Optimal. Leaf size=194 \[ -\frac {2 a^2 e^{5/2} \tan ^{-1}\left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d}+\frac {2 a^2 e^{5/2} \tanh ^{-1}\left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d}-\frac {9 a^2 e^2 E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{5 d \sqrt {\sin (c+d x)}}-\frac {4 a^2 e (e \sin (c+d x))^{3/2}}{3 d}-\frac {2 a^2 e \cos (c+d x) (e \sin (c+d x))^{3/2}}{5 d}+\frac {a^2 e \sec (c+d x) (e \sin (c+d x))^{3/2}}{d} \]
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Rubi [A] time = 0.38, antiderivative size = 194, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 12, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.480, Rules used = {3872, 2873, 2635, 2640, 2639, 2564, 321, 329, 298, 203, 206, 2566} \[ -\frac {2 a^2 e^{5/2} \tan ^{-1}\left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d}+\frac {2 a^2 e^{5/2} \tanh ^{-1}\left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d}-\frac {9 a^2 e^2 E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{5 d \sqrt {\sin (c+d x)}}-\frac {4 a^2 e (e \sin (c+d x))^{3/2}}{3 d}-\frac {2 a^2 e \cos (c+d x) (e \sin (c+d x))^{3/2}}{5 d}+\frac {a^2 e \sec (c+d x) (e \sin (c+d x))^{3/2}}{d} \]
Antiderivative was successfully verified.
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Rule 203
Rule 206
Rule 298
Rule 321
Rule 329
Rule 2564
Rule 2566
Rule 2635
Rule 2639
Rule 2640
Rule 2873
Rule 3872
Rubi steps
\begin {align*} \int (a+a \sec (c+d x))^2 (e \sin (c+d x))^{5/2} \, dx &=\int (-a-a \cos (c+d x))^2 \sec ^2(c+d x) (e \sin (c+d x))^{5/2} \, dx\\ &=\int \left (a^2 (e \sin (c+d x))^{5/2}+2 a^2 \sec (c+d x) (e \sin (c+d x))^{5/2}+a^2 \sec ^2(c+d x) (e \sin (c+d x))^{5/2}\right ) \, dx\\ &=a^2 \int (e \sin (c+d x))^{5/2} \, dx+a^2 \int \sec ^2(c+d x) (e \sin (c+d x))^{5/2} \, dx+\left (2 a^2\right ) \int \sec (c+d x) (e \sin (c+d x))^{5/2} \, dx\\ &=-\frac {2 a^2 e \cos (c+d x) (e \sin (c+d x))^{3/2}}{5 d}+\frac {a^2 e \sec (c+d x) (e \sin (c+d x))^{3/2}}{d}+\frac {\left (2 a^2\right ) \operatorname {Subst}\left (\int \frac {x^{5/2}}{1-\frac {x^2}{e^2}} \, dx,x,e \sin (c+d x)\right )}{d e}+\frac {1}{5} \left (3 a^2 e^2\right ) \int \sqrt {e \sin (c+d x)} \, dx-\frac {1}{2} \left (3 a^2 e^2\right ) \int \sqrt {e \sin (c+d x)} \, dx\\ &=-\frac {4 a^2 e (e \sin (c+d x))^{3/2}}{3 d}-\frac {2 a^2 e \cos (c+d x) (e \sin (c+d x))^{3/2}}{5 d}+\frac {a^2 e \sec (c+d x) (e \sin (c+d x))^{3/2}}{d}+\frac {\left (2 a^2 e\right ) \operatorname {Subst}\left (\int \frac {\sqrt {x}}{1-\frac {x^2}{e^2}} \, dx,x,e \sin (c+d x)\right )}{d}+\frac {\left (3 a^2 e^2 \sqrt {e \sin (c+d x)}\right ) \int \sqrt {\sin (c+d x)} \, dx}{5 \sqrt {\sin (c+d x)}}-\frac {\left (3 a^2 e^2 \sqrt {e \sin (c+d x)}\right ) \int \sqrt {\sin (c+d x)} \, dx}{2 \sqrt {\sin (c+d x)}}\\ &=-\frac {9 a^2 e^2 E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{5 d \sqrt {\sin (c+d x)}}-\frac {4 a^2 e (e \sin (c+d x))^{3/2}}{3 d}-\frac {2 a^2 e \cos (c+d x) (e \sin (c+d x))^{3/2}}{5 d}+\frac {a^2 e \sec (c+d x) (e \sin (c+d x))^{3/2}}{d}+\frac {\left (4 a^2 e\right ) \operatorname {Subst}\left (\int \frac {x^2}{1-\frac {x^4}{e^2}} \, dx,x,\sqrt {e \sin (c+d x)}\right )}{d}\\ &=-\frac {9 a^2 e^2 E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{5 d \sqrt {\sin (c+d x)}}-\frac {4 a^2 e (e \sin (c+d x))^{3/2}}{3 d}-\frac {2 a^2 e \cos (c+d x) (e \sin (c+d x))^{3/2}}{5 d}+\frac {a^2 e \sec (c+d x) (e \sin (c+d x))^{3/2}}{d}+\frac {\left (2 a^2 