Integrand size = 22, antiderivative size = 63 \[ \int x^{5/2} \left (a+b x^2\right )^2 \left (c+d x^2\right ) \, dx=\frac {2}{7} a^2 c x^{7/2}+\frac {2}{11} a (2 b c+a d) x^{11/2}+\frac {2}{15} b (b c+2 a d) x^{15/2}+\frac {2}{19} b^2 d x^{19/2} \]
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Time = 0.02 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {459} \[ \int x^{5/2} \left (a+b x^2\right )^2 \left (c+d x^2\right ) \, dx=\frac {2}{7} a^2 c x^{7/2}+\frac {2}{15} b x^{15/2} (2 a d+b c)+\frac {2}{11} a x^{11/2} (a d+2 b c)+\frac {2}{19} b^2 d x^{19/2} \]
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Rule 459
Rubi steps \begin{align*} \text {integral}& = \int \left (a^2 c x^{5/2}+a (2 b c+a d) x^{9/2}+b (b c+2 a d) x^{13/2}+b^2 d x^{17/2}\right ) \, dx \\ & = \frac {2}{7} a^2 c x^{7/2}+\frac {2}{11} a (2 b c+a d) x^{11/2}+\frac {2}{15} b (b c+2 a d) x^{15/2}+\frac {2}{19} b^2 d x^{19/2} \\ \end{align*}
Time = 0.05 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.95 \[ \int x^{5/2} \left (a+b x^2\right )^2 \left (c+d x^2\right ) \, dx=\frac {2 x^{7/2} \left (285 a^2 \left (11 c+7 d x^2\right )+266 a b x^2 \left (15 c+11 d x^2\right )+77 b^2 x^4 \left (19 c+15 d x^2\right )\right )}{21945} \]
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Time = 2.75 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.83
method | result | size |
derivativedivides | \(\frac {2 b^{2} d \,x^{\frac {19}{2}}}{19}+\frac {2 \left (2 a b d +b^{2} c \right ) x^{\frac {15}{2}}}{15}+\frac {2 \left (a^{2} d +2 a b c \right ) x^{\frac {11}{2}}}{11}+\frac {2 a^{2} c \,x^{\frac {7}{2}}}{7}\) | \(52\) |
default | \(\frac {2 b^{2} d \,x^{\frac {19}{2}}}{19}+\frac {2 \left (2 a b d +b^{2} c \right ) x^{\frac {15}{2}}}{15}+\frac {2 \left (a^{2} d +2 a b c \right ) x^{\frac {11}{2}}}{11}+\frac {2 a^{2} c \,x^{\frac {7}{2}}}{7}\) | \(52\) |
gosper | \(\frac {2 x^{\frac {7}{2}} \left (1155 b^{2} d \,x^{6}+2926 a b d \,x^{4}+1463 b^{2} c \,x^{4}+1995 a^{2} d \,x^{2}+3990 a b c \,x^{2}+3135 a^{2} c \right )}{21945}\) | \(56\) |
trager | \(\frac {2 x^{\frac {7}{2}} \left (1155 b^{2} d \,x^{6}+2926 a b d \,x^{4}+1463 b^{2} c \,x^{4}+1995 a^{2} d \,x^{2}+3990 a b c \,x^{2}+3135 a^{2} c \right )}{21945}\) | \(56\) |
risch | \(\frac {2 x^{\frac {7}{2}} \left (1155 b^{2} d \,x^{6}+2926 a b d \,x^{4}+1463 b^{2} c \,x^{4}+1995 a^{2} d \,x^{2}+3990 a b c \,x^{2}+3135 a^{2} c \right )}{21945}\) | \(56\) |
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Time = 0.24 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.89 \[ \int x^{5/2} \left (a+b x^2\right )^2 \left (c+d x^2\right ) \, dx=\frac {2}{21945} \, {\left (1155 \, b^{2} d x^{9} + 1463 \, {\left (b^{2} c + 2 \, a b d\right )} x^{7} + 3135 \, a^{2} c x^{3} + 1995 \, {\left (2 \, a b c + a^{2} d\right )} x^{5}\right )} \sqrt {x} \]
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Time = 0.61 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.27 \[ \int x^{5/2} \left (a+b x^2\right )^2 \left (c+d x^2\right ) \, dx=\frac {2 a^{2} c x^{\frac {7}{2}}}{7} + \frac {2 a^{2} d x^{\frac {11}{2}}}{11} + \frac {4 a b c x^{\frac {11}{2}}}{11} + \frac {4 a b d x^{\frac {15}{2}}}{15} + \frac {2 b^{2} c x^{\frac {15}{2}}}{15} + \frac {2 b^{2} d x^{\frac {19}{2}}}{19} \]
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Time = 0.22 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.81 \[ \int x^{5/2} \left (a+b x^2\right )^2 \left (c+d x^2\right ) \, dx=\frac {2}{19} \, b^{2} d x^{\frac {19}{2}} + \frac {2}{15} \, {\left (b^{2} c + 2 \, a b d\right )} x^{\frac {15}{2}} + \frac {2}{7} \, a^{2} c x^{\frac {7}{2}} + \frac {2}{11} \, {\left (2 \, a b c + a^{2} d\right )} x^{\frac {11}{2}} \]
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Time = 0.28 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.84 \[ \int x^{5/2} \left (a+b x^2\right )^2 \left (c+d x^2\right ) \, dx=\frac {2}{19} \, b^{2} d x^{\frac {19}{2}} + \frac {2}{15} \, b^{2} c x^{\frac {15}{2}} + \frac {4}{15} \, a b d x^{\frac {15}{2}} + \frac {4}{11} \, a b c x^{\frac {11}{2}} + \frac {2}{11} \, a^{2} d x^{\frac {11}{2}} + \frac {2}{7} \, a^{2} c x^{\frac {7}{2}} \]
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Time = 0.05 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.81 \[ \int x^{5/2} \left (a+b x^2\right )^2 \left (c+d x^2\right ) \, dx=x^{11/2}\,\left (\frac {2\,d\,a^2}{11}+\frac {4\,b\,c\,a}{11}\right )+x^{15/2}\,\left (\frac {2\,c\,b^2}{15}+\frac {4\,a\,d\,b}{15}\right )+\frac {2\,a^2\,c\,x^{7/2}}{7}+\frac {2\,b^2\,d\,x^{19/2}}{19} \]
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