Optimal. Leaf size=380 \[ -\frac {a^2 \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{b^{3/2} \left (-a^2+b^2\right )^{3/4} f \sqrt {g}}-\frac {a^2 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{b^{3/2} \left (-a^2+b^2\right )^{3/4} f \sqrt {g}}-\frac {2 \sqrt {g \cos (e+f x)}}{b f g}-\frac {2 a \sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{b^2 f \sqrt {g \cos (e+f x)}}+\frac {a^3 \sqrt {\cos (e+f x)} \Pi \left (\frac {2 b}{b-\sqrt {-a^2+b^2}};\left .\frac {1}{2} (e+f x)\right |2\right )}{b^2 \left (a^2-b \left (b-\sqrt {-a^2+b^2}\right )\right ) f \sqrt {g \cos (e+f x)}}+\frac {a^3 \sqrt {\cos (e+f x)} \Pi \left (\frac {2 b}{b+\sqrt {-a^2+b^2}};\left .\frac {1}{2} (e+f x)\right |2\right )}{b^2 \left (a^2-b \left (b+\sqrt {-a^2+b^2}\right )\right ) f \sqrt {g \cos (e+f x)}} \]
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Rubi [A]
time = 0.59, antiderivative size = 380, normalized size of antiderivative = 1.00, number of steps
used = 15, number of rules used = 13, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.394, Rules used = {2988, 2645,
30, 2946, 2721, 2720, 2781, 2886, 2884, 335, 218, 214, 211} \begin {gather*} -\frac {a^2 \text {ArcTan}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{b^{3/2} f \sqrt {g} \left (b^2-a^2\right )^{3/4}}-\frac {a^2 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{b^{3/2} f \sqrt {g} \left (b^2-a^2\right )^{3/4}}+\frac {a^3 \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^2 f \left (a^2-b \left (b-\sqrt {b^2-a^2}\right )\right ) \sqrt {g \cos (e+f x)}}+\frac {a^3 \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^2 f \left (a^2-b \left (\sqrt {b^2-a^2}+b\right )\right ) \sqrt {g \cos (e+f x)}}-\frac {2 a \sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{b^2 f \sqrt {g \cos (e+f x)}}-\frac {2 \sqrt {g \cos (e+f x)}}{b f g} \end {gather*}
Antiderivative was successfully verified.
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Rule 30
Rule 211
Rule 214
Rule 218
Rule 335
Rule 2645
Rule 2720
Rule 2721
Rule 2781
Rule 2884
Rule 2886
Rule 2946
Rule 2988
Rubi steps
\begin {align*} \int \frac {\sin ^2(e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx &=\frac {\int \frac {\sin (e+f x)}{\sqrt {g \cos (e+f x)}} \, dx}{b}-\frac {a \int \frac {\sin (e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx}{b}\\ &=-\frac {a \int \frac {1}{\sqrt {g \cos (e+f x)}} \, dx}{b^2}+\frac {a^2 \int \frac {1}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx}{b^2}-\frac {\text {Subst}\left (\int \frac {1}{\sqrt {x}} \, dx,x,g \cos (e+f x)\right )}{b f g}\\ &=-\frac {2 \sqrt {g \cos (e+f x)}}{b f g}-\frac {a^3 \int \frac {1}{\sqrt {g \cos (e+f x)} \left (\sqrt {-a^2+b^2}-b \cos (e+f x)\right )} \, dx}{2 b^2 \sqrt {-a^2+b^2}}-\frac {a^3 \int \frac {1}{\sqrt {g \cos (e+f x)} \left (\sqrt {-a^2+b^2}+b \cos (e+f x)\right )} \, dx}{2 b^2 \sqrt {-a^2+b^2}}+\frac {\left (a^2 g\right ) \text {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 )}{b f}-\frac {\left (a \sqrt {\cos (e+f x)}\right ) \int \frac {1}{\sqrt {\cos (e+f x)}} \, dx}{b^2 \sqrt {g \cos (e+f x)}}\\ &=-\frac {2 \sqrt {g \cos (e+f x)}}{b f g}-\frac {2 a \sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{b^2 f \sqrt {g \cos (e+f x)}}+\frac {\left (2 a^2 g\right ) \text {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 )}{b f}-\frac {\left (a^3 \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^2 \sqrt {-a^2+b^2} \sqrt {g \cos (e+f x)}}-\frac {\left (a^3 \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^2 \sqrt {-a^2+b^2} \sqrt {g \cos (e+f x)}}\\ &=-\frac {2 \sqrt {g \cos (e+f x)}}{b f g}-\frac {2 a \sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{b^2 f \sqrt {g \cos (e+f x)}}+\frac {a^3 \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^2 \left (a^2-b \left (b-\sqrt {-a^2+b^2}\right )\right ) f \sqrt {g \cos (e+f x)}}-\frac {a^3 \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^2 \sqrt {-a^2+b^2} \left (b+\sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}}-\frac {a^2 \text {Subst}\left (\int \frac {1}{\sqrt {-a^2+b^2} g-b x^2} \, dx,x,\sqrt {g \cos (e+f x)}\right )}{b \sqrt {-a^2+b^2} f}-\frac {a^2 \text {Subst}\left (\int \frac {1}{\sqrt {-a^2+b^2} g+b x^2} \, dx,x,\sqrt {g \cos (e+f x)}\right )}{b \sqrt {-a^2+b^2} f}\\ &=-\frac {a^2 \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{b^{3/2} \left (-a^2+b^2\right )^{3/4} f \sqrt {g}}-\frac {a^2 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{b^{3/2} \left (-a^2+b^2\right )^{3/4} f \sqrt {g}}-\frac {2 \sqrt {g \cos (e+f x)}}{b f g}-\frac {2 a \sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{b^2 f \sqrt {g \cos (e+f x)}}+\frac {a^3 \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^2 \left (a^2-b \left (b-\sqrt {-a^2+b^2}\right )\right ) f \sqrt {g \cos (e+f x)}}-\frac {a^3 \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^2 \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] Result contains higher order function than in optimal. Order 5 vs. order 4 in
optimal.
time = 38.27, size = 572, normalized size = 1.51 \begin {gather*} -\frac {\left (a^2-b^2\right ) \left (a^2-b^2+b^2 \cos ^2(e+f x)\right ) \sec (e+f x) \sqrt [4]{\sec ^2(e+f x)} \tan ^3(e+f x) \left (\frac {a^2 \tan ^{-1}\left (\frac {\sqrt [4]{-a^2+b^2} \sqrt [4]{1+\tan ^2(e+f x)}}{\sqrt {b}}\right )}{b^{3/2} \left (-a^2+b^2\right )^{3/4}}-\frac {a^2 \tanh ^{-1}\left (\frac {\sqrt [4]{-a^2+b^2} \sqrt [4]{1+\tan ^2(e+f x)}}{\sqrt {b}}\right )}{b^{3/2} \left (-a^2+b^2\right )^{3/4}}+\frac {a \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {3}{2};-\tan ^2(e+f x)\right ) \tan (e+f x)}{a^2-b^2}-\frac {a^3 \cot (e+f x) \Pi \left (-\frac {\sqrt {-a^2+b^2}}{b};\left .\sin ^{-1}\left (\sqrt [4]{1+\tan ^2(e+f x)}\right )\right |-1\right ) \sqrt {-\tan ^2(e+f x)}}{b^2 \left (a^2-b^2\right )}-\frac {a^3 \cot (e+f x) \Pi \left (\frac {\sqrt {-a^2+b^2}}{b};\left .\sin ^{-1}\left (\sqrt [4]{1+\tan ^2(e+f x)}\right )\right |-1\right ) \sqrt {-\tan ^2(e+f x)}}{b^2 \left (a^2-b^2\right )}-\frac {2}{b \sqrt [4]{1+\tan ^2(e+f x)}}\right )}{f \sqrt {g \cos (e+f x)} (b+a \csc (e+f x)) \left (b \left (a^2-b^2\right ) \tan ^5(e+f x)+a^3 \sqrt {-\tan ^2(e+f x)} \sqrt {-\sec ^2(e+f x) \tan ^2(e+f x)}+\tan ^2(e+f x) \left (-a b^2 \sqrt {-\tan ^2(e+f x)} \sqrt {-\sec ^2(e+f x) \tan ^2(e+f x)}+a^3 \left (\sqrt {\sec ^2(e+f x)}+\sqrt {-\tan ^2(e+f x)} \sqrt {-\sec ^2(e+f x) \tan ^2(e+f x)}\right )\right )\right )} \end {gather*}
Warning: Unable to verify antiderivative.
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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order
4.
time = 39.48, size = 821, normalized size = 2.16
method | result | size |
default | \(\text {Expression too large to display}\) | \(821\) |
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(-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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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 {{\sin \left (e+f\,x\right )}^2}{\sqrt {g\,\cos \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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