Integrand size = 25, antiderivative size = 204 \[ \int \frac {\sqrt {e \tan (c+d x)}}{a+b \cos (c+d x)} \, dx=-\frac {2 \sqrt {2} \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (-\frac {\sqrt {-a+b}}{\sqrt {a+b}},\arcsin \left (\frac {\sqrt {\sin (c+d x)}}{\sqrt {1+\cos (c+d x)}}\right ),-1\right ) \sqrt {e \tan (c+d x)}}{\sqrt {-a+b} \sqrt {a+b} d \sqrt {\sin (c+d x)}}+\frac {2 \sqrt {2} \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {\sqrt {-a+b}}{\sqrt {a+b}},\arcsin \left (\frac {\sqrt {\sin (c+d x)}}{\sqrt {1+\cos (c+d x)}}\right ),-1\right ) \sqrt {e \tan (c+d x)}}{\sqrt {-a+b} \sqrt {a+b} d \sqrt {\sin (c+d x)}} \]
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Time = 0.63 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {2812, 2809, 2985, 2984, 504, 1227, 551} \[ \int \frac {\sqrt {e \tan (c+d x)}}{a+b \cos (c+d x)} \, dx=\frac {2 \sqrt {2} \sqrt {\cos (c+d x)} \sqrt {e \tan (c+d x)} \operatorname {EllipticPi}\left (\frac {\sqrt {b-a}}{\sqrt {a+b}},\arcsin \left (\frac {\sqrt {\sin (c+d x)}}{\sqrt {\cos (c+d x)+1}}\right ),-1\right )}{d \sqrt {b-a} \sqrt {a+b} \sqrt {\sin (c+d x)}}-\frac {2 \sqrt {2} \sqrt {\cos (c+d x)} \sqrt {e \tan (c+d x)} \operatorname {EllipticPi}\left (-\frac {\sqrt {b-a}}{\sqrt {a+b}},\arcsin \left (\frac {\sqrt {\sin (c+d x)}}{\sqrt {\cos (c+d x)+1}}\right ),-1\right )}{d \sqrt {b-a} \sqrt {a+b} \sqrt {\sin (c+d x)}} \]
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Rule 504
Rule 551
Rule 1227
Rule 2809
Rule 2812
Rule 2984
Rule 2985
Rubi steps \begin{align*} \text {integral}& = \left (\sqrt {e \cot (c+d x)} \sqrt {e \tan (c+d x)}\right ) \int \frac {1}{(a+b \cos (c+d x)) \sqrt {e \cot (c+d x)}} \, dx \\ & = \frac {\left (\sqrt {-\cos (c+d x)} \sqrt {e \tan (c+d x)}\right ) \int \frac {\sqrt {\sin (c+d x)}}{\sqrt {-\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{\sqrt {\sin (c+d x)}} \\ & = \frac {\left (\sqrt {\cos (c+d x)} \sqrt {e \tan (c+d x)}\right ) \int \frac {\sqrt {\sin (c+d x)}}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{\sqrt {\sin (c+d x)}} \\ & = \frac {\left (4 \sqrt {2} \sqrt {\cos (c+d x)} \sqrt {e \tan (c+d x)}\right ) \text {Subst}\left (\int \frac {x^2}{\sqrt {1-x^4} \left (a+b+(a-b) x^4\right )} \, dx,x,\frac {\sqrt {\sin (c+d x)}}{\sqrt {1+\cos (c+d x)}}\right )}{d \sqrt {\sin (c+d x)}} \\ & = \frac {\left (2 \sqrt {2} \sqrt {\cos (c+d x)} \sqrt {e \tan (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{\left (\sqrt {a+b}-\sqrt {-a+b} x^2\right ) \sqrt {1-x^4}} \, dx,x,\frac {\sqrt {\sin (c+d x)}}{\sqrt {1+\cos (c+d x)}}\right )}{\sqrt {-a+b} d \sqrt {\sin (c+d x)}}-\frac {\left (2 \sqrt {2} \sqrt {\cos (c+d x)} \sqrt {e \tan (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{\left (\sqrt {a+b}+\sqrt {-a+b} x^2\right ) \sqrt {1-x^4}} \, dx,x,\frac {\sqrt {\sin (c+d x)}}{\sqrt {1+\cos (c+d x)}}\right )}{\sqrt {-a+b} d \sqrt {\sin (c+d x)}} \\ & = \frac {\left (2 \sqrt {2} \sqrt {\cos (c+d x)} \sqrt {e \tan (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-x^2} \sqrt {1+x^2} \left (\sqrt {a+b}-\sqrt {-a+b} x^2\right )} \, dx,x,\frac {\sqrt {\sin (c+d x)}}{\sqrt {1+\cos (c+d x)}}\right )}{\sqrt {-a+b} d \sqrt {\sin (c+d x)}}-\frac {\left (2 \sqrt {2} \sqrt {\cos (c+d x)} \sqrt {e \tan (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-x^2} \sqrt {1+x^2} \left (\sqrt {a+b}+\sqrt {-a+b} x^2\right )} \, dx,x,\frac {\sqrt {\sin (c+d x)}}{\sqrt {1+\cos (c+d x)}}\right )}{\sqrt {-a+b} d \sqrt {\sin (c+d x)}} \\ & = -\frac {2 \sqrt {2} \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (-\frac {\sqrt {-a+b}}{\sqrt {a+b}},\arcsin \left (\frac {\sqrt {\sin (c+d x)}}{\sqrt {1+\cos (c+d x)}}\right ),-1\right ) \sqrt {e \tan (c+d x)}}{\sqrt {-a+b} \sqrt {a+b} d \sqrt {\sin (c+d x)}}+\frac {2 \sqrt {2} \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {\sqrt {-a+b}}{\sqrt {a+b}},\arcsin \left (\frac {\sqrt {\sin (c+d x)}}{\sqrt {1+\cos (c+d x)}}\right ),-1\right ) \sqrt {e \tan (c+d x)}}{\sqrt {-a+b} \sqrt {a+b} d \sqrt {\sin (c+d x)}} \\ \end{align*}
Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
Time = 14.92 (sec) , antiderivative size = 363, normalized size of antiderivative = 1.78 \[ \int \frac {\sqrt {e \tan (c+d x)}}{a+b \cos (c+d x)} \, dx=\frac {2 \left (b+a \sqrt {\sec ^2(c+d x)}\right ) \sqrt {e \tan (c+d x)} \left (\frac {-2 \arctan \left (1-\frac {\sqrt {2} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt [4]{a^2-b^2}}\right )+2 \arctan \left (1+\frac {\sqrt {2} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt [4]{a^2-b^2}}\right )+\log \left (\sqrt {a^2-b^2}-\sqrt {2} \sqrt {a} \sqrt [4]{a^2-b^2} \sqrt {\tan (c+d x)}+a \tan (c+d x)\right )-\log \left (\sqrt {a^2-b^2}+\sqrt {2} \sqrt {a} \sqrt [4]{a^2-b^2} \sqrt {\tan (c+d x)}+a \tan (c+d x)\right )}{4 \sqrt {2} \sqrt {a} \sqrt [4]{a^2-b^2}}+\frac {b \operatorname {AppellF1}\left (\frac {3}{4},\frac {1}{2},1,\frac {7}{4},-\tan ^2(c+d x),-\frac {a^2 \tan ^2(c+d x)}{a^2-b^2}\right ) \tan ^{\frac {3}{2}}(c+d x)}{3 \left (-a^2+b^2\right )}\right )}{d (a+b \cos (c+d x)) \sqrt {\sec ^2(c+d x)} \sqrt {\tan (c+d x)}} \]
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Leaf count of result is larger than twice the leaf count of optimal. \(453\) vs. \(2(164)=328\).
Time = 3.76 (sec) , antiderivative size = 454, normalized size of antiderivative = 2.23
method | result | size |
default | \(-\frac {\left (\sqrt {-a^{2}+b^{2}}\, \Pi \left (\sqrt {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}, \frac {a -b}{a -b +\sqrt {-\left (a -b \right ) \left (a +b \right )}}, \frac {\sqrt {2}}{2}\right )-\sqrt {-a^{2}+b^{2}}\, \Pi \left (\sqrt {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}, -\frac {a -b}{-a +b +\sqrt {-\left (a -b \right ) \left (a +b \right )}}, \frac {\sqrt {2}}{2}\right )-a \Pi \left (\sqrt {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}, \frac {a -b}{a -b +\sqrt {-\left (a -b \right ) \left (a +b \right )}}, \frac {\sqrt {2}}{2}\right )+\Pi \left (\sqrt {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}, \frac {a -b}{a -b +\sqrt {-\left (a -b \right ) \left (a +b \right )}}, \frac {\sqrt {2}}{2}\right ) b -a \Pi \left (\sqrt {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}, -\frac {a -b}{-a +b +\sqrt {-\left (a -b \right ) \left (a +b \right )}}, \frac {\sqrt {2}}{2}\right )+\Pi \left (\sqrt {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}, -\frac {a -b}{-a +b +\sqrt {-\left (a -b \right ) \left (a +b \right )}}, \frac {\sqrt {2}}{2}\right ) b \right ) \sqrt {\cot \left (d x +c \right )-\csc \left (d x +c \right )}\, \sqrt {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}\, \sqrt {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}\, \sqrt {e \tan \left (d x +c \right )}\, \left (\cot \left (d x +c \right )+\csc \left (d x +c \right )\right ) \sqrt {2}}{d \left (b +\sqrt {-a^{2}+b^{2}}-a \right ) \left (-b +\sqrt {-a^{2}+b^{2}}+a \right )}\) | \(454\) |
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Timed out. \[ \int \frac {\sqrt {e \tan (c+d x)}}{a+b \cos (c+d x)} \, dx=\text {Timed out} \]
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\[ \int \frac {\sqrt {e \tan (c+d x)}}{a+b \cos (c+d x)} \, dx=\int \frac {\sqrt {e \tan {\left (c + d x \right )}}}{a + b \cos {\left (c + d x \right )}}\, dx \]
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\[ \int \frac {\sqrt {e \tan (c+d x)}}{a+b \cos (c+d x)} \, dx=\int { \frac {\sqrt {e \tan \left (d x + c\right )}}{b \cos \left (d x + c\right ) + a} \,d x } \]
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\[ \int \frac {\sqrt {e \tan (c+d x)}}{a+b \cos (c+d x)} \, dx=\int { \frac {\sqrt {e \tan \left (d x + c\right )}}{b \cos \left (d x + c\right ) + a} \,d x } \]
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Timed out. \[ \int \frac {\sqrt {e \tan (c+d x)}}{a+b \cos (c+d x)} \, dx=\int \frac {\sqrt {e\,\mathrm {tan}\left (c+d\,x\right )}}{a+b\,\cos \left (c+d\,x\right )} \,d x \]
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