Optimal. Leaf size=369 \[ -\frac {b \sqrt {\cos (e+f x)} \Pi \left (\frac {2 b}{b-\sqrt {b^2-a^2}};\left .\frac {1}{2} (e+f x)\right |2\right )}{f \left (a^2-b \left (b-\sqrt {b^2-a^2}\right )\right ) \sqrt {g \cos (e+f x)}}-\frac {b \sqrt {\cos (e+f x)} \Pi \left (\frac {2 b}{b+\sqrt {b^2-a^2}};\left .\frac {1}{2} (e+f x)\right |2\right )}{f \left (a^2-b \left (\sqrt {b^2-a^2}+b\right )\right ) \sqrt {g \cos (e+f x)}}+\frac {b^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{a f \sqrt {g} \left (b^2-a^2\right )^{3/4}}+\frac {b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{a f \sqrt {g} \left (b^2-a^2\right )^{3/4}}-\frac {\tan ^{-1}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a f \sqrt {g}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a f \sqrt {g}} \]
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Rubi [A] time = 0.82, antiderivative size = 369, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 11, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.355, Rules used = {2898, 2565, 329, 212, 206, 203, 2702, 2807, 2805, 208, 205} \[ \frac {b^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{a f \sqrt {g} \left (b^2-a^2\right )^{3/4}}+\frac {b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{a f \sqrt {g} \left (b^2-a^2\right )^{3/4}}-\frac {b \sqrt {\cos (e+f x)} \Pi \left (\frac {2 b}{b-\sqrt {b^2-a^2}};\left .\frac {1}{2} (e+f x)\right |2\right )}{f \left (a^2-b \left (b-\sqrt {b^2-a^2}\right )\right ) \sqrt {g \cos (e+f x)}}-\frac {b \sqrt {\cos (e+f x)} \Pi \left (\frac {2 b}{b+\sqrt {b^2-a^2}};\left .\frac {1}{2} (e+f x)\right |2\right )}{f \left (a^2-b \left (\sqrt {b^2-a^2}+b\right )\right ) \sqrt {g \cos (e+f x)}}-\frac {\tan ^{-1}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a f \sqrt {g}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a f \sqrt {g}} \]
Antiderivative was successfully verified.
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Rule 203
Rule 205
Rule 206
Rule 208
Rule 212
Rule 329
Rule 2565
Rule 2702
Rule 2805
Rule 2807
Rule 2898
Rubi steps
\begin {align*} \int \frac {\csc (e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx &=\int \left (\frac {\csc (e+f x)}{a \sqrt {g \cos (e+f x)}}-\frac {b}{a \sqrt {g \cos (e+f x)} (a+b \sin (e+f x))}\right ) \, dx\\ &=\frac {\int \frac {\csc (e+f x)}{\sqrt {g \cos (e+f x)}} \, dx}{a}-\frac {b \int \frac {1}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx}{a}\\ &=\frac {b \int \frac {1}{\sqrt {g \cos (e+f x)} \left (\sqrt {-a^2+b^2}-b \cos (e+f x)\right )} \, dx}{2 \sqrt {-a^2+b^2}}+\frac {b \int \frac {1}{\sqrt {g \cos (e+f x)} \left (\sqrt {-a^2+b^2}+b \cos (e+f x)\right )} \, dx}{2 \sqrt {-a^2+b^2}}-\frac {\operatorname {Subst}\left (\int \frac {1}{\sqrt {x} \left (1-\frac {x^2}{g^2}\right )} \, dx,x,g \cos (e+f x)\right )}{a f g}-\frac {\left (b^2 g\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {x} \left (\left (a^2-b^2\right ) g^2+b^2 x^2\right )} \, dx,x,g \cos (e+f x)\right )}{a f}\\ &=-\frac {2 \operatorname {Subst}\left (\int \frac {1}{1-\frac {x^4}{g^2}} \, dx,x,\sqrt {g \cos (e+f x)}\right )}{a f g}-\frac {\left (2 b^2 g\right ) \operatorname {Subst}\left (\int \frac {1}{\left (a^2-b^2\right ) g^2+b^2 x^4} \, dx,x,\sqrt {g \cos (e+f x)}\right )}{a f}+\frac {\left (b \sqrt {\cos (e+f x)}\right ) \int \frac {1}{\sqrt {\cos (e+f x)} \left (\sqrt {-a^2+b^2}-b \cos (e+f x)\right )} \, dx}{2 \sqrt {-a^2+b^2} \sqrt {g \cos (e+f x)}}+\frac {\left (b \sqrt {\cos (e+f x)}\right ) \int \frac {1}{\sqrt {\cos (e+f x)} \left (\sqrt {-a^2+b^2}+b \cos (e+f x)\right )} \, dx}{2 \sqrt {-a^2+b^2} \sqrt {g \cos (e+f x)}}\\ &=-\frac {b \sqrt {\cos (e+f x)} \Pi \left (\frac {2 b}{b-\sqrt {-a^2+b^2}};\left .\frac {1}{2} (e+f x)\right |2\right )}{\left (a^2-b \left (b-\sqrt {-a^2+b^2}\right )\right ) f \sqrt {g \cos (e+f x)}}-\frac {b \sqrt {\cos (e+f x)} \Pi \left (\frac {2 b}{b+\sqrt {-a^2+b^2}};\left .\frac {1}{2} (e+f x)\right |2\right )}{\left (a^2-b \left (b+\sqrt {-a^2+b^2}\right )\right ) f \sqrt {g \cos (e+f x)}}-\frac {\operatorname {Subst}\left (\int \frac {1}{g-x^2} \, dx,x,\sqrt {g \cos (e+f x)}\right )}{a f}-\frac {\operatorname {Subst}\left (\int \frac {1}{g+x^2} \, dx,x,\sqrt {g \cos (e+f x)}\right )}{a f}+\frac {b^2 \operatorname {Subst}\left (\int \frac {1}{\sqrt {-a^2+b^2} g-b x^2} \, dx,x,\sqrt {g \cos (e+f x)}\right )}{a \sqrt {-a^2+b^2} f}+\frac {b^2 \operatorname {Subst}\left (\int \frac {1}{\sqrt {-a^2+b^2} g+b x^2} \, dx,x,\sqrt {g \cos (e+f x)}\right )}{a \sqrt {-a^2+b^2} f}\\ &=-\frac {\tan ^{-1}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a f \sqrt {g}}+\frac {b^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{a \left (-a^2+b^2\right )^{3/4} f \sqrt {g}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a f \sqrt {g}}+\frac {b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{a \left (-a^2+b^2\right )^{3/4} f \sqrt {g}}-\frac {b \sqrt {\cos (e+f x)} \Pi \left (\frac {2 b}{b-\sqrt {-a^2+b^2}};\left .\frac {1}{2} (e+f x)\right |2\right )}{\left (a^2-b \left (b-\sqrt {-a^2+b^2}\right )\right ) f \sqrt {g \cos (e+f x)}}-\frac {b \sqrt {\cos (e+f x)} \Pi \left (\frac {2 b}{b+\sqrt {-a^2+b^2}};\left .\frac {1}{2} (e+f x)\right |2\right )}{\left (a^2-b \left (b+\sqrt {-a^2+b^2}\right )\right ) f \sqrt {g \cos (e+f x)}}\\ \end {align*}
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Mathematica [C] time = 20.62, size = 698, normalized size = 1.89 \[ -\frac {2 \sqrt {\cos (e+f x)} \left (\cos ^2(e+f x)-1\right ) \csc (e+f x) \left (a+b \sqrt {1-\cos ^2(e+f x)}\right ) \left (\frac {5 b \left (a^2-b^2\right ) \sqrt {\cos (e+f x)} F_1\left (\frac {1}{4};\frac {1}{2},1;\frac {5}{4};\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{b^2-a^2}\right )}{\sqrt {1-\cos ^2(e+f x)} \left (a^2+b^2 \left (\cos ^2(e+f x)-1\right )\right ) \left (5 \left (a^2-b^2\right ) F_1\left (\frac {1}{4};\frac {1}{2},1;\frac {5}{4};\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{b^2-a^2}\right )-2 \cos ^2(e+f x) \left (2 b^2 F_1\left (\frac {5}{4};\frac {1}{2},2;\frac {9}{4};\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{b^2-a^2}\right )+\left (b^2-a^2\right ) F_1\left (\frac {5}{4};\frac {3}{2},1;\frac {9}{4};\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{b^2-a^2}\right )\right )\right )}-\frac {-2 \left (a^2-b^2\right )^{3/4} \log \left (1-\sqrt {\cos (e+f x)}\right )+2 \left (a^2-b^2\right )^{3/4} \log \left (\sqrt {\cos (e+f x)}+1\right )+4 \left (a^2-b^2\right )^{3/4} \tan ^{-1}\left (\sqrt {\cos (e+f x)}\right )-\sqrt {2} b^{3/2} \log \left (-\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\cos (e+f x)}+\sqrt {a^2-b^2}+b \cos (e+f x)\right )+\sqrt {2} b^{3/2} \log \left (\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\cos (e+f x)}+\sqrt {a^2-b^2}+b \cos (e+f x)\right )-2 \sqrt {2} b^{3/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {b} \sqrt {\cos (e+f x)}}{\sqrt [4]{a^2-b^2}}\right )+2 \sqrt {2} b^{3/2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {b} \sqrt {\cos (e+f x)}}{\sqrt [4]{a^2-b^2}}+1\right )}{8 a \left (a^2-b^2\right )^{3/4}}\right )}{f \left (1-\cos ^2(e+f x)\right ) \sqrt {g \cos (e+f x)} (a \csc (e+f x)+b)} \]
Warning: Unable to verify antiderivative.
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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 \[ \int \frac {\csc \left (f x + e\right )}{\sqrt {g \cos \left (f x + e\right )} {\left (b \sin \left (f x + e\right ) + a\right )}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 3.06, size = 186, normalized size = 0.50 \[ \frac {\ln \left (\frac {2 \sqrt {-g}\, \sqrt {-2 \left (\sin ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) g +g}-2 g}{\cos \left (\frac {f x}{2}+\frac {e}{2}\right )}\right )}{a \sqrt {-g}\, f}-\frac {\ln \left (\frac {2 \sqrt {g}\, \sqrt {-2 \left (\sin ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) g +g}-4 g \cos \left (\frac {f x}{2}+\frac {e}{2}\right )-2 g}{\cos \left (\frac {f x}{2}+\frac {e}{2}\right )+1}\right )}{2 a \sqrt {g}\, f}-\frac {\ln \left (\frac {2 \sqrt {g}\, \sqrt {-2 \left (\sin ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) g +g}+4 g \cos \left (\frac {f x}{2}+\frac {e}{2}\right )-2 g}{-1+\cos \left (\frac {f x}{2}+\frac {e}{2}\right )}\right )}{2 a \sqrt {g}\, f} \]
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 \[ \int \frac {\csc \left (f x + e\right )}{\sqrt {g \cos \left (f x + e\right )} {\left (b \sin \left (f x + e\right ) + a\right )}}\,{d x} \]
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 \[ \int \frac {1}{\sin \left (e+f\,x\right )\,\sqrt {g\,\cos \left (e+f\,x\right )}\,\left (a+b\,\sin \left (e+f\,x\right )\right )} \,d x \]
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 \[ \int \frac {\csc {\left (e + f x \right )}}{\sqrt {g \cos {\left (e + f x \right )}} \left (a + b \sin {\left (e + f x \right )}\right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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