Optimal. Leaf size=76 \[ \frac {(a+b x) (b d-a e) \left (a^2+2 a b x+b^2 x^2\right )^p}{b^2 (2 p+1)}+\frac {e \left (a^2+2 a b x+b^2 x^2\right )^{p+1}}{2 b^2 (p+1)} \]
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Rubi [A] time = 0.02, antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {640, 609} \[ \frac {(a+b x) (b d-a e) \left (a^2+2 a b x+b^2 x^2\right )^p}{b^2 (2 p+1)}+\frac {e \left (a^2+2 a b x+b^2 x^2\right )^{p+1}}{2 b^2 (p+1)} \]
Antiderivative was successfully verified.
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Rule 609
Rule 640
Rubi steps
\begin {align*} \int (d+e x) \left (a^2+2 a b x+b^2 x^2\right )^p \, dx &=\frac {e \left (a^2+2 a b x+b^2 x^2\right )^{1+p}}{2 b^2 (1+p)}+\frac {\left (2 b^2 d-2 a b e\right ) \int \left (a^2+2 a b x+b^2 x^2\right )^p \, dx}{2 b^2}\\ &=\frac {(b d-a e) (a+b x) \left (a^2+2 a b x+b^2 x^2\right )^p}{b^2 (1+2 p)}+\frac {e \left (a^2+2 a b x+b^2 x^2\right )^{1+p}}{2 b^2 (1+p)}\\ \end {align*}
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Mathematica [A] time = 0.04, size = 54, normalized size = 0.71 \[ \frac {(a+b x) \left ((a+b x)^2\right )^p (-a e+2 b d (p+1)+b e (2 p+1) x)}{2 b^2 (p+1) (2 p+1)} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.77, size = 96, normalized size = 1.26 \[ \frac {{\left (2 \, a b d p + 2 \, a b d - a^{2} e + {\left (2 \, b^{2} e p + b^{2} e\right )} x^{2} + 2 \, {\left (b^{2} d + {\left (b^{2} d + a b e\right )} p\right )} x\right )} {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p}}{2 \, {\left (2 \, b^{2} p^{2} + 3 \, b^{2} p + b^{2}\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.20, size = 228, normalized size = 3.00 \[ \frac {2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} b^{2} p x^{2} e + 2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} b^{2} d p x + 2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} a b p x e + {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} b^{2} x^{2} e + 2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} a b d p + 2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} b^{2} d x + 2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} a b d - {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} a^{2} e}{2 \, {\left (2 \, b^{2} p^{2} + 3 \, b^{2} p + b^{2}\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.05, size = 65, normalized size = 0.86 \[ -\frac {\left (-2 b e p x -2 b d p -b e x +a e -2 b d \right ) \left (b x +a \right ) \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{p}}{2 \left (2 p^{2}+3 p +1\right ) b^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 1.20, size = 78, normalized size = 1.03 \[ \frac {{\left (b x + a\right )} {\left (b x + a\right )}^{2 \, p} d}{b {\left (2 \, p + 1\right )}} + \frac {{\left (b^{2} {\left (2 \, p + 1\right )} x^{2} + 2 \, a b p x - a^{2}\right )} {\left (b x + a\right )}^{2 \, p} e}{2 \, {\left (2 \, p^{2} + 3 \, p + 1\right )} b^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.62, size = 112, normalized size = 1.47 \[ \left (\frac {x\,\left (2\,b^2\,d+2\,b^2\,d\,p+2\,a\,b\,e\,p\right )}{2\,b^2\,\left (2\,p^2+3\,p+1\right )}+\frac {a\,\left (2\,b\,d-a\,e+2\,b\,d\,p\right )}{2\,b^2\,\left (2\,p^2+3\,p+1\right )}+\frac {e\,x^2\,\left (2\,p+1\right )}{2\,\left (2\,p^2+3\,p+1\right )}\right )\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^p \]
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 \[ \begin {cases} \left (d x + \frac {e x^{2}}{2}\right ) \left (a^{2}\right )^{p} & \text {for}\: b = 0 \\\frac {a e \log {\left (\frac {a}{b} + x \right )}}{a b^{2} + b^{3} x} + \frac {a e}{a b^{2} + b^{3} x} - \frac {b d}{a b^{2} + b^{3} x} + \frac {b e x \log {\left (\frac {a}{b} + x \right )}}{a b^{2} + b^{3} x} & \text {for}\: p = -1 \\\int \frac {d + e x}{\sqrt {\left (a + b x\right )^{2}}}\, dx & \text {for}\: p = - \frac {1}{2} \\- \frac {a^{2} e \left (a^{2} + 2 a b x + b^{2} x^{2}\right )^{p}}{4 b^{2} p^{2} + 6 b^{2} p + 2 b^{2}} + \frac {2 a b d p \left (a^{2} + 2 a b x + b^{2} x^{2}\right )^{p}}{4 b^{2} p^{2} + 6 b^{2} p + 2 b^{2}} + \frac {2 a b d \left (a^{2} + 2 a b x + b^{2} x^{2}\right )^{p}}{4 b^{2} p^{2} + 6 b^{2} p + 2 b^{2}} + \frac {2 a b e p x \left (a^{2} + 2 a b x + b^{2} x^{2}\right )^{p}}{4 b^{2} p^{2} + 6 b^{2} p + 2 b^{2}} + \frac {2 b^{2} d p x \left (a^{2} + 2 a b x + b^{2} x^{2}\right )^{p}}{4 b^{2} p^{2} + 6 b^{2} p + 2 b^{2}} + \frac {2 b^{2} d x \left (a^{2} + 2 a b x + b^{2} x^{2}\right )^{p}}{4 b^{2} p^{2} + 6 b^{2} p + 2 b^{2}} + \frac {2 b^{2} e p x^{2} \left (a^{2} + 2 a b x + b^{2} x^{2}\right )^{p}}{4 b^{2} p^{2} + 6 b^{2} p + 2 b^{2}} + \frac {b^{2} e x^{2} \left (a^{2} + 2 a b x + b^{2} x^{2}\right )^{p}}{4 b^{2} p^{2} + 6 b^{2} p + 2 b^{2}} & \text {otherwise} \end {cases} \]
Verification of antiderivative is not currently implemented for this CAS.
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