Optimal. Leaf size=160 \[ \frac{\sqrt{a^2+2 a b x+b^2 x^2} (2 a+b x) \sqrt{c+d x^2}}{2 (a+b x)}+\frac{b c \sqrt{a^2+2 a b x+b^2 x^2} \tanh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c+d x^2}}\right )}{2 \sqrt{d} (a+b x)}-\frac{a \sqrt{c} \sqrt{a^2+2 a b x+b^2 x^2} \tanh ^{-1}\left (\frac{\sqrt{c+d x^2}}{\sqrt{c}}\right )}{a+b x} \]
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Rubi [A] time = 0.119417, antiderivative size = 160, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.229, Rules used = {1001, 815, 844, 217, 206, 266, 63, 208} \[ \frac{\sqrt{a^2+2 a b x+b^2 x^2} (2 a+b x) \sqrt{c+d x^2}}{2 (a+b x)}+\frac{b c \sqrt{a^2+2 a b x+b^2 x^2} \tanh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c+d x^2}}\right )}{2 \sqrt{d} (a+b x)}-\frac{a \sqrt{c} \sqrt{a^2+2 a b x+b^2 x^2} \tanh ^{-1}\left (\frac{\sqrt{c+d x^2}}{\sqrt{c}}\right )}{a+b x} \]
Antiderivative was successfully verified.
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Rule 1001
Rule 815
Rule 844
Rule 217
Rule 206
Rule 266
Rule 63
Rule 208
Rubi steps
\begin{align*} \int \frac{\sqrt{a^2+2 a b x+b^2 x^2} \sqrt{c+d x^2}}{x} \, dx &=\frac{\sqrt{a^2+2 a b x+b^2 x^2} \int \frac{\left (2 a b+2 b^2 x\right ) \sqrt{c+d x^2}}{x} \, dx}{2 a b+2 b^2 x}\\ &=\frac{(2 a+b x) \sqrt{a^2+2 a b x+b^2 x^2} \sqrt{c+d x^2}}{2 (a+b x)}+\frac{\sqrt{a^2+2 a b x+b^2 x^2} \int \frac{4 a b c d+2 b^2 c d x}{x \sqrt{c+d x^2}} \, dx}{2 d \left (2 a b+2 b^2 x\right )}\\ &=\frac{(2 a+b x) \sqrt{a^2+2 a b x+b^2 x^2} \sqrt{c+d x^2}}{2 (a+b x)}+\frac{\left (2 a b c \sqrt{a^2+2 a b x+b^2 x^2}\right ) \int \frac{1}{x \sqrt{c+d x^2}} \, dx}{2 a b+2 b^2 x}+\frac{\left (b^2 c \sqrt{a^2+2 a b x+b^2 x^2}\right ) \int \frac{1}{\sqrt{c+d x^2}} \, dx}{2 a b+2 b^2 x}\\ &=\frac{(2 a+b x) \sqrt{a^2+2 a b x+b^2 x^2} \sqrt{c+d x^2}}{2 (a+b x)}+\frac{\left (a b c \sqrt{a^2+2 a b x+b^2 x^2}\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{c+d x}} \, dx,x,x^2\right )}{2 a b+2 b^2 x}+\frac{\left (b^2 c \sqrt{a^2+2 a b x+b^2 x^2}\right ) \operatorname{Subst}\left (\int \frac{1}{1-d x^2} \, dx,x,\frac{x}{\sqrt{c+d x^2}}\right )}{2 a b+2 b^2 x}\\ &=\frac{(2 a+b x) \sqrt{a^2+2 a b x+b^2 x^2} \sqrt{c+d x^2}}{2 (a+b x)}+\frac{b c \sqrt{a^2+2 a b x+b^2 x^2} \tanh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c+d x^2}}\right )}{2 \sqrt{d} (a+b x)}+\frac{\left (2 a b c \sqrt{a^2+2 a b x+b^2 x^2}\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{c}{d}+\frac{x^2}{d}} \, dx,x,\sqrt{c+d x^2}\right )}{d \left (2 a b+2 b^2 x\right )}\\ &=\frac{(2 a+b x) \sqrt{a^2+2 a b x+b^2 x^2} \sqrt{c+d x^2}}{2 (a+b x)}+\frac{b c \sqrt{a^2+2 a b x+b^2 x^2} \tanh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c+d x^2}}\right )}{2 \sqrt{d} (a+b x)}-\frac{a \sqrt{c} \sqrt{a^2+2 a b x+b^2 x^2} \tanh ^{-1}\left (\frac{\sqrt{c+d x^2}}{\sqrt{c}}\right )}{a+b x}\\ \end{align*}
Mathematica [A] time = 0.146834, size = 139, normalized size = 0.87 \[ \frac{\sqrt{(a+b x)^2} \left (\sqrt{d} \sqrt{\frac{d x^2}{c}+1} \left ((2 a+b x) \sqrt{c+d x^2}-2 a \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c+d x^2}}{\sqrt{c}}\right )\right )+b \sqrt{c} \sqrt{c+d x^2} \sinh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )\right )}{2 \sqrt{d} (a+b x) \sqrt{\frac{d x^2}{c}+1}} \]
Antiderivative was successfully verified.
