Integrand size = 25, antiderivative size = 139 \[ \int \frac {(a+a \sec (c+d x))^2}{\sqrt {e \sin (c+d x)}} \, dx=\frac {2 a^2 \arctan \left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d \sqrt {e}}+\frac {2 a^2 \text {arctanh}\left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d \sqrt {e}}+\frac {3 a^2 \operatorname {EllipticF}\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{d \sqrt {e \sin (c+d x)}}+\frac {a^2 \sec (c+d x) \sqrt {e \sin (c+d x)}}{d e} \]
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Time = 0.53 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3957, 2952, 2721, 2720, 2644, 335, 218, 212, 209, 2651} \[ \int \frac {(a+a \sec (c+d x))^2}{\sqrt {e \sin (c+d x)}} \, dx=\frac {2 a^2 \arctan \left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d \sqrt {e}}+\frac {2 a^2 \text {arctanh}\left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d \sqrt {e}}+\frac {a^2 \sec (c+d x) \sqrt {e \sin (c+d x)}}{d e}+\frac {3 a^2 \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{d \sqrt {e \sin (c+d x)}} \]
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Rule 209
Rule 212
Rule 218
Rule 335
Rule 2644
Rule 2651
Rule 2720
Rule 2721
Rule 2952
Rule 3957
Rubi steps \begin{align*} \text {integral}& = \int \frac {(-a-a \cos (c+d x))^2 \sec ^2(c+d x)}{\sqrt {e \sin (c+d x)}} \, dx \\ & = \int \left (\frac {a^2}{\sqrt {e \sin (c+d x)}}+\frac {2 a^2 \sec (c+d x)}{\sqrt {e \sin (c+d x)}}+\frac {a^2 \sec ^2(c+d x)}{\sqrt {e \sin (c+d x)}}\right ) \, dx \\ & = a^2 \int \frac {1}{\sqrt {e \sin (c+d x)}} \, dx+a^2 \int \frac {\sec ^2(c+d x)}{\sqrt {e \sin (c+d x)}} \, dx+\left (2 a^2\right ) \int \frac {\sec (c+d x)}{\sqrt {e \sin (c+d x)}} \, dx \\ & = \frac {a^2 \sec (c+d x) \sqrt {e \sin (c+d x)}}{d e}+\frac {1}{2} a^2 \int \frac {1}{\sqrt {e \sin (c+d x)}} \, dx+\frac {\left (2 a^2\right ) \text {Subst}\left (\int \frac {1}{\sqrt {x} \left (1-\frac {x^2}{e^2}\right )} \, dx,x,e \sin (c+d x)\right )}{d e}+\frac {\left (a^2 \sqrt {\sin (c+d x)}\right ) \int \frac {1}{\sqrt {\sin (c+d x)}} \, dx}{\sqrt {e \sin (c+d x)}} \\ & = \frac {2 a^2 \operatorname {EllipticF}\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{d \sqrt {e \sin (c+d x)}}+\frac {a^2 \sec (c+d x) \sqrt {e \sin (c+d x)}}{d e}+\frac {\left (4 a^2\right ) \text {Subst}\left (\int \frac {1}{1-\frac {x^4}{e^2}} \, dx,x,\sqrt {e \sin (c+d x)}\right )}{d e}+\frac {\left (a^2 \sqrt {\sin (c+d x)}\right ) \int \frac {1}{\sqrt {\sin (c+d x)}} \, dx}{2 \sqrt {e \sin (c+d x)}} \\ & = \frac {3 a^2 \operatorname {EllipticF}\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{d \sqrt {e \sin (c+d x)}}+\frac {a^2 \sec (c+d x) \sqrt {e \sin (c+d x)}}{d e}+\frac {\left (2 a^2\right ) \text {Subst}\left (\int \frac {1}{e-x^2} \, dx,x,\sqrt {e \sin (c+d x)}\right )}{d}+\frac {\left (2 a^2\right ) \text {Subst}\left (\int \frac {1}{e+x^2} \, dx,x,\sqrt {e \sin (c+d x)}\right )}{d} \\ & = \frac {2 a^2 \arctan \left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d \sqrt {e}}+\frac {2 a^2 \text {arctanh}\left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d \sqrt {e}}+\frac {3 a^2 \operatorname {EllipticF}\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{d \sqrt {e \sin (c+d x)}}+\frac {a^2 \sec (c+d x) \sqrt {e \sin (c+d x)}}{d e} \\ \end{align*}
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 19.81 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.18 \[ \int \frac {(a+a \sec (c+d x))^2}{\sqrt {e \sin (c+d x)}} \, dx=\frac {a^2 \cos ^4\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) \sec ^4\left (\frac {1}{2} \arcsin (\sin (c+d x))\right ) \left (2 \arctan \left (\sqrt {\sin (c+d x)}\right ) \sqrt {\cos ^2(c+d x)}+2 \text {arctanh}\left (\sqrt {\sin (c+d x)}\right ) \sqrt {\cos ^2(c+d x)}+\sqrt {\sin (c+d x)}+3 \sqrt {\cos ^2(c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},\sin ^2(c+d x)\right ) \sqrt {\sin (c+d x)}\right ) \sqrt {\sin (c+d x)}}{d \sqrt {e \sin (c+d x)}} \]
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Time = 11.95 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.17
method | result | size |
default | \(\frac {a^{2} \left (-3 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sqrt {\sin \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) \sqrt {e}+4 \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, \arctan \left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\sqrt {e}}\right )+4 \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, \operatorname {arctanh}\left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\sqrt {e}}\right )+2 \sqrt {e}\, \sin \left (d x +c \right )\right )}{2 \sqrt {e}\, \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d}\) | \(163\) |
