Optimal. Leaf size=150 \[ \frac{b^2 \sqrt{x} \sqrt{a+b x} (5 a B+2 A b)}{a}+b^{3/2} (5 a B+2 A b) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a+b x}}\right )-\frac{2 (a+b x)^{5/2} (5 a B+2 A b)}{15 a x^{3/2}}-\frac{2 b (a+b x)^{3/2} (5 a B+2 A b)}{3 a \sqrt{x}}-\frac{2 A (a+b x)^{7/2}}{5 a x^{5/2}} \]
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Rubi [A] time = 0.0654343, antiderivative size = 150, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {78, 47, 50, 63, 217, 206} \[ \frac{b^2 \sqrt{x} \sqrt{a+b x} (5 a B+2 A b)}{a}+b^{3/2} (5 a B+2 A b) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a+b x}}\right )-\frac{2 (a+b x)^{5/2} (5 a B+2 A b)}{15 a x^{3/2}}-\frac{2 b (a+b x)^{3/2} (5 a B+2 A b)}{3 a \sqrt{x}}-\frac{2 A (a+b x)^{7/2}}{5 a x^{5/2}} \]
Antiderivative was successfully verified.
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Rule 78
Rule 47
Rule 50
Rule 63
Rule 217
Rule 206
Rubi steps
\begin{align*} \int \frac{(a+b x)^{5/2} (A+B x)}{x^{7/2}} \, dx &=-\frac{2 A (a+b x)^{7/2}}{5 a x^{5/2}}+\frac{\left (2 \left (A b+\frac{5 a B}{2}\right )\right ) \int \frac{(a+b x)^{5/2}}{x^{5/2}} \, dx}{5 a}\\ &=-\frac{2 (2 A b+5 a B) (a+b x)^{5/2}}{15 a x^{3/2}}-\frac{2 A (a+b x)^{7/2}}{5 a x^{5/2}}+\frac{(b (2 A b+5 a B)) \int \frac{(a+b x)^{3/2}}{x^{3/2}} \, dx}{3 a}\\ &=-\frac{2 b (2 A b+5 a B) (a+b x)^{3/2}}{3 a \sqrt{x}}-\frac{2 (2 A b+5 a B) (a+b x)^{5/2}}{15 a x^{3/2}}-\frac{2 A (a+b x)^{7/2}}{5 a x^{5/2}}+\frac{\left (b^2 (2 A b+5 a B)\right ) \int \frac{\sqrt{a+b x}}{\sqrt{x}} \, dx}{a}\\ &=\frac{b^2 (2 A b+5 a B) \sqrt{x} \sqrt{a+b x}}{a}-\frac{2 b (2 A b+5 a B) (a+b x)^{3/2}}{3 a \sqrt{x}}-\frac{2 (2 A b+5 a B) (a+b x)^{5/2}}{15 a x^{3/2}}-\frac{2 A (a+b x)^{7/2}}{5 a x^{5/2}}+\frac{1}{2} \left (b^2 (2 A b+5 a B)\right ) \int \frac{1}{\sqrt{x} \sqrt{a+b x}} \, dx\\ &=\frac{b^2 (2 A b+5 a B) \sqrt{x} \sqrt{a+b x}}{a}-\frac{2 b (2 A b+5 a B) (a+b x)^{3/2}}{3 a \sqrt{x}}-\frac{2 (2 A b+5 a B) (a+b x)^{5/2}}{15 a x^{3/2}}-\frac{2 A (a+b x)^{7/2}}{5 a x^{5/2}}+\left (b^2 (2 A b+5 a B)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^2}} \, dx,x,\sqrt{x}\right )\\ &=\frac{b^2 (2 A b+5 a B) \sqrt{x} \sqrt{a+b x}}{a}-\frac{2 b (2 A b+5 a B) (a+b x)^{3/2}}{3 a \sqrt{x}}-\frac{2 (2 A b+5 a B) (a+b x)^{5/2}}{15 a x^{3/2}}-\frac{2 A (a+b x)^{7/2}}{5 a x^{5/2}}+\left (b^2 (2 A b+5 a B)\right ) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{\sqrt{x}}{\sqrt{a+b x}}\right )\\ &=\frac{b^2 (2 A b+5 a B) \sqrt{x} \sqrt{a+b x}}{a}-\frac{2 b (2 A b+5 a B) (a+b x)^{3/2}}{3 a \sqrt{x}}-\frac{2 (2 A b+5 a B) (a+b x)^{5/2}}{15 a x^{3/2}}-\frac{2 A (a+b x)^{7/2}}{5 a x^{5/2}}+b^{3/2} (2 A b+5 a B) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a+b x}}\right )\\ \end{align*}
