Optimal. Leaf size=273 \[ \frac {2 \sqrt {2} b \sqrt {\cos (e+f x)} \Pi \left (-\frac {a}{b-\sqrt {b^2-a^2}};\left .\sin ^{-1}\left (\frac {\sqrt {d \sin (e+f x)}}{\sqrt {d} \sqrt {\cos (e+f x)+1}}\right )\right |-1\right )}{a \sqrt {d} f \sqrt {b^2-a^2} \sqrt {g \cos (e+f x)}}-\frac {2 \sqrt {2} b \sqrt {\cos (e+f x)} \Pi \left (-\frac {a}{b+\sqrt {b^2-a^2}};\left .\sin ^{-1}\left (\frac {\sqrt {d \sin (e+f x)}}{\sqrt {d} \sqrt {\cos (e+f x)+1}}\right )\right |-1\right )}{a \sqrt {d} f \sqrt {b^2-a^2} \sqrt {g \cos (e+f x)}}+\frac {\sqrt {\sin (2 e+2 f x)} F\left (\left .e+f x-\frac {\pi }{4}\right |2\right )}{a f \sqrt {d \sin (e+f x)} \sqrt {g \cos (e+f x)}} \]
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Rubi [A] time = 0.59, antiderivative size = 273, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.162, Rules used = {2910, 2573, 2641, 2908, 2907, 1218} \[ \frac {2 \sqrt {2} b \sqrt {\cos (e+f x)} \Pi \left (-\frac {a}{b-\sqrt {b^2-a^2}};\left .\sin ^{-1}\left (\frac {\sqrt {d \sin (e+f x)}}{\sqrt {d} \sqrt {\cos (e+f x)+1}}\right )\right |-1\right )}{a \sqrt {d} f \sqrt {b^2-a^2} \sqrt {g \cos (e+f x)}}-\frac {2 \sqrt {2} b \sqrt {\cos (e+f x)} \Pi \left (-\frac {a}{b+\sqrt {b^2-a^2}};\left .\sin ^{-1}\left (\frac {\sqrt {d \sin (e+f x)}}{\sqrt {d} \sqrt {\cos (e+f x)+1}}\right )\right |-1\right )}{a \sqrt {d} f \sqrt {b^2-a^2} \sqrt {g \cos (e+f x)}}+\frac {\sqrt {\sin (2 e+2 f x)} F\left (\left .e+f x-\frac {\pi }{4}\right |2\right )}{a f \sqrt {d \sin (e+f x)} \sqrt {g \cos (e+f x)}} \]
Antiderivative was successfully verified.
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Rule 1218
Rule 2573
Rule 2641
Rule 2907
Rule 2908
Rule 2910
Rubi steps
\begin {align*} \int \frac {1}{\sqrt {g \cos (e+f x)} \sqrt {d \sin (e+f x)} (a+b \sin (e+f x))} \, dx &=\frac {\int \frac {1}{\sqrt {g \cos (e+f x)} \sqrt {d \sin (e+f x)}} \, dx}{a}-\frac {b \int \frac {\sqrt {d \sin (e+f x)}}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx}{a d}\\ &=-\frac {\left (b \sqrt {\cos (e+f x)}\right ) \int \frac {\sqrt {d \sin (e+f x)}}{\sqrt {\cos (e+f x)} (a+b \sin (e+f x))} \, dx}{a d \sqrt {g \cos (e+f x)}}+\frac {\sqrt {\sin (2 e+2 f x)} \int \frac {1}{\sqrt {\sin (2 e+2 f x)}} \, dx}{a \sqrt {g \cos (e+f x)} \sqrt {d \sin (e+f x)}}\\ &=\frac {F\left (\left .e-\frac {\pi }{4}+f x\right |2\right ) \sqrt {\sin (2 e+2 f x)}}{a f \sqrt {g \cos (e+f x)} \sqrt {d \sin (e+f x)}}-\frac {\left (2 \sqrt {2} b \left (1-\frac {b}{\sqrt {-a^2+b^2}}\right ) \sqrt {\cos (e+f x)}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (\left (b-\sqrt {-a^2+b^2}\right ) d+a x^2\right ) \sqrt {1-\frac {x^4}{d^2}}} \, dx,x,\frac {\sqrt {d \sin (e+f x)}}{\sqrt {1+\cos (e+f x)}}\right )}{a f \sqrt {g \cos (e+f x)}}-\frac {\left (2 \sqrt {2} b \left (1+\frac {b}{\sqrt {-a^2+b^2}}\right ) \sqrt {\cos (e+f x)}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (\left (b+\sqrt {-a^2+b^2}\right ) d+a x^2\right ) \sqrt {1-\frac {x^4}{d^2}}} \, dx,x,\frac {\sqrt {d \sin (e+f x)}}{\sqrt {1+\cos (e+f x)}}\right )}{a f \sqrt {g \cos (e+f x)}}\\ &=\frac {2 \sqrt {2} b \sqrt {\cos (e+f x)} \Pi \left (-\frac {a}{b-\sqrt {-a^2+b^2}};\left .\sin ^{-1}\left (\frac {\sqrt {d \sin (e+f x)}}{\sqrt {d} \sqrt {1+\cos (e+f x)}}\right )\right |-1\right )}{a \sqrt {-a^2+b^2} \sqrt {d} f \sqrt {g \cos (e+f x)}}-\frac {2 \sqrt {2} b \sqrt {\cos (e+f x)} \Pi \left (-\frac {a}{b+\sqrt {-a^2+b^2}};\left .\sin ^{-1}\left (\frac {\sqrt {d \sin (e+f x)}}{\sqrt {d} \sqrt {1+\cos (e+f x)}}\right )\right |-1\right )}{a \sqrt {-a^2+b^2} \sqrt {d} f \sqrt {g \cos (e+f x)}}+\frac {F\left (\left .e-\frac {\pi }{4}+f x\right |2\right ) \sqrt {\sin (2 e+2 f x)}}{a f \sqrt {g \cos (e+f x)} \sqrt {d \sin (e+f x)}}\\ \end {align*}
