Integrand size = 43, antiderivative size = 229 \[ \int \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {64 a^3 (165 A+143 B+125 C) \tan (c+d x)}{3465 d \sqrt {a+a \sec (c+d x)}}+\frac {16 a^2 (165 A+143 B+125 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{3465 d}+\frac {2 a (165 A+143 B+125 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{1155 d}+\frac {2 (99 A-22 B+26 C) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{693 d}+\frac {2 C \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{11 d}+\frac {2 (11 B+5 C) (a+a \sec (c+d x))^{7/2} \tan (c+d x)}{99 a d} \]
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Time = 0.71 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.116, Rules used = {4173, 4095, 4086, 3878, 3877} \[ \int \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {64 a^3 (165 A+143 B+125 C) \tan (c+d x)}{3465 d \sqrt {a \sec (c+d x)+a}}+\frac {16 a^2 (165 A+143 B+125 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3465 d}+\frac {2 (99 A-22 B+26 C) \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{693 d}+\frac {2 a (165 A+143 B+125 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{1155 d}+\frac {2 (11 B+5 C) \tan (c+d x) (a \sec (c+d x)+a)^{7/2}}{99 a d}+\frac {2 C \tan (c+d x) \sec ^2(c+d x) (a \sec (c+d x)+a)^{5/2}}{11 d} \]
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Rule 3877
Rule 3878
Rule 4086
Rule 4095
Rule 4173
Rubi steps \begin{align*} \text {integral}& = \frac {2 C \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{11 d}+\frac {2 \int \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} \left (\frac {1}{2} a (11 A+4 C)+\frac {1}{2} a (11 B+5 C) \sec (c+d x)\right ) \, dx}{11 a} \\ & = \frac {2 C \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{11 d}+\frac {2 (11 B+5 C) (a+a \sec (c+d x))^{7/2} \tan (c+d x)}{99 a d}+\frac {4 \int \sec (c+d x) (a+a \sec (c+d x))^{5/2} \left (\frac {7}{4} a^2 (11 B+5 C)+\frac {1}{4} a^2 (99 A-22 B+26 C) \sec (c+d x)\right ) \, dx}{99 a^2} \\ & = \frac {2 (99 A-22 B+26 C) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{693 d}+\frac {2 C \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{11 d}+\frac {2 (11 B+5 C) (a+a \sec (c+d x))^{7/2} \tan (c+d x)}{99 a d}+\frac {1}{231} (165 A+143 B+125 C) \int \sec (c+d x) (a+a \sec (c+d x))^{5/2} \, dx \\ & = \frac {2 a (165 A+143 B+125 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{1155 d}+\frac {2 (99 A-22 B+26 C) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{693 d}+\frac {2 C \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{11 d}+\frac {2 (11 B+5 C) (a+a \sec (c+d x))^{7/2} \tan (c+d x)}{99 a d}+\frac {(8 a (165 A+143 B+125 C)) \int \sec (c+d x) (a+a \sec (c+d x))^{3/2} \, dx}{1155} \\ & = \frac {16 a^2 (165 A+143 B+125 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{3465 d}+\frac {2 a (165 A+143 B+125 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{1155 d}+\frac {2 (99 A-22 B+26 C) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{693 d}+\frac {2 C \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{11 d}+\frac {2 (11 B+5 C) (a+a \sec (c+d x))^{7/2} \tan (c+d x)}{99 a d}+\frac {\left (32 a^2 (165 A+143 B+125 C)\right ) \int \sec (c+d x) \sqrt {a+a \sec (c+d x)} \, dx}{3465} \\ & = \frac {64 a^3 (165 A+143 B+125 C) \tan (c+d x)}{3465 d \sqrt {a+a \sec (c+d x)}}+\frac {16 a^2 (165 A+143 B+125 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{3465 d}+\frac {2 a (165 A+143 B+125 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{1155 d}+\frac {2 (99 A-22 B+26 C) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{693 d}+\frac {2 C \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{11 d}+\frac {2 (11 B+5 C) (a+a \sec (c+d x))^{7/2} \tan (c+d x)}{99 a d} \\ \end{align*}
