Optimal. Leaf size=173 \[ \frac{i b c^3 \text{PolyLog}\left (2,-i e^{i \sin ^{-1}(c x)}\right )}{d}-\frac{i b c^3 \text{PolyLog}\left (2,i e^{i \sin ^{-1}(c x)}\right )}{d}-\frac{c^2 \left (a+b \sin ^{-1}(c x)\right )}{d x}-\frac{2 i c^3 \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d}-\frac{a+b \sin ^{-1}(c x)}{3 d x^3}-\frac{b c \sqrt{1-c^2 x^2}}{6 d x^2}-\frac{7 b c^3 \tanh ^{-1}\left (\sqrt{1-c^2 x^2}\right )}{6 d} \]
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Rubi [A] time = 0.243657, antiderivative size = 173, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 9, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.36, Rules used = {4701, 4657, 4181, 2279, 2391, 266, 63, 208, 51} \[ \frac{i b c^3 \text{PolyLog}\left (2,-i e^{i \sin ^{-1}(c x)}\right )}{d}-\frac{i b c^3 \text{PolyLog}\left (2,i e^{i \sin ^{-1}(c x)}\right )}{d}-\frac{c^2 \left (a+b \sin ^{-1}(c x)\right )}{d x}-\frac{2 i c^3 \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d}-\frac{a+b \sin ^{-1}(c x)}{3 d x^3}-\frac{b c \sqrt{1-c^2 x^2}}{6 d x^2}-\frac{7 b c^3 \tanh ^{-1}\left (\sqrt{1-c^2 x^2}\right )}{6 d} \]
Antiderivative was successfully verified.
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Rule 4701
Rule 4657
Rule 4181
Rule 2279
Rule 2391
Rule 266
Rule 63
Rule 208
Rule 51
Rubi steps
\begin{align*} \int \frac{a+b \sin ^{-1}(c x)}{x^4 \left (d-c^2 d x^2\right )} \, dx &=-\frac{a+b \sin ^{-1}(c x)}{3 d x^3}+c^2 \int \frac{a+b \sin ^{-1}(c x)}{x^2 \left (d-c^2 d x^2\right )} \, dx+\frac{(b c) \int \frac{1}{x^3 \sqrt{1-c^2 x^2}} \, dx}{3 d}\\ &=-\frac{a+b \sin ^{-1}(c x)}{3 d x^3}-\frac{c^2 \left (a+b \sin ^{-1}(c x)\right )}{d x}+c^4 \int \frac{a+b \sin ^{-1}(c x)}{d-c^2 d x^2} \, dx+\frac{(b c) \operatorname{Subst}\left (\int \frac{1}{x^2 \sqrt{1-c^2 x}} \, dx,x,x^2\right )}{6 d}+\frac{\left (b c^3\right ) \int \frac{1}{x \sqrt{1-c^2 x^2}} \, dx}{d}\\ &=-\frac{b c \sqrt{1-c^2 x^2}}{6 d x^2}-\frac{a+b \sin ^{-1}(c x)}{3 d x^3}-\frac{c^2 \left (a+b \sin ^{-1}(c x)\right )}{d x}+\frac{c^3 \operatorname{Subst}\left (\int (a+b x) \sec (x) \, dx,x,\sin ^{-1}(c x)\right )}{d}+\frac{\left (b c^3\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{1-c^2 x}} \, dx,x,x^2\right )}{12 d}+\frac{\left (b c^3\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{1-c^2 x}} \, dx,x,x^2\right )}{2 d}\\ &=-\frac{b c \sqrt{1-c^2 x^2}}{6 d x^2}-\frac{a+b \sin ^{-1}(c x)}{3 d x^3}-\frac{c^2 \left (a+b \sin ^{-1}(c x)\right )}{d x}-\frac{2 i c^3 \left (a+b \sin ^{-1}(c x)\right ) \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{d}-\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 )}{6 d}-\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 )}{d}-\frac{\left (b c^3\right ) \operatorname{Subst}\left (\int \log \left (1-i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{d}+\frac{\left (b c^3\right ) \operatorname{Subst}\left (\int \log \left (1+i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{d}\\ &=-\frac{b c \sqrt{1-c^2 x^2}}{6 d x^2}-\frac{a+b \sin ^{-1}(c x)}{3 d x^3}-\frac{c^2 \left (a+b \sin ^{-1}(c x)\right )}{d x}-\frac{2 i c^3 \left (a+b \sin ^{-1}(c x)\right ) \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{d}-\frac{7 b c^3 \tanh ^{-1}\left (\sqrt{1-c^2 x^2}\right )}{6 d}+\frac{\left (i b c^3\right ) \operatorname{Subst}\left (\int \frac{\log (1-i x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{d}-\frac{\left (i b c^3\right ) \operatorname{Subst}\left (\int \frac{\log (1+i x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{d}\\ &=-\frac{b c \sqrt{1-c^2 x^2}}{6 d x^2}-\frac{a+b \sin ^{-1}(c x)}{3 d x^3}-\frac{c^2 \left (a+b \sin ^{-1}(c x)\right )}{d x}-\frac{2 i c^3 \left (a+b \sin ^{-1}(c x)\right ) \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{d}-\frac{7 b c^3 \tanh ^{-1}\left (\sqrt{1-c^2 x^2}\right )}{6 d}+\frac{i b c^3 \text{Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )}{d}-\frac{i b c^3 \text{Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )}{d}\\ \end{align*}
