Integrand size = 12, antiderivative size = 115 \[ \int e^{\text {arcsinh}(a+b x)} x^2 \, dx=-\frac {e^{-2 \text {arcsinh}(a+b x)}}{16 b^3}-\frac {a e^{-\text {arcsinh}(a+b x)}}{2 b^3}-\frac {\left (1-4 a^2\right ) e^{2 \text {arcsinh}(a+b x)}}{16 b^3}-\frac {a e^{3 \text {arcsinh}(a+b x)}}{6 b^3}+\frac {e^{4 \text {arcsinh}(a+b x)}}{32 b^3}-\frac {\left (1-4 a^2\right ) \text {arcsinh}(a+b x)}{8 b^3} \]
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Time = 0.10 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {5873, 2320, 12, 1642} \[ \int e^{\text {arcsinh}(a+b x)} x^2 \, dx=-\frac {\left (1-4 a^2\right ) e^{2 \text {arcsinh}(a+b x)}}{16 b^3}-\frac {\left (1-4 a^2\right ) \text {arcsinh}(a+b x)}{8 b^3}-\frac {a e^{-\text {arcsinh}(a+b x)}}{2 b^3}-\frac {a e^{3 \text {arcsinh}(a+b x)}}{6 b^3}-\frac {e^{-2 \text {arcsinh}(a+b x)}}{16 b^3}+\frac {e^{4 \text {arcsinh}(a+b x)}}{32 b^3} \]
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Rule 12
Rule 1642
Rule 2320
Rule 5873
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int e^x \cosh (x) \left (-\frac {a}{b}+\frac {\sinh (x)}{b}\right )^2 \, dx,x,\text {arcsinh}(a+b x)\right )}{b} \\ & = \frac {\text {Subst}\left (\int \frac {\left (1+2 a x-x^2\right )^2 \left (1+x^2\right )}{8 b^2 x^3} \, dx,x,e^{\text {arcsinh}(a+b x)}\right )}{b} \\ & = \frac {\text {Subst}\left (\int \frac {\left (1+2 a x-x^2\right )^2 \left (1+x^2\right )}{x^3} \, dx,x,e^{\text {arcsinh}(a+b x)}\right )}{8 b^3} \\ & = \frac {\text {Subst}\left (\int \left (\frac {1}{x^3}+\frac {4 a}{x^2}+\frac {-1+4 a^2}{x}+\left (-1+4 a^2\right ) x-4 a x^2+x^3\right ) \, dx,x,e^{\text {arcsinh}(a+b x)}\right )}{8 b^3} \\ & = -\frac {e^{-2 \text {arcsinh}(a+b x)}}{16 b^3}-\frac {a e^{-\text {arcsinh}(a+b x)}}{2 b^3}-\frac {\left (1-4 a^2\right ) e^{2 \text {arcsinh}(a+b x)}}{16 b^3}-\frac {a e^{3 \text {arcsinh}(a+b x)}}{6 b^3}+\frac {e^{4 \text {arcsinh}(a+b x)}}{32 b^3}-\frac {\left (1-4 a^2\right ) \text {arcsinh}(a+b x)}{8 b^3} \\ \end{align*}
Time = 0.08 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.89 \[ \int e^{\text {arcsinh}(a+b x)} x^2 \, dx=\frac {8 a b^3 x^3+6 b^4 x^4+\sqrt {1+a^2+2 a b x+b^2 x^2} \left (2 a^3+3 b x-2 a^2 b x+6 b^3 x^3+a \left (-13+2 b^2 x^2\right )\right )+3 (-1+2 a) (1+2 a) \text {arcsinh}(a+b x)}{24 b^3} \]
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Leaf count of result is larger than twice the leaf count of optimal. \(287\) vs. \(2(143)=286\).
