Optimal. Leaf size=344 \[ -\frac {\cos (e+f x) \sqrt {c+d \sin (e+f x)}}{5 (c-d) f (a+a \sin (e+f x))^3}-\frac {2 (c-3 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{15 a (c-d)^2 f (a+a \sin (e+f x))^2}-\frac {\left (4 c^2-15 c d+27 d^2\right ) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{30 (c-d)^3 f \left (a^3+a^3 \sin (e+f x)\right )}-\frac {\left (4 c^2-15 c d+27 d^2\right ) E\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )|\frac {2 d}{c+d}\right ) \sqrt {c+d \sin (e+f x)}}{30 a^3 (c-d)^3 f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {\left (4 c^2-11 c d+15 d^2\right ) F\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )|\frac {2 d}{c+d}\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}{30 a^3 (c-d)^2 f \sqrt {c+d \sin (e+f x)}} \]
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Rubi [A]
time = 0.50, antiderivative size = 344, normalized size of antiderivative = 1.00, number of steps
used = 8, number of rules used = 7, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {2845, 3057,
2831, 2742, 2740, 2734, 2732} \begin {gather*} -\frac {\left (4 c^2-15 c d+27 d^2\right ) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{30 f (c-d)^3 \left (a^3 \sin (e+f x)+a^3\right )}+\frac {\left (4 c^2-11 c d+15 d^2\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}} F\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{30 a^3 f (c-d)^2 \sqrt {c+d \sin (e+f x)}}-\frac {\left (4 c^2-15 c d+27 d^2\right ) \sqrt {c+d \sin (e+f x)} E\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{30 a^3 f (c-d)^3 \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}-\frac {2 (c-3 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{15 a f (c-d)^2 (a \sin (e+f x)+a)^2}-\frac {\cos (e+f x) \sqrt {c+d \sin (e+f x)}}{5 f (c-d) (a \sin (e+f x)+a)^3} \end {gather*}
Antiderivative was successfully verified.
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Rule 2732
Rule 2734
Rule 2740
Rule 2742
Rule 2831
Rule 2845
Rule 3057
Rubi steps
\begin {align*} \int \frac {1}{(a+a \sin (e+f x))^3 \sqrt {c+d \sin (e+f x)}} \, dx &=-\frac {\cos (e+f x) \sqrt {c+d \sin (e+f x)}}{5 (c-d) f (a+a \sin (e+f x))^3}-\frac {\int \frac {-\frac {1}{2} a (4 c-9 d)-\frac {3}{2} a d \sin (e+f x)}{(a+a \sin (e+f x))^2 \sqrt {c+d \sin (e+f x)}} \, dx}{5 a^2 (c-d)}\\ &=-\frac {\cos (e+f x) \sqrt {c+d \sin (e+f x)}}{5 (c-d) f (a+a \sin (e+f x))^3}-\frac {2 (c-3 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{15 a (c-d)^2 f (a+a \sin (e+f x))^2}+\frac {\int \frac {\frac {1}{2} a^2 \left (4 c^2-13 c d+21 d^2\right )+a^2 (c-3 d) d \sin (e+f x)}{(a+a \sin (e+f x)) \sqrt {c+d \sin (e+f x)}} \, dx}{15 a^4 (c-d)^2}\\ &=-\frac {\cos (e+f x) \sqrt {c+d \sin (e+f x)}}{5 (c-d) f (a+a \sin (e+f x))^3}-\frac {2 (c-3 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{15 a (c-d)^2 f (a+a \sin (e+f x))^2}-\frac {\left (4 c^2-15 c d+27 d^2\right ) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{30 (c-d)^3 f \left (a^3+a^3 \sin (e+f x)\right )}-\frac {\int \frac {\frac {1}{4} a^3 d^2 (c+15 d)+\frac {1}{4} a^3 d \left (4 c^2-15 c d+27 d^2\right ) \sin (e+f x)}{\sqrt {c+d \sin (e+f x)}} \, dx}{15 a^6 (c-d)^3}\\ &=-\frac {\cos (e+f x) \sqrt {c+d \sin (e+f x)}}{5 (c-d) f (a+a \sin (e+f x))^3}-\frac {2 (c-3 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{15 a (c-d)^2 f (a+a \sin (e+f x))^2}-\frac {\left (4 c^2-15 c d+27 d^2\right ) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{30 (c-d)^3 f \left (a^3+a^3 \sin (e+f x)\right )}+\frac {\left (4 c^2-11 c d+15 d^2\right ) \int \frac {1}{\sqrt {c+d \sin (e+f x)}} \, dx}{60 a^3 (c-d)^2}-\frac {\left (4 c^2-15 c d+27 d^2\right ) \int \sqrt {c+d \sin (e+f x)} \, dx}{60 a^3 (c-d)^3}\\ &=-\frac {\cos (e+f x) \sqrt {c+d \sin (e+f x)}}{5 (c-d) f (a+a \sin (e+f x))^3}-\frac {2 (c-3 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{15 a (c-d)^2 f (a+a \sin (e+f x))^2}-\frac {\left (4 c^2-15 c d+27 d^2\right ) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{30 (c-d)^3 f \left (a^3+a^3 \sin (e+f x)\right )}-\frac {\left (\left (4 c^2-15 c d+27 d^2\right ) \sqrt {c+d \sin (e+f x)}\right ) \int \sqrt {\frac {c}{c+d}+\frac {d \sin (e+f x)}{c+d}} \, dx}{60 a^3 (c-d)^3 \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {\left (\left (4 c^2-11 c d+15 d^2\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}\right ) \int \frac {1}{\sqrt {\frac {c}{c+d}+\frac {d \sin (e+f x)}{c+d}}} \, dx}{60 a^3 (c-d)^2 \sqrt {c+d \sin (e+f x)}}\\ &=-\frac {\cos (e+f x) \sqrt {c+d \sin (e+f x)}}{5 (c-d) f (a+a \sin (e+f x))^3}-\frac {2 (c-3 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{15 a (c-d)^2 f (a+a \sin (e+f x))^2}-\frac {\left (4 c^2-15 c d+27 d^2\right ) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{30 (c-d)^3 f \left (a^3+a^3 \sin (e+f x)\right )}-\frac {\left (4 c^2-15 c d+27 d^2\right ) E\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )|\frac {2 d}{c+d}\right ) \sqrt {c+d \sin (e+f x)}}{30 a^3 (c-d)^3 f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {\left (4 c^2-11 c d+15 d^2\right ) F\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )|\frac {2 d}{c+d}\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}{30 a^3 (c-d)^2 f \sqrt {c+d \sin (e+f x)}}\\ \end {align*}
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Mathematica [A]
time = 4.19, size = 445, normalized size = 1.29 \begin {gather*} \frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^6 \left (-\left (\left (4 c^2-15 c d+27 d^2\right ) (c+d \sin (e+f x))\right )+\frac {2 \left (6 (c-d)^2 \sin \left (\frac {1}{2} (e+f x)\right )-3 (c-d)^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )+4 (c-3 d) (c-d) \sin \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2-2 (c-3 d) (c-d) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3+\left (4 c^2-15 c d+27 d^2\right ) \sin \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^4\right ) (c+d \sin (e+f x))}{\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^5}+d^2 (c+15 d) F\left (\frac {1}{4} (-2 e+\pi -2 f x)|\frac {2 d}{c+d}\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}+\left (4 c^2-15 c d+27 d^2\right ) \left ((c+d) E\left (\frac {1}{4} (-2 e+\pi -2 f x)|\frac {2 d}{c+d}\right )-c F\left (\frac {1}{4} (-2 e+\pi -2 f x)|\frac {2 d}{c+d}\right )\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}\right )}{30 a^3 (c-d)^3 f (1+\sin (e+f x))^3 \sqrt {c+d \sin (e+f x)}} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 17.37, size = 593, normalized size = 1.72
