Optimal. Leaf size=124 \[ -\frac{i b c^2 \text{PolyLog}\left (2,-e^{2 i \cos ^{-1}(c x)}\right )}{2 d}+\frac{i b c^2 \text{PolyLog}\left (2,e^{2 i \cos ^{-1}(c x)}\right )}{2 d}+\frac{2 c^2 \tanh ^{-1}\left (e^{2 i \cos ^{-1}(c x)}\right ) \left (a+b \cos ^{-1}(c x)\right )}{d}-\frac{a+b \cos ^{-1}(c x)}{2 d x^2}+\frac{b c \sqrt{1-c^2 x^2}}{2 d x} \]
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Rubi [A] time = 0.185964, antiderivative size = 124, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.28, Rules used = {4702, 4680, 4419, 4183, 2279, 2391, 264} \[ -\frac{i b c^2 \text{PolyLog}\left (2,-e^{2 i \cos ^{-1}(c x)}\right )}{2 d}+\frac{i b c^2 \text{PolyLog}\left (2,e^{2 i \cos ^{-1}(c x)}\right )}{2 d}+\frac{2 c^2 \tanh ^{-1}\left (e^{2 i \cos ^{-1}(c x)}\right ) \left (a+b \cos ^{-1}(c x)\right )}{d}-\frac{a+b \cos ^{-1}(c x)}{2 d x^2}+\frac{b c \sqrt{1-c^2 x^2}}{2 d x} \]
Antiderivative was successfully verified.
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Rule 4702
Rule 4680
Rule 4419
Rule 4183
Rule 2279
Rule 2391
Rule 264
Rubi steps
\begin{align*} \int \frac{a+b \cos ^{-1}(c x)}{x^3 \left (d-c^2 d x^2\right )} \, dx &=-\frac{a+b \cos ^{-1}(c x)}{2 d x^2}+c^2 \int \frac{a+b \cos ^{-1}(c x)}{x \left (d-c^2 d x^2\right )} \, dx-\frac{(b c) \int \frac{1}{x^2 \sqrt{1-c^2 x^2}} \, dx}{2 d}\\ &=\frac{b c \sqrt{1-c^2 x^2}}{2 d x}-\frac{a+b \cos ^{-1}(c x)}{2 d x^2}-\frac{c^2 \operatorname{Subst}\left (\int (a+b x) \csc (x) \sec (x) \, dx,x,\cos ^{-1}(c x)\right )}{d}\\ &=\frac{b c \sqrt{1-c^2 x^2}}{2 d x}-\frac{a+b \cos ^{-1}(c x)}{2 d x^2}-\frac{\left (2 c^2\right ) \operatorname{Subst}\left (\int (a+b x) \csc (2 x) \, dx,x,\cos ^{-1}(c x)\right )}{d}\\ &=\frac{b c \sqrt{1-c^2 x^2}}{2 d x}-\frac{a+b \cos ^{-1}(c x)}{2 d x^2}+\frac{2 c^2 \left (a+b \cos ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{2 i \cos ^{-1}(c x)}\right )}{d}+\frac{\left (b c^2\right ) \operatorname{Subst}\left (\int \log \left (1-e^{2 i x}\right ) \, dx,x,\cos ^{-1}(c x)\right )}{d}-\frac{\left (b c^2\right ) \operatorname{Subst}\left (\int \log \left (1+e^{2 i x}\right ) \, dx,x,\cos ^{-1}(c x)\right )}{d}\\ &=\frac{b c \sqrt{1-c^2 x^2}}{2 d x}-\frac{a+b \cos ^{-1}(c x)}{2 d x^2}+\frac{2 c^2 \left (a+b \cos ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{2 i \cos ^{-1}(c x)}\right )}{d}-\frac{\left (i b c^2\right ) \operatorname{Subst}\left (\int \frac{\log (1-x)}{x} \, dx,x,e^{2 i \cos ^{-1}(c x)}\right )}{2 d}+\frac{\left (i b c^2\right ) \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{2 i \cos ^{-1}(c x)}\right )}{2 d}\\ &=\frac{b c \sqrt{1-c^2 x^2}}{2 d x}-\frac{a+b \cos ^{-1}(c x)}{2 d x^2}+\frac{2 c^2 \left (a+b \cos ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{2 i \cos ^{-1}(c x)}\right )}{d}-\frac{i b c^2 \text{Li}_2\left (-e^{2 i \cos ^{-1}(c x)}\right )}{2 d}+\frac{i b c^2 \text{Li}_2\left (e^{2 i \cos ^{-1}(c x)}\right )}{2 d}\\ \end{align*}
