Optimal. Leaf size=355 \[ \frac {i a h \text {Li}_2\left (-i e^{i (e+f x)}\right ) \cos (e+f x)}{f^2 \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c}}-\frac {i a h \text {Li}_2\left (i e^{i (e+f x)}\right ) \cos (e+f x)}{f^2 \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c}}-\frac {i a h \text {Li}_2\left (-e^{2 i (e+f x)}\right ) \cos (e+f x)}{2 f^2 \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c}}-\frac {i a (g+h x)^2 \cos (e+f x)}{2 h \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c}}+\frac {a (g+h x) \log \left (1+e^{2 i (e+f x)}\right ) \cos (e+f x)}{f \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c}}-\frac {2 i a (g+h x) \cos (e+f x) \tan ^{-1}\left (e^{i (e+f x)}\right )}{f \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c}} \]
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Rubi [A] time = 0.51, antiderivative size = 355, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 9, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.257, Rules used = {4604, 6741, 12, 6742, 4181, 2279, 2391, 3719, 2190} \[ \frac {i a h \cos (e+f x) \text {PolyLog}\left (2,-i e^{i (e+f x)}\right )}{f^2 \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c}}-\frac {i a h \cos (e+f x) \text {PolyLog}\left (2,i e^{i (e+f x)}\right )}{f^2 \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c}}-\frac {i a h \cos (e+f x) \text {PolyLog}\left (2,-e^{2 i (e+f x)}\right )}{2 f^2 \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c}}-\frac {i a (g+h x)^2 \cos (e+f x)}{2 h \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c}}+\frac {a (g+h x) \log \left (1+e^{2 i (e+f x)}\right ) \cos (e+f x)}{f \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c}}-\frac {2 i a (g+h x) \cos (e+f x) \tan ^{-1}\left (e^{i (e+f x)}\right )}{f \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c}} \]
Antiderivative was successfully verified.
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Rule 12
Rule 2190
Rule 2279
Rule 2391
Rule 3719
Rule 4181
Rule 4604
Rule 6741
Rule 6742
Rubi steps
\begin {align*} \int \frac {(g+h x) \sqrt {a-a \sin (e+f x)}}{\sqrt {c+c \sin (e+f x)}} \, dx &=\frac {\cos (e+f x) \int (g+h x) \sec (e+f x) (a-a \sin (e+f x)) \, dx}{\sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}\\ &=\frac {\cos (e+f x) \int a (g+h x) \sec (e+f x) (1-\sin (e+f x)) \, dx}{\sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}\\ &=\frac {(a \cos (e+f x)) \int (g+h x) \sec (e+f x) (1-\sin (e+f x)) \, dx}{\sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}\\ &=\frac {(a \cos (e+f x)) \int ((g+h x) \sec (e+f x)-(g+h x) \tan (e+f x)) \, dx}{\sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}\\ &=\frac {(a \cos (e+f x)) \int (g+h x) \sec (e+f x) \, dx}{\sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}-\frac {(a \cos (e+f x)) \int (g+h x) \tan (e+f x) \, dx}{\sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}\\ &=-\frac {i a (g+h x)^2 \cos (e+f x)}{2 h \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}-\frac {2 i a (g+h x) \tan ^{-1}\left (e^{i (e+f x)}\right ) \cos (e+f x)}{f \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}+\frac {(2 i a \cos (e+f x)) \int \frac {e^{2 i (e+f x)} (g+h x)}{1+e^{2 i (e+f x)}} \, dx}{\sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}-\frac {(a h \cos (e+f x)) \int \log \left (1-i e^{i (e+f x)}\right ) \, dx}{f \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}+\frac {(a h \cos (e+f x)) \int \log \left (1+i e^{i (e+f x)}\right ) \, dx}{f \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}\\ &=-\frac {i a (g+h x)^2 \cos (e+f x)}{2 h \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}-\frac {2 i a (g+h x) \tan ^{-1}\left (e^{i (e+f x)}\right ) \cos (e+f x)}{f \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}+\frac {a (g+h x) \cos (e+f x) \log \left (1+e^{2 i (e+f x)}\right )}{f \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}+\frac {(i a h \cos (e+f x)) \operatorname {Subst}\left (\int \frac {\log (1-i x)}{x} \, dx,x,e^{i (e+f x)}\right )}{f^2 \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}-\frac {(i a h \cos (e+f x)) \operatorname {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{i (e+f x)}\right )}{f^2 \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}-\frac {(a h \cos (e+f x)) \int \log \left (1+e^{2 i (e+f x)}\right ) \, dx}{f \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}\\ &=-\frac {i a (g+h x)^2 \cos (e+f x)}{2 h \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}-\frac {2 i a (g+h x) \tan ^{-1}\left (e^{i (e+f x)}\right ) \cos (e+f x)}{f \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}+\frac {a (g+h x) \cos (e+f x) \log \left (1+e^{2 i (e+f x)}\right )}{f \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}+\frac {i a h \cos (e+f x) \text {Li}_2\left (-i e^{i (e+f x)}\right )}{f^2 \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}-\frac {i a h \cos (e+f x) \text {Li}_2\left (i e^{i (e+f x)}\right )}{f^2 \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}+\frac {(i a h \cos (e+f x)) \operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{2 i (e+f x)}\right )}{2 f^2 \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}\\ &=-\frac {i a (g+h x)^2 \cos (e+f x)}{2 h \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}-\frac {2 i a (g+h x) \tan ^{-1}\left (e^{i (e+f x)}\right ) \cos (e+f x)}{f \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}+\frac {a (g+h x) \cos (e+f x) \log \left (1+e^{2 i (e+f x)}\right )}{f \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}+\frac {i a h \cos (e+f x) \text {Li}_2\left (-i e^{i (e+f x)}\right )}{f^2 \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}-\frac {i a h \cos (e+f x) \text {Li}_2\left (i e^{i (e+f x)}\right )}{f^2 \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}-\frac {i a h \cos (e+f x) \text {Li}_2\left (-e^{2 i (e+f x)}\right )}{2 f^2 \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}\\ \end {align*}
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Mathematica [A] time = 1.22, size = 154, normalized size = 0.43 \[ \frac {\left (e^{i (e+f x)}+i\right ) \sqrt {a-a \sin (e+f x)} \left (f \left (f x (2 g+h x)-4 i (g+h x) \log \left (1+i e^{-i (e+f x)}\right )\right )+4 h \text {Li}_2\left (-i e^{-i (e+f x)}\right )\right )}{\sqrt {2} f^2 \left (e^{i (e+f x)}-i\right ) \sqrt {-i c e^{-i (e+f x)} \left (e^{i (e+f x)}+i\right )^2}} \]
Antiderivative was successfully verified.
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fricas [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]
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 \[ \int \frac {{\left (h x + g\right )} \sqrt {-a \sin \left (f x + e\right ) + a}}{\sqrt {c \sin \left (f x + e\right ) + c}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.35, size = 0, normalized size = 0.00 \[ \int \frac {\left (h x +g \right ) \sqrt {a -a \sin \left (f x +e \right )}}{\sqrt {c +c \sin \left (f x +e \right )}}\, dx \]
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 \[ \int \frac {{\left (h x + g\right )} \sqrt {-a \sin \left (f x + e\right ) + a}}{\sqrt {c \sin \left (f x + e\right ) + c}}\,{d x} \]
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 \[ \int \frac {\left (g+h\,x\right )\,\sqrt {a-a\,\sin \left (e+f\,x\right )}}{\sqrt {c+c\,\sin \left (e+f\,x\right )}} \,d x \]
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 \[ \int \frac {\sqrt {- a \left (\sin {\left (e + f x \right )} - 1\right )} \left (g + h x\right )}{\sqrt {c \left (\sin {\left (e + f x \right )} + 1\right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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