Integrand size = 25, antiderivative size = 154 \[ \int x^3 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {2 b d^2 n \sqrt {d+e x^2}}{15 e^2}+\frac {2 b d n \left (d+e x^2\right )^{3/2}}{45 e^2}-\frac {b n \left (d+e x^2\right )^{5/2}}{25 e^2}-\frac {2 b d^{5/2} n \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{15 e^2}-\frac {d \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2} \]
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Time = 0.13 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {272, 45, 2392, 12, 457, 81, 52, 65, 214} \[ \int x^3 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {d \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}-\frac {2 b d^{5/2} n \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{15 e^2}+\frac {2 b d^2 n \sqrt {d+e x^2}}{15 e^2}+\frac {2 b d n \left (d+e x^2\right )^{3/2}}{45 e^2}-\frac {b n \left (d+e x^2\right )^{5/2}}{25 e^2} \]
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Rule 12
Rule 45
Rule 52
Rule 65
Rule 81
Rule 214
Rule 272
Rule 457
Rule 2392
Rubi steps \begin{align*} \text {integral}& = -\frac {d \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}-(b n) \int \frac {\left (d+e x^2\right )^{3/2} \left (-2 d+3 e x^2\right )}{15 e^2 x} \, dx \\ & = -\frac {d \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}-\frac {(b n) \int \frac {\left (d+e x^2\right )^{3/2} \left (-2 d+3 e x^2\right )}{x} \, dx}{15 e^2} \\ & = -\frac {d \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}-\frac {(b n) \text {Subst}\left (\int \frac {(d+e x)^{3/2} (-2 d+3 e x)}{x} \, dx,x,x^2\right )}{30 e^2} \\ & = -\frac {b n \left (d+e x^2\right )^{5/2}}{25 e^2}-\frac {d \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}+\frac {(b d n) \text {Subst}\left (\int \frac {(d+e x)^{3/2}}{x} \, dx,x,x^2\right )}{15 e^2} \\ & = \frac {2 b d n \left (d+e x^2\right )^{3/2}}{45 e^2}-\frac {b n \left (d+e x^2\right )^{5/2}}{25 e^2}-\frac {d \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}+\frac {\left (b d^2 n\right ) \text {Subst}\left (\int \frac {\sqrt {d+e x}}{x} \, dx,x,x^2\right )}{15 e^2} \\ & = \frac {2 b d^2 n \sqrt {d+e x^2}}{15 e^2}+\frac {2 b d n \left (d+e x^2\right )^{3/2}}{45 e^2}-\frac {b n \left (d+e x^2\right )^{5/2}}{25 e^2}-\frac {d \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}+\frac {\left (b d^3 n\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {d+e x}} \, dx,x,x^2\right )}{15 e^2} \\ & = \frac {2 b d^2 n \sqrt {d+e x^2}}{15 e^2}+\frac {2 b d n \left (d+e x^2\right )^{3/2}}{45 e^2}-\frac {b n \left (d+e x^2\right )^{5/2}}{25 e^2}-\frac {d \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}+\frac {\left (2 b d^3 n\right ) \text {Subst}\left (\int \frac {1}{-\frac {d}{e}+\frac {x^2}{e}} \, dx,x,\sqrt {d+e x^2}\right )}{15 e^3} \\ & = \frac {2 b d^2 n \sqrt {d+e x^2}}{15 e^2}+\frac {2 b d n \left (d+e x^2\right )^{3/2}}{45 e^2}-\frac {b n \left (d+e x^2\right )^{5/2}}{25 e^2}-\frac {2 b d^{5/2} n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{15 e^2}-\frac {d \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2} \\ \end{align*}
Time = 0.10 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.32 \[ \int x^3 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {2 b d^{5/2} n \log (x)}{15 e^2}-\frac {b n \sqrt {d+e x^2} \left (2 d^2-d e x^2-3 e^2 x^4\right ) \log (x)}{15 e^2}+\sqrt {d+e x^2} \left (\frac {1}{25} x^4 \left (5 a-b n+5 b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )+\frac {d x^2 \left (15 a-8 b n+15 b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )}{225 e}-\frac {d^2 \left (30 a-31 b n+30 b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )}{225 e^2}\right )-\frac {2 b d^{5/2} n \log \left (d+\sqrt {d} \sqrt {d+e x^2}\right )}{15 e^2} \]
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\[\int x^{3} \left (a +b \ln \left (c \,x^{n}\right )\right ) \sqrt {e \,x^{2}+d}d x\]
