Optimal. Leaf size=327 \[ \frac {2}{15} \left (-\frac {15 d^{5/2} \tanh ^{-1}\left (\frac {\sqrt {d+e x^r}}{\sqrt {d}}\right )}{r}+\frac {15 d^2 \sqrt {d+e x^r}}{r}+\frac {5 d \left (d+e x^r\right )^{3/2}}{r}+\frac {3 \left (d+e x^r\right )^{5/2}}{r}\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {2 b d^{5/2} n \text {Li}_2\left (1-\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {e x^r+d}}\right )}{r^2}+\frac {2 b d^{5/2} n \tanh ^{-1}\left (\frac {\sqrt {d+e x^r}}{\sqrt {d}}\right )^2}{r^2}+\frac {92 b d^{5/2} n \tanh ^{-1}\left (\frac {\sqrt {d+e x^r}}{\sqrt {d}}\right )}{15 r^2}-\frac {4 b d^{5/2} n \log \left (\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x^r}}\right ) \tanh ^{-1}\left (\frac {\sqrt {d+e x^r}}{\sqrt {d}}\right )}{r^2}-\frac {92 b d^2 n \sqrt {d+e x^r}}{15 r^2}-\frac {32 b d n \left (d+e x^r\right )^{3/2}}{45 r^2}-\frac {4 b n \left (d+e x^r\right )^{5/2}}{25 r^2} \]
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Rubi [A] time = 0.48, antiderivative size = 327, normalized size of antiderivative = 1.00, number of steps used = 23, number of rules used = 9, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {266, 50, 63, 208, 2348, 5984, 5918, 2402, 2315} \[ -\frac {2 b d^{5/2} n \text {PolyLog}\left (2,1-\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x^r}}\right )}{r^2}+\frac {2}{15} \left (\frac {15 d^2 \sqrt {d+e x^r}}{r}-\frac {15 d^{5/2} \tanh ^{-1}\left (\frac {\sqrt {d+e x^r}}{\sqrt {d}}\right )}{r}+\frac {5 d \left (d+e x^r\right )^{3/2}}{r}+\frac {3 \left (d+e x^r\right )^{5/2}}{r}\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {92 b d^2 n \sqrt {d+e x^r}}{15 r^2}+\frac {2 b d^{5/2} n \tanh ^{-1}\left (\frac {\sqrt {d+e x^r}}{\sqrt {d}}\right )^2}{r^2}+\frac {92 b d^{5/2} n \tanh ^{-1}\left (\frac {\sqrt {d+e x^r}}{\sqrt {d}}\right )}{15 r^2}-\frac {4 b d^{5/2} n \log \left (\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x^r}}\right ) \tanh ^{-1}\left (\frac {\sqrt {d+e x^r}}{\sqrt {d}}\right )}{r^2}-\frac {32 b d n \left (d+e x^r\right )^{3/2}}{45 r^2}-\frac {4 b n \left (d+e x^r\right )^{5/2}}{25 r^2} \]
Antiderivative was successfully verified.