parts | \(-\frac {a^{2} \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sqrt {\sin \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )}{\cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d}-\frac {a^{2} \sqrt {\cos \left (d x +c \right )^{2} e \sin \left (d x +c \right )}\, \left (\sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sqrt {\sin \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-2 \sin \left (d x +c \right )\right )}{2 \sqrt {-e \sin \left (d x +c \right ) \left (\sin \left (d x +c \right )-1\right ) \left (1+\sin \left (d x +c \right )\right )}\, \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d}+\frac {2 a^{2} \left (\arctan \left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\sqrt {e}}\right )+\operatorname {arctanh}\left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\sqrt {e}}\right )\right )}{\sqrt {e}\, d}\) | \(251\) |
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Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.19 (sec) , antiderivative size = 660, normalized size of antiderivative = 4.75 \[ \int \frac {(a+a \sec (c+d x))^2}{\sqrt {e \sin (c+d x)}} \, dx=\left [\frac {6 \, \sqrt {2} a^{2} \sqrt {-i \, e} \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 6 \, \sqrt {2} a^{2} \sqrt {i \, e} \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 2 \, a^{2} \sqrt {-e} \arctan \left (\frac {{\left (\cos \left (d x + c\right )^{2} - 6 \, \sin \left (d x + c\right ) - 2\right )} \sqrt {e \sin \left (d x + c\right )} \sqrt {-e}}{4 \, {\left (e \cos \left (d x + c\right )^{2} - e \sin \left (d x + c\right ) - e\right )}}\right ) \cos \left (d x + c\right ) - a^{2} \sqrt {-e} \cos \left (d x + c\right ) \log \left (\frac {e \cos \left (d x + c\right )^{4} - 72 \, e \cos \left (d x + c\right )^{2} - 8 \, {\left (7 \, \cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right )^{2} - 8\right )} \sin \left (d x + c\right ) - 8\right )} \sqrt {e \sin \left (d x + c\right )} \sqrt {-e} + 28 \, {\left (e \cos \left (d x + c\right )^{2} - 2 \, e\right )} \sin \left (d x + c\right ) + 72 \, e}{\cos \left (d x + c\right )^{4} - 8 \, \cos \left (d x + c\right )^{2} - 4 \, {\left (\cos \left (d x + c\right )^{2} - 2\right )} \sin \left (d x + c\right ) + 8}\right ) + 4 \, \sqrt {e \sin \left (d x + c\right )} a^{2}}{4 \, d e \cos \left (d x + c\right )}, \frac {6 \, \sqrt {2} a^{2} \sqrt {-i \, e} \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 6 \, \sqrt {2} a^{2} \sqrt {i \, e} \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 2 \, a^{2} \sqrt {e} \arctan \left (\frac {{\left (\cos \left (d x + c\right )^{2} + 6 \, \sin \left (d x + c\right ) - 2\right )} \sqrt {e \sin \left (d x + c\right )} \sqrt {e}}{4 \, {\left (e \cos \left (d x + c\right )^{2} + e \sin \left (d x + c\right ) - e\right )}}\right ) \cos \left (d x + c\right ) + a^{2} \sqrt {e} \cos \left (d x + c\right ) \log \left (\frac {e \cos \left (d x + c\right )^{4} - 72 \, e \cos \left (d x + c\right )^{2} - 8 \, {\left (7 \, \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 8\right )} \sin \left (d x + c\right ) - 8\right )} \sqrt {e \sin \left (d x + c\right )} \sqrt {e} - 28 \, {\left (e \cos \left (d x + c\right )^{2} - 2 \, e\right )} \sin \left (d x + c\right ) + 72 \, e}{\cos \left (d x + c\right )^{4} - 8 \, \cos \left (d x + c\right )^{2} + 4 \, {\left (\cos \left (d x + c\right )^{2} - 2\right )} \sin \left (d x + c\right ) + 8}\right ) + 4 \, \sqrt {e \sin \left (d x + c\right )} a^{2}}{4 \, d e \cos \left (d x + c\right )}\right ] \]
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\[ \int \frac {(a+a \sec (c+d x))^2}{\sqrt {e \sin (c+d x)}} \, dx=a^{2} \left (\int \frac {1}{\sqrt {e \sin {\left (c + d x \right )}}}\, dx + \int \frac {2 \sec {\left (c + d x \right )}}{\sqrt {e \sin {\left (c + d x \right )}}}\, dx + \int \frac {\sec ^{2}{\left (c + d x \right )}}{\sqrt {e \sin {\left (c + d x \right )}}}\, dx\right ) \]
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\[ \int \frac {(a+a \sec (c+d x))^2}{\sqrt {e \sin (c+d x)}} \, dx=\int { \frac {{\left (a \sec \left (d x + c\right ) + a\right )}^{2}}{\sqrt {e \sin \left (d x + c\right )}} \,d x } \]
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\[ \int \frac {(a+a \sec (c+d x))^2}{\sqrt {e \sin (c+d x)}} \, dx=\int { \frac {{\left (a \sec \left (d x + c\right ) + a\right )}^{2}}{\sqrt {e \sin \left (d x + c\right )}} \,d x } \]
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Timed out. \[ \int \frac {(a+a \sec (c+d x))^2}{\sqrt {e \sin (c+d x)}} \, dx=\int \frac {{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^2}{\sqrt {e\,\sin \left (c+d\,x\right )}} \,d x \]
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