Optimal. Leaf size=171 \[ \frac{\sqrt{b} \left (15 a^2 d^2-10 a b c d+3 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{4 d^{5/2}}-\frac{2 a^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{\sqrt{c}}-\frac{b \sqrt{a+b x} \sqrt{c+d x} (3 b c-7 a d)}{4 d^2}+\frac{b (a+b x)^{3/2} \sqrt{c+d x}}{2 d} \]
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Rubi [A] time = 0.149323, antiderivative size = 171, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.364, Rules used = {102, 154, 157, 63, 217, 206, 93, 208} \[ \frac{\sqrt{b} \left (15 a^2 d^2-10 a b c d+3 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{4 d^{5/2}}-\frac{2 a^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{\sqrt{c}}-\frac{b \sqrt{a+b x} \sqrt{c+d x} (3 b c-7 a d)}{4 d^2}+\frac{b (a+b x)^{3/2} \sqrt{c+d x}}{2 d} \]
Antiderivative was successfully verified.
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Rule 102
Rule 154
Rule 157
Rule 63
Rule 217
Rule 206
Rule 93
Rule 208
Rubi steps
\begin{align*} \int \frac{(a+b x)^{5/2}}{x \sqrt{c+d x}} \, dx &=\frac{b (a+b x)^{3/2} \sqrt{c+d x}}{2 d}+\frac{\int \frac{\sqrt{a+b x} \left (2 a^2 d-\frac{1}{2} b (3 b c-7 a d) x\right )}{x \sqrt{c+d x}} \, dx}{2 d}\\ &=-\frac{b (3 b c-7 a d) \sqrt{a+b x} \sqrt{c+d x}}{4 d^2}+\frac{b (a+b x)^{3/2} \sqrt{c+d x}}{2 d}+\frac{\int \frac{2 a^3 d^2+\frac{1}{4} b \left (3 b^2 c^2-10 a b c d+15 a^2 d^2\right ) x}{x \sqrt{a+b x} \sqrt{c+d x}} \, dx}{2 d^2}\\ &=-\frac{b (3 b c-7 a d) \sqrt{a+b x} \sqrt{c+d x}}{4 d^2}+\frac{b (a+b x)^{3/2} \sqrt{c+d x}}{2 d}+a^3 \int \frac{1}{x \sqrt{a+b x} \sqrt{c+d x}} \, dx+\frac{\left (b \left (3 b^2 c^2-10 a b c d+15 a^2 d^2\right )\right ) \int \frac{1}{\sqrt{a+b x} \sqrt{c+d x}} \, dx}{8 d^2}\\ &=-\frac{b (3 b c-7 a d) \sqrt{a+b x} \sqrt{c+d x}}{4 d^2}+\frac{b (a+b x)^{3/2} \sqrt{c+d x}}{2 d}+\left (2 a^3\right ) \operatorname{Subst}\left (\int \frac{1}{-a+c x^2} \, dx,x,\frac{\sqrt{a+b x}}{\sqrt{c+d x}}\right )+\frac{\left (3 b^2 c^2-10 a b c d+15 a^2 d^2\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{c-\frac{a d}{b}+\frac{d x^2}{b}}} \, dx,x,\sqrt{a+b x}\right )}{4 d^2}\\ &=-\frac{b (3 b c-7 a d) \sqrt{a+b x} \sqrt{c+d x}}{4 d^2}+\frac{b (a+b x)^{3/2} \sqrt{c+d x}}{2 d}-\frac{2 a^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{\sqrt{c}}+\frac{\left (3 b^2 c^2-10 a b c d+15 a^2 d^2\right ) \operatorname{Subst}\left (\int \frac{1}{1-\frac{d x^2}{b}} \, dx,x,\frac{\sqrt{a+b x}}{\sqrt{c+d x}}\right )}{4 d^2}\\ &=-\frac{b (3 b c-7 a d) \sqrt{a+b x} \sqrt{c+d x}}{4 d^2}+\frac{b (a+b x)^{3/2} \sqrt{c+d x}}{2 d}-\frac{2 a^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{\sqrt{c}}+\frac{\sqrt{b} \left (3 b^2 c^2-10 a b c d+15 a^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{4 d^{5/2}}\\ \end{align*}
Mathematica [A] time = 0.593526, size = 197, normalized size = 1.15 \[ \frac{1}{4} \left (\frac{\left (25 a^2 b c d^2-15 a^3 d^3-13 a b^2 c^2 d+3 b^3 c^3\right ) \sqrt{\frac{b (c+d x)}{b c-a d}} \sinh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b c-a d}}\right )}{d^{5/2} \sqrt{c+d x} \sqrt{b c-a d}}-\frac{8 a^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{\sqrt{c}}+\frac{b \sqrt{a+b x} \sqrt{c+d x} (9 a d-3 b c+2 b d x)}{d^2}\right ) \]
Antiderivative was successfully verified.
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Maple [B] time = 0.017, size = 342, normalized size = 2. \begin{align*}{\frac{1}{8\,{d}^{2}}\sqrt{bx+a}\sqrt{dx+c} \left ( 15\,\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ){a}^{2}b{d}^{2}\sqrt{ac}-10\,\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ) a{b}^{2}cd\sqrt{ac}+3\,\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ){b}^{3}{c}^{2}\sqrt{ac}-8\,\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+2\,ac}{x}} \right ){a}^{3}{d}^{2}\sqrt{bd}+4\,x{b}^{2}d\sqrt{bd}\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+18\,abd\sqrt{bd}\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }-6\,{b}^{2}c\sqrt{bd}\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) } \right ){\frac{1}{\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }}}{\frac{1}{\sqrt{bd}}}{\frac{1}{\sqrt{ac}}}} \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 = 17.4571, size = 2240, normalized size = 13.1 \begin{align*} \text{result too large to display} \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 (a + b x\right )^{\frac{5}{2}}}{x \sqrt{c + d x}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.3727, size = 335, normalized size = 1.96 \begin{align*} -\frac{{\left (\frac{16 \, \sqrt{b d} a^{3} \arctan \left (-\frac{b^{2} c + a b d -{\left (\sqrt{b d} \sqrt{b x + a} - \sqrt{b^{2} c +{\left (b x + a\right )} b d - a b d}\right )}^{2}}{2 \, \sqrt{-a b c d} b}\right )}{\sqrt{-a b c d} b} - 2 \, \sqrt{b^{2} c +{\left (b x + a\right )} b d - a b d} \sqrt{b x + a}{\left (\frac{2 \,{\left (b x + a\right )}}{b d} - \frac{3 \, b^{2} c d - 7 \, a b d^{2}}{b^{2} d^{3}}\right )} + \frac{{\left (3 \, \sqrt{b d} b^{2} c^{2} - 10 \, \sqrt{b d} a b c d + 15 \, \sqrt{b d} a^{2} d^{2}\right )} \log \left ({\left (\sqrt{b d} \sqrt{b x + a} - \sqrt{b^{2} c +{\left (b x + a\right )} b d - a b d}\right )}^{2}\right )}{b d^{3}}\right )} b^{2}}{8 \,{\left | b \right |}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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