Integrand size = 35, antiderivative size = 160 \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2}}{x} \, dx=\frac {(2 a+b x) \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2}}{2 (a+b x)}+\frac {b c \sqrt {a^2+2 a b x+b^2 x^2} \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{2 \sqrt {d} (a+b x)}-\frac {a \sqrt {c} \sqrt {a^2+2 a b x+b^2 x^2} \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{a+b x} \]
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Time = 0.07 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.229, Rules used = {1015, 829, 858, 223, 212, 272, 65, 214} \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2}}{x} \, dx=\frac {b c \sqrt {a^2+2 a b x+b^2 x^2} \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{2 \sqrt {d} (a+b x)}-\frac {a \sqrt {c} \sqrt {a^2+2 a b x+b^2 x^2} \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{a+b x}+\frac {\sqrt {a^2+2 a b x+b^2 x^2} (2 a+b x) \sqrt {c+d x^2}}{2 (a+b x)} \]
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Rule 65
Rule 212
Rule 214
Rule 223
Rule 272
Rule 829
Rule 858
Rule 1015
Rubi steps \begin{align*} \text {integral}& = \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 ) \text {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 ) \text {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 ) \text {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*}
Time = 0.17 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.74 \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2}}{x} \, dx=\frac {\sqrt {(a+b x)^2} \left (\sqrt {d} (2 a+b x) \sqrt {c+d x^2}+4 a \sqrt {c} \sqrt {d} \text {arctanh}\left (\frac {\sqrt {d} x-\sqrt {c+d x^2}}{\sqrt {c}}\right )-b c \log \left (-\sqrt {d} x+\sqrt {c+d x^2}\right )\right )}{2 \sqrt {d} (a+b x)} \]
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Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.49 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.58
method | result | size |
default | \(\frac {\operatorname {csgn}\left (b x +a \right ) \left (\sqrt {d \,x^{2}+c}\, \sqrt {d}\, b x -2 \sqrt {d}\, \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {d \,x^{2}+c}}{x}\right ) \sqrt {c}\, a +2 \sqrt {d \,x^{2}+c}\, \sqrt {d}\, a +\ln \left (\sqrt {d}\, x +\sqrt {d \,x^{2}+c}\right ) b c \right )}{2 \sqrt {d}}\) | \(92\) |
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none
Time = 0.30 (sec) , antiderivative size = 341, normalized size of antiderivative = 2.13 \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2}}{x} \, dx=\left [\frac {b c \sqrt {d} \log \left (-2 \, d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) + 2 \, a \sqrt {c} d \log \left (-\frac {d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right ) + 2 \, {\left (b d x + 2 \, a d\right )} \sqrt {d x^{2} + c}}{4 \, d}, -\frac {b c \sqrt {-d} \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) - a \sqrt {c} d \log \left (-\frac {d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right ) - {\left (b d x + 2 \, a d\right )} \sqrt {d x^{2} + c}}{2 \, d}, \frac {4 \, a \sqrt {-c} d \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{2} + c}}\right ) + b c \sqrt {d} \log \left (-2 \, d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) + 2 \, {\left (b d x + 2 \, a d\right )} \sqrt {d x^{2} + c}}{4 \, d}, -\frac {b c \sqrt {-d} \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) - 2 \, a \sqrt {-c} d \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{2} + c}}\right ) - {\left (b d x + 2 \, a d\right )} \sqrt {d x^{2} + c}}{2 \, d}\right ] \]
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\[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2}}{x} \, dx=\int \frac {\sqrt {c + d x^{2}} \sqrt {\left (a + b x\right )^{2}}}{x}\, dx \]
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\[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2}}{x} \, dx=\int { \frac {\sqrt {d x^{2} + c} \sqrt {{\left (b x + a\right )}^{2}}}{x} \,d x } \]
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Exception generated. \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2}}{x} \, dx=\text {Exception raised: TypeError} \]
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Timed out. \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2}}{x} \, dx=\int \frac {\sqrt {{\left (a+b\,x\right )}^2}\,\sqrt {d\,x^2+c}}{x} \,d x \]
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