Optimal. Leaf size=143 \[ -\frac{\left (a+b x^2\right )^{5/2} (4 a B+3 A b)}{8 a x^2}+\frac{5 b \left (a+b x^2\right )^{3/2} (4 a B+3 A b)}{24 a}+\frac{5}{8} b \sqrt{a+b x^2} (4 a B+3 A b)-\frac{5}{8} \sqrt{a} b (4 a B+3 A b) \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )-\frac{A \left (a+b x^2\right )^{7/2}}{4 a x^4} \]
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Rubi [A] time = 0.10115, antiderivative size = 143, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {446, 78, 47, 50, 63, 208} \[ -\frac{\left (a+b x^2\right )^{5/2} (4 a B+3 A b)}{8 a x^2}+\frac{5 b \left (a+b x^2\right )^{3/2} (4 a B+3 A b)}{24 a}+\frac{5}{8} b \sqrt{a+b x^2} (4 a B+3 A b)-\frac{5}{8} \sqrt{a} b (4 a B+3 A b) \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )-\frac{A \left (a+b x^2\right )^{7/2}}{4 a x^4} \]
Antiderivative was successfully verified.
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Rule 446
Rule 78
Rule 47
Rule 50
Rule 63
Rule 208
Rubi steps
\begin{align*} \int \frac{\left (a+b x^2\right )^{5/2} \left (A+B x^2\right )}{x^5} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{(a+b x)^{5/2} (A+B x)}{x^3} \, dx,x,x^2\right )\\ &=-\frac{A \left (a+b x^2\right )^{7/2}}{4 a x^4}+\frac{\left (\frac{3 A b}{2}+2 a B\right ) \operatorname{Subst}\left (\int \frac{(a+b x)^{5/2}}{x^2} \, dx,x,x^2\right )}{4 a}\\ &=-\frac{(3 A b+4 a B) \left (a+b x^2\right )^{5/2}}{8 a x^2}-\frac{A \left (a+b x^2\right )^{7/2}}{4 a x^4}+\frac{(5 b (3 A b+4 a B)) \operatorname{Subst}\left (\int \frac{(a+b x)^{3/2}}{x} \, dx,x,x^2\right )}{16 a}\\ &=\frac{5 b (3 A b+4 a B) \left (a+b x^2\right )^{3/2}}{24 a}-\frac{(3 A b+4 a B) \left (a+b x^2\right )^{5/2}}{8 a x^2}-\frac{A \left (a+b x^2\right )^{7/2}}{4 a x^4}+\frac{1}{16} (5 b (3 A b+4 a B)) \operatorname{Subst}\left (\int \frac{\sqrt{a+b x}}{x} \, dx,x,x^2\right )\\ &=\frac{5}{8} b (3 A b+4 a B) \sqrt{a+b x^2}+\frac{5 b (3 A b+4 a B) \left (a+b x^2\right )^{3/2}}{24 a}-\frac{(3 A b+4 a B) \left (a+b x^2\right )^{5/2}}{8 a x^2}-\frac{A \left (a+b x^2\right )^{7/2}}{4 a x^4}+\frac{1}{16} (5 a b (3 A b+4 a B)) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,x^2\right )\\ &=\frac{5}{8} b (3 A b+4 a B) \sqrt{a+b x^2}+\frac{5 b (3 A b+4 a B) \left (a+b x^2\right )^{3/2}}{24 a}-\frac{(3 A b+4 a B) \left (a+b x^2\right )^{5/2}}{8 a x^2}-\frac{A \left (a+b x^2\right )^{7/2}}{4 a x^4}+\frac{1}{8} (5 a (3 A b+4 a B)) \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b x^2}\right )\\ &=\frac{5}{8} b (3 A b+4 a B) \sqrt{a+b x^2}+\frac{5 b (3 A b+4 a B) \left (a+b x^2\right )^{3/2}}{24 a}-\frac{(3 A b+4 a B) \left (a+b x^2\right )^{5/2}}{8 a x^2}-\frac{A \left (a+b x^2\right )^{7/2}}{4 a x^4}-\frac{5}{8} \sqrt{a} b (3 A b+4 a B) \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )\\ \end{align*}
