Optimal. Leaf size=346 \[ -\frac{\left (-5 a^2 B+3 a A b+2 b^2 B\right ) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{3 b^2 d \left (a^2-b^2\right )}-\frac{\left (3 a^2 A b-5 a^3 B+4 a b^2 B-2 A b^3\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{b^3 d \left (a^2-b^2\right )}-\frac{a \left (3 a^2 A b-5 a^3 B+7 a b^2 B-5 A b^3\right ) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{b^3 d (a-b) (a+b)^2}+\frac{a (A b-a B) \sin (c+d x)}{b d \left (a^2-b^2\right ) \cos ^{\frac{3}{2}}(c+d x) (a \cos (c+d x)+b)}-\frac{\left (-5 a^2 B+3 a A b+2 b^2 B\right ) \sin (c+d x)}{3 b^2 d \left (a^2-b^2\right ) \cos ^{\frac{3}{2}}(c+d x)}+\frac{\left (3 a^2 A b-5 a^3 B+4 a b^2 B-2 A b^3\right ) \sin (c+d x)}{b^3 d \left (a^2-b^2\right ) \sqrt{\cos (c+d x)}} \]
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Rubi [A] time = 1.31062, antiderivative size = 346, normalized size of antiderivative = 1., 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 2954
Rule 3000
Rule 3055
Rule 3059
Rule 2639
Rule 3002
Rule 2641
Rule 2805
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*}
Mathematica [A] time = 6.90735, size = 429, normalized size = 1.24 \[ \frac{\frac{\left (-24 a^2 A b^2+40 a^3 b B-28 a b^3 B+12 A b^4\right ) \left (2 \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )-\frac{2 b \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a+b}\right )}{a}+\frac{\left (-9 a^3 A b-12 a^2 b^2 B+15 a^4 B+6 a A b^3\right ) \sin (c+d x) \cos (2 (c+d x)) \left (4 b (a+b) \text{EllipticF}\left (\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right ),-1\right )-2 \left (a^2-2 b^2\right ) \Pi \left (-\frac{a}{b};\left .-\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )-4 a b E\left (\left .\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )\right )}{a^2 b \sqrt{1-\cos ^2(c+d x)} \left (2 \cos ^2(c+d x)-1\right )}+\frac{2 \left (-27 a^3 A b-44 a^2 b^2 B+45 a^4 B+30 a A b^3-4 b^4 B\right ) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a+b}}{12 b^3 d (a-b) (a+b)}+\frac{\sqrt{\cos (c+d x)} \left (\frac{a^4 B \sin (c+d x)-a^3 A b \sin (c+d x)}{b^3 \left (b^2-a^2\right ) (a \cos (c+d x)+b)}+\frac{2 \sec (c+d x) (A b \sin (c+d x)-2 a B \sin (c+d x))}{b^3}+\frac{2 B \tan (c+d x) \sec (c+d x)}{3 b^2}\right )}{d} \]
Antiderivative was successfully verified.
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Maple [B] time = 9.885, size = 1024, normalized size = 3. \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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