Optimal. Leaf size=227 \[ -\frac{\sqrt{a^2+2 a b x+b^2 x^2} (2 d x+e) \sqrt{c+d x^2+e x} \left (8 a d e+4 b c d-5 b e^2\right )}{64 d^3 (a+b x)}-\frac{\sqrt{a^2+2 a b x+b^2 x^2} \left (4 c d-e^2\right ) \left (8 a d e+4 b c d-5 b e^2\right ) \tanh ^{-1}\left (\frac{2 d x+e}{2 \sqrt{d} \sqrt{c+d x^2+e x}}\right )}{128 d^{7/2} (a+b x)}+\frac{\sqrt{a^2+2 a b x+b^2 x^2} \left (c+d x^2+e x\right )^{3/2} (8 a d+6 b d x-5 b e)}{24 d^2 (a+b x)} \]
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Rubi [A] time = 0.126839, antiderivative size = 227, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.139, Rules used = {1000, 779, 612, 621, 206} \[ -\frac{\sqrt{a^2+2 a b x+b^2 x^2} (2 d x+e) \sqrt{c+d x^2+e x} \left (8 a d e+4 b c d-5 b e^2\right )}{64 d^3 (a+b x)}-\frac{\sqrt{a^2+2 a b x+b^2 x^2} \left (4 c d-e^2\right ) \left (8 a d e+4 b c d-5 b e^2\right ) \tanh ^{-1}\left (\frac{2 d x+e}{2 \sqrt{d} \sqrt{c+d x^2+e x}}\right )}{128 d^{7/2} (a+b x)}+\frac{\sqrt{a^2+2 a b x+b^2 x^2} \left (c+d x^2+e x\right )^{3/2} (8 a d+6 b d x-5 b e)}{24 d^2 (a+b x)} \]
Antiderivative was successfully verified.
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Rule 1000
Rule 779
Rule 612
Rule 621
Rule 206
Rubi steps
\begin{align*} \int x \sqrt{a^2+2 a b x+b^2 x^2} \sqrt{c+e x+d x^2} \, dx &=\frac{\sqrt{a^2+2 a b x+b^2 x^2} \int x \left (2 a b+2 b^2 x\right ) \sqrt{c+e x+d x^2} \, dx}{2 a b+2 b^2 x}\\ &=\frac{(8 a d-5 b e+6 b d x) \sqrt{a^2+2 a b x+b^2 x^2} \left (c+e x+d x^2\right )^{3/2}}{24 d^2 (a+b x)}-\frac{\left (b \left (4 b c d+8 a d e-5 b e^2\right ) \sqrt{a^2+2 a b x+b^2 x^2}\right ) \int \sqrt{c+e x+d x^2} \, dx}{8 d^2 \left (2 a b+2 b^2 x\right )}\\ &=-\frac{\left (4 b c d+8 a d e-5 b e^2\right ) (e+2 d x) \sqrt{a^2+2 a b x+b^2 x^2} \sqrt{c+e x+d x^2}}{64 d^3 (a+b x)}+\frac{(8 a d-5 b e+6 b d x) \sqrt{a^2+2 a b x+b^2 x^2} \left (c+e x+d x^2\right )^{3/2}}{24 d^2 (a+b x)}-\frac{\left (b \left (4 c d-e^2\right ) \left (4 b c d+8 a d e-5 b e^2\right ) \sqrt{a^2+2 a b x+b^2 x^2}\right ) \int \frac{1}{\sqrt{c+e x+d x^2}} \, dx}{64 d^3 \left (2 a b+2 b^2 x\right )}\\ &=-\frac{\left (4 b c d+8 a d e-5 b e^2\right ) (e+2 d x) \sqrt{a^2+2 a b x+b^2 x^2} \sqrt{c+e x+d x^2}}{64 d^3 (a+b x)}+\frac{(8 a d-5 b e+6 b d x) \sqrt{a^2+2 a b x+b^2 x^2} \left (c+e x+d x^2\right )^{3/2}}{24 d^2 (a+b x)}-\frac{\left (b \left (4 c d-e^2\right ) \left (4 b c d+8 a d e-5 b e^2\right ) \sqrt{a^2+2 a b x+b^2 x^2}\right ) \operatorname{Subst}\left (\int \frac{1}{4 d-x^2} \, dx,x,\frac{e+2 d x}{\sqrt{c+e x+d x^2}}\right )}{32 d^3 \left (2 a b+2 b^2 x\right )}\\ &=-\frac{\left (4 b c d+8 a d e-5 b e^2\right ) (e+2 d x) \sqrt{a^2+2 a b x+b^2 x^2} \sqrt{c+e x+d x^2}}{64 d^3 (a+b x)}+\frac{(8 a d-5 b e+6 b d x) \sqrt{a^2+2 a b x+b^2 x^2} \left (c+e x+d x^2\right )^{3/2}}{24 d^2 (a+b x)}-\frac{\left (4 c d-e^2\right ) \left (4 b c d+8 a d e-5 b e^2\right ) \sqrt{a^2+2 a b x+b^2 x^2} \tanh ^{-1}\left (\frac{e+2 d x}{2 \sqrt{d} \sqrt{c+e x+d x^2}}\right )}{128 d^{7/2} (a+b x)}\\ \end{align*}
Mathematica [A] time = 0.136698, size = 147, normalized size = 0.65 \[ \frac{\sqrt{(a+b x)^2} \left ((c+x (d x+e))^{3/2} (8 a d+6 b d x-5 b e)-\frac{3 \left (8 a d e+4 b c d-5 b e^2\right ) \left (\left (4 c d-e^2\right ) \tanh ^{-1}\left (\frac{2 d x+e}{2 \sqrt{d} \sqrt{c+x (d x+e)}}\right )+2 \sqrt{d} (2 d x+e) \sqrt{c+x (d x+e)}\right )}{16 d^{3/2}}\right )}{24 d^2 (a+b x)} \]
Antiderivative was successfully verified.
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Maple [C] time = 0.213, size = 381, normalized size = 1.7 \begin{align*}{\frac{{\it csgn} \left ( bx+a \right ) }{384} \left ( 96\,{d}^{7/2} \left ( d{x}^{2}+ex+c \right ) ^{3/2}xb+128\,{d}^{7/2} \left ( d{x}^{2}+ex+c \right ) ^{3/2}a-80\,{d}^{5/2} \left ( d{x}^{2}+ex+c \right ) ^{3/2}be-96\,{d}^{7/2}\sqrt{d{x}^{2}+ex+c}xae-48\,{d}^{7/2}\sqrt{d{x}^{2}+ex+c}xbc+60\,{d}^{5/2}\sqrt{d{x}^{2}+ex+c}xb{e}^{2}-48\,{d}^{5/2}\sqrt{d{x}^{2}+ex+c}a{e}^{2}-24\,{d}^{5/2}\sqrt{d{x}^{2}+ex+c}bce+30\,{d}^{3/2}\sqrt{d{x}^{2}+ex+c}b{e}^{3}-96\,\ln \left ( 1/2\,{\frac{2\,\sqrt{d{x}^{2}+ex+c}\sqrt{d}+2\,dx+e}{\sqrt{d}}} \right ) ac{d}^{3}e+24\,\ln \left ( 1/2\,{\frac{2\,\sqrt{d{x}^{2}+ex+c}\sqrt{d}+2\,dx+e}{\sqrt{d}}} \right ) a{d}^{2}{e}^{3}-48\,\ln \left ( 1/2\,{\frac{2\,\sqrt{d{x}^{2}+ex+c}\sqrt{d}+2\,dx+e}{\sqrt{d}}} \right ) b{c}^{2}{d}^{3}+72\,\ln \left ( 1/2\,{\frac{2\,\sqrt{d{x}^{2}+ex+c}\sqrt{d}+2\,dx+e}{\sqrt{d}}} \right ) bc{d}^{2}{e}^{2}-15\,\ln \left ( 1/2\,{\frac{2\,\sqrt{d{x}^{2}+ex+c}\sqrt{d}+2\,dx+e}{\sqrt{d}}} \right ) bd{e}^{4} \right ){d}^{-{\frac{9}{2}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{d x^{2} + e x + c} \sqrt{{\left (b x + a\right )}^{2}} x\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.78244, size = 930, normalized size = 4.1 \begin{align*} \left [\frac{3 \,{\left (16 \, b c^{2} d^{2} + 32 \, a c d^{2} e - 24 \, b c d e^{2} - 8 \, a d e^{3} + 5 \, b e^{4}\right )} \sqrt{d} \log \left (8 \, d^{2} x^{2} + 8 \, d e x - 4 \, \sqrt{d x^{2} + e x + c}{\left (2 \, d x + e\right )} \sqrt{d} + 4 \, c d + e^{2}\right ) + 4 \,{\left (48 \, b d^{4} x^{3} + 64 \, a c d^{3} - 52 \, b c d^{2} e - 24 \, a d^{2} e^{2} + 15 \, b d e^{3} + 8 \,{\left (8 \, a d^{4} + b d^{3} e\right )} x^{2} + 2 \,{\left (12 \, b c d^{3} + 8 \, a d^{3} e - 5 \, b d^{2} e^{2}\right )} x\right )} \sqrt{d x^{2} + e x + c}}{768 \, d^{4}}, \frac{3 \,{\left (16 \, b c^{2} d^{2} + 32 \, a c d^{2} e - 24 \, b c d e^{2} - 8 \, a d e^{3} + 5 \, b e^{4}\right )} \sqrt{-d} \arctan \left (\frac{\sqrt{d x^{2} + e x + c}{\left (2 \, d x + e\right )} \sqrt{-d}}{2 \,{\left (d^{2} x^{2} + d e x + c d\right )}}\right ) + 2 \,{\left (48 \, b d^{4} x^{3} + 64 \, a c d^{3} - 52 \, b c d^{2} e - 24 \, a d^{2} e^{2} + 15 \, b d e^{3} + 8 \,{\left (8 \, a d^{4} + b d^{3} e\right )} x^{2} + 2 \,{\left (12 \, b c d^{3} + 8 \, a d^{3} e - 5 \, b d^{2} e^{2}\right )} x\right )} \sqrt{d x^{2} + e x + c}}{384 \, d^{4}}\right ] \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 x \sqrt{c + d x^{2} + e x} \sqrt{\left (a + b x\right )^{2}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.19218, size = 362, normalized size = 1.59 \begin{align*} \frac{1}{192} \, \sqrt{d x^{2} + x e + c}{\left (2 \,{\left (4 \,{\left (6 \, b x \mathrm{sgn}\left (b x + a\right ) + \frac{8 \, a d^{3} \mathrm{sgn}\left (b x + a\right ) + b d^{2} e \mathrm{sgn}\left (b x + a\right )}{d^{3}}\right )} x + \frac{12 \, b c d^{2} \mathrm{sgn}\left (b x + a\right ) + 8 \, a d^{2} e \mathrm{sgn}\left (b x + a\right ) - 5 \, b d e^{2} \mathrm{sgn}\left (b x + a\right )}{d^{3}}\right )} x + \frac{64 \, a c d^{2} \mathrm{sgn}\left (b x + a\right ) - 52 \, b c d e \mathrm{sgn}\left (b x + a\right ) - 24 \, a d e^{2} \mathrm{sgn}\left (b x + a\right ) + 15 \, b e^{3} \mathrm{sgn}\left (b x + a\right )}{d^{3}}\right )} + \frac{{\left (16 \, b c^{2} d^{2} \mathrm{sgn}\left (b x + a\right ) + 32 \, a c d^{2} e \mathrm{sgn}\left (b x + a\right ) - 24 \, b c d e^{2} \mathrm{sgn}\left (b x + a\right ) - 8 \, a d e^{3} \mathrm{sgn}\left (b x + a\right ) + 5 \, b e^{4} \mathrm{sgn}\left (b x + a\right )\right )} \log \left ({\left | -2 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + x e + c}\right )} \sqrt{d} - e \right |}\right )}{128 \, d^{\frac{7}{2}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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