Integrand size = 23, antiderivative size = 217 \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d+e x}} \, dx=\frac {64 b d^3 n \sqrt {d+e x}}{35 e^4}-\frac {76 b d^2 n (d+e x)^{3/2}}{105 e^4}+\frac {64 b d n (d+e x)^{5/2}}{175 e^4}-\frac {4 b n (d+e x)^{7/2}}{49 e^4}-\frac {64 b d^{7/2} n \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{35 e^4}-\frac {2 d^3 \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^4}+\frac {2 d^2 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{e^4}-\frac {6 d (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^4}+\frac {2 (d+e x)^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^4} \]
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Time = 0.15 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {45, 2392, 12, 1634, 52, 65, 214} \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d+e x}} \, dx=-\frac {2 d^3 \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^4}+\frac {2 d^2 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{e^4}-\frac {6 d (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^4}+\frac {2 (d+e x)^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^4}-\frac {64 b d^{7/2} n \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{35 e^4}+\frac {64 b d^3 n \sqrt {d+e x}}{35 e^4}-\frac {76 b d^2 n (d+e x)^{3/2}}{105 e^4}+\frac {64 b d n (d+e x)^{5/2}}{175 e^4}-\frac {4 b n (d+e x)^{7/2}}{49 e^4} \]
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Rule 12
Rule 45
Rule 52
Rule 65
Rule 214
Rule 1634
Rule 2392
Rubi steps \begin{align*} \text {integral}& = -\frac {2 d^3 \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^4}+\frac {2 d^2 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{e^4}-\frac {6 d (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^4}+\frac {2 (d+e x)^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^4}-(b n) \int \frac {2 \sqrt {d+e x} \left (-16 d^3+8 d^2 e x-6 d e^2 x^2+5 e^3 x^3\right )}{35 e^4 x} \, dx \\ & = -\frac {2 d^3 \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^4}+\frac {2 d^2 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{e^4}-\frac {6 d (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^4}+\frac {2 (d+e x)^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^4}-\frac {(2 b n) \int \frac {\sqrt {d+e x} \left (-16 d^3+8 d^2 e x-6 d e^2 x^2+5 e^3 x^3\right )}{x} \, dx}{35 e^4} \\ & = -\frac {2 d^3 \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^4}+\frac {2 d^2 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{e^4}-\frac {6 d (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^4}+\frac {2 (d+e x)^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^4}-\frac {(2 b n) \int \left (19 d^2 e \sqrt {d+e x}-\frac {16 d^3 \sqrt {d+e x}}{x}-16 d e (d+e x)^{3/2}+5 e (d+e x)^{5/2}\right ) \, dx}{35 e^4} \\ & = -\frac {76 b d^2 n (d+e x)^{3/2}}{105 e^4}+\frac {64 b d n (d+e x)^{5/2}}{175 e^4}-\frac {4 b n (d+e x)^{7/2}}{49 e^4}-\frac {2 d^3 \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^4}+\frac {2 d^2 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{e^4}-\frac {6 d (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^4}+\frac {2 (d+e x)^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^4}+\frac {\left (32 b d^3 n\right ) \int \frac {\sqrt {d+e x}}{x} \, dx}{35 e^4} \\ & = \frac {64 b d^3 n \sqrt {d+e x}}{35 e^4}-\frac {76 b d^2 n (d+e x)^{3/2}}{105 e^4}+\frac {64 b d n (d+e x)^{5/2}}{175 e^4}-\frac {4 b n (d+e x)^{7/2}}{49 e^4}-\frac {2 d^3 \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^4}+\frac {2 d^2 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{e^4}-\frac {6 d (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^4}+\frac {2 (d+e x)^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^4}+\frac {\left (32 b d^4 n\right ) \int \frac {1}{x \sqrt {d+e x}} \, dx}{35 e^4} \\ & = \frac {64 b d^3 n \sqrt {d+e x}}{35 e^4}-\frac {76 b d^2 n (d+e x)^{3/2}}{105 