Integrand size = 29, antiderivative size = 84 \[ \int \frac {\left (a+b x^2\right ) \left (A+B x^2\right ) \sqrt {c+d x^2}}{x} \, dx=a A \sqrt {c+d x^2}-\frac {\left (c+d x^2\right )^{3/2} \left (2 b B c-5 (A b+a B) d-3 b B d x^2\right )}{15 d^2}-a A \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right ) \]
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Time = 0.06 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {587, 152, 52, 65, 214} \[ \int \frac {\left (a+b x^2\right ) \left (A+B x^2\right ) \sqrt {c+d x^2}}{x} \, dx=-a A \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )-\frac {\left (c+d x^2\right )^{3/2} \left (-5 d (a B+A b)+2 b B c-3 b B d x^2\right )}{15 d^2}+a A \sqrt {c+d x^2} \]
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Rule 52
Rule 65
Rule 152
Rule 214
Rule 587
Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {(a+b x) (A+B x) \sqrt {c+d x}}{x} \, dx,x,x^2\right ) \\ & = -\frac {\left (c+d x^2\right )^{3/2} \left (2 b B c-5 (A b+a B) d-3 b B d x^2\right )}{15 d^2}+\frac {1}{2} (a A) \text {Subst}\left (\int \frac {\sqrt {c+d x}}{x} \, dx,x,x^2\right ) \\ & = a A \sqrt {c+d x^2}-\frac {\left (c+d x^2\right )^{3/2} \left (2 b B c-5 (A b+a B) d-3 b B d x^2\right )}{15 d^2}+\frac {1}{2} (a A c) \text {Subst}\left (\int \frac {1}{x \sqrt {c+d x}} \, dx,x,x^2\right ) \\ & = a A \sqrt {c+d x^2}-\frac {\left (c+d x^2\right )^{3/2} \left (2 b B c-5 (A b+a B) d-3 b B d x^2\right )}{15 d^2}+\frac {(a A c) \text {Subst}\left (\int \frac {1}{-\frac {c}{d}+\frac {x^2}{d}} \, dx,x,\sqrt {c+d x^2}\right )}{d} \\ & = a A \sqrt {c+d x^2}-\frac {\left (c+d x^2\right )^{3/2} \left (2 b B c-5 (A b+a B) d-3 b B d x^2\right )}{15 d^2}-a A \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right ) \\ \end{align*}
Time = 0.12 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.08 \[ \int \frac {\left (a+b x^2\right ) \left (A+B x^2\right ) \sqrt {c+d x^2}}{x} \, dx=\frac {\sqrt {c+d x^2} \left (-b \left (c+d x^2\right ) \left (2 B c-5 A d-3 B d x^2\right )+5 a d \left (3 A d+B \left (c+d x^2\right )\right )\right )}{15 d^2}-a A \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right ) \]
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Time = 3.41 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.13
method | result | size |
pseudoelliptic | \(\frac {-A a \sqrt {c}\, d^{2} \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{2}+c}}{\sqrt {c}}\right )+\sqrt {d \,x^{2}+c}\, \left (\left (\frac {\left (\frac {3 b \,x^{2}}{5}+a \right ) x^{2} B}{3}+A \left (\frac {b \,x^{2}}{3}+a \right )\right ) d^{2}+\frac {c \left (\left (\frac {b \,x^{2}}{5}+a \right ) B +A b \right ) d}{3}-\frac {2 B b \,c^{2}}{15}\right )}{d^{2}}\) | \(95\) |
default | \(B b \left (\frac {x^{2} \left (d \,x^{2}+c \right )^{\frac {3}{2}}}{5 d}-\frac {2 c \left (d \,x^{2}+c \right )^{\frac {3}{2}}}{15 d^{2}}\right )+\frac {A b \left (d \,x^{2}+c \right )^{\frac {3}{2}}}{3 d}+\frac {B a \left (d \,x^{2}+c \right )^{\frac {3}{2}}}{3 d}+A a \left (\sqrt {d \,x^{2}+c}-\sqrt {c}\, \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {d \,x^{2}+c}}{x}\right )\right )\) | \(111\) |
