Integrand size = 43, antiderivative size = 447 \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=-\frac {\left (3 a^2 b B-2 b^3 B-a b^2 (A-4 C)-5 a^3 C\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{b^3 \left (a^2-b^2\right ) d}+\frac {\left (3 A b^2-3 a b B+5 a^2 C-2 b^2 C\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 b^2 \left (a^2-b^2\right ) d}-\frac {\left (3 A b^4+3 a^3 b B-5 a b^3 B-a^2 b^2 (A-7 C)-5 a^4 C\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{(a-b) b^3 (a+b)^2 d}+\frac {\left (3 a^2 b B-2 b^3 B-a b^2 (A-4 C)-5 a^3 C\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{b^3 \left (a^2-b^2\right ) d}+\frac {\left (3 A b^2-3 a b B+5 a^2 C-2 b^2 C\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d}-\frac {\left (A b^2-a (b B-a C)\right ) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))} \]
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Time = 1.55 (sec) , antiderivative size = 447, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.209, Rules used = {4183, 4187, 4191, 3934, 2884, 3872, 3856, 2719, 2720} \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=-\frac {\sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}+\frac {\sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \left (5 a^2 C-3 a b B+3 A b^2-2 b^2 C\right )}{3 b^2 d \left (a^2-b^2\right )}+\frac {\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (5 a^2 C-3 a b B+3 A b^2-2 b^2 C\right )}{3 b^2 d \left (a^2-b^2\right )}+\frac {\sin (c+d x) \sqrt {\sec (c+d x)} \left (-5 a^3 C+3 a^2 b B-a b^2 (A-4 C)-2 b^3 B\right )}{b^3 d \left (a^2-b^2\right )}-\frac {\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (-5 a^3 C+3 a^2 b B-a b^2 (A-4 C)-2 b^3 B\right )}{b^3 d \left (a^2-b^2\right )}-\frac {\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-5 a^4 C+3 a^3 b B-a^2 b^2 (A-7 C)-5 a b^3 B+3 A b^4\right ) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{b^3 d (a-b) (a+b)^2} \]
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Rule 2719
Rule 2720
Rule 2884
Rule 3856
Rule 3872
Rule 3934
Rule 4183
Rule 4187
Rule 4191
Rubi steps \begin{align*} \text {integral}& = -\frac {\left (A b^2-a (b B-a C)\right ) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac {\int \frac {\sec ^{\frac {3}{2}}(c+d x) \left (\frac {3}{2} \left (A b^2-a (b B-a C)\right )+b (b B-a (A+C)) \sec (c+d x)-\frac {1}{2} \left (3 A b^2-3 a b B+5 a^2 C-2 b^2 C\right ) \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{b \left (a^2-b^2\right )} \\ & = \frac {\left (3 A b^2-3 a b B+5 a^2 C-2 b^2 C\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d}-\frac {\left (A b^2-a (b B-a C)\right ) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac {2 \int \frac {\sqrt {\sec (c+d x)} \left (-\frac {1}{4} a \left (3 A b^2-3 a b B+5 a^2 C-2 b^2 C\right )+\frac {1}{2} b \left (3 A b^2-3 a b B+2 a^2 C+b^2 C\right ) \sec (c+d x)-\frac {3}{4} \left (3 a^2 b B-2 b^3 B-a b^2 (A-4 C)-5 a^3 C\right ) \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{3 b^2 \left (a^2-b^2\right )} \\ & = \frac {\left (3 a^2 b B-2 b^3 B-a b^2 (A-4 C)-5 a^3 C\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{b^3 \left (a^2-b^2\right ) d}+\frac {\left (3 A b^2-3 a b B+5 a^2 C-2 b^2 C\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d}-\frac {\left (A b^2-a (b B-a C)\right ) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac {4 \int \frac {\frac {3}{8} a \left (3 a^2 b B-2 b^3 B-a b^2 (A-4 C)-5 a^3 C\right )+\frac {1}{4} b \left (6 a^2 b B-3 b^3 B-a b^2 (3 A-7 C)-10 a^3 C\right ) \sec (c+d x)+\frac {1}{8} \left (9 a^3 b B-12 a b^3 B-a^2 b^2 (3 A-16 C)-15 a^4 C+2 b^4 (3 A+C)\right ) \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (a+b \sec (c+d x))} \, dx}{3 b^3 \left (a^2-b^2\right )} \\ & = \frac {\left (3 a^2 b B-2 b^3 B-a b^2 (A-4 C)-5 a^3 C\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{b^3 \left (a^2-b^2\right ) d}+\frac {\left (3 A b^2-3 a b B+5 a^2 C-2 b^2 C\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d}-\frac {\left (A b^2-a (b B-a C)\right ) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac {4 \int \frac {\frac {3}{8} a^2 \left (3 a^2 b B-2 b^3 B-a b^2 (A-4 C)-5 a^3 C\right )-\left (-\frac {1}{4} a b \left (6 a^2 b B-3 b^3 B-a b^2 (3 A-7 C)-10 a^3 C\right )+\frac {3}{8} a b \left (3 a^2 b B-2 b^3 B-a b^2 (A-4 C)-5 a^3 C\right )\right ) \sec (c+d x)}{\sqrt {\sec (c+d x)}} \, dx}{3 a^2 b^3 \left (a^2-b^2\right )}-\frac {\left (3 A b^4+3 a^3 b B-5 a b^3 B-a^2 b^2 (A-7 C)-5 a^4 C\right ) \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{a+b \sec (c+d x)} \, dx}{2 b^3 \left (a^2-b^2\right )} \\ & = \frac {\left (3 a^2 b B-2 b^3 B-a b^2 (A-4 C)-5 a^3 C\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{b^3 \left (a^2-b^2\right ) d}+\frac {\left (3 A b^2-3 a b B+5 a^2 C-2 b^2 C\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d}-\frac {\left (A b^2-a (b B-a C)\right ) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac {\left (3 a^2 b B-2 b^3 B-a b^2 (A-4 C)-5 a^3 C\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx}{2 