Optimal. Leaf size=79 \[ \frac {\sqrt {\frac {(b \sin (e+f x)+c \cos (e+f x))^2}{a}+1} F\left (e+f x+\tan ^{-1}(b,c)|-\frac {b^2+c^2}{a}\right )}{f \sqrt {a+(b \sin (e+f x)+c \cos (e+f x))^2}} \]
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Rubi [F] time = 0.70, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \[ \int \frac {1}{\sqrt {a+(c \cos (e+f x)+b \sin (e+f x))^2}} \, dx \]
Verification is Not applicable to the result.
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Rubi steps
\begin {align*} \int \frac {1}{\sqrt {a+(c \cos (e+f x)+b \sin (e+f x))^2}} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {1}{\left (1+x^2\right ) \sqrt {a+\frac {(c+b x)^2}{1+x^2}}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {\operatorname {Subst}\left (\int \left (\frac {i}{2 (i-x) \sqrt {a+\frac {(c+b x)^2}{1+x^2}}}+\frac {i}{2 (i+x) \sqrt {a+\frac {(c+b x)^2}{1+x^2}}}\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {i \operatorname {Subst}\left (\int \frac {1}{(i-x) \sqrt {a+\frac {(c+b x)^2}{1+x^2}}} \, dx,x,\tan (e+f x)\right )}{2 f}+\frac {i \operatorname {Subst}\left (\int \frac {1}{(i+x) \sqrt {a+\frac {(c+b x)^2}{1+x^2}}} \, dx,x,\tan (e+f x)\right )}{2 f}\\ \end {align*}
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Mathematica [C] time = 1.59, size = 529, normalized size = 6.70 \[ \frac {\sqrt {2} \sec \left (\tan ^{-1}\left (\frac {c^2-b^2}{2 b c}\right )+2 (e+f x)\right ) \sqrt {-\frac {b c \sqrt {\frac {\left (b^2+c^2\right )^2}{b^2 c^2}} \left (\sin \left (\tan ^{-1}\left (\frac {c^2-b^2}{2 b c}\right )+2 (e+f x)\right )-1\right )}{2 a+b c \sqrt {\frac {\left (b^2+c^2\right )^2}{b^2 c^2}}+b^2+c^2}} \sqrt {-\frac {b c \sqrt {\frac {\left (b^2+c^2\right )^2}{b^2 c^2}} \left (\sin \left (\tan ^{-1}\left (\frac {c^2-b^2}{2 b c}\right )+2 (e+f x)\right )+1\right )}{2 a-b c \sqrt {\frac {\left (b^2+c^2\right )^2}{b^2 c^2}}+b^2+c^2}} \sqrt {2 a+b c \sqrt {\frac {\left (b^2+c^2\right )^2}{b^2 c^2}} \sin \left (\tan ^{-1}\left (\frac {c^2-b^2}{2 b c}\right )+2 (e+f x)\right )+b^2+c^2} F_1\left (\frac {1}{2};\frac {1}{2},\frac {1}{2};\frac {3}{2};\frac {b^2+c \sqrt {\frac {\left (b^2+c^2\right )^2}{b^2 c^2}} \sin \left (2 (e+f x)+\tan ^{-1}\left (\frac {c^2-b^2}{2 b c}\right )\right ) b+c^2+2 a}{b^2-c \sqrt {\frac {\left (b^2+c^2\right )^2}{b^2 c^2}} b+c^2+2 a},\frac {b^2+c \sqrt {\frac {\left (b^2+c^2\right )^2}{b^2 c^2}} \sin \left (2 (e+f x)+\tan ^{-1}\left (\frac {c^2-b^2}{2 b c}\right )\right ) b+c^2+2 a}{b^2+c \sqrt {\frac {\left (b^2+c^2\right )^2}{b^2 c^2}} b+c^2+2 a}\right )}{b c f \sqrt {\frac {\left (b^2+c^2\right )^2}{b^2 c^2}}} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {1}{\sqrt {2 \, b c \cos \left (f x + e\right ) \sin \left (f x + e\right ) - {\left (b^{2} - c^{2}\right )} \cos \left (f x + e\right )^{2} + b^{2} + a}}, x\right ) \]
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 \[ \int \frac {1}{\sqrt {{\left (c \cos \left (f x + e\right ) + b \sin \left (f x + e\right )\right )}^{2} + a}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 3.76, size = 258179, normalized size = 3268.09 \[ \text {output too large to display} \]
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 \[ \int \frac {1}{\sqrt {{\left (c \cos \left (f x + e\right ) + b \sin \left (f x + e\right )\right )}^{2} + a}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{\sqrt {a+{\left (c\,\cos \left (e+f\,x\right )+b\,\sin \left (e+f\,x\right )\right )}^2}} \,d x \]
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 \[ \int \frac {1}{\sqrt {a + \left (b \sin {\left (e + f x \right )} + c \cos {\left (e + f x \right )}\right )^{2}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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