Optimal. Leaf size=356 \[ -\frac {b^2 e \sqrt {\sin (c+d x)} \Pi \left (\frac {2 a}{a-\sqrt {a^2-b^2}};\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{a^2 d \left (a-\sqrt {a^2-b^2}\right ) \sqrt {e \sin (c+d x)}}-\frac {b^2 e \sqrt {\sin (c+d x)} \Pi \left (\frac {2 a}{a+\sqrt {a^2-b^2}};\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{a^2 d \left (\sqrt {a^2-b^2}+a\right ) \sqrt {e \sin (c+d x)}}+\frac {b \sqrt {e} \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt {e} \sqrt [4]{a^2-b^2}}\right )}{a^{3/2} d \sqrt [4]{a^2-b^2}}-\frac {b \sqrt {e} \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt {e} \sqrt [4]{a^2-b^2}}\right )}{a^{3/2} d \sqrt [4]{a^2-b^2}}+\frac {2 E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{a d \sqrt {\sin (c+d x)}} \]
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Rubi [A] time = 0.76, antiderivative size = 356, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 11, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {3872, 2867, 2640, 2639, 2701, 2807, 2805, 329, 298, 205, 208} \[ \frac {b \sqrt {e} \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt {e} \sqrt [4]{a^2-b^2}}\right )}{a^{3/2} d \sqrt [4]{a^2-b^2}}-\frac {b \sqrt {e} \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt {e} \sqrt [4]{a^2-b^2}}\right )}{a^{3/2} d \sqrt [4]{a^2-b^2}}-\frac {b^2 e \sqrt {\sin (c+d x)} \Pi \left (\frac {2 a}{a-\sqrt {a^2-b^2}};\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{a^2 d \left (a-\sqrt {a^2-b^2}\right ) \sqrt {e \sin (c+d x)}}-\frac {b^2 e \sqrt {\sin (c+d x)} \Pi \left (\frac {2 a}{a+\sqrt {a^2-b^2}};\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{a^2 d \left (\sqrt {a^2-b^2}+a\right ) \sqrt {e \sin (c+d x)}}+\frac {2 E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{a d \sqrt {\sin (c+d x)}} \]
Antiderivative was successfully verified.
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Rule 205
Rule 208
Rule 298
Rule 329
Rule 2639
Rule 2640
Rule 2701
Rule 2805
Rule 2807
Rule 2867
Rule 3872
Rubi steps
\begin {align*} \int \frac {\sqrt {e \sin (c+d x)}}{a+b \sec (c+d x)} \, dx &=-\int \frac {\cos (c+d x) \sqrt {e \sin (c+d x)}}{-b-a \cos (c+d x)} \, dx\\ &=\frac {\int \sqrt {e \sin (c+d x)} \, dx}{a}+\frac {b \int \frac {\sqrt {e \sin (c+d x)}}{-b-a \cos (c+d x)} \, dx}{a}\\ &=\frac {\left (b^2 e\right ) \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {a^2-b^2}-a \sin (c+d x)\right )} \, dx}{2 a^2}-\frac {\left (b^2 e\right ) \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {a^2-b^2}+a \sin (c+d x)\right )} \, dx}{2 a^2}+\frac {(b e) \operatorname {Subst}\left (\int \frac {\sqrt {x}}{\left (-a^2+b^2\right ) e^2+a^2 x^2} \, dx,x,e \sin (c+d x)\right )}{d}+\frac {\sqrt {e \sin (c+d x)} \int \sqrt {\sin (c+d x)} \, dx}{a \sqrt {\sin (c+d x)}}\\ &=\frac {2 E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{a d \sqrt {\sin (c+d x)}}+\frac {(2 b e) \operatorname {Subst}\left (\int \frac {x^2}{\left (-a^2+b^2\right ) e^2+a^2 x^4} \, dx,x,\sqrt {e \sin (c+d x)}\right )}{d}+\frac {\left (b^2 e \sqrt {\sin (c+d x)}\right ) \int \frac {1}{\sqrt {\sin (c+d x)} \left (\sqrt {a^2-b^2}-a \sin (c+d x)\right )} \, dx}{2 a^2 \sqrt {e \sin (c+d x)}}-\frac {\left (b^2 e \sqrt {\sin (c+d x)}\right ) \int \frac {1}{\sqrt {\sin (c+d x)} \left (\sqrt {a^2-b^2}+a \sin (c+d x)\right )} \, dx}{2 a^2 \sqrt {e \sin (c+d x)}}\\ &=-\frac {b^2 e \Pi \left (\frac {2 a}{a-\sqrt {a^2-b^2}};\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{a^2 \left (a-\sqrt {a^2-b^2}\right ) d \sqrt {e \sin (c+d x)}}-\frac {b^2 e \Pi \left (\frac {2 a}{a+\sqrt {a^2-b^2}};\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{a^2 \left (a+\sqrt {a^2-b^2}\right ) d \sqrt {e \sin (c+d x)}}+\frac {2 E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{a d \sqrt {\sin (c+d x)}}-\frac {(b e) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a^2-b^2} e-a x^2} \, dx,x,\sqrt {e \sin (c+d x)}\right )}{a d}+\frac {(b e) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a^2-b^2} e+a x^2} \, dx,x,\sqrt {e \sin (c+d x)}\right )}{a d}\\ &=\frac {b \sqrt {e} \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt [4]{a^2-b^2} \sqrt {e}}\right )}{a^{3/2} \sqrt [4]{a^2-b^2} d}-\frac {b \sqrt {e} \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt [4]{a^2-b^2} \sqrt {e}}\right )}{a^{3/2} \sqrt [4]{a^2-b^2} d}-\frac {b^2 e \Pi \left (\frac {2 a}{a-\sqrt {a^2-b^2}};\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{a^2 \left (a-\sqrt {a^2-b^2}\right ) d \sqrt {e \sin (c+d x)}}-\frac {b^2 e \Pi \left (\frac {2 a}{a+\sqrt {a^2-b^2}};\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{a^2 \left (a+\sqrt {a^2-b^2}\right ) d \sqrt {e \sin (c+d x)}}+\frac {2 E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{a d \sqrt {\sin (c+d x)}}\\ \end {align*}
