Integrand size = 22, antiderivative size = 145 \[ \int \frac {x^4 \left (a+b x^2\right )^2}{\left (c+d x^2\right )^2} \, dx=\frac {(7 b c-3 a d) (b c-a d) x}{2 d^4}-\frac {(7 b c-3 a d) (b c-a d) x^3}{6 c d^3}+\frac {b^2 x^5}{5 d^2}+\frac {(b c-a d)^2 x^5}{2 c d^2 \left (c+d x^2\right )}-\frac {\sqrt {c} (7 b c-3 a d) (b c-a d) \arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{2 d^{9/2}} \]
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Time = 0.10 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {474, 470, 308, 211} \[ \int \frac {x^4 \left (a+b x^2\right )^2}{\left (c+d x^2\right )^2} \, dx=-\frac {\sqrt {c} (7 b c-3 a d) (b c-a d) \arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{2 d^{9/2}}+\frac {x (7 b c-3 a d) (b c-a d)}{2 d^4}-\frac {x^3 (7 b c-3 a d) (b c-a d)}{6 c d^3}+\frac {x^5 (b c-a d)^2}{2 c d^2 \left (c+d x^2\right )}+\frac {b^2 x^5}{5 d^2} \]
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Rule 211
Rule 308
Rule 470
Rule 474
Rubi steps \begin{align*} \text {integral}& = \frac {(b c-a d)^2 x^5}{2 c d^2 \left (c+d x^2\right )}-\frac {\int \frac {x^4 \left (-2 a^2 d^2+5 (b c-a d)^2-2 b^2 c d x^2\right )}{c+d x^2} \, dx}{2 c d^2} \\ & = \frac {b^2 x^5}{5 d^2}+\frac {(b c-a d)^2 x^5}{2 c d^2 \left (c+d x^2\right )}-\frac {((7 b c-3 a d) (b c-a d)) \int \frac {x^4}{c+d x^2} \, dx}{2 c d^2} \\ & = \frac {b^2 x^5}{5 d^2}+\frac {(b c-a d)^2 x^5}{2 c d^2 \left (c+d x^2\right )}-\frac {((7 b c-3 a d) (b c-a d)) \int \left (-\frac {c}{d^2}+\frac {x^2}{d}+\frac {c^2}{d^2 \left (c+d x^2\right )}\right ) \, dx}{2 c d^2} \\ & = \frac {(7 b c-3 a d) (b c-a d) x}{2 d^4}-\frac {(7 b c-3 a d) (b c-a d) x^3}{6 c d^3}+\frac {b^2 x^5}{5 d^2}+\frac {(b c-a d)^2 x^5}{2 c d^2 \left (c+d x^2\right )}-\frac {(c (7 b c-3 a d) (b c-a d)) \int \frac {1}{c+d x^2} \, dx}{2 d^4} \\ & = \frac {(7 b c-3 a d) (b c-a d) x}{2 d^4}-\frac {(7 b c-3 a d) (b c-a d) x^3}{6 c d^3}+\frac {b^2 x^5}{5 d^2}+\frac {(b c-a d)^2 x^5}{2 c d^2 \left (c+d x^2\right )}-\frac {\sqrt {c} (7 b c-3 a d) (b c-a d) \tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{2 d^{9/2}} \\ \end{align*}
Time = 0.06 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.95 \[ \int \frac {x^4 \left (a+b x^2\right )^2}{\left (c+d x^2\right )^2} \, dx=\frac {\left (3 b^2 c^2-4 a b c d+a^2 d^2\right ) x}{d^4}-\frac {2 b (b c-a d) x^3}{3 d^3}+\frac {b^2 x^5}{5 d^2}+\frac {c (b c-a d)^2 x}{2 d^4 \left (c+d x^2\right )}-\frac {\sqrt {c} \left (7 b^2 c^2-10 a b c d+3 a^2 d^2\right ) \arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{2 d^{9/2}} \]
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Time = 2.68 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.97
method | result | size |
