Optimal. Leaf size=98 \[ \frac{\text{Si}\left (\cos ^{-1}(a x)\right )}{16 a^5}+\frac{27 \text{Si}\left (3 \cos ^{-1}(a x)\right )}{32 a^5}+\frac{25 \text{Si}\left (5 \cos ^{-1}(a x)\right )}{32 a^5}+\frac{x^4 \sqrt{1-a^2 x^2}}{2 a \cos ^{-1}(a x)^2}-\frac{2 x^3}{a^2 \cos ^{-1}(a x)}+\frac{5 x^5}{2 \cos ^{-1}(a x)} \]
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Rubi [A] time = 0.344457, antiderivative size = 98, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 5, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {4634, 4720, 4636, 4406, 3299} \[ \frac{\text{Si}\left (\cos ^{-1}(a x)\right )}{16 a^5}+\frac{27 \text{Si}\left (3 \cos ^{-1}(a x)\right )}{32 a^5}+\frac{25 \text{Si}\left (5 \cos ^{-1}(a x)\right )}{32 a^5}+\frac{x^4 \sqrt{1-a^2 x^2}}{2 a \cos ^{-1}(a x)^2}-\frac{2 x^3}{a^2 \cos ^{-1}(a x)}+\frac{5 x^5}{2 \cos ^{-1}(a x)} \]
Antiderivative was successfully verified.
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Rule 4634
Rule 4720
Rule 4636
Rule 4406
Rule 3299
Rubi steps
\begin{align*} \int \frac{x^4}{\cos ^{-1}(a x)^3} \, dx &=\frac{x^4 \sqrt{1-a^2 x^2}}{2 a \cos ^{-1}(a x)^2}-\frac{2 \int \frac{x^3}{\sqrt{1-a^2 x^2} \cos ^{-1}(a x)^2} \, dx}{a}+\frac{1}{2} (5 a) \int \frac{x^5}{\sqrt{1-a^2 x^2} \cos ^{-1}(a x)^2} \, dx\\ &=\frac{x^4 \sqrt{1-a^2 x^2}}{2 a \cos ^{-1}(a x)^2}-\frac{2 x^3}{a^2 \cos ^{-1}(a x)}+\frac{5 x^5}{2 \cos ^{-1}(a x)}-\frac{25}{2} \int \frac{x^4}{\cos ^{-1}(a x)} \, dx+\frac{6 \int \frac{x^2}{\cos ^{-1}(a x)} \, dx}{a^2}\\ &=\frac{x^4 \sqrt{1-a^2 x^2}}{2 a \cos ^{-1}(a x)^2}-\frac{2 x^3}{a^2 \cos ^{-1}(a x)}+\frac{5 x^5}{2 \cos ^{-1}(a x)}-\frac{6 \operatorname{Subst}\left (\int \frac{\cos ^2(x) \sin (x)}{x} \, dx,x,\cos ^{-1}(a x)\right )}{a^5}+\frac{25 \operatorname{Subst}\left (\int \frac{\cos ^4(x) \sin (x)}{x} \, dx,x,\cos ^{-1}(a x)\right )}{2 a^5}\\ &=\frac{x^4 \sqrt{1-a^2 x^2}}{2 a \cos ^{-1}(a x)^2}-\frac{2 x^3}{a^2 \cos ^{-1}(a x)}+\frac{5 x^5}{2 \cos ^{-1}(a x)}-\frac{6 \operatorname{Subst}\left (\int \left (\frac{\sin (x)}{4 x}+\frac{\sin (3 x)}{4 x}\right ) \, dx,x,\cos ^{-1}(a x)\right )}{a^5}+\frac{25 \operatorname{Subst}\left (\int \left (\frac{\sin (x)}{8 x}+\frac{3 \sin (3 x)}{16 x}+\frac{\sin (5 x)}{16 x}\right ) \, dx,x,\cos ^{-1}(a x)\right )}{2 a^5}\\ &=\frac{x^4 \sqrt{1-a^2 x^2}}{2 a \cos ^{-1}(a x)^2}-\frac{2 x^3}{a^2 \cos ^{-1}(a x)}+\frac{5 x^5}{2 \cos ^{-1}(a x)}+\frac{25 \operatorname{Subst}\left (\int \frac{\sin (5 x)}{x} \, dx,x,\cos ^{-1}(a x)\right )}{32 a^5}-\frac{3 \operatorname{Subst}\left (\int \frac{\sin (x)}{x} \, dx,x,\cos ^{-1}(a x)\right )}{2 a^5}-\frac{3 \operatorname{Subst}\left (\int \frac{\sin (3 x)}{x} \, dx,x,\cos ^{-1}(a x)\right )}{2 a^5}+\frac{25 \operatorname{Subst}\left (\int \frac{\sin (x)}{x} \, dx,x,\cos ^{-1}(a x)\right )}{16 a^5}+\frac{75 \operatorname{Subst}\left (\int \frac{\sin (3 x)}{x} \, dx,x,\cos ^{-1}(a x)\right )}{32 a^5}\\ &=\frac{x^4 \sqrt{1-a^2 x^2}}{2 a \cos ^{-1}(a x)^2}-\frac{2 x^3}{a^2 \cos ^{-1}(a x)}+\frac{5 x^5}{2 \cos ^{-1}(a x)}+\frac{\text{Si}\left (\cos ^{-1}(a x)\right )}{16 a^5}+\frac{27 \text{Si}\left (3 \cos ^{-1}(a x)\right )}{32 a^5}+\frac{25 \text{Si}\left (5 \cos ^{-1}(a x)\right )}{32 a^5}\\ \end{align*}
