Integrand size = 28, antiderivative size = 149 \[ \int x^2 \left (a+b x^2\right )^3 \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {1}{3} a^3 A x^3+\frac {1}{4} a^3 B x^4+\frac {1}{5} a^2 (3 A b+a C) x^5+\frac {1}{6} a^2 (3 b B+a D) x^6+\frac {3}{7} a b (A b+a C) x^7+\frac {3}{8} a b (b B+a D) x^8+\frac {1}{9} b^2 (A b+3 a C) x^9+\frac {1}{10} b^2 (b B+3 a D) x^{10}+\frac {1}{11} b^3 C x^{11}+\frac {1}{12} b^3 D x^{12} \]
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Time = 0.09 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.036, Rules used = {1816} \[ \int x^2 \left (a+b x^2\right )^3 \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {1}{3} a^3 A x^3+\frac {1}{4} a^3 B x^4+\frac {1}{5} a^2 x^5 (a C+3 A b)+\frac {1}{6} a^2 x^6 (a D+3 b B)+\frac {1}{9} b^2 x^9 (3 a C+A b)+\frac {3}{7} a b x^7 (a C+A b)+\frac {1}{10} b^2 x^{10} (3 a D+b B)+\frac {3}{8} a b x^8 (a D+b B)+\frac {1}{11} b^3 C x^{11}+\frac {1}{12} b^3 D x^{12} \]
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Rule 1816
Rubi steps \begin{align*} \text {integral}& = \int \left (a^3 A x^2+a^3 B x^3+a^2 (3 A b+a C) x^4+a^2 (3 b B+a D) x^5+3 a b (A b+a C) x^6+3 a b (b B+a D) x^7+b^2 (A b+3 a C) x^8+b^2 (b B+3 a D) x^9+b^3 C x^{10}+b^3 D x^{11}\right ) \, dx \\ & = \frac {1}{3} a^3 A x^3+\frac {1}{4} a^3 B x^4+\frac {1}{5} a^2 (3 A b+a C) x^5+\frac {1}{6} a^2 (3 b B+a D) x^6+\frac {3}{7} a b (A b+a C) x^7+\frac {3}{8} a b (b B+a D) x^8+\frac {1}{9} b^2 (A b+3 a C) x^9+\frac {1}{10} b^2 (b B+3 a D) x^{10}+\frac {1}{11} b^3 C x^{11}+\frac {1}{12} b^3 D x^{12} \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.84 \[ \int x^2 \left (a+b x^2\right )^3 \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {14 b^3 x^9 \left (220 A+3 x \left (66 B+60 C x+55 D x^2\right )\right )+462 a^3 x^3 (20 A+x (15 B+2 x (6 C+5 D x)))+99 a^2 b x^5 (168 A+5 x (28 B+3 x (8 C+7 D x)))+33 a b^2 x^7 (360 A+7 x (45 B+4 x (10 C+9 D x)))}{27720} \]
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Time = 3.43 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.99
method | result | size |
norman | \(\frac {b^{3} D x^{12}}{12}+\frac {b^{3} C \,x^{11}}{11}+\left (\frac {1}{10} B \,b^{3}+\frac {3}{10} a \,b^{2} D\right ) x^{10}+\left (\frac {1}{9} b^{3} A +\frac {1}{3} C \,b^{2} a \right ) x^{9}+\left (\frac {3}{8} a \,b^{2} B +\frac {3}{8} D a^{2} b \right ) x^{8}+\left (\frac {3}{7} a \,b^{2} A +\frac {3}{7} C \,a^{2} b \right ) x^{7}+\left (\frac {1}{2} a^{2} b B +\frac {1}{6} D a^{3}\right ) x^{6}+\left (\frac {3}{5} a^{2} b A +\frac {1}{5} C \,a^{3}\right ) x^{5}+\frac {a^{3} B \,x^{4}}{4}+\frac {a^{3} A \,x^{3}}{3}\) | \(148\) |
