Optimal. Leaf size=218 \[ a^{3/2} (-A) \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+b x+c x^2}}\right )-\frac{\sqrt{a+b x+c x^2} \left (2 c x \left (-12 a B c-8 A b c+3 b^2 B\right )-64 a A c^2-12 a b B c-8 A b^2 c+3 b^3 B\right )}{64 c^2}+\frac{\left (\left (b^2-4 a c\right ) \left (-12 a B c-8 A b c+3 b^2 B\right )+64 a A b c^2\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{128 c^{5/2}}+\frac{\left (a+b x+c x^2\right )^{3/2} (8 A c+3 b B+6 B c x)}{24 c} \]
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Rubi [A] time = 0.25405, antiderivative size = 218, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {814, 843, 621, 206, 724} \[ a^{3/2} (-A) \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+b x+c x^2}}\right )-\frac{\sqrt{a+b x+c x^2} \left (2 c x \left (-12 a B c-8 A b c+3 b^2 B\right )-64 a A c^2-12 a b B c-8 A b^2 c+3 b^3 B\right )}{64 c^2}+\frac{\left (\left (b^2-4 a c\right ) \left (-12 a B c-8 A b c+3 b^2 B\right )+64 a A b c^2\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{128 c^{5/2}}+\frac{\left (a+b x+c x^2\right )^{3/2} (8 A c+3 b B+6 B c x)}{24 c} \]
Antiderivative was successfully verified.
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Rule 814
Rule 843
Rule 621
Rule 206
Rule 724
Rubi steps
\begin{align*} \int \frac{(A+B x) \left (a+b x+c x^2\right )^{3/2}}{x} \, dx &=\frac{(3 b B+8 A c+6 B c x) \left (a+b x+c x^2\right )^{3/2}}{24 c}-\frac{\int \frac{\left (-8 a A c-\frac{1}{2} \left (8 A b c-3 B \left (b^2-4 a c\right )\right ) x\right ) \sqrt{a+b x+c x^2}}{x} \, dx}{8 c}\\ &=-\frac{\left (3 b^3 B-8 A b^2 c-12 a b B c-64 a A c^2+2 c \left (3 b^2 B-8 A b c-12 a B c\right ) x\right ) \sqrt{a+b x+c x^2}}{64 c^2}+\frac{(3 b B+8 A c+6 B c x) \left (a+b x+c x^2\right )^{3/2}}{24 c}+\frac{\int \frac{32 a^2 A c^2+\frac{1}{4} \left (64 a A b c^2-\left (b^2-4 a c\right ) \left (8 A b c-3 B \left (b^2-4 a c\right )\right )\right ) x}{x \sqrt{a+b x+c x^2}} \, dx}{32 c^2}\\ &=-\frac{\left (3 b^3 B-8 A b^2 c-12 a b B c-64 a A c^2+2 c \left (3 b^2 B-8 A b c-12 a B c\right ) x\right ) \sqrt{a+b x+c x^2}}{64 c^2}+\frac{(3 b B+8 A c+6 B c x) \left (a+b x+c x^2\right )^{3/2}}{24 c}+\left (a^2 A\right ) \int \frac{1}{x \sqrt{a+b x+c x^2}} \, dx+\frac{1}{128} \left (64 a A b+\frac{\left (b^2-4 a c\right ) \left (3 b^2 B-8 A b c-12 a B c\right )}{c^2}\right ) \int \frac{1}{\sqrt{a+b x+c x^2}} \, dx\\ &=-\frac{\left (3 b^3 B-8 A b^2 c-12 a b B c-64 a A c^2+2 c \left (3 b^2 B-8 A b c-12 a B c\right ) x\right ) \sqrt{a+b x+c x^2}}{64 c^2}+\frac{(3 b B+8 A c+6 B c x) \left (a+b x+c x^2\right )^{3/2}}{24 c}-\left (2 a^2 A\right ) \operatorname{Subst}\left (\int \frac{1}{4 a-x^2} \, dx,x,\frac{2 a+b x}{\sqrt{a+b