Optimal. Leaf size=370 \[ -\frac {b \text {ArcTan}\left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt [4]{a^2-b^2} \sqrt {e}}\right )}{\sqrt {a} \left (a^2-b^2\right )^{3/4} d \sqrt {e}}-\frac {b \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt [4]{a^2-b^2} \sqrt {e}}\right )}{\sqrt {a} \left (a^2-b^2\right )^{3/4} d \sqrt {e}}+\frac {2 F\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{a d \sqrt {e \sin (c+d x)}}+\frac {b^2 \Pi \left (\frac {2 a}{a-\sqrt {a^2-b^2}};\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{a \left (a^2-b^2-a \sqrt {a^2-b^2}\right ) d \sqrt {e \sin (c+d x)}}+\frac {b^2 \Pi \left (\frac {2 a}{a+\sqrt {a^2-b^2}};\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{a \left (a^2-b^2+a \sqrt {a^2-b^2}\right ) d \sqrt {e \sin (c+d x)}} \]
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Rubi [A]
time = 0.55, antiderivative size = 370, 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 = {3957, 2946,
2721, 2720, 2781, 2886, 2884, 335, 218, 214, 211} \begin {gather*} -\frac {b \text {ArcTan}\left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt {e} \sqrt [4]{a^2-b^2}}\right )}{\sqrt {a} d \sqrt {e} \left (a^2-b^2\right )^{3/4}}-\frac {b \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt {e} \sqrt [4]{a^2-b^2}}\right )}{\sqrt {a} d \sqrt {e} \left (a^2-b^2\right )^{3/4}}+\frac {b^2 \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 d \left (-a \sqrt {a^2-b^2}+a^2-b^2\right ) \sqrt {e \sin (c+d x)}}+\frac {b^2 \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 d \left (a \sqrt {a^2-b^2}+a^2-b^2\right ) \sqrt {e \sin (c+d x)}}+\frac {2 \sqrt {\sin (c+d x)} F\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{a d \sqrt {e \sin (c+d x)}} \end {gather*}
Antiderivative was successfully verified.
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Rule 211
Rule 214
Rule 218
Rule 335
Rule 2720
Rule 2721
Rule 2781
Rule 2884
Rule 2886
Rule 2946
Rule 3957
Rubi steps
\begin {align*} \int \frac {1}{(a+b \sec (c+d x)) \sqrt {e \sin (c+d x)}} \, dx &=-\int \frac {\cos (c+d x)}{(-b-a \cos (c+d x)) \sqrt {e \sin (c+d x)}} \, dx\\ &=\frac {\int \frac {1}{\sqrt {e \sin (c+d x)}} \, dx}{a}+\frac {b \int \frac {1}{(-b-a \cos (c+d x)) \sqrt {e \sin (c+d x)}} \, dx}{a}\\ &=\frac {b^2 \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {a^2-b^2}-a \sin (c+d x)\right )} \, dx}{2 a \sqrt {a^2-b^2}}+\frac {b^2 \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {a^2-b^2}+a \sin (c+d x)\right )} \, dx}{2 a \sqrt {a^2-b^2}}+\frac {(b e) \text {Subst}\left (\int \frac {1}{\sqrt {x} \left (\left (-a^2+b^2\right ) e^2+a^2 x^2\right )} \, dx,x,e \sin (c+d x)\right )}{d}+\frac {\sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)}} \, dx}{a \sqrt {e \sin (c+d x)}}\\ &=\frac {2 F\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{a d \sqrt {e \sin (c+d x)}}+\frac {(2 b e) \text {Subst}\left (\int \frac {1}{\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 \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 \sqrt {a^2-b^2} \sqrt {e \sin (c+d x)}}+\frac {\left (b^2 \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 \sqrt {a^2-b^2} \sqrt {e \sin (c+d x)}}\\ &=\frac {2 F\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{a d \sqrt {e \sin (c+d x)}}+\frac {b^2 \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 \left (a^2-b^2-a \sqrt {a^2-b^2}\right ) d \sqrt {e \sin (c+d x)}}+\frac {b^2 \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 \sqrt {a^2-b^2} \left (a+\sqrt {a^2-b^2}\right ) d \sqrt {e \sin (c+d x)}}-\frac {b \text {Subst}\left (\int \frac {1}{\sqrt {a^2-b^2} e-a x^2} \, dx,x,\sqrt {e \sin (c+d x)}\right )}{\sqrt {a^2-b^2} d}-\frac {b \text {Subst}\left (\int \frac {1}{\sqrt {a^2-b^2} e+a x^2} \, dx,x,\sqrt {e \sin (c+d x)}\right )}{\sqrt {a^2-b^2} d}\\ &=-\frac {b \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt [4]{a^2-b^2} \sqrt {e}}\right )}{\sqrt {a} \left (a^2-b^2\right )^{3/4} d \sqrt {e}}-\frac {b \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt [4]{a^2-b^2} \sqrt {e}}\right )}{\sqrt {a} \left (a^2-b^2\right )^{3/4} d \sqrt {e}}+\frac {2 F\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{a d \sqrt {e \sin (c+d x)}}+\frac {b^2 \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 \left (a^2-b^2-a \sqrt {a^2-b^2}\right ) d \sqrt {e \sin (c+d x)}}+\frac {b^2 \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 \sqrt {a^2-b^2} \left (a+\sqrt {a^2-b^2}\right ) d \sqrt {e \sin (c+d 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 = 14.00, size = 546, normalized size = 1.48 \begin {gather*} \frac {2 \left (b+a \sqrt {\cos ^2(c+d x)}\right ) \sqrt {\sin (c+d x)} \left (\frac {b \left (-2 \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt {a} \sqrt {\sin (c+d x)}}{\sqrt [4]{-a^2+b^2}}\right )+2 \text {ArcTan}\left (1+\frac {\sqrt {2} \sqrt {a} \sqrt {\sin (c+d 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 {\sin (c+d x)}+a \sin (c+d x)\right )+\log \left (\sqrt {-a^2+b^2}+\sqrt {2} \sqrt {a} \sqrt [4]{-a^2+b^2} \sqrt {\sin (c+d x)}+a \sin (c+d x)\right )\right )}{4 \sqrt {2} \sqrt {a} \left (-a^2+b^2\right )^{3/4}}-\frac {5 a \left (a^2-b^2\right ) F_1\left (\frac {1}{4};-\frac {1}{2},1;\frac {5}{4};\sin ^2(c+d x),\frac {a^2 \sin ^2(c+d x)}{a^2-b^2}\right ) \sqrt {\cos ^2(c+d x)} \sqrt {\sin (c+d x)}}{\left (-a^2+b^2+a^2 \sin ^2(c+d x)\right ) \left (5 \left (a^2-b^2\right ) F_1\left (\frac {1}{4};-\frac {1}{2},1;\frac {5}{4};\sin ^2(c+d x),\frac {a^2 \sin ^2(c+d x)}{a^2-b^2}\right )+2 \left (2 a^2 F_1\left (\frac {5}{4};-\frac {1}{2},2;\frac {9}{4};\sin ^2(c+d x),\frac {a^2 \sin ^2(c+d x)}{a^2-b^2}\right )+\left (-a^2+b^2\right ) F_1\left (\frac {5}{4};\frac {1}{2},1;\frac {9}{4};\sin ^2(c+d x),\frac {a^2 \sin ^2(c+d x)}{a^2-b^2}\right )\right ) \sin ^2(c+d x)\right )}\right )}{d (b+a \cos (c+d x)) \sqrt {e \sin (c+d x)}} \end {gather*}
Warning: Unable to verify antiderivative.
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Maple [A]
time = 0.25, size = 530, normalized size = 1.43
method | result | size |
default | \(\frac {\frac {b e \left (\frac {e^{2} \left (a^{2}-b^{2}\right )}{a^{2}}\right )^{\frac {1}{4}} \ln \left (\frac {\sqrt {e \sin \left (d x +c \right )}+\left (\frac {e^{2} \left (a^{2}-b^{2}\right )}{a^{2}}\right )^{\frac {1}{4}}}{\sqrt {e \sin \left (d x +c \right )}-\left (\frac {e^{2} \left (a^{2}-b^{2}\right )}{a^{2}}\right )^{\frac {1}{4}}}\right )}{-2 a^{2} e^{2}+2 b^{2} e^{2}}+\frac {b e \left (\frac {e^{2} \left (a^{2}-b^{2}\right )}{a^{2}}\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\left (\frac {e^{2} \left (a^{2}-b^{2}\right )}{a^{2}}\right )^{\frac {1}{4}}}\right )}{-a^{2} e^{2}+b^{2} e^{2}}-\frac {\sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) \left (2 \EllipticF \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) \left (a^{2}-b^{2}\right )^{\frac {3}{2}}-2 \EllipticF \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) \sqrt {a^{2}-b^{2}}\, a^{2}+\EllipticPi \left (\sqrt {-\sin \left (d x +c \right )+1}, -\frac {a}{\sqrt {a^{2}-b^{2}}-a}, \frac {\sqrt {2}}{2}\right ) \sqrt {a^{2}-b^{2}}\, b^{2}+\EllipticPi \left (\sqrt {-\sin \left (d x +c \right )+1}, -\frac {a}{\sqrt {a^{2}-b^{2}}-a}, \frac {\sqrt {2}}{2}\right ) a \,b^{2}+\EllipticPi \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {a}{a +\sqrt {a^{2}-b^{2}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {a^{2}-b^{2}}\, b^{2}-\EllipticPi \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {a}{a +\sqrt {a^{2}-b^{2}}}, \frac {\sqrt {2}}{2}\right ) a \,b^{2}\right )}{2 a \sqrt {a^{2}-b^{2}}\, \left (\sqrt {a^{2}-b^{2}}-a \right ) \left (a +\sqrt {a^{2}-b^{2}}\right ) \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}}}{d}\) | \(530\) |
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 {1}{\sqrt {e \sin {\left (c + d x \right )}} \left (a + b \sec {\left (c + d 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 {\cos \left (c+d\,x\right )}{\sqrt {e\,\sin \left (c+d\,x\right )}\,\left (b+a\,\cos \left (c+d\,x\right )\right )} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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