Optimal. Leaf size=239 \[ -\frac{5 i b c^3 \text{PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right )}{2 d^2}+\frac{5 i b c^3 \text{PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right )}{2 d^2}+\frac{5 c^4 x \left (a+b \sinh ^{-1}(c x)\right )}{2 d^2 \left (c^2 x^2+1\right )}+\frac{5 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 d^2 x \left (c^2 x^2+1\right )}-\frac{a+b \sinh ^{-1}(c x)}{3 d^2 x^3 \left (c^2 x^2+1\right )}+\frac{5 c^3 \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{d^2}+\frac{b c^3}{3 d^2 \sqrt{c^2 x^2+1}}-\frac{b c}{6 d^2 x^2 \sqrt{c^2 x^2+1}}+\frac{13 b c^3 \tanh ^{-1}\left (\sqrt{c^2 x^2+1}\right )}{6 d^2} \]
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Rubi [A] time = 0.309059, antiderivative size = 264, normalized size of antiderivative = 1.1, number of steps used = 19, number of rules used = 11, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.458, Rules used = {5747, 5690, 5693, 4180, 2279, 2391, 261, 266, 51, 63, 208} \[ -\frac{5 i b c^3 \text{PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right )}{2 d^2}+\frac{5 i b c^3 \text{PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right )}{2 d^2}+\frac{5 c^4 x \left (a+b \sinh ^{-1}(c x)\right )}{2 d^2 \left (c^2 x^2+1\right )}+\frac{5 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 d^2 x \left (c^2 x^2+1\right )}-\frac{a+b \sinh ^{-1}(c x)}{3 d^2 x^3 \left (c^2 x^2+1\right )}+\frac{5 c^3 \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{d^2}+\frac{5 b c^3}{6 d^2 \sqrt{c^2 x^2+1}}-\frac{b c \sqrt{c^2 x^2+1}}{2 d^2 x^2}+\frac{b c}{3 d^2 x^2 \sqrt{c^2 x^2+1}}+\frac{13 b c^3 \tanh ^{-1}\left (\sqrt{c^2 x^2+1}\right )}{6 d^2} \]
Antiderivative was successfully verified.
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Rule 5747
Rule 5690
Rule 5693
Rule 4180
Rule 2279
Rule 2391
Rule 261
Rule 266
Rule 51
Rule 63
Rule 208
Rubi steps
\begin{align*} \int \frac{a+b \sinh ^{-1}(c x)}{x^4 \left (d+c^2 d x^2\right )^2} \, dx &=-\frac{a+b \sinh ^{-1}(c x)}{3 d^2 x^3 \left (1+c^2 x^2\right )}-\frac{1}{3} \left (5 c^2\right ) \int \frac{a+b \sinh ^{-1}(c x)}{x^2 \left (d+c^2 d x^2\right )^2} \, dx+\frac{(b c) \int \frac{1}{x^3 \left (1+c^2 x^2\right )^{3/2}} \, dx}{3 d^2}\\ &=-\frac{a+b \sinh ^{-1}(c x)}{3 d^2 x^3 \left (1+c^2 x^2\right )}+\frac{5 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 d^2 x \left (1+c^2 x^2\right )}+\left (5 c^4\right ) \int \frac{a+b \sinh ^{-1}(c x)}{\left (d+c^2 d x^2\right )^2} \, dx+\frac{(b c) \operatorname{Subst}\left (\int \frac{1}{x^2 \left (1+c^2 x\right )^{3/2}} \, dx,x,x^2\right )}{6 d^2}-\frac{\left (5 b c^3\right ) \int \frac{1}{x \left (1+c^2 x^2\right )^{3/2}} \, dx}{3 d^2}\\ &=\frac{b c}{3 d^2 x^2 \sqrt{1+c^2 x^2}}-\frac{a+b \sinh ^{-1}(c x)}{3 d^2 x^3 \left (1+c^2 x^2\right )}+\frac{5 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 d^2 x \left (1+c^2 x^2\right )}+\frac{5 c^4 x \left (a+b \sinh ^{-1}(c x)\right )}{2 d^2 \left (1+c^2 x^2\right )}+\frac{(b c) \operatorname{Subst}\left (\int \frac{1}{x^2 \sqrt{1+c^2 x}} \, dx,x,x^2\right )}{2 d^2}-\frac{\left (5 b c^3\right ) \operatorname{Subst}\left (\int \frac{1}{x \left (1+c^2 x\right )^{3/2}} \, dx,x,x^2\right )}{6 d^2}-\frac{\left (5 b c^5\right ) \int \frac{x}{\left (1+c^2 x^2\right )^{3/2}} \, dx}{2 d^2}+\frac{\left (5 c^4\right ) \int \frac{a+b \sinh ^{-1}(c x)}{d+c^2 d x^2} \, dx}{2 