Integrand size = 21, antiderivative size = 86 \[ \int \frac {x^4 \text {arcsinh}(a x)}{\sqrt {1+a^2 x^2}} \, dx=\frac {3 x^2}{16 a^3}-\frac {x^4}{16 a}-\frac {3 x \sqrt {1+a^2 x^2} \text {arcsinh}(a x)}{8 a^4}+\frac {x^3 \sqrt {1+a^2 x^2} \text {arcsinh}(a x)}{4 a^2}+\frac {3 \text {arcsinh}(a x)^2}{16 a^5} \]
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Time = 0.11 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {5812, 5783, 30} \[ \int \frac {x^4 \text {arcsinh}(a x)}{\sqrt {1+a^2 x^2}} \, dx=\frac {3 \text {arcsinh}(a x)^2}{16 a^5}+\frac {3 x^2}{16 a^3}+\frac {x^3 \sqrt {a^2 x^2+1} \text {arcsinh}(a x)}{4 a^2}-\frac {3 x \sqrt {a^2 x^2+1} \text {arcsinh}(a x)}{8 a^4}-\frac {x^4}{16 a} \]
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Rule 30
Rule 5783
Rule 5812
Rubi steps \begin{align*} \text {integral}& = \frac {x^3 \sqrt {1+a^2 x^2} \text {arcsinh}(a x)}{4 a^2}-\frac {3 \int \frac {x^2 \text {arcsinh}(a x)}{\sqrt {1+a^2 x^2}} \, dx}{4 a^2}-\frac {\int x^3 \, dx}{4 a} \\ & = -\frac {x^4}{16 a}-\frac {3 x \sqrt {1+a^2 x^2} \text {arcsinh}(a x)}{8 a^4}+\frac {x^3 \sqrt {1+a^2 x^2} \text {arcsinh}(a x)}{4 a^2}+\frac {3 \int \frac {\text {arcsinh}(a x)}{\sqrt {1+a^2 x^2}} \, dx}{8 a^4}+\frac {3 \int x \, dx}{8 a^3} \\ & = \frac {3 x^2}{16 a^3}-\frac {x^4}{16 a}-\frac {3 x \sqrt {1+a^2 x^2} \text {arcsinh}(a x)}{8 a^4}+\frac {x^3 \sqrt {1+a^2 x^2} \text {arcsinh}(a x)}{4 a^2}+\frac {3 \text {arcsinh}(a x)^2}{16 a^5} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.73 \[ \int \frac {x^4 \text {arcsinh}(a x)}{\sqrt {1+a^2 x^2}} \, dx=\frac {3 a^2 x^2-a^4 x^4+2 a x \sqrt {1+a^2 x^2} \left (-3+2 a^2 x^2\right ) \text {arcsinh}(a x)+3 \text {arcsinh}(a x)^2}{16 a^5} \]
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Time = 0.21 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.86
method | result | size |
default | \(\frac {4 a^{3} x^{3} \operatorname {arcsinh}\left (a x \right ) \sqrt {a^{2} x^{2}+1}-a^{4} x^{4}-6 \,\operatorname {arcsinh}\left (a x \right ) \sqrt {a^{2} x^{2}+1}\, a x +3 a^{2} x^{2}+3 \operatorname {arcsinh}\left (a x \right )^{2}+3}{16 a^{5}}\) | \(74\) |
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Time = 0.27 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.97 \[ \int \frac {x^4 \text {arcsinh}(a x)}{\sqrt {1+a^2 x^2}} \, dx=-\frac {a^{4} x^{4} - 3 \, a^{2} x^{2} - 2 \, {\left (2 \, a^{3} x^{3} - 3 \, a x\right )} \sqrt {a^{2} x^{2} + 1} \log \left (a x + \sqrt {a^{2} x^{2} + 1}\right ) - 3 \, \log \left (a x + \sqrt {a^{2} x^{2} + 1}\right )^{2}}{16 \, a^{5}} \]
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Time = 0.42 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.95 \[ \int \frac {x^4 \text {arcsinh}(a x)}{\sqrt {1+a^2 x^2}} \, dx=\begin {cases} - \frac {x^{4}}{16 a} + \frac {x^{3} \sqrt {a^{2} x^{2} + 1} \operatorname {asinh}{\left (a x \right )}}{4 a^{2}} + \frac {3 x^{2}}{16 a^{3}} - \frac {3 x \sqrt {a^{2} x^{2} + 1} \operatorname {asinh}{\left (a x \right )}}{8 a^{4}} + \frac {3 \operatorname {asinh}^{2}{\left (a x \right )}}{16 a^{5}} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \]
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Time = 0.20 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.97 \[ \int \frac {x^4 \text {arcsinh}(a x)}{\sqrt {1+a^2 x^2}} \, dx=-\frac {1}{16} \, {\left (\frac {x^{4}}{a^{2}} - \frac {3 \, x^{2}}{a^{4}} + \frac {3 \, \operatorname {arsinh}\left (a x\right )^{2}}{a^{6}}\right )} a + \frac {1}{8} \, {\left (\frac {2 \, \sqrt {a^{2} x^{2} + 1} x^{3}}{a^{2}} - \frac {3 \, \sqrt {a^{2} x^{2} + 1} x}{a^{4}} + \frac {3 \, \operatorname {arsinh}\left (a x\right )}{a^{5}}\right )} \operatorname {arsinh}\left (a x\right ) \]
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\[ \int \frac {x^4 \text {arcsinh}(a x)}{\sqrt {1+a^2 x^2}} \, dx=\int { \frac {x^{4} \operatorname {arsinh}\left (a x\right )}{\sqrt {a^{2} x^{2} + 1}} \,d x } \]
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Timed out. \[ \int \frac {x^4 \text {arcsinh}(a x)}{\sqrt {1+a^2 x^2}} \, dx=\int \frac {x^4\,\mathrm {asinh}\left (a\,x\right )}{\sqrt {a^2\,x^2+1}} \,d x \]
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