default | \(\frac {b^{3} D x^{12}}{12}+\frac {b^{3} C \,x^{11}}{11}+\frac {\left (B \,b^{3}+3 a \,b^{2} D\right ) x^{10}}{10}+\frac {\left (b^{3} A +3 C \,b^{2} a \right ) x^{9}}{9}+\frac {\left (3 a \,b^{2} B +3 D a^{2} b \right ) x^{8}}{8}+\frac {\left (3 a \,b^{2} A +3 C \,a^{2} b \right ) x^{7}}{7}+\frac {\left (3 a^{2} b B +D a^{3}\right ) x^{6}}{6}+\frac {\left (3 a^{2} b A +C \,a^{3}\right ) x^{5}}{5}+\frac {a^{3} B \,x^{4}}{4}+\frac {a^{3} A \,x^{3}}{3}\) | \(150\) |
gosper | \(\frac {1}{12} b^{3} D x^{12}+\frac {1}{11} b^{3} C \,x^{11}+\frac {1}{10} x^{10} B \,b^{3}+\frac {3}{10} x^{10} a \,b^{2} D+\frac {1}{9} x^{9} b^{3} A +\frac {1}{3} x^{9} C \,b^{2} a +\frac {3}{8} x^{8} a \,b^{2} B +\frac {3}{8} x^{8} D a^{2} b +\frac {3}{7} x^{7} a \,b^{2} A +\frac {3}{7} x^{7} C \,a^{2} b +\frac {1}{2} x^{6} a^{2} b B +\frac {1}{6} x^{6} D a^{3}+\frac {3}{5} x^{5} a^{2} b A +\frac {1}{5} x^{5} C \,a^{3}+\frac {1}{4} a^{3} B \,x^{4}+\frac {1}{3} a^{3} A \,x^{3}\) | \(154\) |
parallelrisch | \(\frac {1}{12} b^{3} D x^{12}+\frac {1}{11} b^{3} C \,x^{11}+\frac {1}{10} x^{10} B \,b^{3}+\frac {3}{10} x^{10} a \,b^{2} D+\frac {1}{9} x^{9} b^{3} A +\frac {1}{3} x^{9} C \,b^{2} a +\frac {3}{8} x^{8} a \,b^{2} B +\frac {3}{8} x^{8} D a^{2} b +\frac {3}{7} x^{7} a \,b^{2} A +\frac {3}{7} x^{7} C \,a^{2} b +\frac {1}{2} x^{6} a^{2} b B +\frac {1}{6} x^{6} D a^{3}+\frac {3}{5} x^{5} a^{2} b A +\frac {1}{5} x^{5} C \,a^{3}+\frac {1}{4} a^{3} B \,x^{4}+\frac {1}{3} a^{3} A \,x^{3}\) | \(154\) |
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Time = 0.27 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.97 \[ \int x^2 \left (a+b x^2\right )^3 \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {1}{12} \, D b^{3} x^{12} + \frac {1}{11} \, C b^{3} x^{11} + \frac {1}{10} \, {\left (3 \, D a b^{2} + B b^{3}\right )} x^{10} + \frac {1}{9} \, {\left (3 \, C a b^{2} + A b^{3}\right )} x^{9} + \frac {3}{8} \, {\left (D a^{2} b + B a b^{2}\right )} x^{8} + \frac {1}{4} \, B a^{3} x^{4} + \frac {3}{7} \, {\left (C a^{2} b + A a b^{2}\right )} x^{7} + \frac {1}{3} \, A a^{3} x^{3} + \frac {1}{6} \, {\left (D a^{3} + 3 \, B a^{2} b\right )} x^{6} + \frac {1}{5} \, {\left (C a^{3} + 3 \, A a^{2} b\right )} x^{5} \]