d}\\ &=\frac{5 b c^3}{6 d^2 \sqrt{1+c^2 x^2}}+\frac{b c}{3 d^2 x^2 \sqrt{1+c^2 x^2}}-\frac{b c \sqrt{1+c^2 x^2}}{2 d^2 x^2}-\frac{a+b \sinh ^{-1}(c x)}{3 d^2 x^3 \left (1+c^2 x^2\right )}+\frac{5 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 d^2 x \left (1+c^2 x^2\right )}+\frac{5 c^4 x \left (a+b \sinh ^{-1}(c x)\right )}{2 d^2 \left (1+c^2 x^2\right )}+\frac{\left (5 c^3\right ) \operatorname{Subst}\left (\int (a+b x) \text{sech}(x) \, dx,x,\sinh ^{-1}(c x)\right )}{2 d^2}-\frac{\left (b c^3\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{1+c^2 x}} \, dx,x,x^2\right )}{4 d^2}-\frac{\left (5 b c^3\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{1+c^2 x}} \, dx,x,x^2\right )}{6 d^2}\\ &=\frac{5 b c^3}{6 d^2 \sqrt{1+c^2 x^2}}+\frac{b c}{3 d^2 x^2 \sqrt{1+c^2 x^2}}-\frac{b c \sqrt{1+c^2 x^2}}{2 d^2 x^2}-\frac{a+b \sinh ^{-1}(c x)}{3 d^2 x^3 \left (1+c^2 x^2\right )}+\frac{5 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 d^2 x \left (1+c^2 x^2\right )}+\frac{5 c^4 x \left (a+b \sinh ^{-1}(c x)\right )}{2 d^2 \left (1+c^2 x^2\right )}+\frac{5 c^3 \left (a+b \sinh ^{-1}(c x)\right ) \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{d^2}-\frac{(b c) \operatorname{Subst}\left (\int \frac{1}{-\frac{1}{c^2}+\frac{x^2}{c^2}} \, dx,x,\sqrt{1+c^2 x^2}\right )}{2 d^2}-\frac{(5 b c) \operatorname{Subst}\left (\int \frac{1}{-\frac{1}{c^2}+\frac{x^2}{c^2}} \, dx,x,\sqrt{1+c^2 x^2}\right )}{3 d^2}-\frac{\left (5 i b c^3\right ) \operatorname{Subst}\left (\int \log \left (1-i e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{2 d^2}+\frac{\left (5 i b c^3\right ) \operatorname{Subst}\left (\int \log \left (1+i e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{2 d^2}\\ &=\frac{5 b c^3}{6 d^2 \sqrt{1+c^2 x^2}}+\frac{b c}{3 d^2 x^2 \sqrt{1+c^2 x^2}}-\frac{b c \sqrt{1+c^2 x^2}}{2 d^2 x^2}-\frac{a+b \sinh ^{-1}(c x)}{3 d^2 x^3 \left (1+c^2 x^2\right )}+\frac{5 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 d^2 x \left (1+c^2 x^2\right )}+\frac{5 c^4 x \left (a+b \sinh ^{-1}(c x)\right )}{2 d^2 \left (1+c^2 x^2\right )}+\frac{5 c^3 \left (a+b \sinh ^{-1}(c x)\right ) \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{d^2}+\frac{13 b c^3 \tanh ^{-1}\left (\sqrt{1+c^2 x^2}\right )}{6 d^2}-\frac{\left (5 i b c^3\right ) \operatorname{Subst}\left (\int \frac{\log (1-i x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{2 d^2}+\frac{\left (5 i b c^3\right ) \operatorname{Subst}\left (\int \frac{\log (1+i x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{2 d^2}\\ &=\frac{5 b c^3}{6 d^2 \sqrt{1+c^2 x^2}}+\frac{b c}{3 d^2 x^2 \sqrt{1+c^2 x^2}}-\frac{b c \sqrt{1+c^2 x^2}}{2 d^2 x^2}-\frac{a+b \sinh ^{-1}(c x)}{3 d^2 x^3 \left (1+c^2 x^2\right )}+\frac{5 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 d^2 x \left (1+c^2 x^2\right )}+\frac{5 c^4 x \left (a+b \sinh ^{-1}(c x)\right )}{2 d^2 \left (1+c^2 x^2\right )}+\frac{5 c^3 \left (a+b \sinh ^{-1}(c x)\right ) \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{d^2}+\frac{13 b c^3 \tanh ^{-1}\left (\sqrt{1+c^2 x^2}\right )}{6 d^2}-\frac{5 i b c^3 \text{Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )}{2 d^2}+\frac{5 i b c^3 \text{Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )}{2 d^2}\\ \end{align*}
