Optimal. Leaf size=207 \[ -\frac{\sqrt{c+d x} \left (11 a^2 d^2-18 a b c d+8 b^2 c^2\right )}{8 a^3 x}+\frac{\left (30 a^2 b c d^2-5 a^3 d^3-40 a b^2 c^2 d+16 b^3 c^3\right ) \tanh ^{-1}\left (\frac{\sqrt{c+d x}}{\sqrt{c}}\right )}{8 a^4 \sqrt{c}}+\frac{c \sqrt{c+d x} (2 b c-3 a d)}{4 a^2 x^2}-\frac{2 \sqrt{b} (b c-a d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{a^4}-\frac{c (c+d x)^{3/2}}{3 a x^3} \]
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Rubi [A] time = 0.26857, antiderivative size = 207, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {98, 149, 151, 156, 63, 208} \[ -\frac{\sqrt{c+d x} \left (11 a^2 d^2-18 a b c d+8 b^2 c^2\right )}{8 a^3 x}+\frac{\left (30 a^2 b c d^2-5 a^3 d^3-40 a b^2 c^2 d+16 b^3 c^3\right ) \tanh ^{-1}\left (\frac{\sqrt{c+d x}}{\sqrt{c}}\right )}{8 a^4 \sqrt{c}}+\frac{c \sqrt{c+d x} (2 b c-3 a d)}{4 a^2 x^2}-\frac{2 \sqrt{b} (b c-a d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{a^4}-\frac{c (c+d x)^{3/2}}{3 a x^3} \]
Antiderivative was successfully verified.
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Rule 98
Rule 149
Rule 151
Rule 156
Rule 63
Rule 208
Rubi steps
\begin{align*} \int \frac{(c+d x)^{5/2}}{x^4 (a+b x)} \, dx &=-\frac{c (c+d x)^{3/2}}{3 a x^3}-\frac{\int \frac{\sqrt{c+d x} \left (\frac{3}{2} c (2 b c-3 a d)+\frac{3}{2} d (b c-2 a d) x\right )}{x^3 (a+b x)} \, dx}{3 a}\\ &=\frac{c (2 b c-3 a d) \sqrt{c+d x}}{4 a^2 x^2}-\frac{c (c+d x)^{3/2}}{3 a x^3}-\frac{\int \frac{-\frac{3}{4} c \left (8 b^2 c^2-18 a b c d+11 a^2 d^2\right )-\frac{3}{4} d \left (6 b^2 c^2-13 a b c d+8 a^2 d^2\right ) x}{x^2 (a+b x) \sqrt{c+d x}} \, dx}{6 a^2}\\ &=\frac{c (2 b c-3 a d) \sqrt{c+d x}}{4 a^2 x^2}-\frac{\left (8 b^2 c^2-18 a b c d+11 a^2 d^2\right ) \sqrt{c+d x}}{8 a^3 x}-\frac{c (c+d x)^{3/2}}{3 a x^3}+\frac{\int \frac{-\frac{3}{8} c \left (16 b^3 c^3-40 a b^2 c^2 d+30 a^2 b c d^2-5 a^3 d^3\right )-\frac{3}{8} b c d \left (8 b^2 c^2-18 a b c d+11 a^2 d^2\right ) x}{x (a+b x) \sqrt{c+d x}} \, dx}{6 a^3 c}\\ &=\frac{c (2 b c-3 a d) \sqrt{c+d x}}{4 a^2 x^2}-\frac{\left (8 b^2 c^2-18 a b c d+11 a^2 d^2\right ) \sqrt{c+d x}}{8 a^3 x}-\frac{c (c+d x)^{3/2}}{3 a x^3}+\frac{\left (b (b c-a d)^3\right ) \int \frac{1}{(a+b x) \sqrt{c+d x}} \, dx}{a^4}-\frac{\left (16 b^3 c^3-40 a b^2 c^2 d+30 a^2 b c d^2-5 a^3 d^3\right ) \int \frac{1}{x \sqrt{c+d x}} \, dx}{16 a^4}\\ &=\frac{c (2 b c-3 a d) \sqrt{c+d x}}{4 a^2 x^2}-\frac{\left (8 b^2 c^2-18 a b c d+11 a^2 d^2\right ) \sqrt{c+d x}}{8 a^3 x}-\frac{c (c+d x)^{3/2}}{3 a x^3}+\frac{\left (2 b (b c-a d)^3\right ) \operatorname{Subst}\left (\int \frac{1}{a-\frac{b c}{d}+\frac{b x^2}{d}} \, dx,x,\sqrt{c+d x}\right )}{a^4 d}-\frac{\left (16 b^3 c^3-40 a b^2 c^2 d+30 a^2 b c d^2-5 a^3 d^3\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{c}{d}+\frac{x^2}{d}} \, dx,x,\sqrt{c+d x}\right )}{8 a^4 d}\\ &=\frac{c (2 b c-3 a d) \sqrt{c+d x}}{4 a^2 x^2}-\frac{\left (8 b^2 c^2-18 a b c d+11 a^2 d^2\right ) \sqrt{c+d x}}{8 a^3 x}-\frac{c (c+d x)^{3/2}}{3 a x^3}+\frac{\left (16 b^3 c^3-40 a b^2 c^2 d+30 a^2 b c d^2-5 a^3 d^3\right ) \tanh ^{-1}\left (\frac{\sqrt{c+d x}}{\sqrt{c}}\right )}{8 a^4 \sqrt{c}}-\frac{2 \sqrt{b} (b c-a d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{a^4}\\ \end{align*}
