Integrand size = 34, antiderivative size = 446 \[ \int \frac {B \sec (c+d x)+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^{7/2}} \, dx=-\frac {2 \left (23 a^2 b B+9 b^3 B-3 a^3 C-29 a b^2 C\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{15 (a-b)^2 b^2 (a+b)^{5/2} d}+\frac {2 \left (3 a^2 (5 B+C)-8 a b (B+3 C)+b^2 (9 B+5 C)\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{15 b \sqrt {a+b} \left (a^2-b^2\right )^2 d}-\frac {2 (b B-a C) \tan (c+d x)}{5 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{5/2}}-\frac {2 \left (8 a b B-3 a^2 C-5 b^2 C\right ) \tan (c+d x)}{15 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^{3/2}}-\frac {2 \left (23 a^2 b B+9 b^3 B-3 a^3 C-29 a b^2 C\right ) \tan (c+d x)}{15 \left (a^2-b^2\right )^3 d \sqrt {a+b \sec (c+d x)}} \]
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Time = 0.95 (sec) , antiderivative size = 446, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.147, Rules used = {4145, 4143, 12, 3917, 4089} \[ \int \frac {B \sec (c+d x)+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^{7/2}} \, dx=\frac {2 \left (3 a^2 (5 B+C)-8 a b (B+3 C)+b^2 (9 B+5 C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{15 b d \sqrt {a+b} \left (a^2-b^2\right )^2}-\frac {2 \left (-3 a^2 C+8 a b B-5 b^2 C\right ) \tan (c+d x)}{15 d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))^{3/2}}-\frac {2 (b B-a C) \tan (c+d x)}{5 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{5/2}}-\frac {2 \left (-3 a^3 C+23 a^2 b B-29 a b^2 C+9 b^3 B\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{15 b^2 d (a-b)^2 (a+b)^{5/2}}-\frac {2 \left (-3 a^3 C+23 a^2 b B-29 a b^2 C+9 b^3 B\right ) \tan (c+d x)}{15 d \left (a^2-b^2\right )^3 \sqrt {a+b \sec (c+d x)}} \]
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Rule 12
Rule 3917
Rule 4089
Rule 4143
Rule 4145
Rubi steps \begin{align*} \text {integral}& = -\frac {2 (b B-a C) \tan (c+d x)}{5 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{5/2}}-\frac {2 \int \frac {-\frac {5}{2} a (a B-b C) \sec (c+d x)+\frac {3}{2} a (b B-a C) \sec ^2(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx}{5 a \left (a^2-b^2\right )} \\ & = -\frac {2 (b B-a C) \tan (c+d x)}{5 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{5/2}}-\frac {2 \left (8 a b B-3 a^2 C-5 b^2 C\right ) \tan (c+d x)}{15 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^{3/2}}+\frac {4 \int \frac {\frac {3}{4} a^2 \left (5 a^2 B+3 b^2 B-8 a b C\right ) \sec (c+d x)-\frac {1}{4} a^2 \left (8 a b B-3 a^2 C-5 b^2 C\right ) \sec ^2(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx}{15 a^2 \left (a^2-b^2\right )^2} \\ & = -\frac {2 (b B-a C) \tan (c+d x)}{5 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{5/2}}-\frac {2 \left (8 a b B-3 a^2 C-5 b^2 C\right ) \tan (c+d x)}{15 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^{3/2}}-\frac {2 \left (23 a^2 b B+9 b^3 B-3 a^3 C-29 a b^2 C\right ) \tan (c+d x)}{15 \left (a^2-b^2\right )^3 d \sqrt {a+b \sec (c+d x)}}-\frac {8 \int \frac {-\frac {1}{8} a^3 \left (15 a^3 B+17 a b^2 B-27 a^2 b C-5 b^3 C\right ) \sec (c+d x)-\frac {1}{8} a^3 \left (23 a^2 b B+9 b^3 B-3 a^3 C-29 a b^2 C\right ) \sec ^2(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{15 a^3 \left (a^2-b^2\right )^3} \\ & = -\frac {2 (b B-a C) \tan (c+d x)}{5 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{5/2}}-\frac {2 \left (8 a b B-3 a^2 C-5 b^2 C\right ) \tan (c+d x)}{15 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^{3/2}}-\frac {2 \left (23 a^2 b B+9 b^3 B-3 a^3 C-29 a b^2 C\right ) \tan (c+d x)}{15 \left (a^2-b^2\right )^3 d \sqrt {a+b \sec (c+d x)}}-\frac {8 \int \frac {\left (\frac {1}{8} a^3 \left (23 a^2 b B+9 b^3 B-3 a^3 C-29 a b^2 C\right )-\frac {1}{8} a^3 \left (15 a^3 B+17 a b^2 B-27 a^2 b C-5 b^3 C\right )\right ) \sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{15 a^3 \left (a^2-b^2\right )^3}+\frac {\left (23 a^2 b B+9 b^3 B-3 a^3 C-29 a b^2 C\right ) \int \frac {\sec (c+d x) (1+\sec (c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx}{15 \left (a^2-b^2\right )^3} \\ & = -\frac {2 \left (23 a^2 b B+9 b^3 B-3 a^3 C-29 a b^2 C\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{15 (a-b)^2 b^2 (a+b)^{5/2} d}-\frac {2 (b B-a C) \tan (c+d x)}{5 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{5/2}}-\frac {2 \left (8 a b B-3 a^2 C-5 b^2 C\right ) \tan (c+d x)}{15 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^{3/2}}-\frac {2 \left (23 a^2 b B+9 b^3 B-3 a^3 C-29 a b^2 C\right ) \tan (c+d x)}{15 \left (a^2-b^2\right )^3 d \sqrt {a+b \sec (c+d x)}}+\frac {\left (3 a^2 (5 B+C)-8 a b (B+3 C)+b^2 (9 B+5 C)\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{15 (a-b)^2 (a+b)^3} \\ & = -\frac {2 \left (23 a^2 b B+9 b^3 B-3 a^3 C-29 a b^2 C\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{15 (a-b)^2 b^2 (a+b)^{5/2} d}+\frac {2 \left (3 a^2 (5 B+C)-8 a b (B+3 C)+b^2 (9 B+5 C)\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{15 (a-b)^2 b (a+b)^{5/2} d}-\frac {2 (b B-a C) \tan (c+d x)}{5 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{5/2}}-\frac {2 \left (8 a b B-3 a^2 C-5 b^2 C\right ) \tan (c+d x)}{15 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^{3/2}}-\frac {2 \left (23 a^2 b B+9 b^3 B-3 a^3 C-29 a b^2 C\right ) \tan (c+d x)}{15 \left (a^2-b^2\right )^3 d \sqrt {a+b \sec (c+d x)}} \\ \end{align*}
Time = 22.17 (sec) , antiderivative size = 723, normalized size of antiderivative = 1.62 \[ \int \frac {B \sec (c+d x)+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^{7/2}} \, dx=\frac {(b+a \cos (c+d x))^4 \sec ^4(c+d x) \left (-\frac {2 \left (23 a^2 b B+9 b^3 B-3 a^3 C-29 a b^2 C\right ) \sin (c+d x)}{15 b \left (-a^2+b^2\right )^3}-\frac {2 \left (b^3 B \sin (c+d x)-a b^2 C \sin (c+d x)\right )}{5 a^2 \left (a^2-b^2\right ) (b+a \cos (c+d x))^3}-\frac {2 \left (-14 a^2 b^2 B \sin (c+d x)+6 b^4 B \sin (c+d x)+9 a^3 b C \sin (c+d x)-a b^3 C \sin (c+d x)\right )}{15 a^2 \left (a^2-b^2\right )^2 (b+a \cos (c+d x))^2}+\frac {2 \left (-34 a^4 b B \sin (c+d x)+5 a^2 b^3 B \sin (c+d x)-3 b^5 B \sin (c+d x)+9 a^5 C \sin (c+d x)+25 a^3 b^2 C \sin (c+d x)-2 a b^4 C \sin (c+d x)\right )}{15 a^2 \left (a^2-b^2\right )^3 (b+a \cos (c+d x))}\right )}{d (a+b \sec (c+d x))^{7/2}}-\frac {2 (b+a \cos (c+d x))^3 \sec ^{\frac {7}{2}}(c+d x) \sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)} \left (2 (a+b) \left (-23 a^2 b B-9 b^3 B+3 a^3 C+29 a b^2 C\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right )+2 b (a+b) \left (b^2 (9 B-5 C)+8 a b (B-3 C)+3 a^2 (5 B-C)\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right )+\left (-23 a^2 b B-9 b^3 B+3 a^3 C+29 a b^2 C\right ) \cos (c+d x) (b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{15 b \left (-a^2+b^2\right )^3 d \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )} (a+b \sec (c+d x))^{7/2}} \]
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Leaf count of result is larger than twice the leaf count of optimal. \(9223\) vs. \(2(412)=824\).
Time = 17.39 (sec) , antiderivative size = 9224, normalized size of antiderivative = 20.68
method | result | size |
parts | \(\text {Expression too large to display}\) | \(9224\) |
default | \(\text {Expression too large to display}\) | \(9320\) |
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\[ \int \frac {B \sec (c+d x)+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^{7/2}} \, dx=\int { \frac {C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right )}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {7}{2}}} \,d x } \]
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\[ \int \frac {B \sec (c+d x)+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^{7/2}} \, dx=\int \frac {\left (B + C \sec {\left (c + d x \right )}\right ) \sec {\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{\frac {7}{2}}}\, dx \]
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Timed out. \[ \int \frac {B \sec (c+d x)+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^{7/2}} \, dx=\text {Timed out} \]
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\[ \int \frac {B \sec (c+d x)+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^{7/2}} \, dx=\int { \frac {C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right )}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {7}{2}}} \,d x } \]
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Timed out. \[ \int \frac {B \sec (c+d x)+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^{7/2}} \, dx=\int \frac {\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{7/2}} \,d x \]
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