} \end {gather*}
Antiderivative was successfully verified.
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Rule 15
Rule 30
Rule 2092
Rubi steps
\begin {align*} \int \left (a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3\right )^p \, dx &=\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*}
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Mathematica [A]
time = 0.02, size = 23, normalized size = 0.77 \begin {gather*} \frac {(a+b x) \left ((a+b x)^3\right )^p}{b+3 b p} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.03, size = 26, normalized size = 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 | \(\frac {\left (b x +a \right ) \left (b^{3} x^{3}+3 a \,b^{2} x^{2}+3 a^{2} b x +a^{3}\right )^{p}}{b \left (1+3 p \right )}\) | \(46\) |
norman | \(\frac {x \,{\mathrm e}^{p \ln \left (b^{3} x^{3}+3 a \,b^{2} x^{2}+3 a^{2} b x +a^{3}\right )}}{1+3 p}+\frac {a \,{\mathrm e}^{p \ln \left (b^{3} x^{3}+3 a \,b^{2} x^{2}+3 a^{2} b x +a^{3}\right )}}{b \left (1+3 p \right )}\) | \(85\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.26, size = 25, normalized size = 0.83 \begin {gather*} \frac {{\left (b x + a\right )} {\left (b x + a\right )}^{3 \, p}}{b {\left (3 \, p + 1\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.38, size = 43, normalized size = 1.43 \begin {gather*} \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} \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*} \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} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 73 vs.
\(2 (30) = 60\).
time = 3.90, size = 73, normalized size = 2.43 \begin {gather*} \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} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 2.13, size = 52, normalized size = 1.73 \begin {gather*} \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 \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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