Mathematica [C] time = 0.0402359, size = 76, normalized size = 0.51 \[ \frac{2 \sqrt{a+b x} \left (-\frac{a^2 x (5 a B+2 A b) \, _2F_1\left (-\frac{5}{2},-\frac{3}{2};-\frac{1}{2};-\frac{b x}{a}\right )}{\sqrt{\frac{b x}{a}+1}}-3 A (a+b x)^3\right )}{15 a x^{5/2}} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.012, size = 206, normalized size = 1.4 \begin{align*}{\frac{1}{30}\sqrt{bx+a} \left ( 30\,A\ln \left ( 1/2\,{\frac{2\,\sqrt{x \left ( bx+a \right ) }\sqrt{b}+2\,bx+a}{\sqrt{b}}} \right ){x}^{3}{b}^{3}+75\,B\ln \left ( 1/2\,{\frac{2\,\sqrt{x \left ( bx+a \right ) }\sqrt{b}+2\,bx+a}{\sqrt{b}}} \right ){x}^{3}a{b}^{2}+30\,B\sqrt{x \left ( bx+a \right ) }{b}^{5/2}{x}^{3}-92\,A\sqrt{x \left ( bx+a \right ) }{b}^{5/2}{x}^{2}-140\,B\sqrt{x \left ( bx+a \right ) }{b}^{3/2}{x}^{2}a-44\,Axa{b}^{3/2}\sqrt{x \left ( bx+a \right ) }-20\,Bx{a}^{2}\sqrt{x \left ( bx+a \right ) }\sqrt{b}-12\,A{a}^{2}\sqrt{x \left ( bx+a \right ) }\sqrt{b} \right ){x}^{-{\frac{5}{2}}}{\frac{1}{\sqrt{x \left ( bx+a \right ) }}}{\frac{1}{\sqrt{b}}}} \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 = 2.67244, size = 548, normalized size = 3.65 \begin{align*} \left [\frac{15 \,{\left (5 \, B a b + 2 \, A b^{2}\right )} \sqrt{b} x^{3} \log \left (2 \, b x + 2 \, \sqrt{b x + a} \sqrt{b} \sqrt{x} + a\right ) + 2 \,{\left (15 \, B b^{2} x^{3} - 6 \, A a^{2} - 2 \,{\left (35 \, B a b + 23 \, A b^{2}\right )} x^{2} - 2 \,{\left (5 \, B a^{2} + 11 \, A a b\right )} x\right )} \sqrt{b x + a} \sqrt{x}}{30 \, x^{3}}, -\frac{15 \,{\left (5 \, B a b + 2 \, A b^{2}\right )} \sqrt{-b} x^{3} \arctan \left (\frac{\sqrt{b x + a} \sqrt{-b}}{b \sqrt{x}}\right ) -{\left (15 \, B b^{2} x^{3} - 6 \, A a^{2} - 2 \,{\left (35 \, B a b + 23 \, A b^{2}\right )} x^{2} - 2 \,{\left (5 \, B a^{2} + 11 \, A a b\right )} x\right )} \sqrt{b x + a} \sqrt{x}}{15 \, x^{3}}\right ] \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A] time = 138.969, size = 201, normalized size = 1.34 \begin{align*} A \left (- \frac{2 a^{2} \sqrt{b} \sqrt{\frac{a}{b x} + 1}}{5 x^{2}} - \frac{22 a b^{\frac{3}{2}} \sqrt{\frac{a}{b x} + 1}}{15 x} - \frac{46 b^{\frac{5}{2}} \sqrt{\frac{a}{b x} + 1}}{15} - b^{\frac{5}{2}} \log{\left (\frac{a}{b x} \right )} + 2 b^{\frac{5}{2}} \log{\left (\sqrt{\frac{a}{b x} + 1} + 1 \right )}\right ) + B \left (- \frac{2 a^{2} \sqrt{b} \sqrt{\frac{a}{b x} + 1}}{3 x} - \frac{14 a b^{\frac{3}{2}} \sqrt{\frac{a}{b x} + 1}}{3} - \frac{5 a b^{\frac{3}{2}} \log{\left (\frac{a}{b x} \right )}}{2} + 5 a b^{\frac{3}{2}} \log{\left (\sqrt{\frac{a}{b x} + 1} + 1 \right )} + b^{\frac{5}{2}} x \sqrt{\frac{a}{b x} + 1}\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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