Optimal. Leaf size=40 \[ \frac{2 a (a+b x)^{m+2}}{b (m+2)}-\frac{(a+b x)^{m+3}}{b (m+3)} \]
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Rubi [A] time = 0.0558223, antiderivative size = 40, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1 \[ \frac{2 a (a+b x)^{m+2}}{b (m+2)}-\frac{(a+b x)^{m+3}}{b (m+3)} \]
Antiderivative was successfully verified.
[In] Int[(a + b*x)^m*(a^2 - b^2*x^2),x]
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Rubi in Sympy [A] time = 10.776, size = 29, normalized size = 0.72 \[ \frac{2 a \left (a + b x\right )^{m + 2}}{b \left (m + 2\right )} - \frac{\left (a + b x\right )^{m + 3}}{b \left (m + 3\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
[In] rubi_integrate((b*x+a)**m*(-b**2*x**2+a**2),x)
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Mathematica [A] time = 0.0401319, size = 36, normalized size = 0.9 \[ \frac{(a+b x)^{m+2} (a (m+4)-b (m+2) x)}{b (m+2) (m+3)} \]
Antiderivative was successfully verified.
[In] Integrate[(a + b*x)^m*(a^2 - b^2*x^2),x]
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Maple [A] time = 0.004, size = 40, normalized size = 1. \[{\frac{ \left ( bx+a \right ) ^{2+m} \left ( -bmx+am-2\,bx+4\,a \right ) }{b \left ({m}^{2}+5\,m+6 \right ) }} \]
Verification of antiderivative is not currently implemented for this CAS.
[In] int((b*x+a)^m*(-b^2*x^2+a^2),x)
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Maxima [F] time = 0., size = 0, normalized size = 0. \[ \text{Exception raised: ValueError} \]
Verification of antiderivative is not currently implemented for this CAS.
[In] integrate(-(b^2*x^2 - a^2)*(b*x + a)^m,x, algorithm="maxima")
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Fricas [A] time = 0.232417, size = 103, normalized size = 2.58 \[ -\frac{{\left (a b^{2} m x^{2} - a^{3} m +{\left (b^{3} m + 2 \, b^{3}\right )} x^{3} - 4 \, a^{3} -{\left (a^{2} b m + 6 \, a^{2} b\right )} x\right )}{\left (b x + a\right )}^{m}}{b m^{2} + 5 \, b m + 6 \, b} \]
Verification of antiderivative is not currently implemented for this CAS.
[In] integrate(-(b^2*x^2 - a^2)*(b*x + a)^m,x, algorithm="fricas")
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Sympy [A] time = 2.5727, size = 275, normalized size = 6.88 \[ \begin{cases} a^{2} a^{m} x & \text{for}\: b = 0 \\- \frac{a \log{\left (\frac{a}{b} + x \right )}}{a b + b^{2} x} - \frac{a}{a b + b^{2} x} - \frac{b x \log{\left (\frac{a}{b} + x \right )}}{a b + b^{2} x} + \frac{b x}{a b + b^{2} x} & \text{for}\: m = -3 \\\frac{2 a \log{\left (\frac{a}{b} + x \right )}}{b} - x & \text{for}\: m = -2 \\\frac{a^{3} m \left (a + b x\right )^{m}}{b m^{2} + 5 b m + 6 b} + \frac{4 a^{3} \left (a + b x\right )^{m}}{b m^{2} + 5 b m + 6 b} + \frac{a^{2} b m x \left (a + b x\right )^{m}}{b m^{2} + 5 b m + 6 b} + \frac{6 a^{2} b x \left (a + b x\right )^{m}}{b m^{2} + 5 b m + 6 b} - \frac{a b^{2} m x^{2} \left (a + b x\right )^{m}}{b m^{2} + 5 b m + 6 b} - \frac{b^{3} m x^{3} \left (a + b x\right )^{m}}{b m^{2} + 5 b m + 6 b} - \frac{2 b^{3} x^{3} \left (a + b x\right )^{m}}{b m^{2} + 5 b m + 6 b} & \text{otherwise} \end{cases} \]
Verification of antiderivative is not currently implemented for this CAS.
[In] integrate((b*x+a)**m*(-b**2*x**2+a**2),x)
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GIAC/XCAS [A] time = 0.212177, size = 178, normalized size = 4.45 \[ -\frac{b^{3} m x^{3} e^{\left (m{\rm ln}\left (b x + a\right )\right )} + a b^{2} m x^{2} e^{\left (m{\rm ln}\left (b x + a\right )\right )} + 2 \, b^{3} x^{3} e^{\left (m{\rm ln}\left (b x + a\right )\right )} - a^{2} b m x e^{\left (m{\rm ln}\left (b x + a\right )\right )} - a^{3} m e^{\left (m{\rm ln}\left (b x + a\right )\right )} - 6 \, a^{2} b x e^{\left (m{\rm ln}\left (b x + a\right )\right )} - 4 \, a^{3} e^{\left (m{\rm ln}\left (b x + a\right )\right )}}{b m^{2} + 5 \, b m + 6 \, b} \]
Verification of antiderivative is not currently implemented for this CAS.
[In] integrate(-(b^2*x^2 - a^2)*(b*x + a)^m,x, algorithm="giac")
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