Optimal. Leaf size=154 \[ -\frac{\left (3 b^2-8 a c\right ) \sqrt{a x+b x^3+c x^5}}{2 a^2 x^{5/2} \left (b^2-4 a c\right )}+\frac{3 b \tanh ^{-1}\left (\frac{\sqrt{x} \left (2 a+b x^2\right )}{2 \sqrt{a} \sqrt{a x+b x^3+c x^5}}\right )}{4 a^{5/2}}+\frac{-2 a c+b^2+b c x^2}{a x^{3/2} \left (b^2-4 a c\right ) \sqrt{a x+b x^3+c x^5}} \]
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Rubi [A] time = 0.174953, antiderivative size = 154, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {1924, 1951, 12, 1913, 206} \[ -\frac{\left (3 b^2-8 a c\right ) \sqrt{a x+b x^3+c x^5}}{2 a^2 x^{5/2} \left (b^2-4 a c\right )}+\frac{3 b \tanh ^{-1}\left (\frac{\sqrt{x} \left (2 a+b x^2\right )}{2 \sqrt{a} \sqrt{a x+b x^3+c x^5}}\right )}{4 a^{5/2}}+\frac{-2 a c+b^2+b c x^2}{a x^{3/2} \left (b^2-4 a c\right ) \sqrt{a x+b x^3+c x^5}} \]
Antiderivative was successfully verified.
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Rule 1924
Rule 1951
Rule 12
Rule 1913
Rule 206
Rubi steps
\begin{align*} \int \frac{1}{x^{3/2} \left (a x+b x^3+c x^5\right )^{3/2}} \, dx &=\frac{b^2-2 a c+b c x^2}{a \left (b^2-4 a c\right ) x^{3/2} \sqrt{a x+b x^3+c x^5}}-\frac{\int \frac{-3 b^2+8 a c-2 b c x^2}{x^{5/2} \sqrt{a x+b x^3+c x^5}} \, dx}{a \left (b^2-4 a c\right )}\\ &=\frac{b^2-2 a c+b c x^2}{a \left (b^2-4 a c\right ) x^{3/2} \sqrt{a x+b x^3+c x^5}}-\frac{\left (3 b^2-8 a c\right ) \sqrt{a x+b x^3+c x^5}}{2 a^2 \left (b^2-4 a c\right ) x^{5/2}}+\frac{\int -\frac{3 b \left (b^2-4 a c\right )}{\sqrt{x} \sqrt{a x+b x^3+c x^5}} \, dx}{2 a^2 \left (b^2-4 a c\right )}\\ &=\frac{b^2-2 a c+b c x^2}{a \left (b^2-4 a c\right ) x^{3/2} \sqrt{a x+b x^3+c x^5}}-\frac{\left (3 b^2-8 a c\right ) \sqrt{a x+b x^3+c x^5}}{2 a^2 \left (b^2-4 a c\right ) x^{5/2}}-\frac{(3 b) \int \frac{1}{\sqrt{x} \sqrt{a x+b x^3+c x^5}} \, dx}{2 a^2}\\ &=\frac{b^2-2 a c+b c x^2}{a \left (b^2-4 a c\right ) x^{3/2} \sqrt{a x+b x^3+c x^5}}-\frac{\left (3 b^2-8 a c\right ) \sqrt{a x+b x^3+c x^5}}{2 a^2 \left (b^2-4 a c\right ) x^{5/2}}+\frac{(3 b) \operatorname{Subst}\left (\int \frac{1}{4 a-x^2} \, dx,x,\frac{\sqrt{x} \left (2 a+b x^2\right )}{\sqrt{a x+b x^3+c x^5}}\right )}{2 a^2}\\ &=\frac{b^2-2 a c+b c x^2}{a \left (b^2-4 a c\right ) x^{3/2} \sqrt{a x+b x^3+c x^5}}-\frac{\left (3 b^2-8 a c\right ) \sqrt{a x+b x^3+c x^5}}{2 a^2 \left (b^2-4 a c\right ) x^{5/2}}+\frac{3 b \tanh ^{-1}\left (\frac{\sqrt{x} \left (2 a+b x^2\right )}{2 \sqrt{a} \sqrt{a x+b x^3+c x^5}}\right )}{4 a^{5/2}}\\ \end{align*}
