Integrand size = 32, antiderivative size = 37 \[ \int \frac {1+k x}{(-1+k x) \sqrt {(1-x) x \left (1-k^2 x\right )}} \, dx=-\frac {2 \arctan \left (\frac {(-1+k) x}{\sqrt {x+\left (-1-k^2\right ) x^2+k^2 x^3}}\right )}{-1+k} \]
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Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
Time = 0.33 (sec) , antiderivative size = 159, normalized size of antiderivative = 4.30, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {6850, 1621, 174, 552, 551, 116} \[ \int \frac {1+k x}{(-1+k x) \sqrt {(1-x) x \left (1-k^2 x\right )}} \, dx=\frac {2 \sqrt {1-x} \sqrt {x} \sqrt {1-k^2 x} \operatorname {EllipticF}\left (\arcsin \left (\sqrt {x}\right ),k^2\right )}{\sqrt {(1-x) x \left (1-k^2 x\right )}}+\frac {4 \sqrt {1-x} \sqrt {x} \sqrt {\frac {k^2 (1-x)}{1-k^2}+1} \operatorname {EllipticPi}\left (-\frac {k}{1-k},\arcsin \left (\sqrt {1-x}\right ),-\frac {k^2}{1-k^2}\right )}{(1-k) \sqrt {(1-x) x \left (1-k^2 x\right )}} \]
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Rule 116
Rule 174
Rule 551
Rule 552
Rule 1621
Rule 6850
Rubi steps \begin{align*} \text {integral}& = \frac {\left (\sqrt {1-x} \sqrt {x} \sqrt {1-k^2 x}\right ) \int \frac {1+k x}{\sqrt {1-x} \sqrt {x} (-1+k x) \sqrt {1-k^2 x}} \, dx}{\sqrt {(1-x) x \left (1-k^2 x\right )}} \\ & = \frac {\left (\sqrt {1-x} \sqrt {x} \sqrt {1-k^2 x}\right ) \int \frac {1}{\sqrt {1-x} \sqrt {x} \sqrt {1-k^2 x}} \, dx}{\sqrt {(1-x) x \left (1-k^2 x\right )}}+\frac {\left (2 \sqrt {1-x} \sqrt {x} \sqrt {1-k^2 x}\right ) \int \frac {1}{\sqrt {1-x} \sqrt {x} (-1+k x) \sqrt {1-k^2 x}} \, dx}{\sqrt {(1-x) x \left (1-k^2 x\right )}} \\ & = \frac {2 \sqrt {1-x} \sqrt {x} \sqrt {1-k^2 x} \operatorname {EllipticF}\left (\arcsin \left (\sqrt {x}\right ),k^2\right )}{\sqrt {(1-x) x \left (1-k^2 x\right )}}-\frac {\left (4 \sqrt {1-x} \sqrt {x} \sqrt {1-k^2 x}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-x^2} \left (-1+k-k x^2\right ) \sqrt {1-k^2+k^2 x^2}} \, dx,x,\sqrt {1-x}\right )}{\sqrt {(1-x) x \left (1-k^2 x\right )}} \\ & = \frac {2 \sqrt {1-x} \sqrt {x} \sqrt {1-k^2 x} \operatorname {EllipticF}\left (\arcsin \left (\sqrt {x}\right ),k^2\right )}{\sqrt {(1-x) x \left (1-k^2 x\right )}}-\frac {\left (4 \sqrt {1+\frac {k^2 (-1+x)}{-1+k^2}} \sqrt {1-x} \sqrt {x}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-x^2} \left (-1+k-k x^2\right ) \sqrt {1+\frac {k^2 x^2}{1-k^2}}} \, dx,x,\sqrt {1-x}\right )}{\sqrt {(1-x) x \left (1-k^2 x\right )}} \\ & = \frac {2 \sqrt {1-x} \sqrt {x} \sqrt {1-k^2 x} \operatorname {EllipticF}\left (\arcsin \left (\sqrt {x}\right ),k^2\right )}{\sqrt {(1-x) x \left (1-k^2 x\right )}}+\frac {4 \sqrt {1+\frac {k^2 (1-x)}{1-k^2}} \sqrt {1-x} \sqrt {x} \operatorname {EllipticPi}\left (-\frac {k}{1-k},\arcsin \left (\sqrt {1-x}\right ),-\frac {k^2}{1-k^2}\right )}{(1-k) \sqrt {(1-x) x \left (1-k^2 x\right )}} \\ \end{align*}