method | result | size |
default | \(\frac {\sqrt {-\left (-d \sin \left (f x +e \right )-c \right ) \left (\cos ^{2}\left (f x +e \right )\right )}\, \left (-\frac {\sqrt {-\left (-d \sin \left (f x +e \right )-c \right ) \left (\cos ^{2}\left (f x +e \right )\right )}}{5 \left (c -d \right ) \left (1+\sin \left (f x +e \right )\right )^{3}}-\frac {2 \left (c -3 d \right ) \sqrt {-\left (-d \sin \left (f x +e \right )-c \right ) \left (\cos ^{2}\left (f x +e \right )\right )}}{15 \left (c -d \right )^{2} \left (1+\sin \left (f x +e \right )\right )^{2}}-\frac {\left (-\left (\sin ^{2}\left (f x +e \right )\right ) d -c \sin \left (f x +e \right )+d \sin \left (f x +e \right )+c \right ) \left (4 c^{2}-15 c d +27 d^{2}\right )}{30 \left (c -d \right )^{3} \sqrt {\left (-d \sin \left (f x +e \right )-c \right ) \left (\sin \left (f x +e \right )-1\right ) \left (1+\sin \left (f x +e \right )\right )}}+\frac {2 \left (-c \,d^{2}-15 d^{3}\right ) \left (\frac {c}{d}-1\right ) \sqrt {\frac {c +d \sin \left (f x +e \right )}{c -d}}\, \sqrt {\frac {d \left (1-\sin \left (f x +e \right )\right )}{c +d}}\, \sqrt {\frac {\left (-1-\sin \left (f x +e \right )\right ) d}{c -d}}\, \EllipticF \left (\sqrt {\frac {c +d \sin \left (f x +e \right )}{c -d}}, \sqrt {\frac {c -d}{c +d}}\right )}{\left (60 c^{3}-180 c^{2} d +180 c \,d^{2}-60 d^{3}\right ) \sqrt {-\left (-d \sin \left (f x +e \right )-c \right ) \left (\cos ^{2}\left (f x +e \right )\right )}}-\frac {d \left (4 c^{2}-15 c d +27 d^{2}\right ) \left (\frac {c}{d}-1\right ) \sqrt {\frac {c +d \sin \left (f x +e \right )}{c -d}}\, \sqrt {\frac {d \left (1-\sin \left (f x +e \right )\right )}{c +d}}\, \sqrt {\frac {\left (-1-\sin \left (f x +e \right )\right ) d}{c -d}}\, \left (\left (-\frac {c}{d}-1\right ) \EllipticE \left (\sqrt {\frac {c +d \sin \left (f x +e \right )}{c -d}}, \sqrt {\frac {c -d}{c +d}}\right )+\EllipticF \left (\sqrt {\frac {c +d \sin \left (f x +e \right )}{c -d}}, \sqrt {\frac {c -d}{c +d}}\right )\right )}{30 \left (c -d \right )^{3} \sqrt {-\left (-d \sin \left (f x +e \right )-c \right ) \left (\cos ^{2}\left (f x +e \right )\right )}}\right )}{a^{3} \cos \left (f x +e \right ) \sqrt {c +d \sin \left (f x +e \right )}\, f}\) | \(593\) |
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 [C] Result contains higher order function than in optimal. Order 9 vs. order
4.
time = 0.19, size = 1789, normalized size = 5.20 \begin {gather*} \text {Too large to display} \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*} \frac {\int \frac {1}{\sqrt {c + d \sin {\left (e + f x \right )}} \sin ^{3}{\left (e + f x \right )} + 3 \sqrt {c + d \sin {\left (e + f x \right )}} \sin ^{2}{\left (e + f x \right )} + 3 \sqrt {c + d \sin {\left (e + f x \right )}} \sin {\left (e + f x \right )} + \sqrt {c + d \sin {\left (e + f x \right )}}}\, dx}{a^{3}} \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 {1}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^3\,\sqrt {c+d\,\sin \left (e+f\,x\right )}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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