Integrand size = 24, antiderivative size = 195 \[ \int \frac {x^2 \left (c+d x^2\right )^{5/2}}{\left (a+b x^2\right )^2} \, dx=\frac {d (11 b c-12 a d) x \sqrt {c+d x^2}}{8 b^3}+\frac {3 d x \left (c+d x^2\right )^{3/2}}{4 b^2}-\frac {x \left (c+d x^2\right )^{5/2}}{2 b \left (a+b x^2\right )}+\frac {(b c-6 a d) (b c-a d)^{3/2} \arctan \left (\frac {\sqrt {b c-a d} x}{\sqrt {a} \sqrt {c+d x^2}}\right )}{2 \sqrt {a} b^4}+\frac {\sqrt {d} \left (15 b^2 c^2-40 a b c d+24 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{8 b^4} \]
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Time = 0.17 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {478, 542, 537, 223, 212, 385, 211} \[ \int \frac {x^2 \left (c+d x^2\right )^{5/2}}{\left (a+b x^2\right )^2} \, dx=\frac {\sqrt {d} \left (24 a^2 d^2-40 a b c d+15 b^2 c^2\right ) \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{8 b^4}+\frac {(b c-6 a d) (b c-a d)^{3/2} \arctan \left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{2 \sqrt {a} b^4}+\frac {d x \sqrt {c+d x^2} (11 b c-12 a d)}{8 b^3}-\frac {x \left (c+d x^2\right )^{5/2}}{2 b \left (a+b x^2\right )}+\frac {3 d x \left (c+d x^2\right )^{3/2}}{4 b^2} \]
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Rule 211
Rule 212
Rule 223
Rule 385
Rule 478
Rule 537
Rule 542
Rubi steps \begin{align*} \text {integral}& = -\frac {x \left (c+d x^2\right )^{5/2}}{2 b \left (a+b x^2\right )}+\frac {\int \frac {\left (c+d x^2\right )^{3/2} \left (c+6 d x^2\right )}{a+b x^2} \, dx}{2 b} \\ & = \frac {3 d x \left (c+d x^2\right )^{3/2}}{4 b^2}-\frac {x \left (c+d x^2\right )^{5/2}}{2 b \left (a+b x^2\right )}+\frac {\int \frac {\sqrt {c+d x^2} \left (2 c (2 b c-3 a d)+2 d (11 b c-12 a d) x^2\right )}{a+b x^2} \, dx}{8 b^2} \\ & = \frac {d (11 b c-12 a d) x \sqrt {c+d x^2}}{8 b^3}+\frac {3 d x \left (c+d x^2\right )^{3/2}}{4 b^2}-\frac {x \left (c+d x^2\right )^{5/2}}{2 b \left (a+b x^2\right )}+\frac {\int \frac {2 c \left (4 b^2 c^2-17 a b c d+12 a^2 d^2\right )+2 d \left (15 b^2 c^2-40 a b c d+24 a^2 d^2\right ) x^2}{\left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx}{16 b^3} \\ & = \frac {d (11 b c-12 a d) x \sqrt {c+d x^2}}{8 b^3}+\frac {3 d x \left (c+d x^2\right )^{3/2}}{4 b^2}-\frac {x \left (c+d x^2\right )^{5/2}}{2 b \left (a+b x^2\right )}+\frac {\left ((b c-6 a d) (b c-a d)^2\right ) \int \frac {1}{\left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx}{2 b^4}+\frac {\left (d \left (15 b^2 c^2-40 a b c d+24 a^2 d^2\right )\right ) \int \frac {1}{\sqrt {c+d x^2}} \, dx}{8 b^4} \\ & = \frac {d (11 b c-12 a d) x \sqrt {c+d x^2}}{8 b^3}+\frac {3 d x \left (c+d x^2\right )^{3/2}}{4 b^2}-\frac {x \left (c+d x^2\right )^{5/2}}{2 b \left (a+b x^2\right )}+\frac {\left ((b c-6 a d) (b c-a d)^2\right ) \text {Subst}\left (\int \frac {1}{a-(-b c+a d) x^2} \, dx,x,\frac {x}{\sqrt {c+d x^2}}\right )}{2 b^4}+\frac {\left (d \left (15 b^2 c^2-40 a b c d+24 a^2 d^2\right )\right ) \text {Subst}\left (\int \frac {1}{1-d x^2} \, dx,x,\frac {x}{\sqrt {c+d x^2}}\right )}{8 b^4} \\ & = \frac {d (11 b c-12 a d) x \sqrt {c+d x^2}}{8 b^3}+\frac {3 d x \left (c+d x^2\right )^{3/2}}{4 b^2}-\frac {x \left (c+d x^2\right )^{5/2}}{2 b \left (a+b x^2\right )}+\frac {(b c-6 a d) (b c-a d)^{3/2} \tan ^{-1}\left (\frac {\sqrt {b c-a d} x}{\sqrt {a} \sqrt {c+d x^2}}\right )}{2 \sqrt {a} b^4}+\frac {\sqrt {d} \left (15 b^2 c^2-40 a b c d+24 a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{8 b^4} \\ \end{align*}
Time = 10.15 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.89 \[ \int \frac {x^2 \left (c+d x^2\right )^{5/2}}{\left (a+b x^2\right )^2} \, dx=\frac {b x \sqrt {c+d x^2} \left (d (9 b c-8 a d)+2 b d^2 x^2-\frac {4 (b c-a d)^2}{a+b x^2}\right )+\frac {4 (b c-6 a d) (b c-a d)^{3/2} \arctan \left (\frac {\sqrt {b c-a d} x}{\sqrt {a} \sqrt {c+d x^2}}\right )}{\sqrt {a}}+\sqrt {d} \left (15 b^2 c^2-40 a b c d+24 a^2 d^2\right ) \log \left (d x+\sqrt {d} \sqrt {c+d x^2}\right )}{8 b^4} \]