b^3 \left (a^2-b^2\right )}+\frac {\left (3 A b^2-3 a b B+5 a^2 C-2 b^2 C\right ) \int \sqrt {\sec (c+d x)} \, dx}{6 b^2 \left (a^2-b^2\right )}-\frac {\left (\left (3 A b^4+3 a^3 b B-5 a b^3 B-a^2 b^2 (A-7 C)-5 a^4 C\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{2 b^3 \left (a^2-b^2\right )} \\ & = -\frac {\left (3 A b^4+3 a^3 b B-5 a b^3 B-a^2 b^2 (A-7 C)-5 a^4 C\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{(a-b) b^3 (a+b)^2 d}+\frac {\left (3 a^2 b B-2 b^3 B-a b^2 (A-4 C)-5 a^3 C\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{b^3 \left (a^2-b^2\right ) d}+\frac {\left (3 A b^2-3 a b B+5 a^2 C-2 b^2 C\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d}-\frac {\left (A b^2-a (b B-a C)\right ) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac {\left (\left (3 a^2 b B-2 b^3 B-a b^2 (A-4 C)-5 a^3 C\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{2 b^3 \left (a^2-b^2\right )}+\frac {\left (\left (3 A b^2-3 a b B+5 a^2 C-2 b^2 C\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{6 b^2 \left (a^2-b^2\right )} \\ & = -\frac {\left (3 a^2 b B-2 b^3 B-a b^2 (A-4 C)-5 a^3 C\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{b^3 \left (a^2-b^2\right ) d}+\frac {\left (3 A b^2-3 a b B+5 a^2 C-2 b^2 C\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 b^2 \left (a^2-b^2\right ) d}-\frac {\left (3 A b^4+3 a^3 b B-5 a b^3 B-a^2 b^2 (A-7 C)-5 a^4 C\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{(a-b) b^3 (a+b)^2 d}+\frac {\left (3 a^2 b B-2 b^3 B-a b^2 (A-4 C)-5 a^3 C\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{b^3 \left (a^2-b^2\right ) d}+\frac {\left (3 A b^2-3 a b B+5 a^2 C-2 b^2 C\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d}-\frac {\left (A b^2-a (b B-a C)\right ) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))} \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(926\) vs. \(2(447)=894\).
Time = 12.81 (sec) , antiderivative size = 926, normalized size of antiderivative = 2.07 \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=\frac {(b+a \cos (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (\frac {2 \left (9 a^2 A b^2-12 A b^4-27 a^3 b B+30 a b^3 B+45 a^4 C-44 a^2 b^2 C-4 b^4 C\right ) \cos ^2(c+d x) \left (\operatorname {EllipticF}\left (\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right )-\operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right )\right ) (a+b \sec (c+d x)) \sqrt {1-\sec ^2(c+d x)} \sin (c+d x)}{b (b+a \cos (c+d x)) \left (1-\cos ^2(c+d x)\right )}+\frac {2 \left (12 a A b^3-24 a^2 b^2 B+12 b^4 B+40 a^3 b C-28 a b^3 C\right ) \cos ^2(c+d x) \operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) (a+b \sec (c+d x)) \sqrt {1-\sec ^2(c+d x)} \sin (c+d x)}{a (b+a \cos (c+d x)) \left (1-\cos ^2(c+d x)\right )}+\frac {\left (3 a^2 A b^2-9 a^3 b B+6 a b^3 B+15 a^4 C-12 a^2 b^2 C\right ) \cos (2 (c+d x)) (a+b \sec (c+d x)) \left (-4 a b+4 a b \sec ^2(c+d x)-4 a b E\left (\left .\arcsin \left (\sqrt {\sec (c+d x)}\right )\right |-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}-2 a (a-2 b) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}+2 a^2 \operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}-4 b^2 \operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}\right ) \sin (c+d x)}{a^2 b (b+a \cos (c+d x)) \left (1-\cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)} \left (2-\sec ^2(c+d x)\right )}\right )}{6 (a-b) b^3 (a+b) d (A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 d x)) (a+b \sec (c+d x))^2}+\frac {(b+a \cos (c+d x))^2 \sqrt {\sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (\frac {2 \left (a A b^2-3 a^2 b B+2 b^3 B+5 a^3 C-4 a b^2 C\right ) \sin (c+d x)}{b^3 \left (-a^2+b^2\right )}-\frac {2 \left (a A b^2 \sin (c+d x)-a^2 b B \sin (c+d x)+a^3 C \sin (c+d x)\right )}{b^2 \left (-a^2+b^2\right ) (b+a \cos (c+d x))}+\frac {4 C \tan (c+d x)}{3 b^2}\right )}{d (A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 d x)) (a+b \sec (c+d x))^2} \]
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Time = 32.81 (sec) , antiderivative size = 1004, normalized size of antiderivative = 2.25
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Timed out. \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=\text {Timed out} \]
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Timed out. \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=\text {Timed out} \]
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Timed out. \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=\text {Timed out} \]
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\[ \int \frac {\sec ^{\frac {5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{\frac {5}{2}}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
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Timed out. \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=\int \frac {{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^2} \,d x \]
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