Optimal. Leaf size=133 \[ -\frac {x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (a e^2-b d^2 (2 p+3)\right ) \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};-\frac {b x^2}{a}\right )}{b (2 p+3)}+\frac {e (d+e x) \left (a+b x^2\right )^{p+1}}{b (2 p+3)}+\frac {d e (p+2) \left (a+b x^2\right )^{p+1}}{b (p+1) (2 p+3)} \]
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Rubi [A] time = 0.08, antiderivative size = 125, normalized size of antiderivative = 0.94, number of steps used = 4, number of rules used = 4, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {743, 641, 246, 245} \[ x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (d^2-\frac {a e^2}{2 b p+3 b}\right ) \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};-\frac {b x^2}{a}\right )+\frac {e (d+e x) \left (a+b x^2\right )^{p+1}}{b (2 p+3)}+\frac {d e (p+2) \left (a+b x^2\right )^{p+1}}{b (p+1) (2 p+3)} \]
Antiderivative was successfully verified.
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Rule 245
Rule 246
Rule 641
Rule 743
Rubi steps
\begin {align*} \int (d+e x)^2 \left (a+b x^2\right )^p \, dx &=\frac {e (d+e x) \left (a+b x^2\right )^{1+p}}{b (3+2 p)}+\frac {\int \left (-a e^2+b d^2 (3+2 p)+2 b d e (2+p) x\right ) \left (a+b x^2\right )^p \, dx}{b (3+2 p)}\\ &=\frac {d e (2+p) \left (a+b x^2\right )^{1+p}}{b (1+p) (3+2 p)}+\frac {e (d+e x) \left (a+b x^2\right )^{1+p}}{b (3+2 p)}+\left (d^2-\frac {a e^2}{3 b+2 b p}\right ) \int \left (a+b x^2\right )^p \, dx\\ &=\frac {d e (2+p) \left (a+b x^2\right )^{1+p}}{b (1+p) (3+2 p)}+\frac {e (d+e x) \left (a+b x^2\right )^{1+p}}{b (3+2 p)}+\left (\left (d^2-\frac {a e^2}{3 b+2 b p}\right ) \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p}\right ) \int \left (1+\frac {b x^2}{a}\right )^p \, dx\\ &=\frac {d e (2+p) \left (a+b x^2\right )^{1+p}}{b (1+p) (3+2 p)}+\frac {e (d+e x) \left (a+b x^2\right )^{1+p}}{b (3+2 p)}+\left (d^2-\frac {a e^2}{3 b+2 b p}\right ) x \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};-\frac {b x^2}{a}\right )\\ \end {align*}
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Mathematica [A] time = 0.12, size = 133, normalized size = 1.00 \[ \frac {\left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (3 b d^2 (p+1) x \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};-\frac {b x^2}{a}\right )+e \left (3 d \left (b x^2 \left (\frac {b x^2}{a}+1\right )^p+a \left (\left (\frac {b x^2}{a}+1\right )^p-1\right )\right )+b e (p+1) x^3 \, _2F_1\left (\frac {3}{2},-p;\frac {5}{2};-\frac {b x^2}{a}\right )\right )\right )}{3 b (p+1)} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.70, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (e^{2} x^{2} + 2 \, d e x + d^{2}\right )} {\left (b x^{2} + a\right )}^{p}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (e x + d\right )}^{2} {\left (b x^{2} + a\right )}^{p}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.04, size = 0, normalized size = 0.00 \[ \int \left (e x +d \right )^{2} \left (b \,x^{2}+a \right )^{p}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (e x + d\right )}^{2} {\left (b x^{2} + a\right )}^{p}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (b\,x^2+a\right )}^p\,{\left (d+e\,x\right )}^2 \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 11.61, size = 97, normalized size = 0.73 \[ a^{p} d^{2} x {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, - p \\ \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )} + \frac {a^{p} e^{2} x^{3} {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, - p \\ \frac {5}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{3} + 2 d e \left (\begin {cases} \frac {a^{p} x^{2}}{2} & \text {for}\: b = 0 \\\frac {\begin {cases} \frac {\left (a + b x^{2}\right )^{p + 1}}{p + 1} & \text {for}\: p \neq -1 \\\log {\left (a + b x^{2} \right )} & \text {otherwise} \end {cases}}{2 b} & \text {otherwise} \end {cases}\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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