Integrand size = 24, antiderivative size = 80 \[ \int \frac {1}{x \left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{a \sqrt {c}}+\frac {\sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{a \sqrt {b c-a d}} \]
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Time = 0.05 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {457, 88, 65, 214} \[ \int \frac {1}{x \left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx=\frac {\sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{a \sqrt {b c-a d}}-\frac {\text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{a \sqrt {c}} \]
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Rule 65
Rule 88
Rule 214
Rule 457
Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {1}{x (a+b x) \sqrt {c+d x}} \, dx,x,x^2\right ) \\ & = \frac {\text {Subst}\left (\int \frac {1}{x \sqrt {c+d x}} \, dx,x,x^2\right )}{2 a}-\frac {b \text {Subst}\left (\int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx,x,x^2\right )}{2 a} \\ & = \frac {\text {Subst}\left (\int \frac {1}{-\frac {c}{d}+\frac {x^2}{d}} \, dx,x,\sqrt {c+d x^2}\right )}{a d}-\frac {b \text {Subst}\left (\int \frac {1}{a-\frac {b c}{d}+\frac {b x^2}{d}} \, dx,x,\sqrt {c+d x^2}\right )}{a d} \\ & = -\frac {\tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{a \sqrt {c}}+\frac {\sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{a \sqrt {b c-a d}} \\ \end{align*}
Time = 0.11 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.98 \[ \int \frac {1}{x \left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx=-\frac {\frac {\sqrt {b} \arctan \left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {-b c+a d}}\right )}{\sqrt {-b c+a d}}+\frac {\text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{\sqrt {c}}}{a} \]
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Time = 3.01 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.98
method | result | size |
pseudoelliptic | \(-\frac {b \arctan \left (\frac {b \sqrt {d \,x^{2}+c}}{\sqrt {\left (a d -b c \right ) b}}\right ) \sqrt {c}+\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{2}+c}}{\sqrt {c}}\right ) \sqrt {\left (a d -b c \right ) b}}{a \sqrt {\left (a d -b c \right ) b}\, \sqrt {c}}\) | \(78\) |
default | \(-\frac {\ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {d \,x^{2}+c}}{x}\right )}{a \sqrt {c}}+\frac {\ln \left (\frac {-\frac {2 \left (a d -b c \right )}{b}+\frac {2 d \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {d \left (x -\frac {\sqrt {-a b}}{b}\right )^{2}+\frac {2 d \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}{b}-\frac {a d -b c}{b}}}{x -\frac {\sqrt {-a b}}{b}}\right )}{2 a \sqrt {-\frac {a d -b c}{b}}}+\frac {\ln \left (\frac {-\frac {2 \left (a d -b c \right )}{b}-\frac {2 d \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {d \left (x +\frac {\sqrt {-a b}}{b}\right )^{2}-\frac {2 d \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}{b}-\frac {a d -b c}{b}}}{x +\frac {\sqrt {-a b}}{b}}\right )}{2 a \sqrt {-\frac {a d -b c}{b}}}\) | \(331\) |
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Time = 0.32 (sec) , antiderivative size = 603, normalized size of antiderivative = 7.54 \[ \int \frac {1}{x \left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx=\left [\frac {c \sqrt {\frac {b}{b c - a d}} \log \left (\frac {b^{2} d^{2} x^{4} + 8 \, b^{2} c^{2} - 8 \, a b c d + a^{2} d^{2} + 2 \, {\left (4 \, b^{2} c d - 3 \, a b d^{2}\right )} x^{2} + 4 \, {\left (2 \, b^{2} c^{2} - 3 \, a b c d + a^{2} d^{2} + {\left (b^{2} c d - a b d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c} \sqrt {\frac {b}{b c - a d}}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right ) + 2 \, \sqrt {c} \log \left (-\frac {d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right )}{4 \, a c}, \frac {c \sqrt {\frac {b}{b c - a d}} \log \left (\frac {b^{2} d^{2} x^{4} + 8 \, b^{2} c^{2} - 8 \, a b c d + a^{2} d^{2} + 2 \, {\left (4 \, b^{2} c d - 3 \, a b d^{2}\right )} x^{2} + 4 \, {\left (2 \, b^{2} c^{2} - 3 \, a b c d + a^{2} d^{2} + {\left (b^{2} c d - a b d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c} \sqrt {\frac {b}{b c - a d}}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right ) + 4 \, \sqrt {-c} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{2} + c}}\right )}{4 \, a c}, -\frac {c \sqrt {-\frac {b}{b c - a d}} \arctan \left (\frac {{\left (b d x^{2} + 2 \, b c - a d\right )} \sqrt {d x^{2} + c} \sqrt {-\frac {b}{b c - a d}}}{2 \, {\left (b d x^{2} + b c\right )}}\right ) - \sqrt {c} \log \left (-\frac {d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right )}{2 \, a c}, -\frac {c \sqrt {-\frac {b}{b c - a d}} \arctan \left (\frac {{\left (b d x^{2} + 2 \, b c - a d\right )} \sqrt {d x^{2} + c} \sqrt {-\frac {b}{b c - a d}}}{2 \, {\left (b d x^{2} + b c\right )}}\right ) - 2 \, \sqrt {-c} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{2} + c}}\right )}{2 \, a c}\right ] \]
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Time = 4.10 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.36 \[ \int \frac {1}{x \left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx=\begin {cases} \frac {2 \left (- \frac {d \operatorname {atan}{\left (\frac {\sqrt {c + d x^{2}}}{\sqrt {\frac {a d - b c}{b}}} \right )}}{2 a \sqrt {\frac {a d - b c}{b}}} + \frac {d \operatorname {atan}{\left (\frac {\sqrt {c + d x^{2}}}{\sqrt {- c}} \right )}}{2 a \sqrt {- c}}\right )}{d} & \text {for}\: d \neq 0 \\\frac {\operatorname {atan}{\left (\frac {2 \left (\frac {a}{2 b} + x^{2}\right )}{\sqrt {- \frac {a^{2}}{b^{2}}}} \right )}}{b \sqrt {c} \sqrt {- \frac {a^{2}}{b^{2}}}} & \text {otherwise} \end {cases} \]
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\[ \int \frac {1}{x \left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )} \sqrt {d x^{2} + c} x} \,d x } \]
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Time = 0.28 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.88 \[ \int \frac {1}{x \left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx=-\frac {b \arctan \left (\frac {\sqrt {d x^{2} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{\sqrt {-b^{2} c + a b d} a} + \frac {\arctan \left (\frac {\sqrt {d x^{2} + c}}{\sqrt {-c}}\right )}{a \sqrt {-c}} \]
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Time = 5.52 (sec) , antiderivative size = 651, normalized size of antiderivative = 8.14 \[ \int \frac {1}{x \left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx=-\frac {\mathrm {atanh}\left (\frac {\sqrt {d\,x^2+c}}{\sqrt {c}}\right )}{a\,\sqrt {c}}-\frac {\mathrm {atan}\left (\frac {\frac {\sqrt {b^2\,c-a\,b\,d}\,\left (2\,b^3\,d^2\,\sqrt {d\,x^2+c}-\frac {\sqrt {b^2\,c-a\,b\,d}\,\left (2\,a^2\,b^2\,d^3-\frac {\left (8\,a^3\,b^2\,d^3-16\,a^2\,b^3\,c\,d^2\right )\,\sqrt {d\,x^2+c}\,\sqrt {b^2\,c-a\,b\,d}}{4\,\left (a^2\,d-a\,b\,c\right )}\right )}{2\,\left (a^2\,d-a\,b\,c\right )}\right )\,1{}\mathrm {i}}{a^2\,d-a\,b\,c}+\frac {\sqrt {b^2\,c-a\,b\,d}\,\left (2\,b^3\,d^2\,\sqrt {d\,x^2+c}+\frac {\sqrt {b^2\,c-a\,b\,d}\,\left (2\,a^2\,b^2\,d^3+\frac {\left (8\,a^3\,b^2\,d^3-16\,a^2\,b^3\,c\,d^2\right )\,\sqrt {d\,x^2+c}\,\sqrt {b^2\,c-a\,b\,d}}{4\,\left (a^2\,d-a\,b\,c\right )}\right )}{2\,\left (a^2\,d-a\,b\,c\right )}\right )\,1{}\mathrm {i}}{a^2\,d-a\,b\,c}}{\frac {\sqrt {b^2\,c-a\,b\,d}\,\left (2\,b^3\,d^2\,\sqrt {d\,x^2+c}-\frac {\sqrt {b^2\,c-a\,b\,d}\,\left (2\,a^2\,b^2\,d^3-\frac {\left (8\,a^3\,b^2\,d^3-16\,a^2\,b^3\,c\,d^2\right )\,\sqrt {d\,x^2+c}\,\sqrt {b^2\,c-a\,b\,d}}{4\,\left (a^2\,d-a\,b\,c\right )}\right )}{2\,\left (a^2\,d-a\,b\,c\right )}\right )}{a^2\,d-a\,b\,c}-\frac {\sqrt {b^2\,c-a\,b\,d}\,\left (2\,b^3\,d^2\,\sqrt {d\,x^2+c}+\frac {\sqrt {b^2\,c-a\,b\,d}\,\left (2\,a^2\,b^2\,d^3+\frac {\left (8\,a^3\,b^2\,d^3-16\,a^2\,b^3\,c\,d^2\right )\,\sqrt {d\,x^2+c}\,\sqrt {b^2\,c-a\,b\,d}}{4\,\left (a^2\,d-a\,b\,c\right )}\right )}{2\,\left (a^2\,d-a\,b\,c\right )}\right )}{a^2\,d-a\,b\,c}}\right )\,\sqrt {b^2\,c-a\,b\,d}\,1{}\mathrm {i}}{a^2\,d-a\,b\,c} \]
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