e^3\right ) \operatorname {Subst}\left (\int \frac {1}{e-x^2} \, dx,x,\sqrt {e \sin (c+d x)}\right )}{d}-\frac {\left (2 a^2 e^3\right ) \operatorname {Subst}\left (\int \frac {1}{e+x^2} \, dx,x,\sqrt {e \sin (c+d x)}\right )}{d}\\ &=-\frac {2 a^2 e^{5/2} \tan ^{-1}\left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d}+\frac {2 a^2 e^{5/2} \tanh ^{-1}\left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d}-\frac {9 a^2 e^2 E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{5 d \sqrt {\sin (c+d x)}}-\frac {4 a^2 e (e \sin (c+d x))^{3/2}}{3 d}-\frac {2 a^2 e \cos (c+d x) (e \sin (c+d x))^{3/2}}{5 d}+\frac {a^2 e \sec (c+d x) (e \sin (c+d x))^{3/2}}{d}\\ \end {align*}
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Mathematica [C] time = 17.79, size = 205, normalized size = 1.06 \[ \frac {2 a^2 \cos ^4\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) (e \sin (c+d x))^{5/2} \sec ^4\left (\frac {1}{2} \sin ^{-1}(\sin (c+d x))\right ) \left (9 \sin ^{\frac {3}{2}}(c+d x) \sqrt {\cos ^2(c+d x)} \, _2F_1\left (\frac {3}{4},\frac {3}{2};\frac {7}{4};\sin ^2(c+d x)\right )+3 \sin ^{\frac {7}{2}}(c+d x)-9 \sin ^{\frac {3}{2}}(c+d x)-10 \sin ^{\frac {3}{2}}(c+d x) \sqrt {\cos ^2(c+d x)}-15 \sqrt {\cos ^2(c+d x)} \tan ^{-1}\left (\sqrt {\sin (c+d x)}\right )+15 \sqrt {\cos ^2(c+d x)} \tanh ^{-1}\left (\sqrt {\sin (c+d x)}\right )\right )}{15 d \sin ^{\frac {5}{2}}(c+d x)} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 1.91, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-{\left (a^{2} e^{2} \cos \left (d x + c\right )^{2} - a^{2} e^{2} + {\left (a^{2} e^{2} \cos \left (d x + c\right )^{2} - a^{2} e^{2}\right )} \sec \left (d x + c\right )^{2} + 2 \, {\left (a^{2} e^{2} \cos \left (d x + c\right )^{2} - a^{2} e^{2}\right )} \sec \left (d x + c\right )\right )} \sqrt {e \sin \left (d x + c\right )}, x\right ) \]
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 \[ \int {\left (a \sec \left (d x + c\right ) + a\right )}^{2} \left (e \sin \left (d x + c\right )\right )^{\frac {5}{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 6.14, size = 265, normalized size = 1.37 \[ -\frac {a^{2} \left (60 \arctan \left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\sqrt {e}}\right ) \sqrt {e \sin \left (d x +c \right )}\, e^{\frac {5}{2}} \cos \left (d x +c \right )-60 \sqrt {e \sin \left (d x +c \right )}\, \arctanh \left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\sqrt {e}}\right ) e^{\frac {5}{2}} \cos \left (d x +c \right )-54 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) \EllipticE \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) e^{3}+27 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) \EllipticF \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) e^{3}-12 e^{3} \left (\cos ^{4}\left (d x +c \right )\right )-40 e^{3} \left (\cos ^{3}\left (d x +c \right )\right )+42 e^{3} \left (\cos ^{2}\left (d x +c \right )\right )+40 e^{3} \cos \left (d x +c \right )-30 e^{3}\right )}{30 \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [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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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (e\,\sin \left (c+d\,x\right )\right )}^{5/2}\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^2 \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [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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