Optimal. Leaf size=326 \[ \frac {\sinh ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {2}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{\sqrt {g}}-\frac {b n \text {Li}_2\left (-\frac {\sqrt {2} e e^{\sinh ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {2}}\right )}}{d \sqrt {g}-\sqrt {g d^2+2 e^2}}\right )}{\sqrt {g}}-\frac {b n \text {Li}_2\left (-\frac {\sqrt {2} e e^{\sinh ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {2}}\right )}}{\sqrt {g} d+\sqrt {g d^2+2 e^2}}\right )}{\sqrt {g}}-\frac {b n \sinh ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {2}}\right ) \log \left (\frac {\sqrt {2} e e^{\sinh ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {2}}\right )}}{d \sqrt {g}-\sqrt {d^2 g+2 e^2}}+1\right )}{\sqrt {g}}-\frac {b n \sinh ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {2}}\right ) \log \left (\frac {\sqrt {2} e e^{\sinh ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {2}}\right )}}{\sqrt {d^2 g+2 e^2}+d \sqrt {g}}+1\right )}{\sqrt {g}}+\frac {b n \sinh ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {2}}\right )^2}{2 \sqrt {g}} \]
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Rubi [A] time = 0.42, antiderivative size = 326, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {215, 2404, 12, 5799, 5561, 2190, 2279, 2391} \[ -\frac {b n \text {PolyLog}\left (2,-\frac {\sqrt {2} e e^{\sinh ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {2}}\right )}}{d \sqrt {g}-\sqrt {d^2 g+2 e^2}}\right )}{\sqrt {g}}-\frac {b n \text {PolyLog}\left (2,-\frac {\sqrt {2} e e^{\sinh ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {2}}\right )}}{\sqrt {d^2 g+2 e^2}+d \sqrt {g}}\right )}{\sqrt {g}}+\frac {\sinh ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {2}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{\sqrt {g}}-\frac {b n \sinh ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {2}}\right ) \log \left (\frac {\sqrt {2} e e^{\sinh ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {2}}\right )}}{d \sqrt {g}-\sqrt {d^2 g+2 e^2}}+1\right )}{\sqrt {g}}-\frac {b n \sinh ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {2}}\right ) \log \left (\frac {\sqrt {2} e e^{\sinh ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {2}}\right )}}{\sqrt {d^2 g+2 e^2}+d \sqrt {g}}+1\right )}{\sqrt {g}}+\frac {b n \sinh ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {2}}\right )^2}{2 \sqrt {g}} \]
Antiderivative was successfully verified.
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Rule 12
Rule 215
Rule 2190
Rule 2279
Rule 2391
Rule 2404
Rule 5561
Rule 5799
Rubi steps
\begin {align*} \int \frac {a+b \log \left (c (d+e x)^n\right )}{\sqrt {2+g x^2}} \, dx &=\frac {\sinh ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {2}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{\sqrt {g}}-(b e n) \int \frac {\sinh ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {2}}\right )}{\sqrt {g} (d+e x)} \, dx\\ &=\frac {\sinh ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {2}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{\sqrt {g}}-\frac {(b e n) \int \frac {\sinh ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {2}}\right )}{d+e x} \, dx}{\sqrt {g}}\\ &=\frac {\sinh ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {2}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{\sqrt {g}}-\frac {(b e n) \operatorname {Subst}\left (\int \frac {x \cosh (x)}{\frac {d \sqrt {g}}{\sqrt {2}}+e \sinh (x)} \, dx,x,\sinh ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {2}}\right )\right )}{\sqrt {g}}\\ &=\frac {b n \sinh ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {2}}\right )^2}{2 \sqrt {g}}+\frac {\sinh ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {2}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{\sqrt {g}}-\frac {(b e n) \operatorname {Subst}\left (\int \frac {e^x x}{e e^x+\frac {d \sqrt {g}}{\sqrt {2}}-\frac {\sqrt {2 e^2+d^2 g}}{\sqrt {2}}} \, dx,x,\sinh ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {2}}\right )\right )}{\sqrt {g}}-\frac {(b e n) \operatorname {Subst}\left (\int \frac {e^x x}{e e^x+\frac {d \sqrt {g}}{\sqrt {2}}+\frac {\sqrt {2 e^2+d^2 g}}{\sqrt {2}}} \, dx,x,\sinh ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {2}}\right )\right )}{\sqrt {g}}\\ &=\frac {b n \sinh ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {2}}\right )^2}{2 \sqrt {g}}-\frac {b n \sinh ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {2}}\right ) \log \left (1+\frac {\sqrt {2} e e^{\sinh ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {2}}\right )}}{d \sqrt {g}-\sqrt {2 e^2+d^2 g}}\right )}{\sqrt {g}}-\frac {b n \sinh ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {2}}\right ) \log \left (1+\frac {\sqrt {2} e e^{\sinh ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {2}}\right )}}{d \sqrt {g}+\sqrt {2 e^2+d^2 g}}\right )}{\sqrt {g}}+\frac {\sinh ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {2}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{\sqrt {g}}+\frac {(b n) \operatorname {Subst}\left (\int \log \left (1+\frac {e e^x}{\frac {d \sqrt {g}}{\sqrt {2}}-\frac {\sqrt {2 e^2+d^2 g}}{\sqrt {2}}}\right ) \, dx,x,\sinh ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {2}}\right )\right )}{\sqrt {g}}+\frac {(b n) \operatorname {Subst}\left (\int \log \left (1+\frac {e e^x}{\frac {d \sqrt {g}}{\sqrt {2}}+\frac {\sqrt {2 e^2+d^2 g}}{\sqrt {2}}}\right ) \, dx,x,\sinh ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {2}}\right )\right )}{\sqrt {g}}\\ &=\frac {b n \sinh ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {2}}\right )^2}{2 \sqrt {g}}-\frac {b n \sinh ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {2}}\right ) \log \left (1+\frac {\sqrt {2} e e^{\sinh ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {2}}\right )}}{d \sqrt {g}-\sqrt {2 e^2+d^2 g}}\right )}{\sqrt {g}}-\frac {b n \sinh ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {2}}\right ) \log \left (1+\frac {\sqrt {2} e e^{\sinh ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {2}}\right )}}{d \sqrt {g}+\sqrt {2 e^2+d^2 g}}\right )}{\sqrt {g}}+\frac {\sinh ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {2}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{\sqrt {g}}+\frac {(b n) \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {e x}{\frac {d \sqrt {g}}{\sqrt {2}}-\frac {\sqrt {2 e^2+d^2 g}}{\sqrt {2}}}\right )}{x} \, dx,x,e^{\sinh ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {2}}\right )}\right )}{\sqrt {g}}+\frac {(b n) \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {e x}{\frac {d \sqrt {g}}{\sqrt {2}}+\frac {\sqrt {2 e^2+d^2 g}}{\sqrt {2}}}\right )}{x} \, dx,x,e^{\sinh ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {2}}\right )}\right )}{\sqrt {g}}\\ &=\frac {b n \sinh ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {2}}\right )^2}{2 \sqrt {g}}-\frac {b n \sinh ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {2}}\right ) \log \left (1+\frac {\sqrt {2} e e^{\sinh ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {2}}\right )}}{d \sqrt {g}-\sqrt {2 e^2+d^2 g}}\right )}{\sqrt {g}}-\frac {b n \sinh ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {2}}\right ) \log \left (1+\frac {\sqrt {2} e e^{\sinh ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {2}}\right )}}{d \sqrt {g}+\sqrt {2 e^2+d^2 g}}\right )}{\sqrt {g}}+\frac {\sinh ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {2}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{\sqrt {g}}-\frac {b n \text {Li}_2\left (-\frac {\sqrt {2} e e^{\sinh ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {2}}\right )}}{d \sqrt {g}-\sqrt {2 e^2+d^2 g}}\right )}{\sqrt {g}}-\frac {b n \text {Li}_2\left (-\frac {\sqrt {2} e e^{\sinh ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {2}}\right )}}{d \sqrt {g}+\sqrt {2 e^2+d^2 g}}\right )}{\sqrt {g}}\\ \end {align*}
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Mathematica [A] time = 0.25, size = 275, normalized size = 0.84 \[ \frac {\sinh ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {2}}\right ) \left (2 a+2 b \log \left (c (d+e x)^n\right )-2 b n \log \left (\frac {\sqrt {2} e e^{\sinh ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {2}}\right )}}{d \sqrt {g}-\sqrt {d^2 g+2 e^2}}+1\right )-2 b n \log \left (\frac {\sqrt {2} e e^{\sinh ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {2}}\right )}}{\sqrt {d^2 g+2 e^2}+d \sqrt {g}}+1\right )+b n \sinh ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {2}}\right )\right )-2 b n \text {Li}_2\left (\frac {\sqrt {2} e e^{\sinh ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {2}}\right )}}{\sqrt {g d^2+2 e^2}-d \sqrt {g}}\right )-2 b n \text {Li}_2\left (-\frac {\sqrt {2} e e^{\sinh ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {2}}\right )}}{\sqrt {g} d+\sqrt {g d^2+2 e^2}}\right )}{2 \sqrt {g}} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {g x^{2} + 2} b \log \left ({\left (e x + d\right )}^{n} c\right ) + \sqrt {g x^{2} + 2} a}{g x^{2} + 2}, x\right ) \]
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 \[ \int \frac {b \log \left ({\left (e x + d\right )}^{n} c\right ) + a}{\sqrt {g x^{2} + 2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.45, size = 0, normalized size = 0.00 \[ \int \frac {b \ln \left (c \left (e x +d \right )^{n}\right )+a}{\sqrt {g \,x^{2}+2}}\, dx \]
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 \[ b \int \frac {\log \left ({\left (e x + d\right )}^{n}\right ) + \log \relax (c)}{\sqrt {g x^{2} + 2}}\,{d x} + \frac {a \operatorname {arsinh}\left (\frac {1}{2} \, \sqrt {2} \sqrt {g} x\right )}{\sqrt {g}} \]
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 \[ \int \frac {a+b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )}{\sqrt {g\,x^2+2}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {a + b \log {\left (c \left (d + e x\right )^{n} \right )}}{\sqrt {g x^{2} + 2}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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