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Rule 50
Rule 63
Rule 208
Rule 266
Rule 2315
Rule 2348
Rule 2402
Rule 5918
Rule 5984
Rubi steps
\begin {align*} \int \frac {\left (d+e x^r\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{x} \, dx &=\frac {2}{15} \left (\frac {15 d^2 \sqrt {d+e x^r}}{r}+\frac {5 d \left (d+e x^r\right )^{3/2}}{r}+\frac {3 \left (d+e x^r\right )^{5/2}}{r}-\frac {15 d^{5/2} \tanh ^{-1}\left (\frac {\sqrt {d+e x^r}}{\sqrt {d}}\right )}{r}\right ) \left (a+b \log \left (c x^n\right )\right )-(b n) \int \left (\frac {2 d^2 \sqrt {d+e x^r}}{r x}+\frac {2 d \left (d+e x^r\right )^{3/2}}{3 r x}+\frac {2 \left (d+e x^r\right )^{5/2}}{5 r x}-\frac {2 d^{5/2} \tanh ^{-1}\left (\frac {\sqrt {d+e x^r}}{\sqrt {d}}\right )}{r x}\right ) \, dx\\ &=\frac {2}{15} \left (\frac {15 d^2 \sqrt {d+e x^r}}{r}+\frac {5 d \left (d+e x^r\right )^{3/2}}{r}+\frac {3 \left (d+e x^r\right )^{5/2}}{r}-\frac {15 d^{5/2} \tanh ^{-1}\left (\frac {\sqrt {d+e x^r}}{\sqrt {d}}\right )}{r}\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {(2 b n) \int \frac {\left (d+e x^r\right )^{5/2}}{x} \, dx}{5 r}-\frac {(2 b d n) \int \frac {\left (d+e x^r\right )^{3/2}}{x} \, dx}{3 r}-\frac {\left (2 b d^2 n\right ) \int \frac {\sqrt {d+e x^r}}{x} \, dx}{r}+\frac {\left (2 b d^{5/2} n\right ) \int \frac {\tanh ^{-1}\left (\frac {\sqrt {d+e x^r}}{\sqrt {d}}\right )}{x} \, dx}{r}\\ &=\frac {2}{15} \left (\frac {15 d^2 \sqrt {d+e x^r}}{r}+\frac {5 d \left (d+e x^r\right )^{3/2}}{r}+\frac {3 \left (d+e x^r\right )^{5/2}}{r}-\frac {15 d^{5/2} \tanh ^{-1}\left (\frac {\sqrt {d+e x^r}}{\sqrt {d}}\right )}{r}\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {(2 b n) \operatorname {Subst}\left (\int \frac {(d+e x)^{5/2}}{x} \, dx,x,x^r\right )}{5 r^2}-\frac {(2 b d n) \operatorname {Subst}\left (\int \frac {(d+e x)^{3/2}}{x} \, dx,x,x^r\right )}{3 r^2}-\frac {\left (2 b d^2 n\right ) \operatorname {Subst}\left (\int \frac {\sqrt {d+e x}}{x} \, dx,x,x^r\right )}{r^2}+\frac {\left (2 b d^{5/2} n\right ) \operatorname {Subst}\left (\int \frac {\tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{x} \, dx,x,x^r\right )}{r^2}\\ &=-\frac {4 b d^2 n \sqrt {d+e x^r}}{r^2}-\frac {4 b d n \left (d+e x^r\right )^{3/2}}{9 r^2}-\frac {4 b n \left (d+e x^r\right )^{5/2}}{25 r^2}+\frac {2}{15} \left (\frac {15 d^2 \sqrt {d+e x^r}}{r}+\frac {5 d \left (d+e x^r\right )^{3/2}}{r}+\frac {3 \left (d+e x^r\right )^{5/2}}{r}-\frac {15 d^{5/2} \tanh ^{-1}\left (\frac {\sqrt {d+e x^r}}{\sqrt {d}}\right )}{r}\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {(2 b d n) \operatorname {Subst}\left (\int \frac {(d+e x)^{3/2}}{x} \, dx,x,x^r\right )}{5 r^2}-\frac {\left (2 b d^2 n\right ) \operatorname {Subst}\left (\int \frac {\sqrt {d+e x}}{x} \, dx,x,x^r\right )}{3 r^2}+\frac {\left (4 b d^{5/2} n\right ) \operatorname {Subst}\left (\int \frac {x \tanh ^{-1}\left (\frac {x}{\sqrt {d}}\right )}{-d+x^2} \, dx,x,\sqrt {d+e