Mathematica [A] time = 0.130407, size = 103, normalized size = 1.05 \[ \frac{16 a^4 x^4 \sqrt{1-a^2 x^2}+80 a^5 x^5 \cos ^{-1}(a x)-64 a^3 x^3 \cos ^{-1}(a x)+2 \cos ^{-1}(a x)^2 \text{Si}\left (\cos ^{-1}(a x)\right )+27 \cos ^{-1}(a x)^2 \text{Si}\left (3 \cos ^{-1}(a x)\right )+25 \cos ^{-1}(a x)^2 \text{Si}\left (5 \cos ^{-1}(a x)\right )}{32 a^5 \cos ^{-1}(a x)^2} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.059, size = 121, normalized size = 1.2 \begin{align*}{\frac{1}{{a}^{5}} \left ({\frac{3\,\sin \left ( 3\,\arccos \left ( ax \right ) \right ) }{32\, \left ( \arccos \left ( ax \right ) \right ) ^{2}}}+{\frac{9\,\cos \left ( 3\,\arccos \left ( ax \right ) \right ) }{32\,\arccos \left ( ax \right ) }}+{\frac{27\,{\it Si} \left ( 3\,\arccos \left ( ax \right ) \right ) }{32}}+{\frac{\sin \left ( 5\,\arccos \left ( ax \right ) \right ) }{32\, \left ( \arccos \left ( ax \right ) \right ) ^{2}}}+{\frac{5\,\cos \left ( 5\,\arccos \left ( ax \right ) \right ) }{32\,\arccos \left ( ax \right ) }}+{\frac{25\,{\it Si} \left ( 5\,\arccos \left ( ax \right ) \right ) }{32}}+{\frac{1}{16\, \left ( \arccos \left ( ax \right ) \right ) ^{2}}\sqrt{-{a}^{2}{x}^{2}+1}}+{\frac{ax}{16\,\arccos \left ( ax \right ) }}+{\frac{{\it Si} \left ( \arccos \left ( ax \right ) \right ) }{16}} \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{\sqrt{a x + 1} \sqrt{-a x + 1} a x^{4} - \arctan \left (\sqrt{a x + 1} \sqrt{-a x + 1}, a x\right )^{2} \int \frac{{\left (25 \, a^{2} x^{2} - 12\right )} x^{2}}{\arctan \left (\sqrt{a x + 1} \sqrt{-a x + 1}, a x\right )}\,{d x} +{\left (5 \, a^{2} x^{5} - 4 \, x^{3}\right )} \arctan \left (\sqrt{a x + 1} \sqrt{-a x + 1}, a x\right )}{2 \, a^{2} \arctan \left (\sqrt{a x + 1} \sqrt{-a x + 1}, a x\right )^{2}} \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{x^{4}}{\arccos \left (a x\right )^{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*} \int \frac{x^{4}}{\operatorname{acos}^{3}{\left (a x \right )}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.23122, size = 116, normalized size = 1.18 \begin{align*} \frac{5 \, x^{5}}{2 \, \arccos \left (a x\right )} + \frac{\sqrt{-a^{2} x^{2} + 1} x^{4}}{2 \, a \arccos \left (a x\right )^{2}} - \frac{2 \, x^{3}}{a^{2} \arccos \left (a x\right )} + \frac{25 \, \operatorname{Si}\left (5 \, \arccos \left (a x\right )\right )}{32 \, a^{5}} + \frac{27 \, \operatorname{Si}\left (3 \, \arccos \left (a x\right )\right )}{32 \, a^{5}} + \frac{\operatorname{Si}\left (\arccos \left (a x\right )\right )}{16 \, a^{5}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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