Integrand size = 24, antiderivative size = 92 \[ \int \frac {\left (a+b x^2\right )^2 \sqrt {c+d x^2}}{x} \, dx=a^2 \sqrt {c+d x^2}-\frac {b (b c-2 a d) \left (c+d x^2\right )^{3/2}}{3 d^2}+\frac {b^2 \left (c+d x^2\right )^{5/2}}{5 d^2}-a^2 \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right ) \]
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Time = 0.06 (sec) , antiderivative size = 92, 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, 90, 52, 65, 214} \[ \int \frac {\left (a+b x^2\right )^2 \sqrt {c+d x^2}}{x} \, dx=-a^2 \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )+a^2 \sqrt {c+d x^2}-\frac {b \left (c+d x^2\right )^{3/2} (b c-2 a d)}{3 d^2}+\frac {b^2 \left (c+d x^2\right )^{5/2}}{5 d^2} \]
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Rule 52
Rule 65
Rule 90
Rule 214
Rule 457
Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {(a+b x)^2 \sqrt {c+d x}}{x} \, dx,x,x^2\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \left (-\frac {b (b c-2 a d) \sqrt {c+d x}}{d}+\frac {a^2 \sqrt {c+d x}}{x}+\frac {b^2 (c+d x)^{3/2}}{d}\right ) \, dx,x,x^2\right ) \\ & = -\frac {b (b c-2 a d) \left (c+d x^2\right )^{3/2}}{3 d^2}+\frac {b^2 \left (c+d x^2\right )^{5/2}}{5 d^2}+\frac {1}{2} a^2 \text {Subst}\left (\int \frac {\sqrt {c+d x}}{x} \, dx,x,x^2\right ) \\ & = a^2 \sqrt {c+d x^2}-\frac {b (b c-2 a d) \left (c+d x^2\right )^{3/2}}{3 d^2}+\frac {b^2 \left (c+d x^2\right )^{5/2}}{5 d^2}+\frac {1}{2} \left (a^2 c\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {c+d x}} \, dx,x,x^2\right ) \\ & = a^2 \sqrt {c+d x^2}-\frac {b (b c-2 a d) \left (c+d x^2\right )^{3/2}}{3 d^2}+\frac {b^2 \left (c+d x^2\right )^{5/2}}{5 d^2}+\frac {\left (a^2 c\right ) \text {Subst}\left (\int \frac {1}{-\frac {c}{d}+\frac {x^2}{d}} \, dx,x,\sqrt {c+d x^2}\right )}{d} \\ & = a^2 \sqrt {c+d x^2}-\frac {b (b c-2 a d) \left (c+d x^2\right )^{3/2}}{3 d^2}+\frac {b^2 \left (c+d x^2\right )^{5/2}}{5 d^2}-a^2 \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right ) \\ \end{align*}
Time = 0.10 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+b x^2\right )^2 \sqrt {c+d x^2}}{x} \, dx=\frac {\sqrt {c+d x^2} \left (15 a^2 d^2+10 a b d \left (c+d x^2\right )+b^2 \left (-2 c^2+c d x^2+3 d^2 x^4\right )\right )}{15 d^2}-a^2 \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right ) \]
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Time = 2.90 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.93
method | result | size |
pseudoelliptic | \(\frac {-\sqrt {c}\, a^{2} d^{2} \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{2}+c}}{\sqrt {c}}\right )+\left (\left (\frac {1}{5} b^{2} x^{4}+\frac {2}{3} a b \,x^{2}+a^{2}\right ) d^{2}+\frac {2 b \left (\frac {b \,x^{2}}{10}+a \right ) c d}{3}-\frac {2 b^{2} c^{2}}{15}\right ) \sqrt {d \,x^{2}+c}}{d^{2}}\) | \(86\) |
default | \(b^{2} \left (\frac {x^{2} \left (d \,x^{2}+c \right )^{\frac {3}{2}}}{5 d}-\frac {2 c \left (d \,x^{2}+c \right )^{\frac {3}{2}}}{15 d^{2}}\right )+a^{2} \left (\sqrt {d \,x^{2}+c}-\sqrt {c}\, \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {d \,x^{2}+c}}{x}\right )\right )+\frac {2 a b \left (d \,x^{2}+c \right )^{\frac {3}{2}}}{3 d}\) | \(97\) |