x^r}\right )}{r^2}-\frac {\left (2 b d^3 n\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {d+e x}} \, dx,x,x^r\right )}{r^2}\\ &=-\frac {16 b d^2 n \sqrt {d+e x^r}}{3 r^2}-\frac {32 b d n \left (d+e x^r\right )^{3/2}}{45 r^2}-\frac {4 b n \left (d+e x^r\right )^{5/2}}{25 r^2}+\frac {2 b d^{5/2} n \tanh ^{-1}\left (\frac {\sqrt {d+e x^r}}{\sqrt {d}}\right )^2}{r^2}+\frac {2}{15} \left (\frac {15 d^2 \sqrt {d+e x^r}}{r}+\frac {5 d \left (d+e x^r\right )^{3/2}}{r}+\frac {3 \left (d+e x^r\right )^{5/2}}{r}-\frac {15 d^{5/2} \tanh ^{-1}\left (\frac {\sqrt {d+e x^r}}{\sqrt {d}}\right )}{r}\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {\left (2 b d^2 n\right ) \operatorname {Subst}\left (\int \frac {\sqrt {d+e x}}{x} \, dx,x,x^r\right )}{5 r^2}-\frac {\left (4 b d^2 n\right ) \operatorname {Subst}\left (\int \frac {\tanh ^{-1}\left (\frac {x}{\sqrt {d}}\right )}{1-\frac {x}{\sqrt {d}}} \, dx,x,\sqrt {d+e x^r}\right )}{r^2}-\frac {\left (2 b d^3 n\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {d+e x}} \, dx,x,x^r\right )}{3 r^2}-\frac {\left (4 b d^3 n\right ) \operatorname {Subst}\left (\int \frac {1}{-\frac {d}{e}+\frac {x^2}{e}} \, dx,x,\sqrt {d+e x^r}\right )}{e r^2}\\ &=-\frac {92 b d^2 n \sqrt {d+e x^r}}{15 r^2}-\frac {32 b d n \left (d+e x^r\right )^{3/2}}{45 r^2}-\frac {4 b n \left (d+e x^r\right )^{5/2}}{25 r^2}+\frac {4 b d^{5/2} n \tanh ^{-1}\left (\frac {\sqrt {d+e x^r}}{\sqrt {d}}\right )}{r^2}+\frac {2 b d^{5/2} n \tanh ^{-1}\left (\frac {\sqrt {d+e x^r}}{\sqrt {d}}\right )^2}{r^2}+\frac {2}{15} \left (\frac {15 d^2 \sqrt {d+e x^r}}{r}+\frac {5 d \left (d+e x^r\right )^{3/2}}{r}+\frac {3 \left (d+e x^r\right )^{5/2}}{r}-\frac {15 d^{5/2} \tanh ^{-1}\left (\frac {\sqrt {d+e x^r}}{\sqrt {d}}\right )}{r}\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {4 b d^{5/2} n \tanh ^{-1}\left (\frac {\sqrt {d+e x^r}}{\sqrt {d}}\right ) \log \left (\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x^r}}\right )}{r^2}+\frac {\left (4 b d^2 n\right ) \operatorname {Subst}\left (\int \frac {\log \left (\frac {2}{1-\frac {x}{\sqrt {d}}}\right )}{1-\frac {x^2}{d}} \, dx,x,\sqrt {d+e x^r}\right )}{r^2}-\frac {\left (2 b d^3 n\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {d+e x}} \, dx,x,x^r\right )}{5 r^2}-\frac {\left (4 b d^3 n\right ) \operatorname {Subst}\left (\int \frac {1}{-\frac {d}{e}+\frac {x^2}{e}} \, dx,x,\sqrt {d+e x^r}\right )}{3 e r^2}\\ &=-\frac {92 b d^2 n \sqrt {d+e x^r}}{15 r^2}-\frac {32 b d n \left (d+e x^r\right )^{3/2}}{45 r^2}-\frac {4 b n \left (d+e x^r\right )^{5/2}}{25 r^2}+\frac {16 b d^{5/2} n \tanh ^{-1}\left (\frac {\sqrt {d+e x^r}}{\sqrt {d}}\right )}{3 r^2}+\frac {2 b d^{5/2} n \tanh ^{-1}\left (\frac {\sqrt {d+e x^r}}{\sqrt {d}}\right )^2}{r^2}+\frac {2}{15} \left (\frac {15 d^2 \sqrt {d+e x^r}}{r}+\frac {5 d \left (d+e x^r\right )^{3/2}}{r}+\frac {3 \left (d+e x^r\right )^{5/2}}{r}-\frac {15 d^{5/2} \tanh ^{-1}\left (\frac {\sqrt {d+e