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Mathematica [A] time = 10.10, size = 209, normalized size = 0.77 \[ -\frac {4 \sqrt {2} \cos ^2\left (\frac {1}{2} (e+f x)\right ) \sqrt {\frac {\cos (e+f x)}{\cos (e+f x)-1}} \tan ^{\frac {3}{2}}\left (\frac {1}{2} (e+f x)\right ) \left (\sqrt {b^2-a^2} F\left (\left .\sin ^{-1}\left (\frac {1}{\sqrt {\tan \left (\frac {1}{2} (e+f x)\right )}}\right )\right |-1\right )+b \left (\Pi \left (\frac {a}{\sqrt {b^2-a^2}-b};\left .\sin ^{-1}\left (\frac {1}{\sqrt {\tan \left (\frac {1}{2} (e+f x)\right )}}\right )\right |-1\right )-\Pi \left (-\frac {a}{b+\sqrt {b^2-a^2}};\left .\sin ^{-1}\left (\frac {1}{\sqrt {\tan \left (\frac {1}{2} (e+f x)\right )}}\right )\right |-1\right )\right )\right )}{a f \sqrt {b^2-a^2} \sqrt {d \sin (e+f x)} \sqrt {g \cos (e+f x)}} \]
Antiderivative was successfully verified.
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fricas [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {g \cos \left (f x + e\right )} {\left (b \sin \left (f x + e\right ) + a\right )} \sqrt {d \sin \left (f x + e\right )}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.68, size = 631, normalized size = 2.31 \[ \frac {\left (2 \EllipticF \left (\sqrt {-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {\sqrt {2}}{2}\right ) a \sqrt {-a^{2}+b^{2}}-2 \EllipticF \left (\sqrt {-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {\sqrt {2}}{2}\right ) \sqrt {-a^{2}+b^{2}}\, b +\EllipticPi \left (\sqrt {-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {a}{a -b +\sqrt {-a^{2}+b^{2}}}, \frac {\sqrt {2}}{2}\right ) a b -\EllipticPi \left (\sqrt {-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {a}{a -b +\sqrt {-a^{2}+b^{2}}}, \frac {\sqrt {2}}{2}\right ) b^{2}+\EllipticPi \left (\sqrt {-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {a}{a -b +\sqrt {-a^{2}+b^{2}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {-a^{2}+b^{2}}\, b -\EllipticPi \left (\sqrt {-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, -\frac {a}{b +\sqrt {-a^{2}+b^{2}}-a}, \frac {\sqrt {2}}{2}\right ) a b +\EllipticPi \left (\sqrt {-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, -\frac {a}{b +\sqrt {-a^{2}+b^{2}}-a}, \frac {\sqrt {2}}{2}\right ) b^{2}+\EllipticPi \left (\sqrt {-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, -\frac {a}{b +\sqrt {-a^{2}+b^{2}}-a}, \frac {\sqrt {2}}{2}\right ) \sqrt {-a^{2}+b^{2}}\, b \right ) \sqrt {\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \left (\sin ^{2}\left (f x +e \right )\right ) \sqrt {2}}{f \sqrt {d \sin \left (f x +e \right )}\, \left (-1+\cos \left (f x +e \right )\right ) \sqrt {g \cos \left (f x +e \right )}\, \sqrt {-a^{2}+b^{2}}\, \left (a -b +\sqrt {-a^{2}+b^{2}}\right ) \left (b +\sqrt {-a^{2}+b^{2}}-a \right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {g \cos \left (f x + e\right )} {\left (b \sin \left (f x + e\right ) + a\right )} \sqrt {d \sin \left (f x + e\right )}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {1}{\sqrt {g\,\cos \left (e+f\,x\right )}\,\sqrt {d\,\sin \left (e+f\,x\right )}\,\left (a+b\,\sin \left (e+f\,x\right )\right )} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {d \sin {\left (e + f x \right )}} \sqrt {g \cos {\left (e + f x \right )}} \left (a + b \sin {\left (e + f x \right )}\right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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