Integrand size = 38, antiderivative size = 202 \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+e x+d x^2}}{x^2} \, dx=-\frac {(a-b x) \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+e x+d x^2}}{x (a+b x)}+\frac {(2 a d+b e) \sqrt {a^2+2 a b x+b^2 x^2} \text {arctanh}\left (\frac {e+2 d x}{2 \sqrt {d} \sqrt {c+e x+d x^2}}\right )}{2 \sqrt {d} (a+b x)}-\frac {(2 b c+a e) \sqrt {a^2+2 a b x+b^2 x^2} \text {arctanh}\left (\frac {2 c+e x}{2 \sqrt {c} \sqrt {c+e x+d x^2}}\right )}{2 \sqrt {c} (a+b x)} \]
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Time = 0.10 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {1014, 826, 857, 635, 212, 738} \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+e x+d x^2}}{x^2} \, dx=\frac {\sqrt {a^2+2 a b x+b^2 x^2} (2 a d+b e) \text {arctanh}\left (\frac {2 d x+e}{2 \sqrt {d} \sqrt {c+d x^2+e x}}\right )}{2 \sqrt {d} (a+b x)}-\frac {\sqrt {a^2+2 a b x+b^2 x^2} (a e+2 b c) \text {arctanh}\left (\frac {2 c+e x}{2 \sqrt {c} \sqrt {c+d x^2+e x}}\right )}{2 \sqrt {c} (a+b x)}-\frac {\sqrt {a^2+2 a b x+b^2 x^2} (a-b x) \sqrt {c+d x^2+e x}}{x (a+b x)} \]
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Rule 212
Rule 635
Rule 738
Rule 826
Rule 857
Rule 1014
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+e x+d x^2}}{x^2} \, dx}{2 a b+2 b^2 x} \\ & = -\frac {(a-b x) \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+e x+d x^2}}{x (a+b x)}-\frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {-2 b (2 b c+a e)-2 b (2 a d+b e) x}{x \sqrt {c+e x+d x^2}} \, dx}{2 \left (2 a b+2 b^2 x\right )} \\ & = -\frac {(a-b x) \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+e x+d x^2}}{x (a+b x)}+\frac {\left (b (2 b c+a e) \sqrt {a^2+2 a b x+b^2 x^2}\right ) \int \frac {1}{x \sqrt {c+e x+d x^2}} \, dx}{2 a b+2 b^2 x}+\frac {\left (b (2 a d+b e) \sqrt {a^2+2 a b x+b^2 x^2}\right ) \int \frac {1}{\sqrt {c+e x+d x^2}} \, dx}{2 a b+2 b^2 x} \\ & = -\frac {(a-b x) \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+e x+d x^2}}{x (a+b x)}-\frac {\left (2 b (2 b c+a e) \sqrt {a^2+2 a b x+b^2 x^2}\right ) \text {Subst}\left (\int \frac {1}{4 c-x^2} \, dx,x,\frac {2 c+e x}{\sqrt {c+e x+d x^2}}\right )}{2 a b+2 b^2 x}+\frac {\left (2 b (2 a d+b e) \sqrt {a^2+2 a b x+b^2 x^2}\right ) \text {Subst}\left (\int \frac {1}{4 d-x^2} \, dx,x,\frac {e+2 d x}{\sqrt {c+e x+d x^2}}\right )}{2 a b+2 b^2 x} \\ & = -\frac {(a-b x) \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+e x+d x^2}}{x (a+b x)}+\frac {(2 a d+b e) \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 )}{2 \sqrt {d} (a+b x)}-\frac {(2 b c+a e) \sqrt {a^2+2 a b x+b^2 x^2} \tanh ^{-1}\left (\frac {2 c+e x}{2 \sqrt {c} \sqrt {c+e x+d x^2}}\right )}{2 \sqrt {c} (a+b x)} \\ \end{align*}
Time = 0.45 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.75 \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+e x+d x^2}}{x^2} \, dx=-\frac {\sqrt {(a+b x)^2} \left (2 \sqrt {d} (2 b c+a e) x \text {arctanh}\left (\frac {-\sqrt {d} x+\sqrt {c+x (e+d x)}}{\sqrt {c}}\right )+\sqrt {c} \left (2 \sqrt {d} (a-b x) \sqrt {c+x (e+d x)}+(2 a d+b e) x \log \left (e+2 d x-2 \sqrt {d} \sqrt {c+x (e+d x)}\right )\right )\right )}{2 \sqrt {c} \sqrt {d} x (a+b x)} \]
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Time = 0.59 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.00
method | result | size |
risch | \(-\frac {a \sqrt {d \,x^{2}+e x +c}\, \sqrt {\left (b x +a \right )^{2}}}{x \left (b x +a \right )}+\frac {\left (a \sqrt {d}\, \ln \left (\frac {\frac {e}{2}+d x}{\sqrt {d}}+\sqrt {d \,x^{2}+e x +c}\right )+\frac {b e \ln \left (\frac {\frac {e}{2}+d x}{\sqrt {d}}+\sqrt {d \,x^{2}+e x +c}\right )}{2 \sqrt {d}}+b \sqrt {d \,x^{2}+e x +c}-\frac {\ln \left (\frac {2 c +e x +2 \sqrt {c}\, \sqrt {d \,x^{2}+e x +c}}{x}\right ) a e}{2 \sqrt {c}}-\sqrt {c}\, \ln \left (\frac {2 c +e x +2 \sqrt {c}\, \sqrt {d \,x^{2}+e x +c}}{x}\right ) b \right ) \sqrt {\left (b x +a \right )^{2}}}{b x +a}\) | \(201\) |