x^r}}{\sqrt {d}}\right )}{r}\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {4 b d^{5/2} n \tanh ^{-1}\left (\frac {\sqrt {d+e x^r}}{\sqrt {d}}\right ) \log \left (\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x^r}}\right )}{r^2}-\frac {\left (4 b d^{5/2} n\right ) \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1-\frac {\sqrt {d+e x^r}}{\sqrt {d}}}\right )}{r^2}-\frac {\left (4 b d^3 n\right ) \operatorname {Subst}\left (\int \frac {1}{-\frac {d}{e}+\frac {x^2}{e}} \, dx,x,\sqrt {d+e x^r}\right )}{5 e r^2}\\ &=-\frac {92 b d^2 n \sqrt {d+e x^r}}{15 r^2}-\frac {32 b d n \left (d+e x^r\right )^{3/2}}{45 r^2}-\frac {4 b n \left (d+e x^r\right )^{5/2}}{25 r^2}+\frac {92 b d^{5/2} n \tanh ^{-1}\left (\frac {\sqrt {d+e x^r}}{\sqrt {d}}\right )}{15 r^2}+\frac {2 b d^{5/2} n \tanh ^{-1}\left (\frac {\sqrt {d+e x^r}}{\sqrt {d}}\right )^2}{r^2}+\frac {2}{15} \left (\frac {15 d^2 \sqrt {d+e x^r}}{r}+\frac {5 d \left (d+e x^r\right )^{3/2}}{r}+\frac {3 \left (d+e x^r\right )^{5/2}}{r}-\frac {15 d^{5/2} \tanh ^{-1}\left (\frac {\sqrt {d+e x^r}}{\sqrt {d}}\right )}{r}\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {4 b d^{5/2} n \tanh ^{-1}\left (\frac {\sqrt {d+e x^r}}{\sqrt {d}}\right ) \log \left (\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x^r}}\right )}{r^2}-\frac {2 b d^{5/2} n \text {Li}_2\left (1-\frac {2}{1-\frac {\sqrt {d+e x^r}}{\sqrt {d}}}\right )}{r^2}\\ \end {align*}
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Mathematica [F] time = 0.55, size = 0, normalized size = 0.00 \[ \int \frac {\left (d+e x^r\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{x} \, dx \]
Verification is Not applicable to the result.
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fricas [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (e x^{r} + d\right )}^{\frac {5}{2}} {\left (b \log \left (c x^{n}\right ) + a\right )}}{x}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.59, size = 0, normalized size = 0.00 \[ \int \frac {\left (e \,x^{r}+d \right )^{\frac {5}{2}} \left (b \ln \left (c \,x^{n}\right )+a \right )}{x}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{15} \, {\left (\frac {15 \, d^{\frac {5}{2}} \log \left (\frac {\sqrt {e x^{r} + d} - \sqrt {d}}{\sqrt {e x^{r} + d} + \sqrt {d}}\right )}{r} + \frac {2 \, {\left (3 \, {\left (e x^{r} + d\right )}^{\frac {5}{2}} + 5 \, {\left (e x^{r} + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {e x^{r} + d} d^{2}\right )}}{r}\right )} a + b \int \frac {{\left (e^{2} x^{2 \, r} \log \relax (c) + 2 \, d e x^{r} \log \relax (c) + d^{2} \log \relax (c) + {\left (e^{2} x^{2 \, r} + 2 \, d e x^{r} + d^{2}\right )} \log \left (x^{n}\right )\right )} \sqrt {e x^{r} + d}}{x}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (d+e\,x^r\right )}^{5/2}\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{x} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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