Optimal. Leaf size=405 \[ \frac {b (A b-a B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{a d \left (a^2-b^2\right ) (a \sec (c+d x)+b)}+\frac {\left (2 a^2 A+3 a b B-5 A b^2\right ) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 a^2 d \left (a^2-b^2\right )}+\frac {\left (2 a^2 A+3 a b B-5 A b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 a^2 d \left (a^2-b^2\right )}-\frac {\left (-2 a^3 B+4 a^2 A b+3 a b^2 B-5 A b^3\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{a^3 d \left (a^2-b^2\right )}+\frac {\left (-2 a^3 B+4 a^2 A b+3 a b^2 B-5 A b^3\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^3 d \left (a^2-b^2\right )}+\frac {b \left (-5 a^3 B+7 a^2 A b+3 a b^2 B-5 A b^3\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a^3 d (a-b) (a+b)^2} \]
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Rubi [A] time = 1.29, antiderivative size = 405, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 10, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.303, Rules used = {2960, 4029, 4102, 4106, 3849, 2805, 3787, 3771, 2639, 2641} \[ \frac {b (A b-a B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{a d \left (a^2-b^2\right ) (a \sec (c+d x)+b)}+\frac {\left (2 a^2 A+3 a b B-5 A b^2\right ) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 a^2 d \left (a^2-b^2\right )}-\frac {\left (4 a^2 A b-2 a^3 B+3 a b^2 B-5 A b^3\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{a^3 d \left (a^2-b^2\right )}+\frac {\left (2 a^2 A+3 a b B-5 A b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 a^2 d \left (a^2-b^2\right )}+\frac {\left (4 a^2 A b-2 a^3 B+3 a b^2 B-5 A b^3\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^3 d \left (a^2-b^2\right )}+\frac {b \left (7 a^2 A b-5 a^3 B+3 a b^2 B-5 A b^3\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a^3 d (a-b) (a+b)^2} \]
Antiderivative was successfully verified.
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Rule 2639
Rule 2641
Rule 2805
Rule 2960
Rule 3771
Rule 3787
Rule 3849
Rule 4029
Rule 4102
Rule 4106
Rubi steps
\begin {align*} \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x)}{(a+b \cos (c+d x))^2} \, dx &=\int \frac {\sec ^{\frac {7}{2}}(c+d x) (B+A \sec (c+d x))}{(b+a \sec (c+d x))^2} \, dx\\ &=\frac {b (A b-a B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (b+a \sec (c+d x))}-\frac {\int \frac {\sec ^{\frac {3}{2}}(c+d x) \left (-\frac {3}{2} b (A b-a B)+a (A b-a B) \sec (c+d x)-\frac {1}{2} \left (2 a^2 A-5 A b^2+3 a b B\right ) \sec ^2(c+d x)\right )}{b+a \sec (c+d x)} \, dx}{a \left (a^2-b^2\right )}\\ &=\frac {\left (2 a^2 A-5 A b^2+3 a b B\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d}+\frac {b (A b-a B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (b+a \sec (c+d x))}-\frac {2 \int \frac {\sqrt {\sec (c+d x)} \left (-\frac {1}{4} b \left (2 a^2 A-5 A b^2+3 a b B\right )-\frac {1}{2} a \left (a^2 A+2 A b^2-3 a b B\right ) \sec (c+d x)+\frac {3}{4} \left (4 a^2 A b-5 A b^3-2 a^3 B+3 a b^2 B\right ) \sec ^2(c+d x)\right )}{b+a \sec (c+d x)} \, dx}{3 a^2 \left (a^2-b^2\right )}\\ &=-\frac {\left (4 a^2 A b-5 A b^3-2 a^3 B+3 a b^2 B\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{a^3 \left (a^2-b^2\right ) d}+\frac {\left (2 a^2 A-5 A b^2+3 a b B\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d}+\frac {b (A b-a B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (b+a \sec (c+d x))}-\frac {4 \int \frac {-\frac {3}{8} b \left (4 a^2 A b-5 A b^3-2 a^3 B+3 a b^2 B\right )-\frac {1}{4} a \left (7 a^2 A b-10 A b^3-3 a^3 B+6 a b^2 B\right ) \sec (c+d x)-\frac {1}{8} \left (2 a^4 A+16 a^2 A b^2-15 A b^4-12 a^3 b B+9 a b^3 B\right ) \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (b+a \sec (c+d x))} \, dx}{3 a^3 \left (a^2-b^2\right )}\\ &=-\frac {\left (4 a^2 A b-5 A b^3-2 a^3 B+3 a b^2 B\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{a^3 \left (a^2-b^2\right ) d}+\frac {\left (2 a^2 A-5 A b^2+3 a b B\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d}+\frac {b (A b-a B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (b+a \sec (c+d x))}-\frac {4 \int \frac {-\frac {3}{8} b^2 \left (4 a^2 A b-5 A b^3-2 a^3 B+3 a b^2 B\right )-\left (-\frac {3}{8} a b \left (4 a^2 A b-5 A b^3-2 a^3 B+3 a b^2 B\right )+\frac {1}{4} a b \left (7 a^2 A b-10 A b^3-3 a^3 B+6 a b^2 B\right )\right ) \sec (c+d x)}{\sqrt {\sec (c+d x)}} \, dx}{3 a^3 b^2 \left (a^2-b^2\right )}+\frac {\left (b \left (7 a^2 A b-5 A b^3-5 a^3 B+3 a b^2 B\right )\right ) \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{b+a \sec (c+d x)} \, dx}{2 a^3 \left (a^2-b^2\right )}\\ &=-\frac {\left (4 a^2 A b-5 A b^3-2 a^3 B+3 a b^2 B\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{a^3 \left (a^2-b^2\right ) d}+\frac {\left (2 a^2 A-5 A b^2+3 a b