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Mathematica [C] time = 20.75, size = 351, normalized size = 0.99 \[ \frac {\sqrt {e \sin (c+d x)} \left (a \sqrt {\cos ^2(c+d x)}+b\right ) \left (3 \sqrt {2} b \left (b^2-a^2\right )^{3/4} \left (-\log \left (-\sqrt {2} \sqrt {a} \sqrt [4]{b^2-a^2} \sqrt {\sin (c+d x)}+\sqrt {b^2-a^2}+a \sin (c+d x)\right )+\log \left (\sqrt {2} \sqrt {a} \sqrt [4]{b^2-a^2} \sqrt {\sin (c+d x)}+\sqrt {b^2-a^2}+a \sin (c+d x)\right )+2 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {a} \sqrt {\sin (c+d x)}}{\sqrt [4]{b^2-a^2}}\right )-2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {\sin (c+d x)}}{\sqrt [4]{b^2-a^2}}+1\right )\right )+8 a^{5/2} \sin ^{\frac {3}{2}}(c+d x) F_1\left (\frac {3}{4};-\frac {1}{2},1;\frac {7}{4};\sin ^2(c+d x),\frac {a^2 \sin ^2(c+d x)}{a^2-b^2}\right )\right )}{12 a^{3/2} d \left (a^2-b^2\right ) \sqrt {\sin (c+d x)} (a \cos (c+d x)+b)} \]
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 \[ \int \frac {\sqrt {e \sin \left (d x + c\right )}}{b \sec \left (d x + c\right ) + a}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 6.69, size = 919, normalized size = 2.58 \[ \frac {e b \arctan \left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\left (\frac {e^{2} \left (a^{2}-b^{2}\right )}{a^{2}}\right )^{\frac {1}{4}}}\right )}{d \,a^{2} \left (\frac {e^{2} \left (a^{2}-b^{2}\right )}{a^{2}}\right )^{\frac {1}{4}}}-\frac {e b \ln \left (\frac {\sqrt {e \sin \left (d x +c \right )}+\left (\frac {e^{2} \left (a^{2}-b^{2}\right )}{a^{2}}\right )^{\frac {1}{4}}}{\sqrt {e \sin \left (d x +c \right )}-\left (\frac {e^{2} \left (a^{2}-b^{2}\right )}{a^{2}}\right )^{\frac {1}{4}}}\right )}{2 d \,a^{2} \left (\frac {e^{2} \left (a^{2}-b^{2}\right )}{a^{2}}\right )^{\frac {1}{4}}}-\frac {e \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) b^{2} \EllipticPi \left (\sqrt {-\sin \left (d x +c \right )+1}, -\frac {a}{-a +\sqrt {a^{2}-b^{2}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {a^{2}-b^{2}}}{2 d \,a^{2} \left (-a +\sqrt {a^{2}-b^{2}}\right ) \left (a +\sqrt {a^{2}-b^{2}}\right ) \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}}+\frac {e \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) b^{2} \EllipticPi \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {a}{a +\sqrt {a^{2}-b^{2}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {a^{2}-b^{2}}}{2 d \,a^{2} \left (-a +\sqrt {a^{2}-b^{2}}\right ) \left (a +\sqrt {a^{2}-b^{2}}\right ) \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}}+\frac {2 e \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) b^{2} \EllipticE \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )}{d a \left (-a +\sqrt {a^{2}-b^{2}}\right ) \left (a +\sqrt {a^{2}-b^{2}}\right ) \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}}-\frac {e \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) b^{2} \EllipticF \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )}{d a \left (-a +\sqrt {a^{2}-b^{2}}\right ) \left (a +\sqrt {a^{2}-b^{2}}\right ) \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}}-\frac {e \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) b^{2} \EllipticPi \left (\sqrt {-\sin \left (d x +c \right )+1}, -\frac {a}{-a +\sqrt {a^{2}-b^{2}}}, \frac {\sqrt {2}}{2}\right )}{2 d a \left (-a +\sqrt {a^{2}-b^{2}}\right ) \left (a +\sqrt {a^{2}-b^{2}}\right ) \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}}-\frac {e \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) b^{2} \EllipticPi \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {a}{a +\sqrt {a^{2}-b^{2}}}, \frac {\sqrt {2}}{2}\right )}{2 d a \left (-a +\sqrt {a^{2}-b^{2}}\right ) \left (a +\sqrt {a^{2}-b^{2}}\right ) \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}} \]
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 \[ \int \frac {\sqrt {e \sin \left (d x + c\right )}}{b \sec \left (d x + c\right ) + a}\,{d x} \]
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 \[ \int \frac {\cos \left (c+d\,x\right )\,\sqrt {e\,\sin \left (c+d\,x\right )}}{b+a\,\cos \left (c+d\,x\right )} \,d x \]
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 \[ \int \frac {\sqrt {e \sin {\left (c + d x \right )}}}{a + b \sec {\left (c + d x \right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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