Mathematica [A] time = 0.42693, size = 150, normalized size = 1.21 \[ -\frac{b c^2 \left (i \text{PolyLog}\left (2,-e^{2 i \cos ^{-1}(c x)}\right )-i \text{PolyLog}\left (2,e^{2 i \cos ^{-1}(c x)}\right )-\frac{\sqrt{1-c^2 x^2}}{c x}+\frac{\cos ^{-1}(c x)}{c^2 x^2}+2 \cos ^{-1}(c x) \log \left (1-e^{2 i \cos ^{-1}(c x)}\right )-2 \cos ^{-1}(c x) \log \left (1+e^{2 i \cos ^{-1}(c x)}\right )\right )+a c^2 \log \left (1-c^2 x^2\right )-2 a c^2 \log (x)+\frac{a}{x^2}}{2 d} \]
Warning: Unable to verify antiderivative.
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Maple [B] time = 0.177, size = 301, normalized size = 2.4 \begin{align*} -{\frac{{c}^{2}a\ln \left ( cx-1 \right ) }{2\,d}}-{\frac{{c}^{2}a\ln \left ( cx+1 \right ) }{2\,d}}-{\frac{a}{2\,d{x}^{2}}}+{\frac{{c}^{2}a\ln \left ( cx \right ) }{d}}+{\frac{{\frac{i}{2}}{c}^{2}b}{d}}+{\frac{bc}{2\,dx}\sqrt{-{c}^{2}{x}^{2}+1}}-{\frac{b\arccos \left ( cx \right ) }{2\,d{x}^{2}}}-{\frac{{c}^{2}b\arccos \left ( cx \right ) }{d}\ln \left ( 1+cx+i\sqrt{-{c}^{2}{x}^{2}+1} \right ) }+{\frac{i{c}^{2}b}{d}{\it polylog} \left ( 2,-cx-i\sqrt{-{c}^{2}{x}^{2}+1} \right ) }-{\frac{{c}^{2}b\arccos \left ( cx \right ) }{d}\ln \left ( 1-cx-i\sqrt{-{c}^{2}{x}^{2}+1} \right ) }+{\frac{i{c}^{2}b}{d}{\it polylog} \left ( 2,cx+i\sqrt{-{c}^{2}{x}^{2}+1} \right ) }+{\frac{{c}^{2}b\arccos \left ( cx \right ) }{d}\ln \left ( 1+ \left ( cx+i\sqrt{-{c}^{2}{x}^{2}+1} \right ) ^{2} \right ) }-{\frac{{\frac{i}{2}}{c}^{2}b}{d}{\it polylog} \left ( 2,- \left ( cx+i\sqrt{-{c}^{2}{x}^{2}+1} \right ) ^{2} \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}{2} \,{\left (\frac{c^{2} \log \left (c x + 1\right )}{d} + \frac{c^{2} \log \left (c x - 1\right )}{d} - \frac{2 \, c^{2} \log \left (x\right )}{d} + \frac{1}{d x^{2}}\right )} a - b \int \frac{\arctan \left (\sqrt{c x + 1} \sqrt{-c x + 1}, c x\right )}{c^{2} d x^{5} - d x^{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 \arccos \left (c x\right ) + a}{c^{2} d x^{5} - d x^{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}{c^{2} x^{5} - x^{3}}\, dx + \int \frac{b \operatorname{acos}{\left (c x \right )}}{c^{2} x^{5} - x^{3}}\, dx}{d} \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 \arccos \left (c x\right ) + a}{{\left (c^{2} d x^{2} - d\right )} x^{3}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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