Mathematica [B] time = 0.147467, size = 350, normalized size = 2.02 \[ -\frac{-6 i b c^3 x^3 \text{PolyLog}\left (2,-i e^{i \sin ^{-1}(c x)}\right )+6 i b c^3 x^3 \text{PolyLog}\left (2,i e^{i \sin ^{-1}(c x)}\right )+6 a c^2 x^2+3 a c^3 x^3 \log (1-c x)-3 a c^3 x^3 \log (c x+1)+2 a+b c x \sqrt{1-c^2 x^2}+3 i \pi b c^3 x^3 \sin ^{-1}(c x)+6 b c^2 x^2 \sin ^{-1}(c x)+7 b c^3 x^3 \tanh ^{-1}\left (\sqrt{1-c^2 x^2}\right )-6 b c^3 x^3 \sin ^{-1}(c x) \log \left (1-i e^{i \sin ^{-1}(c x)}\right )-3 \pi b c^3 x^3 \log \left (1-i e^{i \sin ^{-1}(c x)}\right )+6 b c^3 x^3 \sin ^{-1}(c x) \log \left (1+i e^{i \sin ^{-1}(c x)}\right )-3 \pi b c^3 x^3 \log \left (1+i e^{i \sin ^{-1}(c x)}\right )+3 \pi b c^3 x^3 \log \left (\sin \left (\frac{1}{4} \left (2 \sin ^{-1}(c x)+\pi \right )\right )\right )+3 \pi b c^3 x^3 \log \left (-\cos \left (\frac{1}{4} \left (2 \sin ^{-1}(c x)+\pi \right )\right )\right )+2 b \sin ^{-1}(c x)}{6 d x^3} \]
Warning: Unable to verify antiderivative.
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Maple [A] time = 0.177, size = 303, normalized size = 1.8 \begin{align*} -{\frac{{c}^{3}a\ln \left ( cx-1 \right ) }{2\,d}}+{\frac{{c}^{3}a\ln \left ( cx+1 \right ) }{2\,d}}-{\frac{a}{3\,d{x}^{3}}}-{\frac{{c}^{2}a}{dx}}-{\frac{{c}^{2}b\arcsin \left ( cx \right ) }{dx}}-{\frac{bc}{6\,d{x}^{2}}\sqrt{-{c}^{2}{x}^{2}+1}}-{\frac{b\arcsin \left ( cx \right ) }{3\,d{x}^{3}}}+{\frac{7\,b{c}^{3}}{6\,d}\ln \left ( icx+\sqrt{-{c}^{2}{x}^{2}+1}-1 \right ) }-{\frac{7\,b{c}^{3}}{6\,d}\ln \left ( 1+icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) }+{\frac{b{c}^{3}\arcsin \left ( cx \right ) }{d}\ln \left ( 1-i \left ( icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) \right ) }-{\frac{b{c}^{3}\arcsin \left ( cx \right ) }{d}\ln \left ( 1+i \left ( icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) \right ) }-{\frac{i{c}^{3}b}{d}{\it dilog} \left ( 1-i \left ( icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) \right ) }+{\frac{i{c}^{3}b}{d}{\it dilog} \left ( 1+i \left ( icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) \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{3 \, c^{3} \log \left (c x + 1\right )}{d} - \frac{3 \, c^{3} \log \left (c x - 1\right )}{d} - \frac{2 \,{\left (3 \, c^{2} x^{2} + 1\right )}}{d x^{3}}\right )} a + \frac{{\left (3 \, c^{3} x^{3} \arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right ) \log \left (c x + 1\right ) - 3 \, c^{3} x^{3} \arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right ) \log \left (-c x + 1\right ) + d x^{3} \int \frac{{\left (3 \, c^{4} x^{3} \log \left (c x + 1\right ) - 3 \, c^{4} x^{3} \log \left (-c x + 1\right ) - 6 \, c^{3} x^{2} - 2 \, c\right )} \sqrt{c x + 1} \sqrt{-c x + 1}}{c^{2} d x^{5} - d x^{3}}\,{d x} - 2 \,{\left (3 \, c^{2} x^{2} + 1\right )} \arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right )\right )} b}{6 \, d x^{3}} \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 \arcsin \left (c x\right ) + a}{c^{2} d x^{6} - d 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^{2} x^{6} - x^{4}}\, dx + \int \frac{b \operatorname{asin}{\left (c x \right )}}{c^{2} x^{6} - x^{4}}\, 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 \arcsin \left (c x\right ) + a}{{\left (c^{2} d x^{2} - d\right )} x^{4}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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