Integrand size = 22, antiderivative size = 98 \[ \int \frac {x}{\left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx=\frac {1}{3 (b c-a d) \left (c+d x^2\right )^{3/2}}+\frac {b}{(b c-a d)^2 \sqrt {c+d x^2}}-\frac {b^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{(b c-a d)^{5/2}} \]
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Time = 0.05 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {455, 53, 65, 214} \[ \int \frac {x}{\left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx=-\frac {b^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{(b c-a d)^{5/2}}+\frac {b}{\sqrt {c+d x^2} (b c-a d)^2}+\frac {1}{3 \left (c+d x^2\right )^{3/2} (b c-a d)} \]
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Rule 53
Rule 65
Rule 214
Rule 455
Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {1}{(a+b x) (c+d x)^{5/2}} \, dx,x,x^2\right ) \\ & = \frac {1}{3 (b c-a d) \left (c+d x^2\right )^{3/2}}+\frac {b \text {Subst}\left (\int \frac {1}{(a+b x) (c+d x)^{3/2}} \, dx,x,x^2\right )}{2 (b c-a d)} \\ & = \frac {1}{3 (b c-a d) \left (c+d x^2\right )^{3/2}}+\frac {b}{(b c-a d)^2 \sqrt {c+d x^2}}+\frac {b^2 \text {Subst}\left (\int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx,x,x^2\right )}{2 (b c-a d)^2} \\ & = \frac {1}{3 (b c-a d) \left (c+d x^2\right )^{3/2}}+\frac {b}{(b c-a d)^2 \sqrt {c+d x^2}}+\frac {b^2 \text {Subst}\left (\int \frac {1}{a-\frac {b c}{d}+\frac {b x^2}{d}} \, dx,x,\sqrt {c+d x^2}\right )}{d (b c-a d)^2} \\ & = \frac {1}{3 (b c-a d) \left (c+d x^2\right )^{3/2}}+\frac {b}{(b c-a d)^2 \sqrt {c+d x^2}}-\frac {b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{(b c-a d)^{5/2}} \\ \end{align*}
Time = 0.14 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.92 \[ \int \frac {x}{\left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx=\frac {4 b c-a d+3 b d x^2}{3 (b c-a d)^2 \left (c+d x^2\right )^{3/2}}+\frac {b^{3/2} \arctan \left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {-b c+a d}}\right )}{(-b c+a d)^{5/2}} \]
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Time = 2.96 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.04
method | result | size |
pseudoelliptic | \(-\frac {-3 b^{2} \arctan \left (\frac {b \sqrt {d \,x^{2}+c}}{\sqrt {\left (a d -b c \right ) b}}\right ) \left (d \,x^{2}+c \right )^{\frac {3}{2}}+\sqrt {\left (a d -b c \right ) b}\, \left (\left (-3 d \,x^{2}-4 c \right ) b +a d \right )}{3 \left (d \,x^{2}+c \right )^{\frac {3}{2}} \sqrt {\left (a d -b c \right ) b}\, \left (a d -b c \right )^{2}}\) | \(102\) |
default | \(\text {Expression too large to display}\) | \(1390\) |
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Leaf count of result is larger than twice the leaf count of optimal. 213 vs. \(2 (82) = 164\).
