Optimal. Leaf size=100 \[ -\frac{2 b^2 \text{PolyLog}\left (2,-e^{\sinh ^{-1}(c+d x)}\right )}{d e^2}+\frac{2 b^2 \text{PolyLog}\left (2,e^{\sinh ^{-1}(c+d x)}\right )}{d e^2}-\frac{\left (a+b \sinh ^{-1}(c+d x)\right )^2}{d e^2 (c+d x)}-\frac{4 b \tanh ^{-1}\left (e^{\sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )}{d e^2} \]
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Rubi [A] time = 0.175678, antiderivative size = 100, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.304, Rules used = {5865, 12, 5661, 5760, 4182, 2279, 2391} \[ -\frac{2 b^2 \text{PolyLog}\left (2,-e^{\sinh ^{-1}(c+d x)}\right )}{d e^2}+\frac{2 b^2 \text{PolyLog}\left (2,e^{\sinh ^{-1}(c+d x)}\right )}{d e^2}-\frac{\left (a+b \sinh ^{-1}(c+d x)\right )^2}{d e^2 (c+d x)}-\frac{4 b \tanh ^{-1}\left (e^{\sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )}{d e^2} \]
Antiderivative was successfully verified.
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Rule 5865
Rule 12
Rule 5661
Rule 5760
Rule 4182
Rule 2279
Rule 2391
Rubi steps
\begin{align*} \int \frac{\left (a+b \sinh ^{-1}(c+d x)\right )^2}{(c e+d e x)^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b \sinh ^{-1}(x)\right )^2}{e^2 x^2} \, dx,x,c+d x\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b \sinh ^{-1}(x)\right )^2}{x^2} \, dx,x,c+d x\right )}{d e^2}\\ &=-\frac{\left (a+b \sinh ^{-1}(c+d x)\right )^2}{d e^2 (c+d x)}+\frac{(2 b) \operatorname{Subst}\left (\int \frac{a+b \sinh ^{-1}(x)}{x \sqrt{1+x^2}} \, dx,x,c+d x\right )}{d e^2}\\ &=-\frac{\left (a+b \sinh ^{-1}(c+d x)\right )^2}{d e^2 (c+d x)}+\frac{(2 b) \operatorname{Subst}\left (\int (a+b x) \text{csch}(x) \, dx,x,\sinh ^{-1}(c+d x)\right )}{d e^2}\\ &=-\frac{\left (a+b \sinh ^{-1}(c+d x)\right )^2}{d e^2 (c+d x)}-\frac{4 b \left (a+b \sinh ^{-1}(c+d x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c+d x)}\right )}{d e^2}-\frac{\left (2 b^2\right ) \operatorname{Subst}\left (\int \log \left (1-e^x\right ) \, dx,x,\sinh ^{-1}(c+d x)\right )}{d e^2}+\frac{\left (2 b^2\right ) \operatorname{Subst}\left (\int \log \left (1+e^x\right ) \, dx,x,\sinh ^{-1}(c+d x)\right )}{d e^2}\\ &=-\frac{\left (a+b \sinh ^{-1}(c+d x)\right )^2}{d e^2 (c+d x)}-\frac{4 b \left (a+b \sinh ^{-1}(c+d x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c+d x)}\right )}{d e^2}-\frac{\left (2 b^2\right ) \operatorname{Subst}\left (\int \frac{\log (1-x)}{x} \, dx,x,e^{\sinh ^{-1}(c+d x)}\right )}{d e^2}+\frac{\left (2 b^2\right ) \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{\sinh ^{-1}(c+d x)}\right )}{d e^2}\\ &=-\frac{\left (a+b \sinh ^{-1}(c+d x)\right )^2}{d e^2 (c+d x)}-\frac{4 b \left (a+b \sinh ^{-1}(c+d x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c+d x)}\right )}{d e^2}-\frac{2 b^2 \text{Li}_2\left (-e^{\sinh ^{-1}(c+d x)}\right )}{d e^2}+\frac{2 b^2 \text{Li}_2\left (e^{\sinh ^{-1}(c+d x)}\right )}{d e^2}\\ \end{align*}
Mathematica [A] time = 0.590954, size = 154, normalized size = 1.54 \[ \frac{b^2 \left (2 \text{PolyLog}\left (2,-e^{-\sinh ^{-1}(c+d x)}\right )-2 \text{PolyLog}\left (2,e^{-\sinh ^{-1}(c+d x)}\right )+\sinh ^{-1}(c+d x) \left (-\frac{\sinh ^{-1}(c+d x)}{c+d x}+2 \log \left (1-e^{-\sinh ^{-1}(c+d x)}\right )-2 \log \left (e^{-\sinh ^{-1}(c+d x)}+1\right )\right )\right )-\frac{a^2}{c+d x}+a b \left (2 \log \left (\frac{2 \sinh ^2\left (\frac{1}{2} \sinh ^{-1}(c+d x)\right )}{c+d x}\right )-\frac{2 \sinh ^{-1}(c+d x)}{c+d x}\right )}{d e^2} \]
Warning: Unable to verify antiderivative.
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Maple [A] time = 0.046, size = 229, normalized size = 2.3 \begin{align*} -{\frac{{a}^{2}}{d{e}^{2} \left ( dx+c \right ) }}-{\frac{{b}^{2} \left ({\it Arcsinh} \left ( dx+c \right ) \right ) ^{2}}{d{e}^{2} \left ( dx+c \right ) }}-2\,{\frac{{b}^{2}{\it Arcsinh} \left ( dx+c \right ) \ln \left ( 1+dx+c+\sqrt{1+ \left ( dx+c \right ) ^{2}} \right ) }{d{e}^{2}}}-2\,{\frac{{b}^{2}{\it polylog} \left ( 2,-dx-c-\sqrt{1+ \left ( dx+c \right ) ^{2}} \right ) }{d{e}^{2}}}+2\,{\frac{{b}^{2}{\it Arcsinh} \left ( dx+c \right ) \ln \left ( 1-dx-c-\sqrt{1+ \left ( dx+c \right ) ^{2}} \right ) }{d{e}^{2}}}+2\,{\frac{{b}^{2}{\it polylog} \left ( 2,dx+c+\sqrt{1+ \left ( dx+c \right ) ^{2}} \right ) }{d{e}^{2}}}-2\,{\frac{ab{\it Arcsinh} \left ( dx+c \right ) }{d{e}^{2} \left ( dx+c \right ) }}-2\,{\frac{ab{\it Artanh} \left ({\frac{1}{\sqrt{1+ \left ( dx+c \right ) ^{2}}}} \right ) }{d{e}^{2}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \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{b^{2} \operatorname{arsinh}\left (d x + c\right )^{2} + 2 \, a b \operatorname{arsinh}\left (d x + c\right ) + a^{2}}{d^{2} e^{2} x^{2} + 2 \, c d e^{2} x + c^{2} e^{2}}, 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*} \frac{\int \frac{a^{2}}{c^{2} + 2 c d x + d^{2} x^{2}}\, dx + \int \frac{b^{2} \operatorname{asinh}^{2}{\left (c + d x \right )}}{c^{2} + 2 c d x + d^{2} x^{2}}\, dx + \int \frac{2 a b \operatorname{asinh}{\left (c + d x \right )}}{c^{2} + 2 c d x + d^{2} x^{2}}\, dx}{e^{2}} \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 (d x + c\right ) + a\right )}^{2}}{{\left (d e x + c e\right )}^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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