Optimal. Leaf size=203 \[ \frac {\left (2 a^2 A-a b B-A b^2\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d \left (a^2-b^2\right )}+\frac {(A b-a B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d \left (a^2-b^2\right )}-\frac {(A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}-\frac {\left (a^3 (-B)+3 a^2 A b-a b^2 B-A b^3\right ) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d (a-b) (a+b)^2} \]
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Rubi [A] time = 0.61, antiderivative size = 203, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {2954, 2999, 3059, 2639, 3002, 2641, 2805} \[ \frac {\left (2 a^2 A-a b B-A b^2\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d \left (a^2-b^2\right )}+\frac {(A b-a B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d \left (a^2-b^2\right )}-\frac {\left (3 a^2 A b+a^3 (-B)-a b^2 B-A b^3\right ) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d (a-b) (a+b)^2}-\frac {(A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \left (a^2-b^2\right ) (a \cos (c+d x)+b)} \]
Antiderivative was successfully verified.
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Rule 2639
Rule 2641
Rule 2805
Rule 2954
Rule 2999
Rule 3002
Rule 3059
Rubi steps
\begin {align*} \int \frac {A+B \sec (c+d x)}{\sqrt {\cos (c+d x)} (a+b \sec (c+d x))^2} \, dx &=\int \frac {\sqrt {\cos (c+d x)} (B+A \cos (c+d x))}{(b+a \cos (c+d x))^2} \, dx\\ &=-\frac {(A b-a B) \sqrt {\cos (c+d x)} \sin (c+d x)}{\left (a^2-b^2\right ) d (b+a \cos (c+d x))}+\frac {\int \frac {\frac {1}{2} (-A b+a B)+(a A-b B) \cos (c+d x)+\frac {1}{2} (A b-a B) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{a^2-b^2}\\ &=-\frac {(A b-a B) \sqrt {\cos (c+d x)} \sin (c+d x)}{\left (a^2-b^2\right ) d (b+a \cos (c+d x))}-\frac {\int \frac {\frac {1}{2} a (A b-a B)-\frac {1}{2} \left (2 a^2 A-A b^2-a b B\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{a \left (a^2-b^2\right )}+\frac {(A b-a B) \int \sqrt {\cos (c+d x)} \, dx}{2 a \left (a^2-b^2\right )}\\ &=\frac {(A b-a B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a \left (a^2-b^2\right ) d}-\frac {(A b-a B) \sqrt {\cos (c+d x)} \sin (c+d x)}{\left (a^2-b^2\right ) d (b+a \cos (c+d x))}+\frac {\left (2 a^2 A-A b^2-a b B\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{2 a^2 \left (a^2-b^2\right )}-\frac {\left (3 a^2 A b-A b^3-a^3 B-a b^2 B\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{2 a^2 \left (a^2-b^2\right )}\\ &=\frac {(A b-a B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a \left (a^2-b^2\right ) d}+\frac {\left (2 a^2 A-A b^2-a b B\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 \left (a^2-b^2\right ) d}-\frac {\left (3 a^2 A b-A b^3-a^3 B-a b^2 B\right ) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 (a-b) (a+b)^2 d}-\frac {(A b-a B) \sqrt {\cos (c+d x)} \sin (c+d x)}{\left (a^2-b^2\right ) d (b+a \cos (c+d x))}\\ \end {align*}
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Mathematica [A] time = 2.71, size = 260, normalized size = 1.28 \[ \frac {\frac {4 (a B-A b) \sin (c+d x) \sqrt {\cos (c+d x)}}{\left (a^2-b^2\right ) (a \cos (c+d x)+b)}-\frac {\frac {2 (A b-a B) \sin (c+d x) \left (\left (a^2-2 b^2\right ) \Pi \left (-\frac {a}{b};\left .\sin ^{-1}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 b (a+b) F\left (\left .\sin ^{-1}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )-2 a b E\left (\left .\sin ^{-1}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )\right )}{a^2 b \sqrt {\sin ^2(c+d x)}}+\frac {2 (a B-A b) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a+b}+\frac {(4 a A-4 b B) \left (2 F\left (\left .\frac {1}{2} (c+d x)\right |2\right )-\frac {2 b \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a+b}\right )}{a}}{(b-a) (a+b)}}{4 d} \]
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 \[ \int \frac {B \sec \left (d x + c\right ) + A}{{\left (b \sec \left (d x + c\right ) + a\right )}^{2} \sqrt {\cos \left (d x + c\right )}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 11.82, size = 802, normalized size = 3.95 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {B \sec \left (d x + c\right ) + A}{{\left (b \sec \left (d x + c\right ) + a\right )}^{2} \sqrt {\cos \left (d x + c\right )}}\,{d x} \]
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 \[ \int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}}{\sqrt {\cos \left (c+d\,x\right )}\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^2} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {A + B \sec {\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{2} \sqrt {\cos {\left (c + d x \right )}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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