Integrand size = 24, antiderivative size = 115 \[ \int \frac {x^3 \left (c+d x^2\right )^{3/2}}{a+b x^2} \, dx=-\frac {a (b c-a d) \sqrt {c+d x^2}}{b^3}-\frac {a \left (c+d x^2\right )^{3/2}}{3 b^2}+\frac {\left (c+d x^2\right )^{5/2}}{5 b d}+\frac {a (b c-a d)^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{b^{7/2}} \]
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Time = 0.08 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {457, 81, 52, 65, 214} \[ \int \frac {x^3 \left (c+d x^2\right )^{3/2}}{a+b x^2} \, dx=\frac {a (b c-a d)^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{b^{7/2}}-\frac {a \sqrt {c+d x^2} (b c-a d)}{b^3}-\frac {a \left (c+d x^2\right )^{3/2}}{3 b^2}+\frac {\left (c+d x^2\right )^{5/2}}{5 b d} \]
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Rule 52
Rule 65
Rule 81
Rule 214
Rule 457
Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {x (c+d x)^{3/2}}{a+b x} \, dx,x,x^2\right ) \\ & = \frac {\left (c+d x^2\right )^{5/2}}{5 b d}-\frac {a \text {Subst}\left (\int \frac {(c+d x)^{3/2}}{a+b x} \, dx,x,x^2\right )}{2 b} \\ & = -\frac {a \left (c+d x^2\right )^{3/2}}{3 b^2}+\frac {\left (c+d x^2\right )^{5/2}}{5 b d}-\frac {(a (b c-a d)) \text {Subst}\left (\int \frac {\sqrt {c+d x}}{a+b x} \, dx,x,x^2\right )}{2 b^2} \\ & = -\frac {a (b c-a d) \sqrt {c+d x^2}}{b^3}-\frac {a \left (c+d x^2\right )^{3/2}}{3 b^2}+\frac {\left (c+d x^2\right )^{5/2}}{5 b d}-\frac {\left (a (b c-a d)^2\right ) \text {Subst}\left (\int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx,x,x^2\right )}{2 b^3} \\ & = -\frac {a (b c-a d) \sqrt {c+d x^2}}{b^3}-\frac {a \left (c+d x^2\right )^{3/2}}{3 b^2}+\frac {\left (c+d x^2\right )^{5/2}}{5 b d}-\frac {\left (a (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 )}{b^3 d} \\ & = -\frac {a (b c-a d) \sqrt {c+d x^2}}{b^3}-\frac {a \left (c+d x^2\right )^{3/2}}{3 b^2}+\frac {\left (c+d x^2\right )^{5/2}}{5 b d}+\frac {a (b c-a d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{b^{7/2}} \\ \end{align*}
Time = 0.20 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.95 \[ \int \frac {x^3 \left (c+d x^2\right )^{3/2}}{a+b x^2} \, dx=\frac {\sqrt {c+d x^2} \left (15 a^2 d^2+3 b^2 \left (c+d x^2\right )^2-5 a b d \left (4 c+d x^2\right )\right )}{15 b^3 d}-\frac {a (-b c+a d)^{3/2} \arctan \left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {-b c+a d}}\right )}{b^{7/2}} \]
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Time = 3.09 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.02
method | result | size |
pseudoelliptic | \(-\frac {-\left (\frac {b^{2} \left (d \,x^{2}+c \right )^{2}}{5}-\frac {4 \left (\frac {d \,x^{2}}{4}+c \right ) d a b}{3}+a^{2} d^{2}\right ) \sqrt {\left (a d -b c \right ) b}\, \sqrt {d \,x^{2}+c}+a d \left (a d -b c \right )^{2} \arctan \left (\frac {b \sqrt {d \,x^{2}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{\sqrt {\left (a d -b c \right ) b}\, d \,b^{3}}\) | \(117\) |
risch | \(\frac {\left (3 b^{2} d^{2} x^{4}-5 x^{2} a b \,d^{2}+6 x^{2} b^{2} c d +15 a^{2} d^{2}-20 a b c d +3 b^{2} c^{2}\right ) \sqrt {d \,x^{2}+c}}{15 d \,b^{3}}-\frac {a \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \left (-\frac {\ln \left (\frac {-\frac {2 \left (a d -b c \right )}{b}+\frac {2 d \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {d \left (x -\frac {\sqrt {-a b}}{b}\right )^{2}+\frac {2 d \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}{b}-\frac {a d -b c}{b}}}{x -\frac {\sqrt {-a b}}{b}}\right )}{2 b \sqrt {-\frac {a d -b c}{b}}}-\frac {\ln \left (\frac {-\frac {2 \left (a d -b c \right )}{b}-\frac {2 d \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {d \left (x +\frac {\sqrt {-a b}}{b}\right )^{2}-\frac {2 d \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}{b}-\frac {a d -b c}{b}}}{x +\frac {\sqrt {-a b}}{b}}\right )}{2 b \sqrt {-\frac {a d -b c}{b}}}\right )}{b^{3}}\) | \(399\) |
