Optimal. Leaf size=332 \[ -\frac{m \left (a+b \sinh ^{-1}(c x)\right ) \text{PolyLog}\left (2,-\frac{g e^{\sinh ^{-1}(c x)}}{c f-\sqrt{c^2 f^2+g^2}}\right )}{c}-\frac{m \left (a+b \sinh ^{-1}(c x)\right ) \text{PolyLog}\left (2,-\frac{g e^{\sinh ^{-1}(c x)}}{\sqrt{c^2 f^2+g^2}+c f}\right )}{c}+\frac{b m \text{PolyLog}\left (3,-\frac{g e^{\sinh ^{-1}(c x)}}{c f-\sqrt{c^2 f^2+g^2}}\right )}{c}+\frac{b m \text{PolyLog}\left (3,-\frac{g e^{\sinh ^{-1}(c x)}}{\sqrt{c^2 f^2+g^2}+c f}\right )}{c}+\frac{m \left (a+b \sinh ^{-1}(c x)\right )^3}{6 b^2 c}-\frac{m \left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (\frac{g e^{\sinh ^{-1}(c x)}}{c f-\sqrt{c^2 f^2+g^2}}+1\right )}{2 b c}-\frac{m \left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (\frac{g e^{\sinh ^{-1}(c x)}}{\sqrt{c^2 f^2+g^2}+c f}+1\right )}{2 b c}+\frac{\left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (h (f+g x)^m\right )}{2 b c} \]
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Rubi [A] time = 0.558748, antiderivative size = 332, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 8, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {5675, 5838, 5799, 5561, 2190, 2531, 2282, 6589} \[ -\frac{m \left (a+b \sinh ^{-1}(c x)\right ) \text{PolyLog}\left (2,-\frac{g e^{\sinh ^{-1}(c x)}}{c f-\sqrt{c^2 f^2+g^2}}\right )}{c}-\frac{m \left (a+b \sinh ^{-1}(c x)\right ) \text{PolyLog}\left (2,-\frac{g e^{\sinh ^{-1}(c x)}}{\sqrt{c^2 f^2+g^2}+c f}\right )}{c}+\frac{b m \text{PolyLog}\left (3,-\frac{g e^{\sinh ^{-1}(c x)}}{c f-\sqrt{c^2 f^2+g^2}}\right )}{c}+\frac{b m \text{PolyLog}\left (3,-\frac{g e^{\sinh ^{-1}(c x)}}{\sqrt{c^2 f^2+g^2}+c f}\right )}{c}+\frac{m \left (a+b \sinh ^{-1}(c x)\right )^3}{6 b^2 c}-\frac{m \left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (\frac{g e^{\sinh ^{-1}(c x)}}{c f-\sqrt{c^2 f^2+g^2}}+1\right )}{2 b c}-\frac{m \left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (\frac{g e^{\sinh ^{-1}(c x)}}{\sqrt{c^2 f^2+g^2}+c f}+1\right )}{2 b c}+\frac{\left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (h (f+g x)^m\right )}{2 b c} \]
Antiderivative was successfully verified.