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Time = 0.34 (sec) , antiderivative size = 309, normalized size of antiderivative = 2.01 \[ \int x^3 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right ) \, dx=\left [\frac {15 \, b d^{\frac {5}{2}} n \log \left (-\frac {e x^{2} - 2 \, \sqrt {e x^{2} + d} \sqrt {d} + 2 \, d}{x^{2}}\right ) - {\left (9 \, {\left (b e^{2} n - 5 \, a e^{2}\right )} x^{4} - 31 \, b d^{2} n + 30 \, a d^{2} + {\left (8 \, b d e n - 15 \, a d e\right )} x^{2} - 15 \, {\left (3 \, b e^{2} x^{4} + b d e x^{2} - 2 \, b d^{2}\right )} \log \left (c\right ) - 15 \, {\left (3 \, b e^{2} n x^{4} + b d e n x^{2} - 2 \, b d^{2} n\right )} \log \left (x\right )\right )} \sqrt {e x^{2} + d}}{225 \, e^{2}}, \frac {30 \, b \sqrt {-d} d^{2} n \arctan \left (\frac {\sqrt {-d}}{\sqrt {e x^{2} + d}}\right ) - {\left (9 \, {\left (b e^{2} n - 5 \, a e^{2}\right )} x^{4} - 31 \, b d^{2} n + 30 \, a d^{2} + {\left (8 \, b d e n - 15 \, a d e\right )} x^{2} - 15 \, {\left (3 \, b e^{2} x^{4} + b d e x^{2} - 2 \, b d^{2}\right )} \log \left (c\right ) - 15 \, {\left (3 \, b e^{2} n x^{4} + b d e n x^{2} - 2 \, b d^{2} n\right )} \log \left (x\right )\right )} \sqrt {e x^{2} + d}}{225 \, e^{2}}\right ] \]
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Time = 14.27 (sec) , antiderivative size = 343, normalized size of antiderivative = 2.23 \[ \int x^3 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right ) \, dx=a \left (\begin {cases} - \frac {2 d^{2} \sqrt {d + e x^{2}}}{15 e^{2}} + \frac {d x^{2} \sqrt {d + e x^{2}}}{15 e} + \frac {x^{4} \sqrt {d + e x^{2}}}{5} & \text {for}\: e \neq 0 \\\frac {\sqrt {d} x^{4}}{4} & \text {otherwise} \end {cases}\right ) - b n \left (\begin {cases} \frac {2 d^{\frac {5}{2}} \operatorname {asinh}{\left (\frac {\sqrt {d}}{\sqrt {e} x} \right )}}{15 e^{2}} - \frac {2 d^{3}}{15 e^{\frac {5}{2}} x \sqrt {\frac {d}{e x^{2}} + 1}} - \frac {2 d^{2} x}{15 e^{\frac {3}{2}} \sqrt {\frac {d}{e x^{2}} + 1}} + \frac {d \left (\begin {cases} \frac {d \sqrt {d + e x^{2}}}{3 e} + \frac {x^{2} \sqrt {d + e x^{2}}}{3} & \text {for}\: e \neq 0 \\\frac {\sqrt {d} x^{2}}{2} & \text {otherwise} \end {cases}\right )}{15 e} + \frac {\begin {cases} - \frac {2 d^{2} \sqrt {d + e x^{2}}}{15 e^{2}} + \frac {d x^{2} \sqrt {d + e x^{2}}}{15 e} + \frac {x^{4} \sqrt {d + e x^{2}}}{5} & \text {for}\: e \neq 0 \\\frac {\sqrt {d} x^{4}}{4} & \text {otherwise} \end {cases}}{5} & \text {for}\: e > -\infty \wedge e < \infty \wedge e \neq 0 \\\frac {\sqrt {d} x^{4}}{16} & \text {otherwise} \end {cases}\right ) + b \left (\begin {cases} - \frac {2 d^{2} \sqrt {d + e x^{2}}}{15 e^{2}} + \frac {d x^{2} \sqrt {d + e x^{2}}}{15 e} + \frac {x^{4} \sqrt {d + e x^{2}}}{5} & \text {for}\: e \neq 0 \\\frac {\sqrt {d} x^{4}}{4} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )} \]
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Exception generated. \[ \int x^3 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right ) \, dx=\text {Exception raised: ValueError} \]
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Time = 0.35 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.40 \[ \int x^3 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {1}{5} \, \sqrt {e x^{2} + d} b x^{4} \log \left (c\right ) + \frac {1}{5} \, \sqrt {e x^{2} + d} a x^{4} + \frac {\sqrt {e x^{2} + d} b d x^{2} \log \left (c\right )}{15 \, e} + \frac {\sqrt {e x^{2} + d} a d x^{2}}{15 \, e} + \frac {1}{225} \, b n {\left (\frac {15 \, {\left (3 \, {\left (e x^{2} + d\right )}^{\frac {5}{2}} - 5 \, {\left (e x^{2} + d\right )}^{\frac {3}{2}} d\right )} \log \left (x\right )}{e^{2}} + \frac {\frac {30 \, d^{3} \arctan \left (\frac {\sqrt {e x^{2} + d}}{\sqrt {-d}}\right )}{\sqrt {-d}} - 9 \, {\left (e x^{2} + d\right )}^{\frac {5}{2}} + 10 \, {\left (e x^{2} + d\right )}^{\frac {3}{2}} d + 30 \, \sqrt {e x^{2} + d} d^{2}}{e^{2}}\right )} - \frac {2 \, \sqrt {e x^{2} + d} b d^{2} \log \left (c\right )}{15 \, e^{2}} - \frac {2 \, \sqrt {e x^{2} + d} a d^{2}}{15 \, e^{2}} \]
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Timed out. \[ \int x^3 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right ) \, dx=\int x^3\,\sqrt {e\,x^2+d}\,\left (a+b\,\ln \left (c\,x^n\right )\right ) \,d x \]
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