Integrand size = 25, antiderivative size = 495 \[ \int \frac {A+B \sec (c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx=\frac {2 \left (7 a^2 A b-3 A b^3-4 a^3 B\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 a^2 (a-b) b (a+b)^{3/2} d}-\frac {2 \left (6 a^2 A b-a A b^2-3 A b^3-3 a^3 B+a^2 b B\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 a^2 (a-b) b (a+b)^{3/2} d}-\frac {2 A \sqrt {a+b} \cot (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{a^3 d}+\frac {2 b (A b-a B) \tan (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}+\frac {2 b \left (7 a^2 A b-3 A b^3-4 a^3 B\right ) \tan (c+d x)}{3 a^2 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}} \]
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Time = 0.92 (sec) , antiderivative size = 495, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {4008, 4145, 4143, 4006, 3869, 3917, 4089} \[ \int \frac {A+B \sec (c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx=-\frac {2 A \sqrt {a+b} \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{a^3 d}+\frac {2 b (A b-a B) \tan (c+d x)}{3 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}+\frac {2 \left (-4 a^3 B+7 a^2 A b-3 A b^3\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{3 a^2 b d (a-b) (a+b)^{3/2}}-\frac {2 \left (-3 a^3 B+6 a^2 A b+a^2 b B-a A b^2-3 A b^3\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{3 a^2 b d (a-b) (a+b)^{3/2}}+\frac {2 b \left (-4 a^3 B+7 a^2 A b-3 A b^3\right ) \tan (c+d x)}{3 a^2 d \left (a^2-b^2\right )^2 \sqrt {a+b \sec (c+d x)}} \]
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Rule 3869
Rule 3917
Rule 4006
Rule 4008
Rule 4089
Rule 4143
Rule 4145
Rubi steps \begin{align*} \text {integral}& = \frac {2 b (A b-a B) \tan (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {2 \int \frac {-\frac {3}{2} A \left (a^2-b^2\right )+\frac {3}{2} a (A b-a B) \sec (c+d x)-\frac {1}{2} b (A b-a B) \sec ^2(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx}{3 a \left (a^2-b^2\right )} \\ & = \frac {2 b (A b-a B) \tan (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}+\frac {2 b \left (7 a^2 A b-3 A b^3-4 a^3 B\right ) \tan (c+d x)}{3 a^2 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}}+\frac {4 \int \frac {\frac {3}{4} A \left (a^2-b^2\right )^2-\frac {1}{4} a \left (6 a^2 A b-2 A b^3-3 a^3 B-a b^2 B\right ) \sec (c+d x)-\frac {1}{4} b \left (7 a^2 A b-3 A b^3-4 a^3 B\right ) \sec ^2(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{3 a^2 \left (a^2-b^2\right )^2} \\ & = \frac {2 b (A b-a B) \tan (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}+\frac {2 b \left (7 a^2 A b-3 A b^3-4 a^3 B\right ) \tan (c+d x)}{3 a^2 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}}+\frac {4 \int \frac {\frac {3}{4} A \left (a^2-b^2\right )^2+\left (\frac {1}{4} b \left (7 a^2 A b-3 A b^3-4 a^3 B\right )-\frac {1}{4} a \left (6 a^2 A b-2 A b^3-3 a^3 B-a b^2 B\right )\right ) \sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{3 a^2 \left (a^2-b^2\right )^2}-\frac {\left (b \left (7 a^2 A b-3 A b^3-4 a^3 B\right )\right ) \int \frac {\sec (c+d x) (1+\sec (c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx}{3 a^2 \left (a^2-b^2\right )^2} \\ & = \frac {2 \left (7 a^2 A b-3 A b^3-4 a^3 B\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}}}{3 a^2 (a-b) b (a+b)^{3/2} d}+\frac {2 b (A b-a B) \tan (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}+\frac {2 b \left (7 a^2 A b-3 A b^3-4 a^3 B\right ) \tan (c+d x)}{3 a^2 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}}+\frac {A \int \frac {1}{\sqrt {a+b \sec (c+d x)}} \, dx}{a^2}+\frac {\left (a A b^2+3 A b^3+3 a^3 B-a^2 b (6 A+B)\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{3 a^2 (a-b) (a+b)^2} \\ & = \frac {2 \left (7 a^2 A b-3 A b^3-4 a^3 B\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}}}{3 a^2 (a-b) b (a+b)^{3/2} d}+\frac {2 \left (a A b^2+3 A b^3+3 a^3 B-a^2 b (6 A+B)\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}}}{3 a^2 (a-b) b (a+b)^{3/2} d}-\frac {2 A \sqrt {a+b} \cot (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{a},\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}}}{a^3 d}+\frac {2 b (A b-a B) \tan (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}+\frac {2 b \left (7 a^2 A b-3 A b^3-4 a^3 B\right ) \tan (c+d x)}{3 a^2 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}} \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(1589\) vs. \(2(495)=990\).