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Time = 0.03 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.11 \[ \int x^2 \left (a+b x^2\right )^3 \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {A a^{3} x^{3}}{3} + \frac {B a^{3} x^{4}}{4} + \frac {C b^{3} x^{11}}{11} + \frac {D b^{3} x^{12}}{12} + x^{10} \left (\frac {B b^{3}}{10} + \frac {3 D a b^{2}}{10}\right ) + x^{9} \left (\frac {A b^{3}}{9} + \frac {C a b^{2}}{3}\right ) + x^{8} \cdot \left (\frac {3 B a b^{2}}{8} + \frac {3 D a^{2} b}{8}\right ) + x^{7} \cdot \left (\frac {3 A a b^{2}}{7} + \frac {3 C a^{2} b}{7}\right ) + x^{6} \left (\frac {B a^{2} b}{2} + \frac {D a^{3}}{6}\right ) + x^{5} \cdot \left (\frac {3 A a^{2} b}{5} + \frac {C a^{3}}{5}\right ) \]
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Time = 0.20 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.97 \[ \int x^2 \left (a+b x^2\right )^3 \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {1}{12} \, D b^{3} x^{12} + \frac {1}{11} \, C b^{3} x^{11} + \frac {1}{10} \, {\left (3 \, D a b^{2} + B b^{3}\right )} x^{10} + \frac {1}{9} \, {\left (3 \, C a b^{2} + A b^{3}\right )} x^{9} + \frac {3}{8} \, {\left (D a^{2} b + B a b^{2}\right )} x^{8} + \frac {1}{4} \, B a^{3} x^{4} + \frac {3}{7} \, {\left (C a^{2} b + A a b^{2}\right )} x^{7} + \frac {1}{3} \, A a^{3} x^{3} + \frac {1}{6} \, {\left (D a^{3} + 3 \, B a^{2} b\right )} x^{6} + \frac {1}{5} \, {\left (C a^{3} + 3 \, A a^{2} b\right )} x^{5} \]
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Time = 0.30 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.03 \[ \int x^2 \left (a+b x^2\right )^3 \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {1}{12} \, D b^{3} x^{12} + \frac {1}{11} \, C b^{3} x^{11} + \frac {3}{10} \, D a b^{2} x^{10} + \frac {1}{10} \, B b^{3} x^{10} + \frac {1}{3} \, C a b^{2} x^{9} + \frac {1}{9} \, A b^{3} x^{9} + \frac {3}{8} \, D a^{2} b x^{8} + \frac {3}{8} \, B a b^{2} x^{8} + \frac {3}{7} \, C a^{2} b x^{7} + \frac {3}{7} \, A a b^{2} x^{7} + \frac {1}{6} \, D a^{3} x^{6} + \frac {1}{2} \, B a^{2} b x^{6} + \frac {1}{5} \, C a^{3} x^{5} + \frac {3}{5} \, A a^{2} b x^{5} + \frac {1}{4} \, B a^{3} x^{4} + \frac {1}{3} \, A a^{3} x^{3} \]
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Time = 5.85 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.03 \[ \int x^2 \left (a+b x^2\right )^3 \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {A\,a^3\,x^3}{3}+\frac {B\,a^3\,x^4}{4}+\frac {A\,b^3\,x^9}{9}+\frac {C\,a^3\,x^5}{5}+\frac {B\,b^3\,x^{10}}{10}+\frac {C\,b^3\,x^{11}}{11}+\frac {a^3\,x^6\,D}{6}+\frac {b^3\,x^{12}\,D}{12}+\frac {3\,a^2\,b\,x^8\,D}{8}+\frac {3\,a\,b^2\,x^{10}\,D}{10}+\frac {3\,A\,a^2\,b\,x^5}{5}+\frac {3\,A\,a\,b^2\,x^7}{7}+\frac {B\,a^2\,b\,x^6}{2}+\frac {3\,B\,a\,b^2\,x^8}{8}+\frac {3\,C\,a^2\,b\,x^7}{7}+\frac {C\,a\,b^2\,x^9}{3} \]
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