default | \(-\frac {\operatorname {csgn}\left (b x +a \right ) \left (-2 \sqrt {d \,x^{2}+e x +c}\, d^{\frac {5}{2}} a \,x^{2}+2 d^{\frac {3}{2}} c^{\frac {3}{2}} \ln \left (\frac {2 c +e x +2 \sqrt {c}\, \sqrt {d \,x^{2}+e x +c}}{x}\right ) b x +d^{\frac {3}{2}} \sqrt {c}\, \ln \left (\frac {2 c +e x +2 \sqrt {c}\, \sqrt {d \,x^{2}+e x +c}}{x}\right ) a e x +2 \left (d \,x^{2}+e x +c \right )^{\frac {3}{2}} d^{\frac {3}{2}} a -2 \sqrt {d \,x^{2}+e x +c}\, d^{\frac {3}{2}} a e x -2 \sqrt {d \,x^{2}+e x +c}\, d^{\frac {3}{2}} b c x -2 \ln \left (\frac {2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {d}+2 d x +e}{2 \sqrt {d}}\right ) a c \,d^{2} x -\ln \left (\frac {2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {d}+2 d x +e}{2 \sqrt {d}}\right ) d b c e x \right )}{2 c x \,d^{\frac {3}{2}}}\) | \(249\) |
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Time = 0.45 (sec) , antiderivative size = 647, normalized size of antiderivative = 3.20 \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+e x+d x^2}}{x^2} \, dx=\left [\frac {{\left (2 \, a c d + b c e\right )} \sqrt {d} x \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 ) + {\left (2 \, b c d + a d e\right )} \sqrt {c} x \log \left (\frac {8 \, c e x + {\left (4 \, c d + e^{2}\right )} x^{2} - 4 \, \sqrt {d x^{2} + e x + c} {\left (e x + 2 \, c\right )} \sqrt {c} + 8 \, c^{2}}{x^{2}}\right ) + 4 \, {\left (b c d x - a c d\right )} \sqrt {d x^{2} + e x + c}}{4 \, c d x}, -\frac {2 \, {\left (2 \, a c d + b c e\right )} \sqrt {-d} x \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 ) - {\left (2 \, b c d + a d e\right )} \sqrt {c} x \log \left (\frac {8 \, c e x + {\left (4 \, c d + e^{2}\right )} x^{2} - 4 \, \sqrt {d x^{2} + e x + c} {\left (e x + 2 \, c\right )} \sqrt {c} + 8 \, c^{2}}{x^{2}}\right ) - 4 \, {\left (b c d x - a c d\right )} \sqrt {d x^{2} + e x + c}}{4 \, c d x}, \frac {2 \, {\left (2 \, b c d + a d e\right )} \sqrt {-c} x \arctan \left (\frac {\sqrt {d x^{2} + e x + c} {\left (e x + 2 \, c\right )} \sqrt {-c}}{2 \, {\left (c d x^{2} + c e x + c^{2}\right )}}\right ) + {\left (2 \, a c d + b c e\right )} \sqrt {d} x \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 (b c d x - a c d\right )} \sqrt {d x^{2} + e x + c}}{4 \, c d x}, \frac {{\left (2 \, b c d + a d e\right )} \sqrt {-c} x \arctan \left (\frac {\sqrt {d x^{2} + e x + c} {\left (e x + 2 \, c\right )} \sqrt {-c}}{2 \, {\left (c d x^{2} + c e x + c^{2}\right )}}\right ) - {\left (2 \, a c d + b c e\right )} \sqrt {-d} x \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 (b c d x - a c d\right )} \sqrt {d x^{2} + e x + c}}{2 \, c d x}\right ] \]
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\[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+e x+d x^2}}{x^2} \, dx=\int \frac {\sqrt {c + d x^{2} + e x} \sqrt {\left (a + b x\right )^{2}}}{x^{2}}\, dx \]
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\[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+e x+d x^2}}{x^2} \, dx=\int { \frac {\sqrt {d x^{2} + e x + c} \sqrt {{\left (b x + a\right )}^{2}}}{x^{2}} \,d x } \]
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Time = 0.66 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.00 \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+e x+d x^2}}{x^2} \, dx=\sqrt {d x^{2} + e x + c} b \mathrm {sgn}\left (b x + a\right ) + \frac {{\left (2 \, b c \mathrm {sgn}\left (b x + a\right ) + a e \mathrm {sgn}\left (b x + a\right )\right )} \arctan \left (-\frac {\sqrt {d} x - \sqrt {d x^{2} + e x + c}}{\sqrt {-c}}\right )}{\sqrt {-c}} - \frac {{\left (2 \, a d \mathrm {sgn}\left (b x + a\right ) + b e \mathrm {sgn}\left (b x + a\right )\right )} \log \left ({\left | -2 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + e x + c}\right )} \sqrt {d} - e \right |}\right )}{2 \, \sqrt {d}} + \frac {{\left (\sqrt {d} x - \sqrt {d x^{2} + e x + c}\right )} a e \mathrm {sgn}\left (b x + a\right ) + 2 \, a c \sqrt {d} \mathrm {sgn}\left (b x + a\right )}{{\left (\sqrt {d} x - \sqrt {d x^{2} + e x + c}\right )}^{2} - c} \]
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Timed out. \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+e x+d x^2}}{x^2} \, dx=\int \frac {\sqrt {{\left (a+b\,x\right )}^2}\,\sqrt {d\,x^2+e\,x+c}}{x^2} \,d x \]
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