Time = 0.63 (sec) , antiderivative size = 288, normalized size of antiderivative = 2.50
method | result | size |
default | \(\frac {x \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )^{\frac {3}{2}}}{4 b^{2}}-\frac {5 a \left (\frac {\left (b^{2} x^{2}+2 a b x +a^{2}+1\right )^{\frac {3}{2}}}{3 b^{2}}-\frac {a \left (\frac {\left (2 b^{2} x +2 a b \right ) \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{4 b^{2}}+\frac {\left (4 b^{2} \left (a^{2}+1\right )-4 a^{2} b^{2}\right ) \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{8 b^{2} \sqrt {b^{2}}}\right )}{b}\right )}{4 b}-\frac {\left (a^{2}+1\right ) \left (\frac {\left (2 b^{2} x +2 a b \right ) \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{4 b^{2}}+\frac {\left (4 b^{2} \left (a^{2}+1\right )-4 a^{2} b^{2}\right ) \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{8 b^{2} \sqrt {b^{2}}}\right )}{4 b^{2}}+\frac {b \,x^{4}}{4}+\frac {a \,x^{3}}{3}\) | \(288\) |
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Time = 0.26 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.02 \[ \int e^{\text {arcsinh}(a+b x)} x^2 \, dx=\frac {6 \, b^{4} x^{4} + 8 \, a b^{3} x^{3} - 3 \, {\left (4 \, a^{2} - 1\right )} \log \left (-b x - a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) + {\left (6 \, b^{3} x^{3} + 2 \, a b^{2} x^{2} + 2 \, a^{3} - {\left (2 \, a^{2} - 3\right )} b x - 13 \, a\right )} \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{24 \, b^{3}} \]
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Time = 0.81 (sec) , antiderivative size = 292, normalized size of antiderivative = 2.54 \[ \int e^{\text {arcsinh}(a+b x)} x^2 \, dx=\frac {a x^{3}}{3} + \frac {b x^{4}}{4} + \begin {cases} \frac {\left (- \frac {a \left (- \frac {3 a \left (\frac {1}{4} - \frac {a^{2}}{6}\right )}{2 b} - \frac {a \left (2 a^{2} + 2\right )}{12 b}\right )}{b} - \frac {\left (\frac {1}{4} - \frac {a^{2}}{6}\right ) \left (a^{2} + 1\right )}{2 b^{2}}\right ) \log {\left (2 a b + 2 b^{2} x + 2 \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1} \sqrt {b^{2}} \right )}}{\sqrt {b^{2}}} + \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1} \left (\frac {a x^{2}}{12 b} + \frac {x^{3}}{4} + \frac {x \left (\frac {1}{4} - \frac {a^{2}}{6}\right )}{2 b^{2}} + \frac {- \frac {3 a \left (\frac {1}{4} - \frac {a^{2}}{6}\right )}{2 b} - \frac {a \left (2 a^{2} + 2\right )}{12 b}}{b^{2}}\right ) & \text {for}\: b^{2} \neq 0 \\\frac {\frac {\left (- 2 a^{2} - 2\right ) \left (a^{2} + 2 a b x + 1\right )^{\frac {5}{2}}}{5} + \frac {\left (a^{2} + 2 a b x + 1\right )^{\frac {7}{2}}}{7} + \frac {\left (a^{2} + 2 a b x + 1\right )^{\frac {3}{2}} \left (a^{4} + 2 a^{2} + 1\right )}{3}}{4 a^{3} b^{3}} & \text {for}\: a b \neq 0 \\\frac {x^{3} \sqrt {a^{2} + 1}}{3} & \text {otherwise} \end {cases} \]
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Time = 0.25 (sec) , antiderivative size = 273, normalized size of antiderivative = 2.37 \[ \int e^{\text {arcsinh}(a+b x)} x^2 \, dx=\frac {1}{4} \, b x^{4} + \frac {1}{3} \, a x^{3} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}^{\frac {3}{2}} x}{4 \, b^{2}} - \frac {5 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}^{\frac {3}{2}} a}{12 \, b^{3}} - \frac {{\left (5 \, a^{2} b^{2} - {\left (a^{2} + 1\right )} b^{2}\right )} a^{2} \operatorname {arsinh}\left (\frac {2 \, {\left (b^{2} x + a b\right )}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}}}\right )}{8 \, b^{5}} + \frac {{\left (5 \, a^{2} b^{2} - {\left (a^{2} + 1\right )} b^{2}\right )} \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} x}{8 \, b^{4}} + \frac {{\left (5 \, a^{2} b^{2} - {\left (a^{2} + 1\right )} b^{2}\right )} {\left (a^{2} + 1\right )} \operatorname {arsinh}\left (\frac {2 \, {\left (b^{2} x + a b\right )}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}}}\right )}{8 \, b^{5}} + \frac {{\left (5 \, a^{2} b^{2} - {\left (a^{2} + 1\right )} b^{2}\right )} \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} a}{8 \, b^{5}} \]
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Time = 0.27 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.22 \[ \int e^{\text {arcsinh}(a+b x)} x^2 \, dx=\frac {1}{4} \, b x^{4} + \frac {1}{3} \, a x^{3} + \frac {1}{24} \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left ({\left (2 \, {\left (3 \, x + \frac {a}{b}\right )} x - \frac {2 \, a^{2} b^{3} - 3 \, b^{3}}{b^{5}}\right )} x + \frac {2 \, a^{3} b^{2} - 13 \, a b^{2}}{b^{5}}\right )} - \frac {{\left (4 \, a^{2} - 1\right )} \log \left (-a b - {\left (x {\left | b \right |} - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )} {\left | b \right |}\right )}{8 \, b^{2} {\left | b \right |}} \]
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Timed out. \[ \int e^{\text {arcsinh}(a+b x)} x^2 \, dx=\int x^2\,\left (a+\sqrt {{\left (a+b\,x\right )}^2+1}+b\,x\right ) \,d x \]
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