Optimal. Leaf size=214 \[ \frac{3 c^3 \sqrt{b x^2+c x^4} (2 b B-A c)}{256 b^3 x^3}-\frac{c^2 \sqrt{b x^2+c x^4} (2 b B-A c)}{128 b^2 x^5}-\frac{3 c^4 (2 b B-A c) \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{b x^2+c x^4}}\right )}{256 b^{7/2}}-\frac{c \sqrt{b x^2+c x^4} (2 b B-A c)}{32 b x^7}-\frac{\left (b x^2+c x^4\right )^{3/2} (2 b B-A c)}{16 b x^{11}}-\frac{A \left (b x^2+c x^4\right )^{5/2}}{10 b x^{15}} \]
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Rubi [A] time = 0.339416, antiderivative size = 214, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192, Rules used = {2038, 2020, 2025, 2008, 206} \[ \frac{3 c^3 \sqrt{b x^2+c x^4} (2 b B-A c)}{256 b^3 x^3}-\frac{c^2 \sqrt{b x^2+c x^4} (2 b B-A c)}{128 b^2 x^5}-\frac{3 c^4 (2 b B-A c) \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{b x^2+c x^4}}\right )}{256 b^{7/2}}-\frac{c \sqrt{b x^2+c x^4} (2 b B-A c)}{32 b x^7}-\frac{\left (b x^2+c x^4\right )^{3/2} (2 b B-A c)}{16 b x^{11}}-\frac{A \left (b x^2+c x^4\right )^{5/2}}{10 b x^{15}} \]
Antiderivative was successfully verified.
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Rule 2038
Rule 2020
Rule 2025
Rule 2008
Rule 206
Rubi steps
\begin{align*} \int \frac{\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{14}} \, dx &=-\frac{A \left (b x^2+c x^4\right )^{5/2}}{10 b x^{15}}-\frac{(-10 b B+5 A c) \int \frac{\left (b x^2+c x^4\right )^{3/2}}{x^{12}} \, dx}{10 b}\\ &=-\frac{(2 b B-A c) \left (b x^2+c x^4\right )^{3/2}}{16 b x^{11}}-\frac{A \left (b x^2+c x^4\right )^{5/2}}{10 b x^{15}}+\frac{(3 c (2 b B-A c)) \int \frac{\sqrt{b x^2+c x^4}}{x^8} \, dx}{16 b}\\ &=-\frac{c (2 b B-A c) \sqrt{b x^2+c x^4}}{32 b x^7}-\frac{(2 b B-A c) \left (b x^2+c x^4\right )^{3/2}}{16 b x^{11}}-\frac{A \left (b x^2+c x^4\right )^{5/2}}{10 b x^{15}}+\frac{\left (c^2 (2 b B-A c)\right ) \int \frac{1}{x^4 \sqrt{b x^2+c x^4}} \, dx}{32 b}\\ &=-\frac{c (2 b B-A c) \sqrt{b x^2+c x^4}}{32 b x^7}-\frac{c^2 (2 b B-A c) \sqrt{b x^2+c x^4}}{128 b^2 x^5}-\frac{(2 b B-A c) \left (b x^2+c x^4\right )^{3/2}}{16 b x^{11}}-\frac{A \left (b x^2+c x^4\right )^{5/2}}{10 b x^{15}}-\frac{\left (3 c^3 (2 b B-A c)\right ) \int \frac{1}{x^2 \sqrt{b x^2+c x^4}} \, dx}{128 b^2}\\ &=-\frac{c (2 b B-A c) \sqrt{b x^2+c x^4}}{32 b x^7}-\frac{c^2 (2 b B-A c) \sqrt{b x^2+c x^4}}{128 b^2 x^5}+\frac{3 c^3 (2 b B-A c) \sqrt{b x^2+c x^4}}{256 b^3 x^3}-\frac{(2 b B-A c) \left (b x^2+c x^4\right )^{3/2}}{16 b x^{11}}-\frac{A \left (b x^2+c x^4\right )^{5/2}}{10 b x^{15}}+\frac{\left (3 c^4 (2 b B-A c)\right ) \int \frac{1}{\sqrt{b x^2+c x^4}} \, dx}{256 b^3}\\ &=-\frac{c (2 b B-A c) \sqrt{b x^2+c x^4}}{32 b x^7}-\frac{c^2 (2 b B-A c) \sqrt{b x^2+c x^4}}{128 b^2 x^5}+\frac{3 c^3 (2 b B-A c) \sqrt{b x^2+c x^4}}{256 b^3 x^3}-\frac{(2 b B-A c) \left (b x^2+c x^4\right )^{3/2}}{16 b x^{11}}-\frac{A \left (b x^2+c x^4\right )^{5/2}}{10 b x^{15}}-\frac{\left (3 c^4 (2 b B-A c)\right ) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{x}{\sqrt{b x^2+c x^4}}\right )}{256 b^3}\\ &=-\frac{c (2 b B-A c) \sqrt{b x^2+c x^4}}{32 b x^7}-\frac{c^2 (2 b B-A c) \sqrt{b x^2+c x^4}}{128 b^2 x^5}+\frac{3 c^3 (2 b B-A c) \sqrt{b x^2+c x^4}}{256 b^3 x^3}-\frac{(2 b B-A c) \left (b x^2+c x^4\right )^{3/2}}{16 b x^{11}}-\frac{A \left (b x^2+c x^4\right )^{5/2}}{10 b x^{15}}-\frac{3 c^4 (2 b B-A c) \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{b x^2+c x^4}}\right )}{256 b^{7/2}}\\ \end{align*}
Mathematica [C] time = 0.0329536, size = 65, normalized size = 0.3 \[ -\frac{\left (x^2 \left (b+c x^2\right )\right )^{5/2} \left (A b^5+c^4 x^{10} (2 b B-A c) \, _2F_1\left (\frac{5}{2},5;\frac{7}{2};\frac{c x^2}{b}+1\right )\right )}{10 b^6 x^{15}} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.033, size = 344, normalized size = 1.6 \begin{align*}{\frac{1}{1280\,{x}^{13}{b}^{5}} \left ( c{x}^{4}+b{x}^{2} \right ) ^{{\frac{3}{2}}} \left ( -5\,A \left ( c{x}^{2}+b \right ) ^{3/2}{x}^{10}{c}^{5}+15\,A{b}^{3/2}\ln \left ( 2\,{\frac{\sqrt{b}\sqrt{c{x}^{2}+b}+b}{x}} \right ){x}^{10}{c}^{5}+10\,B \left ( c{x}^{2}+b \right ) ^{3/2}{x}^{10}b{c}^{4}-30\,B{b}^{5/2}\ln \left ( 2\,{\frac{\sqrt{b}\sqrt{c{x}^{2}+b}+b}{x}} \right ){x}^{10}{c}^{4}+5\,A \left ( c{x}^{2}+b \right ) ^{5/2}{x}^{8}{c}^{4}-15\,A\sqrt{c{x}^{2}+b}{x}^{10}b{c}^{5}-10\,B \left ( c{x}^{2}+b \right ) ^{5/2}{x}^{8}b{c}^{3}+30\,B\sqrt{c{x}^{2}+b}{x}^{10}{b}^{2}{c}^{4}+10\,A \left ( c{x}^{2}+b \right ) ^{5/2}{x}^{6}b{c}^{3}-20\,B \left ( c{x}^{2}+b \right ) ^{5/2}{x}^{6}{b}^{2}{c}^{2}-40\,A \left ( c{x}^{2}+b \right ) ^{5/2}{x}^{4}{b}^{2}{c}^{2}+80\,B \left ( c{x}^{2}+b \right ) ^{5/2}{x}^{4}{b}^{3}c+80\,A \left ( c{x}^{2}+b \right ) ^{5/2}{x}^{2}{b}^{3}c-160\,B \left ( c{x}^{2}+b \right ) ^{5/2}{x}^{2}{b}^{4}-128\,A \left ( c{x}^{2}+b \right ) ^{5/2}{b}^{4} \right ) \left ( c{x}^{2}+b \right ) ^{-{\frac{3}{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{{\left (c x^{4} + b x^{2}\right )}^{\frac{3}{2}}{\left (B x^{2} + A\right )}}{x^{14}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.4328, size = 765, normalized size = 3.57 \begin{align*} \left [-\frac{15 \,{\left (2 \, B b c^{4} - A c^{5}\right )} \sqrt{b} x^{11} \log \left (-\frac{c x^{3} + 2 \, b x + 2 \, \sqrt{c x^{4} + b x^{2}} \sqrt{b}}{x^{3}}\right ) - 2 \,{\left (15 \,{\left (2 \, B b^{2} c^{3} - A b c^{4}\right )} x^{8} - 10 \,{\left (2 \, B b^{3} c^{2} - A b^{2} c^{3}\right )} x^{6} - 128 \, A b^{5} - 8 \,{\left (30 \, B b^{4} c + A b^{3} c^{2}\right )} x^{4} - 16 \,{\left (10 \, B b^{5} + 11 \, A b^{4} c\right )} x^{2}\right )} \sqrt{c x^{4} + b x^{2}}}{2560 \, b^{4} x^{11}}, \frac{15 \,{\left (2 \, B b c^{4} - A c^{5}\right )} \sqrt{-b} x^{11} \arctan \left (\frac{\sqrt{c x^{4} + b x^{2}} \sqrt{-b}}{c x^{3} + b x}\right ) +{\left (15 \,{\left (2 \, B b^{2} c^{3} - A b c^{4}\right )} x^{8} - 10 \,{\left (2 \, B b^{3} c^{2} - A b^{2} c^{3}\right )} x^{6} - 128 \, A b^{5} - 8 \,{\left (30 \, B b^{4} c + A b^{3} c^{2}\right )} x^{4} - 16 \,{\left (10 \, B b^{5} + 11 \, A b^{4} c\right )} x^{2}\right )} \sqrt{c x^{4} + b x^{2}}}{1280 \, b^{4} x^{11}}\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{\left (x^{2} \left (b + c x^{2}\right )\right )^{\frac{3}{2}} \left (A + B x^{2}\right )}{x^{14}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.29525, size = 316, normalized size = 1.48 \begin{align*} \frac{\frac{15 \,{\left (2 \, B b c^{5} \mathrm{sgn}\left (x\right ) - A c^{6} \mathrm{sgn}\left (x\right )\right )} \arctan \left (\frac{\sqrt{c x^{2} + b}}{\sqrt{-b}}\right )}{\sqrt{-b} b^{3}} + \frac{30 \,{\left (c x^{2} + b\right )}^{\frac{9}{2}} B b c^{5} \mathrm{sgn}\left (x\right ) - 140 \,{\left (c x^{2} + b\right )}^{\frac{7}{2}} B b^{2} c^{5} \mathrm{sgn}\left (x\right ) + 140 \,{\left (c x^{2} + b\right )}^{\frac{3}{2}} B b^{4} c^{5} \mathrm{sgn}\left (x\right ) - 30 \, \sqrt{c x^{2} + b} B b^{5} c^{5} \mathrm{sgn}\left (x\right ) - 15 \,{\left (c x^{2} + b\right )}^{\frac{9}{2}} A c^{6} \mathrm{sgn}\left (x\right ) + 70 \,{\left (c x^{2} + b\right )}^{\frac{7}{2}} A b c^{6} \mathrm{sgn}\left (x\right ) - 128 \,{\left (c x^{2} + b\right )}^{\frac{5}{2}} A b^{2} c^{6} \mathrm{sgn}\left (x\right ) - 70 \,{\left (c x^{2} + b\right )}^{\frac{3}{2}} A b^{3} c^{6} \mathrm{sgn}\left (x\right ) + 15 \, \sqrt{c x^{2} + b} A b^{4} c^{6} \mathrm{sgn}\left (x\right )}{b^{3} c^{5} x^{10}}}{1280 \, c} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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