Integrand size = 24, antiderivative size = 198 \[ \int \frac {x^3 \left (c+d x^2\right )^{5/2}}{\left (a+b x^2\right )^2} \, dx=\frac {(2 b c-7 a d) (b c-a d) \sqrt {c+d x^2}}{2 b^4}+\frac {(2 b c-7 a d) \left (c+d x^2\right )^{3/2}}{6 b^3}+\frac {(2 b c-7 a d) \left (c+d x^2\right )^{5/2}}{10 b^2 (b c-a d)}+\frac {a \left (c+d x^2\right )^{7/2}}{2 b (b c-a d) \left (a+b x^2\right )}-\frac {(2 b c-7 a d) (b c-a d)^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{2 b^{9/2}} \]
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Time = 0.13 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {457, 79, 52, 65, 214} \[ \int \frac {x^3 \left (c+d x^2\right )^{5/2}}{\left (a+b x^2\right )^2} \, dx=-\frac {(2 b c-7 a d) (b c-a d)^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{2 b^{9/2}}+\frac {\sqrt {c+d x^2} (2 b c-7 a d) (b c-a d)}{2 b^4}+\frac {\left (c+d x^2\right )^{3/2} (2 b c-7 a d)}{6 b^3}+\frac {\left (c+d x^2\right )^{5/2} (2 b c-7 a d)}{10 b^2 (b c-a d)}+\frac {a \left (c+d x^2\right )^{7/2}}{2 b \left (a+b x^2\right ) (b c-a d)} \]
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Rule 52
Rule 65
Rule 79
Rule 214
Rule 457
Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {x (c+d x)^{5/2}}{(a+b x)^2} \, dx,x,x^2\right ) \\ & = \frac {a \left (c+d x^2\right )^{7/2}}{2 b (b c-a d) \left (a+b x^2\right )}+\frac {(2 b c-7 a d) \text {Subst}\left (\int \frac {(c+d x)^{5/2}}{a+b x} \, dx,x,x^2\right )}{4 b (b c-a d)} \\ & = \frac {(2 b c-7 a d) \left (c+d x^2\right )^{5/2}}{10 b^2 (b c-a d)}+\frac {a \left (c+d x^2\right )^{7/2}}{2 b (b c-a d) \left (a+b x^2\right )}+\frac {(2 b c-7 a d) \text {Subst}\left (\int \frac {(c+d x)^{3/2}}{a+b x} \, dx,x,x^2\right )}{4 b^2} \\ & = \frac {(2 b c-7 a d) \left (c+d x^2\right )^{3/2}}{6 b^3}+\frac {(2 b c-7 a d) \left (c+d x^2\right )^{5/2}}{10 b^2 (b c-a d)}+\frac {a \left (c+d x^2\right )^{7/2}}{2 b (b c-a d) \left (a+b x^2\right )}+\frac {((2 b c-7 a d) (b c-a d)) \text {Subst}\left (\int \frac {\sqrt {c+d x}}{a+b x} \, dx,x,x^2\right )}{4 b^3} \\ & = \frac {(2 b c-7 a d) (b c-a d) \sqrt {c+d x^2}}{2 b^4}+\frac {(2 b c-7 a d) \left (c+d x^2\right )^{3/2}}{6 b^3}+\frac {(2 b c-7 a d) \left (c+d x^2\right )^{5/2}}{10 b^2 (b c-a d)}+\frac {a \left (c+d x^2\right )^{7/2}}{2 b (b c-a d) \left (a+b x^2\right )}+\frac {\left ((2 b c-7 a d) (b c-a d)^2\right ) \text {Subst}\left (\int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx,x,x^2\right )}{4 b^4} \\ & = \frac {(2 b c-7 a d) (b c-a d) \sqrt {c+d x^2}}{2 b^4}+\frac {(2 b c-7 a d) \left (c+d x^2\right )^{3/2}}{6 b^3}+\frac {(2 b c-7 a d) \left (c+d x^2\right )^{5/2}}{10 b^2 (b c-a d)}+\frac {a \left (c+d x^2\right )^{7/2}}{2 b (b c-a d) \left (a+b x^2\right )}+\frac {\left ((2 b c-7 a d) (b c-a d)^2\right ) \text {Subst}\left (\int \frac {1}{a-\frac {b c}{d}+\frac {b x^2}{d}} \, dx,x,\sqrt {c+d x^2}\right )}{2 b^4 d} \\ & = \frac {(2 b c-7 a d) (b c-a d) \sqrt {c+d x^2}}{2 b^4}+\frac {(2 b c-7 a d) \left (c+d x^2\right )^{3/2}}{6 b^3}+\frac {(2 b c-7 a d) \left (c+d x^2\right )^{5/2}}{10 b^2 (b c-a d)}+\frac {a \left (c+d x^2\right )^{7/2}}{2 b (b c-a d) \left (a+b x^2\right )}-\frac {(2 b c-7 a d) (b c-a d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{2 b^{9/2}} \\ \end{align*}