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Rule 5675
Rule 5838
Rule 5799
Rule 5561
Rule 2190
Rule 2531
Rule 2282
Rule 6589
Rubi steps
\begin{align*} \int \frac{\left (a+b \sinh ^{-1}(c x)\right ) \log \left (h (f+g x)^m\right )}{\sqrt{1+c^2 x^2}} \, dx &=\frac{\left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (h (f+g x)^m\right )}{2 b c}-\frac{(g m) \int \frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{f+g x} \, dx}{2 b c}\\ &=\frac{\left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (h (f+g x)^m\right )}{2 b c}-\frac{(g m) \operatorname{Subst}\left (\int \frac{(a+b x)^2 \cosh (x)}{c f+g \sinh (x)} \, dx,x,\sinh ^{-1}(c x)\right )}{2 b c}\\ &=\frac{m \left (a+b \sinh ^{-1}(c x)\right )^3}{6 b^2 c}+\frac{\left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (h (f+g x)^m\right )}{2 b c}-\frac{(g m) \operatorname{Subst}\left (\int \frac{e^x (a+b x)^2}{c f+e^x g-\sqrt{c^2 f^2+g^2}} \, dx,x,\sinh ^{-1}(c x)\right )}{2 b c}-\frac{(g m) \operatorname{Subst}\left (\int \frac{e^x (a+b x)^2}{c f+e^x g+\sqrt{c^2 f^2+g^2}} \, dx,x,\sinh ^{-1}(c x)\right )}{2 b c}\\ &=\frac{m \left (a+b \sinh ^{-1}(c x)\right )^3}{6 b^2 c}-\frac{m \left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (1+\frac{e^{\sinh ^{-1}(c x)} g}{c f-\sqrt{c^2 f^2+g^2}}\right )}{2 b c}-\frac{m \left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (1+\frac{e^{\sinh ^{-1}(c x)} g}{c f+\sqrt{c^2 f^2+g^2}}\right )}{2 b c}+\frac{\left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (h (f+g x)^m\right )}{2 b c}+\frac{m \operatorname{Subst}\left (\int (a+b x) \log \left (1+\frac{e^x g}{c f-\sqrt{c^2 f^2+g^2}}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{c}+\frac{m \operatorname{Subst}\left (\int (a+b x) \log \left (1+\frac{e^x g}{c f+\sqrt{c^2 f^2+g^2}}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{c}\\ &=\frac{m \left (a+b \sinh ^{-1}(c x)\right )^3}{6 b^2 c}-\frac{m \left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (1+\frac{e^{\sinh ^{-1}(c x)} g}{c f-\sqrt{c^2 f^2+g^2}}\right )}{2 b c}-\frac{m \left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (1+\frac{e^{\sinh ^{-1}(c x)} g}{c f+\sqrt{c^2 f^2+g^2}}\right )}{2 b c}+\frac{\left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (h (f+g x)^m\right )}{2 b c}-\frac{m \left (a+b \sinh ^{-1}(c x)\right ) \text{Li}_2\left (-\frac{e^{\sinh ^{-1}(c x)} g}{c f-\sqrt{c^2 f^2+g^2}}\right )}{c}-\frac{m \left (a+b \sinh ^{-1}(c x)\right ) \text{Li}_2\left (-\frac{e^{\sinh ^{-1}(c x)} g}{c f+\sqrt{c^2 f^2+g^2}}\right )}{c}+\frac{(b m) \operatorname{Subst}\left (\int \text{Li}_2\left (-\frac{e^x g}{c f-\sqrt{c^2 f^2+g^2}}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{c}+\frac{(b m) \operatorname{Subst}\left (\int \text{Li}_2\left (-\frac{e^x g}{c f+\sqrt{c^2 f^2+g^2}}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{c}\\ &=\frac{m \left (a+b \sinh ^{-1}(c x)\right )^3}{6 b^2 c}-\frac{m \left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (1+\frac{e^{\sinh ^{-1}(c x)} g}{c f-\sqrt{c^2 f^2+g^2}}\right )}{2 b c}-\frac{m \left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (1+\frac{e^{\sinh ^{-1}(c x)} g}{c f+\sqrt{c^2 f^2+g^2}}\right )}{2 b c}+\frac{\left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (h (f+g x)^m\right )}{2 b c}-\frac{m \left (a+b \sinh ^{-1}(c x)\right ) \text{Li}_2\left (-\frac{e^{\sinh ^{-1}(c x)} g}{c f-\sqrt{c^2 f^2+g^2}}\right )}{c}-\frac{m \left (a+b \sinh ^{-1}(c x)\right ) \text{Li}_2\left (-\frac{e^{\sinh ^{-1}(c x)} g}{c f+\sqrt{c^2 f^2+g^2}}\right )}{c}+\frac{(b m) \operatorname{Subst}\left (\int \frac{\text{Li}_2\left (\frac{g x}{-c f+\sqrt{c^2 f^2+g^2}}\right )}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{c}+\frac{(b m) \operatorname{Subst}\left (\int \frac{\text{Li}_2\left (-\frac{g x}{c f+\sqrt{c^2 f^2+g^2}}\right )}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{c}\\ &=\frac{m \left (a+b \sinh ^{-1}(c x)\right )^3}{6 b^2 c}-\frac{m \left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (1+\frac{e^{\sinh ^{-1}(c x)} g}{c f-\sqrt{c^2 f^2+g^2}}\right )}{2 b c}-\frac{m \left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (1+\frac{e^{\sinh ^{-1}(c x)} g}{c f+\sqrt{c^2 f^2+g^2}}\right )}{2 b c}+\frac{\left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (h (f+g x)^m\right )}{2 b c}-\frac{m \left (a+b \sinh ^{-1}(c x)\right ) \text{Li}_2\left (-\frac{e^{\sinh ^{-1}(c x)} g}{c f-\sqrt{c^2 f^2+g^2}}\right )}{c}-\frac{m \left (a+b \sinh ^{-1}(c x)\right ) \text{Li}_2\left (-\frac{e^{\sinh ^{-1}(c x)} g}{c f+\sqrt{c^2 f^2+g^2}}\right )}{c}+\frac{b m \text{Li}_3\left (-\frac{e^{\sinh ^{-1}(c x)} g}{c f-\sqrt{c^2 f^2+g^2}}\right )}{c}+\frac{b m \text{Li}_3\left (-\frac{e^{\sinh ^{-1}(c x)} g}{c f+\sqrt{c^2 f^2+g^2}}\right )}{c}\\ \end{align*}
Mathematica [A] time = 0.30484, size = 303, normalized size = 0.91 \[ -\frac{2 b m \left (\left (a+b \sinh ^{-1}(c x)\right ) \text{PolyLog}\left (2,\frac{g e^{\sinh ^{-1}(c x)}}{\sqrt{c^2 f^2+g^2}-c f}\right )-b \text{PolyLog}\left (3,\frac{g e^{\sinh ^{-1}(c x)}}{\sqrt{c^2 f^2+g^2}-c f}\right )\right )+2 b m \left (\left (a+b \sinh ^{-1}(c x)\right ) \text{PolyLog}\left (2,-\frac{g e^{\sinh ^{-1}(c x)}}{\sqrt{c^2 f^2+g^2}+c f}\right )-b \text{PolyLog}\left (3,-\frac{g e^{\sinh ^{-1}(c x)}}{\sqrt{c^2 f^2+g^2}+c f}\right )\right )+m \left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (\frac{g e^{\sinh ^{-1}(c x)}}{c f-\sqrt{c^2 f^2+g^2}}+1\right )+m \left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (\frac{g e^{\sinh ^{-1}(c x)}}{\sqrt{c^2 f^2+g^2}+c f}+1\right )-\left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (h (f+g x)^m\right )-\frac{m \left (a+b \sinh ^{-1}(c x)\right )^3}{3 b}}{2 b c} \]
Antiderivative was successfully verified.
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Maple [F] time = 1.241, size = 0, normalized size = 0. \begin{align*} \int{ \left ( a+b{\it Arcsinh} \left ( cx \right ) \right ) \ln \left ( h \left ( gx+f \right ) ^{m} \right ){\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \operatorname{arsinh}\left (c x\right ) + a\right )} \log \left ({\left (g x + f\right )}^{m} h\right )}{\sqrt{c^{2} x^{2} + 1}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (b \operatorname{arsinh}\left (c x\right ) + a\right )} \log \left ({\left (g x + f\right )}^{m} h\right )}{\sqrt{c^{2} x^{2} + 1}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b \operatorname{asinh}{\left (c x \right )}\right ) \log{\left (h \left (f + g x\right )^{m} \right )}}{\sqrt{c^{2} x^{2} + 1}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \operatorname{arsinh}\left (c x\right ) + a\right )} \log \left ({\left (g x + f\right )}^{m} h\right )}{\sqrt{c^{2} x^{2} + 1}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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