Optimal. Leaf size=352 \[ \frac {a \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{\sqrt {b} f \sqrt {g} \left (b^2-a^2\right )^{3/4}}+\frac {a \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{\sqrt {b} f \sqrt {g} \left (b^2-a^2\right )^{3/4}}-\frac {a^2 \sqrt {\cos (e+f x)} \Pi \left (\frac {2 b}{b-\sqrt {b^2-a^2}};\left .\frac {1}{2} (e+f x)\right |2\right )}{b f \left (b \sqrt {b^2-a^2}+a^2-b^2\right ) \sqrt {g \cos (e+f x)}}-\frac {a^2 \sqrt {\cos (e+f x)} \Pi \left (\frac {2 b}{b+\sqrt {b^2-a^2}};\left .\frac {1}{2} (e+f x)\right |2\right )}{b f \left (a^2-b \left (\sqrt {b^2-a^2}+b\right )\right ) \sqrt {g \cos (e+f x)}}+\frac {2 \sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{b f \sqrt {g \cos (e+f x)}} \]
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Rubi [A] time = 0.73, antiderivative size = 352, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 10, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.323, Rules used = {2867, 2642, 2641, 2702, 2807, 2805, 329, 212, 208, 205} \[ \frac {a \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{\sqrt {b} f \sqrt {g} \left (b^2-a^2\right )^{3/4}}+\frac {a \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{\sqrt {b} f \sqrt {g} \left (b^2-a^2\right )^{3/4}}-\frac {a^2 \sqrt {\cos (e+f x)} \Pi \left (\frac {2 b}{b-\sqrt {b^2-a^2}};\left .\frac {1}{2} (e+f x)\right |2\right )}{b f \left (b \sqrt {b^2-a^2}+a^2-b^2\right ) \sqrt {g \cos (e+f x)}}-\frac {a^2 \sqrt {\cos (e+f x)} \Pi \left (\frac {2 b}{b+\sqrt {b^2-a^2}};\left .\frac {1}{2} (e+f x)\right |2\right )}{b f \left (a^2-b \left (\sqrt {b^2-a^2}+b\right )\right ) \sqrt {g \cos (e+f x)}}+\frac {2 \sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{b f \sqrt {g \cos (e+f x)}} \]
Antiderivative was successfully verified.
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Rule 205
Rule 208
Rule 212
Rule 329
Rule 2641
Rule 2642
Rule 2702
Rule 2805
Rule 2807
Rule 2867
Rubi steps
\begin {align*} \int \frac {\sin (e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx &=\frac {\int \frac {1}{\sqrt {g \cos (e+f x)}} \, dx}{b}-\frac {a \int \frac {1}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx}{b}\\ &=\frac {a^2 \int \frac {1}{\sqrt {g \cos (e+f x)} \left (\sqrt {-a^2+b^2}-b \cos (e+f x)\right )} \, dx}{2 b \sqrt {-a^2+b^2}}+\frac {a^2 \int \frac {1}{\sqrt {g \cos (e+f x)} \left (\sqrt {-a^2+b^2}+b \cos (e+f x)\right )} \, dx}{2 b \sqrt {-a^2+b^2}}-\frac {(a g) \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 )}{f}+\frac {\sqrt {\cos (e+f x)} \int \frac {1}{\sqrt {\cos (e+f x)}} \, dx}{b \sqrt {g \cos (e+f x)}}\\ &=\frac {2 \sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{b f \sqrt {g \cos (e+f x)}}-\frac {(2 a g) \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 )}{f}+\frac {\left (a^2 \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 b \sqrt {-a^2+b^2} \sqrt {g \cos (e+f x)}}+\frac {\left (a^2 \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 b \sqrt {-a^2+b^2} \sqrt {g \cos (e+f x)}}\\ &=\frac {2 \sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{b f \sqrt {g \cos (e+f x)}}-\frac {a^2 \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 )}{b \left (a^2-b^2+b \sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}}+\frac {a^2 \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 )}{b \sqrt {-a^2+b^2} \left (b+\sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}}+\frac {a \operatorname {Subst}\left (\int \frac {1}{\sqrt {-a^2+b^2} g-b x^2} \, dx,x,\sqrt {g \cos (e+f x)}\right )}{\sqrt {-a^2+b^2} f}+\frac {a \operatorname {Subst}\left (\int \frac {1}{\sqrt {-a^2+b^2} g+b x^2} \, dx,x,\sqrt {g \cos (e+f x)}\right )}{\sqrt {-a^2+b^2} f}\\ &=\frac {a \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{\sqrt {b} \left (-a^2+b^2\right )^{3/4} f \sqrt {g}}+\frac {a \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{\sqrt {b} \left (-a^2+b^2\right )^{3/4} f \sqrt {g}}+\frac {2 \sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{b f \sqrt {g \cos (e+f x)}}-\frac {a^2 \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 )}{b \left (a^2-b^2+b \sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}}+\frac {a^2 \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 )}{b \sqrt {-a^2+b^2} \left (b+\sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}}\\ \end {align*}
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Mathematica [C] time = 6.20, size = 546, normalized size = 1.55 \[ -\frac {2 \sqrt {\cos (e+f x)} \left (a+b \sqrt {\sin ^2(e+f x)}\right ) \left (\frac {5 b \left (a^2-b^2\right ) \sqrt {\sin ^2(e+f x)} \sqrt {\cos (e+f x)} F_1\left (\frac {1}{4};-\frac {1}{2},1;\frac {5}{4};\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{b^2-a^2}\right )}{\left (a^2+b^2 \cos ^2(e+f x)-b^2\right ) \left (2 \cos ^2(e+f x) \left (2 b^2 F_1\left (\frac {5}{4};-\frac {1}{2},2;\frac {9}{4};\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{b^2-a^2}\right )+\left (a^2-b^2\right ) F_1\left (\frac {5}{4};\frac {1}{2},1;\frac {9}{4};\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{b^2-a^2}\right )\right )-5 \left (a^2-b^2\right ) F_1\left (\frac {1}{4};-\frac {1}{2},1;\frac {5}{4};\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{b^2-a^2}\right )\right )}+\frac {a \left (-\log \left (-\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\cos (e+f x)}+\sqrt {a^2-b^2}+b \cos (e+f x)\right )+\log \left (\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\cos (e+f x)}+\sqrt {a^2-b^2}+b \cos (e+f x)\right )-2 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {b} \sqrt {\cos (e+f x)}}{\sqrt [4]{a^2-b^2}}\right )+2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {b} \sqrt {\cos (e+f x)}}{\sqrt [4]{a^2-b^2}}+1\right )\right )}{4 \sqrt {2} \sqrt {b} \left (a^2-b^2\right )^{3/4}}\right )}{f \sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \]
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 \[ \int \frac {\sin \left (f x + e\right )}{\sqrt {g \cos \left (f x + e\right )} {\left (b \sin \left (f x + e\right ) + a\right )}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 6.05, size = 1181, normalized size = 3.36 \[ \text {result too large to display} \]
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 \[ \int \frac {\sin \left (f x + e\right )}{\sqrt {g \cos \left (f x + e\right )} {\left (b \sin \left (f x + e\right ) + a\right )}}\,{d x} \]
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 \[ \int \frac {\sin \left (e+f\,x\right )}{\sqrt {g\,\cos \left (e+f\,x\right )}\,\left (a+b\,\sin \left (e+f\,x\right )\right )} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-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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