x+c x^2}}\right )+\frac{1}{64} \left (64 a A b+\frac{\left (b^2-4 a c\right ) \left (3 b^2 B-8 A b c-12 a B c\right )}{c^2}\right ) \operatorname{Subst}\left (\int \frac{1}{4 c-x^2} \, dx,x,\frac{b+2 c x}{\sqrt{a+b x+c x^2}}\right )\\ &=-\frac{\left (3 b^3 B-8 A b^2 c-12 a b B c-64 a A c^2+2 c \left (3 b^2 B-8 A b c-12 a B c\right ) x\right ) \sqrt{a+b x+c x^2}}{64 c^2}+\frac{(3 b B+8 A c+6 B c x) \left (a+b x+c x^2\right )^{3/2}}{24 c}-a^{3/2} A \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+b x+c x^2}}\right )+\frac{\left (64 a A b+\frac{\left (b^2-4 a c\right ) \left (3 b^2 B-8 A b c-12 a B c\right )}{c^2}\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{128 \sqrt{c}}\\ \end{align*}
Mathematica [A] time = 0.366382, size = 206, normalized size = 0.94 \[ \frac{\left (48 a^2 B c^2+96 a A b c^2-24 a b^2 B c-8 A b^3 c+3 b^4 B\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+x (b+c x)}}\right )}{128 c^{5/2}}+a^{3/2} (-A) \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+x (b+c x)}}\right )+\frac{\sqrt{a+x (b+c x)} \left (4 b c (15 a B+2 c x (14 A+9 B x))+8 c^2 \left (32 a A+15 a B x+8 A c x^2+6 B c x^3\right )+6 b^2 c (4 A+B x)-9 b^3 B\right )}{192 c^2} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.007, size = 390, normalized size = 1.8 \begin{align*}{\frac{Bx}{4} \left ( c{x}^{2}+bx+a \right ) ^{{\frac{3}{2}}}}+{\frac{bB}{8\,c} \left ( c{x}^{2}+bx+a \right ) ^{{\frac{3}{2}}}}+{\frac{3\,aBx}{8}\sqrt{c{x}^{2}+bx+a}}-{\frac{3\,{b}^{2}Bx}{32\,c}\sqrt{c{x}^{2}+bx+a}}+{\frac{3\,abB}{16\,c}\sqrt{c{x}^{2}+bx+a}}-{\frac{3\,{b}^{3}B}{64\,{c}^{2}}\sqrt{c{x}^{2}+bx+a}}+{\frac{3\,B{a}^{2}}{8}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){\frac{1}{\sqrt{c}}}}-{\frac{3\,Ba{b}^{2}}{16}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){c}^{-{\frac{3}{2}}}}+{\frac{3\,{b}^{4}B}{128}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){c}^{-{\frac{5}{2}}}}+{\frac{A}{3} \left ( c{x}^{2}+bx+a \right ) ^{{\frac{3}{2}}}}+{\frac{Abx}{4}\sqrt{c{x}^{2}+bx+a}}+{\frac{A{b}^{2}}{8\,c}\sqrt{c{x}^{2}+bx+a}}+{\frac{3\,Aab}{4}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){\frac{1}{\sqrt{c}}}}-{\frac{A{b}^{3}}{16}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){c}^{-{\frac{3}{2}}}}+Aa\sqrt{c{x}^{2}+bx+a}-A{a}^{{\frac{3}{2}}}\ln \left ({\frac{1}{x} \left ( 2\,a+bx+2\,\sqrt{a}\sqrt{c{x}^{2}+bx+a} \right ) } \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 = 31.5857, size = 2435, normalized size = 11.17 \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 ) \left (a + b x + c x^{2}\right )^{\frac{3}{2}}}{x}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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