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Time = 0.27 (sec) , antiderivative size = 207, normalized size of antiderivative = 2.25 \[ \int \frac {\left (a+b x^2\right )^2 \sqrt {c+d x^2}}{x} \, dx=\left [\frac {15 \, a^{2} \sqrt {c} d^{2} \log \left (-\frac {d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right ) + 2 \, {\left (3 \, b^{2} d^{2} x^{4} - 2 \, b^{2} c^{2} + 10 \, a b c d + 15 \, a^{2} d^{2} + {\left (b^{2} c d + 10 \, a b d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{30 \, d^{2}}, \frac {15 \, a^{2} \sqrt {-c} d^{2} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{2} + c}}\right ) + {\left (3 \, b^{2} d^{2} x^{4} - 2 \, b^{2} c^{2} + 10 \, a b c d + 15 \, a^{2} d^{2} + {\left (b^{2} c d + 10 \, a b d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{15 \, d^{2}}\right ] \]
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Time = 7.22 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.50 \[ \int \frac {\left (a+b x^2\right )^2 \sqrt {c+d x^2}}{x} \, dx=\frac {\begin {cases} \frac {2 a^{2} c \operatorname {atan}{\left (\frac {\sqrt {c + d x^{2}}}{\sqrt {- c}} \right )}}{\sqrt {- c}} + 2 a^{2} \sqrt {c + d x^{2}} + \frac {2 b^{2} \left (c + d x^{2}\right )^{\frac {5}{2}}}{5 d^{2}} + \frac {2 \left (c + d x^{2}\right )^{\frac {3}{2}} \cdot \left (2 a b d - b^{2} c\right )}{3 d^{2}} & \text {for}\: d \neq 0 \\a^{2} \sqrt {c} \log {\left (x^{2} \right )} + 2 a b \sqrt {c} x^{2} + \frac {b^{2} \sqrt {c} x^{4}}{2} & \text {otherwise} \end {cases}}{2} \]
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Time = 0.21 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.96 \[ \int \frac {\left (a+b x^2\right )^2 \sqrt {c+d x^2}}{x} \, dx=\frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} x^{2}}{5 \, d} - a^{2} \sqrt {c} \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right ) + \sqrt {d x^{2} + c} a^{2} - \frac {2 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} c}{15 \, d^{2}} + \frac {2 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a b}{3 \, d} \]
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Time = 0.30 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.10 \[ \int \frac {\left (a+b x^2\right )^2 \sqrt {c+d x^2}}{x} \, dx=\frac {a^{2} c \arctan \left (\frac {\sqrt {d x^{2} + c}}{\sqrt {-c}}\right )}{\sqrt {-c}} + \frac {3 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} b^{2} d^{8} - 5 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} c d^{8} + 10 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a b d^{9} + 15 \, \sqrt {d x^{2} + c} a^{2} d^{10}}{15 \, d^{10}} \]
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Time = 5.28 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.47 \[ \int \frac {\left (a+b x^2\right )^2 \sqrt {c+d x^2}}{x} \, dx=\sqrt {d\,x^2+c}\,\left (\frac {{\left (a\,d-b\,c\right )}^2}{d^2}-c\,\left (\frac {2\,b^2\,c-2\,a\,b\,d}{d^2}-\frac {b^2\,c}{d^2}\right )\right )-\left (\frac {2\,b^2\,c-2\,a\,b\,d}{3\,d^2}-\frac {b^2\,c}{3\,d^2}\right )\,{\left (d\,x^2+c\right )}^{3/2}+\frac {b^2\,{\left (d\,x^2+c\right )}^{5/2}}{5\,d^2}+a^2\,\sqrt {c}\,\mathrm {atan}\left (\frac {\sqrt {d\,x^2+c}\,1{}\mathrm {i}}{\sqrt {c}}\right )\,1{}\mathrm {i} \]
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