Integrand size = 25, antiderivative size = 292 \[ \int \frac {\sqrt {e \cos (c+d x)}}{a+b \sin (c+d x)} \, dx=\frac {\sqrt {e} \arctan \left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{\sqrt {b} \sqrt [4]{-a^2+b^2} d}-\frac {\sqrt {e} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{\sqrt {b} \sqrt [4]{-a^2+b^2} d}+\frac {a e \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} (c+d x),2\right )}{b \left (b-\sqrt {-a^2+b^2}\right ) d \sqrt {e \cos (c+d x)}}+\frac {a e \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} (c+d x),2\right )}{b \left (b+\sqrt {-a^2+b^2}\right ) d \sqrt {e \cos (c+d x)}} \]
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Time = 0.39 (sec) , antiderivative size = 292, 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 = {2780, 2886, 2884, 335, 304, 211, 214} \[ \int \frac {\sqrt {e \cos (c+d x)}}{a+b \sin (c+d x)} \, dx=\frac {\sqrt {e} \arctan \left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{\sqrt {b} d \sqrt [4]{b^2-a^2}}-\frac {\sqrt {e} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{\sqrt {b} d \sqrt [4]{b^2-a^2}}+\frac {a e \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {b^2-a^2}},\frac {1}{2} (c+d x),2\right )}{b d \left (b-\sqrt {b^2-a^2}\right ) \sqrt {e \cos (c+d x)}}+\frac {a e \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {b^2-a^2}},\frac {1}{2} (c+d x),2\right )}{b d \left (\sqrt {b^2-a^2}+b\right ) \sqrt {e \cos (c+d x)}} \]
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Rule 211
Rule 214
Rule 304
Rule 335
Rule 2780
Rule 2884
Rule 2886
Rubi steps \begin{align*} \text {integral}& = -\frac {(a e) \int \frac {1}{\sqrt {e \cos (c+d x)} \left (\sqrt {-a^2+b^2}-b \cos (c+d x)\right )} \, dx}{2 b}+\frac {(a e) \int \frac {1}{\sqrt {e \cos (c+d x)} \left (\sqrt {-a^2+b^2}+b \cos (c+d x)\right )} \, dx}{2 b}+\frac {(b e) \text {Subst}\left (\int \frac {\sqrt {x}}{\left (a^2-b^2\right ) e^2+b^2 x^2} \, dx,x,e \cos (c+d x)\right )}{d} \\ & = \frac {(2 b e) \text {Subst}\left (\int \frac {x^2}{\left (a^2-b^2\right ) e^2+b^2 x^4} \, dx,x,\sqrt {e \cos (c+d x)}\right )}{d}-\frac {\left (a e \sqrt {\cos (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)} \left (\sqrt {-a^2+b^2}-b \cos (c+d x)\right )} \, dx}{2 b \sqrt {e \cos (c+d x)}}+\frac {\left (a e \sqrt {\cos (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)} \left (\sqrt {-a^2+b^2}+b \cos (c+d x)\right )} \, dx}{2 b \sqrt {e \cos (c+d x)}} \\ & = \frac {a e \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} (c+d x),2\right )}{b \left (b-\sqrt {-a^2+b^2}\right ) d \sqrt {e \cos (c+d x)}}+\frac {a e \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} (c+d x),2\right )}{b \left (b+\sqrt {-a^2+b^2}\right ) d \sqrt {e \cos (c+d x)}}-\frac {e \text {Subst}\left (\int \frac {1}{\sqrt {-a^2+b^2} e-b x^2} \, dx,x,\sqrt {e \cos (c+d x)}\right )}{d}+\frac {e \text {Subst}\left (\int \frac {1}{\sqrt {-a^2+b^2} e+b x^2} \, dx,x,\sqrt {e \cos (c+d x)}\right )}{d} \\ & = \frac {\sqrt {e} \arctan \left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{\sqrt {b} \sqrt [4]{-a^2+b^2} d}-\frac {\sqrt {e} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{\sqrt {b} \sqrt [4]{-a^2+b^2} d}+\frac {a e \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} (c+d x),2\right )}{b \left (b-\sqrt {-a^2+b^2}\right ) d \sqrt {e \cos (c+d x)}}+\frac {a e \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} (c+d x),2\right )}{b \left (b+\sqrt {-a^2+b^2}\right ) d \sqrt {e \cos (c+d x)}} \\ \end{align*}
Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
Time = 13.01 (sec) , antiderivative size = 361, normalized size of antiderivative = 1.24 \[ \int \frac {\sqrt {e \cos (c+d x)}}{a+b \sin (c+d x)} \, dx=-\frac {2 \sqrt {e \cos (c+d x)} \left (\frac {a \operatorname {AppellF1}\left (\frac {3}{4},\frac {1}{2},1,\frac {7}{4},\cos ^2(c+d x),\frac {b^2 \cos ^2(c+d x)}{-a^2+b^2}\right ) \cos ^{\frac {3}{2}}(c+d x)}{3 \left (a^2-b^2\right )}+\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) \left (2 \arctan \left (1-\frac {(1+i) \sqrt {b} \sqrt {\cos (c+d x)}}{\sqrt [4]{-a^2+b^2}}\right )-2 \arctan \left (1+\frac {(1+i) \sqrt {b} \sqrt {\cos (c+d x)}}{\sqrt [4]{-a^2+b^2}}\right )-\log \left (\sqrt {-a^2+b^2}-(1+i) \sqrt {b} \sqrt [4]{-a^2+b^2} \sqrt {\cos (c+d x)}+i b \cos (c+d x)\right )+\log \left (\sqrt {-a^2+b^2}+(1+i) \sqrt {b} \sqrt [4]{-a^2+b^2} \sqrt {\cos (c+d x)}+i b \cos (c+d x)\right )\right )}{\sqrt {b} \sqrt [4]{-a^2+b^2}}\right ) \sin (c+d x) \left (a+b \sqrt {\sin ^2(c+d x)}\right )}{d \sqrt {\cos (c+d x)} \sqrt {\sin ^2(c+d x)} (a+b \sin (c+d x))} \]
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Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 2.39 (sec) , antiderivative size = 825, normalized size of antiderivative = 2.83
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\[ \int \frac {\sqrt {e \cos (c+d x)}}{a+b \sin (c+d x)} \, dx=\int { \frac {\sqrt {e \cos \left (d x + c\right )}}{b \sin \left (d x + c\right ) + a} \,d x } \]
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Timed out. \[ \int \frac {\sqrt {e \cos (c+d x)}}{a+b \sin (c+d x)} \, dx=\text {Timed out} \]
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\[ \int \frac {\sqrt {e \cos (c+d x)}}{a+b \sin (c+d x)} \, dx=\int { \frac {\sqrt {e \cos \left (d x + c\right )}}{b \sin \left (d x + c\right ) + a} \,d x } \]
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\[ \int \frac {\sqrt {e \cos (c+d x)}}{a+b \sin (c+d x)} \, dx=\int { \frac {\sqrt {e \cos \left (d x + c\right )}}{b \sin \left (d x + c\right ) + a} \,d x } \]
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Timed out. \[ \int \frac {\sqrt {e \cos (c+d x)}}{a+b \sin (c+d x)} \, dx=\int \frac {\sqrt {e\,\cos \left (c+d\,x\right )}}{a+b\,\sin \left (c+d\,x\right )} \,d x \]
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