Time = 0.35 (sec) , antiderivative size = 183, normalized size of antiderivative = 0.92 \[ \int \frac {x^3 \left (c+d x^2\right )^{5/2}}{\left (a+b x^2\right )^2} \, dx=\frac {\sqrt {c+d x^2} \left (105 a^3 d^2+10 a^2 b d \left (-17 c+7 d x^2\right )+a b^2 \left (61 c^2-118 c d x^2-14 d^2 x^4\right )+2 b^3 x^2 \left (23 c^2+11 c d x^2+3 d^2 x^4\right )\right )}{30 b^4 \left (a+b x^2\right )}-\frac {\sqrt {-b c+a d} \left (2 b^2 c^2-9 a b c d+7 a^2 d^2\right ) \arctan \left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {-b c+a d}}\right )}{2 b^{9/2}} \]
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Time = 3.09 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.89
method | result | size |
pseudoelliptic | \(-\frac {7 \left (\left (b \,x^{2}+a \right ) \left (a d -\frac {2 b c}{7}\right ) \left (a d -b c \right )^{2} \arctan \left (\frac {b \sqrt {d \,x^{2}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )-\left (\frac {46 x^{2} \left (\frac {3}{23} d^{2} x^{4}+\frac {11}{23} c d \,x^{2}+c^{2}\right ) b^{3}}{105}+\frac {61 \left (-\frac {14}{61} d^{2} x^{4}-\frac {118}{61} c d \,x^{2}+c^{2}\right ) a \,b^{2}}{105}-\frac {34 \left (-\frac {7 d \,x^{2}}{17}+c \right ) d \,a^{2} b}{21}+a^{3} d^{2}\right ) \sqrt {d \,x^{2}+c}\, \sqrt {\left (a d -b c \right ) b}\right )}{2 \sqrt {\left (a d -b c \right ) b}\, b^{4} \left (b \,x^{2}+a \right )}\) | \(176\) |
risch | \(\text {Expression too large to display}\) | \(1036\) |
default | \(\text {Expression too large to display}\) | \(5284\) |
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Time = 0.30 (sec) , antiderivative size = 573, normalized size of antiderivative = 2.89 \[ \int \frac {x^3 \left (c+d x^2\right )^{5/2}}{\left (a+b x^2\right )^2} \, dx=\left [\frac {15 \, {\left (2 \, a b^{2} c^{2} - 9 \, a^{2} b c d + 7 \, a^{3} d^{2} + {\left (2 \, b^{3} c^{2} - 9 \, a b^{2} c d + 7 \, a^{2} b d^{2}\right )} x^{2}\right )} \sqrt {\frac {b c - a d}{b}} \log \left (\frac {b^{2} d^{2} x^{4} + 8 \, b^{2} c^{2} - 8 \, a b c d + a^{2} d^{2} + 2 \, {\left (4 \, b^{2} c d - 3 \, a b d^{2}\right )} x^{2} - 4 \, {\left (b^{2} d x^{2} + 2 \, b^{2} c - a b d\right )} \sqrt {d x^{2} + c} \sqrt {\frac {b c - a d}{b}}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right ) + 4 \, {\left (6 \, b^{3} d^{2} x^{6} + 61 \, a b^{2} c^{2} - 170 \, a^{2} b c d + 105 \, a^{3} d^{2} + 2 \, {\left (11 \, b^{3} c d - 7 \, a b^{2} d^{2}\right )} x^{4} + 2 \, {\left (23 \, b^{3} c^{2} - 59 \, a b^{2} c d + 35 \, a^{2} b d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{120 \, {\left (b^{5} x^{2} + a b^{4}\right )}}, -\frac {15 \, {\left (2 \, a b^{2} c^{2} - 9 \, a^{2} b c d + 7 \, a^{3} d^{2} + {\left (2 \, b^{3} c^{2} - 9 \, a b^{2} c d + 7 \, a^{2} b d^{2}\right )} x^{2}\right )} \sqrt {-\frac {b c - a d}{b}} \arctan \left (-\frac {{\left (b d x^{2} + 2 \, b c - a d\right )} \sqrt {d x^{2} + c} \sqrt {-\frac {b c - a d}{b}}}{2 \, {\left (b c^{2} - a c d + {\left (b c d - a d^{2}\right )} x^{2}\right )}}\right ) - 2 \, {\left (6 \, b^{3} d^{2} x^{6} + 61 \, a b^{2} c^{2} - 170 \, a^{2} b c d + 105 \, a^{3} d^{2} + 2 \, {\left (11 \, b^{3} c d - 7 \, a b^{2} d^{2}\right )} x^{4} + 2 \, {\left (23 \, b^{3} c^{2} - 59 \, a b^{2} c d + 35 \, a^{2} b d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{60 \, {\left (b^{5} x^{2} + a b^{4}\right )}}\right ] \]
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Timed out. \[ \int \frac {x^3 \left (c+d x^2\right )^{5/2}}{\left (a+b x^2\right )^2} \, dx=\text {Timed out} \]
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Exception generated. \[ \int \frac {x^3 \left (c+d x^2\right )^{5/2}}{\left (a+b x^2\right )^2} \, dx=\text {Exception raised: ValueError} \]
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Time = 0.30 (sec) , antiderivative size = 264, normalized size of antiderivative = 1.33 \[ \int \frac {x^3 \left (c+d x^2\right )^{5/2}}{\left (a+b x^2\right )^2} \, dx=\frac {{\left (2 \, b^{3} c^{3} - 11 \, a b^{2} c^{2} d + 16 \, a^{2} b c d^{2} - 7 \, a^{3} d^{3}\right )} \arctan \left (\frac {\sqrt {d x^{2} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{2 \, \sqrt {-b^{2} c + a b d} b^{4}} + \frac {\sqrt {d x^{2} + c} a b^{2} c^{2} d - 2 \, \sqrt {d x^{2} + c} a^{2} b c d^{2} + \sqrt {d x^{2} + c} a^{3} d^{3}}{2 \, {\left ({\left (d x^{2} + c\right )} b - b c + a d\right )} b^{4}} + \frac {3 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} b^{8} + 5 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{8} c + 15 \, \sqrt {d x^{2} + c} b^{8} c^{2} - 10 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a b^{7} d - 60 \, \sqrt {d x^{2} + c} a b^{7} c d + 45 \, \sqrt {d x^{2} + c} a^{2} b^{6} d^{2}}{15 \, b^{10}} \]
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Time = 5.81 (sec) , antiderivative size = 276, normalized size of antiderivative = 1.39 \[ \int \frac {x^3 \left (c+d x^2\right )^{5/2}}{\left (a+b x^2\right )^2} \, dx=\frac {{\left (d\,x^2+c\right )}^{5/2}}{5\,b^2}-\sqrt {d\,x^2+c}\,\left (\frac {{\left (a\,d-b\,c\right )}^2}{b^4}+\frac {\left (2\,b^2\,c-2\,a\,b\,d\right )\,\left (\frac {c}{b^2}-\frac {2\,b^2\,c-2\,a\,b\,d}{b^4}\right )}{b^2}\right )-{\left (d\,x^2+c\right )}^{3/2}\,\left (\frac {c}{3\,b^2}-\frac {2\,b^2\,c-2\,a\,b\,d}{3\,b^4}\right )+\frac {\sqrt {d\,x^2+c}\,\left (\frac {a^3\,d^3}{2}-a^2\,b\,c\,d^2+\frac {a\,b^2\,c^2\,d}{2}\right )}{b^5\,\left (d\,x^2+c\right )-b^5\,c+a\,b^4\,d}-\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {d\,x^2+c}\,{\left (a\,d-b\,c\right )}^{3/2}\,\left (7\,a\,d-2\,b\,c\right )}{7\,a^3\,d^3-16\,a^2\,b\,c\,d^2+11\,a\,b^2\,c^2\,d-2\,b^3\,c^3}\right )\,{\left (a\,d-b\,c\right )}^{3/2}\,\left (7\,a\,d-2\,b\,c\right )}{2\,b^{9/2}} \]
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