Mathematica [C] time = 0.634604, size = 311, normalized size = 1.3 \[ \frac{\frac{b c^3 \text{Hypergeometric2F1}\left (-\frac{1}{2},2,\frac{1}{2},c^2 x^2+1\right )}{\sqrt{c^2 x^2+1}}+5 b \left (-c^2\right )^{3/2} \text{PolyLog}\left (2,\frac{c e^{\sinh ^{-1}(c x)}}{\sqrt{-c^2}}\right )-5 b \left (-c^2\right )^{3/2} \text{PolyLog}\left (2,\frac{\sqrt{-c^2} e^{\sinh ^{-1}(c x)}}{c}\right )+\frac{a}{c^2 x^5+x^3}+\frac{5 a c^2}{x}+5 a c^3 \tan ^{-1}(c x)-\frac{5 a}{3 x^3}-\frac{5 b c \sqrt{c^2 x^2+1}}{6 x^2}+\frac{b \sinh ^{-1}(c x)}{c^2 x^5+x^3}+\frac{35}{6} b c^3 \tanh ^{-1}\left (\sqrt{c^2 x^2+1}\right )+\frac{5 b c^2 \sinh ^{-1}(c x)}{x}-5 b \left (-c^2\right )^{3/2} \sinh ^{-1}(c x) \log \left (\frac{c e^{\sinh ^{-1}(c x)}}{\sqrt{-c^2}}+1\right )+5 b \left (-c^2\right )^{3/2} \sinh ^{-1}(c x) \log \left (\frac{\sqrt{-c^2} e^{\sinh ^{-1}(c x)}}{c}+1\right )-\frac{5 b \sinh ^{-1}(c x)}{3 x^3}}{2 d^2} \]
Warning: Unable to verify antiderivative.
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Maple [A] time = 0.021, size = 332, normalized size = 1.4 \begin{align*} -{\frac{a}{3\,{d}^{2}{x}^{3}}}+2\,{\frac{{c}^{2}a}{{d}^{2}x}}+{\frac{{c}^{4}ax}{2\,{d}^{2} \left ({c}^{2}{x}^{2}+1 \right ) }}+{\frac{5\,{c}^{3}a\arctan \left ( cx \right ) }{2\,{d}^{2}}}-{\frac{b{\it Arcsinh} \left ( cx \right ) }{3\,{d}^{2}{x}^{3}}}+2\,{\frac{{c}^{2}b{\it Arcsinh} \left ( cx \right ) }{{d}^{2}x}}+{\frac{{c}^{4}b{\it Arcsinh} \left ( cx \right ) x}{2\,{d}^{2} \left ({c}^{2}{x}^{2}+1 \right ) }}+{\frac{5\,b{c}^{3}{\it Arcsinh} \left ( cx \right ) \arctan \left ( cx \right ) }{2\,{d}^{2}}}+{\frac{b{c}^{3}}{3\,{d}^{2}}{\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}}+{\frac{13\,b{c}^{3}}{6\,{d}^{2}}{\it Artanh} \left ({\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}} \right ) }-{\frac{bc}{6\,{d}^{2}{x}^{2}}{\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}}+{\frac{5\,b{c}^{3}\arctan \left ( cx \right ) }{2\,{d}^{2}}\ln \left ( 1+{i \left ( 1+icx \right ){\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}} \right ) }-{\frac{5\,b{c}^{3}\arctan \left ( cx \right ) }{2\,{d}^{2}}\ln \left ( 1-{i \left ( 1+icx \right ){\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}} \right ) }-{\frac{{\frac{5\,i}{2}}{c}^{3}b}{{d}^{2}}{\it dilog} \left ( 1+{i \left ( 1+icx \right ){\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}} \right ) }+{\frac{{\frac{5\,i}{2}}{c}^{3}b}{{d}^{2}}{\it dilog} \left ( 1-{i \left ( 1+icx \right ){\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}} \right ) } \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}{6} \,{\left (\frac{15 \, c^{3} \arctan \left (c x\right )}{d^{2}} + \frac{15 \, c^{4} x^{4} + 10 \, c^{2} x^{2} - 2}{c^{2} d^{2} x^{5} + d^{2} x^{3}}\right )} a + b \int \frac{\log \left (c x + \sqrt{c^{2} x^{2} + 1}\right )}{c^{4} d^{2} x^{8} + 2 \, c^{2} d^{2} x^{6} + d^{2} x^{4}}\,{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 \operatorname{arsinh}\left (c x\right ) + a}{c^{4} d^{2} x^{8} + 2 \, c^{2} d^{2} x^{6} + d^{2} x^{4}}, 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}{c^{4} x^{8} + 2 c^{2} x^{6} + x^{4}}\, dx + \int \frac{b \operatorname{asinh}{\left (c x \right )}}{c^{4} x^{8} + 2 c^{2} x^{6} + x^{4}}\, dx}{d^{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{b \operatorname{arsinh}\left (c x\right ) + a}{{\left (c^{2} d x^{2} + d\right )}^{2} x^{4}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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