Optimal. Leaf size=355 \[ \frac {\sqrt {g} \tan ^{-1}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a f}-\frac {\sqrt {b} \sqrt {g} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{a \sqrt [4]{-a^2+b^2} f}-\frac {\sqrt {g} \tanh ^{-1}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a f}+\frac {\sqrt {b} \sqrt {g} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{a \sqrt [4]{-a^2+b^2} f}-\frac {g \sqrt {\cos (e+f x)} \Pi \left (\frac {2 b}{b-\sqrt {-a^2+b^2}};\left .\frac {1}{2} (e+f x)\right |2\right )}{\left (b-\sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}}-\frac {g \sqrt {\cos (e+f x)} \Pi \left (\frac {2 b}{b+\sqrt {-a^2+b^2}};\left .\frac {1}{2} (e+f x)\right |2\right )}{\left (b+\sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}} \]
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Rubi [A]
time = 0.54, antiderivative size = 355, 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 = {2977, 2645,
335, 304, 209, 212, 2780, 2886, 2884, 211, 214} \begin {gather*} -\frac {\sqrt {b} \sqrt {g} \text {ArcTan}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{a f \sqrt [4]{b^2-a^2}}+\frac {\sqrt {b} \sqrt {g} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{a f \sqrt [4]{b^2-a^2}}-\frac {g \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 (b-\sqrt {b^2-a^2}\right ) \sqrt {g \cos (e+f x)}}-\frac {g \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 (\sqrt {b^2-a^2}+b\right ) \sqrt {g \cos (e+f x)}}+\frac {\sqrt {g} \text {ArcTan}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a f}-\frac {\sqrt {g} \tanh ^{-1}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a f} \end {gather*}
Antiderivative was successfully verified.
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Rule 209
Rule 211
Rule 212
Rule 214
Rule 304
Rule 335
Rule 2645
Rule 2780
Rule 2884
Rule 2886
Rule 2977
Rubi steps
\begin {align*} \int \frac {\sqrt {g \cos (e+f x)} \csc (e+f x)}{a+b \sin (e+f x)} \, dx &=\int \left (\frac {\sqrt {g \cos (e+f x)} \csc (e+f x)}{a}-\frac {b \sqrt {g \cos (e+f x)}}{a (a+b \sin (e+f x))}\right ) \, dx\\ &=\frac {\int \sqrt {g \cos (e+f x)} \csc (e+f x) \, dx}{a}-\frac {b \int \frac {\sqrt {g \cos (e+f x)}}{a+b \sin (e+f x)} \, dx}{a}\\ &=-\frac {\text {Subst}\left (\int \frac {\sqrt {x}}{1-\frac {x^2}{g^2}} \, dx,x,g \cos (e+f x)\right )}{a f g}+\frac {1}{2} g \int \frac {1}{\sqrt {g \cos (e+f x)} \left (\sqrt {-a^2+b^2}-b \cos (e+f x)\right )} \, dx-\frac {1}{2} g \int \frac {1}{\sqrt {g \cos (e+f x)} \left (\sqrt {-a^2+b^2}+b \cos (e+f x)\right )} \, dx-\frac {\left (b^2 g\right ) \text {Subst}\left (\int \frac {\sqrt {x}}{\left (a^2-b^2\right ) g^2+b^2 x^2} \, dx,x,g \cos (e+f x)\right )}{a f}\\ &=-\frac {2 \text {Subst}\left (\int \frac {x^2}{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 ) \text {Subst}\left (\int \frac {x^2}{\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 (g \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 {g \cos (e+f x)}}-\frac {\left (g \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 {g \cos (e+f x)}}\\ &=-\frac {g \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 (b-\sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}}-\frac {g \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 (b+\sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}}-\frac {g \text {Subst}\left (\int \frac {1}{g-x^2} \, dx,x,\sqrt {g \cos (e+f x)}\right )}{a f}+\frac {g \text {Subst}\left (\int \frac {1}{g+x^2} \, dx,x,\sqrt {g \cos (e+f x)}\right )}{a f}+\frac {(b g) \text {Subst}\left (\int \frac {1}{\sqrt {-a^2+b^2} g-b x^2} \, dx,x,\sqrt {g \cos (e+f x)}\right )}{a f}-\frac {(b g) \text {Subst}\left (\int \frac {1}{\sqrt {-a^2+b^2} g+b x^2} \, dx,x,\sqrt {g \cos (e+f x)}\right )}{a f}\\ &=\frac {\sqrt {g} \tan ^{-1}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a f}-\frac {\sqrt {b} \sqrt {g} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{a \sqrt [4]{-a^2+b^2} f}-\frac {\sqrt {g} \tanh ^{-1}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a f}+\frac {\sqrt {b} \sqrt {g} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{a \sqrt [4]{-a^2+b^2} f}-\frac {g \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 (b-\sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}}-\frac {g \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 (b+\sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}}\\ \end {align*}
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Mathematica [C] Result contains higher order function than in optimal. Order 6 vs. order 4 in
optimal.