Mathematica [C] time = 0.0307477, size = 60, normalized size = 0.42 \[ \frac{\left (a+b x^2\right )^{7/2} \left (b x^4 (4 a B+3 A b) \, _2F_1\left (2,\frac{7}{2};\frac{9}{2};\frac{b x^2}{a}+1\right )-7 a^2 A\right )}{28 a^3 x^4} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.009, size = 213, normalized size = 1.5 \begin{align*} -{\frac{A}{4\,a{x}^{4}} \left ( b{x}^{2}+a \right ) ^{{\frac{7}{2}}}}-{\frac{3\,Ab}{8\,{a}^{2}{x}^{2}} \left ( b{x}^{2}+a \right ) ^{{\frac{7}{2}}}}+{\frac{3\,A{b}^{2}}{8\,{a}^{2}} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}}+{\frac{5\,A{b}^{2}}{8\,a} \left ( b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}-{\frac{15\,A{b}^{2}}{8}\sqrt{a}\ln \left ({\frac{1}{x} \left ( 2\,a+2\,\sqrt{a}\sqrt{b{x}^{2}+a} \right ) } \right ) }+{\frac{15\,A{b}^{2}}{8}\sqrt{b{x}^{2}+a}}-{\frac{B}{2\,a{x}^{2}} \left ( b{x}^{2}+a \right ) ^{{\frac{7}{2}}}}+{\frac{Bb}{2\,a} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}}+{\frac{5\,Bb}{6} \left ( b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}-{\frac{5\,Bb}{2}{a}^{{\frac{3}{2}}}\ln \left ({\frac{1}{x} \left ( 2\,a+2\,\sqrt{a}\sqrt{b{x}^{2}+a} \right ) } \right ) }+{\frac{5\,Bba}{2}\sqrt{b{x}^{2}+a}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.72509, size = 516, normalized size = 3.61 \begin{align*} \left [\frac{15 \,{\left (4 \, B a b + 3 \, A b^{2}\right )} \sqrt{a} x^{4} \log \left (-\frac{b x^{2} - 2 \, \sqrt{b x^{2} + a} \sqrt{a} + 2 \, a}{x^{2}}\right ) + 2 \,{\left (8 \, B b^{2} x^{6} + 8 \,{\left (7 \, B a b + 3 \, A b^{2}\right )} x^{4} - 6 \, A a^{2} - 3 \,{\left (4 \, B a^{2} + 9 \, A a b\right )} x^{2}\right )} \sqrt{b x^{2} + a}}{48 \, x^{4}}, \frac{15 \,{\left (4 \, B a b + 3 \, A b^{2}\right )} \sqrt{-a} x^{4} \arctan \left (\frac{\sqrt{-a}}{\sqrt{b x^{2} + a}}\right ) +{\left (8 \, B b^{2} x^{6} + 8 \,{\left (7 \, B a b + 3 \, A b^{2}\right )} x^{4} - 6 \, A a^{2} - 3 \,{\left (4 \, B a^{2} + 9 \, A a b\right )} x^{2}\right )} \sqrt{b x^{2} + a}}{24 \, x^{4}}\right ] \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A] time = 61.9685, size = 279, normalized size = 1.95 \begin{align*} - \frac{15 A \sqrt{a} b^{2} \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} x} \right )}}{8} - \frac{A a^{3}}{4 \sqrt{b} x^{5} \sqrt{\frac{a}{b x^{2}} + 1}} - \frac{3 A a^{2} \sqrt{b}}{8 x^{3} \sqrt{\frac{a}{b x^{2}} + 1}} - \frac{A a b^{\frac{3}{2}} \sqrt{\frac{a}{b x^{2}} + 1}}{x} + \frac{7 A a b^{\frac{3}{2}}}{8 x \sqrt{\frac{a}{b x^{2}} + 1}} + \frac{A b^{\frac{5}{2}} x}{\sqrt{\frac{a}{b x^{2}} + 1}} - \frac{5 B a^{\frac{3}{2}} b \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} x} \right )}}{2} - \frac{B a^{2} \sqrt{b} \sqrt{\frac{a}{b x^{2}} + 1}}{2 x} + \frac{2 B a^{2} \sqrt{b}}{x \sqrt{\frac{a}{b x^{2}} + 1}} + \frac{2 B a b^{\frac{3}{2}} x}{\sqrt{\frac{a}{b x^{2}} + 1}} + B b^{2} \left (\begin{cases} \frac{\sqrt{a} x^{2}}{2} & \text{for}\: b = 0 \\\frac{\left (a + b x^{2}\right )^{\frac{3}{2}}}{3 b} & \text{otherwise} \end{cases}\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.15734, size = 231, normalized size = 1.62 \begin{align*} \frac{8 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}} B b^{2} + 48 \, \sqrt{b x^{2} + a} B a b^{2} + 24 \, \sqrt{b x^{2} + a} A b^{3} + \frac{15 \,{\left (4 \, B a^{2} b^{2} + 3 \, A a b^{3}\right )} \arctan \left (\frac{\sqrt{b x^{2} + a}}{\sqrt{-a}}\right )}{\sqrt{-a}} - \frac{3 \,{\left (4 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}} B a^{2} b^{2} - 4 \, \sqrt{b x^{2} + a} B a^{3} b^{2} + 9 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}} A a b^{3} - 7 \, \sqrt{b x^{2} + a} A a^{2} b^{3}\right )}}{b^{2} x^{4}}}{24 \, b} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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