Time = 10.32 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.78 \[ \int \frac {1+k x}{(-1+k x) \sqrt {(1-x) x \left (1-k^2 x\right )}} \, dx=-\frac {2 \arctan \left (\frac {(-1+k) x}{\sqrt {(-1+x) x \left (-1+k^2 x\right )}}\right )}{-1+k} \]
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Time = 2.00 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.86
method | result | size |
default | \(\frac {2 \arctan \left (\frac {\sqrt {\left (x -1\right ) x \left (k^{2} x -1\right )}}{\left (-1+k \right ) x}\right )}{-1+k}\) | \(32\) |
pseudoelliptic | \(\frac {2 \arctan \left (\frac {\sqrt {\left (x -1\right ) x \left (k^{2} x -1\right )}}{\left (-1+k \right ) x}\right )}{-1+k}\) | \(32\) |
elliptic | \(-\frac {2 \sqrt {-\left (x -\frac {1}{k^{2}}\right ) k^{2}}\, \sqrt {\frac {x -1}{\frac {1}{k^{2}}-1}}\, \sqrt {k^{2} x}\, \operatorname {EllipticF}\left (\sqrt {-\left (x -\frac {1}{k^{2}}\right ) k^{2}}, \sqrt {\frac {1}{k^{2} \left (\frac {1}{k^{2}}-1\right )}}\right )}{k^{2} \sqrt {k^{2} x^{3}-k^{2} x^{2}-x^{2}+x}}-\frac {4 \sqrt {-\left (x -\frac {1}{k^{2}}\right ) k^{2}}\, \sqrt {\frac {x -1}{\frac {1}{k^{2}}-1}}\, \sqrt {k^{2} x}\, \operatorname {EllipticPi}\left (\sqrt {-\left (x -\frac {1}{k^{2}}\right ) k^{2}}, \frac {1}{k^{2} \left (\frac {1}{k^{2}}-\frac {1}{k}\right )}, \sqrt {\frac {1}{k^{2} \left (\frac {1}{k^{2}}-1\right )}}\right )}{k^{3} \sqrt {k^{2} x^{3}-k^{2} x^{2}-x^{2}+x}\, \left (\frac {1}{k^{2}}-\frac {1}{k}\right )}\) | \(210\) |
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Leaf count of result is larger than twice the leaf count of optimal. 87 vs. \(2 (34) = 68\).
Time = 0.28 (sec) , antiderivative size = 87, normalized size of antiderivative = 2.35 \[ \int \frac {1+k x}{(-1+k x) \sqrt {(1-x) x \left (1-k^2 x\right )}} \, dx=\frac {\arctan \left (\frac {\sqrt {k^{2} x^{3} - {\left (k^{2} + 1\right )} x^{2} + x} {\left (k^{2} x^{2} - 2 \, {\left (k^{2} - k + 1\right )} x + 1\right )}}{2 \, {\left ({\left (k^{3} - k^{2}\right )} x^{3} - {\left (k^{3} - k^{2} + k - 1\right )} x^{2} + {\left (k - 1\right )} x\right )}}\right )}{k - 1} \]
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\[ \int \frac {1+k x}{(-1+k x) \sqrt {(1-x) x \left (1-k^2 x\right )}} \, dx=\int \frac {k x + 1}{\sqrt {x \left (x - 1\right ) \left (k^{2} x - 1\right )} \left (k x - 1\right )}\, dx \]
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\[ \int \frac {1+k x}{(-1+k x) \sqrt {(1-x) x \left (1-k^2 x\right )}} \, dx=\int { \frac {k x + 1}{\sqrt {{\left (k^{2} x - 1\right )} {\left (x - 1\right )} x} {\left (k x - 1\right )}} \,d x } \]
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\[ \int \frac {1+k x}{(-1+k x) \sqrt {(1-x) x \left (1-k^2 x\right )}} \, dx=\int { \frac {k x + 1}{\sqrt {{\left (k^{2} x - 1\right )} {\left (x - 1\right )} x} {\left (k x - 1\right )}} \,d x } \]
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Timed out. \[ \int \frac {1+k x}{(-1+k x) \sqrt {(1-x) x \left (1-k^2 x\right )}} \, dx=\text {Hanged} \]
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