Time = 2.42 (sec) , antiderivative size = 184, normalized size of antiderivative = 0.80 \[ \int \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a^3 (13365 A+15356 B+18140 C+(49830 A+49654 B+50140 C) \cos (c+d x)+4 (4290 A+4642 B+4615 C) \cos (2 (c+d x))+22935 A \cos (3 (c+d x))+20878 B \cos (3 (c+d x))+18460 C \cos (3 (c+d x))+3795 A \cos (4 (c+d x))+3212 B \cos (4 (c+d x))+2840 C \cos (4 (c+d x))+3795 A \cos (5 (c+d x))+3212 B \cos (5 (c+d x))+2840 C \cos (5 (c+d x))) \sec ^5(c+d x) \tan (c+d x)}{13860 d \sqrt {a (1+\sec (c+d x))}} \]
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Time = 198.26 (sec) , antiderivative size = 199, normalized size of antiderivative = 0.87
method | result | size |
default | \(\frac {2 a^{2} \left (7590 A \cos \left (d x +c \right )^{5}+6424 B \cos \left (d x +c \right )^{5}+5680 C \cos \left (d x +c \right )^{5}+3795 A \cos \left (d x +c \right )^{4}+3212 B \cos \left (d x +c \right )^{4}+2840 C \cos \left (d x +c \right )^{4}+1980 A \cos \left (d x +c \right )^{3}+2409 B \cos \left (d x +c \right )^{3}+2130 C \cos \left (d x +c \right )^{3}+495 A \cos \left (d x +c \right )^{2}+1430 B \cos \left (d x +c \right )^{2}+1775 C \cos \left (d x +c \right )^{2}+385 B \cos \left (d x +c \right )+1120 C \cos \left (d x +c \right )+315 C \right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right ) \sec \left (d x +c \right )^{4}}{3465 d \left (\cos \left (d x +c \right )+1\right )}\) | \(199\) |
parts | \(\frac {2 A \,a^{2} \left (46 \cos \left (d x +c \right )^{3}+23 \cos \left (d x +c \right )^{2}+12 \cos \left (d x +c \right )+3\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right ) \sec \left (d x +c \right )^{2}}{21 d \left (\cos \left (d x +c \right )+1\right )}+\frac {2 B \,a^{2} \left (584 \cos \left (d x +c \right )^{4}+292 \cos \left (d x +c \right )^{3}+219 \cos \left (d x +c \right )^{2}+130 \cos \left (d x +c \right )+35\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right ) \sec \left (d x +c \right )^{3}}{315 d \left (\cos \left (d x +c \right )+1\right )}+\frac {2 C \,a^{2} \left (1136 \cos \left (d x +c \right )^{5}+568 \cos \left (d x +c \right )^{4}+426 \cos \left (d x +c \right )^{3}+355 \cos \left (d x +c \right )^{2}+224 \cos \left (d x +c \right )+63\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right ) \sec \left (d x +c \right )^{4}}{693 d \left (\cos \left (d x +c \right )+1\right )}\) | \(257\) |
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Time = 0.27 (sec) , antiderivative size = 168, normalized size of antiderivative = 0.73 \[ \int \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 \, {\left (2 \, {\left (3795 \, A + 3212 \, B + 2840 \, C\right )} a^{2} \cos \left (d x + c\right )^{5} + {\left (3795 \, A + 3212 \, B + 2840 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + 3 \, {\left (660 \, A + 803 \, B + 710 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 5 \, {\left (99 \, A + 286 \, B + 355 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 35 \, {\left (11 \, B + 32 \, C\right )} a^{2} \cos \left (d x + c\right ) + 315 \, C a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{3465 \, {\left (d \cos \left (d x + c\right )^{6} + d \cos \left (d x + c\right )^{5}\right )}} \]
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Timed out. \[ \int \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]
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Timed out. \[ \int \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]
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\[ \int \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sec \left (d x + c\right )^{2} \,d x } \]
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Time = 28.70 (sec) , antiderivative size = 1034, normalized size of antiderivative = 4.52 \[ \int \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Too large to display} \]
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