Optimal. Leaf size=251 \[ \frac {1}{3} \left (-\frac {3 \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{d^{5/2}}+\frac {3}{d^2 \sqrt {d+e x^2}}+\frac {1}{d \left (d+e x^2\right )^{3/2}}\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {b n \text {Li}_2\left (1-\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {e x^2+d}}\right )}{2 d^{5/2}}+\frac {b n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )^2}{2 d^{5/2}}+\frac {4 b n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{3 d^{5/2}}-\frac {b n \log \left (\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x^2}}\right ) \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{d^{5/2}}-\frac {b n}{3 d^2 \sqrt {d+e x^2}} \]
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Rubi [A] time = 0.40, antiderivative size = 251, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 9, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {266, 51, 63, 208, 2348, 5984, 5918, 2402, 2315} \[ -\frac {b n \text {PolyLog}\left (2,1-\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x^2}}\right )}{2 d^{5/2}}+\frac {1}{3} \left (\frac {3}{d^2 \sqrt {d+e x^2}}-\frac {3 \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{d^{5/2}}+\frac {1}{d \left (d+e x^2\right )^{3/2}}\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {b n}{3 d^2 \sqrt {d+e x^2}}+\frac {b n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )^2}{2 d^{5/2}}+\frac {4 b n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{3 d^{5/2}}-\frac {b n \log \left (\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x^2}}\right ) \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{d^{5/2}} \]
Antiderivative was successfully verified.
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Rule 51
Rule 63
Rule 208
Rule 266
Rule 2315
Rule 2348
Rule 2402
Rule 5918
Rule 5984
Rubi steps
\begin {align*} \int \frac {a+b \log \left (c x^n\right )}{x \left (d+e x^2\right )^{5/2}} \, dx &=\frac {1}{3} \left (\frac {1}{d \left (d+e x^2\right )^{3/2}}+\frac {3}{d^2 \sqrt {d+e x^2}}-\frac {3 \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{d^{5/2}}\right ) \left (a+b \log \left (c x^n\right )\right )-(b n) \int \left (\frac {1}{3 d x \left (d+e x^2\right )^{3/2}}+\frac {1}{d^2 x \sqrt {d+e x^2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{d^{5/2} x}\right ) \, dx\\ &=\frac {1}{3} \left (\frac {1}{d \left (d+e x^2\right )^{3/2}}+\frac {3}{d^2 \sqrt {d+e x^2}}-\frac {3 \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{d^{5/2}}\right ) \left (a+b \log \left (c x^n\right )\right )+\frac {(b n) \int \frac {\tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{x} \, dx}{d^{5/2}}-\frac {(b n) \int \frac {1}{x \sqrt {d+e x^2}} \, dx}{d^2}-\frac {(b n) \int \frac {1}{x \left (d+e x^2\right )^{3/2}} \, dx}{3 d}\\ &=\frac {1}{3} \left (\frac {1}{d \left (d+e x^2\right )^{3/2}}+\frac {3}{d^2 \sqrt {d+e x^2}}-\frac {3 \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{d^{5/2}}\right ) \left (a+b \log \left (c x^n\right )\right )+\frac {(b n) \operatorname {Subst}\left (\int \frac {\tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{x} \, dx,x,x^2\right )}{2 d^{5/2}}-\frac {(b n) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {d+e x}} \, dx,x,x^2\right )}{2 d^2}-\frac {(b n) \operatorname {Subst}\left (\int \frac {1}{x (d+e x)^{3/2}} \, dx,x,x^2\right )}{6 d}\\ &=-\frac {b n}{3 d^2 \sqrt {d+e x^2}}+\frac {1}{3} \left (\frac {1}{d \left (d+e x^2\right )^{3/2}}+\frac {3}{d^2 \sqrt {d+e x^2}}-\frac {3 \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{d^{5/2}}\right ) \left (a+b \log \left (c x^n\right )\right )+\frac {(b n) \operatorname {Subst}\left (\int \frac {x \tanh ^{-1}\left (\frac {x}{\sqrt {d}}\right )}{-d+x^2} \, dx,x,\sqrt {d+e x^2}\right )}{d^{5/2}}-\frac {(b n) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {d+e x}} \, dx,x,x^2\right )}{6 d^2}-\frac {(b n) \operatorname {Subst}\left (\int \frac {1}{-\frac {d}{e}+\frac {x^2}{e}} \, dx,x,\sqrt {d+e x^2}\right )}{d^2 e}\\ &=-\frac {b n}{3 d^2 \sqrt {d+e x^2}}+\frac {b n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{d^{5/2}}+\frac {b n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )^2}{2 d^{5/2}}+\frac {1}{3} \left (\frac {1}{d \left (d+e x^2\right )^{3/2}}+\frac {3}{d^2 \sqrt {d+e x^2}}-\frac {3 \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{d^{5/2}}\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {(b n) \operatorname {Subst}\left (\int \frac {\tanh ^{-1}\left (\frac {x}{\sqrt {d}}\right )}{1-\frac {x}{\sqrt {d}}} \, dx,x,\sqrt {d+e x^2}\right )}{d^3}-\frac {(b n) \operatorname {Subst}\left (\int \frac {1}{-\frac {d}{e}+\frac {x^2}{e}} \, dx,x,\sqrt {d+e x^2}\right )}{3 d^2 e}\\ &=-\frac {b n}{3 d^2 \sqrt {d+e x^2}}+\frac {4 b n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{3 d^{5/2}}+\frac {b n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )^2}{2 d^{5/2}}+\frac {1}{3} \left (\frac {1}{d \left (d+e x^2\right )^{3/2}}+\frac {3}{d^2 \sqrt {d+e x^2}}-\frac {3 \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{d^{5/2}}\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {b n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right ) \log \left (\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x^2}}\right )}{d^{5/2}}+\frac {(b n) \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^2}\right )}{d^3}\\ &=-\frac {b n}{3 d^2 \sqrt {d+e x^2}}+\frac {4 b n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{3 d^{5/2}}+\frac {b n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )^2}{2 d^{5/2}}+\frac {1}{3} \left (\frac {1}{d \left (d+e x^2\right )^{3/2}}+\frac {3}{d^2 \sqrt {d+e x^2}}-\frac {3 \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{d^{5/2}}\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {b n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right ) \log \left (\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x^2}}\right )}{d^{5/2}}-\frac {(b n) \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1-\frac {\sqrt {d+e x^2}}{\sqrt {d}}}\right )}{d^{5/2}}\\ &=-\frac {b n}{3 d^2 \sqrt {d+e x^2}}+\frac {4 b n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{3 d^{5/2}}+\frac {b n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )^2}{2 d^{5/2}}+\frac {1}{3} \left (\frac {1}{d \left (d+e x^2\right )^{3/2}}+\frac {3}{d^2 \sqrt {d+e x^2}}-\frac {3 \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{d^{5/2}}\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {b n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right ) \log \left (\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x^2}}\right )}{d^{5/2}}-\frac {b n \text {Li}_2\left (1-\frac {2}{1-\frac {\sqrt {d+e x^2}}{\sqrt {d}}}\right )}{2 d^{5/2}}\\ \end {align*}
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Mathematica [C] time = 0.45, size = 273, normalized size = 1.09 \[ \frac {b n \sqrt {\frac {d}{e x^2}+1} \left (-3 d^{5/2} \left (d+e x^2\right )^2 \, _3F_2\left (\frac {5}{2},\frac {5}{2},\frac {5}{2};\frac {7}{2},\frac {7}{2};-\frac {d}{e x^2}\right )-75 e^{5/2} x^5 \log (x) \left (d+e x^2\right )^2 \sinh ^{-1}\left (\frac {\sqrt {d}}{\sqrt {e} x}\right )+25 \sqrt {d} e^3 x^6 \log (x) \sqrt {\frac {d}{e x^2}+1} \left (4 d+3 e x^2\right )\right )}{75 d^{5/2} e^2 x^4 \left (d+e x^2\right )^{5/2}}-\frac {\log \left (\sqrt {d} \sqrt {d+e x^2}+d\right ) \left (a+b \log \left (c x^n\right )-b n \log (x)\right )}{d^{5/2}}+\frac {\log (x) \left (a+b \log \left (c x^n\right )-b n \log (x)\right )}{d^{5/2}}+\frac {\left (4 d+3 e x^2\right ) \left (a+b \log \left (c x^n\right )-b n \log (x)\right )}{3 d^2 \left (d+e x^2\right )^{3/2}} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {e x^{2} + d} b \log \left (c x^{n}\right ) + \sqrt {e x^{2} + d} a}{e^{3} x^{7} + 3 \, d e^{2} x^{5} + 3 \, d^{2} e x^{3} + d^{3} x}, x\right ) \]
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 \[ \int \frac {b \log \left (c x^{n}\right ) + a}{{\left (e x^{2} + d\right )}^{\frac {5}{2}} x}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.35, size = 0, normalized size = 0.00 \[ \int \frac {b \ln \left (c \,x^{n}\right )+a}{\left (e \,x^{2}+d \right )^{\frac {5}{2}} x}\, dx \]
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 \[ -\frac {1}{3} \, a {\left (\frac {3 \, \operatorname {arsinh}\left (\frac {d}{\sqrt {d e} {\left | x \right |}}\right )}{d^{\frac {5}{2}}} - \frac {3}{\sqrt {e x^{2} + d} d^{2}} - \frac {1}{{\left (e x^{2} + d\right )}^{\frac {3}{2}} d}\right )} + b \int \frac {\log \relax (c) + \log \left (x^{n}\right )}{{\left (e^{2} x^{5} + 2 \, d e x^{3} + d^{2} x\right )} \sqrt {e x^{2} + d}}\,{d x} \]
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 \[ \int \frac {a+b\,\ln \left (c\,x^n\right )}{x\,{\left (e\,x^2+d\right )}^{5/2}} \,d x \]
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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