Optimal. Leaf size=272 \[ -\frac {a \sqrt {e} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} d}+\frac {a \sqrt {e} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}+1\right )}{\sqrt {2} d}+\frac {a \sqrt {e} \log \left (\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} d}-\frac {a \sqrt {e} \log \left (\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} d}+\frac {2 a \cos (c+d x) (e \tan (c+d x))^{3/2}}{d e}-\frac {2 a \cos (c+d x) E\left (\left .c+d x-\frac {\pi }{4}\right |2\right ) \sqrt {e \tan (c+d x)}}{d \sqrt {\sin (2 c+2 d x)}} \]
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Rubi [A] time = 0.24, antiderivative size = 272, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 13, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.565, Rules used = {3884, 3476, 329, 297, 1162, 617, 204, 1165, 628, 2613, 2615, 2572, 2639} \[ -\frac {a \sqrt {e} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} d}+\frac {a \sqrt {e} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}+1\right )}{\sqrt {2} d}+\frac {a \sqrt {e} \log \left (\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} d}-\frac {a \sqrt {e} \log \left (\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} d}+\frac {2 a \cos (c+d x) (e \tan (c+d x))^{3/2}}{d e}-\frac {2 a \cos (c+d x) E\left (\left .c+d x-\frac {\pi }{4}\right |2\right ) \sqrt {e \tan (c+d x)}}{d \sqrt {\sin (2 c+2 d x)}} \]
Antiderivative was successfully verified.
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Rule 204
Rule 297
Rule 329
Rule 617
Rule 628
Rule 1162
Rule 1165
Rule 2572
Rule 2613
Rule 2615
Rule 2639
Rule 3476
Rule 3884
Rubi steps
\begin {align*} \int (a+a \sec (c+d x)) \sqrt {e \tan (c+d x)} \, dx &=a \int \sqrt {e \tan (c+d x)} \, dx+a \int \sec (c+d x) \sqrt {e \tan (c+d x)} \, dx\\ &=\frac {2 a \cos (c+d x) (e \tan (c+d x))^{3/2}}{d e}-(2 a) \int \cos (c+d x) \sqrt {e \tan (c+d x)} \, dx+\frac {(a e) \operatorname {Subst}\left (\int \frac {\sqrt {x}}{e^2+x^2} \, dx,x,e \tan (c+d x)\right )}{d}\\ &=\frac {2 a \cos (c+d x) (e \tan (c+d x))^{3/2}}{d e}+\frac {(2 a e) \operatorname {Subst}\left (\int \frac {x^2}{e^2+x^4} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{d}-\frac {\left (2 a \sqrt {\cos (c+d x)} \sqrt {e \tan (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \sqrt {\sin (c+d x)} \, dx}{\sqrt {\sin (c+d x)}}\\ &=\frac {2 a \cos (c+d x) (e \tan (c+d x))^{3/2}}{d e}-\frac {(a e) \operatorname {Subst}\left (\int \frac {e-x^2}{e^2+x^4} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{d}+\frac {(a e) \operatorname {Subst}\left (\int \frac {e+x^2}{e^2+x^4} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{d}-\frac {\left (2 a \cos (c+d x) \sqrt {e \tan (c+d x)}\right ) \int \sqrt {\sin (2 c+2 d x)} \, dx}{\sqrt {\sin (2 c+2 d x)}}\\ &=-\frac {2 a \cos (c+d x) E\left (\left .c-\frac {\pi }{4}+d x\right |2\right ) \sqrt {e \tan (c+d x)}}{d \sqrt {\sin (2 c+2 d x)}}+\frac {2 a \cos (c+d x) (e \tan (c+d x))^{3/2}}{d e}+\frac {\left (a \sqrt {e}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {2} \sqrt {e}+2 x}{-e-\sqrt {2} \sqrt {e} x-x^2} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} d}+\frac {\left (a \sqrt {e}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {2} \sqrt {e}-2 x}{-e+\sqrt {2} \sqrt {e} x-x^2} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} d}+\frac {(a e) \operatorname {Subst}\left (\int \frac {1}{e-\sqrt {2} \sqrt {e} x+x^2} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{2 d}+\frac {(a e) \operatorname {Subst}\left (\int \frac {1}{e+\sqrt {2} \sqrt {e} x+x^2} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{2 d}\\ &=\frac {a \sqrt {e} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} d}-\frac {a \sqrt {e} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} d}-\frac {2 a \cos (c+d x) E\left (\left .c-\frac {\pi }{4}+d x\right |2\right ) \sqrt {e \tan (c+d x)}}{d \sqrt {\sin (2 c+2 d x)}}+\frac {2 a \cos (c+d x) (e \tan (c+d x))^{3/2}}{d e}+\frac {\left (a \sqrt {e}\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} d}-\frac {\left (a \sqrt {e}\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} d}\\ &=-\frac {a \sqrt {e} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} d}+\frac {a \sqrt {e} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} d}+\frac {a \sqrt {e} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} d}-\frac {a \sqrt {e} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} d}-\frac {2 a \cos (c+d x) E\left (\left .c-\frac {\pi }{4}+d x\right |2\right ) \sqrt {e \tan (c+d x)}}{d \sqrt {\sin (2 c+2 d x)}}+\frac {2 a \cos (c+d x) (e \tan (c+d x))^{3/2}}{d e}\\ \end {align*}
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Mathematica [C] time = 1.57, size = 182, normalized size = 0.67 \[ -\frac {a (\cos (c+d x)+1) \csc (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {e \tan (c+d x)} \left (8 \tan ^2(c+d x) \, _2F_1\left (\frac {3}{4},\frac {3}{2};\frac {7}{4};-\tan ^2(c+d x)\right )+3 \sqrt {\sec ^2(c+d x)} \left (-4 \sin ^2(c+d x)+\sqrt {\sin (2 (c+d x))} \sin ^{-1}(\cos (c+d x)-\sin (c+d x))+\sqrt {\sin (2 (c+d x))} \log \left (\sin (c+d x)+\sqrt {\sin (2 (c+d x))}+\cos (c+d x)\right )\right )\right )}{12 d \sqrt {\sec ^2(c+d x)}} \]
Antiderivative was successfully verified.
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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 \[ \int {\left (a \sec \left (d x + c\right ) + a\right )} \sqrt {e \tan \left (d x + c\right )}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 1.97, size = 1409, normalized size = 5.18 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.46, size = 154, normalized size = 0.57 \[ \frac {a e {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {e} + 2 \, \sqrt {e \tan \left (d x + c\right )}\right )}}{2 \, \sqrt {e}}\right )}{\sqrt {e}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {e} - 2 \, \sqrt {e \tan \left (d x + c\right )}\right )}}{2 \, \sqrt {e}}\right )}{\sqrt {e}} - \frac {\sqrt {2} \log \left (e \tan \left (d x + c\right ) + \sqrt {2} \sqrt {e \tan \left (d x + c\right )} \sqrt {e} + e\right )}{\sqrt {e}} + \frac {\sqrt {2} \log \left (e \tan \left (d x + c\right ) - \sqrt {2} \sqrt {e \tan \left (d x + c\right )} \sqrt {e} + e\right )}{\sqrt {e}}\right )}}{4 \, d} \]
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 \[ \int \sqrt {e\,\mathrm {tan}\left (c+d\,x\right )}\,\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right ) \,d x \]
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 \[ a \left (\int \sqrt {e \tan {\left (c + d x \right )}}\, dx + \int \sqrt {e \tan {\left (c + d x \right )}} \sec {\left (c + d x \right )}\, dx\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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