Optimal. Leaf size=184 \[ -\frac{i b \text{PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right )}{8 c^3 d^3}+\frac{i b \text{PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right )}{8 c^3 d^3}+\frac{x \left (a+b \sinh ^{-1}(c x)\right )}{8 c^2 d^3 \left (c^2 x^2+1\right )}-\frac{x \left (a+b \sinh ^{-1}(c x)\right )}{4 c^2 d^3 \left (c^2 x^2+1\right )^2}+\frac{\tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{4 c^3 d^3}+\frac{b}{8 c^3 d^3 \sqrt{c^2 x^2+1}}-\frac{b}{12 c^3 d^3 \left (c^2 x^2+1\right )^{3/2}} \]
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Rubi [A] time = 0.175743, antiderivative size = 184, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 7, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.292, Rules used = {5751, 5690, 5693, 4180, 2279, 2391, 261} \[ -\frac{i b \text{PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right )}{8 c^3 d^3}+\frac{i b \text{PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right )}{8 c^3 d^3}+\frac{x \left (a+b \sinh ^{-1}(c x)\right )}{8 c^2 d^3 \left (c^2 x^2+1\right )}-\frac{x \left (a+b \sinh ^{-1}(c x)\right )}{4 c^2 d^3 \left (c^2 x^2+1\right )^2}+\frac{\tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{4 c^3 d^3}+\frac{b}{8 c^3 d^3 \sqrt{c^2 x^2+1}}-\frac{b}{12 c^3 d^3 \left (c^2 x^2+1\right )^{3/2}} \]
Antiderivative was successfully verified.
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Rule 5751
Rule 5690
Rule 5693
Rule 4180
Rule 2279
Rule 2391
Rule 261
Rubi steps
\begin{align*} \int \frac{x^2 \left (a+b \sinh ^{-1}(c x)\right )}{\left (d+c^2 d x^2\right )^3} \, dx &=-\frac{x \left (a+b \sinh ^{-1}(c x)\right )}{4 c^2 d^3 \left (1+c^2 x^2\right )^2}+\frac{b \int \frac{x}{\left (1+c^2 x^2\right )^{5/2}} \, dx}{4 c d^3}+\frac{\int \frac{a+b \sinh ^{-1}(c x)}{\left (d+c^2 d x^2\right )^2} \, dx}{4 c^2 d}\\ &=-\frac{b}{12 c^3 d^3 \left (1+c^2 x^2\right )^{3/2}}-\frac{x \left (a+b \sinh ^{-1}(c x)\right )}{4 c^2 d^3 \left (1+c^2 x^2\right )^2}+\frac{x \left (a+b \sinh ^{-1}(c x)\right )}{8 c^2 d^3 \left (1+c^2 x^2\right )}-\frac{b \int \frac{x}{\left (1+c^2 x^2\right )^{3/2}} \, dx}{8 c d^3}+\frac{\int \frac{a+b \sinh ^{-1}(c x)}{d+c^2 d x^2} \, dx}{8 c^2 d^2}\\ &=-\frac{b}{12 c^3 d^3 \left (1+c^2 x^2\right )^{3/2}}+\frac{b}{8 c^3 d^3 \sqrt{1+c^2 x^2}}-\frac{x \left (a+b \sinh ^{-1}(c x)\right )}{4 c^2 d^3 \left (1+c^2 x^2\right )^2}+\frac{x \left (a+b \sinh ^{-1}(c x)\right )}{8 c^2 d^3 \left (1+c^2 x^2\right )}+\frac{\operatorname{Subst}\left (\int (a+b x) \text{sech}(x) \, dx,x,\sinh ^{-1}(c x)\right )}{8 c^3 d^3}\\ &=-\frac{b}{12 c^3 d^3 \left (1+c^2 x^2\right )^{3/2}}+\frac{b}{8 c^3 d^3 \sqrt{1+c^2 x^2}}-\frac{x \left (a+b \sinh ^{-1}(c x)\right )}{4 c^2 d^3 \left (1+c^2 x^2\right )^2}+\frac{x \left (a+b \sinh ^{-1}(c x)\right )}{8 c^2 d^3 \left (1+c^2 x^2\right )}+\frac{\left (a+b \sinh ^{-1}(c x)\right ) \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{4 c^3 d^3}-\frac{(i b) \operatorname{Subst}\left (\int \log \left (1-i e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{8 c^3 d^3}+\frac{(i b) \operatorname{Subst}\left (\int \log \left (1+i e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{8 c^3 d^3}\\ &=-\frac{b}{12 c^3 d^3 \left (1+c^2 x^2\right )^{3/2}}+\frac{b}{8 c^3 d^3 \sqrt{1+c^2 x^2}}-\frac{x \left (a+b \sinh ^{-1}(c x)\right )}{4 c^2 d^3 \left (1+c^2 x^2\right )^2}+\frac{x \left (a+b \sinh ^{-1}(c x)\right )}{8 c^2 d^3 \left (1+c^2 x^2\right )}+\frac{\left (a+b \sinh ^{-1}(c x)\right ) \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{4 c^3 d^3}-\frac{(i b) \operatorname{Subst}\left (\int \frac{\log (1-i x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{8 c^3 d^3}+\frac{(i b) \operatorname{Subst}\left (\int \frac{\log (1+i x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{8 c^3 d^3}\\ &=-\frac{b}{12 c^3 d^3 \left (1+c^2 x^2\right )^{3/2}}+\frac{b}{8 c^3 d^3 \sqrt{1+c^2 x^2}}-\frac{x \left (a+b \sinh ^{-1}(c x)\right )}{4 c^2 d^3 \left (1+c^2 x^2\right )^2}+\frac{x \left (a+b \sinh ^{-1}(c x)\right )}{8 c^2 d^3 \left (1+c^2 x^2\right )}+\frac{\left (a+b \sinh ^{-1}(c x)\right ) \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{4 c^3 d^3}-\frac{i b \text{Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )}{8 c^3 d^3}+\frac{i b \text{Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )}{8 c^3 d^3}\\ \end{align*}
