Optimal. Leaf size=209 \[ \frac {2 \sqrt {2} \sqrt {d} \Pi \left (-\frac {a}{b-\sqrt {-a^2+b^2}};\left .\text {ArcSin}\left (\frac {\sqrt {d \cos (e+f x)}}{\sqrt {d} \sqrt {1+\sin (e+f x)}}\right )\right |-1\right ) \sqrt {\sin (e+f x)}}{\sqrt {-a^2+b^2} f \sqrt {g \sin (e+f x)}}-\frac {2 \sqrt {2} \sqrt {d} \Pi \left (-\frac {a}{b+\sqrt {-a^2+b^2}};\left .\text {ArcSin}\left (\frac {\sqrt {d \cos (e+f x)}}{\sqrt {d} \sqrt {1+\sin (e+f x)}}\right )\right |-1\right ) \sqrt {\sin (e+f x)}}{\sqrt {-a^2+b^2} f \sqrt {g \sin (e+f x)}} \]
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Rubi [A]
time = 0.29, antiderivative size = 209, normalized size of antiderivative = 1.00, number of steps
used = 4, number of rules used = 3, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.081, Rules used = {2987, 2986,
1232} \begin {gather*} \frac {2 \sqrt {2} \sqrt {d} \sqrt {\sin (e+f x)} \Pi \left (-\frac {a}{b-\sqrt {b^2-a^2}};\left .\text {ArcSin}\left (\frac {\sqrt {d \cos (e+f x)}}{\sqrt {d} \sqrt {\sin (e+f x)+1}}\right )\right |-1\right )}{f \sqrt {b^2-a^2} \sqrt {g \sin (e+f x)}}-\frac {2 \sqrt {2} \sqrt {d} \sqrt {\sin (e+f x)} \Pi \left (-\frac {a}{b+\sqrt {b^2-a^2}};\left .\text {ArcSin}\left (\frac {\sqrt {d \cos (e+f x)}}{\sqrt {d} \sqrt {\sin (e+f x)+1}}\right )\right |-1\right )}{f \sqrt {b^2-a^2} \sqrt {g \sin (e+f x)}} \end {gather*}
Antiderivative was successfully verified.
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Rule 1232
Rule 2986
Rule 2987
Rubi steps
\begin {align*} \int \frac {\sqrt {d \cos (e+f x)}}{(a+b \cos (e+f x)) \sqrt {g \sin (e+f x)}} \, dx &=\frac {\sqrt {\sin (e+f x)} \int \frac {\sqrt {d \cos (e+f x)}}{(a+b \cos (e+f x)) \sqrt {\sin (e+f x)}} \, dx}{\sqrt {g \sin (e+f x)}}\\ &=-\frac {\left (2 \sqrt {2} \left (1-\frac {b}{\sqrt {-a^2+b^2}}\right ) d \sqrt {\sin (e+f x)}\right ) \text {Subst}\left (\int \frac {1}{\left (-\left (-b+\sqrt {-a^2+b^2}\right ) d+a x^2\right ) \sqrt {1-\frac {x^4}{d^2}}} \, dx,x,\frac {\sqrt {d \cos (e+f x)}}{\sqrt {1+\sin (e+f x)}}\right )}{f \sqrt {g \sin (e+f x)}}-\frac {\left (2 \sqrt {2} \left (1+\frac {b}{\sqrt {-a^2+b^2}}\right ) d \sqrt {\sin (e+f x)}\right ) \text {Subst}\left (\int \frac {1}{\left (-\left (-b-\sqrt {-a^2+b^2}\right ) d+a x^2\right ) \sqrt {1-\frac {x^4}{d^2}}} \, dx,x,\frac {\sqrt {d \cos (e+f x)}}{\sqrt {1+\sin (e+f x)}}\right )}{f \sqrt {g \sin (e+f x)}}\\ &=\frac {2 \sqrt {2} \sqrt {d} \Pi \left (-\frac {a}{b-\sqrt {-a^2+b^2}};\left .\sin ^{-1}\left (\frac {\sqrt {d \cos (e+f x)}}{\sqrt {d} \sqrt {1+\sin (e+f x)}}\right )\right |-1\right ) \sqrt {\sin (e+f x)}}{\sqrt {-a^2+b^2} f \sqrt {g \sin (e+f x)}}-\frac {2 \sqrt {2} \sqrt {d} \Pi \left (-\frac {a}{b+\sqrt {-a^2+b^2}};\left .\sin ^{-1}\left (\frac {\sqrt {d \cos (e+f x)}}{\sqrt {d} \sqrt {1+\sin (e+f x)}}\right )\right |-1\right ) \sqrt {\sin (e+f x)}}{\sqrt {-a^2+b^2} f \sqrt {g \sin (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 = 19.63, size = 594, normalized size = 2.84 \begin {gather*} \frac {2 \sqrt {d \cos (e+f x)} \sec ^2(e+f x) \sqrt {\tan (e+f x)} \left (b+a \sqrt {1+\tan ^2(e+f x)}\right ) \left (\frac {\sqrt {a} \left (-2 \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt {a} \sqrt {\tan (e+f x)}}{\sqrt [4]{a^2-b^2}}\right )+2 \text {ArcTan}\left (1+\frac {\sqrt {2} \sqrt {a} \sqrt {\tan (e+f x)}}{\sqrt [4]{a^2-b^2}}\right )-\log \left (\sqrt {a^2-b^2}-\sqrt {2} \sqrt {a} \sqrt [4]{a^2-b^2} \sqrt {\tan (e+f x)}+a \tan (e+f x)\right )+\log \left (\sqrt {a^2-b^2}+\sqrt {2} \sqrt {a} \sqrt [4]{a^2-b^2} \sqrt {\tan (e+f x)}+a \tan (e+f x)\right )\right )}{4 \sqrt {2} \left (a^2-b^2\right )^{3/4}}+\frac {5 b \left (a^2-b^2\right ) F_1\left (\frac {1}{4};\frac {1}{2},1;\frac {5}{4};-\tan ^2(e+f x),-\frac {a^2 \tan ^2(e+f x)}{a^2-b^2}\right ) \sqrt {\tan (e+f x)}}{\sqrt {1+\tan ^2(e+f x)} \left (-5 \left (a^2-b^2\right ) F_1\left (\frac {1}{4};\frac {1}{2},1;\frac {5}{4};-\tan ^2(e+f x),-\frac {a^2 \tan ^2(e+f x)}{a^2-b^2}\right )+2 \left (2 a^2 F_1\left (\frac {5}{4};\frac {1}{2},2;\frac {9}{4};-\tan ^2(e+f x),-\frac {a^2 \tan ^2(e+f x)}{a^2-b^2}\right )+\left (a^2-b^2\right ) F_1\left (\frac {5}{4};\frac {3}{2},1;\frac {9}{4};-\tan ^2(e+f x),-\frac {a^2 \tan ^2(e+f x)}{a^2-b^2}\right )\right ) \tan ^2(e+f x)\right ) \left (-b^2+a^2 \left (1+\tan ^2(e+f x)\right )\right )}\right )}{f (a+b \cos (e+f x)) \sqrt {g \sin (e+f x)} \left (1+\tan ^2(e+f x)\right )^{3/2}} \end {gather*}
Warning: Unable to verify antiderivative.
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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(615\) vs.
\(2(173)=346\).
time = 0.29, size = 616, normalized size = 2.95
method | result | size |
default | \(-\frac {\sqrt {d \cos \left (f x +e \right )}\, \left (2 \EllipticF \left (\sqrt {-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {\sqrt {2}}{2}\right ) \sqrt {-a^{2}+b^{2}}+a \EllipticPi \left (\sqrt {-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {a -b}{a -b +\sqrt {-\left (a -b \right ) \left (a +b \right )}}, \frac {\sqrt {2}}{2}\right )-\EllipticPi \left (\sqrt {-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {a -b}{a -b +\sqrt {-\left (a -b \right ) \left (a +b \right )}}, \frac {\sqrt {2}}{2}\right ) b -\sqrt {-a^{2}+b^{2}}\, \EllipticPi \left (\sqrt {-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {a -b}{a -b +\sqrt {-\left (a -b \right ) \left (a +b \right )}}, \frac {\sqrt {2}}{2}\right )-a \EllipticPi \left (\sqrt {-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, -\frac {a -b}{-a +b +\sqrt {-\left (a -b \right ) \left (a +b \right )}}, \frac {\sqrt {2}}{2}\right )+\EllipticPi \left (\sqrt {-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, -\frac {a -b}{-a +b +\sqrt {-\left (a -b \right ) \left (a +b \right )}}, \frac {\sqrt {2}}{2}\right ) b -\sqrt {-a^{2}+b^{2}}\, \EllipticPi \left (\sqrt {-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, -\frac {a -b}{-a +b +\sqrt {-\left (a -b \right ) \left (a +b \right )}}, \frac {\sqrt {2}}{2}\right )\right ) \sqrt {\frac {\cos \left (f x +e \right )-1}{\sin \left (f x +e \right )}}\, \sqrt {\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \left (\sin ^{2}\left (f x +e \right )\right ) \sqrt {2}\, a}{f \sqrt {g \sin \left (f x +e \right )}\, \cos \left (f x +e \right ) \left (\cos \left (f x +e \right )-1\right ) \sqrt {-a^{2}+b^{2}}\, \left (\sqrt {-a^{2}+b^{2}}+a -b \right ) \left (\sqrt {-a^{2}+b^{2}}-a +b \right )}\) | \(616\) |
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 {d \cos {\left (e + f x \right )}}}{\sqrt {g \sin {\left (e + f x \right )}} \left (a + b \cos {\left (e + f x \right )}\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 {d\,\cos \left (e+f\,x\right )}}{\sqrt {g\,\sin \left (e+f\,x\right )}\,\left (a+b\,\cos \left (e+f\,x\right )\right )} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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