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Time = 0.26 (sec) , antiderivative size = 217, normalized size of antiderivative = 2.58 \[ \int \frac {\left (a+b x^2\right ) \left (A+B x^2\right ) \sqrt {c+d x^2}}{x} \, dx=\left [\frac {15 \, A a \sqrt {c} d^{2} \log \left (-\frac {d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right ) + 2 \, {\left (3 \, B b d^{2} x^{4} - 2 \, B b c^{2} + 15 \, A a d^{2} + 5 \, {\left (B a + A b\right )} c d + {\left (B b c d + 5 \, {\left (B a + A b\right )} d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{30 \, d^{2}}, \frac {15 \, A a \sqrt {-c} d^{2} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{2} + c}}\right ) + {\left (3 \, B b d^{2} x^{4} - 2 \, B b c^{2} + 15 \, A a d^{2} + 5 \, {\left (B a + A b\right )} c d + {\left (B b c d + 5 \, {\left (B a + A b\right )} d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{15 \, d^{2}}\right ] \]
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Time = 9.67 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.80 \[ \int \frac {\left (a+b x^2\right ) \left (A+B x^2\right ) \sqrt {c+d x^2}}{x} \, dx=\frac {\begin {cases} \frac {2 A a c \operatorname {atan}{\left (\frac {\sqrt {c + d x^{2}}}{\sqrt {- c}} \right )}}{\sqrt {- c}} + 2 A a \sqrt {c + d x^{2}} + \frac {2 B b \left (c + d x^{2}\right )^{\frac {5}{2}}}{5 d^{2}} + \frac {2 \left (c + d x^{2}\right )^{\frac {3}{2}} \left (A b d + B a d - B b c\right )}{3 d^{2}} & \text {for}\: d \neq 0 \\A a \sqrt {c} \log {\left (x^{2} \right )} + A b \sqrt {c} x^{2} + B a \sqrt {c} x^{2} + \frac {B b \sqrt {c} x^{4}}{2} & \text {otherwise} \end {cases}}{2} \]
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Time = 0.22 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.19 \[ \int \frac {\left (a+b x^2\right ) \left (A+B x^2\right ) \sqrt {c+d x^2}}{x} \, dx=\frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} B b x^{2}}{5 \, d} - A a \sqrt {c} \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right ) + \sqrt {d x^{2} + c} A a - \frac {2 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} B b c}{15 \, d^{2}} + \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} B a}{3 \, d} + \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} A b}{3 \, d} \]
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Time = 0.29 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.35 \[ \int \frac {\left (a+b x^2\right ) \left (A+B x^2\right ) \sqrt {c+d x^2}}{x} \, dx=\frac {A a c \arctan \left (\frac {\sqrt {d x^{2} + c}}{\sqrt {-c}}\right )}{\sqrt {-c}} + \frac {3 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} B b d^{8} - 5 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} B b c d^{8} + 5 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} B a d^{9} + 5 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} A b d^{9} + 15 \, \sqrt {d x^{2} + c} A a d^{10}}{15 \, d^{10}} \]
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Time = 5.94 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.20 \[ \int \frac {\left (a+b x^2\right ) \left (A+B x^2\right ) \sqrt {c+d x^2}}{x} \, dx=\sqrt {d\,x^2+c}\,\left (\frac {B\,b\,x^4}{5}+\frac {B\,c\,\left (5\,a\,d-2\,b\,c\right )}{15\,d^2}+\frac {B\,x^2\,\left (5\,a\,d+b\,c\right )}{15\,d}\right )+A\,a\,\sqrt {d\,x^2+c}-A\,a\,\sqrt {c}\,\mathrm {atanh}\left (\frac {\sqrt {d\,x^2+c}}{\sqrt {c}}\right )+\frac {A\,b\,{\left (d\,x^2+c\right )}^{3/2}}{3\,d} \]
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