B\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d}+\frac {b (A b-a B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (b+a \sec (c+d x))}+\frac {\left (2 a^2 A-5 A b^2+3 a b B\right ) \int \sqrt {\sec (c+d x)} \, dx}{6 a^2 \left (a^2-b^2\right )}+\frac {\left (4 a^2 A b-5 A b^3-2 a^3 B+3 a b^2 B\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx}{2 a^3 \left (a^2-b^2\right )}+\frac {\left (b \left (7 a^2 A b-5 A b^3-5 a^3 B+3 a b^2 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{2 a^3 \left (a^2-b^2\right )}\\ &=\frac {b \left (7 a^2 A b-5 A b^3-5 a^3 B+3 a b^2 B\right ) \sqrt {\cos (c+d x)} \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{a^3 (a-b) (a+b)^2 d}-\frac {\left (4 a^2 A b-5 A b^3-2 a^3 B+3 a b^2 B\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{a^3 \left (a^2-b^2\right ) d}+\frac {\left (2 a^2 A-5 A b^2+3 a b B\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d}+\frac {b (A b-a B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (b+a \sec (c+d x))}+\frac {\left (\left (2 a^2 A-5 A b^2+3 a b B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{6 a^2 \left (a^2-b^2\right )}+\frac {\left (\left (4 a^2 A b-5 A b^3-2 a^3 B+3 a b^2 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{2 a^3 \left (a^2-b^2\right )}\\ &=\frac {\left (4 a^2 A b-5 A b^3-2 a^3 B+3 a b^2 B\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{a^3 \left (a^2-b^2\right ) d}+\frac {\left (2 a^2 A-5 A b^2+3 a b B\right ) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{3 a^2 \left (a^2-b^2\right ) d}+\frac {b \left (7 a^2 A b-5 A b^3-5 a^3 B+3 a b^2 B\right ) \sqrt {\cos (c+d x)} \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{a^3 (a-b) (a+b)^2 d}-\frac {\left (4 a^2 A b-5 A b^3-2 a^3 B+3 a b^2 B\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{a^3 \left (a^2-b^2\right ) d}+\frac {\left (2 a^2 A-5 A b^2+3 a b B\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d}+\frac {b (A b-a B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (b+a \sec (c+d x))}\\ \end {align*}
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Mathematica [A] time = 7.11, size = 735, normalized size = 1.81 \[ \frac {\sqrt {\sec (c+d x)} \left (\frac {a b^2 B \sin (c+d x)-A b^3 \sin (c+d x)}{a^2 \left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac {2 A \tan (c+d x)}{3 a^2}+\frac {\left (2 a^3 B-4 a^2 A b-3 a b^2 B+5 A b^3\right ) \sin (c+d x)}{a^3 \left (a^2-b^2\right )}\right )}{d}+\frac {\frac {\left (6 a^3 b B-12 a^2 A b^2-9 a b^3 B+15 A b^4\right ) \sin (c+d x) \cos (2 (c+d x)) (a \sec (c+d x)+b) \left (-4 a^2 \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)} \Pi \left (-\frac {a}{b};\left .\sin ^{-1}\left (\sqrt {\sec (c+d x)}\right )\right |-1\right )+2 b^2 \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)} \Pi \left (-\frac {a}{b};\left .\sin ^{-1}\left (\sqrt {\sec (c+d x)}\right )\right |-1\right )+4 a b \sec ^2(c+d x)+2 b (2 a-b) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)} F\left (\left .\sin ^{-1}\left (\sqrt {\sec (c+d x)}\right )\right |-1\right )-4 a b \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)} E\left (\left .\sin ^{-1}\left (\sqrt {\sec (c+d x)}\right )\right |-1\right )-4 a b\right )}{a b^2 \left (1-\cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)} \left (2-\sec ^2(c+d x)\right ) (a+b \cos (c+d x))}+\frac {2 \left (12 a^4 B-28 a^3 A b-24 a^2 b^2 B+40 a A b^3\right ) \sin (c+d x) \cos ^2(c+d x) \sqrt {1-\sec ^2(c+d x)} (a \sec (c+d x)+b) \Pi \left (-\frac {a}{b};\left .\sin ^{-1}\left (\sqrt {\sec (c+d x)}\right )\right |-1\right )}{b \left (1-\cos ^2(c+d x)\right ) (a+b \cos (c+d x))}+\frac {2 \left (-4 a^4 A+30 a^3 b B-44 a^2 A b^2-27 a b^3 B+45 A b^4\right ) \sin (c+d x) \cos ^2(c+d x) \sqrt {1-\sec ^2(c+d x)} (a \sec (c+d x)+b) \left (F\left (\left .\sin ^{-1}\left (\sqrt {\sec (c+d x)}\right )\right |-1\right )-\Pi \left (-\frac {a}{b};\left .\sin ^{-1}\left (\sqrt {\sec (c+d x)}\right )\right |-1\right )\right )}{a \left (1-\cos ^2(c+d x)\right ) (a+b \cos (c+d x))}}{12 a^3 d (b-a) (a+b)} \]
Warning: Unable to verify antiderivative.
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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 \[ \int \frac {{\left (B \cos \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{\frac {5}{2}}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 6.90, size = 1031, normalized size = 2.55 \[ \text {result too large to display} \]
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 \[ \int \frac {\left (A+B\,\cos \left (c+d\,x\right )\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}}{{\left (a+b\,\cos \left (c+d\,x\right )\right )}^2} \,d x \]
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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