Optimal. Leaf size=346 \[ -\frac {\left (-5 a^2 B+3 a A b+2 b^2 B\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 b^2 d \left (a^2-b^2\right )}+\frac {a (A b-a B) \sin (c+d x)}{b d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+b)}-\frac {\left (-5 a^2 B+3 a A b+2 b^2 B\right ) \sin (c+d x)}{3 b^2 d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x)}-\frac {\left (-5 a^3 B+3 a^2 A b+4 a b^2 B-2 A b^3\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b^3 d \left (a^2-b^2\right )}-\frac {a \left (-5 a^3 B+3 a^2 A b+7 a b^2 B-5 A b^3\right ) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{b^3 d (a-b) (a+b)^2}+\frac {\left (-5 a^3 B+3 a^2 A b+4 a b^2 B-2 A b^3\right ) \sin (c+d x)}{b^3 d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)}} \]
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Rubi [A] time = 1.31, antiderivative size = 346, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.242, Rules used = {2954, 3000, 3055, 3059, 2639, 3002, 2641, 2805} \[ -\frac {\left (-5 a^2 B+3 a A b+2 b^2 B\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 b^2 d \left (a^2-b^2\right )}-\frac {\left (3 a^2 A b-5 a^3 B+4 a b^2 B-2 A b^3\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b^3 d \left (a^2-b^2\right )}-\frac {a \left (3 a^2 A b-5 a^3 B+7 a b^2 B-5 A b^3\right ) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{b^3 d (a-b) (a+b)^2}+\frac {a (A b-a B) \sin (c+d x)}{b d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+b)}-\frac {\left (-5 a^2 B+3 a A b+2 b^2 B\right ) \sin (c+d x)}{3 b^2 d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x)}+\frac {\left (3 a^2 A b-5 a^3 B+4 a b^2 B-2 A b^3\right ) \sin (c+d x)}{b^3 d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)}} \]
Antiderivative was successfully verified.
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Rule 2639
Rule 2641
Rule 2805
Rule 2954
Rule 3000
Rule 3002
Rule 3055
Rule 3059
Rubi steps
\begin {align*} \int \frac {A+B \sec (c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+b \sec (c+d x))^2} \, dx &=\int \frac {B+A \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (b+a \cos (c+d x))^2} \, dx\\ &=\frac {a (A b-a B) \sin (c+d x)}{b \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x) (b+a \cos (c+d x))}-\frac {\int \frac {\frac {1}{2} \left (3 a A b-5 a^2 B+2 b^2 B\right )+b (A b-a B) \cos (c+d x)-\frac {3}{2} a (A b-a B) \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (b+a \cos (c+d x))} \, dx}{b \left (a^2-b^2\right )}\\ &=-\frac {\left (3 a A b-5 a^2 B+2 b^2 B\right ) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x)}+\frac {a (A b-a B) \sin (c+d x)}{b \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x) (b+a \cos (c+d x))}-\frac {2 \int \frac {-\frac {3}{4} \left (3 a^2 A b-2 A b^3-5 a^3 B+4 a b^2 B\right )-\frac {1}{2} b \left (3 a A b-2 a^2 B-b^2 B\right ) \cos (c+d x)+\frac {1}{4} a \left (3 a A b-5 a^2 B+2 b^2 B\right ) \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (b+a \cos (c+d x))} \, dx}{3 b^2 \left (a^2-b^2\right )}\\ &=-\frac {\left (3 a A b-5 a^2 B+2 b^2 B\right ) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x)}+\frac {\left (3 a^2 A b-2 A b^3-5 a^3 B+4 a b^2 B\right ) \sin (c+d x)}{b^3 \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)}}+\frac {a (A b-a B) \sin (c+d x)}{b \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x) (b+a \cos (c+d x))}-\frac {4 \int \frac {\frac {1}{8} \left (9 a^3 A b-12 a A b^3-15 a^4 B+16 a^2 b^2 B+2 b^4 B\right )+\frac {1}{4} b \left (6 a^2 A b-3 A b^3-10 a^3 B+7 a b^2 B\right ) \cos (c+d x)+\frac {3}{8} a \left (3 a^2 A b-2 A b^3-5 a^3 B+4 a b^2 B\right ) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{3 b^3 \left (a^2-b^2\right )}\\ &=-\frac {\left (3 a A b-5 a^2 B+2 b^2 B\right ) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x)}+\frac {\left (3 a^2 A b-2 A b^3-5 a^3 B+4 a b^2 