default | \(\frac {\frac {1}{5} b^{2} d^{2} x^{5}+\frac {2}{3} x^{3} a b \,d^{2}-\frac {2}{3} x^{3} b^{2} c d +a^{2} d^{2} x -4 a b c d x +3 b^{2} c^{2} x}{d^{4}}-\frac {c \left (\frac {\left (-\frac {1}{2} a^{2} d^{2}+a b c d -\frac {1}{2} b^{2} c^{2}\right ) x}{d \,x^{2}+c}+\frac {\left (3 a^{2} d^{2}-10 a b c d +7 b^{2} c^{2}\right ) \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{2 \sqrt {c d}}\right )}{d^{4}}\) | \(141\) |
risch | \(\frac {b^{2} x^{5}}{5 d^{2}}+\frac {2 x^{3} a b}{3 d^{2}}-\frac {2 x^{3} b^{2} c}{3 d^{3}}+\frac {a^{2} x}{d^{2}}-\frac {4 a b c x}{d^{3}}+\frac {3 b^{2} c^{2} x}{d^{4}}+\frac {\left (\frac {1}{2} c \,a^{2} d^{2}-a b \,c^{2} d +\frac {1}{2} b^{2} c^{3}\right ) x}{d^{4} \left (d \,x^{2}+c \right )}+\frac {3 \sqrt {-c d}\, \ln \left (-\sqrt {-c d}\, x -c \right ) a^{2}}{4 d^{3}}-\frac {5 \sqrt {-c d}\, \ln \left (-\sqrt {-c d}\, x -c \right ) a b c}{2 d^{4}}+\frac {7 \sqrt {-c d}\, \ln \left (-\sqrt {-c d}\, x -c \right ) b^{2} c^{2}}{4 d^{5}}-\frac {3 \sqrt {-c d}\, \ln \left (\sqrt {-c d}\, x -c \right ) a^{2}}{4 d^{3}}+\frac {5 \sqrt {-c d}\, \ln \left (\sqrt {-c d}\, x -c \right ) a b c}{2 d^{4}}-\frac {7 \sqrt {-c d}\, \ln \left (\sqrt {-c d}\, x -c \right ) b^{2} c^{2}}{4 d^{5}}\) | \(275\) |
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Time = 0.27 (sec) , antiderivative size = 400, normalized size of antiderivative = 2.76 \[ \int \frac {x^4 \left (a+b x^2\right )^2}{\left (c+d x^2\right )^2} \, dx=\left [\frac {12 \, b^{2} d^{3} x^{7} - 4 \, {\left (7 \, b^{2} c d^{2} - 10 \, a b d^{3}\right )} x^{5} + 20 \, {\left (7 \, b^{2} c^{2} d - 10 \, a b c d^{2} + 3 \, a^{2} d^{3}\right )} x^{3} + 15 \, {\left (7 \, b^{2} c^{3} - 10 \, a b c^{2} d + 3 \, a^{2} c d^{2} + {\left (7 \, b^{2} c^{2} d - 10 \, a b c d^{2} + 3 \, a^{2} d^{3}\right )} x^{2}\right )} \sqrt {-\frac {c}{d}} \log \left (\frac {d x^{2} - 2 \, d x \sqrt {-\frac {c}{d}} - c}{d x^{2} + c}\right ) + 30 \, {\left (7 \, b^{2} c^{3} - 10 \, a b c^{2} d + 3 \, a^{2} c d^{2}\right )} x}{60 \, {\left (d^{5} x^{2} + c d^{4}\right )}}, \frac {6 \, b^{2} d^{3} x^{7} - 2 \, {\left (7 \, b^{2} c d^{2} - 10 \, a b d^{3}\right )} x^{5} + 10 \, {\left (7 \, b^{2} c^{2} d - 10 \, a b c d^{2} + 3 \, a^{2} d^{3}\right )} x^{3} - 15 \, {\left (7 \, b^{2} c^{3} - 10 \, a b c^{2} d + 3 \, a^{2} c d^{2} + {\left (7 \, b^{2} c^{2} d - 10 \, a b c d^{2} + 3 \, a^{2} d^{3}\right )} x^{2}\right )} \sqrt {\frac {c}{d}} \arctan \left (\frac {d x \sqrt {\frac {c}{d}}}{c}\right ) + 15 \, {\left (7 \, b^{2} c^{3} - 10 \, a b c^{2} d + 3 \, a^{2} c d^{2}\right )} x}{30 \, {\left (d^{5} x^{2} + c d^{4}\right )}}\right ] \]
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Leaf count of result is larger than twice the leaf count of optimal. 286 vs. \(2 (133) = 266\).