Time = 0.33 (sec) , antiderivative size = 511, normalized size of antiderivative = 5.21 \[ \int \frac {x}{\left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx=\left [\frac {3 \, {\left (b d^{2} x^{4} + 2 \, b c d x^{2} + b c^{2}\right )} \sqrt {\frac {b}{b c - a d}} \log \left (\frac {b^{2} d^{2} x^{4} + 8 \, b^{2} c^{2} - 8 \, a b c d + a^{2} d^{2} + 2 \, {\left (4 \, b^{2} c d - 3 \, a b d^{2}\right )} x^{2} - 4 \, {\left (2 \, b^{2} c^{2} - 3 \, a b c d + a^{2} d^{2} + {\left (b^{2} c d - a b d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c} \sqrt {\frac {b}{b c - a d}}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right ) + 4 \, {\left (3 \, b d x^{2} + 4 \, b c - a d\right )} \sqrt {d x^{2} + c}}{12 \, {\left (b^{2} c^{4} - 2 \, a b c^{3} d + a^{2} c^{2} d^{2} + {\left (b^{2} c^{2} d^{2} - 2 \, a b c d^{3} + a^{2} d^{4}\right )} x^{4} + 2 \, {\left (b^{2} c^{3} d - 2 \, a b c^{2} d^{2} + a^{2} c d^{3}\right )} x^{2}\right )}}, \frac {3 \, {\left (b d^{2} x^{4} + 2 \, b c d x^{2} + b c^{2}\right )} \sqrt {-\frac {b}{b c - a d}} \arctan \left (\frac {{\left (b d x^{2} + 2 \, b c - a d\right )} \sqrt {d x^{2} + c} \sqrt {-\frac {b}{b c - a d}}}{2 \, {\left (b d x^{2} + b c\right )}}\right ) + 2 \, {\left (3 \, b d x^{2} + 4 \, b c - a d\right )} \sqrt {d x^{2} + c}}{6 \, {\left (b^{2} c^{4} - 2 \, a b c^{3} d + a^{2} c^{2} d^{2} + {\left (b^{2} c^{2} d^{2} - 2 \, a b c d^{3} + a^{2} d^{4}\right )} x^{4} + 2 \, {\left (b^{2} c^{3} d - 2 \, a b c^{2} d^{2} + a^{2} c d^{3}\right )} x^{2}\right )}}\right ] \]
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Time = 7.41 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.42 \[ \int \frac {x}{\left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx=\begin {cases} \frac {2 \left (\frac {b d}{2 \sqrt {c + d x^{2}} \left (a d - b c\right )^{2}} + \frac {b d \operatorname {atan}{\left (\frac {\sqrt {c + d x^{2}}}{\sqrt {\frac {a d - b c}{b}}} \right )}}{2 \sqrt {\frac {a d - b c}{b}} \left (a d - b c\right )^{2}} - \frac {d}{6 \left (c + d x^{2}\right )^{\frac {3}{2}} \left (a d - b c\right )}\right )}{d} & \text {for}\: d \neq 0 \\\begin {cases} \frac {x^{2}}{2 a c^{\frac {5}{2}}} & \text {for}\: b = 0 \\\tilde {\infty } x^{2} & \text {for}\: c^{\frac {5}{2}} = 0 \\\frac {\log {\left (2 a c^{\frac {5}{2}} + 2 b c^{\frac {5}{2}} x^{2} \right )}}{2 b c^{\frac {5}{2}}} & \text {otherwise} \end {cases} & \text {otherwise} \end {cases} \]
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Exception generated. \[ \int \frac {x}{\left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx=\text {Exception raised: ValueError} \]
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none
Time = 0.28 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.20 \[ \int \frac {x}{\left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx=\frac {b^{2} \arctan \left (\frac {\sqrt {d x^{2} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt {-b^{2} c + a b d}} + \frac {3 \, {\left (d x^{2} + c\right )} b + b c - a d}{3 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} {\left (d x^{2} + c\right )}^{\frac {3}{2}}} \]
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Time = 5.61 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.05 \[ \int \frac {x}{\left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx=\frac {b^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {d\,x^2+c}\,\left (a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2\right )}{{\left (a\,d-b\,c\right )}^{5/2}}\right )}{{\left (a\,d-b\,c\right )}^{5/2}}-\frac {\frac {1}{3\,\left (a\,d-b\,c\right )}-\frac {b\,\left (d\,x^2+c\right )}{{\left (a\,d-b\,c\right )}^2}}{{\left (d\,x^2+c\right )}^{3/2}} \]
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