Time = 17.19 (sec) , antiderivative size = 1589, normalized size of antiderivative = 3.21 \[ \int \frac {A+B \sec (c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx=\frac {(b+a \cos (c+d x))^3 \sec ^2(c+d x) (A+B \sec (c+d x)) \left (\frac {2 \left (-7 a^2 A b+3 A b^3+4 a^3 B\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right )^2}-\frac {2 \left (A b^3 \sin (c+d x)-a b^2 B \sin (c+d x)\right )}{3 a^2 \left (a^2-b^2\right ) (b+a \cos (c+d x))^2}-\frac {2 \left (-8 a^2 A b^2 \sin (c+d x)+4 A b^4 \sin (c+d x)+5 a^3 b B \sin (c+d x)-a b^3 B \sin (c+d x)\right )}{3 a^2 \left (a^2-b^2\right )^2 (b+a \cos (c+d x))}\right )}{d (B+A \cos (c+d x)) (a+b \sec (c+d x))^{5/2}}-\frac {2 (b+a \cos (c+d x))^{5/2} \sec ^{\frac {3}{2}}(c+d x) (A+B \sec (c+d x)) \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{1+\tan ^2\left (\frac {1}{2} (c+d x)\right )}} \left (-7 a^3 A b \tan \left (\frac {1}{2} (c+d x)\right )-7 a^2 A b^2 \tan \left (\frac {1}{2} (c+d x)\right )+3 a A b^3 \tan \left (\frac {1}{2} (c+d x)\right )+3 A b^4 \tan \left (\frac {1}{2} (c+d x)\right )+4 a^4 B \tan \left (\frac {1}{2} (c+d x)\right )+4 a^3 b B \tan \left (\frac {1}{2} (c+d x)\right )+14 a^3 A b \tan ^3\left (\frac {1}{2} (c+d x)\right )-6 a A b^3 \tan ^3\left (\frac {1}{2} (c+d x)\right )-8 a^4 B \tan ^3\left (\frac {1}{2} (c+d x)\right )-7 a^3 A b \tan ^5\left (\frac {1}{2} (c+d x)\right )+7 a^2 A b^2 \tan ^5\left (\frac {1}{2} (c+d x)\right )+3 a A b^3 \tan ^5\left (\frac {1}{2} (c+d x)\right )-3 A b^4 \tan ^5\left (\frac {1}{2} (c+d x)\right )+4 a^4 B \tan ^5\left (\frac {1}{2} (c+d x)\right )-4 a^3 b B \tan ^5\left (\frac {1}{2} (c+d x)\right )-6 a^4 A \operatorname {EllipticPi}\left (-1,\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}+12 a^2 A b^2 \operatorname {EllipticPi}\left (-1,\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}-6 A b^4 \operatorname {EllipticPi}\left (-1,\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}-6 a^4 A \operatorname {EllipticPi}\left (-1,\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}+12 a^2 A b^2 \operatorname {EllipticPi}\left (-1,\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}-6 A b^4 \operatorname {EllipticPi}\left (-1,\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}+(a+b) \left (-7 a^2 A b+3 A b^3+4 a^3 B\right ) E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}+a (a+b) \left (-2 A b^2+3 a^2 (A-B)+a b (3 A-B)\right ) \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}\right )}{3 a^2 \left (a^2-b^2\right )^2 d (B+A \cos (c+d x)) (a+b \sec (c+d x))^{5/2} \left (-1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \sqrt {\frac {1+\tan ^2\left (\frac {1}{2} (c+d x)\right )}{1-\tan ^2\left (\frac {1}{2} (c+d x)\right )}} \left (a \left (-1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )-b \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )\right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. \(7174\) vs. \(2(456)=912\).
Time = 18.08 (sec) , antiderivative size = 7175, normalized size of antiderivative = 14.49
method | result | size |
parts | \(\text {Expression too large to display}\) | \(7175\) |
default | \(\text {Expression too large to display}\) | \(7241\) |
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Timed out. \[ \int \frac {A+B \sec (c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx=\text {Timed out} \]
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\[ \int \frac {A+B \sec (c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx=\int \frac {A + B \sec {\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx \]
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\[ \int \frac {A+B \sec (c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx=\int { \frac {B \sec \left (d x + c\right ) + A}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
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\[ \int \frac {A+B \sec (c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx=\int { \frac {B \sec \left (d x + c\right ) + A}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
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Timed out. \[ \int \frac {A+B \sec (c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx=\int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}}{{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]
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