Time = 0.55 (sec) , antiderivative size = 286, normalized size of antiderivative = 1.97 \[ \int \frac {x^4 \left (a+b x^2\right )^2}{\left (c+d x^2\right )^2} \, dx=\frac {b^{2} x^{5}}{5 d^{2}} + x^{3} \cdot \left (\frac {2 a b}{3 d^{2}} - \frac {2 b^{2} c}{3 d^{3}}\right ) + x \left (\frac {a^{2}}{d^{2}} - \frac {4 a b c}{d^{3}} + \frac {3 b^{2} c^{2}}{d^{4}}\right ) + \frac {x \left (a^{2} c d^{2} - 2 a b c^{2} d + b^{2} c^{3}\right )}{2 c d^{4} + 2 d^{5} x^{2}} + \frac {\sqrt {- \frac {c}{d^{9}}} \left (a d - b c\right ) \left (3 a d - 7 b c\right ) \log {\left (- \frac {d^{4} \sqrt {- \frac {c}{d^{9}}} \left (a d - b c\right ) \left (3 a d - 7 b c\right )}{3 a^{2} d^{2} - 10 a b c d + 7 b^{2} c^{2}} + x \right )}}{4} - \frac {\sqrt {- \frac {c}{d^{9}}} \left (a d - b c\right ) \left (3 a d - 7 b c\right ) \log {\left (\frac {d^{4} \sqrt {- \frac {c}{d^{9}}} \left (a d - b c\right ) \left (3 a d - 7 b c\right )}{3 a^{2} d^{2} - 10 a b c d + 7 b^{2} c^{2}} + x \right )}}{4} \]
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Time = 0.29 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.03 \[ \int \frac {x^4 \left (a+b x^2\right )^2}{\left (c+d x^2\right )^2} \, dx=\frac {{\left (b^{2} c^{3} - 2 \, a b c^{2} d + a^{2} c d^{2}\right )} x}{2 \, {\left (d^{5} x^{2} + c d^{4}\right )}} - \frac {{\left (7 \, b^{2} c^{3} - 10 \, a b c^{2} d + 3 \, a^{2} c d^{2}\right )} \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{2 \, \sqrt {c d} d^{4}} + \frac {3 \, b^{2} d^{2} x^{5} - 10 \, {\left (b^{2} c d - a b d^{2}\right )} x^{3} + 15 \, {\left (3 \, b^{2} c^{2} - 4 \, a b c d + a^{2} d^{2}\right )} x}{15 \, d^{4}} \]
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Time = 0.28 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.08 \[ \int \frac {x^4 \left (a+b x^2\right )^2}{\left (c+d x^2\right )^2} \, dx=-\frac {{\left (7 \, b^{2} c^{3} - 10 \, a b c^{2} d + 3 \, a^{2} c d^{2}\right )} \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{2 \, \sqrt {c d} d^{4}} + \frac {b^{2} c^{3} x - 2 \, a b c^{2} d x + a^{2} c d^{2} x}{2 \, {\left (d x^{2} + c\right )} d^{4}} + \frac {3 \, b^{2} d^{8} x^{5} - 10 \, b^{2} c d^{7} x^{3} + 10 \, a b d^{8} x^{3} + 45 \, b^{2} c^{2} d^{6} x - 60 \, a b c d^{7} x + 15 \, a^{2} d^{8} x}{15 \, d^{10}} \]
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Time = 0.09 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.38 \[ \int \frac {x^4 \left (a+b x^2\right )^2}{\left (c+d x^2\right )^2} \, dx=x\,\left (\frac {a^2}{d^2}+\frac {2\,c\,\left (\frac {2\,b^2\,c}{d^3}-\frac {2\,a\,b}{d^2}\right )}{d}-\frac {b^2\,c^2}{d^4}\right )-x^3\,\left (\frac {2\,b^2\,c}{3\,d^3}-\frac {2\,a\,b}{3\,d^2}\right )+\frac {b^2\,x^5}{5\,d^2}+\frac {x\,\left (\frac {a^2\,c\,d^2}{2}-a\,b\,c^2\,d+\frac {b^2\,c^3}{2}\right )}{d^5\,x^2+c\,d^4}-\frac {\sqrt {c}\,\mathrm {atan}\left (\frac {\sqrt {c}\,\sqrt {d}\,x\,\left (a\,d-b\,c\right )\,\left (3\,a\,d-7\,b\,c\right )}{3\,a^2\,c\,d^2-10\,a\,b\,c^2\,d+7\,b^2\,c^3}\right )\,\left (a\,d-b\,c\right )\,\left (3\,a\,d-7\,b\,c\right )}{2\,d^{9/2}} \]
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