Optimal. Leaf size=297 \[ \frac {2 b^5 (d x)^{27/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{27 d^{11} \left (a+b x^2\right )}+\frac {10 a b^4 (d x)^{23/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{23 d^9 \left (a+b x^2\right )}+\frac {20 a^2 b^3 (d x)^{19/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{19 d^7 \left (a+b x^2\right )}+\frac {2 a^5 (d x)^{7/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{7 d \left (a+b x^2\right )}+\frac {10 a^4 b (d x)^{11/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{11 d^3 \left (a+b x^2\right )}+\frac {4 a^3 b^2 (d x)^{15/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 d^5 \left (a+b x^2\right )} \]
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Rubi [A] time = 0.08, antiderivative size = 297, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {1112, 270} \begin {gather*} \frac {2 b^5 (d x)^{27/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{27 d^{11} \left (a+b x^2\right )}+\frac {10 a b^4 (d x)^{23/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{23 d^9 \left (a+b x^2\right )}+\frac {20 a^2 b^3 (d x)^{19/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{19 d^7 \left (a+b x^2\right )}+\frac {4 a^3 b^2 (d x)^{15/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 d^5 \left (a+b x^2\right )}+\frac {10 a^4 b (d x)^{11/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{11 d^3 \left (a+b x^2\right )}+\frac {2 a^5 (d x)^{7/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{7 d \left (a+b x^2\right )} \end {gather*}
Antiderivative was successfully verified.
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Rule 270
Rule 1112
Rubi steps
\begin {align*} \int (d x)^{5/2} \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx &=\frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int (d x)^{5/2} \left (a b+b^2 x^2\right )^5 \, dx}{b^4 \left (a b+b^2 x^2\right )}\\ &=\frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \left (a^5 b^5 (d x)^{5/2}+\frac {5 a^4 b^6 (d x)^{9/2}}{d^2}+\frac {10 a^3 b^7 (d x)^{13/2}}{d^4}+\frac {10 a^2 b^8 (d x)^{17/2}}{d^6}+\frac {5 a b^9 (d x)^{21/2}}{d^8}+\frac {b^{10} (d x)^{25/2}}{d^{10}}\right ) \, dx}{b^4 \left (a b+b^2 x^2\right )}\\ &=\frac {2 a^5 (d x)^{7/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{7 d \left (a+b x^2\right )}+\frac {10 a^4 b (d x)^{11/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{11 d^3 \left (a+b x^2\right )}+\frac {4 a^3 b^2 (d x)^{15/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 d^5 \left (a+b x^2\right )}+\frac {20 a^2 b^3 (d x)^{19/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{19 d^7 \left (a+b x^2\right )}+\frac {10 a b^4 (d x)^{23/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{23 d^9 \left (a+b x^2\right )}+\frac {2 b^5 (d x)^{27/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{27 d^{11} \left (a+b x^2\right )}\\ \end {align*}
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Mathematica [A] time = 0.04, size = 88, normalized size = 0.30 \begin {gather*} \frac {2 x (d x)^{5/2} \sqrt {\left (a+b x^2\right )^2} \left (129789 a^5+412965 a^4 b x^2+605682 a^3 b^2 x^4+478170 a^2 b^3 x^6+197505 a b^4 x^8+33649 b^5 x^{10}\right )}{908523 \left (a+b x^2\right )} \end {gather*}
Antiderivative was successfully verified.
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IntegrateAlgebraic [A] time = 126.12, size = 141, normalized size = 0.47 \begin {gather*} \frac {2 \left (a d^2+b d^2 x^2\right ) \left (129789 a^5 d^{10} (d x)^{7/2}+412965 a^4 b d^8 (d x)^{11/2}+605682 a^3 b^2 d^6 (d x)^{15/2}+478170 a^2 b^3 d^4 (d x)^{19/2}+197505 a b^4 d^2 (d x)^{23/2}+33649 b^5 (d x)^{27/2}\right )}{908523 d^{13} \sqrt {\frac {\left (a d^2+b d^2 x^2\right )^2}{d^4}}} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 1.58, size = 82, normalized size = 0.28 \begin {gather*} \frac {2}{908523} \, {\left (33649 \, b^{5} d^{2} x^{13} + 197505 \, a b^{4} d^{2} x^{11} + 478170 \, a^{2} b^{3} d^{2} x^{9} + 605682 \, a^{3} b^{2} d^{2} x^{7} + 412965 \, a^{4} b d^{2} x^{5} + 129789 \, a^{5} d^{2} x^{3}\right )} \sqrt {d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.17, size = 153, normalized size = 0.52 \begin {gather*} \frac {2}{27} \, \sqrt {d x} b^{5} d^{2} x^{13} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {10}{23} \, \sqrt {d x} a b^{4} d^{2} x^{11} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {20}{19} \, \sqrt {d x} a^{2} b^{3} d^{2} x^{9} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {4}{3} \, \sqrt {d x} a^{3} b^{2} d^{2} x^{7} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {10}{11} \, \sqrt {d x} a^{4} b d^{2} x^{5} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {2}{7} \, \sqrt {d x} a^{5} d^{2} x^{3} \mathrm {sgn}\left (b x^{2} + a\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.01, size = 83, normalized size = 0.28 \begin {gather*} \frac {2 \left (33649 b^{5} x^{10}+197505 a \,b^{4} x^{8}+478170 a^{2} b^{3} x^{6}+605682 a^{3} b^{2} x^{4}+412965 a^{4} b \,x^{2}+129789 a^{5}\right ) \left (d x \right )^{\frac {5}{2}} \left (\left (b \,x^{2}+a \right )^{2}\right )^{\frac {5}{2}} x}{908523 \left (b \,x^{2}+a \right )^{5}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 1.47, size = 147, normalized size = 0.49 \begin {gather*} \frac {2}{621} \, {\left (23 \, b^{5} d^{\frac {5}{2}} x^{3} + 27 \, a b^{4} d^{\frac {5}{2}} x\right )} x^{\frac {21}{2}} + \frac {8}{437} \, {\left (19 \, a b^{4} d^{\frac {5}{2}} x^{3} + 23 \, a^{2} b^{3} d^{\frac {5}{2}} x\right )} x^{\frac {17}{2}} + \frac {4}{95} \, {\left (15 \, a^{2} b^{3} d^{\frac {5}{2}} x^{3} + 19 \, a^{3} b^{2} d^{\frac {5}{2}} x\right )} x^{\frac {13}{2}} + \frac {8}{165} \, {\left (11 \, a^{3} b^{2} d^{\frac {5}{2}} x^{3} + 15 \, a^{4} b d^{\frac {5}{2}} x\right )} x^{\frac {9}{2}} + \frac {2}{77} \, {\left (7 \, a^{4} b d^{\frac {5}{2}} x^{3} + 11 \, a^{5} d^{\frac {5}{2}} x\right )} x^{\frac {5}{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int {\left (d\,x\right )}^{5/2}\,{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{5/2} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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