B\right ) \sin (c+d x)}{b^3 \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)}}+\frac {a (A b-a B) \sin (c+d x)}{b \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x) (b+a \cos (c+d x))}+\frac {4 \int \frac {-\frac {1}{8} a \left (9 a^3 A b-12 a A b^3-15 a^4 B+16 a^2 b^2 B+2 b^4 B\right )-\frac {1}{8} a^2 b \left (3 a A b-5 a^2 B+2 b^2 B\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{3 a b^3 \left (a^2-b^2\right )}-\frac {\left (3 a^2 A b-2 A b^3-5 a^3 B+4 a b^2 B\right ) \int \sqrt {\cos (c+d x)} \, dx}{2 b^3 \left (a^2-b^2\right )}\\ &=-\frac {\left (3 a^2 A b-2 A b^3-5 a^3 B+4 a b^2 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b^3 \left (a^2-b^2\right ) d}-\frac {\left (3 a A b-5 a^2 B+2 b^2 B\right ) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x)}+\frac {\left (3 a^2 A b-2 A b^3-5 a^3 B+4 a b^2 B\right ) \sin (c+d x)}{b^3 \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)}}+\frac {a (A b-a B) \sin (c+d x)}{b \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x) (b+a \cos (c+d x))}-\frac {\left (3 a A b-5 a^2 B+2 b^2 B\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{6 b^2 \left (a^2-b^2\right )}-\frac {\left (a \left (3 a^2 A b-5 A b^3-5 a^3 B+7 a b^2 B\right )\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^2 A b-2 A b^3-5 a^3 B+4 a b^2 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b^3 \left (a^2-b^2\right ) d}-\frac {\left (3 a A b-5 a^2 B+2 b^2 B\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 b^2 \left (a^2-b^2\right ) d}-\frac {a \left (3 a^2 A b-5 A b^3-5 a^3 B+7 a b^2 B\right ) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{(a-b) b^3 (a+b)^2 d}-\frac {\left (3 a A b-5 a^2 B+2 b^2 B\right ) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x)}+\frac {\left (3 a^2 A b-2 A b^3-5 a^3 B+4 a b^2 B\right ) \sin (c+d x)}{b^3 \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)}}+\frac {a (A b-a B) \sin (c+d x)}{b \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x) (b+a \cos (c+d x))}\\ \end {align*}
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Mathematica [A] time = 6.94, size = 427, normalized size = 1.23 \[ \frac {\sqrt {\cos (c+d x)} \left (\frac {a^4 B \sin (c+d x)-a^3 A b \sin (c+d x)}{b^3 \left (b^2-a^2\right ) (a \cos (c+d x)+b)}+\frac {2 \sec (c+d x) (A b \sin (c+d x)-2 a B \sin (c+d x))}{b^3}+\frac {2 B \tan (c+d x) \sec (c+d x)}{3 b^2}\right )}{d}+\frac {\frac {\left (40 a^3 b B-24 a^2 A b^2-28 a b^3 B+12 A b^4\right ) \left (2 F\left (\left .\frac {1}{2} (c+d x)\right |2\right )-\frac {2 b \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a+b}\right )}{a}+\frac {2 \left (15 a^4 B-9 a^3 A b-12 a^2 b^2 B+6 a A b^3\right ) \sin (c+d x) \cos (2 (c+d x)) \left (\left (a^2-2 b^2\right ) \Pi \left (-\frac {a}{b};\left .\sin ^{-1}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 b (a+b) F\left (\left .\sin ^{-1}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )-2 a b E\left (\left .\sin ^{-1}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )\right )}{a^2 b \sqrt {1-\cos ^2(c+d x)} \left (2 \cos ^2(c+d x)-1\right )}+\frac {2 \left (45 a^4 B-27 a^3 A b-44 a^2 b^2 B+30 a A b^3-4 b^4 B\right ) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a+b}}{12 b^3 d (a-b) (a+b)} \]
Antiderivative was successfully verified.
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fricas [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {B \sec \left (d x + c\right ) + A}{{\left (b \sec \left (d x + c\right ) + a\right )}^{2} \cos \left (d x + c\right )^{\frac {7}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 20.45, size = 1024, normalized size = 2.96 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}}{{\cos \left (c+d\,x\right )}^{7/2}\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^2} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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