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Maple [C] time = 0.205, size = 94, normalized size = 0.6 \begin{align*} -{\frac{{\it csgn} \left ( bx+a \right ) }{2} \left ( 2\,\sqrt{d}\ln \left ( 2\,{\frac{\sqrt{c}\sqrt{d{x}^{2}+c}+c}{x}} \right ) \sqrt{c}a-\sqrt{d}\sqrt{d{x}^{2}+c}xb-2\,\sqrt{d}\sqrt{d{x}^{2}+c}a-\ln \left ( x\sqrt{d}+\sqrt{d{x}^{2}+c} \right ) bc \right ){\frac{1}{\sqrt{d}}}} \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 \frac{\sqrt{d x^{2} + 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 = 1.85328, size = 859, normalized size = 5.37 \begin{align*} \left [\frac{b c \sqrt{d} \log \left (-2 \, d x^{2} - 2 \, \sqrt{d x^{2} + c} \sqrt{d} x - c\right ) + 2 \, a \sqrt{c} d \log \left (-\frac{d x^{2} - 2 \, \sqrt{d x^{2} + c} \sqrt{c} + 2 \, c}{x^{2}}\right ) + 2 \,{\left (b d x + 2 \, a d\right )} \sqrt{d x^{2} + c}}{4 \, d}, -\frac{b c \sqrt{-d} \arctan \left (\frac{\sqrt{-d} x}{\sqrt{d x^{2} + c}}\right ) - a \sqrt{c} d \log \left (-\frac{d x^{2} - 2 \, \sqrt{d x^{2} + c} \sqrt{c} + 2 \, c}{x^{2}}\right ) -{\left (b d x + 2 \, a d\right )} \sqrt{d x^{2} + c}}{2 \, d}, \frac{4 \, a \sqrt{-c} d \arctan \left (\frac{\sqrt{-c}}{\sqrt{d x^{2} + c}}\right ) + b c \sqrt{d} \log \left (-2 \, d x^{2} - 2 \, \sqrt{d x^{2} + c} \sqrt{d} x - c\right ) + 2 \,{\left (b d x + 2 \, a d\right )} \sqrt{d x^{2} + c}}{4 \, d}, -\frac{b c \sqrt{-d} \arctan \left (\frac{\sqrt{-d} x}{\sqrt{d x^{2} + c}}\right ) - 2 \, a \sqrt{-c} d \arctan \left (\frac{\sqrt{-c}}{\sqrt{d x^{2} + c}}\right ) -{\left (b d x + 2 \, a d\right )} \sqrt{d x^{2} + c}}{2 \, d}\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 \frac{\sqrt{c + d x^{2}} \sqrt{\left (a + b x\right )^{2}}}{x}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.20625, size = 138, normalized size = 0.86 \begin{align*} \frac{2 \, a c \arctan \left (-\frac{\sqrt{d} x - \sqrt{d x^{2} + c}}{\sqrt{-c}}\right ) \mathrm{sgn}\left (b x + a\right )}{\sqrt{-c}} - \frac{b c \log \left ({\left | -\sqrt{d} x + \sqrt{d x^{2} + c} \right |}\right ) \mathrm{sgn}\left (b x + a\right )}{2 \, \sqrt{d}} + \frac{1}{2} \, \sqrt{d x^{2} + c}{\left (b x \mathrm{sgn}\left (b x + a\right ) + 2 \, a \mathrm{sgn}\left (b x + a\right )\right )} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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