default | \(\text {Expression too large to display}\) | \(1256\) |
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Time = 0.29 (sec) , antiderivative size = 397, normalized size of antiderivative = 3.45 \[ \int \frac {x^3 \left (c+d x^2\right )^{3/2}}{a+b x^2} \, dx=\left [-\frac {15 \, {\left (a b c d - a^{2} d^{2}\right )} \sqrt {\frac {b c - a d}{b}} \log \left (\frac {b^{2} d^{2} x^{4} + 8 \, b^{2} c^{2} - 8 \, a b c d + a^{2} d^{2} + 2 \, {\left (4 \, b^{2} c d - 3 \, a b d^{2}\right )} x^{2} - 4 \, {\left (b^{2} d x^{2} + 2 \, b^{2} c - a b d\right )} \sqrt {d x^{2} + c} \sqrt {\frac {b c - a d}{b}}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right ) - 4 \, {\left (3 \, b^{2} d^{2} x^{4} + 3 \, b^{2} c^{2} - 20 \, a b c d + 15 \, a^{2} d^{2} + {\left (6 \, b^{2} c d - 5 \, a b d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{60 \, b^{3} d}, \frac {15 \, {\left (a b c d - a^{2} d^{2}\right )} \sqrt {-\frac {b c - a d}{b}} \arctan \left (-\frac {{\left (b d x^{2} + 2 \, b c - a d\right )} \sqrt {d x^{2} + c} \sqrt {-\frac {b c - a d}{b}}}{2 \, {\left (b c^{2} - a c d + {\left (b c d - a d^{2}\right )} x^{2}\right )}}\right ) + 2 \, {\left (3 \, b^{2} d^{2} x^{4} + 3 \, b^{2} c^{2} - 20 \, a b c d + 15 \, a^{2} d^{2} + {\left (6 \, b^{2} c d - 5 \, a b d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{30 \, b^{3} d}\right ] \]
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Time = 7.89 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.24 \[ \int \frac {x^3 \left (c+d x^2\right )^{3/2}}{a+b x^2} \, dx=\begin {cases} \frac {2 \left (- \frac {a d \left (c + d x^{2}\right )^{\frac {3}{2}}}{6 b^{2}} - \frac {a d \left (a d - b c\right )^{2} \operatorname {atan}{\left (\frac {\sqrt {c + d x^{2}}}{\sqrt {\frac {a d - b c}{b}}} \right )}}{2 b^{4} \sqrt {\frac {a d - b c}{b}}} + \frac {\left (c + d x^{2}\right )^{\frac {5}{2}}}{10 b} + \frac {\sqrt {c + d x^{2}} \left (a^{2} d^{2} - a b c d\right )}{2 b^{3}}\right )}{d} & \text {for}\: d \neq 0 \\c^{\frac {3}{2}} \left (- \frac {a \left (\begin {cases} \frac {x^{2}}{a} & \text {for}\: b = 0 \\\frac {\log {\left (a + b x^{2} \right )}}{b} & \text {otherwise} \end {cases}\right )}{2 b} + \frac {x^{2}}{2 b}\right ) & \text {otherwise} \end {cases} \]
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Exception generated. \[ \int \frac {x^3 \left (c+d x^2\right )^{3/2}}{a+b x^2} \, dx=\text {Exception raised: ValueError} \]
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Time = 0.28 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.31 \[ \int \frac {x^3 \left (c+d x^2\right )^{3/2}}{a+b x^2} \, dx=-\frac {{\left (a b^{2} c^{2} - 2 \, a^{2} b c d + a^{3} d^{2}\right )} \arctan \left (\frac {\sqrt {d x^{2} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{\sqrt {-b^{2} c + a b d} b^{3}} + \frac {3 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} b^{4} d^{4} - 5 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a b^{3} d^{5} - 15 \, \sqrt {d x^{2} + c} a b^{3} c d^{5} + 15 \, \sqrt {d x^{2} + c} a^{2} b^{2} d^{6}}{15 \, b^{5} d^{5}} \]
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Time = 5.29 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.56 \[ \int \frac {x^3 \left (c+d x^2\right )^{3/2}}{a+b x^2} \, dx=\frac {{\left (d\,x^2+c\right )}^{5/2}}{5\,b\,d}-{\left (d\,x^2+c\right )}^{3/2}\,\left (\frac {c}{3\,b\,d}+\frac {a\,d^2-b\,c\,d}{3\,b^2\,d^2}\right )-\frac {a\,\mathrm {atan}\left (\frac {a\,\sqrt {b}\,\sqrt {d\,x^2+c}\,{\left (a\,d-b\,c\right )}^{3/2}}{a^3\,d^2-2\,a^2\,b\,c\,d+a\,b^2\,c^2}\right )\,{\left (a\,d-b\,c\right )}^{3/2}}{b^{7/2}}+\frac {\sqrt {d\,x^2+c}\,\left (a\,d^2-b\,c\,d\right )\,\left (\frac {c}{b\,d}+\frac {a\,d^2-b\,c\,d}{b^2\,d^2}\right )}{b\,d} \]
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