Mathematica [A] time = 0.0820778, size = 160, normalized size = 1.04 \[ \frac{2 \sqrt{a} \left (-4 a^2 c+a \left (b^2-10 b c x^2-8 c^2 x^4\right )+3 b^2 x^2 \left (b+c x^2\right )\right )-3 b x^2 \left (b^2-4 a c\right ) \sqrt{a+b x^2+c x^4} \tanh ^{-1}\left (\frac{2 a+b x^2}{2 \sqrt{a} \sqrt{a+b x^2+c x^4}}\right )}{4 a^{5/2} x^{3/2} \left (4 a c-b^2\right ) \sqrt{x \left (a+b x^2+c x^4\right )}} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.019, size = 220, normalized size = 1.4 \begin{align*}{\frac{1}{ \left ( 4\,c{x}^{4}+4\,b{x}^{2}+4\,a \right ) \left ( 4\,ac-{b}^{2} \right ) }\sqrt{x \left ( c{x}^{4}+b{x}^{2}+a \right ) } \left ( -16\,{x}^{4}{a}^{3/2}{c}^{2}+6\,{x}^{4}{b}^{2}c\sqrt{a}+12\,\ln \left ({\frac{2\,a+b{x}^{2}+2\,\sqrt{a}\sqrt{c{x}^{4}+b{x}^{2}+a}}{{x}^{2}}} \right ){x}^{2}abc\sqrt{c{x}^{4}+b{x}^{2}+a}-3\,\ln \left ({\frac{2\,a+b{x}^{2}+2\,\sqrt{a}\sqrt{c{x}^{4}+b{x}^{2}+a}}{{x}^{2}}} \right ){x}^{2}{b}^{3}\sqrt{c{x}^{4}+b{x}^{2}+a}-20\,{a}^{3/2}{x}^{2}bc+6\,{x}^{2}{b}^{3}\sqrt{a}-8\,{a}^{5/2}c+2\,{a}^{3/2}{b}^{2} \right ){a}^{-{\frac{5}{2}}}{x}^{-{\frac{5}{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*} \int \frac{1}{{\left (c x^{5} + b x^{3} + a x\right )}^{\frac{3}{2}} x^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.85444, size = 1084, normalized size = 7.04 \begin{align*} \left [\frac{3 \,{\left ({\left (b^{3} c - 4 \, a b c^{2}\right )} x^{7} +{\left (b^{4} - 4 \, a b^{2} c\right )} x^{5} +{\left (a b^{3} - 4 \, a^{2} b c\right )} x^{3}\right )} \sqrt{a} \log \left (-\frac{{\left (b^{2} + 4 \, a c\right )} x^{5} + 8 \, a b x^{3} + 8 \, a^{2} x + 4 \, \sqrt{c x^{5} + b x^{3} + a x}{\left (b x^{2} + 2 \, a\right )} \sqrt{a} \sqrt{x}}{x^{5}}\right ) - 4 \, \sqrt{c x^{5} + b x^{3} + a x}{\left ({\left (3 \, a b^{2} c - 8 \, a^{2} c^{2}\right )} x^{4} + a^{2} b^{2} - 4 \, a^{3} c +{\left (3 \, a b^{3} - 10 \, a^{2} b c\right )} x^{2}\right )} \sqrt{x}}{8 \,{\left ({\left (a^{3} b^{2} c - 4 \, a^{4} c^{2}\right )} x^{7} +{\left (a^{3} b^{3} - 4 \, a^{4} b c\right )} x^{5} +{\left (a^{4} b^{2} - 4 \, a^{5} c\right )} x^{3}\right )}}, -\frac{3 \,{\left ({\left (b^{3} c - 4 \, a b c^{2}\right )} x^{7} +{\left (b^{4} - 4 \, a b^{2} c\right )} x^{5} +{\left (a b^{3} - 4 \, a^{2} b c\right )} x^{3}\right )} \sqrt{-a} \arctan \left (\frac{\sqrt{c x^{5} + b x^{3} + a x}{\left (b x^{2} + 2 \, a\right )} \sqrt{-a} \sqrt{x}}{2 \,{\left (a c x^{5} + a b x^{3} + a^{2} x\right )}}\right ) + 2 \, \sqrt{c x^{5} + b x^{3} + a x}{\left ({\left (3 \, a b^{2} c - 8 \, a^{2} c^{2}\right )} x^{4} + a^{2} b^{2} - 4 \, a^{3} c +{\left (3 \, a b^{3} - 10 \, a^{2} b c\right )} x^{2}\right )} \sqrt{x}}{4 \,{\left ({\left (a^{3} b^{2} c - 4 \, a^{4} c^{2}\right )} x^{7} +{\left (a^{3} b^{3} - 4 \, a^{4} b c\right )} x^{5} +{\left (a^{4} b^{2} - 4 \, a^{5} c\right )} x^{3}\right )}}\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{1}{x^{\frac{3}{2}} \left (x \left (a + b x^{2} + c x^{4}\right )\right )^{\frac{3}{2}}}\, dx \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{1}{{\left (c x^{5} + b x^{3} + a x\right )}^{\frac{3}{2}} x^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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