time = 23.48, size = 534, normalized size = 1.50 \begin {gather*} \frac {\sqrt {g \cos (e+f x)} \csc (e+f x) \left (8 a b F_1\left (\frac {3}{4};\frac {1}{2},1;\frac {7}{4};\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{-a^2+b^2}\right ) \cos ^{\frac {3}{2}}(e+f x)+3 \left (2 \sqrt {2} \sqrt {b} \left (a^2-b^2\right )^{3/4} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {b} \sqrt {\cos (e+f x)}}{\sqrt [4]{a^2-b^2}}\right )-2 \sqrt {2} \sqrt {b} \left (a^2-b^2\right )^{3/4} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {b} \sqrt {\cos (e+f x)}}{\sqrt [4]{a^2-b^2}}\right )+4 a^2 \tan ^{-1}\left (\sqrt {\cos (e+f x)}\right )-4 b^2 \tan ^{-1}\left (\sqrt {\cos (e+f x)}\right )+2 a^2 \log \left (1-\sqrt {\cos (e+f x)}\right )-2 b^2 \log \left (1-\sqrt {\cos (e+f x)}\right )-2 a^2 \log \left (1+\sqrt {\cos (e+f x)}\right )+2 b^2 \log \left (1+\sqrt {\cos (e+f x)}\right )-\sqrt {2} \sqrt {b} \left (a^2-b^2\right )^{3/4} \log \left (\sqrt {a^2-b^2}-\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\cos (e+f x)}+b \cos (e+f x)\right )+\sqrt {2} \sqrt {b} \left (a^2-b^2\right )^{3/4} \log \left (\sqrt {a^2-b^2}+\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\cos (e+f x)}+b \cos (e+f x)\right )\right )\right ) \left (a+b \sqrt {\sin ^2(e+f x)}\right )}{12 a \left (a^2-b^2\right ) f \sqrt {\cos (e+f x)} (b+a \csc (e+f x))} \end {gather*}
Warning: Unable to verify antiderivative.
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Maple [A]
time = 16.75, size = 186, normalized size = 0.52
method | result | size |
default | \(-\frac {\sqrt {g}\, \ln \left (\frac {2 \sqrt {g}\, \sqrt {-2 \left (\sin ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) g +g}-4 g \cos \left (\frac {f x}{2}+\frac {e}{2}\right )-2 g}{\cos \left (\frac {f x}{2}+\frac {e}{2}\right )+1}\right ) \sqrt {-g}+\sqrt {g}\, \ln \left (\frac {2 \sqrt {g}\, \sqrt {-2 \left (\sin ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) g +g}+4 g \cos \left (\frac {f x}{2}+\frac {e}{2}\right )-2 g}{\cos \left (\frac {f x}{2}+\frac {e}{2}\right )-1}\right ) \sqrt {-g}+2 g \ln \left (\frac {2 \sqrt {-g}\, \sqrt {-2 \left (\sin ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) g +g}-2 g}{\cos \left (\frac {f x}{2}+\frac {e}{2}\right )}\right )}{2 a \sqrt {-g}\, f}\) | \(186\) |
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 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
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 \begin {gather*} \int \frac {\sqrt {g \cos {\left (e + f x \right )}} \csc {\left (e + f x \right )}}{a + b \sin {\left (e + f x \right )}}\, dx \end {gather*}
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 \begin {gather*} \text {could not integrate} \end {gather*}
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 \begin {gather*} \int \frac {\sqrt {g\,\cos \left (e+f\,x\right )}}{\sin \left (e+f\,x\right )\,\left (a+b\,\sin \left (e+f\,x\right )\right )} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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