Mathematica [A] time = 0.327246, size = 178, normalized size = 0.86 \[ -\frac{\frac{a \sqrt{c+d x} \left (a^2 \left (8 c^2+26 c d x+33 d^2 x^2\right )-6 a b c x (2 c+9 d x)+24 b^2 c^2 x^2\right )}{x^3}-\frac{3 \left (30 a^2 b c d^2-5 a^3 d^3-40 a b^2 c^2 d+16 b^3 c^3\right ) \tanh ^{-1}\left (\frac{\sqrt{c+d x}}{\sqrt{c}}\right )}{\sqrt{c}}+48 \sqrt{b} (b c-a d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{24 a^4} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.014, size = 461, normalized size = 2.2 \begin{align*} -{\frac{11}{8\,a{x}^{3}} \left ( dx+c \right ) ^{{\frac{5}{2}}}}+{\frac{9\,bc}{4\,d{a}^{2}{x}^{3}} \left ( dx+c \right ) ^{{\frac{5}{2}}}}-{\frac{{b}^{2}{c}^{2}}{{d}^{2}{a}^{3}{x}^{3}} \left ( dx+c \right ) ^{{\frac{5}{2}}}}+{\frac{5\,c}{3\,a{x}^{3}} \left ( dx+c \right ) ^{{\frac{3}{2}}}}-4\,{\frac{ \left ( dx+c \right ) ^{3/2}b{c}^{2}}{d{a}^{2}{x}^{3}}}+2\,{\frac{ \left ( dx+c \right ) ^{3/2}{b}^{2}{c}^{3}}{{d}^{2}{a}^{3}{x}^{3}}}+{\frac{7\,{c}^{3}b}{4\,d{a}^{2}{x}^{3}}\sqrt{dx+c}}-{\frac{{b}^{2}{c}^{4}}{{d}^{2}{a}^{3}{x}^{3}}\sqrt{dx+c}}-{\frac{5\,{c}^{2}}{8\,a{x}^{3}}\sqrt{dx+c}}-{\frac{5\,{d}^{3}}{8\,a}{\it Artanh} \left ({\sqrt{dx+c}{\frac{1}{\sqrt{c}}}} \right ){\frac{1}{\sqrt{c}}}}+{\frac{15\,{d}^{2}b}{4\,{a}^{2}}\sqrt{c}{\it Artanh} \left ({\sqrt{dx+c}{\frac{1}{\sqrt{c}}}} \right ) }-5\,{\frac{d{c}^{3/2}{b}^{2}}{{a}^{3}}{\it Artanh} \left ({\frac{\sqrt{dx+c}}{\sqrt{c}}} \right ) }+2\,{\frac{{c}^{5/2}{b}^{3}}{{a}^{4}}{\it Artanh} \left ({\frac{\sqrt{dx+c}}{\sqrt{c}}} \right ) }-2\,{\frac{{d}^{3}b}{a\sqrt{ \left ( ad-bc \right ) b}}\arctan \left ({\frac{b\sqrt{dx+c}}{\sqrt{ \left ( ad-bc \right ) b}}} \right ) }+6\,{\frac{{d}^{2}{b}^{2}c}{{a}^{2}\sqrt{ \left ( ad-bc \right ) b}}\arctan \left ({\frac{b\sqrt{dx+c}}{\sqrt{ \left ( ad-bc \right ) b}}} \right ) }-6\,{\frac{d{b}^{3}{c}^{2}}{{a}^{3}\sqrt{ \left ( ad-bc \right ) b}}\arctan \left ({\frac{b\sqrt{dx+c}}{\sqrt{ \left ( ad-bc \right ) b}}} \right ) }+2\,{\frac{{b}^{4}{c}^{3}}{{a}^{4}\sqrt{ \left ( ad-bc \right ) b}}\arctan \left ({\frac{b\sqrt{dx+c}}{\sqrt{ \left ( ad-bc \right ) b}}} \right ) } \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 = 5.71013, size = 2075, normalized size = 10.02 \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(-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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Giac [A] time = 1.17614, size = 405, normalized size = 1.96 \begin{align*} \frac{2 \,{\left (b^{4} c^{3} - 3 \, a b^{3} c^{2} d + 3 \, a^{2} b^{2} c d^{2} - a^{3} b d^{3}\right )} \arctan \left (\frac{\sqrt{d x + c} b}{\sqrt{-b^{2} c + a b d}}\right )}{\sqrt{-b^{2} c + a b d} a^{4}} - \frac{{\left (16 \, b^{3} c^{3} - 40 \, a b^{2} c^{2} d + 30 \, a^{2} b c d^{2} - 5 \, a^{3} d^{3}\right )} \arctan \left (\frac{\sqrt{d x + c}}{\sqrt{-c}}\right )}{8 \, a^{4} \sqrt{-c}} - \frac{24 \,{\left (d x + c\right )}^{\frac{5}{2}} b^{2} c^{2} d - 48 \,{\left (d x + c\right )}^{\frac{3}{2}} b^{2} c^{3} d + 24 \, \sqrt{d x + c} b^{2} c^{4} d - 54 \,{\left (d x + c\right )}^{\frac{5}{2}} a b c d^{2} + 96 \,{\left (d x + c\right )}^{\frac{3}{2}} a b c^{2} d^{2} - 42 \, \sqrt{d x + c} a b c^{3} d^{2} + 33 \,{\left (d x + c\right )}^{\frac{5}{2}} a^{2} d^{3} - 40 \,{\left (d x + c\right )}^{\frac{3}{2}} a^{2} c d^{3} + 15 \, \sqrt{d x + c} a^{2} c^{2} d^{3}}{24 \, a^{3} d^{3} x^{3}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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