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Time = 3.25 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.87
method | result | size |
pseudoelliptic | \(-\frac {\frac {d \left (b \sqrt {d \,x^{2}+c}\, \left (-2 b d \,x^{2}+8 a d -9 b c \right ) x -\frac {\left (24 a^{2} d^{2}-40 a b c d +15 b^{2} c^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{2}+c}}{x \sqrt {d}}\right )}{\sqrt {d}}\right )}{2}-2 \left (a d -b c \right )^{2} \left (-\frac {b \sqrt {d \,x^{2}+c}\, x}{b \,x^{2}+a}-\frac {\left (6 a d -b c \right ) \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{2}+c}\, a}{x \sqrt {\left (a d -b c \right ) a}}\right )}{\sqrt {\left (a d -b c \right ) a}}\right )}{4 b^{4}}\) | \(169\) |
risch | \(\text {Expression too large to display}\) | \(1046\) |
default | \(\text {Expression too large to display}\) | \(5284\) |
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Time = 0.86 (sec) , antiderivative size = 1379, normalized size of antiderivative = 7.07 \[ \int \frac {x^2 \left (c+d x^2\right )^{5/2}}{\left (a+b x^2\right )^2} \, dx=\text {Too large to display} \]
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\[ \int \frac {x^2 \left (c+d x^2\right )^{5/2}}{\left (a+b x^2\right )^2} \, dx=\int \frac {x^{2} \left (c + d x^{2}\right )^{\frac {5}{2}}}{\left (a + b x^{2}\right )^{2}}\, dx \]
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\[ \int \frac {x^2 \left (c+d x^2\right )^{5/2}}{\left (a+b x^2\right )^2} \, dx=\int { \frac {{\left (d x^{2} + c\right )}^{\frac {5}{2}} x^{2}}{{\left (b x^{2} + a\right )}^{2}} \,d x } \]
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Leaf count of result is larger than twice the leaf count of optimal. 446 vs. \(2 (163) = 326\).
Time = 0.33 (sec) , antiderivative size = 446, normalized size of antiderivative = 2.29 \[ \int \frac {x^2 \left (c+d x^2\right )^{5/2}}{\left (a+b x^2\right )^2} \, dx=\frac {1}{8} \, \sqrt {d x^{2} + c} {\left (\frac {2 \, d^{2} x^{2}}{b^{2}} + \frac {9 \, b^{7} c d^{3} - 8 \, a b^{6} d^{4}}{b^{9} d^{2}}\right )} x - \frac {{\left (15 \, b^{2} c^{2} \sqrt {d} - 40 \, a b c d^{\frac {3}{2}} + 24 \, a^{2} d^{\frac {5}{2}}\right )} \log \left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2}\right )}{16 \, b^{4}} - \frac {{\left (b^{3} c^{3} \sqrt {d} - 8 \, a b^{2} c^{2} d^{\frac {3}{2}} + 13 \, a^{2} b c d^{\frac {5}{2}} - 6 \, a^{3} d^{\frac {7}{2}}\right )} \arctan \left (\frac {{\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b - b c + 2 \, a d}{2 \, \sqrt {a b c d - a^{2} d^{2}}}\right )}{2 \, \sqrt {a b c d - a^{2} d^{2}} b^{4}} + \frac {{\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b^{3} c^{3} \sqrt {d} - 4 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a b^{2} c^{2} d^{\frac {3}{2}} + 5 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a^{2} b c d^{\frac {5}{2}} - 2 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a^{3} d^{\frac {7}{2}} - b^{3} c^{4} \sqrt {d} + 2 \, a b^{2} c^{3} d^{\frac {3}{2}} - a^{2} b c^{2} d^{\frac {5}{2}}}{{\left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} b - 2 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b c + 4 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a d + b c^{2}\right )} b^{4}} \]
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Timed out. \[ \int \frac {x^2 \left (c+d x^2\right )^{5/2}}{\left (a+b x^2\right )^2} \, dx=\int \frac {x^2\,{\left (d\,x^2+c\right )}^{5/2}}{{\left (b\,x^2+a\right )}^2} \,d x \]
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