Optimal. Leaf size=86 \[ \frac {\left (a+b x^2\right )^3 \left (a^2+2 a b x^2+b^2 x^4\right )^p}{2 b^2 (2 p+3)}-\frac {a \left (a+b x^2\right )^2 \left (a^2+2 a b x^2+b^2 x^4\right )^p}{4 b^2 (p+1)} \]
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Rubi [A] time = 0.09, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {1249, 770, 21, 43} \[ \frac {\left (a+b x^2\right )^3 \left (a^2+2 a b x^2+b^2 x^4\right )^p}{2 b^2 (2 p+3)}-\frac {a \left (a+b x^2\right )^2 \left (a^2+2 a b x^2+b^2 x^4\right )^p}{4 b^2 (p+1)} \]
Antiderivative was successfully verified.
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Rule 21
Rule 43
Rule 770
Rule 1249
Rubi steps
\begin {align*} \int x^3 \left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int x (a+b x) \left (a^2+2 a b x+b^2 x^2\right )^p \, dx,x,x^2\right )\\ &=\frac {1}{2} \left (\left (b \left (a+b x^2\right )\right )^{-2 p} \left (a^2+2 a b x^2+b^2 x^4\right )^p\right ) \operatorname {Subst}\left (\int x (a+b x) \left (a b+b^2 x\right )^{2 p} \, dx,x,x^2\right )\\ &=\frac {\left (\left (b \left (a+b x^2\right )\right )^{-2 p} \left (a^2+2 a b x^2+b^2 x^4\right )^p\right ) \operatorname {Subst}\left (\int x \left (a b+b^2 x\right )^{1+2 p} \, dx,x,x^2\right )}{2 b}\\ &=\frac {\left (\left (b \left (a+b x^2\right )\right )^{-2 p} \left (a^2+2 a b x^2+b^2 x^4\right )^p\right ) \operatorname {Subst}\left (\int \left (-\frac {a \left (a b+b^2 x\right )^{1+2 p}}{b}+\frac {\left (a b+b^2 x\right )^{2+2 p}}{b^2}\right ) \, dx,x,x^2\right )}{2 b}\\ &=-\frac {a \left (a+b x^2\right )^2 \left (a^2+2 a b x^2+b^2 x^4\right )^p}{4 b^2 (1+p)}+\frac {\left (a+b x^2\right )^3 \left (a^2+2 a b x^2+b^2 x^4\right )^p}{2 b^2 (3+2 p)}\\ \end {align*}
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Mathematica [A] time = 0.02, size = 45, normalized size = 0.52 \[ \frac {\left (\left (a+b x^2\right )^2\right )^{p+1} \left (2 b (p+1) x^2-a\right )}{4 b^2 (p+1) (2 p+3)} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.52, size = 92, normalized size = 1.07 \[ \frac {{\left (2 \, {\left (b^{3} p + b^{3}\right )} x^{6} + 2 \, a^{2} b p x^{2} + {\left (4 \, a b^{2} p + 3 \, a b^{2}\right )} x^{4} - a^{3}\right )} {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p}}{4 \, {\left (2 \, b^{2} p^{2} + 5 \, b^{2} p + 3 \, b^{2}\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.43, size = 196, normalized size = 2.28 \[ \frac {2 \, {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} b^{3} p x^{6} + 2 \, {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} b^{3} x^{6} + 4 \, {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} a b^{2} p x^{4} + 3 \, {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} a b^{2} x^{4} + 2 \, {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} a^{2} b p x^{2} - {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} a^{3}}{4 \, {\left (2 \, b^{2} p^{2} + 5 \, b^{2} p + 3 \, b^{2}\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.01, size = 62, normalized size = 0.72 \[ -\frac {\left (-2 x^{2} p b -2 b \,x^{2}+a \right ) \left (b \,x^{2}+a \right )^{2} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{p}}{4 \left (2 p^{2}+5 p +3\right ) b^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.74, size = 135, normalized size = 1.57 \[ \frac {{\left (b^{2} {\left (2 \, p + 1\right )} x^{4} + 2 \, a b p x^{2} - a^{2}\right )} {\left (b x^{2} + a\right )}^{2 \, p} a}{4 \, {\left (2 \, p^{2} + 3 \, p + 1\right )} b^{2}} + \frac {{\left ({\left (2 \, p^{2} + 3 \, p + 1\right )} b^{3} x^{6} + {\left (2 \, p^{2} + p\right )} a b^{2} x^{4} - 2 \, a^{2} b p x^{2} + a^{3}\right )} {\left (b x^{2} + a\right )}^{2 \, p}}{2 \, {\left (4 \, p^{3} + 12 \, p^{2} + 11 \, p + 3\right )} b^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.17, size = 108, normalized size = 1.26 \[ {\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^p\,\left (\frac {b\,x^6\,\left (p+1\right )}{2\,\left (2\,p^2+5\,p+3\right )}-\frac {a^3}{4\,b^2\,\left (2\,p^2+5\,p+3\right )}+\frac {a\,x^4\,\left (4\,p+3\right )}{4\,\left (2\,p^2+5\,p+3\right )}+\frac {a^2\,p\,x^2}{2\,b\,\left (2\,p^2+5\,p+3\right )}\right ) \]
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 \[ \begin {cases} \frac {a x^{4} \left (a^{2}\right )^{p}}{4} & \text {for}\: b = 0 \\\int \frac {x^{3} \left (a + b x^{2}\right )}{\left (\left (a + b x^{2}\right )^{2}\right )^{\frac {3}{2}}}\, dx & \text {for}\: p = - \frac {3}{2} \\- \frac {a \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + x \right )}}{2 b^{2}} - \frac {a \log {\left (i \sqrt {a} \sqrt {\frac {1}{b}} + x \right )}}{2 b^{2}} + \frac {x^{2}}{2 b} & \text {for}\: p = -1 \\- \frac {a^{3} \left (a^{2} + 2 a b x^{2} + b^{2} x^{4}\right )^{p}}{8 b^{2} p^{2} + 20 b^{2} p + 12 b^{2}} + \frac {2 a^{2} b p x^{2} \left (a^{2} + 2 a b x^{2} + b^{2} x^{4}\right )^{p}}{8 b^{2} p^{2} + 20 b^{2} p + 12 b^{2}} + \frac {4 a b^{2} p x^{4} \left (a^{2} + 2 a b x^{2} + b^{2} x^{4}\right )^{p}}{8 b^{2} p^{2} + 20 b^{2} p + 12 b^{2}} + \frac {3 a b^{2} x^{4} \left (a^{2} + 2 a b x^{2} + b^{2} x^{4}\right )^{p}}{8 b^{2} p^{2} + 20 b^{2} p + 12 b^{2}} + \frac {2 b^{3} p x^{6} \left (a^{2} + 2 a b x^{2} + b^{2} x^{4}\right )^{p}}{8 b^{2} p^{2} + 20 b^{2} p + 12 b^{2}} + \frac {2 b^{3} x^{6} \left (a^{2} + 2 a b x^{2} + b^{2} x^{4}\right )^{p}}{8 b^{2} p^{2} + 20 b^{2} p + 12 b^{2}} & \text {otherwise} \end {cases} \]
Verification of antiderivative is not currently implemented for this CAS.
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