e^4}+\frac {64 b d n (d+e x)^{5/2}}{175 e^4}-\frac {4 b n (d+e x)^{7/2}}{49 e^4}-\frac {2 d^3 \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^4}+\frac {2 d^2 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{e^4}-\frac {6 d (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^4}+\frac {2 (d+e x)^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^4}+\frac {\left (64 b d^4 n\right ) \text {Subst}\left (\int \frac {1}{-\frac {d}{e}+\frac {x^2}{e}} \, dx,x,\sqrt {d+e x}\right )}{35 e^5} \\ & = \frac {64 b d^3 n \sqrt {d+e x}}{35 e^4}-\frac {76 b d^2 n (d+e x)^{3/2}}{105 e^4}+\frac {64 b d n (d+e x)^{5/2}}{175 e^4}-\frac {4 b n (d+e x)^{7/2}}{49 e^4}-\frac {64 b d^{7/2} n \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{35 e^4}-\frac {2 d^3 \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^4}+\frac {2 d^2 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{e^4}-\frac {6 d (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^4}+\frac {2 (d+e x)^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^4} \\ \end{align*}
Time = 0.15 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.69 \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d+e x}} \, dx=-\frac {2 \left (3360 b d^{7/2} n \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )+\sqrt {d+e x} \left (105 a \left (16 d^3-8 d^2 e x+6 d e^2 x^2-5 e^3 x^3\right )+2 b n \left (-1276 d^3+218 d^2 e x-111 d e^2 x^2+75 e^3 x^3\right )+105 b \left (16 d^3-8 d^2 e x+6 d e^2 x^2-5 e^3 x^3\right ) \log \left (c x^n\right )\right )\right )}{3675 e^4} \]
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\[\int \frac {x^{3} \left (a +b \ln \left (c \,x^{n}\right )\right )}{\sqrt {e x +d}}d x\]
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Time = 0.32 (sec) , antiderivative size = 395, normalized size of antiderivative = 1.82 \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d+e x}} \, dx=\left [\frac {2 \, {\left (1680 \, b d^{\frac {7}{2}} n \log \left (\frac {e x - 2 \, \sqrt {e x + d} \sqrt {d} + 2 \, d}{x}\right ) + {\left (2552 \, b d^{3} n - 1680 \, a d^{3} - 75 \, {\left (2 \, b e^{3} n - 7 \, a e^{3}\right )} x^{3} + 6 \, {\left (37 \, b d e^{2} n - 105 \, a d e^{2}\right )} x^{2} - 4 \, {\left (109 \, b d^{2} e n - 210 \, a d^{2} e\right )} x + 105 \, {\left (5 \, b e^{3} x^{3} - 6 \, b d e^{2} x^{2} + 8 \, b d^{2} e x - 16 \, b d^{3}\right )} \log \left (c\right ) + 105 \, {\left (5 \, b e^{3} n x^{3} - 6 \, b d e^{2} n x^{2} + 8 \, b d^{2} e n x - 16 \, b d^{3} n\right )} \log \left (x\right )\right )} \sqrt {e x + d}\right )}}{3675 \, e^{4}}, \frac {2 \, {\left (3360 \, b \sqrt {-d} d^{3} n \arctan \left (\frac {\sqrt {e x + d} \sqrt {-d}}{d}\right ) + {\left (2552 \, b d^{3} n - 1680 \, a d^{3} - 75 \, {\left (2 \, b e^{3} n - 7 \, a e^{3}\right )} x^{3} + 6 \, {\left (37 \, b d e^{2} n - 105 \, a d e^{2}\right )} x^{2} - 4 \, {\left (109 \, b d^{2} e n - 210 \, a d^{2} e\right )} x + 105 \, {\left (5 \, b e^{3} x^{3} - 6 \, b d e^{2} x^{2} + 8 \, b d^{2} e x - 16 \, b d^{3}\right )} \log \left (c\right ) + 105 \, {\left (5 \, b e^{3} n x^{3} - 6 \, b d e^{2} n x^{2} + 8 \, b d^{2} e n x - 16 \, b d^{3} n\right )} \log \left (x\right )\right )} \sqrt {e x + d}\right )}}{3675 \, e^{4}}\right ] \]
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Time = 57.10 (sec) , antiderivative size = 396, normalized size of antiderivative = 1.82 \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d+e x}} \, dx=a \left (\begin {cases} - \frac {2 d^{3} \sqrt {d + e x}}{e^{4}} + \frac {2 d^{2} \left (d + e x\right )^{\frac {3}{2}}}{e^{4}} - \frac {6 d \left (d + e x\right )^{\frac {5}{2}}}{5 e^{4}} + \frac {2 \left (d + e x\right )^{\frac {7}{2}}}{7 e^{4}} & \text {for}\: e \neq 0 \\\frac {x^{4}}{4 \sqrt {d}} & \text {otherwise} \end {cases}\right ) - b n \left (\begin {cases} \frac {9596 d^{\frac {7}{2}} \sqrt {1 + \frac {e x}{d}}}{3675 e^{4}} + \frac {38 d^{\frac {7}{2}} \log {\left (\frac {e x}{d} \right )}}{35 e^{4}} - \frac {76 d^{\frac {7}{2}} \log {\left (\sqrt {1 + \frac {e x}{d}} + 1 \right )}}{35 e^{4}} + \frac {4 d^{\frac {7}{2}} \operatorname {asinh}{\left (\frac {\sqrt {d}}{\sqrt {e} \sqrt {x}} \right )}}{e^{4}} + \frac {872 d^{\frac {5}{2}} x \sqrt {1 + \frac {e x}{d}}}{3675 e^{3}} - \frac {148 d^{\frac {3}{2}} x^{2} \sqrt {1 + \frac {e x}{d}}}{1225 e^{2}} + \frac {4 \sqrt {d} x^{3} \sqrt {1 + \frac {e x}{d}}}{49 e} - \frac {4 d^{4}}{e^{\frac {9}{2}} \sqrt {x} \sqrt {\frac {d}{e x} + 1}} - \frac {4 d^{3} \sqrt {x}}{e^{\frac {7}{2}} \sqrt {\frac {d}{e x} + 1}} & \text {for}\: e > -\infty \wedge e < \infty \wedge e \neq 0 \\\frac {x^{4}}{16 \sqrt {d}} & \text {otherwise} \end {cases}\right ) + b \left (\begin {cases} - \frac {2 d^{3} \sqrt {d + e x}}{e^{4}} + \frac {2 d^{2} \left (d + e x\right )^{\frac {3}{2}}}{e^{4}} - \frac {6 d \left (d + e x\right )^{\frac {5}{2}}}{5 e^{4}} + \frac {2 \left (d + e x\right )^{\frac {7}{2}}}{7 e^{4}} & \text {for}\: e \neq 0 \\\frac {x^{4}}{4 \sqrt {d}} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )} \]
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Time = 0.27 (sec) , antiderivative size = 215, normalized size of antiderivative = 0.99 \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d+e x}} \, dx=\frac {4}{3675} \, b n {\left (\frac {840 \, d^{\frac {7}{2}} \log \left (\frac {\sqrt {e x + d} - \sqrt {d}}{\sqrt {e x + d} + \sqrt {d}}\right )}{e^{4}} - \frac {75 \, {\left (e x + d\right )}^{\frac {7}{2}} - 336 \, {\left (e x + d\right )}^{\frac {5}{2}} d + 665 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{2} - 1680 \, \sqrt {e x + d} d^{3}}{e^{4}}\right )} + \frac {2}{35} \, b {\left (\frac {5 \, {\left (e x + d\right )}^{\frac {7}{2}}}{e^{4}} - \frac {21 \, {\left (e x + d\right )}^{\frac {5}{2}} d}{e^{4}} + \frac {35 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{2}}{e^{4}} - \frac {35 \, \sqrt {e x + d} d^{3}}{e^{4}}\right )} \log \left (c x^{n}\right ) + \frac {2}{35} \, a {\left (\frac {5 \, {\left (e x + d\right )}^{\frac {7}{2}}}{e^{4}} - \frac {21 \, {\left (e x + d\right )}^{\frac {5}{2}} d}{e^{4}} + \frac {35 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{2}}{e^{4}} - \frac {35 \, \sqrt {e x + d} d^{3}}{e^{4}}\right )} \]
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Time = 0.38 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.16 \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d+e x}} \, dx=\frac {64 \, b d^{4} n \arctan \left (\frac {\sqrt {e x + d}}{\sqrt {-d}}\right )}{35 \, \sqrt {-d} e^{4}} + \frac {2}{35} \, {\left (\frac {5 \, {\left (e x + d\right )}^{\frac {7}{2}} b n}{e^{4}} - \frac {21 \, {\left (e x + d\right )}^{\frac {5}{2}} b d n}{e^{4}} + \frac {35 \, {\left (e x + d\right )}^{\frac {3}{2}} b d^{2} n}{e^{4}} - \frac {35 \, \sqrt {e x + d} b d^{3} n}{e^{4}}\right )} \log \left (e x\right ) - \frac {2 \, {\left (7 \, b n \log \left (e\right ) + 2 \, b n - 7 \, b \log \left (c\right ) - 7 \, a\right )} {\left (e x + d\right )}^{\frac {7}{2}}}{49 \, e^{4}} + \frac {2 \, {\left (105 \, b d n \log \left (e\right ) + 32 \, b d n - 105 \, b d \log \left (c\right ) - 105 \, a d\right )} {\left (e x + d\right )}^{\frac {5}{2}}}{175 \, e^{4}} - \frac {2 \, {\left (105 \, b d^{2} n \log \left (e\right ) + 38 \, b d^{2} n - 105 \, b d^{2} \log \left (c\right ) - 105 \, a d^{2}\right )} {\left (e x + d\right )}^{\frac {3}{2}}}{105 \, e^{4}} + \frac {2 \, {\left (35 \, b d^{3} n \log \left (e\right ) + 32 \, b d^{3} n - 35 \, b d^{3} \log \left (c\right ) - 35 \, a d^{3}\right )} \sqrt {e x + d}}{35 \, e^{4}} \]
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Timed out. \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d+e x}} \, dx=\int \frac {x^3\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{\sqrt {d+e\,x}} \,d x \]
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