Mathematica [A] time = 0.24785, size = 340, normalized size = 1.85 \[ \frac{-3 i b \left (c^2 x^2+1\right )^2 \text{PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right )+3 i b \left (c^2 x^2+1\right )^2 \text{PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right )+3 a c^3 x^3+3 a c^4 x^4 \tan ^{-1}(c x)+6 a c^2 x^2 \tan ^{-1}(c x)-3 a c x+3 a \tan ^{-1}(c x)+3 b c^2 x^2 \sqrt{c^2 x^2+1}+b \sqrt{c^2 x^2+1}+3 b c^3 x^3 \sinh ^{-1}(c x)+3 i b c^4 x^4 \sinh ^{-1}(c x) \log \left (1-i e^{\sinh ^{-1}(c x)}\right )-3 i b c^4 x^4 \sinh ^{-1}(c x) \log \left (1+i e^{\sinh ^{-1}(c x)}\right )+6 i b c^2 x^2 \sinh ^{-1}(c x) \log \left (1-i e^{\sinh ^{-1}(c x)}\right )-6 i b c^2 x^2 \sinh ^{-1}(c x) \log \left (1+i e^{\sinh ^{-1}(c x)}\right )-3 b c x \sinh ^{-1}(c x)+3 i b \sinh ^{-1}(c x) \log \left (1-i e^{\sinh ^{-1}(c x)}\right )-3 i b \sinh ^{-1}(c x) \log \left (1+i e^{\sinh ^{-1}(c x)}\right )}{24 c^3 d^3 \left (c^2 x^2+1\right )^2} \]
Warning: Unable to verify antiderivative.
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Maple [A] time = 0.013, size = 307, normalized size = 1.7 \begin{align*}{\frac{a{x}^{3}}{8\,{d}^{3} \left ({c}^{2}{x}^{2}+1 \right ) ^{2}}}-{\frac{ax}{8\,{c}^{2}{d}^{3} \left ({c}^{2}{x}^{2}+1 \right ) ^{2}}}+{\frac{a\arctan \left ( cx \right ) }{8\,{c}^{3}{d}^{3}}}+{\frac{b{\it Arcsinh} \left ( cx \right ){x}^{3}}{8\,{d}^{3} \left ({c}^{2}{x}^{2}+1 \right ) ^{2}}}-{\frac{b{\it Arcsinh} \left ( cx \right ) x}{8\,{c}^{2}{d}^{3} \left ({c}^{2}{x}^{2}+1 \right ) ^{2}}}+{\frac{b{\it Arcsinh} \left ( cx \right ) \arctan \left ( cx \right ) }{8\,{c}^{3}{d}^{3}}}+{\frac{b\arctan \left ( cx \right ) }{8\,{c}^{3}{d}^{3}}\ln \left ( 1+{i \left ( 1+icx \right ){\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}} \right ) }-{\frac{b\arctan \left ( cx \right ) }{8\,{c}^{3}{d}^{3}}\ln \left ( 1-{i \left ( 1+icx \right ){\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}} \right ) }-{\frac{{\frac{i}{8}}b}{{c}^{3}{d}^{3}}{\it dilog} \left ( 1+{i \left ( 1+icx \right ){\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}} \right ) }+{\frac{{\frac{i}{8}}b}{{c}^{3}{d}^{3}}{\it dilog} \left ( 1-{i \left ( 1+icx \right ){\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}} \right ) }+{\frac{b{x}^{2}}{8\,c{d}^{3}} \left ({c}^{2}{x}^{2}+1 \right ) ^{-{\frac{3}{2}}}}+{\frac{b}{24\,{c}^{3}{d}^{3}} \left ({c}^{2}{x}^{2}+1 \right ) ^{-{\frac{3}{2}}}} \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*} \frac{1}{8} \, a{\left (\frac{c^{2} x^{3} - x}{c^{6} d^{3} x^{4} + 2 \, c^{4} d^{3} x^{2} + c^{2} d^{3}} + \frac{\arctan \left (c x\right )}{c^{3} d^{3}}\right )} + b \int \frac{x^{2} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right )}{c^{6} d^{3} x^{6} + 3 \, c^{4} d^{3} x^{4} + 3 \, c^{2} d^{3} x^{2} + d^{3}}\,{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{b x^{2} \operatorname{arsinh}\left (c x\right ) + a x^{2}}{c^{6} d^{3} x^{6} + 3 \, c^{4} d^{3} x^{4} + 3 \, c^{2} d^{3} x^{2} + d^{3}}, 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 x^{2}}{c^{6} x^{6} + 3 c^{4} x^{4} + 3 c^{2} x^{2} + 1}\, dx + \int \frac{b x^{2} \operatorname{asinh}{\left (c x \right )}}{c^{6} x^{6} + 3 c^{4} x^{4} + 3 c^{2} x^{2} + 1}\, dx}{d^{3}} \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 )} x^{2}}{{\left (c^{2} d x^{2} + d\right )}^{3}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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